Calculus II: Integration & Series

Here is a question you already have a feel for. Suppose you want the area of a region that has a flat bottom, straight sides, and a curved top, like the space underneath a graph. For a rectangle you multiply width times height and you are done. A curved top has no single height, so no neat formula applies. The fix is one of the oldest ideas in mathematics: cover the region with shapes you already know how to measure, rectangles, add them up, and then use thinner and thinner rectangles to sharpen the answer.

The big picture

Calculus I was about the derivative, a rate of change. Calculus II opens with the opposite question: how much piles up in total. We want the area between a curve y = f(x) and the x-axis, from x = a to x = b. We slice that region into thin vertical strips, replace each strip with a rectangle, add the rectangle areas, and then imagine the strips getting infinitely thin. That final total is the definite integral.

Key idea: a definite integral is the exact area you get by adding up infinitely many, infinitely thin rectangles.

Building a Riemann sum, one piece at a time

Think of the region as a loaf of bread you are cutting into equal slices. Here is the recipe.

1. Pick how many strips you want, and call that number n. More strips means a better estimate.
2. Find the width of each strip. Cut the interval from a to b into n equal pieces, so each strip has width dx = (b - a)/n. Read that as "the width equals b minus a, divided by n."
3. For each strip, choose one x-value and measure the height of the curve there, which is f at that x.
4. Each rectangle has area height times width. Add up all n rectangle areas. That total is a Riemann sum.

Written out, a Riemann sum looks like this:

S = f(x1) dx + f(x2) dx + ... + f(xn) dx

The only choice left is where in each strip to measure the height. Use the left edge of every strip and you get a left sum. Use the right edge and you get a right sum. Use the middle and you get a midpoint sum. For a curve that is climbing the whole way, the left sum comes out a little low and the right sum a little high, because the left edge of a rising strip is its shortest point and the right edge is its tallest.

Key idea: a Riemann sum has exactly three ingredients - how many strips, how wide each one is, and where inside each strip you sample the height.

Sigma notation, so we can talk about all n strips at once

Writing f(x1) dx + f(x2) dx + ... + f(xn) dx every time is tiring, and the dots hide exactly the thing we want to control. The compact form uses the summation sign:

Sn = sum from i = 1 to n of f(xi) dx

Read it as "add up f of x-sub-i times dx, letting the counter i run from 1 up to n." Nothing new has happened; we have only given the pattern a name. The sample points follow a pattern too: for a right sum the i-th edge is xi = a + i*dx, for a left sum xi = a + (i - 1)*dx, and for a midpoint sum xi = a + (i - 0.5)*dx.

One closed-form sum, provable by induction, lets us collapse the whole thing into a formula in n:

sum from i = 1 to n of i2 = n(n + 1)(2n + 1)/6

Sanity-check it on n = 3: 1 + 4 + 9 = 14, and the formula gives 3*4*7/6 = 84/6 = 14. It holds, and we will use it shortly to get an exact area rather than an estimate.

Key idea: sigma notation plus a closed formula turns "add up n rectangles" into "simplify one expression in n," which is what makes the limit computable.

Let us work one all the way through

Estimate the area under f(x) = x2 on the interval from 0 to 2, using a right sum with n = 4 strips. We will take it one small step at a time.

1. Find the strip width. Here a = 0 and b = 2 and n = 4, so dx = (2 - 0)/4 = 0.5. Each strip is half a unit wide.
2. List the right edges of the four strips. Starting at 0 and stepping by 0.5, the right edges are 0.5, 1.0, 1.5, and 2.0.
3. Find the height at each right edge. Since f(x) = x2 means "square the input," we get f(0.5) = 0.25, f(1) = 1, f(1.5) = 2.25, and f(2) = 4.
4. Add the four heights first. 0.25 + 1 + 2.25 + 4 = 7.5.
5. Multiply that total by the common width. 7.5 * 0.5 = 3.75.

So our estimate is 3.75. The true area, which we will confirm below, is 8/3 ~= 2.667. Our estimate is a bit high, exactly as expected for a right sum of a rising curve. Read that once more if you like: right edges are the tall side of each rising strip, so the rectangles stick out above the curve and the total overshoots. Using more strips would shrink the overshoot.

The same problem with left and midpoint samples

Change only the sampling rule and rerun the identical recipe. The interval, the function, and n = 4 all stay put, so dx = 0.5 throughout.

Left sum. The left edges are 0, 0.5, 1.0, 1.5. Their heights are f(0) = 0, f(0.5) = 0.25, f(1) = 1, f(1.5) = 2.25. Add the heights: 0 + 0.25 + 1 + 2.25 = 3.5. Times the width: 3.5 * 0.5 = 1.75. As promised for a rising curve, the left sum undershoots the true 2.667.

Midpoint sum. The midpoints sit half a step in from each left edge: 0.25, 0.75, 1.25, 1.75. Squaring each gives 0.0625, 0.5625, 1.5625, 3.0625. Add them: 0.0625 + 0.5625 + 1.5625 + 3.0625 = 5.25. Times the width: 5.25 * 0.5 = 2.625.

Now line the three up against the exact answer 8/3 = 2.6667 (four decimals):

Rule	Estimate	Error
Left, n = 4	1.7500	0.9167 low
Right, n = 4	3.7500	1.0833 high
Midpoint, n = 4	2.6250	0.0417 low

The midpoint rule is more than twenty times as accurate here with exactly the same arithmetic, and the reason is visible if you draw one strip. A midpoint rectangle cuts the curve: too short on the right half of the strip, too tall on the left half, and the two errors very nearly cancel. An endpoint rectangle makes its whole error in one direction with nothing to cancel against. For smooth functions the midpoint error shrinks like 1/n² while endpoint error shrinks like 1/n, so doubling n roughly halves an endpoint error but quarters a midpoint error. Averaging the left and right sums, (1.75 + 3.75)/2 = 2.75, gives the trapezoid rule, whose error of 0.0833 is exactly twice the midpoint error.

Key idea: where you sample changes the accuracy far more than it changes the work; midpoint sampling is the cheapest accuracy upgrade available.

Try it

Estimate the area under f(x) = x on the interval from 0 to 4 with a right sum, n = 4.

Answer: the width is dx = (4 - 0)/4 = 1. The right edges are 1, 2, 3, 4, with heights 1, 2, 3, 4. Add the heights, 1 + 2 + 3 + 4 = 10, then times the width, 10 * 1 = 10. The region is a triangle with exact area 8, so the right sum of 10 overestimates. Nice work if you remembered the times-the-width step.

From a sum to the integral sign

Now let the strips get infinitely thin. As n grows without bound, the rectangles hug the curve better and better, and the sum settles on one exact number. We write that limiting total with a new symbol:

integral from a to b of f(x) dx = lim(n -> infinity) [ f(x1) dx + ... + f(xn) dx ]

Read the left side aloud as "the integral from a to b of f of x, dee x." The tall symbol integral is a stretched-out letter S, and it stands for "sum," a reminder that we are still just adding up strips. The little dx is the width of one infinitely thin strip. The numbers a and b, sitting at the bottom and top of the integral sign, are the limits of integration, meaning where the region starts and stops.

One subtlety worth stating plainly: when the curve dips below the x-axis, its strips have negative height, so the integral counts that area as negative. The integral measures signed area: area above the axis adds, area below subtracts.

Key idea: the integral sign is a stretched S for "sum," and dx is the width of one paper-thin strip.

Getting the exact area from the definition, with no shortcuts

We claimed the true area under x2 from 0 to 2 is 8/3. Let us earn that number using nothing but the definition and the sum formula. This is the only time you will do it the hard way, and it is worth doing once so the shortcut in the next lesson feels like the gift it is.

1. Set up the general right sum. With a = 0, b = 2, and n strips, the width is dx = (2 - 0)/n = 2/n.
2. Locate the i-th right edge: xi = 0 + i*(2/n) = 2i/n.
3. Find its height: f(xi) = (2i/n)2 = 4i2/n2.
4. Write the sum: Sn = sum from i = 1 to n of (4i2/n2)*(2/n) = (8/n3) * sum from i = 1 to n of i2. The constants 8 and n³ do not depend on i, so they slide out in front of the sum.
5. Replace the sum with its closed form: Sn = (8/n3) * n(n + 1)(2n + 1)/6.
6. Expand the numerator: (n + 1)(2n + 1) = 2n2 + 3n + 1, so Sn = (8/6) * (2n2 + 3n + 1)/n2 = (4/3)*(2 + 3/n + 1/n2).
7. Take the limit. As n -> infinity, both 3/n and 1/n2 go to 0, leaving Sn -> (4/3)*2 = 8/3.

So integral from 0 to 2 of x2 dx = 8/3 exactly, and 8/3 = 2.6667 to four places. Check the formula against the arithmetic we already did: at n = 4 it predicts (4/3)*(2 + 0.75 + 0.0625) = (4/3)*(2.8125) = 3.75, which is precisely the right sum we computed strip by strip. The formula and the hand count agree, which is the sort of cross-check worth insisting on.

The formula also settles how fast the estimate improves. At n = 8 it gives (4/3)*(2 + 0.375 + 0.015625) = 3.1875, an error of 0.5208 against the earlier 1.0833 - just under half, exactly the 1/n behaviour claimed above. Three correct decimals would need roughly a thousand strips, which is a good reason to want the Fundamental Theorem.

Key idea: the definite integral is defined as a limit, and for simple functions you really can evaluate that limit by hand; every later technique is a way of avoiding this labour.

Where the idea came from

Slicing a curved region into manageable pieces is older than calculus by nineteen centuries: Archimedes found the area of a parabolic segment by exhausting it with triangles and summing a geometric series. What was missing until the nineteenth century was a precise account of which functions the procedure is guaranteed to work for. Bernhard Riemann supplied it in his 1854 Habilitation lecture at Gottingen, defining the integral through sums like ours with arbitrary sample points and arbitrary strip widths, and giving a criterion for when the limit exists no matter how the strips are chosen. That is why the sums carry his name, and why the definition is stated with a general sample point rather than a specific edge.

Four properties that save work

These rules follow straight from thinking of the integral as accumulated area, and they mirror the limit laws you already used in Calculus I.

• integral from a to a of f dx = 0. No width from a to a means no area.
• integral from a to b of f dx = - integral from b to a of f dx. Sweeping the region backward flips the sign.
• integral from a to b of [f + g] dx = integral f dx + integral g dx. The area of a sum is the sum of the areas.
• integral from a to c of f dx + integral from c to b of f dx = integral from a to b of f dx. You can split a region at any inside point c and add the pieces.

Where people get stuck

• Forgetting to multiply by dx. This is far and away the most common slip. It is easy to add up the heights and stop, but a Riemann sum is a sum of areas, and every area is height times width. Skipping dx in our example would have produced 7.5 instead of 3.75. A quick habit: after you add the heights, ask "times the width?" before writing anything down.
• "A Riemann sum is the exact area." A Riemann sum with a fixed number of strips is an estimate, full stop. The exact area is the limit as the number of strips goes to infinity. Our n = 4 right sum of 3.75 misses the true 8/3 by more than a whole unit.
• "Right sums are always too big." They overshoot for a rising curve, but for a falling curve the right edge is the short side, so a right sum undershoots. It is the direction of the function, not the word "right," that decides. On f(x) = 4 - x over [0, 4] the right sum is the low one.
• "The integral is always positive." Only when the curve stays at or above the axis. On [0, 2*pi] the integral of sin(x) is 0, because the hump above the axis and the trough below cancel exactly. The integral tracks signed area, and a question about total distance travelled or total physical area needs the absolute value instead.
• Counting the sample points wrong. With n strips there are n + 1 grid points but only n rectangles. A right sum on [0, 2] with n = 4 uses 0.5, 1.0, 1.5, 2.0 and never uses 0; a left sum uses 0, 0.5, 1.0, 1.5 and never uses 2. Using all five values gives an answer wrong by a whole rectangle. Related: the width is (b - a)/n, not n/(b - a), so if dx comes out bigger than the interval you have flipped the fraction.

Recap

• To find area under a curve, cover it with n rectangles of width dx = (b - a)/n and add their areas. That total is a Riemann sum, written compactly as sum from i = 1 to n of f(xi) dx.
• Left, right, and midpoint sums differ only in where you measure each rectangle's height. For a rising curve the left sum is low and the right sum is high; the midpoint sum splits the difference and is far more accurate.
• Endpoint sums have error proportional to 1/n; midpoint and trapezoid sums have error proportional to 1/n². Doubling n halves the first kind of error and quarters the second.
• With the closed form sum i2 = n(n + 1)(2n + 1)/6 the right sum for x2 on [0, 2] simplifies to (4/3)(2 + 3/n + 1/n2), whose limit is exactly 8/3.
• The definite integral integral from a to b of f(x) dx is the limit of those sums as the strips become infinitely thin, giving exact signed area.
• Handy properties: equal limits give 0, swapping limits flips the sign, integrals of sums split apart, and an interval can be split at an interior point.

Computing every integral from its definition would be unbearable. The next lesson supplies the shortcut that makes 8/3 appear in one line.

Sources

1. OpenStax. (2016). Approximating areas. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). The definite integral. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 5.1: Approximating areas. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). The area problem. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Dawkins, P. (n.d.). Definition of the definite integral. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. O'Connor, J. J., & Robertson, E. F. (n.d.). Bernhard Riemann. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk
7. Sealey, V. (2014). A framework for characterizing student understanding of Riemann sums and definite integrals. The Journal of Mathematical Behavior, 33, 230-245. doi.org/10.1016/j.jmathb.2013.12.002

In the last lesson, finding an exact area meant adding up infinitely many thin strips, which is a lot of work. This lesson is the shortcut that changed mathematics. It says that the two big ideas of calculus, rates of change and running totals, are two sides of the same coin. Once you see that, most definite integrals become a quick plug-and-subtract.

The big picture

Picture a car with an odometer and a speedometer. The speedometer reading is the rate of change of the odometer. Run that backward: if you know the speed at every instant and add it all up, you recover the total distance on the odometer. The Fundamental Theorem of Calculus (FTC) says this always works: integrating a rate gives back the total, and differentiating a running total gives back the rate. Integration and differentiation undo each other. The theorem comes in two parts.

Key idea: integration and differentiation are opposite operations, so an area problem can be solved with an antiderivative instead of an infinite sum.

Part 1: the running-total function

Fix a starting point a, and let x be a movable right edge. Define the accumulation function

g(x) = integral from a to x of f(t) dt

which reads "g of x equals the area under f from a out to x." As you slide x to the right, the area g(x) grows. How fast does it grow? Exactly at the current height of the curve. A taller curve piles on area faster. That single sentence is Part 1:

d/dx integral from a to x of f(t) dt = f(x)

Differentiating the area function hands you back the original curve. When the top edge is itself a function u(x) rather than a plain x, tack on the chain rule: d/dx integral from a to u(x) of f(t) dt = f(u(x)) * u'(x). For instance, d/dx integral from 0 to x2 of cos(t) dt = cos(x2) * 2x, because the top edge x² has derivative 2x.

Why Part 1 is true

The argument is short enough to follow completely, and seeing it makes the theorem stop feeling like magic. Start from the definition of the derivative applied to g:

g'(x) = lim(h -> 0) [ g(x + h) - g(x) ] / h

1. Interpret the numerator. g(x + h) is the area from a out to x + h, and g(x) is the area from a out to x. Subtracting leaves only the sliver between x and x + h, so g(x + h) - g(x) = integral from x to x+h of f(t) dt.
2. Estimate the sliver. It is a strip of width h whose height is f, and over a tiny interval a continuous f barely changes, so its area is close to f(x) * h. Precisely, the Mean Value Theorem for integrals guarantees some c between x and x + h with integral from x to x+h of f(t) dt = f(c) * h exactly.
3. Divide by h: the quotient is exactly f(c).
4. Let h -> 0. Then c is squeezed between x and x + h, so c -> x, and because f is continuous, f(c) -> f(x). Therefore g'(x) = f(x).

Notice where continuity was used, in step 4. That hypothesis is not decoration, and a later example shows what goes wrong when it fails.

Key idea: the area function grows at the rate given by the current height, because a sliver of width h has area about f(x) times h.

Worked Part 1 problems

Three variations, in increasing difficulty. In each case do not integrate anything - just read off the answer.

1. g(x) = integral from 1 to x of the square root of (1 + t3) dt. Nobody can write an elementary antiderivative for that integrand, and it does not matter. Part 1 gives g'(x) = the square root of (1 + x3) immediately: replace t by x and stop.
2. d/dx integral from x to 5 of cos(t2) dt. Here x is the bottom limit. Flip it to the top first, which costs a minus sign: the integral equals - integral from 5 to x of cos(t2) dt, whose derivative is -cos(x2).
3. d/dx integral from x to x2 of ln(t) dt. Both limits move, so split at any convenient constant, say 2: the integral is integral from 2 to x2 of ln(t) dt - integral from 2 to x of ln(t) dt. Differentiate each piece with the chain rule: ln(x2) * 2x - ln(x) * 1 = 4x*ln(x) - ln(x), using ln(x2) = 2 ln(x).

Part 2: the workhorse for evaluating integrals

Part 2 is the one you will use constantly. If F is any antiderivative of f, meaning a function whose derivative is f, then

integral from a to b of f(x) dx = F(b) - F(a)

In plain words: find an antiderivative, plug in the top limit, plug in the bottom limit, and subtract. The bracket shorthand [F(x)] from a to b just means "F at b minus F at a."

Part 2 follows from Part 1 in two lines. The accumulation function g(x) = integral from a to x of f(t) dt has g' = f by Part 1, and F has F' = f by assumption. Two functions with the same derivative differ by a constant, so F(x) = g(x) + C. Then F(b) - F(a) = [g(b) + C] - [g(a) + C] = g(b) - g(a), and since g(a) = 0 (no width, no area), that is exactly integral from a to b of f(x) dx. The constant cancels, which is why definite integrals never carry a "+ C".

Key idea: "any antiderivative" really means any - the arbitrary constant subtracts itself away.

A short table you should be able to reconstruct

Every entry below is verified the same way: differentiate the right column and check you get the left. Do that, out loud, every time you use one.

f(x)	An antiderivative F(x)	Check: F'(x)
xn, n not -1	xn+1/(n + 1)	(n + 1)xn/(n + 1) = xn
1/x	ln|x|	1/x
ekx	ekx/k	k*ekx/k = ekx
sin(x)	-cos(x)	-(-sin x) = sin x
cos(x)	sin(x)	cos x
sec2(x)	tan(x)	sec2 x
1/(1 + x2)	arctan(x)	1/(1 + x2)

The power rule fails at n = -1 because dividing by n + 1 = 0 is illegal, and that gap is exactly where the logarithm lives. The absolute value in ln|x| matters: it makes the formula valid on the negative side of the axis as well, where d/dx ln(-x) = (-1)/(-x) = 1/x still.

Worked example one, step by step

Evaluate integral from 0 to 2 of x2 dx.

1. Find an antiderivative of x², a function whose derivative is x². That is F(x) = x3/3, since the derivative of x³/3 is x².
2. Plug in the top limit, x = 2: F(2) = 23/3 = 8/3.
3. Plug in the bottom limit, x = 0: F(0) = 03/3 = 0.
4. Subtract, top minus bottom: 8/3 - 0 = 8/3.

So the exact area is 8/3 ~= 2.667. That matches the prediction from last lesson, where the right sum of 3.75 was an overestimate. No infinite sum needed this time.

Worked example two, watch the signs

Evaluate integral from 0 to pi of sin(x) dx.

1. Find an antiderivative of sin(x). The derivative of -cos(x) is sin(x), so F(x) = -cos(x).
2. Plug in the top limit, x = pi: -cos(pi) = -(-1) = 1.
3. Plug in the bottom limit, x = 0: -cos(0) = -(1) = -1.
4. Subtract, top minus bottom: 1 - (-1) = 2.

The area under one arch of the sine curve is exactly 2, a strikingly clean result. Verify the antiderivative before trusting it: d/dx[-cos(x)] = -(-sin x) = sin x. It checks.

Worked example three, a fractional power

Evaluate integral from 1 to 4 of 1/(the square root of x) dx.

1. Rewrite the integrand as a power: 1/(the square root of x) = x-1/2.
2. Apply the power rule with n = -1/2, so n + 1 = 1/2: F(x) = x1/2/(1/2) = 2*x1/2.
3. Verify by differentiating back: d/dx[2x1/2] = 2*(1/2)*x-1/2 = x-1/2. That is the integrand, so F is right.
4. Evaluate: F(4) - F(1) = 2*2 - 2*1 = 4 - 2 = 2.

Worked example four, and a trap worth meeting early

Evaluate integral from 0 to 1 of e3x dx. An antiderivative is F(x) = e3x/3; check it, d/dx[e3x/3] = 3e3x/3 = e3x. Then F(1) - F(0) = e3/3 - 1/3 = (e3 - 1)/3, which is about (20.086 - 1)/3 = 6.362. The forgotten 1/3 is the single most common error here; it comes from the chain rule when you differentiate back, which is exactly why you differentiate back.

Now the trap. Try integral from -1 to 1 of 1/x2 dx mechanically. An antiderivative of x-2 is -x-1 = -1/x, so the machine produces [-1/x] from -1 to 1 = (-1/1) - (-1/(-1)) = -1 - 1 = -2. That answer is impossible: the integrand 1/x2 is positive everywhere it is defined, so any honest area must be positive. The error is that 1/x2 is not continuous on [-1, 1] - it blows up at x = 0 - and Part 2 requires continuity on the closed interval. The true value is not -2; the integral does not converge at all, as Lesson 8 will show. Check the hypothesis before you turn the crank.

Key idea: FTC Part 2 applies only when f is continuous on the whole closed interval; a blow-up inside the interval silently produces a meaningless number.

Try it

Evaluate integral from 1 to 3 of 2x dx.

Answer: an antiderivative of 2x is x². Plug in the top, 32 = 9. Plug in the bottom, 12 = 1. Subtract, 9 - 1 = 8. Nice work. Notice you did not need a single rectangle.

Who actually discovered this

Isaac Barrow, Newton's predecessor in the Lucasian chair at Cambridge, published a geometric result in 1670 that amounts to Part 1, though he stated it as a theorem about tangents and areas rather than about functions. Newton had the calculus in workable form by 1666 and Leibniz independently by the mid-1670s, and it was Leibniz who gave us the elongated-S integral sign and the dx notation that made the subject easy to compute with. The bitter priority dispute that followed poisoned English and Continental mathematics for a century, with English mathematicians clinging to Newton's clumsier fluxion notation out of loyalty. The moral is not about credit: it is that good notation is a real mathematical contribution.

Where people get stuck

• Subtracting in the wrong order. It is always F(top) minus F(bottom), never the reverse. Reversing them flips the sign of the whole answer, and because a wrong sign often still looks plausible, nothing warns you.
• Losing a double negative. In the sine example, 1 - (-1) is 1 + 1 = 2, not 0. When a minus meets a negative they combine into a plus. Write the parentheses and slow that line down.
• "I need the +C on a definite integral." No. Any constant cancels in F(b) - F(a), as the two-line derivation above shows. Keep the +C for indefinite integrals, where the answer really is a whole family of functions.
• Forgetting the inner constant. The antiderivative of e3x is e3x/3, not e3x, and the antiderivative of cos(5x) is sin(5x)/5. Differentiating your candidate back catches this in one line, every time.
• Applying Part 2 across a discontinuity. The 1/x2 example produced -2 for a strictly positive integrand. Before evaluating, glance at the integrand for a denominator that vanishes or a logarithm of zero inside the interval.
• "Part 1 and Part 2 are unrelated." They are one fact seen twice, and Part 2 was derived from Part 1 above. Part 1 says differentiating an area returns the curve; Part 2 turns that around to evaluate areas.
• "Every function has an easy antiderivative." Finding one is usually the hard part, and some integrands such as e-x2 have no elementary antiderivative at all. The next several lessons build the techniques; Part 2 is only easy once F is in hand.

Recap

• FTC Part 1: the derivative of the running-area function integral from a to x of f(t) dt is just f(x); with a function on top, multiply by its derivative (chain rule), and with x on the bottom, flip the limits and pick up a minus sign.
• Part 1 holds because the sliver from x to x + h has area f(c)*h for some c in between, and continuity forces f(c) -> f(x).
• FTC Part 2: integral from a to b of f(x) dx = F(b) - F(a) for any antiderivative F, provided f is continuous on the closed interval [a, b].
• The recipe is: antiderivative, verify by differentiating back, plug in top, plug in bottom, subtract in that order.
• The arbitrary constant cancels in the subtraction, so definite integrals carry no "+ C".
• The theorem ties the two halves of calculus together, and almost every integral from here on leans on it.

Sources

1. OpenStax. (2016). The fundamental theorem of calculus. In Calculus volume 2. Rice University. openstax.org
2. Strang, G., & Herman, E. (n.d.). 5.3: The fundamental theorem of calculus. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
3. Dawkins, P. (n.d.). Computing definite integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Jerison, D. (n.d.). 18.01SC Single variable calculus [Course materials, Fall 2010]. MIT OpenCourseWare. ocw.mit.edu
5. O'Connor, J. J., & Robertson, E. F. (n.d.). Isaac Barrow. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk
6. O'Connor, J. J., & Robertson, E. F. (n.d.). The rise of calculus. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk
7. Grabiner, J. V. (1983). Who gave you the epsilon? Cauchy and the origins of rigorous calculus. The American Mathematical Monthly, 90(3), 185-194. doi.org/10.1080/00029890.1983.11971185

Integrals are not only about area on a graph. They answer a very practical kind of question: if you know the rate at which something changes, how much does it change in total? If water flows into a tub at a known rate each second, the integral of that rate is the total water added. This one idea connects integrals to physics, biology, and money.

The big picture

The Fundamental Theorem said integral from a to b of F'(x) dx = F(b) - F(a). Read the left side as "add up the rate of change." Read the right side as "the net change in the quantity." So integrating a rate gives the total change. That is the Net Change Theorem.

Key idea: the integral of a rate of change over an interval is the net change in the quantity across that interval.

Displacement versus total distance

Velocity is the rate of change of position, so integrating velocity gives the change in position, called displacement:

displacement = integral from a to b of v(t) dt

Here is the catch that surprises people. Velocity can be negative when the object moves backward, and those backward stretches subtract from the total. So displacement is "where you ended up compared to where you started," which can be small even after a long trip. To get the actual total distance covered, integrate the speed, which is the absolute value of velocity so every bit of motion counts as positive:

total distance = integral from a to b of |v(t)| dt

Worked example: an object that backs up, then moves forward

An object has velocity v(t) = t - 2 meters per second for t from 0 to 3. Find both its displacement and its total distance.

First the displacement, in tiny steps.

1. An antiderivative of t - 2 is t2/2 - 2t.
2. Plug in the top, t = 3: 32/2 - 2*3 = 4.5 - 6 = -1.5.
3. Plug in the bottom, t = 0: 0 - 0 = 0.
4. Subtract: -1.5 - 0 = -1.5 meters. The object ends up 1.5 meters behind where it started.

Now the total distance. The velocity t - 2 is negative before t = 2 and positive after, so the object first backs up, then goes forward. We split the trip at t = 2 and add the sizes of the two pieces.

1. From 0 to 2: [t2/2 - 2t] from 0 to 2 = (2 - 4) - 0 = -2. That is 2 meters of backward travel, so the distance is 2.
2. From 2 to 3: [t2/2 - 2t] from 2 to 3 = (4.5 - 6) - (2 - 4) = -1.5 - (-2) = 0.5. That is 0.5 meters forward.
3. Add the two distances: 2 + 0.5 = 2.5 meters.

So displacement is -1.5 meters but total distance is 2.5 meters. They differ because the object doubled back. Read that once more: displacement is net position change; distance is how far the object actually traveled.

Key idea: integrate v for displacement, integrate |v| for distance, and the two agree only when v never changes sign.

Second worked example: a trigonometric velocity

A particle has velocity v(t) = sin(t) meters per second for t from 0 to 3*pi/2. Find displacement and total distance.

1. An antiderivative of sin(t) is -cos(t); check by differentiating, d/dt[-cos t] = sin t.
2. Displacement: [-cos t] from 0 to 3pi/2 = -cos(3pi/2) - (-cos 0) = -0 + 1 = 1 meter.
3. Find the sign changes. On [0, pi] the sine is at or above zero; on [pi, 3pi/2] it is at or below zero. So split at t = pi.
4. First piece: [-cos t] from 0 to pi = -(-1) - (-1) = 1 + 1 = 2. Positive, so it contributes 2 meters.
5. Second piece: [-cos t] from pi to 3pi/2 = -cos(3pi/2) - (-cos pi) = 0 - 1 = -1. Its size is 1 meter.
6. Total distance: 2 + 1 = 3 meters.

Displacement 1 metre, distance 3 metres. The particle went 2 metres forward, then 1 metre back, and finished 1 metre ahead - and the arithmetic says exactly that.

Symmetry: two shortcuts you should always look for

Before computing any integral over an interval centred at 0, check the symmetry of the integrand.

• If f is odd, meaning f(-x) = -f(x), then integral from -a to a of f(x) dx = 0. The area on the left is the exact negative of the area on the right, and they annihilate.
• If f is even, meaning f(-x) = f(x), then integral from -a to a of f(x) dx = 2 * integral from 0 to a of f(x) dx. The two halves are mirror images, so compute one and double it.

Test both on cases you can check by hand. For the odd function x3: [x4/4] from -2 to 2 = 16/4 - 16/4 = 0, as promised. For the even function x2: the rule predicts 2*[x3/3] from 0 to 2 = 2*(8/3) = 16/3, and the direct computation gives [x3/3] from -2 to 2 = 8/3 - (-8/3) = 16/3. They agree. On an exam this can turn a page of work into a single word, "odd."

Net change beyond motion

The theorem has nothing to do with physics in particular; it applies wherever a rate is known and a total is wanted.

A filling tank. Water enters at r(t) = 20 - 2t litres per minute. How much enters during the first 5 minutes? Integrate the rate: integral from 0 to 5 of (20 - 2t) dt = [20t - t2] from 0 to 5 = (100 - 25) - 0 = 75 litres. Verify the antiderivative: d/dt[20t - t2] = 20 - 2t. Correct. Note the units multiply properly - litres per minute times minutes gives litres - and units are a free error check on every applied integral.

Marginal cost. In economics, marginal cost C'(x) is the cost of producing one more unit at production level x. The added cost of going from 10 units to 20 units is therefore integral from 10 to 20 of C'(x) dx. If C'(x) = 3x2 + 10 dollars per unit, then [x3 + 10x] from 10 to 20 = (8000 + 200) - (1000 + 100) = 8200 - 1100 = 7100 dollars. Check the antiderivative: d/dx[x3 + 10x] = 3x2 + 10. Correct.

Population growth rates, drug elimination rates, rainfall intensity, and electric current all work the same way: integrate the rate over the interval, and the answer is the total.

Key idea: whenever a problem hands you a "per unit time" or "per unit item" quantity and asks for a total, you are being asked for an integral.

The average value of a function

You know how to average a list of numbers. But a function has infinitely many values across an interval, so we use an integral to add them all and then divide by the width of the interval:

favg = (1/(b - a)) * integral from a to b of f(x) dx

Worked example. Find the average value of f(x) = x2 on the interval from 0 to 3.

1. Integrate: integral from 0 to 3 of x2 dx = [x3/3] from 0 to 3 = 27/3 = 9.
2. Divide by the width, which is 3 - 0 = 3: favg = 9/3 = 3.

The average value is 3. Sanity-check it against the function's range on [0, 3], which runs from 0 up to 9: an average of 3 sits inside that range, low rather than central because x2 spends most of the interval small and only shoots up near the right end. An average outside the range would be an immediate sign of an arithmetic error.

The Mean Value Theorem for Integrals promises that a continuous function actually attains its own average somewhere on the interval. Here we can find the spot: solve x2 = 3 to get x = the square root of 3 ~= 1.732, which does lie in (0, 3). Geometrically the theorem says a rectangle of height 3 and width 3, area 9, has exactly the same area as the region under the parabola - you can always flatten a curved region into a rectangle of the same area, and the required height is a value the function really takes.

Key idea: the average value is the height of the rectangle with the same width and the same area as the region under the curve.

Try it

A particle has velocity v(t) = 3t2 meters per second on the interval from 0 to 2. Find the displacement.

Answer: an antiderivative of 3t2 is t3. Then [t3] from 0 to 2 = 8 - 0 = 8 meters. Since 3t2 is never negative here, the total distance is also 8 meters. Nice work.

Where people get stuck

• Reporting displacement when the question asks for distance. This is the classic error, and the tell is a sign change in velocity. Before integrating, solve v(t) = 0 and check whether any root lies strictly inside the interval. If one does, split there and add the sizes; if none does, the two answers agree in size.
• Taking absolute values at the wrong moment. Total distance is integral |v| dt, not |integral v dt|. In the first worked example those give 2.5 and 1.5 respectively - different numbers. The absolute value has to go inside, before the accumulation, or the cancellation has already happened.
• "Average value is (f(a) + f(b))/2." That averages two points and ignores everything between them. On x2 over [0, 3] the endpoint rule gives (0 + 9)/2 = 4.5, half again as large as the true average of 3.
• Forgetting to divide by the width. The integral alone is the accumulated total, not the average. Dropping the 1/(b - a) in the example would report 9 instead of 3, and 9 is not even in the range of f on that interval - which is exactly the check that catches it.
• "A negative integral is a mistake." Negative results are meaningful: net motion in the negative direction, a tank draining, a population shrinking. Only quantities that cannot be negative, such as total distance or physical area, warrant suspicion.
• Ignoring units. A rate in litres per minute integrated over minutes gives litres. If your answer's units do not match what was asked, you have integrated the wrong thing - often the quantity rather than its rate.

Recap

• Net Change Theorem: integrating a rate of change gives the total change, since integral F' dx = F(b) - F(a).
• Displacement is integral v dt; total distance is integral |v| dt. Split at every sign change of v, then add the sizes of the pieces.
• For v(t) = t - 2 on [0, 3] the displacement is -1.5 m and the distance is 2.5 m; for v(t) = sin t on [0, 3pi/2] they are 1 m and 3 m.
• Odd integrands vanish over intervals centred at 0; even integrands are twice the half-interval, and checking symmetry first can eliminate the work entirely.
• Average value is (1/(b - a)) integral from a to b of f dx, the height of the rectangle with the same area as the region.
• The Mean Value Theorem for Integrals says a continuous function actually attains its average value at some interior point.
• Any "per unit" rate - litres per minute, dollars per unit, people per year - integrates to a total in the matching units.

Sources

1. OpenStax. (2016). Integration formulas and the net change theorem. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Physical applications. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 5.4: Integration formulas and the net change theorem. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). Computing definite integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Guichard, D. (n.d.). Calculus: Early transcendentals. Whitman College. whitman.edu
6. Jerison, D. (n.d.). 18.01SC Single variable calculus [Course materials, Fall 2010]. MIT OpenCourseWare. ocw.mit.edu
7. Khan Academy. (n.d.). Applications of integrals. khanacademy.org

Some integrals look complicated only because of a messy inner piece, like the x² tucked inside cos(x²). Substitution is the move that gives that messy piece a short temporary nickname so the whole integral turns into one you already know. Then you rename it back at the end. It is the single most used integration technique, so it is worth taking slowly.

The big picture

Back in Calculus I, the chain rule differentiated a function-inside-a-function. Substitution runs that rule in reverse. When an integrand contains an inner function and (up to a constant) its own derivative, we rename the inner function u, and the integral collapses into something simple.

Key idea: substitution reverses the chain rule by renaming the inner function u, so that integral f(g(x)) g'(x) dx becomes the easy integral f(u) du.

The method in three moves

1. Choose u to be the inner function, the thing tucked inside something else.
2. Differentiate it to get du = g'(x) dx. Read this as "a small change in u equals the derivative of u, times dx."
3. Swap every x-piece for a u-piece. If the integral now reads entirely in u, you are set. Integrate, then rename u back to x at the end.

Worked example one, a perfect match

Evaluate integral 2x cos(x2) dx.

1. The inner function is x², so let u = x2.
2. Differentiate: du = 2x dx.
3. Notice the integrand already contains 2x dx, so it swaps out cleanly. The integral becomes integral cos(u) du.
4. Integrate: integral cos(u) du = sin(u) + C.
5. Rename u back to x²: sin(x2) + C.

Worked example two, fixing a missing constant

Often the derivative is almost present, off by a number. You patch that with a fraction. Evaluate integral x(x2 + 1)5 dx.

1. Let u = x2 + 1, the inner function.
2. Differentiate: du = 2x dx.
3. The integrand has only x dx, not 2x dx. Solve for what you have: x dx = du/2.
4. Substitute: integral u5 (du/2) = (1/2) integral u5 du.
5. Integrate: (1/2)(u6/6) = u6/12.
6. Rename back: (x2 + 1)6/12 + C.

Verify by differentiating: d/dx[(x2 + 1)6/12] = 6(x2 + 1)5*(2x)/12 = x(x2 + 1)5. That is the integrand, so the answer stands. Do this after every substitution until it becomes automatic - it costs one line and it catches every dropped constant.

How to recognise a substitution problem

Substitution is the right tool exactly when two things are true at once: the integrand contains a composite function f(g(x)), and the derivative g'(x) appears as a factor, up to a constant multiple. Nothing else qualifies. In practice you spot the composite first and then hunt for its derivative:

• Something in parentheses raised to a power, as in (x3 + 4)2. Try u = the thing in parentheses. Its derivative 3x2 should be sitting outside.
• Something under a root, as in the square root of (x2 + 9). Try u = the expression under the root.
• An exponent on e, as in esin x. Try u = the exponent; look for cos x outside.
• The argument of a trig or log function, as in cos(x2) or ln(3x + 1). Try u = the argument.
• A fraction whose numerator looks like the derivative of its denominator. Try u = the denominator; the answer will be a logarithm.

And here is the negative test, which matters just as much. If, after computing du, a stray variable is left over that you cannot express in u, substitution alone will not finish the job. A constant left over is fine, because constants pass through the integral sign; a leftover x is fatal. That is why x*ex resists substitution - the derivative of x is 1, not ex, and the derivative of ex is ex, not x. Neither factor is the derivative of the other, so the next lesson's tool is required.

Key idea: substitution needs an inside function and its derivative; a leftover constant is repairable, a leftover variable means the wrong tool.

Definite integrals: change the limits

For a definite integral you have a clean option: convert the x-limits into u-limits, and then you never need to rename back. Evaluate integral from 0 to 2 of x(x2 + 1)3 dx.

1. Let u = x2 + 1, so du = 2x dx, which gives x dx = du/2.
2. Change the bottom limit: when x = 0, u = 02 + 1 = 1.
3. Change the top limit: when x = 2, u = 22 + 1 = 5.
4. Rewrite the whole integral in u: (1/2) integral from 1 to 5 of u3 du.
5. Integrate: (1/2)[u4/4] from 1 to 5 = (1/8)(54 - 14) = (1/8)(625 - 1) = 624/8 = 78.

The answer is 78. Because we switched the limits to u-values, the 1 and 5 already belong to u, so there is nothing to undo.

A special case worth memorizing

Whenever the top of a fraction is the derivative of the bottom, the answer is a logarithm: integral g'(x)/g(x) dx = ln|g(x)| + C. That is just u = g(x) turning the integral into integral du/u. For example, integral (2x)/(x2 + 1) dx = ln(x2 + 1) + C, since the top 2x is exactly the derivative of the bottom x² + 1.

Worked example three: an exponential composite

Evaluate integral esin x cos(x) dx.

1. The composite is e(something), so let u = sin(x), the exponent.
2. Differentiate: du = cos(x) dx. That factor is present exactly, with no constant to patch.
3. Substitute: integral eu du = eu + C.
4. Rename back: esin x + C.
5. Verify: d/dx[esin x] = esin x * cos(x). Correct.

Worked example four: a definite integral of tan(x)

Evaluate integral from 0 to pi/4 of tan(x) dx. There is no composite in sight until you rewrite the integrand.

1. Rewrite: tan(x) = sin(x)/cos(x). Now it is a fraction whose numerator is nearly the derivative of its denominator.
2. Let u = cos(x), so du = -sin(x) dx, which gives sin(x) dx = -du.
3. Change the limits: when x = 0, u = cos(0) = 1; when x = pi/4, u = cos(pi/4) = (the square root of 2)/2.
4. Rewrite: integral from 1 to (sqrt2)/2 of (-du)/u = -[ln|u|] from 1 to (sqrt2)/2.
5. Evaluate: -[ln((sqrt2)/2) - ln(1)] = -ln((sqrt2)/2). Since (sqrt2)/2 = 1/(sqrt2) = 2-1/2, this is -(-1/2)ln(2) = (1/2)ln(2) ~= 0.3466.

Verify the antiderivative: d/dx[-ln|cos x|] = -(-sin x)/cos(x) = tan(x). Correct. The answer is positive, which it must be, since tan(x) is positive on (0, pi/4).

Worked example five: when you must solve for x too

Sometimes du removes only part of the x's, and you finish by substituting for x itself. Evaluate integral x/(the square root of (x + 1)) dx.

1. Let u = x + 1, the expression under the root. Then du = dx.
2. The numerator x is still there. But u = x + 1 rearranges to x = u - 1, so replace it: the integral becomes integral (u - 1)/(u1/2) du.
3. Split the fraction: integral (u1/2 - u-1/2) du.
4. Integrate term by term: u3/2/(3/2) - u1/2/(1/2) = (2/3)u3/2 - 2u1/2 + C.
5. Rename back: (2/3)(x + 1)3/2 - 2(x + 1)1/2 + C.

Verify: d/dx[(2/3)(x+1)3/2] = (x+1)1/2 and d/dx[-2(x+1)1/2] = -(x+1)-1/2. Adding, (x+1)1/2 - (x+1)-1/2 = [(x + 1) - 1]/(x+1)1/2 = x/(the square root of (x + 1)). That is the integrand, so the answer is right. This "solve for x as well" trick rescues many integrals that look hopeless at step 2.

Key idea: a leftover x is only fatal if you cannot rewrite it in terms of u; when u is a simple shift or scaling, solve for x and continue.

Try it

Evaluate integral 3x2 (x3 + 4)2 dx.

Answer: let u = x3 + 4, so du = 3x2 dx, which is exactly present. The integral becomes integral u2 du = u3/3, so the answer is (x3 + 4)3/3 + C. Nice work spotting that the 3x² was ready-made.

Where people get stuck

• Choosing u badly. The make-or-break step is picking u so that its derivative actually appears. After you write du, look back and ask "is the rest of this du, maybe off by a constant?" If a stray x survives and cannot be rewritten as u - 1 or similar, try a different inner piece or a different technique.
• Pulling a variable outside the integral. You may move only constants across the integral sign. Writing integral x(x2+1)5 dx = x integral (x2+1)5 dx is simply false, because x is not constant with respect to x. Only the constant 1/2 was legitimately moved in the worked example.
• Leaving the answer in u. For an indefinite integral you must rename back to x - the question was about x. For a definite integral you may change the limits and stop in u, but you may not change the limits and rename back, which double-counts the substitution.
• Mixing the two definite-integral habits. Either convert the limits to u-values and evaluate in u, or keep the x-limits and rename back before plugging in. Evaluating [u4/8] at the original x-limits 0 and 2 gives 2, not 78. Pick one habit and never blend them.
• Dropping the minus sign. When du = -sin(x) dx, the substitution introduces a minus that must survive to the end, as in the tan(x) example. Omitting it turns +0.3466 into -0.3466, a negative area under a positive curve.
• Forgetting the absolute value in ln|u|. It matters whenever u can be negative on the interval. Here cos(x) stayed positive on [0, pi/4], but on [0, 3pi/4] it would not, and the integral would in fact diverge at pi/2.
• "Substitution works on every integral." It works only when the inner function's derivative is present up to a constant. Products like x*ex, quotients of polynomials, and expressions like the square root of (1 - x2) need the tools of the next three lessons.

Recap

• Substitution reverses the chain rule: let u be the inner function, compute du, and rewrite the integral entirely in u.
• Recognise it by a composite - a parenthesis raised to a power, a root, an exponent, a trig or log argument, a denominator - whose derivative also appears as a factor.
• If a constant is off, solve du for the piece you have (for example x dx = du/2) and carry the fraction along. A leftover variable means a different u or a different technique, unless u can be solved for x.
• For definite integrals, either change the limits to u-values or rename back to x before plugging in, never a mix.
• Special case: the top being the derivative of the bottom gives integral g'/g dx = ln|g| + C; this is where integral tan(x) dx = -ln|cos x| + C comes from.
• Differentiate every answer back before you write it down. It is one line and it catches missing constants and lost minus signs.

Sources

1. OpenStax. (2016). Substitution. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Integrals involving exponential and logarithmic functions. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 5.5: Substitution. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). Substitution rule for indefinite integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Dawkins, P. (n.d.). Substitution rule for definite integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Guichard, D. (n.d.). Calculus: Early transcendentals. Whitman College. whitman.edu
7. Khan Academy. (n.d.). Integration techniques. khanacademy.org

Substitution handles an inner function and its derivative. But what about a plain product of two unrelated things, like x times e^x, where neither is the derivative of the other? Integration by parts is the tool for that. It has a big reputation, mostly because of one small formula and one choice you have to make, so we will nail down both.

The big picture

Integration by parts reverses the product rule, the same way substitution reverses the chain rule. You split the integrand into two pieces: a part called u that you will differentiate, and a part called dv that you will integrate. The formula trades your hard integral for a hopefully easier one.

integral u dv = u*v - integral v du

Read it aloud: "the integral of u dv equals u times v, minus the integral of v du." The whole goal is that the new integral, integral v du, is simpler than the one you started with.

Key idea: split the product into u (to differentiate) and dv (to integrate), then apply integral u dv = uv - integral v du to turn a hard integral into an easier one.

Choosing u with LIATE

Which piece should be u? Pick whichever comes first in this list, because that type of function gets simpler when you differentiate it:

• L, Logarithmic, like ln(x)
• I, Inverse trig, like arctan(x)
• A, Algebraic, like x or x²
• T, Trig, like sin(x) or cos(x)
• E, Exponential, like e^x

Whatever is left over, together with the dx, becomes dv. Spell the word "LIATE" out and it picks u for you.

LIATE is a heuristic, not a theorem, and it is worth knowing why it usually works. A logarithm differentiates into 1/x, an algebraic power drops by one degree, and both are strict simplifications. Trig and exponential functions differentiate into functions of the same complexity, so they are better used as dv, which you integrate. The list is ordered from "improves most when differentiated" to "does not improve at all."

Where the formula comes from

Start with the product rule, which you already know: d/dx[u*v] = u'*v + u*v'. Integrate both sides with respect to x. The left side undoes itself, giving u*v, so

u*v = integral u'*v dx + integral u*v' dx

Solve for the second integral: integral u*v' dx = u*v - integral u'*v dx. Now rename in differential notation, with dv = v' dx and du = u' dx, and that is exactly integral u dv = u*v - integral v du. The whole formula is one rearrangement of the product rule, which is why the minus sign is there: it is the price of moving a term across the equals sign.

One convenience falls out of the derivation. When you integrate dv to get v, you may drop the constant of integration and take the simplest v available. Any constant you added would appear as +C*u in the first term and -C*integral du = -C*u in the second, and the two cancel exactly.

How to recognise a parts problem

Reach for parts when the integrand is a product of two functions of different families and one of them gets simpler when differentiated. The recurring shapes are:

• Polynomial times exponential, such as x ex or x2e-3x. Take u = the polynomial; each round lowers its degree by one.
• Polynomial times trig, such as x sin(x). Same choice, same reason.
• Polynomial times logarithm, such as x2ln(x). Here u = ln(x), because differentiating the log removes it entirely.
• A lone logarithm or inverse trig function, such as ln(x) or arctan(x). There is no visible product, so manufacture one with dv = dx.
• Exponential times trig, such as exsin(x). Neither ever simplifies, so this one boomerangs and is finished with algebra.

The negative test is just as useful: if the two factors are related by differentiation, as in x cos(x2), you want substitution, not parts. Check for substitution first every time; it is faster when it applies.

Key idea: substitution for a composite and its derivative, parts for a product of unrelated types.

Worked example one, step by step

Evaluate integral x ex dx.

1. Apply LIATE. The algebraic x comes before the exponential e^x, so u = x and dv = ex dx.
2. Differentiate u: du = dx.
3. Integrate dv to get v: v = ex.
4. Put the pieces into the formula: u*v - integral v du = x ex - integral ex dx.
5. Finish the leftover integral: x ex - ex + C.

So integral x ex dx = ex(x - 1) + C. You can check any antiderivative by differentiating it, and the product rule brings you right back to x e^x.

Worked example two, a lone logarithm

Evaluate integral ln(x) dx. There is no obvious antiderivative of ln(x), so let the whole thing be u and let dv be just dx.

1. Set u = ln(x) and dv = dx.
2. Differentiate u: du = (1/x) dx.
3. Integrate dv: v = x.
4. Apply the formula: x ln(x) - integral x*(1/x) dx.
5. The leftover simplifies because x times 1/x is 1: x ln(x) - integral 1 dx = x ln(x) - x + C.

Verify: d/dx[x ln(x) - x] = ln(x) + x*(1/x) - 1 = ln(x) + 1 - 1 = ln(x). Correct. The same manufacture-a-dv trick handles inverse trig. For integral arctan(x) dx, take u = arctan(x) and dv = dx, so du = dx/(1 + x2) and v = x. The formula gives x arctan(x) - integral x/(1 + x2) dx, and that leftover is a substitution with w = 1 + x2, yielding (1/2)ln(1 + x2). So the answer is x arctan(x) - (1/2)ln(1 + x2) + C. Differentiating back gives arctan(x) + x/(1 + x2) - x/(1 + x2) = arctan(x). Correct.

When you need parts more than once

A higher power needs parts repeated, once per degree. Evaluate integral x2 ex dx.

1. Take u = x2 and dv = ex dx, so du = 2x dx and v = ex.
2. Apply the formula: x2ex - integral ex*2x dx = x2ex - 2 integral x ex dx.
3. The leftover is worked example one: integral x ex dx = ex(x - 1).
4. Substitute it in: x2ex - 2ex(x - 1) = x2ex - 2x ex + 2ex.
5. Factor: ex(x2 - 2x + 2) + C.

Verify with the product rule: d/dx[ex(x2 - 2x + 2)] = ex(x2 - 2x + 2) + ex(2x - 2) = ex*x2. Correct. Note the pattern: x3ex would take three rounds, and each round costs one degree.

The boomerang case, worked in full

Evaluate I = integral ex sin(x) dx. Neither factor simplifies under differentiation, so no choice of u makes visible progress - and that is exactly the signal for this method.

1. First round. Take u = sin(x), dv = ex dx, so du = cos(x) dx and v = ex. Then I = exsin(x) - integral excos(x) dx. Call that new integral J.
2. Second round, on J. Take u = cos(x), dv = ex dx, so du = -sin(x) dx and v = ex. Then J = excos(x) - integral ex*(-sin x) dx = excos(x) + I. The original integral has come back.
3. Substitute J into the first line. I = exsin(x) - [excos(x) + I] = exsin(x) - excos(x) - I.
4. Solve algebraically. Add I to both sides: 2I = ex(sin(x) - cos(x)), so I = ex(sin(x) - cos(x))/2 + C.

Verify: d/dx[ex(sin x - cos x)/2] = (1/2)[ex(sin x - cos x) + ex(cos x + sin x)] = (1/2)ex(2 sin x) = exsin(x). Correct. One caution: on the second round you must keep the same kind of choice, integrating the exponential both times. If you switch and integrate the trig function on round two, you simply undo round one and arrive at the true but useless statement I = I.

Key idea: when the original integral reappears, it is not a failure - move it to the left side and divide.

Parts on a definite integral

The formula becomes integral from a to b of u dv = [u*v] from a to b - integral from a to b of v du: the boundary term is evaluated at both limits, and the remaining integral keeps the same limits. Evaluate integral from 0 to 1 of x e2x dx.

1. u = x, dv = e2x dx, so du = dx and v = e2x/2. Note the 1/2 from the chain rule; forgetting it is the classic slip here.
2. Boundary term: [x e2x/2] from 0 to 1 = (1*e2/2) - (0) = e2/2.
3. Remaining integral: integral from 0 to 1 of (e2x/2) dx = [e2x/4] from 0 to 1 = e2/4 - 1/4.
4. Subtract: e2/2 - (e2/4 - 1/4) = e2/4 + 1/4 = (e2 + 1)/4 ~= 2.097.

Verify the underlying antiderivative F(x) = x e2x/2 - e2x/4: F'(x) = e2x/2 + x e2x - e2x/2 = x e2x. Correct. As a rough sanity check, the integrand rises from 0 to e2 ~= 7.39 across [0, 1], so an answer near 2 is plausible.

Try it

Evaluate integral x cos(x) dx.

Answer: by LIATE the algebraic x is u and dv = cos(x) dx. Then du = dx and v = sin(x). The formula gives x sin(x) - integral sin(x) dx = x sin(x) + cos(x) + C. Notice the leftover integral of sin(x) is minus cos(x), and subtracting it flips to plus cos(x). Nice work keeping the sign straight.

Where people get stuck

• The minus sign. The formula is uv - integral v du, and the subtraction applies to the entire remaining integral. Keep it in one set of mental parentheses. In the x cos(x) problem the leftover integrates to -cos(x), and subtracting that flips it to +cos(x); losing track there gives the wrong sign on the final term.
• Differentiating both pieces, or integrating both. You differentiate u and integrate dv, never the same operation twice. If your "easier" integral came out harder than the original, you almost certainly swapped them - switch and start over.
• Choosing dv you cannot integrate. LIATE puts the log first partly because you cannot integrate ln(x) at that stage anyway. dv must be something you can antidifferentiate immediately; if it is not, that choice is unavailable regardless of what the mnemonic says.
• Losing the inner constant in v. The antiderivative of e2x dx is e2x/2, and of cos(3x) dx is sin(3x)/3. This single omission wrecks more definite-integral answers than the minus sign does.
• Undoing your own work in the boomerang case. On the second application you must integrate the same family you integrated the first time. Switching returns you to the starting integral and the equation collapses to I = I.
• Forgetting to evaluate the boundary term at both limits. In a definite integral, [u*v] must be evaluated at b and at a and subtracted, exactly like any other antiderivative. Plugging in only the top limit is a common and invisible error.
• Trying parts when substitution would work. For integral x cos(x2) dx, the x is the derivative of the inside, so substitution finishes it in two lines while parts goes nowhere. Always test for substitution first.

Recap

• Integration by parts reverses the product rule: integral u dv = uv - integral v du, obtained by integrating d/dx[uv] = u'v + uv' and rearranging.
• Choose u by LIATE (Log, Inverse trig, Algebraic, Trig, Exponential); whatever remains, together with dx, is dv. The ordering reflects how much each family simplifies when differentiated.
• Recognise parts by a product of unrelated types, or by a lone log or inverse trig function where you set dv = dx.
• Differentiate u, integrate dv (dropping the constant, which cancels), substitute, and finish the leftover integral.
• A polynomial of degree n needs n rounds; integral x2ex dx = ex(x2 - 2x + 2) + C.
• Exponential-times-trig boomerangs: integral exsin(x) dx = ex(sin x - cos x)/2 + C, found by solving for the integral algebraically.
• For definite integrals, evaluate the [uv] term at both limits and keep the same limits on the remaining integral.

Sources

1. OpenStax. (2016). Integration by parts. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Other strategies for integration. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 7.1: Integration by parts. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). Integration by parts. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Dawkins, P. (n.d.). Integration strategy. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Guichard, D. (n.d.). Calculus: Early transcendentals. Whitman College. whitman.edu
7. Khan Academy. (n.d.). Integration techniques. khanacademy.org

This lesson bundles two techniques that both lean on trig identities you have met before. The first handles integrals built from powers of sine and cosine. The second, called trigonometric substitution, tackles integrals with an awkward square root like the square root of 4 - x². Both turn a hard-looking problem into a familiar trig integral. One note on notation up front: sin³(x) means (sin(x)) cubed, the whole sine value raised to a power.

The big picture

Trig identities let you rewrite an integrand into a shape that substitution can finish. For powers of sine and cosine, the Pythagorean identity sin2 + cos2 = 1 lets you trade one function for the other. For a square root of a quadratic, replacing x with a trig function makes the root simplify to a single clean term.

Key idea: a well-chosen trig identity or trig substitution reshapes a messy integrand into a plain trig integral you already know how to do.

Powers of sine and cosine

For integral sinm(x) cosn(x) dx, look at whether the powers are odd or even.

• If the power of sine is odd: peel off one sin(x) to pair with dx, rewrite the rest using sin2 = 1 - cos2, and let u = cos(x).
• If the power of cosine is odd: peel off one cos(x), rewrite the rest with cos2 = 1 - sin2, and let u = sin(x).
• If both powers are even: use the half-angle identities sin2(x) = (1 - cos 2x)/2 and cos2(x) = (1 + cos 2x)/2 to lower the powers first.

Worked example. Evaluate integral sin3(x) cos2(x) dx.

1. The power of sine is 3, which is odd, so split off one sine: sin3 = sin2 * sin.
2. Rewrite the sin² using the identity: sin2 = 1 - cos2. The integral becomes integral (1 - cos2x) cos2x * sin(x) dx.
3. Let u = cos(x), so du = -sin(x) dx. The leftover sin(x) dx becomes -du.
4. Substitute: -integral (1 - u2) u2 du = -integral (u2 - u4) du.
5. Integrate: -(u3/3 - u5/5).
6. Rename u back to cos(x): -cos3(x)/3 + cos5(x)/5 + C.

Verify: d/dx[-cos3x/3] = -3cos2x*(-sin x)/3 = cos2x sin x, and d/dx[cos5x/5] = 5cos4x*(-sin x)/5 = -cos4x sin x. Their sum is cos2x sin x (1 - cos2x) = cos2x sin x * sin2x = sin3x cos2x. That is the integrand. Correct.

Worked example: both powers even

Evaluate integral sin2(x) cos2(x) dx. Nothing is odd, so there is no factor to peel off and the half-angle route is forced.

1. Group the product as a square: sin2x cos2x = (sin x cos x)2.
2. Use the double-angle identity sin(2x) = 2 sin x cos x, so sin x cos x = sin(2x)/2 and the integrand becomes sin2(2x)/4.
3. Lower the remaining even power with sin2(A) = (1 - cos 2A)/2, taking A = 2x: sin2(2x) = (1 - cos(4x))/2. The integrand is now (1 - cos 4x)/8.
4. Integrate term by term: (1/8)*[x - sin(4x)/4] = x/8 - sin(4x)/32 + C.

Verify: d/dx[x/8 - sin(4x)/32] = 1/8 - 4cos(4x)/32 = (1 - cos 4x)/8, which is what we showed the integrand equals. Correct. Notice the pattern: every even power costs one application of a half-angle identity, and each application doubles the angle inside.

A quick definite version worth remembering: integral from 0 to pi of sin2(x) dx = integral from 0 to pi of (1 - cos 2x)/2 dx = [x/2 - sin(2x)/4] from 0 to pi = pi/2. The average height of sin2 over a half period is exactly 1/2, which is why so many physics formulas carry a factor of one half.

Key idea: odd power means peel and substitute; all even means half-angle identities until something is odd or linear.

Trigonometric substitution

When an integrand hides a square root of a quadratic, a trig substitution replaces x with a trig function so a Pythagorean identity collapses the root. Here are the three patterns.

You see	Substitute	Identity that fires
square root of a2 - x2	x = a sin(theta)	1 - sin2 = cos2
square root of a2 + x2	x = a tan(theta)	1 + tan2 = sec2
square root of x2 - a2	x = a sec(theta)	sec2 - 1 = tan2

Worked example. Evaluate integral dx / (square root of 4 - x2).

1. Match the pattern: 4 - x² is a² - x² with a = 2. So let x = 2 sin(theta).
2. Differentiate: dx = 2 cos(theta) d(theta).
3. Simplify the root: square root of (4 - 4 sin2 theta) = square root of 4(1 - sin2 theta) = 2 cos(theta).
4. Put it together: integral (2 cos theta)/(2 cos theta) d(theta) = integral 1 d(theta) = theta + C.
5. Rename back. Since x = 2 sin(theta), we have sin(theta) = x/2, so theta = arcsin(x/2).

The answer is arcsin(x/2) + C, confirming the standard formula integral dx / (square root of a2 - x2) = arcsin(x/a) + C. Verify: d/dx[arcsin(x/2)] = (1/2)/(the square root of (1 - x2/4)), and since the square root of (1 - x2/4) = (the square root of (4 - x2))/2, that equals 1/(the square root of (4 - x2)). Correct.

How to recognise a trig substitution problem, and when NOT to use it

The signal is a square root of a quadratic - or a quadratic raised to any half-integer power such as (x2 + 9)3/2 - that plain substitution cannot touch. The test that separates the two techniques is short:

• integral x/(the square root of (9 - x2)) dx: the derivative of the inside is -2x and an x is sitting right there. Use plain substitution with u = 9 - x2. Trig substitution would also work but wastes several lines.
• integral (the square root of (9 - x2)) dx: no x factor is available to serve as du. Substitution is dead, so use x = 3 sin(theta).

So: look for a lone x factor first. If it is there, substitute. If it is absent, or if the root sits in a denominator with extra powers of x, go trigonometric. And always match the pattern by sign and order - a2 - x2 uses sine, a2 + x2 uses tangent, x2 - a2 uses secant - because getting this backwards produces an imaginary root and no warning.

Second worked trig substitution: the tangent pattern

Evaluate integral dx/(x2 + 9)3/2. There is no lone x, and the base is a2 + x2 with a = 3.

1. Substitute x = 3 tan(theta), so dx = 3 sec2(theta) d(theta).
2. Simplify the base: x2 + 9 = 9 tan2(theta) + 9 = 9(tan2 + 1) = 9 sec2(theta).
3. Raise it to the 3/2 power: (9 sec2)3/2 = 27 sec3(theta). (Take the 3/2 power of 9 to get 27, and of sec² to get sec³.)
4. Assemble: integral [3 sec2(theta)]/[27 sec3(theta)] d(theta) = (1/9) integral d(theta)/sec(theta) = (1/9) integral cos(theta) d(theta).
5. Integrate: (1/9) sin(theta) + C.
6. Convert back with a reference triangle. From tan(theta) = x/3, draw a right triangle with opposite side x and adjacent side 3; the hypotenuse is the square root of (x2 + 9). Then sin(theta) = opposite/hypotenuse = x/(the square root of (x2 + 9)).
7. Final answer: x/(9 * the square root of (x2 + 9)) + C.

Verify by the quotient rule: differentiating x/(9(x2+9)1/2) gives (1/9)*[(x2+9)1/2 - x2(x2+9)-1/2]/(x2+9). The bracket is [(x2+9) - x2]/(x2+9)1/2 = 9/(x2+9)1/2, so the whole thing collapses to 1/(x2+9)3/2. Correct.

Third worked trig substitution, with a geometric check

Evaluate integral (the square root of (9 - x2)) dx.

1. Pattern a2 - x2 with a = 3, so x = 3 sin(theta) and dx = 3 cos(theta) d(theta).
2. The root collapses: the square root of (9 - 9sin2theta) = 3 cos(theta).
3. Assemble: integral 3cos(theta) * 3cos(theta) d(theta) = 9 integral cos2(theta) d(theta).
4. Both powers even, so half-angle: 9 integral (1 + cos 2theta)/2 d(theta) = (9/2)theta + (9/4)sin(2theta) + C.
5. Convert back. theta = arcsin(x/3). For the second term use sin(2theta) = 2 sin(theta)cos(theta) = 2*(x/3)*((the square root of (9 - x2))/3) = 2x(the square root of (9 - x2))/9, so (9/4)sin(2theta) = x(the square root of (9 - x2))/2.
6. Final answer: (9/2)arcsin(x/3) + x(the square root of (9 - x2))/2 + C.

Two checks. Differentiating gives (9/2)/(the square root of (9-x2)) + (9 - 2x2)/(2 * the square root of (9-x2)), which combines to (18 - 2x2)/(2 * the square root of (9-x2)) = the square root of (9 - x2). Correct. And geometrically, y = the square root of (9 - x2) is the upper half of a circle of radius 3, so the integral from -3 to 3 must equal the semicircle area 9pi/2. The formula gives (9/2)arcsin(1) - (9/2)arcsin(-1) = (9/2)(pi/2) + (9/2)(pi/2) = 9pi/2. It matches exactly.

Completing the square first

A quadratic that is not already in one of the three shapes often becomes one after completing the square. Evaluate integral dx/(the square root of (x2 + 6x + 13)).

1. Complete the square: x2 + 6x + 13 = (x2 + 6x + 9) + 4 = (x + 3)2 + 4.
2. Shift with w = x + 3, dw = dx. The integral becomes integral dw/(the square root of (w2 + 4)), the tangent pattern with a = 2.
3. That standard form integrates to ln|w + the square root of (w2 + 4)| + C.
4. Undo the shift: ln|x + 3 + the square root of (x2 + 6x + 13)| + C.

Verify: the derivative of the log is [1 + (x+3)/R]/(x + 3 + R) where R = the square root of (x2+6x+13). Writing the numerator over R gives (R + x + 3)/R, and the (x + 3 + R) cancels, leaving 1/R. Correct.

Key idea: complete the square whenever the quadratic has a linear term; the shift turns it into one of the three memorised patterns.

Try it

Evaluate integral cos3(x) dx.

Answer: the power of cosine is odd, so write cos3 = (1 - sin2) cos. Let u = sin(x), so du = cos(x) dx. The integral becomes integral (1 - u2) du = u - u3/3 = sin(x) - sin3(x)/3 + C. Nice work matching the odd-power strategy.

Where people get stuck

• Not asking "is either power odd?" first. That single question decides the whole strategy. Odd means peel one factor and substitute; only all-even sends you to the half-angle identities, which take three times as long.
• Peeling from the wrong function. You peel a factor off the function with the odd power and then substitute the other one. In sin3x cos2x you peel a sine and set u = cos x. Peeling a cosine here leaves sin3x cos x with an odd sine power still stuck in the way.
• Losing the minus sign from du. u = cos(x) gives du = -sin(x) dx. That minus propagates through every term of the answer, and dropping it flips the sign of the entire result.
• Grabbing the wrong trig-substitution pattern. Match by sign and order: a2 - x2 uses sine, a2 + x2 uses tangent, x2 - a2 uses secant. Using tangent on 4 - x2 produces the square root of (4 - 4tan2), which no identity simplifies.
• Using trig substitution when plain substitution would do. If a lone factor of x sits outside the root, u = the inside finishes it in two lines. Check for that first.
• Forgetting to convert dx. When x = 3 sin(theta), you must also replace dx with 3 cos(theta) d(theta). Substituting only inside the root and leaving dx alone is the single most common trig-substitution error.
• Leaving the answer in theta. An indefinite integral must come back to x. Draw the reference triangle from the substitution equation - opposite, adjacent, hypotenuse - and read off whichever trig function you need. Do not try to remember conversion formulas; the triangle regenerates them.
• Assuming the root has to be visible. (x2 + 9)3/2 is a root in disguise, and x2 + 6x + 13 becomes one of the three patterns only after completing the square.

Recap

• Odd power of sine: peel one sine, use sin2 = 1 - cos2, let u = cos(x). Odd power of cosine: mirror it with u = sin(x).
• Both powers even: lower them with the half-angle identities sin2x = (1 - cos 2x)/2 and cos2x = (1 + cos 2x)/2, repeating as needed. Worked result: integral sin2x cos2x dx = x/8 - sin(4x)/32 + C.
• Trig substitution: a2 - x2 uses a sin(theta), a2 + x2 uses a tan(theta), x2 - a2 uses a sec(theta).
• Use it only when no lone x factor is available for a plain substitution; the presence of that factor is the whole test.
• Worked results to know: integral dx/(x2+9)3/2 = x/(9 * the square root of (x2+9)) + C, and integral (the square root of (9 - x2)) dx = (9/2)arcsin(x/3) + x(the square root of (9 - x2))/2 + C, whose value from -3 to 3 is the semicircle area 9pi/2.
• Complete the square when the quadratic carries a linear term, then shift to reach a standard pattern.
• Always convert the final answer back to the original variable x, using a reference right triangle.

Sources

1. OpenStax. (2016). Trigonometric integrals. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Trigonometric substitution. In Calculus volume 2. Rice University. openstax.org
3. OpenStax. (2016). Integrals resulting in inverse trigonometric functions. In Calculus volume 2. Rice University. openstax.org
4. Strang, G., & Herman, E. (n.d.). 7.2: Trigonometric integrals. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
5. Strang, G., & Herman, E. (n.d.). 7.3: Trigonometric substitution. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
6. Dawkins, P. (n.d.). Integrals involving trig functions. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
7. Dawkins, P. (n.d.). Trig substitutions. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu

You already know how to add two fractions by finding a common denominator. Partial fractions runs that process backward: it takes a single complicated fraction and un-mixes it into a sum of simple ones. Why bother? Because each simple piece is easy to integrate, while the original lump is not.

The big picture

A rational function is a ratio of two polynomials, like (5x - 4)/(x2 - x - 2). On its own it is hard to integrate. But split into 2/(x - 2) + 3/(x + 1), each piece integrates to a logarithm in one line. Partial fractions is the un-mixing step that makes that possible.

Key idea: break a rational function into a sum of simpler fractions, then integrate each piece, usually into logarithms and arctangents.

Two setup rules before you start

1. The fraction must be proper: the top degree must be less than the bottom degree. If it is not, do polynomial long division first, and work on the leftover proper piece.
2. Factor the bottom completely. Then write one partial fraction for each factor, each with an unknown constant on top. For two different linear factors: (top)/((x - r1)(x - r2)) = A/(x - r1) + B/(x - r2).

Worked example, two different linear factors

Evaluate integral (5x - 4)/(x2 - x - 2) dx.

1. Factor the bottom: x2 - x - 2 = (x - 2)(x + 1).
2. Set up the split with unknowns: (5x - 4)/((x - 2)(x + 1)) = A/(x - 2) + B/(x + 1).
3. Multiply both sides by the bottom to clear fractions: 5x - 4 = A(x + 1) + B(x - 2).
4. Substitute a root to isolate one unknown. Let x = 2: 5*2 - 4 = A(3), so 6 = 3A and A = 2.
5. Substitute the other root. Let x = -1: 5*(-1) - 4 = B(-3), so -9 = -3B and B = 3.
6. Integrate each simple piece: integral [2/(x - 2) + 3/(x + 1)] dx = 2 ln|x - 2| + 3 ln|x + 1| + C.

The trick in steps 4 and 5, plugging in one root at a time, makes the other term vanish, so each unknown pops out immediately. Read that once more: choosing x to be a root zeroes out one factor and hands you the other constant.

Verify the whole answer by differentiating: d/dx[2 ln|x - 2| + 3 ln|x + 1|] = 2/(x - 2) + 3/(x + 1). Put that over the common denominator: [2(x + 1) + 3(x - 2)]/((x-2)(x+1)) = (2x + 2 + 3x - 6)/(x2 - x - 2) = (5x - 4)/(x2 - x - 2). That is the original integrand. Correct.

Key idea: substituting each root of the denominator kills every term but one, which is why linear factors are almost free.

How to recognise a partial-fractions problem

The signal is unambiguous: the integrand is a rational function, a polynomial divided by a polynomial, and plain substitution does not apply because the numerator is not the derivative of the denominator up to a constant. Two quick sub-tests:

• integral (2x)/(x2 + 1) dx: the numerator is the derivative of the denominator. Substitute; do not decompose.
• integral (5x - 4)/(x2 - x - 2) dx: the derivative of the bottom is 2x - 1, which is not a multiple of 5x - 4. Decompose.

The technique always terminates, because every real polynomial factors into linear and irreducible-quadratic pieces, and each of those pieces has a known antiderivative. That is a genuine theorem, not a hope: partial fractions gives an elementary antiderivative for every rational function.

Case one: an improper fraction needs division first

Evaluate integral x3/(x2 - 1) dx. The top has degree 3 and the bottom degree 2, so the fraction is improper and no decomposition is legal yet.

1. Divide. Ask what times x2 - 1 approximates x3: the answer is x, and x(x2 - 1) = x3 - x. Subtracting leaves a remainder of x.
2. Rewrite: x3/(x2 - 1) = x + x/(x2 - 1). The leftover is now proper.
3. Decompose the leftover: x/((x-1)(x+1)) = A/(x - 1) + B/(x + 1), so x = A(x + 1) + B(x - 1). Let x = 1: 1 = 2A, so A = 1/2. Let x = -1: -1 = -2B, so B = 1/2.
4. Integrate everything: integral x dx + (1/2) integral dx/(x-1) + (1/2) integral dx/(x+1) = x2/2 + (1/2)ln|x - 1| + (1/2)ln|x + 1| + C.
5. Combine the logs: x2/2 + (1/2)ln|x2 - 1| + C.

Verify: d/dx[x2/2] = x and d/dx[(1/2)ln|x2-1|] = x/(x2-1). Their sum is [x(x2-1) + x]/(x2-1) = x3/(x2-1). Correct. Skipping the division here would have given a decomposition with no solution, and there is no error message to warn you.

Case two: a repeated linear factor

Evaluate integral (x2 + 2x + 3)/((x - 1)(x + 1)2) dx. The factor (x + 1) appears squared, so it earns two terms, one for each power.

1. Set up: (x2+2x+3)/((x-1)(x+1)2) = A/(x - 1) + B/(x + 1) + C/(x + 1)2.
2. Clear denominators: x2 + 2x + 3 = A(x + 1)2 + B(x - 1)(x + 1) + C(x - 1).
3. Let x = 1: the B and C terms vanish, leaving 1 + 2 + 3 = A(4), so A = 3/2.
4. Let x = -1: the A and B terms vanish, leaving 1 - 2 + 3 = C(-2), so 2 = -2C and C = -1.
5. The roots are used up, so match a coefficient for B. The x2 coefficient on the right is A + B, and on the left it is 1, so B = 1 - 3/2 = -1/2.
6. Check with the constant term: A(1) + B(-1) + C(-1) = 3/2 + 1/2 + 1 = 3, matching the left side. The constants are right.
7. Integrate: (3/2)ln|x - 1| - (1/2)ln|x + 1| + integral (-1)(x+1)-2 dx = (3/2)ln|x - 1| - (1/2)ln|x + 1| + 1/(x + 1) + C.

Note that the squared term did not become a logarithm - (x+1)-2 is an ordinary power, integrating to -(x+1)-1, and the leading -1 makes it +1/(x+1). Verify by recombining the derivative over (x-1)(x+1)2: the numerator is (3/2)(x+1)2 - (1/2)(x-1)(x+1) - (x - 1) = (1.5x2 + 3x + 1.5) - (0.5x2 - 0.5) - x + 1 = x2 + 2x + 3. Correct.

Case three: an irreducible quadratic factor

Evaluate integral (2x + 1)/(x(x2 + 1)) dx. The factor x2 + 1 has no real roots, so it is irreducible and takes a full linear numerator.

1. Set up: (2x + 1)/(x(x2+1)) = A/x + (Bx + C)/(x2 + 1).
2. Clear denominators: 2x + 1 = A(x2 + 1) + (Bx + C)x = (A + B)x2 + Cx + A.
3. Match coefficients. Constant: A = 1. Coefficient of x: C = 2. Coefficient of x²: A + B = 0, so B = -1.
4. Rewrite: integral [1/x + (-x + 2)/(x2 + 1)] dx. Split the second fraction into the two building blocks below.
5. Integrate: ln|x| - (1/2)ln(x2 + 1) + 2 arctan(x) + C.

Verify: 1/x - x/(x2+1) + 2/(x2+1) = 1/x + (2 - x)/(x2+1). Over the common denominator: [(x2 + 1) + x(2 - x)]/(x(x2+1)) = (2x + 1)/(x(x2+1)). Correct.

Two building blocks make every irreducible-quadratic case routine: integral dx/(x2 + 1) = arctan(x) + C, and integral x dx/(x2 + 1) = (1/2)ln(x2 + 1) + C by substitution. Split any (Bx + C)/(x2+1) into a multiple of each. When the quadratic carries a linear term, complete the square first and shift, exactly as in the last lesson.

Key idea: one term per power of a repeated factor, and a linear numerator over every irreducible quadratic.

Which technique do I reach for?

You now own the whole standard toolkit. Run this checklist in order; the first match is almost always right.

1. Simplify algebraically first. Expand a product, split a fraction term by term, cancel a common factor, or apply a trig identity. integral (x2 + 1)/x dx is not a partial-fractions problem; it is integral (x + 1/x) dx.
2. Is it already a known form? Powers, ekx, 1/x, basic trig, 1/(1 + x2), 1/(the square root of (1 - x2)). Write the answer down.
3. Substitution. Is there a composite whose inside's derivative appears as a factor, up to a constant? This is the cheapest technique, so always test it before anything else.
4. Integration by parts. A product of unrelated families - polynomial times exponential, trig, or log - or a lone logarithm or inverse trig function.
5. Rational function? Divide if improper, factor the denominator, then partial fractions.
6. Powers of trig functions? Odd power means peel and substitute; all even means half-angle identities.
7. Square root of a quadratic with no lone x outside? Trig substitution, completing the square first if there is a linear term.
8. Still stuck? Try a clever rewrite - multiply by a conjugate, add and subtract a term - or accept that some integrands such as e-x2, sin(x)/x, and the square root of (1 + x4) have no elementary antiderivative at all, and go numerical.

Two habits make the checklist reliable. First, spend ten seconds classifying before writing anything; most wasted pages come from starting on a technique that was never going to apply. Second, differentiate your answer. Every technique in this module is verifiable in one line, and a verified answer needs no second opinion.

Try it

Decompose 1/((x - 1)(x + 1)) into partial fractions.

Answer: write A/(x - 1) + B/(x + 1), so 1 = A(x + 1) + B(x - 1). Let x = 1: 1 = 2A, so A = 1/2. Let x = -1: 1 = -2B, so B = -1/2. The decomposition is (1/2)/(x - 1) - (1/2)/(x + 1). Nice work using the root-substitution trick.

Where people get stuck

• Skipping the proper-fraction check. This is the top mistake. If the top degree is not strictly smaller than the bottom, divide first. Attempting the decomposition anyway produces a system with no solution, and nothing tells you why.
• Giving a squared factor only one term. (x - r)2 contributes both A/(x - r) and B/(x - r)2, and a cubed factor contributes three terms. Count the powers and give each its own fraction, or the system will be short one unknown.
• Putting a constant over an irreducible quadratic. It needs a full linear numerator (Bx + C). Using just B loses a degree of freedom and the system becomes inconsistent.
• Running out of roots and stopping. Substituting the roots gives you only as many equations as there are distinct roots. With repeated or quadratic factors you must finish by matching coefficients of like powers, as in case two.
• Expecting every piece to become a logarithm. Distinct linear factors give logs, but repeated factors give ordinary powers like 1/(x + 1), and irreducible quadratics give an arctangent as well as a log.
• Dropping the absolute value. Write ln|x - 2|, not ln(x - 2). The bars are what make the antiderivative valid on both sides of the root.
• Factoring the denominator incompletely. x4 - 1 is not (x2-1)(x2+1) for these purposes; push on to (x-1)(x+1)(x2+1). A factor you leave unfactored will not decompose.

Recap

• Make the fraction proper (divide first if needed), then factor the denominator completely into linear and irreducible-quadratic pieces.
• Write one fraction per factor with unknown tops, clear denominators, and solve - by plugging in roots where you can and matching coefficients where you cannot.
• Repeated factors need one term per power; irreducible quadratics take a linear numerator Ax + B.
• Worked results: integral (5x-4)/(x2-x-2) dx = 2ln|x-2| + 3ln|x+1| + C; integral x3/(x2-1) dx = x2/2 + (1/2)ln|x2-1| + C; integral (2x+1)/(x(x2+1)) dx = ln|x| - (1/2)ln(x2+1) + 2arctan(x) + C.
• Integrating the pieces produces logarithms, ordinary powers, and arctangents. Every rational function has an elementary antiderivative for exactly this reason.
• Use the eight-step checklist to classify an integral before you start, and differentiate your answer before you finish.

Sources

1. OpenStax. (2016). Partial fractions. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Other strategies for integration. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 7.4: Partial fractions. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Strang, G., & Herman, E. (n.d.). 7.5: Other strategies for integration. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
5. Dawkins, P. (n.d.). Partial fractions. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Dawkins, P. (n.d.). Integration strategy. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
7. Dawkins, P. (n.d.). Integration techniques (introduction). Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu

Here is a question that sounds impossible at first: can a region that stretches out forever still have a finite area? Sometimes yes. An improper integral is how we make sense of that. The definite integral needed a bounded function on a bounded interval. An improper integral relaxes those conditions, and a limit does the careful work of deciding whether the total is finite.

The big picture

There are two ways an integral becomes improper: the interval runs to infinity, or the function shoots up to infinity somewhere inside. In both cases we do not just plug in infinity. We integrate up to a movable stopping point, then take a limit. If that limit is a finite number, the integral converges. If the limit is infinite or does not exist, the integral diverges.

Key idea: an improper integral is a limit of ordinary integrals; a finite limit means it converges, otherwise it diverges.

Type 1: the interval runs to infinity

We define integral from a to infinity of f(x) dx = lim(t -> infinity) integral from a to t of f(x) dx. Integrate up to t, then let t march off to infinity.

Worked example that converges. Evaluate integral from 1 to infinity of 1/x2 dx.

1. Integrate up to t: integral from 1 to t of x-2 dx = [-1/x] from 1 to t = -1/t + 1.
2. Take the limit as t -> infinity: the term -1/t goes to 0, leaving 1 - 0 = 1.

The integral converges to 1. A region stretching forever to the right can still enclose a finite area of 1.

Worked example that diverges. Evaluate integral from 1 to infinity of 1/x dx.

1. Integrate up to t: [ln x] from 1 to t = ln t - 0 = ln t.
2. Take the limit as t -> infinity: ln t grows without bound, so the limit is infinite.

This integral diverges. The graphs of 1/x and 1/x² look almost the same, yet one traps finite area and the other does not. The power matters.

The p-integral, derived in general

Since the exponent decides everything, let us settle every case at once. Consider integral from 1 to infinity of x-p dx for a positive constant p.

1. Case p is not 1. The power rule applies: integral from 1 to t of x-p dx = [x1-p/(1 - p)] from 1 to t = (t1-p - 1)/(1 - p).
2. Now watch the exponent 1 - p. If p > 1 then 1 - p is negative, so t1-p = 1/tp-1 -> 0, and the limit is (0 - 1)/(1 - p) = 1/(p - 1). Finite: converges.
3. If p < 1 then 1 - p is positive, so t1-p -> infinity and the limit is infinite: diverges.
4. Case p = 1. The power rule breaks down (it would divide by zero), and the antiderivative is ln(t), which grows without bound: diverges.

So integral from 1 to infinity of dx/xp converges exactly when p > 1, with value 1/(p - 1). Check it against the cases we did by hand: at p = 2 the formula gives 1/(2-1) = 1, and at p = 3 it gives 1/(3-1) = 1/2. Both match. Notice that p = 1 is not merely a boundary case to memorise - it is the exact point where the antiderivative changes species, from a power to a logarithm, and a logarithm grows slower than any positive power but still without bound.

Key idea: at infinity, the tail must decay faster than 1/x for the area to be finite, and 1/x itself is the borderline that fails.

Type 2: the function blows up inside

If f shoots up to infinity at an endpoint, say at x = 0, we define integral from 0 to 1 of f dx = lim(t -> 0+) integral from t to 1 of f dx, starting just to the right of the trouble spot and sliding toward it.

Worked example that converges. Evaluate integral from 0 to 1 of 1/(the square root of x) dx.

1. Note the trouble: at x = 0 the integrand is undefined and grows without bound, so this is improper at the left endpoint.
2. Cut short: integral from t to 1 of x-1/2 dx = [2x1/2] from t to 1 = 2 - 2(the square root of t).
3. Let t -> 0+: the square root of t goes to 0, so the limit is 2.

It converges to 2, even though the integrand is infinite at 0 - an infinitely tall spike can still enclose finite area if it is thin enough fast enough.

Worked example that diverges. Evaluate integral from 0 to 1 of dx/x. Cut short: [ln|x|] from t to 1 = ln(1) - ln(t) = -ln(t). As t -> 0+, ln(t) -> -infinity, so -ln(t) -> +infinity. It diverges.

The same power-rule argument as before, run on integral from 0 to 1 of x-p dx, gives (1 - t1-p)/(1 - p), which converges exactly when p < 1, with value 1/(1 - p). Check against the worked example: p = 1/2 gives 1/(1 - 1/2) = 2. Correct.

Put the two rules side by side, because the reversal catches almost everyone:

Integral	Converges when	Value
integral from 1 to infinity of dx/xp	p > 1	1/(p - 1)
integral from 0 to 1 of dx/xp	p < 1	1/(1 - p)

The logic behind the reversal is worth saying plainly. Far out at infinity, a large p makes the function tiny, so the tail is thin and the area is finite. Near zero, a large p makes the function enormous, so the spike is fat and the area blows up. Same p, opposite effect, because x-p is small when x is large and large when x is small.

The trap: a blow-up in the middle

An integrand can be improper at an interior point with nothing in the notation to warn you. Consider integral from -1 to 1 of dx/x2, the example that produced the impossible answer -2 back in Lesson 2.

1. Spot the trouble: the integrand blows up at x = 0, which is inside [-1, 1].
2. Split at the trouble spot: the integral is integral from -1 to 0 of dx/x2 + integral from 0 to 1 of dx/x2, and the whole thing converges only if both pieces do.
3. Test the right piece: it is integral from 0 to 1 of x-p dx with p = 2, and 2 is not less than 1, so it diverges.
4. One divergent piece is enough. The original integral diverges, and the mechanical answer -2 was meaningless.

The habit that prevents this: before evaluating any definite integral, ask where the integrand is undefined, and check whether any such point lies in the interval. Denominators that vanish, logarithms of zero, and tangents at odd multiples of pi/2 are the usual suspects. One caution about splitting: you must let each piece take its limit independently. Letting both sides approach zero at the same rate would give 0 for integral from -1 to 1 of dx/x, which is called the principal value and is not the same thing as convergence.

Key idea: improperness at an interior point is invisible in the notation; find it by inspecting the integrand's domain, then split and test each piece separately.

The comparison test

Sometimes you cannot find an antiderivative, but you can still decide convergence by comparing with a simpler function. If 0 <= f(x) <= g(x), then:

• If the bigger one, integral g, converges, then the smaller integral f converges too (it is trapped underneath something finite).
• If the smaller one, integral f, diverges, then the bigger integral g diverges too.

For instance, integral from 1 to infinity of 1/(x2 + 1) dx converges, because 1/(x2 + 1) <= 1/x2 and the 1/x² integral already converged.

Worked comparison where no antiderivative exists. Decide whether integral from 1 to infinity of e-x2 dx converges. This integrand has no elementary antiderivative at all, so evaluation is off the table and comparison is the only route.

1. Find a bound. For x >= 1 we have x2 >= x, so -x2 <= -x, and since the exponential is increasing, e-x2 <= e-x. Both functions are positive, so the setup of the test is satisfied.
2. Evaluate the bigger one: integral from 1 to t of e-x dx = [-e-x] from 1 to t = e-1 - e-t, and as t -> infinity this tends to 1/e. It converges.
3. Apply the test: the smaller function is squeezed under something with finite area, so integral from 1 to infinity of e-x2 dx converges by the comparison test.

Note what was and was not proved. We showed the value is finite and at most 1/e ~= 0.368; we did not find it. (Its true value is about 0.1394.) Comparison answers the yes-or-no question only, and that is usually the question that matters.

The limit comparison test

Direct comparison needs a clean inequality, which is sometimes fiddly to arrange. The limit comparison test avoids that: if f and g are positive and lim(x -> infinity) f(x)/g(x) = L with 0 < L < infinity, then the two integrals either both converge or both diverge. A finite nonzero ratio means the two functions decay at the same rate, and decay rate is the only thing convergence cares about.

Worked example. Decide whether integral from 1 to infinity of (x + 1)/(x3 + 2) dx converges.

1. Identify the dominant behaviour. For large x the top behaves like x and the bottom like x3, so the whole thing behaves like 1/x2. Take g(x) = 1/x2.
2. Form the ratio: f/g = [(x + 1)/(x3 + 2)] * x2 = (x3 + x2)/(x3 + 2).
3. Take the limit by dividing top and bottom by x3: (1 + 1/x)/(1 + 2/x3) -> 1. The limit is 1, which is finite and positive.
4. Since integral from 1 to infinity of dx/x2 converges (p = 2 > 1), the given integral converges by the limit comparison test.

The practical recipe: strip the integrand down to its highest power on top over its highest power on bottom, and compare with that. It converts almost every rational-looking convergence question into a p-integral lookup.

Why anyone cares: a horn you can fill but not paint

Rotate the curve y = 1/x for x >= 1 about the x-axis and you get a trumpet shape called Gabriel's horn. Its volume, by the disk method of Lesson 10, is pi * integral from 1 to infinity of (1/x)2 dx = pi * 1 = pi, a finite number. Its surface area involves 2pi * integral from 1 to infinity of (1/x) * the square root of (1 + 1/x4) dx, and since the square root is at least 1, that integral is at least 2pi * integral from 1 to infinity of dx/x, which diverges. So the horn holds pi cubic units of paint but has infinite surface area. The paradox dissolves once you notice that "painting" a mathematical surface means covering it with zero thickness, while filling it means something else entirely - but the calculation is exactly the p-integral threshold at work, with p = 2 on one side of the line and p = 1 on the other.

Improper integrals are not a curiosity. Probability densities are integrals over infinite ranges that must converge to 1, the Laplace and Fourier transforms are improper integrals, and the entire next module - infinite series - runs on the same convergence machinery, with the integral test in Lesson 15 tying the two together directly.

Try it

Determine whether integral from 1 to infinity of 1/x3 dx converges.

Answer: integrate to t: [-1/(2x2)] from 1 to t = 1/2 - 1/(2t2). As t -> infinity the second term goes to 0, leaving 1/2. It converges to 1/2. This also matches the p-integral rule, since p = 3 is greater than 1. Nice work.

Where people get stuck

• Plugging infinity into the antiderivative. Infinity is a limit, not a number. Writing [-1/x] from 1 to infinity = 0 + 1 happens to give the right answer here and trains a habit that fails elsewhere. Introduce t, integrate, then take the limit - every time.
• Missing an interior blow-up. Nothing in integral from -1 to 1 of dx/x2 announces that it is improper, and turning the crank gives -2 for a positive integrand. Check the integrand's domain against the interval before you evaluate.
• Mixing up the two p-thresholds. At infinity you need p > 1; at zero you need p < 1. They point opposite ways for a reason: a big p makes the function small far out and huge near zero.
• Reversing the comparison test. Only two conclusions are available: a convergent upper bound forces convergence, and a divergent lower bound forces divergence. Showing your function is smaller than something divergent proves nothing at all, and neither does showing it is bigger than something convergent.
• Using a comparison function that is not positive. Both comparison tests require 0 <= f <= g. On an integrand that changes sign, cancellation can produce convergence that no size comparison would predict.
• Reporting a value from a comparison test. Comparison decides convergence, not the number. Our bound showed integral e-x2 is at most 1/e; the actual value is about 0.139, and the test never claimed otherwise.
• "Infinite extent means infinite area." Not so. 1/x2 over an infinite interval gives area 1, and 1/(the square root of x) under an infinite spike gives area 2. Extent is not the issue; decay rate is.
• Letting both halves approach the singularity together. Each side of an interior blow-up must take its own independent limit. Coupling them computes a principal value, which can be finite even when the integral genuinely diverges.

Recap

• An improper integral is defined as a limit: integrate to a finite endpoint, then take the limit. A finite limit means convergence; an infinite or nonexistent limit means divergence.
• Type 1 is an infinite interval; Type 2 is an infinite integrand. An integrand that blows up inside the interval must be split at the blow-up, and every piece must converge.
• integral from 1 to infinity of dx/xp converges exactly when p > 1, with value 1/(p - 1). integral from 0 to 1 of dx/xp converges exactly when p < 1, with value 1/(1 - p).
• The direct comparison test decides convergence without an antiderivative: a convergent upper bound drags the smaller function down, a divergent lower bound pushes the bigger one up.
• The limit comparison test needs only a finite nonzero ratio; strip the integrand to its dominant power and compare with a p-integral.
• integral from 1 to infinity of e-x2 dx converges by comparison with e-x, even though it has no elementary antiderivative.
• Every convergence conclusion should be stated with the test that produced it: "converges by the p-test with p = 2," "converges by limit comparison with 1/x²."

Sources

1. OpenStax. (2016). Improper integrals. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Numerical integration. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 7.7: Improper integrals. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). Improper integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Dawkins, P. (n.d.). Comparison test for improper integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Guichard, D. (n.d.). Calculus: Early transcendentals. Whitman College. whitman.edu
7. Tall, D., & Vinner, S. (1981). Concept image and concept definition in mathematics with particular reference to limits and continuity. Educational Studies in Mathematics, 12(2), 151-169. doi.org/10.1007/BF00305619

A single integral finds the area under one curve. What if you want the area of the region trapped between two curves, like the sliver between a line and a parabola? The idea is refreshingly simple: it is the area under the top curve minus the area under the bottom curve. One integral does both at once.

The big picture

If the curve f(x) sits on top of g(x) across an interval, the region between them has area equal to the integral of the gap, top minus bottom.

A = integral from a to b of [f(x) - g(x)] dx

The limits a and b are almost always the x-values where the curves cross, which you find by setting the two functions equal. Subtracting bottom from top keeps the gap positive everywhere, so the area comes out positive.

Why the integrand is top minus bottom

Every application in this module is built the same way, and it is worth doing the construction once, slowly, because the pattern repeats for volumes and arc length.

1. Slice. Cut the region with a thin vertical strip at position x. This strip is called the representative rectangle, and everything depends on describing it correctly.
2. Measure one slice. Its width is dx. Its height reaches from the lower boundary up to the upper boundary, so the height is f(x) - g(x).
3. Write its area. Height times width: [f(x) - g(x)] dx.
4. Add and take the limit. Summing over all strips and letting the width shrink turns the sum into integral from a to b of [f(x) - g(x)] dx.

That is the whole derivation, and it explains details that otherwise look arbitrary. The subtraction is there because a height is a difference of two positions, not because of a rule about areas. The order is top minus bottom because heights are positive. And the strip is vertical - width dx - precisely because we chose to integrate with respect to x; a horizontal strip would have thickness dy and force the other variable.

The construction also survives cases that trip people up. If the region dips below the x-axis, nothing changes: the height is still upper minus lower, and both may be negative numbers whose difference is positive. On f(x) = -1 and g(x) = -3 over [0, 2], the height is -1 - (-3) = 2 and the area is 4, exactly as the picture demands.

Key idea: the area between two curves is the integral of the top curve minus the bottom curve, over the interval where they overlap - and the integrand is a difference because it measures the height of one representative strip.

Worked example, step by step

Find the area enclosed between the line y = x + 2 and the parabola y = x2.

1. Find where they cross by setting them equal: x + 2 = x2, which rearranges to x2 - x - 2 = 0, and factors as (x - 2)(x + 1) = 0. So the crossings are x = -1 and x = 2. These are your limits.
2. Decide which curve is on top by testing one inside point. At x = 0 the line gives 2 and the parabola gives 0, so the line is on top.
3. Set up the integral, top minus bottom: A = integral from -1 to 2 of [(x + 2) - x2] dx.
4. Find the antiderivative: x2/2 + 2x - x3/3.
5. Evaluate at the top, x = 2: 2 + 4 - 8/3 = 6 - 8/3 = 10/3.
6. Evaluate at the bottom, x = -1: 1/2 - 2 + 1/3 = -7/6.
7. Subtract, top minus bottom: 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2.

The enclosed area is 9/2. Here is the region, with the line riding above the parabola between the two crossings.

Verify the antiderivative: d/dx[x2/2 + 2x - x3/3] = x + 2 - x2, which is top minus bottom. Correct. As a rough check, the region is roughly 3 units wide and at most about 2.25 units tall, so an area of 4.5 is plausible.

When the curves swap places inside the interval

If the two curves cross between your limits, one integral cannot describe the whole region, because "top" changes identity partway through. Find the area enclosed between y = x and y = x3.

1. Find all crossings: x = x3 gives x3 - x = 0, so x(x - 1)(x + 1) = 0 and the crossings are x = -1, 0, 1. Three crossings mean two separate regions.
2. Test each subinterval. At x = -0.5: x3 = -0.125 and x = -0.5, so the cubic is on top on (-1, 0). At x = 0.5: x3 = 0.125 and x = 0.5, so the line is on top on (0, 1).
3. Left piece: integral from -1 to 0 of (x3 - x) dx = [x4/4 - x2/2] from -1 to 0 = 0 - (1/4 - 1/2) = 1/4.
4. Right piece: integral from 0 to 1 of (x - x3) dx = [x2/2 - x4/4] from 0 to 1 = 1/2 - 1/4 = 1/4.
5. Total area: 1/4 + 1/4 = 1/2.

Now see what happens if you skip the split and write one integral from -1 to 1 of (x - x3): the antiderivative x2/2 - x4/4 takes the value 1/4 at both endpoints, so the answer comes out 0. The two halves cancelled, because on the left half the integrand was negative. Zero for a visibly non-empty region is the loudest possible warning, but a subtler example might have produced a merely wrong positive number with no warning at all. Always find every crossing, not just the outermost two.

Key idea: solve for all intersection points, split there, and integrate each piece with its own top and bottom.

Sometimes it is easier to integrate in y

When the curves are simpler to write as x in terms of y, for example sideways-opening parabolas, slice the region with horizontal strips instead and integrate right curve minus left curve: A = integral from c to d of [xright(y) - xleft(y)] dy. The principle is identical, just rotated: the strip now has thickness dy and length right-minus-left. Pick whichever direction keeps the boundaries as clean functions.

Worked example. Find the area enclosed by the line y = x - 1 and the sideways parabola y2 = 2x + 6.

1. Solve both boundaries for x: the line becomes x = y + 1, and the parabola becomes x = (y2 - 6)/2.
2. Find the crossings in y: y + 1 = (y2 - 6)/2. Multiply by 2: 2y + 2 = y2 - 6, so y2 - 2y - 8 = 0, which factors as (y - 4)(y + 2) = 0. The limits are y = -2 and y = 4.
3. Decide which boundary is on the right by testing an inside value, say y = 0: the line gives x = 1, the parabola gives x = -3. The line is on the right.
4. Set up: A = integral from -2 to 4 of [(y + 1) - (y2 - 6)/2] dy = integral from -2 to 4 of (-y2/2 + y + 4) dy.
5. Antiderivative: -y3/6 + y2/2 + 4y. Check it: d/dy = -y2/2 + y + 4. Correct.
6. At y = 4: -64/6 + 8 + 16 = -32/3 + 24 = 40/3. At y = -2: 8/6 + 2 - 8 = 4/3 - 6 = -14/3.
7. Subtract: 40/3 - (-14/3) = 54/3 = 18.

The area is 18. Now consider the cost of the other choice. Integrating in x, the parabola's upper and lower branches are y = the square root of (2x + 6) and y = -the square root of (2x + 6), and the region's lower boundary switches from the parabola's lower branch to the line at x = -1. You would need two integrals, two square roots, and a careful argument about where the switch happens - to reach the same 18. Choosing the slicing direction is a real decision, and the rule of thumb is simple: slice so that each strip enters and exits through the same pair of boundaries all the way across.

Key idea: pick the direction of slicing that lets a single integral describe the whole region.

Try it

Find the area between y = x and y = x2 where they enclose a region.

Answer: set them equal, x = x2, giving crossings at x = 0 and x = 1. Testing x = 1/2, the line 0.5 beats the parabola 0.25, so the line is on top. Then integral from 0 to 1 of (x - x2) dx = [x2/2 - x3/3] from 0 to 1 = 1/2 - 1/3 = 1/6. Nice work.

Where people get stuck

• Guessing which curve is on top. Always test one point strictly between the crossings and see which function is larger there. Guessing from a rough sketch fails exactly when the curves are close, which is when it matters.
• Missing an interior crossing. Solving f(x) = g(x) and keeping only the outermost roots is the error that turned a region of area 1/2 into an answer of 0 above. Find every root, then split at each one.
• Subtracting in the wrong order. A negative area is the signal to swap top and bottom - not to take an absolute value at the end, which would hide a genuine missing split.
• Skipping the search for limits. Without the intersection points there are no limits of integration. If a problem supplies limits explicitly, still check whether the curves cross inside them.
• Forgetting that both curves can be negative. Height is upper minus lower regardless of sign. A region entirely below the axis has a perfectly positive area.
• Slicing in the harder direction. If a vertical strip enters through one curve at some x-values and a different curve at others, switch to horizontal strips and integrate in y. The parabola example is 18 in one integral that way, and two messy integrals the other way.
• Mixing dx with functions of y. If your integrand is written in y, the differential must be dy and the limits must be y-values. Pairing xright(y) - xleft(y) with x-limits is a common and fatal mismatch.

Recap

• Area between curves is integral from a to b of [top - bottom] dx, and the integrand is a difference because it is the height of one representative strip of width dx.
• Find the limits by setting the curves equal, then test a point in each subinterval to see which curve is on top.
• If the curves cross inside the interval, split there and integrate each piece separately; a single integral would let the pieces cancel.
• Worked results: line y = x + 2 above y = x2 encloses 9/2; y = x and y = x3 enclose 1/2 in two pieces of 1/4 each.
• A negative result means you subtracted in the wrong order; swap top and bottom.
• When the boundaries are cleaner as functions of y, use horizontal strips: integral from c to d of [right - left] dy. The line y = x - 1 and the parabola y2 = 2x + 6 enclose area 18, in one integral this way.
• Choose the slicing direction so that every strip enters and exits through the same pair of boundaries.

Sources

1. OpenStax. (2016). Areas between curves. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Determining volumes by slicing. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 6.1: Areas between curves. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). Area between curves. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Guichard, D. (n.d.). Calculus: Early transcendentals. Whitman College. whitman.edu
6. Jerison, D. (n.d.). 18.01SC Single variable calculus [Course materials, Fall 2010]. MIT OpenCourseWare. ocw.mit.edu
7. Khan Academy. (n.d.). Applications of integrals. khanacademy.org

Spin a flat region around a line and it sweeps out a solid, the way a shape spinning on a potter's wheel or a lathe carves out a vase or a table leg. This lesson finds the volume of such a solid by slicing it into thin coins, measuring each coin, and adding them up with an integral. It is the same slice-and-add idea as area, now in three dimensions.

The big picture

Take the region under y = f(x) and rotate it around the x-axis. Each thin vertical strip sweeps out a flat disk, like a coin, standing on edge. A coin has volume pi * (radius)2 * (thickness). The radius is the height of the curve, f(x), and the thickness is dx. Add up all the coins with an integral and you have the volume.

V = integral from a to b of pi * [f(x)]2 dx

Key idea: a solid of revolution is a stack of thin disks; each disk is pi times radius squared times thickness, and the integral adds them all.

Why the integrand looks like that

Do not memorise the formula. Rebuild it, in four steps that will work for every volume problem in this module.

1. Choose a slicing direction. Cut the solid perpendicular to the axis of rotation. Rotating about a horizontal axis means vertical cuts, so the thickness is dx.
2. Describe one slice. When a vertical strip at position x is spun about the x-axis, it sweeps a full circle. The slice is a disk of thickness dx.
3. Measure it. The radius is the distance from the axis of rotation out to the curve. For rotation about the x-axis that distance is just the height f(x). So the slice has area pi[f(x)]2 and volume pi[f(x)]2 dx.
4. Add and take the limit. Summing over all slices gives V = integral from a to b of pi[f(x)]2 dx.

Every part of the integrand now has a job. The pi and the square come from the area of a circle. The f(x) is a distance from the axis, which is why the formula changes when the axis moves off the x-axis. And the dx is the thickness of one slice, which is why the limits must be x-values.

The washer version comes from the same construction with one change at step 3. If the region does not touch the axis, the spun strip sweeps an annulus rather than a full disk: a big circle of radius R with a small circle of radius r removed. Its area is pi R2 - pi r2 = pi(R2 - r2), so the volume of one slice is pi(R2 - r2) dx. The disk method is simply the washer method with r = 0; there is one formula here, not two.

The disk method, step by step

Rotate y = the square root of x on the interval from 0 to 4 about the x-axis.

1. Identify the radius: it is the height of the curve, f(x) = the square root of x.
2. Square the radius: [f(x)]2 = x. This squaring step is the one students most often forget.
3. Set up the integral: V = integral from 0 to 4 of pi * x dx.
4. Integrate: pi * [x2/2] from 0 to 4 = pi * (16/2).
5. Finish: pi * 8 = 8 pi.

The volume is 8 pi cubic units.

Two checks against formulas you already trust

A method that reproduces geometry you learned in school is a method you can rely on. Both classical volume formulas fall out of the disk integral in three lines.

The cone. A cone of radius r and height h is what you get by rotating the line y = (r/h)x from x = 0 to x = h about the x-axis - a straight line from the tip out to the rim.

1. Radius of a slice: (r/h)x. Square it: (r2/h2)x2.
2. Set up: V = pi(r2/h2) integral from 0 to h of x2 dx.
3. Integrate: pi(r2/h2) * [x3/3] from 0 to h = pi(r2/h2)(h3/3) = pi r2h/3.

That is the cone formula exactly, one third of the cylinder that contains it.

The sphere. Rotate the semicircle y = the square root of (R2 - x2) from x = -R to x = R about the x-axis.

1. Square the radius: [the square root of (R2 - x2)]2 = R2 - x2. The root disappears, which is why sphere problems are easier than they look.
2. Set up and integrate: V = pi integral from -R to R of (R2 - x2) dx = pi[R2x - x3/3] from -R to R.
3. Evaluate: at R this is R3 - R3/3 = 2R3/3; at -R it is -2R3/3. Subtracting gives 4R3/3, so V = 4pi R3/3.

Also exactly right. Archimedes proved this by a far harder argument and asked for a sphere inscribed in a cylinder on his tombstone; the disk method gets there in three lines.

Key idea: when a new method reproduces a formula you already trust, you have earned the right to use it on shapes with no formula.

The washer method, when there is a hole

If the region is rotated about an axis it does not touch, each slice is not a full coin but a washer: a disk with a circular hole punched out. It has an outer radius R reaching to the far curve and an inner radius r reaching to the near curve. The slice area is the big circle minus the small circle, pi(R2 - r2), so:

V = integral from a to b of pi ([R(x)]2 - [r(x)]2) dx

Worked example. Rotate the region between y = x and y = x2 about the x-axis. On the interval from 0 to 1, the line y = x is farther from the axis (outer) and the parabola y = x² is nearer (inner).

1. Outer radius R = x, inner radius r = x2.
2. Square each and subtract: R2 - r2 = x2 - x4.
3. Set up: V = integral from 0 to 1 of pi (x2 - x4) dx.
4. Integrate: pi [x3/3 - x5/5] from 0 to 1 = pi (1/3 - 1/5).
5. Combine the fractions: pi (5/15 - 3/15) = 2 pi/15.

The volume is 2 pi/15. Notice that squaring first was essential: (R - r)2 = (x - x2)2 = x2 - 2x3 + x4, which integrates to pi(1/3 - 1/2 + 1/5) = pi/30 - half the right answer, and wrong for a reason that no arithmetic check would catch.

Rotating about a line that is not an axis

Nothing in the derivation assumed the axis was the x-axis. What the formula actually needs is the distance from the axis of rotation to each boundary. Rotate the region between y = x2 and the x-axis, for x from 0 to 1, about the horizontal line y = -1.

1. Draw the axis and measure distances to it. The far boundary is the curve y = x2, at distance x2 - (-1) = x2 + 1 from the line. The near boundary is the x-axis, at distance 0 - (-1) = 1.
2. So R = x2 + 1 and r = 1. The region does not touch the axis, so there is a hole and this is a washer problem.
3. Square each and subtract: R2 - r2 = (x2 + 1)2 - 1 = x4 + 2x2 + 1 - 1 = x4 + 2x2.
4. Integrate: V = pi integral from 0 to 1 of (x4 + 2x2) dx = pi[x5/5 + 2x3/3] from 0 to 1 = pi(1/5 + 2/3).
5. Combine: pi(3/15 + 10/15) = 13pi/15.

Sanity-check the size. Rotating the same region about the x-axis would give pi integral from 0 to 1 of x4 dx = pi/5 = 3pi/15. Moving the axis one unit away pushes the material further out and should increase the volume, and 13pi/15 is indeed larger. If your relocated-axis answer comes out smaller, you have almost certainly forgotten to shift one of the radii.

Key idea: a radius is always a distance from the axis of rotation - subtract the axis position from the curve, not the other way round, and never assume the radius equals the function value.

Slicing in y: rotating about a vertical axis

Rotating about a vertical line means the slices are horizontal, so the thickness is dy and everything must be written in terms of y. Rotate the region bounded by y = x2, y = 4, and x = 0 about the y-axis.

1. Slice horizontally at height y. Spinning that strip about the y-axis sweeps a disk whose radius is the horizontal distance from the axis to the curve.
2. Solve the boundary for x: y = x2 gives x = the square root of y on the right-hand branch. So the radius is the square root of y.
3. Square it: [the square root of y]2 = y.
4. The limits are now y-values, running from 0 up to 4: V = pi integral from 0 to 4 of y dy = pi[y2/2] from 0 to 4 = 8pi.

The volume is 8 pi. Keep that number - the next lesson computes the same solid by a completely different method and must get 8 pi again. Agreement between two independent setups is the strongest check available in this whole module.

Try it

Rotate y = x on the interval from 0 to 2 about the x-axis.

Answer: the disk radius is x, so V = integral from 0 to 2 of pi x2 dx = pi [x3/3] from 0 to 2 = 8 pi/3. Nice work squaring the radius before integrating.

Where people get stuck

• Forgetting to square the radius. The formula is pi times radius squared. Integrating pi f(x) instead of pi[f(x)]2 is the single most common error, and the answer looks perfectly reasonable.
• Writing (R - r)² for R² - r². These are never equal unless r = 0. In the worked washer they give 2pi/15 and pi/30 - a factor of four apart. Square each radius separately, then subtract.
• Using the function value as the radius when the axis has moved. A radius is a distance to the axis. Rotating about y = -1 makes the radius f(x) + 1, and rotating about y = 5 makes it 5 - f(x). Sketch the axis and measure.
• Mismatching the slice direction and the variable. Slices must be perpendicular to the axis. A horizontal axis forces dx and x-limits; a vertical axis forces dy, y-limits, and boundaries solved for x in terms of y.
• Not noticing a hole. If the region does not touch the axis, there is a hole and you need washers. Rotating the region between y = x and y = x2 about the x-axis has no hole only because both curves meet the axis at x = 0; move the axis to y = -1 and a hole appears.
• Losing the pi. It multiplies the whole integral, so factor it out at the start and it cannot be forgotten mid-computation.
• Never checking the result. Compare against a bounding cylinder: the solid must fit inside pi Rmax2 * (length). The 8pi from y = the square root of x on [0, 4] sits comfortably under the bounding cylinder's pi*4*4 = 16pi. An answer larger than the bounding cylinder is definitely wrong.

Recap

• Build every volume the same way: slice perpendicular to the axis, describe one slice, measure it, integrate.
• Disk method: V = integral pi[radius]2 d(thickness), for a solid with no hole. Washer method: V = integral pi(R2 - r2) d(thickness), when the region does not touch the axis. They are one formula, with r = 0 in the first case.
• The radius is the distance from the axis of rotation to the boundary curve, not the function value, whenever the axis is not a coordinate axis.
• Always square each radius before integrating, and subtract the squares rather than squaring the difference.
• Worked results: y = the square root of x on [0, 4] about the x-axis gives 8 pi; the region between y = x and y = x2 gives 2pi/15; the region under y = x2 on [0, 1] about y = -1 gives 13pi/15.
• The method reproduces the cone volume pi r2h/3 and the sphere volume 4pi R3/3 in three lines each, which is why it can be trusted on shapes with no formula.
• Rotating about a vertical axis means horizontal slices, dy, and boundaries written as x in terms of y.

Sources

1. OpenStax. (2016). Determining volumes by slicing. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Volumes of revolution: Cylindrical shells. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 6.2: Determining volumes by slicing. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). Volumes of solids of revolution, method of rings. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Dawkins, P. (n.d.). More volume problems. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. O'Connor, J. J., & Robertson, E. F. (n.d.). Archimedes of Syracuse. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk
7. Khan Academy. (n.d.). Applications of integrals. khanacademy.org

The disk method slices a solid into flat coins. The shell method builds the same kind of solid a different way: out of thin nested tubes, like the rings of a tree or a set of tin cans that fit one inside the next. This second method shines when you rotate a region about a vertical axis, where disks would be awkward to set up.

The big picture

Imagine the solid made of many thin cylindrical shells nested inside each other. Take one shell at radius x, with height f(x) and thickness dx. Now slit it down the side and unroll it flat. It becomes a thin rectangular sheet: its length is the circumference 2 pi x, its height is f(x), and its thickness is dx. So one shell has volume 2 pi x * f(x) * dx. Add up all the shells with an integral, for rotation about the y-axis:

V = integral from a to b of 2 pi x * f(x) dx

Read the three factors as radius (x), height (f(x)), and thickness (dx).

Key idea: a shell unrolls into a flat sheet of area circumference times height, so its volume is 2 pi x times f(x) times dx.

Why the integrand takes that form, exactly

The unrolling picture is honest but informal, since a real shell has a slightly larger outer circumference than inner. Here is the same result derived without any hand-waving, which also shows why the approximation is harmless.

1. Take a shell of height h whose inner radius is x and outer radius is x + dx. It is the solid between two concentric cylinders.
2. Its exact volume is the big cylinder minus the small one: pi(x + dx)2h - pi x2h = pi h[(x + dx)2 - x2].
3. Expand the bracket: (x + dx)2 - x2 = x2 + 2x dx + (dx)2 - x2 = 2x dx + (dx)2.
4. So the exact volume is 2pi x h dx + pi h (dx)2. The second term carries (dx)2, a squared infinitesimal, which vanishes faster than the first as the shells get thin and contributes nothing in the limit.
5. What survives is dV = 2pi x h dx - circumference times height times thickness - and summing gives the integral.

The three factors now have permanent meanings, and every shell problem is a matter of identifying them correctly:

• 2 pi times radius is the circumference of the shell, where the radius is the distance from the axis of rotation to the strip.
• Height is the length of the strip, which is f(x) for a region above the x-axis and top - bottom when two curves bound the region.
• Thickness is the differential, and it must match the variable everything else is written in.

Worked example, step by step

Rotate the region under y = x2 on the interval from 0 to 2 about the y-axis.

1. Identify the shell radius: the distance from the y-axis to the strip, which is x.
2. Identify the shell height: the height of the region, f(x) = x2.
3. Set up the integral: V = integral from 0 to 2 of 2 pi x * (x2) dx = 2 pi integral from 0 to 2 of x3 dx.
4. Integrate: 2 pi [x4/4] from 0 to 2 = 2 pi (16/4).
5. Finish: 2 pi * 4 = 8 pi.

The volume is 8 pi. Solving this with washers would force you to rewrite y = x2 as x = the square root of y and handle the region in pieces. Shells keep everything in terms of x, which is much cleaner here.

The same solid, both ways

The strongest confidence-builder available is computing one volume by two independent methods. In the last lesson we rotated the region bounded by y = x2, y = 4, and x = 0 about the y-axis and got 8 pi using disks in y. Redo it with shells.

1. Slice vertically. A strip at position x runs from the parabola up to the line y = 4, so the shell height is 4 - x2.
2. The radius is the distance from the y-axis to the strip, which is x.
3. The region spans x from 0 (the y-axis) to 2 (where x2 = 4).
4. Set up: V = 2pi integral from 0 to 2 of x(4 - x2) dx = 2pi integral from 0 to 2 of (4x - x3) dx.
5. Integrate: 2pi[2x2 - x4/4] from 0 to 2 = 2pi[8 - 4] = 8pi.

The two methods agree exactly, having shared no steps at all: one sliced horizontally and squared a radius, the other sliced vertically and multiplied a circumference by a height. When a setup feels shaky, this is the check to run.

A region bounded by two curves

Rotate the region between y = x and y = x2 about the y-axis.

1. Crossings: x = x2 gives x = 0 and x = 1, so those are the limits.
2. Shell height is top minus bottom. Testing x = 0.5 gives 0.5 for the line and 0.25 for the parabola, so the height is x - x2.
3. Radius is x. Set up: V = 2pi integral from 0 to 1 of x(x - x2) dx = 2pi integral from 0 to 1 of (x2 - x3) dx.
4. Integrate: 2pi[x3/3 - x4/4] from 0 to 1 = 2pi(1/3 - 1/4) = 2pi(1/12) = pi/6.

The volume is pi/6. Note that this is a completely different solid from the 2pi/15 in the previous lesson, even though the region is identical - there we rotated about the x-axis, here about the y-axis. The axis is part of the problem, not a detail.

Rotating about a vertical line that is not the y-axis

Just as with washers, only the radius changes. Rotate the region under y = x2 from x = 0 to x = 2 about the line x = -1.

1. The strip at position x is now x - (-1) = x + 1 units from the axis, so the shell radius is x + 1.
2. The height is unchanged at x2, and the limits are still 0 to 2.
3. Set up: V = 2pi integral from 0 to 2 of (x + 1)x2 dx = 2pi integral from 0 to 2 of (x3 + x2) dx.
4. Integrate: 2pi[x4/4 + x3/3] from 0 to 2 = 2pi[4 + 8/3] = 2pi(20/3) = 40pi/3.

Check the size: rotating the same region about the y-axis gave 8 pi, about 25.1, and pushing the axis one unit further away should sweep out more, which 40pi/3 ~= 41.9 does. If a relocated axis makes your volume smaller, the radius was shifted the wrong way.

When shells are the only practical route

Rotate the region under y = sin(x) from x = 0 to x = pi about the y-axis. With washers you would have to invert the sine - two separate branches, x = arcsin(y) and x = pi - arcsin(y) - and integrate their difference in y. With shells the setup is immediate.

1. Radius x, height sin(x), limits 0 to pi: V = 2pi integral from 0 to pi of x sin(x) dx.
2. The remaining integral is integration by parts with u = x, dv = sin(x) dx, so du = dx and v = -cos(x).
3. Apply the formula: [-x cos(x)] from 0 to pi + integral from 0 to pi of cos(x) dx.
4. Boundary term: -pi cos(pi) - 0 = -pi(-1) = pi. Remaining integral: [sin(x)] from 0 to pi = 0 - 0 = 0.
5. So the integral is pi, and V = 2pi * pi = 2pi2 ~= 19.74.

This is where the whole integration toolkit earns its keep: a geometry problem produced an integral that needed a technique from Module 2. Applications are rarely self-contained.

Key idea: choose the method that keeps the region describable in one variable; the arithmetic afterwards is whatever it is.

Choosing between shells and washers

Situation	Often easier
Rotating about the x-axis, region given as y = f(x)	Disks or washers
Rotating about the y-axis, region given as y = f(x)	Shells
A representative strip runs parallel to the axis of rotation	Shells
A representative strip runs perpendicular to the axis	Disks or washers

Both methods always give the same volume. The choice is only about which integral is easier to set up. Here is a reliable rule of thumb: sketch one strip, then look at how it sits relative to the axis. If the strip runs parallel to the axis, use shells. If it runs perpendicular, use disks or washers.

Try it

Rotate the region under y = x on the interval from 0 to 3 about the y-axis, using shells.

Answer: the radius is x and the height is x, so V = integral from 0 to 3 of 2 pi x * (x) dx = 2 pi [x3/3] from 0 to 3 = 2 pi * 9 = 18 pi. Nice work reading off the radius and height.

Where people get stuck

• Confusing radius with height. The radius is how far the strip sits from the axis; the height is how tall the region is at that spot. Label both on a sketch before writing anything. In the y = sin(x) problem the radius is x and the height is sin(x) - swapping them gives 2pi integral sin(x)*x dx, which happens to be the same here but is a coincidence you cannot rely on.
• Squaring the radius. Shells use 2 pi r h, with the radius to the first power. The square belongs to disks and washers. Mixing the formulas is the most damaging error in this lesson.
• Forgetting that height is top minus bottom. When two curves bound the region, the shell height is their difference, exactly as with area between curves. Using only the upper curve silently adds the solid underneath.
• Not shifting the radius for a relocated axis. About x = -1 the radius is x + 1; about x = 5 it is 5 - x. Write the distance, do not guess the sign, and check that it is positive across the whole interval.
• Mismatching the differential. Rotating about a vertical axis and slicing vertically gives shells in dx with x-limits. Rotating about a horizontal axis with shells means horizontal strips, thickness dy, and everything written in y.
• "Shells and washers give different answers." They never do. Getting two different numbers means one setup has an error, and comparing them line by line will find it.
• Choosing the method by habit. Draw one strip and look. Parallel to the axis means shells; perpendicular means disks or washers. That single glance is faster than starting the wrong integral.

Recap

• Shell method: V = integral 2 pi (radius)(height) d(thickness), building the solid from nested cylindrical shells.
• The formula is exact in the limit: the shell between radii x and x + dx has volume 2pi x h dx + pi h (dx)2, and the squared term vanishes.
• Radius is the distance from the axis to the strip; height is the length of the strip, which is top minus bottom when two curves bound the region.
• Worked results: y = x2 on [0, 2] about the y-axis gives 8 pi; the region between y = x and y = x2 about the y-axis gives pi/6; the same parabolic region about x = -1 gives 40pi/3; and y = sin(x) on [0, pi] about the y-axis gives 2pi2.
• Shells and disks always agree - the region under y = x2, y = 4, x = 0 about the y-axis gives 8 pi by either method - so computing both is a genuine check.
• Rule of thumb: strip parallel to the axis means shells; strip perpendicular means disks or washers.
• Some regions force one method: inverting y = sin(x) for washers takes two branches, while shells set it up in a single line.

Sources

1. OpenStax. (2016). Volumes of revolution: Cylindrical shells. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Determining volumes by slicing. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 6.3: Volumes of revolution - Cylindrical shells. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). Volumes of solids of revolution, method of cylinders. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Dawkins, P. (n.d.). More volume problems. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Guichard, D. (n.d.). Calculus: Early transcendentals. Whitman College. whitman.edu
7. Khan Academy. (n.d.). Applications of integrals. khanacademy.org

How long is a curved path, like a winding road or a bent wire? You cannot lay a straight ruler along a curve. But you can approximate the curve with many tiny straight segments, measure each with the Pythagorean theorem, and add them up. Let the segments shrink and the sum becomes an integral. That integral is the arc length.

The big picture

Zoom in on a tiny piece of the curve. It has a small horizontal run dx and a small vertical rise dy, so by the Pythagorean theorem its length is the square root of (dx2 + dy2). Factor out dx and use the fact that the slope dy/dx = f'(x), and you get the arc length of y = f(x) from a to b:

L = integral from a to b of the square root of (1 + [f'(x)]2) dx

The 1 comes from the horizontal run, and the [f'(x)]2 comes from the slope. The whole formula is just the Pythagorean theorem applied to infinitely many infinitely short segments.

Key idea: arc length adds up tiny Pythagorean hypotenuses, giving L = integral the square root of (1 + [f'(x)]2) dx.

Where the formula comes from, step by step

The informal version above is right, but the derivation is worth doing carefully because it shows exactly where the 1 and the square both come from.

1. Approximate with a polygon. Mark n + 1 points along the curve and connect consecutive points with straight segments. The total length of this polygon underestimates the curve, since a straight line is the shortest path between two points, and it improves as n grows.
2. Measure one segment. Between consecutive points the run is dx and the rise is dy, so the segment length is the square root of (dx2 + dy2) by the Pythagorean theorem.
3. Factor out dx. Write dx2 + dy2 = dx2(1 + (dy/dx)2), so the segment length is the square root of (1 + (dy/dx)2) dx. Here is the 1: it is the horizontal run, after the factoring.
4. Use the Mean Value Theorem. On each subinterval there is a point where the segment's slope dy/dx equals the curve's derivative exactly, so we may write f'(x) in place of the ratio.
5. Add and take the limit. The polygon's length becomes L = integral from a to b of the square root of (1 + [f'(x)]2) dx.

Two consequences follow immediately and are useful as checks. First, since 1 + [f'(x)]2 >= 1, the integrand is at least 1, so the arc length is always at least the width of the interval, b - a. Second, the arc length is always at least the straight-line distance between the endpoints. Any answer that violates either bound is wrong.

Worked example, step by step

Find the length of y = (2/3) x3/2 from x = 0 to x = 3. This curve is chosen so the algebra comes out clean.

1. Differentiate: f'(x) = (2/3)*(3/2)*x1/2 = x1/2, which is the square root of x.
2. Square the derivative: [f'(x)]2 = x.
3. Build the integrand: the square root of (1 + x).
4. Set up: L = integral from 0 to 3 of the square root of (1 + x) dx. Let u = 1 + x, giving integral u1/2 du = (2/3) u3/2.
5. Evaluate: (2/3)[(1 + x)3/2] from 0 to 3 = (2/3)(43/2 - 13/2).
6. Simplify the powers: 43/2 = (the square root of 4) cubed = 23 = 8, so (2/3)(8 - 1) = 14/3.

The arc length is 14/3, about 4.667. Check it against the two bounds: the interval width is 3, and the straight-line distance from (0, 0) to (3, 2*33/2/3) = (3, 3.464) is the square root of (9 + 12) = 4.583. Our 4.667 exceeds both, as it must, and only slightly exceeds the chord, which fits a curve this gently bent.

Second worked example: engineering the perfect square

Find the length of y = x3/6 + 1/(2x) from x = 1 to x = 2. The strange-looking function is the point: watch what happens under the root.

1. Differentiate: f'(x) = 3x2/6 - 1/(2x2) = x2/2 - 1/(2x2).
2. Square it, carefully keeping the cross term: [f'(x)]2 = x4/4 - 2*(x2/2)*(1/(2x2)) + 1/(4x4) = x4/4 - 1/2 + 1/(4x4).
3. Add 1. The cross term was -1/2, so adding 1 turns it into +1/2: 1 + [f'(x)]2 = x4/4 + 1/2 + 1/(4x4).
4. Recognise the perfect square: that is (x2/2 + 1/(2x2))2, the same two terms as f'(x) but with a plus sign. The root comes off cleanly: x2/2 + 1/(2x2).
5. Integrate: L = integral from 1 to 2 of (x2/2 + x-2/2) dx = [x3/6 - 1/(2x)] from 1 to 2. Verify the antiderivative: d/dx[x3/6 - (1/2)x-1] = x2/2 + (1/2)x-2. Correct.
6. Evaluate: at x = 2, 8/6 - 1/4 = 4/3 - 1/4 = 13/12; at x = 1, 1/6 - 1/2 = -1/3 = -4/12. Subtract: 13/12 + 4/12 = 17/12.

The length is 17/12 ~= 1.417. Check: the endpoints are (1, 2/3) and (2, 19/12), a chord of the square root of (1 + 0.9172) ~= 1.356. The arc is slightly longer. Correct.

The trick generalises. Any function of the form f'(x) = g(x) - 1/(4g(x)) makes 1 + [f']2 a perfect square, because the cross term is exactly -1/2. Textbook arc-length problems are built backwards from this identity, which is why they always work out and real curves usually do not.

Arc length in terms of y

If a curve is easier to describe as x = g(y), or if it has a vertical tangent (where f'(x) blows up), swap the roles. The identical derivation with dy factored out gives

L = integral from c to d of the square root of (1 + [g'(y)]2) dy

For example, the curve x = y2 from y = 0 to y = 1 has g'(y) = 2y, so L = integral from 0 to 1 of the square root of (1 + 4y2) dy. Written as y = the square root of x instead, the derivative 1/(2 the square root of x) is infinite at x = 0 and the integral is improper - the same curve, but one description is well behaved and the other is not.

When the integral fights back

Arc length integrands often hide a square root with no elementary antiderivative. The length of the parabola y = x2 from 0 to 1 leads to integral from 0 to 1 of the square root of (1 + 4x2) dx, which needs a trig substitution. Doing it that way gives the antiderivative (x/2) the square root of (1 + 4x2) + (1/4)ln(2x + the square root of (1 + 4x2)), and evaluating from 0 to 1 gives (1/2)(the square root of 5) + (1/4)ln(2 + the square root of 5) ~= 1.1180 + 0.3609 = 1.4789.

Confirm it numerically with Simpson's rule on four strips, h = 0.25. The integrand values are 1, 1.1180, 1.4142, 1.8028, 2.2361, and Simpson's weighting gives (0.25/3)[1 + 4(1.1180) + 2(1.4142) + 4(1.8028) + 2.2361] = (0.08333)(17.7477) = 1.4790. The two agree to four figures, so both are almost certainly right. And the chord from (0,0) to (1,1) is the square root of 2 ~= 1.4142, comfortably below.

Most arc lengths in practice - a coastline, a suspension cable, a spacecraft trajectory - are computed numerically. The setup is always the same, and setting it up correctly is the real skill.

The companion formula: surface area of revolution

Spin the curve rather than the region beneath it and you sweep out a surface. One tiny arc of length ds at height f(x) traces a thin band whose circumference is 2 pi f(x), so the band's area is 2 pi f(x) ds. Since ds = the square root of (1 + [f'(x)]2) dx:

S = integral from a to b of 2 pi f(x) * the square root of (1 + [f'(x)]2) dx

Test it on a cone. Rotating y = (r/h)x from 0 to h about the x-axis gives a cone of radius r and height h. Here f' = r/h, so the square root of (1 + r2/h2) = (the square root of (h2 + r2))/h, which is the slant height divided by h. Writing s for the slant height:

S = integral from 0 to h of 2pi(r/h)x * (s/h) dx = (2pi r s/h2)[x2/2] from 0 to h = pi r s

That is precisely the classical lateral-surface formula for a cone, pi times radius times slant height. The construction checks out.

Key idea: arc length uses ds; surface area multiplies ds by the circumference the arc sweeps.

Try it

Set up, but do not evaluate, the arc length of y = x3 from x = 0 to x = 2.

Answer: the derivative is f'(x) = 3x2, so [f'(x)]2 = 9x4 and L = integral from 0 to 2 of the square root of (1 + 9x4) dx. This has no elementary antiderivative, so it would be evaluated numerically. Nice work getting the setup exactly right.

Where people get stuck

• Dropping the 1 under the root. It is not decoration; it is the horizontal run of each tiny segment. Without it the "length" of a horizontal line would come out as zero.
• Forgetting to square the derivative. The integrand is the square root of (1 + [f'(x)]2), in that exact shape. Do four small steps in order - write f'(x), square it, add 1, take the root - and the setup is reliable.
• Integrating f(x) instead of the arc-length integrand. Arc length has nothing to do with the area under the curve. A curve can be long and enclose almost no area, and vice versa.
• Botching the cross term when squaring. (A - B)2 = A2 - 2AB + B2. The middle term is exactly what makes the perfect square appear after adding 1, so losing it destroys the whole problem.
• Missing that the root simplifies. If 1 + [f']2 is a perfect square, the root vanishes and the integral becomes elementary. Before reaching for a technique, try to factor it.
• Expecting every arc length to be elementary. Most are not - even the parabola needs a trig substitution, and y = x3 has no elementary answer at all. Setting up correctly and then integrating numerically is a complete solution, not a failure.
• Not sanity-checking against the chord. The arc length must exceed both the interval width b - a and the straight-line distance between the endpoints. Two seconds of checking catches most algebra errors.
• Confusing arc length with surface area. Surface area carries an extra factor of 2 pi f(x), the circumference the arc sweeps out. Arc length is measured in units of length, surface area in units of length squared.

Recap

• Arc length is L = integral from a to b of the square root of (1 + [f'(x)]2) dx, the Pythagorean theorem applied to infinitely many tiny segments and justified by the Mean Value Theorem.
• Steps: find f'(x), square it, add 1, take the square root, then integrate.
• Worked results: y = (2/3)x3/2 on [0, 3] has length 14/3; y = x3/6 + 1/(2x) on [1, 2] has length 17/12.
• Textbook problems are built so that 1 + [f']2 is a perfect square; real curves usually are not, and y = x2 on [0, 1] gives 1.4789, confirmed to four figures by Simpson's rule.
• When x is the simpler function, integrate in y: L = integral the square root of (1 + [g'(y)]2) dy. This also rescues curves with vertical tangents.
• Every arc length is at least b - a and at least the chord between the endpoints. Use both as checks.
• Surface area of revolution multiplies the arc element by the circumference it sweeps: S = integral 2 pi f(x) the square root of (1 + [f'(x)]2) dx, which reproduces the cone's pi r s.

Sources

1. OpenStax. (2016). Arc length of a curve and surface area. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Numerical integration. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 6.4: Arc length of a curve and surface area. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Dawkins, P. (n.d.). Arc length. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Dawkins, P. (n.d.). Surface area. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Guichard, D. (n.d.). Calculus: Early transcendentals. Whitman College. whitman.edu
7. Khan Academy. (n.d.). Applications of integrals. khanacademy.org

The second half of this course studies the infinite, and it starts with something down to earth: a list of numbers that follows a rule. Think of an account balance printed each month, or a row of steps numbered by position. A sequence is exactly that, an ordered list. The one question we keep asking is: as you go further and further out, where do the numbers head?

The big picture

A sequence is an ordered list a1, a2, a3, ..., produced by a rule for the general term an (read "a sub n," the nth term). For example, an = 1/n gives the list 1, 1/2, 1/3, 1/4, .... The big question is about the far-off behavior: do the terms settle toward a single number, or not?

Key idea: a sequence is a list defined by a rule for its nth term, and the central question is what value the terms approach as n grows.

The limit of a sequence

A sequence converges to a limit L if the terms get and stay arbitrarily close to L as n goes to infinity, written lim(n -> infinity) an = L. If there is no single finite value the terms approach, the sequence diverges. Because n is marching off to infinity, evaluating these limits uses the very same tricks you used for limits at infinity in Calculus I.

Worked example one. Does an = 1/n converge? As n grows, 1/n shrinks toward 0, so lim(n -> infinity) 1/n = 0. It converges to 0.

Worked example two. Does an = (3n2 + 1)/(2n2 - n) converge? Divide top and bottom by the highest power, n²:

1. Top becomes 3 + 1/n2.
2. Bottom becomes 2 - 1/n.
3. As n -> infinity the small pieces 1/n² and 1/n vanish, leaving 3/2.

It converges to 3/2. Check the claim against actual terms: a10 = 301/190 = 1.5842, a100 = 30001/19900 = 1.5076, a1000 = 1.5008. They are closing in on 1.5 from above, exactly as predicted.

What "arbitrarily close" actually means

The phrase "gets and stays close to L" is doing real work, and the precise version is worth seeing once. We say an -> L if for every tolerance epsilon > 0, however small, there is an index N such that |an - L| < epsilon for all n > N. Read it as a challenge and a response: you name a tolerance, and I must produce a point in the list beyond which every single term is within that tolerance.

Worked verification. Prove that an = 1/n converges to 0.

1. Write down what must be shown: |1/n - 0| < epsilon, which is just 1/n < epsilon since 1/n is positive.
2. Solve for n: n > 1/epsilon.
3. So choose N to be any integer at least 1/epsilon. Then every term past N satisfies the requirement.
4. Test with a number. If epsilon = 0.01, take N = 100; every term from a101 = 0.0099 onward is within 0.01 of 0. If you demand epsilon = 0.000001, take N = 1000000. There is always an answer, which is exactly what convergence means.

Notice what the definition rules out. "Gets close once" is not enough - the terms must stay close. That single word is why (-1)n fails: it hits 1 infinitely often, but it never stays near 1.

Key idea: convergence is a promise that can be kept for every tolerance, no matter how strict, from some point onward.

Two patterns that show up constantly

• Powers rⁿ: if |r| < 1, then rn -> 0 (for instance (1/2)n -> 0); if |r| > 1, the terms blow up; if r = 1, they sit at 1.
• Factorial growth: n! (n factorial) grows faster than any power or exponential, so terms like n10/n! and 2n/n! head to 0.

Worked example three. Does an = (-1)n converge? The terms are -1, 1, -1, 1, ... forever, never settling on one value, so the sequence diverges by oscillation.

The growth hierarchy, with real numbers attached

Almost every sequence limit in this course is decided by knowing which of two expressions grows faster. From slowest to fastest, for large n:

ln(n) << np << bn (for b > 1) << n! << nn

The symbol << means "is eventually dwarfed by," so any ratio with a slower item on top goes to 0. Numbers make the gaps vivid. At n = 20: ln(20) ~= 3.0, 202 = 400, 220 = 1048576, 20! ~= 2.4 * 1018, and 2020 ~= 1.0 * 1026. Each step up is not a small edge but a change of scale, and by n = 100 the gaps are astronomically wider.

Use it directly. n10/n! has the slower item on top, so it goes to 0 - even though the first several terms grow enormously before turning around near n = 10. 2n/n100 has the faster item on top, so it diverges, however patient you have to be. (ln n)/n0.001 still goes to 0, because a logarithm loses to every positive power.

A bridge to functions

If an = f(n) for some function f and lim(x -> infinity) f(x) = L, then lim(n -> infinity) an = L as well. This lets you bring in calculus tools, including L'Hopital's rule, to handle a sequence. For example, lim(n -> infinity) (ln n)/n is an infinity-over-infinity form; L'Hopital on the function version gives (1/x)/1 = 1/x -> 0, so the sequence limit is 0.

The bridge runs one way only. If the function limit fails to exist, the sequence may still converge, because the sequence only samples the function at integers. Take f(x) = sin(pi x), which oscillates forever and has no limit, while an = sin(pi n) = 0 for every integer n and converges to 0 immediately.

The squeeze theorem for sequences

When a sequence is hard to evaluate but easy to trap, squeeze it. If bn <= an <= cn and both bn and cn converge to the same L, then an -> L too.

Worked example. Find lim(n -> infinity) sin(n)/n. There is no formula for sin(n) at integer arguments and the values jump around forever, so a direct attack is hopeless.

1. Bound the numerator: -1 <= sin(n) <= 1 for every n.
2. Divide by n, which is positive, so the inequalities keep their direction: -1/n <= sin(n)/n <= 1/n.
3. Both outer sequences converge to 0.
4. By the squeeze theorem, sin(n)/n -> 0.

Monotone and bounded means convergent

One theorem lets you conclude convergence without knowing the limit at all. If a sequence is increasing and bounded above, it converges; likewise if it is decreasing and bounded below. Intuitively the terms keep moving one way but cannot escape past the bound, so they must pile up somewhere.

Worked example. Show that an = n/(n + 1) converges.

1. Rewrite it: n/(n + 1) = 1 - 1/(n + 1).
2. Increasing? As n grows, 1/(n + 1) shrinks, so we are subtracting less each time. Yes, increasing. Terms: 1/2, 2/3, 3/4, 4/5, ....
3. Bounded above? 1/(n + 1) is always positive, so an < 1 always. Yes, bounded by 1.
4. Therefore it converges - and here the limit is visibly 1.

Recursive sequences are where this theorem earns its keep, because no closed formula for an exists. Define a1 = 2 and an+1 = (an + 2/an)/2.

1. Compute terms: a1 = 2; a2 = (2 + 1)/2 = 1.5; a3 = (1.5 + 1.33333)/2 = 1.416667; a4 = (1.416667 + 1.411765)/2 = 1.414216.
2. The terms are decreasing and stay above 1, so by the monotone convergence theorem some limit L exists.
3. Now find it. Since both an and an+1 approach the same L, take the limit of both sides of the recursion: L = (L + 2/L)/2.
4. Solve: 2L = L + 2/L, so L = 2/L, so L2 = 2 and L = the square root of 2 (the positive root, since all terms are positive).

Compare: the square root of 2 = 1.4142136, and a4 = 1.414216 already matches to five decimals. This recursion is Newton's method for the square root, and the same argument - prove a limit exists, then solve for it - is the standard way to analyse any iterative algorithm.

Key idea: you must establish that a limit exists before solving for it; setting L = f(L) on a divergent recursion produces a confident, meaningless number.

Try it

Find lim(n -> infinity) (2n + 1)/(n + 3).

Answer: divide top and bottom by n to get (2 + 1/n)/(1 + 3/n). As n -> infinity the small terms vanish, leaving 2/1 = 2. The sequence converges to 2. Nice work using the divide-by-the-highest-power trick.

Where people get stuck

• Confusing a sequence with a series. This is the biggest source of trouble in the whole module. A sequence is the list of terms; a series is what you get when you add them. The sequence 1/n converges to 0, while the series 1 + 1/2 + 1/3 + ... built from the same terms diverges. Same numbers, opposite answers.
• "Bounded means convergent." Only with monotonicity attached. (-1)n stays inside [-1, 1] forever and still diverges, because it never settles on one value.
• "Monotone means convergent." Only with boundedness attached. an = n is increasing and diverges. You need both halves of the theorem.
• Solving L = f(L) before proving L exists. Applied to an+1 = 2an with a1 = 1, this "gives" L = 2L, so L = 0 - but the terms are 1, 2, 4, 8, ... and diverge. Establish convergence first, then solve.
• Assuming factorials and exponentials are comparable. Factorials win decisively, which is why 2n/n! goes to 0 despite the numerator's early lead. At n = 20 the factorial is over a trillion times larger.
• Reading the first few terms as the trend. n10/n! grows to around 104 before collapsing toward 0. Early terms are not evidence about a limit.
• Applying L'Hopital in reverse. A missing function limit does not imply a missing sequence limit, since the sequence only samples integer inputs - sin(pi x) versus sin(pi n) is the standard counterexample.

Recap

• A sequence is an ordered list with a rule for the nth term an. It converges if the terms approach a single finite limit L, and diverges otherwise, including by oscillation.
• Formally, an -> L means: for every epsilon > 0 there is an N with |an - L| < epsilon for all n > N. For 1/n -> 0, take N = 1/epsilon.
• Divide by the highest power for rational terms; rn -> 0 when |r| < 1; and the growth hierarchy is ln n << np << bn << n! << nn.
• If an = f(n) and the function limit exists, the sequence shares it, so L'Hopital's rule is available. The converse fails.
• Trap hard sequences between two easy ones with the squeeze theorem; sin(n)/n -> 0 this way.
• A monotone bounded sequence converges even when the limit is unknown, which is how recursive sequences are handled: prove a limit exists, then solve L = f(L). The recursion an+1 = (an + 2/an)/2 converges to the square root of 2.

Sources

1. OpenStax. (2016). Sequences. In Calculus volume 2. Rice University. openstax.org
2. Strang, G., & Herman, E. (n.d.). 9.1: Sequences. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
3. Dawkins, P. (n.d.). Sequences. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Dawkins, P. (n.d.). More on sequences. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Rodriguez, C. (n.d.). 18.100A Real analysis [Course materials, Fall 2020]. MIT OpenCourseWare. ocw.mit.edu
6. Guichard, D. (n.d.). Calculus: Early transcendentals. Whitman College. whitman.edu
7. Tall, D., & Vinner, S. (1981). Concept image and concept definition in mathematics with particular reference to limits and continuity. Educational Studies in Mathematics, 12(2), 151-169. doi.org/10.1007/BF00305619

Can you add up infinitely many numbers and still get a finite total? It sounds impossible, yet a bouncing ball that travels half as far on each bounce covers a finite total distance. The secret is that if the terms shrink fast enough, the running total closes in on a number. This lesson makes that precise and gives you the one series you will use most.

The big picture

An infinite series is the sum of every term of a sequence: a1 + a2 + a3 + ..., written sum from n=1 to infinity of an. We make sense of an endless sum by watching its partial sums, the totals you get by stopping after 1 term, 2 terms, 3 terms, and so on.

Key idea: an infinite series is really the limit of its partial sums; if those running totals approach a number, the series converges to it.

Partial sums define the total

The nth partial sum is Sn = a1 + a2 + ... + an, which adds only the first n terms. The series converges to a sum S if the partial sums approach S, that is lim(n -> infinity) Sn = S. If the partial sums have no finite limit, the series diverges. So a series is a limit of ordinary finite sums in disguise.

This is worth stating carefully because it is where the whole subject sits. Two sequences are in play at once and they behave differently. The terms an are the individual numbers being added; the partial sums Sn are the running totals. Convergence of a series is a statement about the second sequence, never the first. Keeping them apart resolves most of the confusion in Module 5.

A worked telescoping series

Very few series let you write down Sn in closed form, but when you can, the definition does all the work. Evaluate sum from n=1 to infinity of 1/(n(n + 1)).

1. Decompose the term by partial fractions: 1/(n(n+1)) = A/n + B/(n+1) gives 1 = A(n+1) + Bn. Setting n = 0 gives A = 1; setting n = -1 gives 1 = -B, so B = -1. The term is 1/n - 1/(n + 1).
2. Write out the partial sum: Sn = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1)).
3. Cancel. Every negative piece is destroyed by the positive piece of the next bracket. Only the very first and very last survive: Sn = 1 - 1/(n + 1). The collapse is why these are called telescoping series.
4. Take the limit: lim(n -> infinity) [1 - 1/(n+1)] = 1.

The series converges to 1. Check the formula at n = 3: it predicts 1 - 1/4 = 3/4, and adding directly gives 1/2 + 1/6 + 1/12 = 6/12 + 2/12 + 1/12 = 9/12 = 3/4. It matches.

Key idea: when a term splits into a difference of consecutive pieces, the partial sum collapses and the series can be evaluated exactly rather than merely classified.

The geometric series, the one to know cold

In a geometric series, each term is a fixed ratio r times the term before it: a + a r + a r2 + a r3 + .... It converges exactly when |r| < 1, and then it has a clean closed-form sum:

sum from n=0 to infinity of a rn = a/(1 - r), valid when |r| < 1

If |r| >= 1, the terms do not shrink and the series diverges.

Worked example. Evaluate 1 + 1/2 + 1/4 + 1/8 + ...

1. Identify the first term and ratio: a = 1 and each term is half the one before, so r = 1/2.
2. Check the condition: |1/2| < 1, so the series converges.
3. Apply the formula: a/(1 - r) = 1/(1 - 1/2) = 1/(1/2) = 2.

The sum is exactly 2. Infinitely many positive numbers added up to a finite total, because the terms shrank quickly enough. As another example, sum from n=1 to infinity of (1/3)n = (1/3)/(1 - 1/3) = (1/3)/(2/3) = 1/2. Note that this one starts at n = 1, so its first term is 1/3, not 1.

Where the geometric formula comes from

The formula is not something to take on faith; it drops out of one algebraic trick.

1. Write the partial sum: Sn = a + ar + ar2 + ... + arn-1.
2. Multiply the whole thing by r: rSn = ar + ar2 + ... + arn-1 + arn. This is the same list shifted one place.
3. Subtract. Every middle term appears in both lines and cancels, leaving Sn - rSn = a - arn.
4. Factor and divide: Sn(1 - r) = a(1 - rn), so Sn = a(1 - rn)/(1 - r) whenever r is not 1.
5. Take the limit. The only part depending on n is rn. If |r| < 1 then rn -> 0 and Sn -> a/(1 - r). If |r| >= 1 then rn does not go to 0, and the partial sums have no finite limit.

Test the partial-sum formula on the example above: with a = 1, r = 1/2, n = 4, it predicts (1 - 1/16)/(1/2) = (15/16)(2) = 15/8 = 1.875. Adding directly: 1 + 0.5 + 0.25 + 0.125 = 1.875. Correct - and you can see the running total creeping up on 2, never reaching it.

Notice also where the condition |r| < 1 comes from. It is not a convention; it is precisely the condition under which rn dies out, and it fails at r = -1 as well as at r = 1, since (-1)n oscillates forever.

Two places geometric series show up immediately

Repeating decimals are geometric series. What rational number is 0.474747...? Write it as 47/100 + 47/10000 + 47/1000000 + ..., which is geometric with a = 47/100 and r = 1/100. The formula gives (47/100)/(1 - 1/100) = (47/100)/(99/100) = 47/99. Dividing 47 by 99 does indeed give 0.474747..., so every repeating decimal is a fraction - and the argument is a geometric series, not a trick.

The same machinery settles a famous argument. Is 0.999... = 1? As a series it is 0.9 + 0.09 + 0.009 + ..., geometric with a = 0.9 and r = 0.1, summing to 0.9/(1 - 0.1) = 0.9/0.9 = 1. Exactly 1, not merely close. The discomfort people feel comes from imagining the sum as a process that never finishes, whereas the definition makes it the limit of the partial sums, and that limit is 1.

A bouncing ball. A ball dropped from 10 metres rebounds to 60 percent of its previous height each bounce. How far does it travel in total? It falls 10 metres, then each subsequent bounce contributes an up-trip and a down-trip of equal length:

1. Rebound heights: 6, 3.6, 2.16, ..., geometric with a = 6 and r = 0.6.
2. Their total is 6/(1 - 0.6) = 6/0.4 = 15 metres.
3. Each of those heights is travelled twice, up and down, so the bouncing contributes 2 * 15 = 30 metres.
4. Add the initial drop: 10 + 30 = 40 metres.

The ball bounces infinitely many times and travels a finite 40 metres - and, since the durations also form a convergent geometric series, it does so in finite time. This is Zeno's dichotomy paradox with the arithmetic filled in: to cross a room you must first cross half, then half the remainder, and so on, an infinite list of tasks whose lengths sum to 1/2 + 1/4 + 1/8 + ... = 1. Zeno's argument that motion is therefore impossible has an unstated premise, that infinitely many positive quantities must add to infinity, and the geometric series shows the premise is false.

The nth term test for divergence

Here is a fast necessary check. If a series converges, its terms must shrink to 0. Turn that around: the nth term test says that if lim(n -> infinity) an is not 0 (or does not exist), the series diverges. For example, sum of n/(n + 1) diverges because an -> 1, not 0.

Two more worked applications. For sum of n/(2n + 1), divide top and bottom by n to get 1/(2 + 1/n) -> 1/2. Since 1/2 is not 0, the series diverges by the nth term test. For sum of cos(n), the terms oscillate without approaching anything, so the limit fails to exist and the series again diverges by the nth term test. Both conclusions took one line, which is the point of running this test first.

One firm warning: this test can only prove divergence. Terms going to 0 does not guarantee convergence, as the harmonic series in the next lesson will show. Reason about it this way: shrinking terms are necessary for the running total to settle, but they are not sufficient, because the terms might shrink too slowly to stop the total climbing. When lim an = 0, the correct thing to say is that the test is inconclusive - not that the series converges, and not that it diverges.

Key idea: the nth term test is a one-way door - it can send a series to "diverges" but it can never send one to "converges".

Try it

Evaluate sum from n=0 to infinity of (2/5)n.

Answer: this is geometric with first term a = 1 and ratio r = 2/5. Since |2/5| < 1, apply the formula: 1/(1 - 2/5) = 1/(3/5) = 5/3. Nice work checking the ratio before summing.

Where people get stuck

• Skipping the ratio condition. The formula a/(1 - r) only applies when |r| < 1. Applied blindly to 1 + 2 + 4 + 8 + ... it produces 1/(1 - 2) = -1, a negative sum for a series of positive terms. Check |r| first, every time.
• Getting the first term wrong. In a/(1 - r), the letter a is the actual first term you are adding. For sum from n=1 to infinity of (1/3)n that is 1/3, not 1, giving 1/2 rather than 3/2. Write out the first two terms before naming a and r.
• Reading the ratio off the exponent. The ratio is an+1/an, so compute it rather than guessing. For sum of 3n/4n+1, the ratio is 3/4 and the first term is 3/16 if the sum starts at n = 1.
• "If the terms go to 0, the series converges." The most consequential error in the module. The nth term test detects divergence only; when the limit is 0 the test says nothing at all, and the harmonic series is the standing counterexample.
• Confusing the terms with the partial sums. For sum 1/(n(n+1)) the terms go to 0 and the partial sums go to 1. Asking "does the sequence converge?" about a series question is a different question with a different answer.
• Expecting to evaluate every convergent series. Geometric and telescoping series can be summed exactly. Almost nothing else can, and the tests in the next three lessons decide convergence without producing a value.
• Treating 0.999... as an approximation. It is a limit, and the limit equals 1 exactly. Nothing is left over.

Recap

• An infinite series converges when its partial sums Sn approach a finite limit. Convergence is a statement about the running totals, not the terms.
• Telescoping series collapse: sum from n=1 to infinity of 1/(n(n+1)) has Sn = 1 - 1/(n+1), so it converges to 1.
• A geometric series sums to a/(1 - r) when |r| < 1, and diverges when |r| >= 1. The formula comes from Sn = a(1 - rn)/(1 - r) and the fact that rn -> 0 exactly when |r| < 1.
• Identify a as the actual first term and r as the common ratio, and mind where the sum begins.
• Repeating decimals are geometric series: 0.474747... = 47/99 and 0.999... = 1 exactly.
• The nth term test: if the terms do not go to 0, the series diverges. If they do go to 0, the test is inconclusive - it proves nothing either way.
• Convergent geometric series dissolve Zeno's dichotomy paradox: infinitely many positive lengths can sum to a finite total.

Sources

1. OpenStax. (2016). Infinite series. In Calculus volume 2. Rice University. openstax.org
2. Strang, G., & Herman, E. (n.d.). 9.2: Infinite series. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
3. Dawkins, P. (n.d.). Series, the basics. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Dawkins, P. (n.d.). Convergence and divergence of series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
5. Huggett, N. (2024). Zeno's paradoxes. In E. N. Zalta (Ed.), The Stanford encyclopedia of philosophy. Stanford University. plato.stanford.edu
6. O'Connor, J. J., & Robertson, E. F. (n.d.). Zeno of Elea. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk
7. Khan Academy. (n.d.). Series. khanacademy.org

The geometric series was special because we could compute its exact sum. Most series are not that friendly, and often you do not need the sum at all, only a yes-or-no answer: does it converge? This lesson gives you a small toolbox of tests that decide convergence for series with positive terms, without ever finding the total.

The big picture

For a series of positive terms, a handful of tests settle convergence by comparing the series to something you already understand, either an integral or a simpler series. Start every problem with the nth term test from last lesson (if the terms do not go to 0, you are done, it diverges), then reach for one of the tools below.

Key idea: convergence tests decide whether a positive series adds up to a finite total by comparing it to a known integral or a known series.

The integral test and the p-series

If an = f(n) for a function f that is positive, decreasing, and continuous, then the series sum of an and the improper integral integral from 1 to infinity of f(x) dx either both converge or both diverge. This ties series back to the improper integrals you already studied.

The most useful consequence is the p-series rule:

sum from n=1 to infinity of 1/np converges if and only if p > 1

So sum of 1/n2 converges (p = 2), while the famous harmonic series sum of 1/n diverges (p = 1), even though its terms go to 0. The harmonic series is the classic reminder that shrinking terms are not enough for convergence.

Why the integral test works

The argument is a picture, and once you see it the test stops feeling like an unrelated coincidence.

1. Draw the curve y = f(x), positive and decreasing.
2. Over each interval [n, n+1], draw a rectangle of width 1 and height f(n), the value at the left end. Because f is decreasing, this rectangle sits entirely above the curve. Adding them: a1 + a2 + ... + aN-1 >= integral from 1 to N of f(x) dx.
3. Now use the right-end value instead, height f(n+1). Those rectangles sit entirely below the curve, giving a2 + a3 + ... + aN <= integral from 1 to N of f(x) dx.
4. Combine. The partial sums and the integral are pinned within one term of each other, forever. If the integral is finite, the second inequality caps the partial sums, and an increasing bounded sequence converges. If the integral is infinite, the first inequality forces the partial sums up with it.

Two conditions did real work. Decreasing is what puts the rectangles on the correct side of the curve. Positive is what makes the partial sums increasing, so that boundedness implies convergence. Drop either and the argument collapses; the test genuinely does not apply to sum of (-1)n/n, which needs Lesson 16.

Note also what the test does not give you. The series and the integral converge together, but they almost never converge to the same number. integral from 1 to infinity of dx/x2 = 1, while sum of 1/n2 = pi2/6 ~= 1.6449 - the value Euler found in 1735, solving a problem that had defeated everyone for decades.

The harmonic series, worked two ways

Why does sum of 1/n diverge when its terms so obviously shrink to zero? Two independent arguments settle it.

Argument one, the integral test.

1. Set f(x) = 1/x. Check the hypotheses: it is positive on [1, infinity), continuous there, and decreasing since its derivative -1/x2 is negative. All three hold.
2. Evaluate the integral: integral from 1 to t of dx/x = [ln x] from 1 to t = ln(t).
3. Take the limit: ln(t) -> infinity, so the improper integral diverges.
4. Conclude: the series sum of 1/n diverges by the integral test.

The rectangle picture also tells you how it diverges: the partial sums track ln(N). Precisely, SN ~= ln(N) + 0.5772, where the constant is the Euler-Mascheroni number. The numbers are startling: S10 ~= 2.93, S100 ~= 5.19, S1000 ~= 7.49, and S1000000 ~= 14.39. A million terms have got you to fourteen. To reach 100 you would need roughly 1043 terms. No amount of numerical evidence would ever reveal this divergence - only the proof does.

Argument two, Oresme's grouping. Nicole Oresme found this around 1350, three centuries before calculus existed. Group the terms in blocks of doubling length:

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + ... + 1/16) + ...

1. The block 1/3 + 1/4 is larger than 1/4 + 1/4 = 1/2, because 1/3 exceeds 1/4.
2. The block 1/5 + 1/6 + 1/7 + 1/8 is larger than four copies of 1/8, which is 1/2.
3. In general the block from 1/(2k+1) to 1/2k+1 has 2k terms, each larger than 1/2k+1, so the block exceeds 2k/2k+1 = 1/2.
4. So the total exceeds 1 + 1/2 + 1/2 + 1/2 + ..., infinitely many halves. It cannot be finite.

This also explains the slowness: you need to double the number of terms just to add another half, which is exactly logarithmic growth.

And why p = 2 converges. Run the identical test on f(x) = 1/x2, which is likewise positive, continuous, and decreasing on [1, infinity). Then integral from 1 to t of x-2 dx = [-1/x] from 1 to t = 1 - 1/t -> 1. The integral converges, so sum of 1/n2 converges by the integral test. The rectangle inequality even bounds it: the sum is at most a1 + integral from 1 to infinity of f = 1 + 1 = 2, and the true value pi2/6 ~= 1.645 obeys that bound.

The whole difference between the two cases is the antiderivative. For 1/x it is ln(x), which grows without bound; for 1/x2 it is -1/x, which levels off at 0. The terms of both series shrink to zero, but only the second shrinks fast enough for the accumulated total to settle.

Key idea: convergence is decided by how fast the terms shrink, not whether they shrink, and 1/n is the exact borderline where "fast enough" fails.

Estimating the sum with the integral test

The same rectangles bound the error left after stopping at N terms. Writing RN = S - SN for the tail,

integral from N+1 to infinity of f(x) dx <= RN <= integral from N to infinity of f(x) dx

For sum of 1/n2 with N = 10, the right bound is integral from 10 to infinity of dx/x2 = 1/10 = 0.1 and the left bound is 1/11 ~= 0.0909. So ten terms leave an error between 0.0909 and 0.1. The true value confirms it: S10 = 1.549768 and pi2/6 = 1.644934, a gap of 0.09517. Inside the predicted window, as promised. This is how you decide in advance how many terms a computation needs.

The comparison test

Compare your mystery series to a known one, term by term. If 0 <= an <= bn, then:

• If the bigger series sum of bn converges, the smaller sum of an converges too.
• If the smaller series sum of an diverges, the bigger sum of bn diverges too.

Worked example one. Does sum of 1/(n2 + 1) converge? Since n2 + 1 > n2, a bigger denominator makes a smaller fraction: 1/(n2 + 1) < 1/n2. The larger series sum of 1/n2 converges as a p-series with p = 2, so the smaller one converges by the direct comparison test.

Worked example two, the other direction. Does sum from n=1 to infinity of ln(n)/n converge? For every n >= 3 we have ln(n) > 1, so ln(n)/n > 1/n. The smaller series sum of 1/n is the harmonic series, which diverges. A divergent lower bound pushes everything above it up, so the series diverges by the direct comparison test. Dropping the first two terms is harmless: adding or removing finitely many terms never changes whether a series converges, only what it converges to.

Watch the direction. Showing 1/(n2 + 1) < 1/n is perfectly true and completely useless, because being smaller than something divergent tells you nothing. The only two valid conclusions are convergent-upper-bound and divergent-lower-bound.

The limit comparison test

Direct comparison can be fussy about which way the inequality points, so the limit comparison test is often smoother. If an and bn are positive and lim(n -> infinity) an/bn = c for some finite positive number c, then the two series do the same thing: converge together or diverge together.

Worked example. Does sum of (2n + 1)/(n3 + 5) converge?

1. Keep only the dominant powers to guess a comparison. For large n it behaves like 2n/n3 = 2/n2.
2. Compare with bn = 1/n2. Form the ratio an/bn = (2n + 1)n2/(n3 + 5), which approaches 2.
3. Since 2 is a finite positive number and sum of 1/n2 converges, the original series converges by the limit comparison test.

Do the ratio carefully, since this is where slips happen. an/bn = [(2n+1)/(n3+5)] * n2 = (2n3 + n2)/(n3 + 5). Divide top and bottom by n3: (2 + 1/n)/(1 + 5/n3) -> 2. Finite and positive, so the test applies.

The limit comparison test is usually the right first choice for anything that looks like a ratio of polynomials, because the comparison series is read straight off the leading powers and no inequality has to be arranged.

Choosing a test: a short decision guide

1. nth term test. Does an fail to approach 0? Then it diverges and you are finished. Otherwise carry on - this tells you nothing more.
2. Is it geometric or telescoping? Then evaluate it exactly rather than testing it.
3. Is it a p-series, or a ratio of polynomials? Use the p-rule directly, or limit comparison against 1/np with p = (bottom degree) minus (top degree).
4. Does the term look like a function you can integrate? Terms with logarithms or 1/(n ln n) shapes are integral-test problems.
5. Can you bound it above by something convergent, or below by something divergent? Direct comparison.
6. Factorials or nth powers present? Save it for the ratio and root tests in the next lesson.

Two habits matter more than the list. First, state the test by name in your conclusion - "converges by limit comparison with 1/n2" is an answer, while "converges" is a guess. Second, check the hypotheses before applying a test, because every one of these tests requires positive terms and the integral test additionally requires a decreasing continuous function.

Try it

Does sum of 1/(n2 + n) converge?

Answer: yes. Since 1/(n2 + n) < 1/n2 and sum of 1/n2 is a convergent p-series (p = 2), the comparison test shows it converges. Nice work spotting the p-series to compare against.

Where people get stuck

• "The harmonic series converges because its terms go to 0." It diverges, by the integral test and by Oresme's grouping. It is the p-series at p = 1, exactly on the boundary where convergence fails. Numerical evidence is no help: a million terms only reach 14.4.
• "Any p-series converges." Only for p > 1. At p = 1 and below it diverges, and p = 0.999 diverges while p = 1.001 converges - the threshold is genuinely sharp.
• Confusing the two p-thresholds. For series sum 1/np and for the improper integral integral from 1 to infinity dx/xp, the condition is the same, p > 1. It is integral from 0 to 1 dx/xp that flips to p < 1. Match the interval before quoting a rule.
• Comparing in a useless direction. The only valid conclusions are convergent-upper-bound and divergent-lower-bound. Showing your series is smaller than a divergent one, or bigger than a convergent one, proves nothing.
• Skipping the hypotheses of the integral test. The function must be positive, continuous, and decreasing. f(x) = (sin2x)/x2 is not decreasing, so the test does not apply even though the series converges for other reasons.
• Believing the integral equals the sum. They converge together and rarely agree. integral from 1 to infinity of dx/x2 = 1 while the series sums to pi2/6 ~= 1.645. The integral test gives a verdict, not a value.
• Getting the limit-comparison ratio upside down. Either orientation works as long as the limit is finite and positive, but a limit of 0 or infinity makes the test inconclusive in one direction, so simplify carefully rather than guessing.
• Announcing a verdict without a test. Always name the test and the comparison series. That habit is what makes an answer checkable.

Recap

• Start with the nth term test, then pick a tool for positive series, and name the test in your conclusion.
• The integral test compares a series with an improper integral using rectangles that straddle the curve; it needs f positive, continuous, and decreasing.
• The p-series sum of 1/np converges exactly when p > 1. The harmonic series (p = 1) diverges because integral dx/x = ln x grows without bound, and Oresme's blocks each exceed 1/2. The p = 2 series converges because integral dx/x2 levels off at 1.
• The harmonic partial sums grow like ln(N) + 0.5772: a million terms reach only 14.4, and 100 would take about 1043 terms.
• The integral test also bounds the tail: integral from N+1 to infinity of f <= RN <= integral from N to infinity of f. For sum 1/n2 at N = 10 that predicts an error between 0.0909 and 0.1, and the true error is 0.0952.
• Comparison test: a convergent upper bound pulls the smaller series down; a divergent lower bound pushes the bigger one up. No other direction concludes anything.
• Limit comparison: if an/bn approaches a finite positive c, both series share the same fate. Read the comparison series off the leading powers.

Sources

1. OpenStax. (2016). The divergence and integral tests. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Comparison tests. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 9.3: The divergence and integral tests. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Strang, G., & Herman, E. (n.d.). 9.4: Comparison tests. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
5. Dawkins, P. (n.d.). Integral test. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Dawkins, P. (n.d.). Strategy for series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
7. O'Connor, J. J., & Robertson, E. F. (n.d.). Nicholas Oresme. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk

Two kinds of series need tools we have not built yet: series whose signs flip back and forth, and series stuffed with factorials or nth powers. This lesson adds an alternating series test and a ratio test to your kit, plus a sharper way to talk about convergence that will matter for the power series coming next.

The big picture

When signs alternate, a series can converge even if the same terms without signs would not, because the plus and minus terms partly cancel. When terms contain factorials or powers, the ratio of consecutive terms reveals whether they shrink fast enough. These two tests, plus the idea of absolute convergence, round out how you classify series.

Key idea: alternating signs can rescue convergence, and the ratio of consecutive terms measures how fast the terms shrink.

The alternating series test

An alternating series flips sign every term, like sum of (-1)n bn with bn > 0. The alternating series test says it converges if two conditions both hold:

1. The sizes bn are eventually decreasing.
2. lim(n -> infinity) bn = 0.

Worked example. The alternating harmonic series 1 - 1/2 + 1/3 - 1/4 + ... has bn = 1/n. Check both conditions: 1/n decreases as n grows, and 1/n -> 0. Both hold, so it converges by the alternating series test. Its sum is ln 2 ~= 0.6931. Compare that with the plain harmonic series, which diverges: the alternating signs are what pull this one back to convergence.

Why the alternating series test works

Watch the partial sums of 1 - 1/2 + 1/3 - 1/4 + ... as they are built:

S1 = 1, S2 = 0.5, S3 = 0.8333, S4 = 0.5833, S5 = 0.7833, S6 = 0.6167

They overshoot, undershoot, overshoot, undershoot - and each swing is shorter than the last, because the terms are shrinking. Precisely: the odd partial sums decrease, the even ones increase, and every even one stays below every odd one. The two subsequences squeeze together, and since the gap between consecutive partial sums is bn, which goes to 0, they must meet at a single value. That value is the sum.

Both hypotheses earn their place in that argument. Decreasing is what makes each swing shorter than the last, so the intervals nest. Terms going to 0 is what shrinks the nested intervals to a point. Drop the first and the argument fails even if the terms tend to 0 - which is why "eventually decreasing" must actually be checked, not assumed from the fact that a formula looks like it should shrink.

The alternating series error bound

The nesting gives a bonus that no other test in this course provides: a guaranteed error bound, for free. Since the true sum always lies between consecutive partial sums,

|S - SN| <= bN+1

The error after N terms is no bigger than the first term you left out. Test it on the alternating harmonic series, whose sum is ln 2 = 0.693147. Stopping at S5 = 0.783333, the bound promises an error at most b6 = 1/6 = 0.1667. The actual error is 0.783333 - 0.693147 = 0.090186, comfortably inside. Correct.

Now use it in reverse, which is the practical direction. How many terms give three-decimal accuracy? Solve bN+1 = 1/(N+1) < 0.0005, giving N + 1 > 2000, so about 2000 terms. That is painfully slow, and it explains why nobody computes ln 2 this way.

Compare a faster series: sum from n=1 to infinity of (-1)n-1/n2 = pi2/12 ~= 0.822467. Here S4 = 1 - 0.25 + 0.111111 - 0.0625 = 0.798611, and the bound is b5 = 1/25 = 0.04. The actual error is 0.023856, again inside the bound. Four terms of the squared version beat five terms of the harmonic one, because its terms shrink faster.

Key idea: for an alternating series meeting both conditions, the error is bounded by the first omitted term - the cheapest error estimate in all of series analysis.

The ratio test

The ratio test is your go-to when terms contain factorials or nth powers. Compute the limit of the size of the ratio of consecutive terms:

L = lim(n -> infinity) |an+1/an|

• If L < 1, the series converges (in fact absolutely).
• If L > 1, the series diverges.
• If L = 1, the test is inconclusive, so try a different test.

Worked example. Does sum of 2n/n! converge?

1. Write the ratio of the next term to the current one: an+1/an = [2n+1/(n+1)!] * [n!/2n].
2. Cancel: the powers give one factor of 2, and the factorials give 1/(n + 1), so the ratio is 2/(n + 1).
3. Take the limit: 2/(n + 1) -> 0, which is less than 1, so the series converges.

Factorials in the denominator almost always win, driving the ratio to 0.

A harder ratio. Does sum of n!/nn converge? Here a factorial fights an nth power, so the answer is not obvious.

1. Form the ratio: an+1/an = [(n+1)!/(n+1)n+1] * [nn/n!].
2. The factorials simplify: (n+1)!/n! = n + 1. So the ratio is (n + 1) * nn/(n+1)n+1.
3. Split the power: (n+1)n+1 = (n+1)(n+1)n, so one factor of n + 1 cancels, leaving nn/(n+1)n = (n/(n+1))n.
4. Rewrite as 1/(1 + 1/n)n, and recall that (1 + 1/n)n -> e. So L = 1/e ~= 0.368.
5. Since L < 1, the series converges by the ratio test. The nth power beats the factorial, matching the growth hierarchy from Lesson 13.

Why L = 1 really is inconclusive. Do not take this on trust; two examples settle it. For sum of 1/n, the ratio is n/(n+1) -> 1, and the series diverges. For sum of 1/n2, the ratio is n2/(n+1)2 -> 1, and the series converges. Same L, opposite outcomes, so no rule could possibly decide from L alone. When L = 1, go to the integral test or a comparison - and note that the ratio test is always inconclusive on p-series, which is exactly the family it should never be used on.

The root test

Where the ratio test compares consecutive terms, the root test looks at the nth root of a single term:

L = lim(n -> infinity) |an|1/n

The conclusions are identical: L < 1 converges absolutely, L > 1 diverges, L = 1 is inconclusive. Reach for it when the whole term is raised to the nth power, since the root then cancels the exponent in one step.

Worked example. Does sum of (n/(2n + 1))n converge? Take the nth root and the outer exponent disappears: |an|1/n = n/(2n + 1). Divide top and bottom by n: 1/(2 + 1/n) -> 1/2. Since 1/2 < 1, the series converges by the root test. The ratio test would have required expanding ((n+1)/(2n+3))n+1 against (n/(2n+1))n, which is genuinely unpleasant.

Second example. Does sum from n=2 to infinity of 1/(ln n)n converge? The nth root is 1/ln(n), which goes to 0 as n grows. Since 0 < 1, it converges by the root test.

Absolute versus conditional convergence

A series sum of an converges absolutely if the series of sizes sum of |an| converges. Absolute convergence is the stronger, better-behaved kind, and it guarantees ordinary convergence. If a series converges but the absolute-value version diverges, it is conditionally convergent. The alternating harmonic series is the star example: sum of (-1)n/n converges, but sum of 1/n diverges, so it converges conditionally. This distinction matters because you can safely rearrange an absolutely convergent series, but rearranging a conditionally convergent one can actually change its sum.

That last sentence deserves more than a mention, because it is one of the genuinely startling facts in mathematics. Riemann's rearrangement theorem says that a conditionally convergent series can be reordered to converge to any real number you choose, or to diverge to plus or minus infinity. Nothing is added or removed; only the order changes.

Here is why it is possible. In the alternating harmonic series the positive terms alone, 1 + 1/3 + 1/5 + ..., sum to infinity, and the negative terms alone, -1/2 - 1/4 - ..., sum to minus infinity. So you can pile up positive terms until you pass any target, then add negative terms until you fall below it, then more positives, and so on. Since all the terms shrink to 0, the overshoots shrink too, and the running total closes in on whatever target you named.

A concrete case: take the alternating harmonic series and rearrange it as 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + ..., two positive terms for every negative one. This rearrangement converges to (3/2)ln(2) ~= 1.0397, not to ln(2) ~= 0.6931. Same terms, same signs, different order, fifty percent larger sum.

For an absolutely convergent series none of this can happen: every rearrangement converges to the same value. That is the practical reason absolute convergence is the property worth wanting, and it is why the next lesson's power series manipulations - differentiating and integrating term by term - are stated inside the interval where convergence is absolute.

Key idea: absolute convergence means the series behaves like a finite sum under reordering; conditional convergence means the order is part of the answer.

Try it

Use the ratio test on sum of n/2n.

Answer: the ratio is an+1/an = [(n + 1)/2n+1] * [2n/n] = (n + 1)/(2n), which approaches 1/2. Since 1/2 is less than 1, the series converges. Nice work simplifying the ratio before taking the limit.

Where people get stuck

• Checking only one condition of the alternating series test. The terms must be decreasing and go to 0. People routinely verify the limit and skip the decreasing check, which is exactly what makes the nested-interval argument work.
• Concluding divergence from L = 1. It is inconclusive, full stop. sum 1/n and sum 1/n2 both give L = 1 and behave oppositely. Switch to the integral test or a comparison.
• Using the ratio test on p-series. It is always inconclusive there, because polynomial terms give a ratio limit of exactly 1. Recognise a p-series and use the p-rule instead.
• Forgetting the absolute values in the ratio test. The test uses |an+1/an|, which is why it detects absolute convergence and works on alternating series without modification.
• Applying the alternating error bound to a series that fails the test. The bound |S - SN| <= bN+1 depends on the nesting, which depends on both hypotheses. If the terms are not decreasing, the bound is not available.
• Confusing absolute with conditional. Absolute means the all-positive version also converges; conditional means only the signed version does. The alternating harmonic series is the standard conditional example, and the alternating 1/n2 series is absolute.
• Assuming a convergent series can be reordered. Only absolutely convergent ones. Riemann's theorem says a conditionally convergent series can be rearranged to hit any target you name.
• Reaching for the root test when nothing is raised to the nth power. Its advantage is cancelling an outer exponent. Without one, the ratio test is usually easier.

Recap

• Alternating series test: sum of (-1)n bn converges if bn is eventually decreasing and tends to 0. The proof nests the odd and even partial sums.
• Error bound, free with the test: |S - SN| <= bN+1, the first omitted term. For the alternating harmonic series, three decimals need about 2000 terms.
• Ratio test: with L = lim |an+1/an|, L < 1 converges absolutely, L > 1 diverges, L = 1 is inconclusive. Best for factorials and exponentials; sum n!/nn gives L = 1/e and converges.
• Root test: with L = lim |an|1/n, same three conclusions. Best when the whole term carries an nth power; sum (n/(2n+1))n gives L = 1/2 and converges.
• Absolute convergence (the sizes converge) implies convergence and allows safe rearrangement.
• Conditional convergence means the signed series converges but the all-positive version diverges. Riemann's rearrangement theorem shows such a series can be reordered to reach any value; regrouping the alternating harmonic series two-positives-to-one-negative gives (3/2)ln 2 instead of ln 2.

Sources

1. OpenStax. (2016). Alternating series. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Ratio and root tests. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 9.5: Alternating series. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Strang, G., & Herman, E. (n.d.). 9.6: Ratio and root tests. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
5. Dawkins, P. (n.d.). Alternating series test. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Dawkins, P. (n.d.). Ratio test. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
7. Dawkins, P. (n.d.). Absolute convergence. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu

Polynomials are the easiest functions in all of mathematics: to evaluate one you only add and multiply. A power series takes that idea to the limit, an "infinite polynomial." The payoff is huge, because many complicated functions can be rewritten as a power series, and then they become as easy to compute as a polynomial. This lesson sets up power series and asks the key question: for which x does the infinite sum actually make sense?

The big picture

A power series is a series whose terms are constants times powers of (x - c). Centered at c, it looks like:

sum from n=0 to infinity of an (x - c)n = a0 + a1(x - c) + a2(x - c)2 + ...

The number c is the center and the an are the coefficients. For each fixed value of x, this is just a series of numbers that either converges or diverges. So a power series quietly defines a function, whose domain is the set of x-values where it converges.

Key idea: a power series is an infinite polynomial in (x - c); plug in a number for x and it becomes an ordinary series that may converge or diverge.

Radius and interval of convergence

A power series converges on an interval centered at c. The radius of convergence R is half the length of that interval: the series converges for |x - c| < R and diverges for |x - c| > R. The two endpoints must be tested separately, because either one might converge or diverge. We almost always find R with the ratio test.

Worked example, an infinite radius. Find where sum from n=0 to infinity of xn/n! converges (this one is centered at 0).

1. Form the ratio of consecutive terms: |an+1/an| = |xn+1/(n+1)! * n!/xn| = |x|/(n + 1).
2. Take the limit: lim(n -> infinity) |x|/(n + 1) = 0 for every fixed x.
3. Since the limit 0 is less than 1 no matter what x is, the series converges for all real x. So R = infinity.

This particular series turns out to equal ex everywhere, which the next lesson explains.

Worked example, a finite radius

Find the radius of convergence of sum from n=0 to infinity of xn.

1. Recognize it: this is geometric with ratio x, so it converges exactly when |x| < 1.
2. Therefore R = 1, and on that interval the sum is 1/(1 - x).

So 1/(1 - x) = 1 + x + x2 + x3 + ... for |x| < 1, our first everyday function written as a power series. Check it at x = 0.5: the left side is 1/0.5 = 2, and the right side is 1 + 0.5 + 0.25 + 0.125 + ... = 2, the geometric series from Lesson 14. They agree.

There is a third possibility besides R = infinity and a finite R. Consider sum of n! xn. The ratio is |(n+1)! xn+1/(n! xn)| = (n + 1)|x|, which tends to infinity for every nonzero x. So the series converges only at x = 0, and R = 0. Those are all the cases: R is 0, a positive number, or infinity.

Finding a full interval of convergence, endpoints included

The radius is only half the job. Here is the complete procedure on sum from n=1 to infinity of (x - 2)n/(n * 3n).

1. Form the ratio. |an+1/an| = |(x-2)n+1/((n+1)3n+1)| * |(n)(3n)/(x-2)n|. The powers of (x-2) leave one factor, the powers of 3 leave 1/3, and the n's leave n/(n+1): the ratio is (|x - 2|/3) * n/(n+1).
2. Take the limit. n/(n+1) -> 1, so L = |x - 2|/3.
3. Impose L < 1. That gives |x - 2| < 3, so R = 3. The centre is 2, so the open interval runs from 2 - 3 = -1 to 2 + 3 = 5.
4. Test the right endpoint, x = 5. Then x - 2 = 3, and the term becomes 3n/(n * 3n) = 1/n. That is the harmonic series, which diverges. So 5 is excluded.
5. Test the left endpoint, x = -1. Then x - 2 = -3, and the term becomes (-3)n/(n * 3n) = (-1)n/n. That is the alternating harmonic series, which converges by the alternating series test. So -1 is included.
6. State the interval: the series converges exactly on [-1, 5).

Look at what happened at the two ends. Same series, same radius, and opposite verdicts - one endpoint in, one out. That is why the ratio test alone never finishes the problem: it is silent precisely when L = 1, which is exactly what happens at both endpoints.

A second interval, with both endpoints in. Take sum from n=1 to infinity of (x + 1)n/n2.

1. Ratio: |x + 1| * n2/(n+1)2 -> |x + 1|. So R = 1, centred at c = -1, giving the open interval from -2 to 0.
2. At x = 0: the term is 1/n2, a convergent p-series with p = 2. Converges.
3. At x = -2: the term is (-1)n/n2, which converges absolutely since sum 1/n2 converges. Converges.
4. Interval of convergence: the closed interval [-2, 0].

Key idea: the ratio test gives you R; only a separate test at each endpoint gives you the interval, and the two ends genuinely can disagree.

Building new series from old ones

Inside the radius of convergence, a power series behaves like a polynomial: you may add, multiply by a constant, substitute, differentiate, and integrate term by term, and the radius does not change (endpoints may). This is what makes the single geometric series worth so much.

Differentiate it. Starting from 1/(1 - x) = 1 + x + x2 + x3 + ... and differentiating both sides:

1/(1 - x)2 = 1 + 2x + 3x2 + 4x3 + ... = sum from n=1 to infinity of n xn-1, for |x| < 1

Check numerically at x = 0.5. The left side is 1/(0.5)2 = 4. Summing the right side: 1 + 1 + 0.75 + 0.5 + 0.3125 + 0.1875 + 0.109 + 0.063 = 3.92 after eight terms, and climbing toward 4. Correct.

Integrate it. Integrating 1/(1 - x) gives -ln(1 - x), and integrating the series term by term gives x + x2/2 + x3/3 + .... So

-ln(1 - x) = sum from n=1 to infinity of xn/n, for |x| < 1

Check at x = 0.5: the left side is -ln(0.5) = ln 2 = 0.6931, and the right side is 0.5 + 0.125 + 0.0417 + 0.0156 + 0.0063 + ... = 0.6931. Correct. Setting x = -1 gives -ln(2) = -1 + 1/2 - 1/3 + ..., which rearranges to the alternating harmonic sum ln 2 we quoted in the last lesson - now derived rather than asserted.

Substitute, then integrate. Replace x by -x2 in the geometric series: 1/(1 + x2) = 1 - x2 + x4 - x6 + ..., valid for |x| < 1. Integrating term by term, and using integral dx/(1 + x2) = arctan(x):

arctan(x) = x - x3/3 + x5/5 - x7/7 + ...

At x = 1 this endpoint converges by the alternating series test, and since arctan(1) = pi/4, we get pi/4 = 1 - 1/3 + 1/5 - 1/7 + .... This is the Gregory-Leibniz series, found by James Gregory in 1671. It is beautiful and useless for computation: by the alternating error bound, ten decimal places would need about 1010 terms.

Three genuinely different functions - a rational function, a logarithm, an inverse trig function - all extracted from one geometric series by calculus. That leverage is the reason power series matter.

Try it

Find the radius of convergence of sum of xn/2n.

Answer: this is geometric with ratio x/2, which converges when |x/2| < 1, that is |x| < 2. So R = 2. Nice work reading off the ratio as x/2.

Where people get stuck

• Confusing radius with interval. The radius R is a single number, the reach from the centre. The interval of convergence is the actual set of x-values, obtained by taking that reach and then testing both endpoints. "R = 3" is not an answer to "find the interval of convergence."
• Skipping the endpoint checks. This is the standard oversight, and the two ends genuinely can disagree - sum (x-2)n/(n 3n) converges at x = -1 and diverges at x = 5.
• Trying to use the ratio test at an endpoint. At an endpoint L = 1 exactly, so the ratio test is silent by construction. Substitute the endpoint value and treat the result as an ordinary numerical series.
• Forgetting the centre when writing the interval. With |x - 2| < 3 the interval is from -1 to 5, not from -3 to 3. Solve the inequality rather than reading R off as if the centre were 0.
• Assuming the radius is always positive. sum n! xn has R = 0 and converges only at its centre; sum xn/n! has R = infinity. Both are legitimate answers.
• Believing endpoint behaviour survives differentiation. Term-by-term differentiation and integration preserve R but can change what happens at the endpoints. Recheck them after any such operation.
• Treating a power series as a formal object with no domain. Outside its interval the sum simply does not exist. Writing 1/(1 - x) = 1 + x + x2 + ... at x = 2 would give -1 = 1 + 2 + 4 + ..., which is nonsense.

Recap

• A power series is sum of an(x - c)n, an infinite polynomial centred at c, and it defines a function on the set of x where it converges.
• It converges for |x - c| < R and diverges for |x - c| > R. R can be 0, a positive number, or infinity.
• Find R with the ratio test, then test each endpoint separately as an ordinary numerical series. sum (x-2)n/(n 3n) has R = 3 and interval [-1, 5); sum (x+1)n/n2 has R = 1 and interval [-2, 0].
• Inside R, a power series can be added, substituted into, differentiated, and integrated term by term, with R unchanged.
• From 1/(1-x) = sum xn alone we obtained 1/(1-x)2 = sum n xn-1, -ln(1-x) = sum xn/n, and arctan(x) = x - x3/3 + x5/5 - ..., whose value at x = 1 gives pi/4.

Sources

1. OpenStax. (2016). Power series and functions. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Properties of power series. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 10.1: Power series and functions. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Strang, G., & Herman, E. (n.d.). 10.2: Properties of power series. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
5. Dawkins, P. (n.d.). Power series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Dawkins, P. (n.d.). Power series and functions. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
7. O'Connor, J. J., & Robertson, E. F. (n.d.). James Gregory. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk

How does a calculator find cos(0.1) or e²? It does not have them stored in a giant table. Instead it adds up a few terms of a Taylor series. Last lesson we wrote a couple of functions as power series almost by luck. Taylor series give a systematic recipe to turn nearly any smooth function into an infinite polynomial, using its derivatives at a single point.

The big picture

The Taylor series of f centered at c is:

f(x) = sum from n=0 to infinity of [f(n)(c)/n!] (x - c)n

Here f(n)(c) means the nth derivative of f, evaluated at c. Written out, that is f(c) + f'(c)(x - c) + [f''(c)/2!](x - c)2 + [f'''(c)/3!](x - c)3 + .... When the center is c = 0, the expansion has its own name, a Maclaurin series, which is the most common case.

Key idea: a Taylor series rebuilds a function from its derivatives at one point, with each term scaled by 1/n!.

Worked example, building e to the x

Find the Maclaurin series of f(x) = ex.

1. Differentiate repeatedly. Every derivative of e^x is again e^x.
2. Evaluate each at 0. Since e⁰ = 1, every derivative gives f(n)(0) = 1.
3. Plug into the formula: sum of [1/n!] xn = 1 + x + x2/2! + x3/3! + ...

So ex = sum from n=0 to infinity of xn/n!, matching the series from last lesson. It converges for all x.

Where the formula comes from

The 1/n! is not a convention to memorise; it is forced. Suppose f can be written as a power series about c, f(x) = a0 + a1(x - c) + a2(x - c)2 + a3(x - c)3 + ..., and hunt for the coefficients.

1. Set x = c. Every term with a factor (x - c) dies, leaving f(c) = a0.
2. Differentiate once: f'(x) = a1 + 2a2(x - c) + 3a3(x - c)2 + .... Set x = c again: f'(c) = a1.
3. Differentiate twice: f''(x) = 2a2 + 6a3(x - c) + .... Setting x = c gives f''(c) = 2a2, so a2 = f''(c)/2.
4. In general, differentiating an(x - c)n exactly n times produces n * (n-1) * ... * 1 * an = n! an, and every other term either has died or still carries a factor of (x - c). So f(n)(c) = n! an, giving an = f(n)(c)/n!.

The factorial is simply what falls out of differentiating a power n times. And notice the logic: if a power series representation exists, its coefficients must be these. Whether the resulting series actually converges back to f is a separate question, settled by the remainder below.

Deriving the sine series from scratch

Do it the long way once, computing every derivative, so the pattern is earned rather than quoted. Take f(x) = sin(x) and c = 0.

1. f(x) = sin(x), so f(0) = 0.
2. f'(x) = cos(x), so f'(0) = 1.
3. f''(x) = -sin(x), so f''(0) = 0.
4. f'''(x) = -cos(x), so f'''(0) = -1.
5. f(4)(x) = sin(x), so f(4)(0) = 0 - and we are back where we started. The derivatives cycle with period 4, and the values at 0 cycle as 0, 1, 0, -1.

Now assemble, term by term, dividing each derivative value by n!:

0 + (1)x + (0)x2/2! + (-1)x3/3! + (0)x4/4! + (1)x5/5! + ... = x - x3/3! + x5/5! - x7/7! + ...

Every even power vanished because the even derivatives of sine at 0 are all 0, which is the analytic reflection of sine being an odd function. Check the first two terms against a calculator at x = 0.2: 0.2 - 0.008/6 = 0.198667, and sin(0.2) = 0.198669. Two terms, five correct digits.

The three Maclaurin series worth memorizing

These come up again and again across mathematics and physics.

Function	Maclaurin series
ex	1 + x + x2/2! + x3/3! + ...
sin(x)	x - x3/3! + x5/5! - x7/7! + ...
cos(x)	1 - x2/2! + x4/4! - x6/6! + ...

Notice sine keeps only odd powers and cosine only even powers, matching the symmetry of each function. All three converge for every real x.

Finding the interval of convergence

"Converges for every real x" is a claim, and the ratio test proves it. For the sine series the terms are (-1)nx2n+1/(2n+1)!, so

|an+1/an| = |x2n+3/(2n+3)!| * |(2n+1)!/x2n+1| = x2/[(2n+2)(2n+3)]

For any fixed x the numerator is a constant while the denominator grows without bound, so the limit is 0, which is less than 1 for every x. Hence R = infinity and the interval of convergence is all real numbers. The same argument works for ex and cos(x): factorials in the denominator always win.

A series with a finite interval. Derive the Maclaurin series of f(x) = ln(1 + x) and find exactly where it is valid.

1. Compute derivatives at 0: f(0) = ln(1) = 0; f'(x) = 1/(1+x) so f'(0) = 1; f''(x) = -1/(1+x)2 so f''(0) = -1; f'''(x) = 2/(1+x)3 so f'''(0) = 2; f(4)(x) = -6/(1+x)4 so f(4)(0) = -6.
2. Divide by the factorials: 1/1! = 1, -1/2! = -1/2, 2/3! = 1/3, -6/4! = -1/4.
3. So ln(1 + x) = x - x2/2 + x3/3 - x4/4 + ..., matching what term-by-term integration gave last lesson.
4. Ratio test: |xn+1/(n+1)| * |n/xn| = |x| * n/(n+1) -> |x|. So R = 1 and the open interval is (-1, 1).
5. Endpoint x = 1: the series is 1 - 1/2 + 1/3 - ..., the alternating harmonic series, which converges - to ln 2, as it must.
6. Endpoint x = -1: the series is -1 - 1/2 - 1/3 - ..., minus the harmonic series, which diverges - as it must, since ln(0) is undefined.
7. Interval of convergence: (-1, 1].

The two endpoints behaving differently is not an accident here; it is the function telling you where it breaks. That is the general pattern - the radius of convergence reaches out to the nearest point where the function misbehaves.

Taylor series about a point other than 0

Expanding ln(x) about c = 0 is impossible, since ln(0) does not exist. Centre it at c = 1 instead.

1. f(1) = ln(1) = 0; f'(x) = 1/x so f'(1) = 1; f''(x) = -1/x2 so f''(1) = -1; f'''(x) = 2/x3 so f'''(1) = 2; f(4)(x) = -6/x4 so f(4)(1) = -6.
2. Assemble in powers of (x - 1): ln(x) = (x - 1) - (x - 1)2/2 + (x - 1)3/3 - (x - 1)4/4 + ...
3. The interval follows from the previous result with u = x - 1: valid for 0 < x <= 2.

Test it at x = 1.5 with four terms: 0.5 - 0.125 + 0.041667 - 0.015625 = 0.401042, against ln(1.5) = 0.405465. Close, and the next term +0.55/5 = 0.00625 would bring it to 0.407292, straddling the answer as an alternating series should.

Key idea: choose the centre where you know the derivatives and near where you want to evaluate; accuracy is best close to c and degrades as you move away.

Taylor polynomials as approximations

Chop the series off after a few terms and you get a Taylor polynomial that approximates f near the center. The more terms you keep, the better the fit.

Worked example. Approximate cos(0.1) with two terms, using cos(x) ~= 1 - x2/2.

1. Substitute x = 0.1: 1 - (0.1)2/2.
2. Square: (0.1)2 = 0.01.
3. Divide and subtract: 1 - 0.01/2 = 1 - 0.005 = 0.995.

The true value is 0.995004..., so just two terms already give four correct digits. This is exactly how calculators evaluate transcendental functions: they sum enough Taylor terms to reach the precision they need.

How wrong can a Taylor polynomial be? The remainder bound

"Close enough" is not a mathematical statement. Write Pn(x) for the Taylor polynomial of degree n and Rn(x) = f(x) - Pn(x) for the error. Taylor's theorem, in Lagrange's form, says there is some z between c and x with

Rn(x) = [f(n+1)(z)/(n+1)!] * (x - c)n+1

You never know z, but you do not need to. Bound |f(n+1)| by some number M on the interval between c and x, and the error obeys

|Rn(x)| <= M|x - c|n+1/(n+1)!

Read the shape of that bound. It is the size of the first omitted term with the unknown derivative replaced by its worst case. It shrinks fast as n grows, because of the factorial, and it grows as x moves away from c, because of the power.

Numerical demonstration one: approximating e^0.5.

1. Use the cubic Maclaurin polynomial P3(x) = 1 + x + x2/2 + x3/6.
2. Evaluate at 0.5: 1 + 0.5 + 0.125 + 0.0208333 = 1.6458333.
3. Bound the error before checking. Here f(4)(z) = ez, and on [0, 0.5] the largest value is e0.5, which is certainly below 1.65. Take M = 1.65.
4. Apply the bound: |R3| <= 1.65 * (0.5)4/4! = 1.65 * 0.0625/24 = 0.103125/24 = 0.004297.
5. Now compare with reality: e0.5 = 1.6487213, so the actual error is 0.0028880. That is indeed less than the promised 0.004297.

The bound was honest and not wasteful - within a factor of 1.5 of the true error. Note the order of operations: the bound was computed before looking up the true value, which is the whole point. In a real computation there is no true value to look up.

Numerical demonstration two: how tight it can get. Return to cos(0.1) ~= 0.995. Since the x3 coefficient of cosine is 0, that estimate is also P3, so use n = 3. The fourth derivative of cosine is cos(z), and |cos(z)| <= 1 always, so M = 1:

|R3(0.1)| <= 1 * (0.1)4/4! = 0.0001/24 = 0.0000041667

The true value is cos(0.1) = 0.9950041653, so the actual error is 0.0000041653. The bound is 0.0000041667. They agree to four significant figures - the estimate is essentially exact.

Using the bound in reverse. The practical question is not "how wrong am I?" but "how many terms do I need?" Suppose you want e1 = e to six decimal places, meaning an error below 0.0000005. On [0, 1] we have ez <= e < 3, so take M = 3 and solve

3 * 1n+1/(n+1)! < 0.0000005, that is (n+1)! > 6000000

Since 10! = 3628800 is too small and 11! = 39916800 is comfortably large, take n + 1 = 11, so a degree-10 polynomial suffices. Eleven terms for six decimals of e, decided in advance with no trial and error. That is what a remainder bound is for.

Key idea: a Taylor approximation without an error bound is a guess; the Lagrange remainder converts it into a guarantee.

Try it

Write the first three nonzero terms of the Maclaurin series for sin(x), then estimate sin(0.2) using the first two terms.

Answer: the series is x - x3/6 + x5/120. Two terms give 0.2 - (0.008)/6 = 0.2 - 0.001333 = 0.198667, very close to the true value 0.19867. Nice work.

Where people get stuck

• Dropping the 1/n!. Write the factorial in before the derivative value, as a fixed habit. It is the single most common omission, and it wrecks every term past the first two.
• Evaluating at c before differentiating. f(n)(c) means differentiate n times and then substitute c. Substituting first turns f into a constant whose derivatives are all zero.
• Thinking Taylor and Maclaurin are different recipes. A Maclaurin series is a Taylor series with c = 0. There is one formula.
• Expanding about a point where the function is undefined. ln(x) has no Maclaurin series because ln(0) does not exist; centre it at 1 instead. Check that f and all its derivatives exist at c before starting.
• Assuming the series converges everywhere. Only some do. ln(1+x) converges on (-1, 1] and nowhere else, and using it at x = 2 produces a divergent series, not a bad approximation.
• Quoting an approximation with no error estimate. "About 1.6458" means nothing on its own. The Lagrange bound turns it into "1.6458, accurate to within 0.0043."
• Choosing M badly. M must bound |f(n+1)| over the whole interval between c and x, not just at one point. Picking too small an M yields a bound that is simply false.
• Expecting accuracy far from the centre. The bound carries |x - c|n+1, so error grows quickly with distance. Approximating e5 with a cubic centred at 0 is hopeless; centre it near 5, or use e5 = (e1)5.

Recap

• Taylor series: f(x) = sum of [f(n)(c)/n!](x - c)n, built from derivatives at c. The factorial is forced, because differentiating (x-c)n n times produces n!.
• A Maclaurin series is the special case centred at c = 0.
• Memorise ex, sin(x), and cos(x); sine has only odd powers, cosine only even, and all three converge for every real x because the factorial drives the ratio to 0.
• Building a series from scratch means computing derivatives at c until the pattern is clear: sine's cycle at 0 is 0, 1, 0, -1, which produces x - x3/3! + x5/5! - ....
• Not every series converges everywhere. ln(1 + x) = x - x2/2 + x3/3 - ... has R = 1, converges at x = 1 to ln 2, diverges at x = -1, so its interval is (-1, 1].
• Truncating gives a Taylor polynomial. The Lagrange remainder bounds its error by M|x - c|n+1/(n+1)!: for e0.5 with a cubic the bound is 0.0043 against a true error of 0.0029, and for cos(0.1) the bound 0.00000417 matches the true error to four figures.
• Run the bound backwards to choose the number of terms in advance: six decimals of e need a degree-10 polynomial.

Sources

1. OpenStax. (2016). Taylor and Maclaurin series. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Working with Taylor series. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 10.3: Taylor and Maclaurin series. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Strang, G., & Herman, E. (n.d.). 10.4: Working with Taylor series. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
5. Dawkins, P. (n.d.). Taylor series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Dawkins, P. (n.d.). Estimating the value of a series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
7. O'Connor, J. J., & Robertson, E. F. (n.d.). Brook Taylor. MacTutor History of Mathematics Archive, University of St Andrews. mathshistory.st-andrews.ac.uk

A plain graph y = f(x) cannot loop around or double back, because each x is allowed only one y. But a moving point can trace any path at all, including circles and spirals. Parametric equations describe a curve by tracking a point over time, recording not just where it is but when it is there. Think of GPS breadcrumbs dropped as you walk.

The big picture

Parametric equations give both coordinates as functions of a third variable, the parameter t (often time):

x = f(t), y = g(t)

As t runs over an interval, the point (x, y) sweeps out the curve. For example, x = cos(t), y = sin(t) for t in [0, 2 pi] traces the unit circle, since x2 + y2 = cos2 t + sin2 t = 1. Parametric form also records the direction and speed of travel, information a plain y = f(x) graph throws away.

Key idea: parametric equations trace a curve as a point moving in time, giving x and y each as a function of the parameter t.

Eliminating the parameter, and what gets lost

Sometimes you can remove t and recover a familiar equation. Take x = t + 1, y = t2. Solve the first for t, getting t = x - 1, and substitute: y = (x - 1)2. The path is a parabola with vertex at (1, 0).

But be careful about what elimination throws away. Two examples make the danger concrete.

• x = cos(t), y = sin(t) for t in [0, pi] satisfies x2 + y2 = 1, the full unit circle - but the actual path is only the upper half, traced right to left. The Cartesian equation describes a superset of the real curve.
• x = t2, y = t4 gives y = x2, but since t2 is never negative, only the right half of the parabola, x >= 0, is traced.

So after eliminating the parameter, always check the range of x and y that the parametrisation actually produces, and note the direction of travel. That direction, called the orientation, is real information: x = cos(t), y = sin(t) and x = cos(-t), y = sin(-t) trace the same circle in opposite directions and are genuinely different parametrisations of the same set of points.

Slope of a parametric curve

You can find the slope dy/dx without eliminating t. By the chain rule, dy/dt = (dy/dx)(dx/dt), so solving for the slope gives:

dy/dx = (dy/dt)/(dx/dt), as long as dx/dt is not 0

Worked example. For x = t2, y = t3 - 3t, find the slope of the tangent at t = 2.

1. Differentiate each coordinate: dx/dt = 2t and dy/dt = 3t2 - 3.
2. Form the ratio: dy/dx = (3t2 - 3)/(2t).
3. Plug in t = 2: (3*4 - 3)/(2*2) = (12 - 3)/4 = 9/4.

The slope at that point is 9/4. A horizontal tangent happens where dy/dt = 0 (here t = 1 or t = -1), and a vertical tangent where dx/dt = 0 (here t = 0).

Notice something a Cartesian graph could never show: at t = 1 the point is (1, -2) and at t = -1 it is (1, 2). The curve passes through x = 1 twice, at different heights, so it is not a function of x at all - and the parametric machinery handled it without complaint.

The second derivative and concavity

To find concavity you need d2y/dx2, and the natural guess - divide the second derivatives - is wrong. The correct rule differentiates the slope with respect to t and then converts:

d2y/dx2 = [d/dt(dy/dx)]/(dx/dt)

The reasoning is the same chain-rule move as before, applied to the function dy/dx instead of to y.

Worked example. Continue with x = t2, y = t3 - 3t.

1. We already have dy/dx = (3t2 - 3)/(2t). Rewrite it for easy differentiation: (3/2)t - (3/2)t-1.
2. Differentiate with respect to t: d/dt(dy/dx) = 3/2 + (3/2)t-2 = (3/2)(1 + 1/t2).
3. Divide by dx/dt = 2t: d2y/dx2 = (3/2)(1 + 1/t2)/(2t) = 3(t2 + 1)/(4t3).
4. Read off the sign. The numerator 3(t2+1) is always positive, so the sign is the sign of t3. At t = 2 the value is 3(5)/32 = 15/32 > 0, so the curve is concave up there; at t = -2 it is -15/32 < 0, concave down.

Compare with the wrong approach: (d2y/dt2)/(d2x/dt2) = 6t/2 = 3t, which at t = 2 gives 6 rather than 15/32. Not a small discrepancy - a different number entirely.

Area under a parametric curve

Area is still integral y dx; you simply express both pieces through t. Since dx = (dx/dt) dt,

A = integral from t=a to t=b of y(t) * (dx/dt) dt

The pattern is the same in every one of these formulas: write the geometric quantity, then convert each differential using the chain rule. Nothing new is being invented.

Arc length in parametric form

The arc length idea carries over directly. A tiny step of the point has length the square root of (dx2 + dy2); factoring out dt gives the parametric arc length from t = a to t = b:

L = integral from a to b of the square root of ((dx/dt)2 + (dy/dt)2) dt

Worked example. Find the circumference of the unit circle x = cos(t), y = sin(t), t in [0, 2 pi].

1. Differentiate: dx/dt = -sin(t) and dy/dt = cos(t).
2. Add the squares: sin2 t + cos2 t = 1, so the integrand is the square root of 1 = 1.
3. Integrate: L = integral from 0 to 2 pi of 1 dt = 2 pi.

The circumference is 2 pi, exactly matching the familiar 2 pi r with r = 1.

The cycloid, a curve only parametric form can handle

Mark a point on the rim of a wheel of radius r and roll the wheel along a straight line. The path the point traces is a cycloid, and it is described by

x = r(t - sin t), y = r(1 - cos t)

where t is the angle the wheel has turned. There is no reasonable Cartesian equation for this curve, which is exactly why parametric form exists. Take r = 1 and find the length of one arch, t from 0 to 2 pi.

1. Differentiate: dx/dt = 1 - cos(t) and dy/dt = sin(t).
2. Add the squares: (1 - cos t)2 + sin2t = 1 - 2cos t + cos2t + sin2t = 2 - 2cos(t), using cos2 + sin2 = 1.
3. Simplify with the half-angle identity 1 - cos(t) = 2sin2(t/2): the sum becomes 4 sin2(t/2).
4. Take the root: 2|sin(t/2)|. On [0, 2pi] the half-angle t/2 lies in [0, pi], where sine is not negative, so the bars come off and the integrand is 2 sin(t/2).
5. Integrate: L = integral from 0 to 2pi of 2 sin(t/2) dt = [-4 cos(t/2)] from 0 to 2pi = -4cos(pi) + 4cos(0) = 4 + 4 = 8.

One arch has length exactly 8, or 8r in general - a whole number with no pi in it, which surprised the seventeenth century considerably. Sanity-check the size: the arch spans 2 pi ~= 6.28 horizontally and rises to height 2, so a path of length 8 is entirely plausible.

The area under one arch is just as clean. With y = 1 - cos(t) and dx = (1 - cos t) dt:

1. A = integral from 0 to 2pi of (1 - cos t)2 dt = integral from 0 to 2pi of (1 - 2cos t + cos2t) dt.
2. Replace cos2t with (1 + cos 2t)/2, giving the integrand 3/2 - 2cos(t) + cos(2t)/2.
3. Antidifferentiate: 3t/2 - 2sin(t) + sin(2t)/4. Both sine terms vanish at 0 and at 2 pi.
4. So A = 3(2pi)/2 = 3 pi, that is 3 pi r2 in general - exactly three times the area of the rolling wheel.

A physical application: projectile motion

Parametric equations are the natural language for motion. A projectile launched at speed v0 and angle theta, ignoring air resistance, follows

x = (v0 cos theta) t, y = (v0 sin theta) t - 4.9 t2

with distances in metres and time in seconds. Take v0 = 20 and theta = 45 degrees, so cos theta = sin theta = 0.7071 and both leading coefficients are 14.142.

1. Landing time: solve y = 0, so t(14.142 - 4.9t) = 0, giving t = 0 at launch and t = 14.142/4.9 = 2.886 seconds at landing.
2. Range: x(2.886) = 14.142 * 2.886 = 40.8 metres. The standard physics formula v02sin(2 theta)/9.8 = 400/9.8 = 40.8 agrees.
3. Peak height: the vertical velocity dy/dt = 14.142 - 9.8t vanishes at t = 1.443 s, where y = 14.142(1.443) - 4.9(1.443)2 = 20.41 - 10.20 = 10.2 metres. Again the textbook formula v02sin2(theta)/(2 * 9.8) = 200/19.6 = 10.2 agrees.

Every one of those questions was answered by differentiating one coordinate function - and the trajectory's actual length would be a parametric arc-length integral, which no Cartesian setup makes any easier.

Key idea: the parameter is not a nuisance to be eliminated; for curves that loop, cross themselves, or describe motion, it is the whole point.

Try it

For x = t2, y = 2t, find dy/dx at t = 3.

Answer: dx/dt = 2t and dy/dt = 2, so dy/dx = 2/(2t) = 1/t, which is 1/3 at t = 3. Nice work keeping dy/dt on top.

Where people get stuck

• Inverting the slope formula. It is dy/dt on top and dx/dt on the bottom, matching rise over run. Flipping it turns every slope into its reciprocal. Quick check: a horizontal tangent must give slope 0, which happens when the top vanishes.
• Computing the second derivative as (d2y/dt2)/(d2x/dt2). That is not a chain rule and is simply wrong. Differentiate dy/dx with respect to t, then divide by dx/dt. On the worked example the two approaches give 15/32 and 6.
• Forgetting to convert dx when integrating. Area is integral y(t)(dx/dt) dt and arc length carries the whole Pythagorean factor. Leaving a bare dx next to a function of t is a type error.
• Assuming elimination is lossless. The Cartesian equation typically describes more of the curve than the parametrisation traces, and it always discards the direction and speed of travel.
• Dropping absolute values inside a root. the square root of (4sin2(t/2)) = 2|sin(t/2)|. It is only safe to drop the bars after checking the sign on the interval, and over two arches of the cycloid the sign really does change.
• Reusing x-limits for a t-integral. Parametric integrals run over parameter values. For one arch of the cycloid the limits are 0 and 2 pi in t, not 0 and 2 pi in x - though here they happen to coincide, which is a coincidence worth not relying on.
• Missing that a point can be visited twice. x = t2, y = t3 - 3t passes through x = 1 at both t = 1 and t = -1, with different slopes. Questions about "the tangent at a point" may need the parameter value, not just the coordinates.

Recap

• Parametric equations give x = f(t) and y = g(t), tracing a curve as t varies, and they record direction and speed as well as position.
• Eliminating the parameter can recover a Cartesian equation but usually describes more of the curve than is actually traced, and always loses the orientation.
• The slope is dy/dx = (dy/dt)/(dx/dt), with dy/dt on top. Horizontal tangents occur where dy/dt = 0; vertical tangents where dx/dt = 0.
• The second derivative is d2y/dx2 = [d/dt(dy/dx)]/(dx/dt), never a ratio of second derivatives.
• Area is integral y(t)(dx/dt) dt; arc length is integral the square root of ((dx/dt)2 + (dy/dt)2) dt.
• One arch of the cycloid x = r(t - sin t), y = r(1 - cos t) has length 8r and encloses area 3 pi r2.
• Projectile motion is naturally parametric, and differentiating each coordinate gives landing time, range, and peak height directly.

Sources

1. OpenStax. (2016). Parametric equations. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Calculus of parametric curves. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 11.1: Parametric equations. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Strang, G., & Herman, E. (n.d.). 11.2: Calculus of parametric curves. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
5. Dawkins, P. (n.d.). Parametric equations and curves. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Dawkins, P. (n.d.). Tangents with parametric equations. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
7. Dawkins, P. (n.d.). Arc length with parametric equations. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu

Usually we locate a point by how far left or right and how far up or down it sits. But sometimes distance and direction are more natural, the way a radar screen reads "2 miles out, at this angle." Polar coordinates use exactly that: how far from the center, and at what angle. For circles and spirals, this description is far simpler than the usual one.

The big picture

Polar coordinates write a point as (r, theta), where r is the distance from the origin (called the pole) and theta is the angle measured counterclockwise from the positive x-axis. Round shapes that are awkward in x and y often become simple in r and theta.

Key idea: a polar point (r, theta) is located by distance r from the pole and angle theta, which makes circles and spirals easy to describe.

Converting between the two systems

Basic right-triangle trigonometry links them:

• Polar to rectangular: x = r cos(theta) and y = r sin(theta).
• Rectangular to polar: r2 = x2 + y2 and tan(theta) = y/x.

Worked example. Convert (r, theta) = (2, pi/3) to rectangular coordinates.

1. Find x: x = 2 cos(pi/3) = 2*(1/2) = 1.
2. Find y: y = 2 sin(pi/3) = 2*(the square root of 3, over 2) = the square root of 3.
3. So the point is (1, the square root of 3) in rectangular form.

Check it: 12 + (the square root of 3)2 = 1 + 3 = 4 = 22, so the point really is 2 units from the origin. Always verify the distance after converting.

Going the other way needs care with the quadrant. Convert (x, y) = (-1, 1). Then r = the square root of (1 + 1) = the square root of 2, and tan(theta) = 1/(-1) = -1. A calculator returns arctan(-1) = -pi/4, which points into the fourth quadrant - but (-1, 1) is in the second. The correct angle is 3pi/4. The arctangent cannot tell the second quadrant from the fourth, so always sketch the point and adjust by pi when x is negative.

Polar coordinates are not unique

This is the structural difference from rectangular coordinates, and it causes most polar confusion. Every point has infinitely many polar names:

• Add a full turn. (2, pi/3), (2, pi/3 + 2pi), and (2, pi/3 - 2pi) are all the same point.
• Use a negative radius. A negative r means "go the opposite way", so (-2, pi/3 + pi) = (-2, 4pi/3) is also that same point.
• The origin has no angle. (0, theta) is the pole for every theta whatsoever.

The practical consequence appears when you look for intersections of polar curves. Setting two equations equal finds points where the curves are at the same place at the same angle, and it can miss crossings that occur at different theta values - including the origin, which is invisible to that algebra entirely. Sketch the curves before trusting an intersection computation.

A few polar curves

A polar curve is usually written r = f(theta). Some standard shapes:

• r = a is a circle of radius a centered at the origin, since the distance stays constant.
• r = 2a cos(theta) is a circle of diameter 2a passing through the origin.
• r = a(1 + cos theta) is a heart-shaped cardioid.
• r = a sin(n theta) is a rose with petals.

The petal count on a rose has a rule worth knowing, and it surprises people: r = a cos(n theta) or a sin(n theta) has n petals when n is odd and 2n petals when n is even. So r = cos(3 theta) has three petals while r = cos(2 theta) has four. The reason is the negative-radius business above: when n is odd, the second half of the sweep retraces petals already drawn, because (-r, theta + pi) is the same point as (r, theta).

Area in polar coordinates

Here the region is swept out by thin pie-slice sectors, not vertical rectangles - a vertical rectangle simply does not fit a region bounded by rays from the origin. To measure one sector, use proportions: a full disk of radius r has area pi r2, and a sector of angle d(theta) is the fraction d(theta)/(2 pi) of the whole, so its area is

[d(theta)/(2 pi)] * pi r2 = (1/2) r2 d(theta)

That is where the unfamiliar one-half and the square come from: they are the sector formula, not a rectangle formula in disguise. Adding the sectors over the swept angle gives:

A = integral from alpha to beta of (1/2) [f(theta)]2 d(theta)

Worked example. Find the area enclosed by the circle r = 2 cos(theta), which is traced as theta runs from -pi/2 to pi/2.

1. Set up: A = integral from -pi/2 to pi/2 of (1/2)(2 cos theta)2 d(theta).
2. Simplify the integrand: (1/2)(4 cos2 theta) = 2 cos2 theta.
3. Apply the half-angle identity cos2 theta = (1 + cos 2theta)/2, giving 1 + cos 2theta.
4. Integrate: [theta + (1/2) sin 2theta] from -pi/2 to pi/2 = (pi/2 + 0) - (-pi/2 + 0) = pi.

The area is pi. As a check, r = 2 cos theta is a circle of radius 1, and a radius-1 circle has area pi r2 = pi. Note also what would have gone wrong with careless limits: integrating from 0 to 2 pi traces this circle twice and reports an area of 2 pi, double the truth.

Second worked area: the cardioid

Find the area enclosed by r = 1 + cos(theta). This heart shape is traced exactly once as theta runs from 0 to 2 pi, since the radius returns to its starting value and the curve closes.

1. Set up: A = (1/2) integral from 0 to 2pi of (1 + cos theta)2 d(theta).
2. Expand the square: (1 + cos theta)2 = 1 + 2cos(theta) + cos2(theta).
3. Lower the even power with cos2(theta) = (1 + cos 2theta)/2. The integrand becomes 1 + 2cos(theta) + 1/2 + cos(2theta)/2 = 3/2 + 2cos(theta) + cos(2theta)/2.
4. Antidifferentiate: 3theta/2 + 2sin(theta) + sin(2theta)/4. Verify by differentiating: 3/2 + 2cos(theta) + cos(2theta)/2. Correct.
5. Evaluate from 0 to 2 pi. Both sine terms vanish at each end, leaving 3(2pi)/2 = 3pi.
6. Multiply by the leading one-half: A = (1/2)(3pi) = 3pi/2.

The area is 3 pi/2, and in general 3 pi a2/2 for r = a(1 + cos theta). Sanity-check the size: the cardioid fits inside a circle of radius 2, whose area is 4 pi ~= 12.6, and 3pi/2 ~= 4.7 is a comfortable fraction of that.

Third worked area: one petal of a rose

Find the area of a single petal of r = cos(2 theta). The hardest part is the limits, so get them first.

1. A petal begins and ends where r = 0. Solve cos(2 theta) = 0: that needs 2theta = -pi/2 or 2theta = pi/2, so theta = -pi/4 and theta = pi/4. Those bound one petal, the one straddling the positive x-axis.
2. Set up: A = (1/2) integral from -pi/4 to pi/4 of cos2(2 theta) d(theta).
3. Half-angle again, with the angle now 2 theta: cos2(2theta) = (1 + cos 4theta)/2. The integral becomes (1/4) integral from -pi/4 to pi/4 of (1 + cos 4theta) d(theta).
4. Antidifferentiate: (1/4)[theta + sin(4theta)/4]. At theta = pi/4 the sine term is sin(pi)/4 = 0; at -pi/4 it is also 0.
5. So A = (1/4)[pi/4 - (-pi/4)] = (1/4)(pi/2) = pi/8.

One petal has area pi/8. Since n = 2 is even there are four petals, so the whole rose encloses 4 * pi/8 = pi/2. Getting these limits wrong is the classic polar-area error: integrating from 0 to 2 pi here would sweep all four petals and give pi/2 when the question asked for one.

Arc length and slope in polar form

Both follow by treating r = f(theta) as the parametric curve x = r cos(theta), y = r sin(theta) with parameter theta. Working through the algebra collapses everything to

L = integral from alpha to beta of the square root of (r2 + (dr/d(theta))2) d(theta)

Test it on the circle r = a: since dr/d(theta) = 0, the integrand is just a, and L = integral from 0 to 2pi of a d(theta) = 2 pi a. The circumference formula, recovered.

Worked example: the cardioid's perimeter. With r = 1 + cos(theta) and dr/d(theta) = -sin(theta):

1. r2 + (dr/d theta)2 = (1 + cos theta)2 + sin2(theta) = 1 + 2cos(theta) + 1 = 2 + 2cos(theta).
2. Use 1 + cos(theta) = 2cos2(theta/2), so the expression is 4cos2(theta/2) and its root is 2|cos(theta/2)|.
3. On [0, pi] the half-angle lies in [0, pi/2], where cosine is not negative, so the bars come off. The curve is symmetric about the x-axis, so compute the top half and double.
4. L = 2 * integral from 0 to pi of 2cos(theta/2) d(theta) = 2 * [4 sin(theta/2)] from 0 to pi = 2 * (4 - 0) = 8.

The cardioid's perimeter is exactly 8, or 8a in general - another whole number where a pi might have been expected, just as with the cycloid arch.

For the slope of a polar curve, the parametric rule gives

dy/dx = [(dr/d theta)sin(theta) + r cos(theta)] / [(dr/d theta)cos(theta) - r sin(theta)]

which is just (dy/d theta)/(dx/d theta) after applying the product rule to each coordinate. Nothing new is required; every polar formula in this lesson is the parametric machinery of the previous lesson with theta playing the role of t.

Key idea: polar calculus is parametric calculus with x = r cos(theta) and y = r sin(theta); the sector area (1/2)r2d(theta) is the only genuinely new ingredient.

This lesson completes your calculus toolkit for curves of every kind, from graphs to parametric paths to polar shapes.

Try it

Convert (r, theta) = (4, pi/2) to rectangular coordinates.

Answer: x = 4 cos(pi/2) = 0 and y = 4 sin(pi/2) = 4, so the point is (0, 4). Nice work; the point sits straight up on the y-axis, exactly where angle pi/2 should place it.

Where people get stuck

• Expecting rectangles. Polar regions are built from pie-slice sectors, so the element is (1/2)r2d(theta), derived from the proportion of a full disk. Trying to force a height-times-width picture onto a region bounded by rays does not work.
• Choosing theta-limits carelessly. This is the dominant error. The limits must trace the region exactly once: r = 2cos(theta) needs [-pi/2, pi/2] and doubles if you use [0, 2pi]; one petal of r = cos(2 theta) needs [-pi/4, pi/4]. Find where r = 0 to locate a petal's boundaries.
• Forgetting the one-half or the square. Both come from the sector formula. Dropping the square is especially damaging because it changes the units of the answer.
• Trusting the arctangent for the angle. arctan(y/x) cannot distinguish the second quadrant from the fourth, or the third from the first. Sketch the point and add pi when x is negative.
• Assuming polar coordinates are unique. Adding 2 pi to theta, or negating r and adding pi, names the same point, and the origin is (0, theta) for every theta. Intersection problems must be checked graphically, not only algebraically.
• Miscounting rose petals. r = a cos(n theta) has n petals when n is odd and 2n when n is even, so cos(3 theta) gives three and cos(2 theta) gives four.
• Dropping absolute values inside a root. the square root of (4cos2(theta/2)) = 2|cos(theta/2)|, and the bars only come off after you have checked the sign on your interval.
• Treating polar formulas as a separate subject. They all come from the parametric formulas with x = r cos(theta) and y = r sin(theta). If you forget one, re-derive it.

Recap

• Polar coordinates (r, theta) give distance from the pole and angle from the positive x-axis, and every point has infinitely many polar names.
• Convert with x = r cos(theta), y = r sin(theta), and r2 = x2 + y2, adjusting the angle by pi when x is negative.
• Polar area is A = integral (1/2)[f(theta)]2 d(theta), built from circular sectors of area (1/2)r2d(theta).
• Choose theta-limits that trace the region exactly once. The circle r = 2cos(theta) encloses pi; the cardioid r = 1 + cos(theta) encloses 3pi/2; one petal of r = cos(2 theta) encloses pi/8.
• Roses have n petals for odd n and 2n petals for even n.
• Polar arc length is integral the square root of (r2 + (dr/d theta)2) d(theta), which returns 2 pi a for a circle and 8 for the cardioid r = 1 + cos(theta).
• Everything here is the parametric calculus of the previous lesson with theta as the parameter; only the sector-area element is new.

Sources

1. OpenStax. (2016). Polar coordinates. In Calculus volume 2. Rice University. openstax.org
2. OpenStax. (2016). Area and arc length in polar coordinates. In Calculus volume 2. Rice University. openstax.org
3. Strang, G., & Herman, E. (n.d.). 11.3: Polar coordinates. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
4. Strang, G., & Herman, E. (n.d.). 11.4: Area and arc length in polar coordinates. In Calculus (OpenStax). Mathematics LibreTexts. math.libretexts.org
5. Dawkins, P. (n.d.). Polar coordinates. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
6. Dawkins, P. (n.d.). Area with polar coordinates. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
7. Dawkins, P. (n.d.). Arc length with polar coordinates. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu