AP-Level Calculus BC

An integral that will not yield

Evaluate ∫ x ex dx. Try substitution three ways. Let u = x: then du = dx and the ex sits there untouched, so nothing simplifies. Let u = ex: then du = ex dx, which consumes the exponential and leaves a lone x with no way to express it in u. Let u = x ex: its derivative is ex + x ex, which is not a factor of the integrand. Every route closes.

The failure is structural, not bad luck. Substitution reverses the chain rule, and x ex is not something a chain rule ever produces. It is a product of two functions that have nothing to do with each other. The rule that produces products is the product rule, so the technique that undoes it is integration by parts.

Key idea: substitution reverses the chain rule; parts reverses the product rule. Between them they cover most of what the AP exam asks you to integrate by hand.

Deriving the formula in four lines

Start with the product rule for two differentiable functions u(x) and v(x):

d/dx [u v] = u' v + u v'

Integrate both sides with respect to x. The left side integrates to u v, because integrating a derivative returns the original function:

u v = ∫ u' v dx + ∫ u v' dx

Now solve for the integral you want:

∫ u v' dx = u v − ∫ v u' dx

Write dv = v' dx and du = u' dx and the formula takes the shape you will use every time:

∫ u dv = u v − ∫ v du

Read it as a trade. You hand over the integral of u dv and receive a product term plus a new integral, ∫ v du. The trade is only worth making if the new integral is easier than the old one. That single sentence is the whole strategy.

Choosing u: LIATE, and why it is a habit rather than a law

The integrand splits into two pieces. One becomes u and gets differentiated; the other becomes dv and gets integrated. You want u to get simpler when differentiated and dv to be something you can actually antidifferentiate. The usual ordering, remembered as LIATE, puts the candidates for u in this order:

• Logarithmic: ln x
• Inverse trigonometric: arctan x, arcsin x
• Algebraic: x, x², polynomials
• Trigonometric: sin x, cos x
• Exponential: ex, 2x

Take u to be whichever factor appears first on that list. The ordering is not a theorem. It is a summary of which functions get worse when you integrate them: logarithms and inverse trigonometric functions have no clean antiderivatives, so they belong in the u slot where they are differentiated instead. Polynomials shrink under differentiation, so they are good u choices whenever nothing further up the list is present. Exponentials and sines are equally happy in either slot, which is exactly why they end up last.

Two integrals, worked

The one that started this. ∫ x ex dx. Algebraic beats exponential on the list, so u = x and dv = ex dx. Then du = dx and v = ex. Substituting:

∫ x ex dx = x ex − ∫ ex dx = x ex − ex + C

Check by differentiating ex(x − 1): the product rule gives ex(x − 1) + ex = x ex. Correct.

Now with a trigonometric partner. ∫ x cos x dx. Again u = x, dv = cos x dx, so du = dx and v = sin x:

∫ x cos x dx = x sin x − ∫ sin x dx = x sin x + cos x + C

Watch the sign. The formula subtracts the new integral, and ∫ sin x dx = −cos x, so subtracting a negative gives the plus. Sign slips at exactly this step are the single most common way this technique goes wrong.

What a bad choice looks like

Go back to ∫ x ex dx and deliberately choose wrongly: u = ex and dv = x dx. Then du = ex dx and v = x²/2, and the formula gives

∫ x ex dx = (x²/2) ex − ∫ (x²/2) ex dx

That statement is true. It is also useless: the new integrand has x² where the old one had x, so the problem grew. This is the diagnostic. If one application of parts produces a harder integral, you did not fail at parts, you picked the wrong split. Swap u and dv and run it again.

ln x, where there appears to be nothing to split

Evaluate ∫ ln x dx. There is only one factor, so parts looks unavailable. But every integrand is secretly a product with 1. Take u = ln x and dv = 1 dx. Then du = (1/x) dx and v = x:

∫ ln x dx = x ln x − ∫ x · (1/x) dx = x ln x − ∫ 1 dx = x ln x − x + C

Differentiate to check: ln x + x(1/x) − 1 = ln x + 1 − 1 = ln x. The same trick evaluates ∫ arctan x dx, giving x arctan x − (1/2)ln(1 + x²) + C, where the leftover integral ∫ x/(1 + x²) dx is an ordinary substitution.

Applying parts twice, and the table that shortens it

Evaluate ∫ x² ex dx. Take u = x², dv = ex dx, so du = 2x dx and v = ex:

∫ x² ex dx = x² ex − 2∫ x ex dx

The remaining integral is the one already solved, x ex − ex, so the answer is x² ex − 2x ex + 2ex + C.

When the polynomial has degree 3 or 4, that bookkeeping gets long. Tabular integration organises it. Differentiate the u column until it reaches 0, antidifferentiate the dv column the same number of times, and alternate signs down the diagonal:

Multiply each entry in the u column by the entry one row down in the dv column, attach the sign on that row, and add: +x²ex − 2x ex + 2ex + C. Same answer, a third of the writing. The method works whenever the u column eventually hits zero, which means whenever u is a polynomial. It is not a different rule, only repeated parts written compactly, and you should be able to say that if asked to justify a step.

The integral that comes back to itself

Evaluate I = ∫ ex sin x dx. Neither factor simplifies under differentiation, so the u column never reaches zero and tabular integration will run forever. Do it anyway, twice, and watch what happens.

First pass: u = sin x, dv = ex dx, giving du = cos x dx and v = ex.

I = ex sin x − ∫ ex cos x dx

Second pass, on the new integral, and here is the one rule that matters: keep the same kind of choice. The trigonometric function was u last time, so it is u again. Take u = cos x, dv = ex dx, so du = −sin x dx and v = ex.

∫ ex cos x dx = ex cos x + ∫ ex sin x dx = ex cos x + I

Substitute that back:

I = ex sin x − (ex cos x + I) = ex sin x − ex cos x − I

The unknown integral appears on both sides, so treat it as an unknown and solve: 2I = ex(sin x − cos x), hence

∫ ex sin x dx = ex(sin x − cos x)/2 + C

If instead you switch roles on the second pass, taking u = ex, everything unwinds and you arrive at I = I, which is true and tells you nothing. Consistency between passes is what makes the algebra close.

The upshot: when parts returns the original integral, you have an equation, not a dead end. Solve it.

Definite integrals

The bracket applies to the product term only, and the limits stay attached to the new integral:

∫(a to b) u dv = [u v](a to b) − ∫(a to b) v du

Evaluate ∫(1 to e) ln x dx. The antiderivative is x ln x − x, so the value is (e · 1 − e) − (1 · 0 − 1) = 0 − (−1) = 1. Exactly 1, which is a pleasant enough fact to remember as a check on the antiderivative.

Evaluate ∫(0 to 1) x ex dx using ex(x − 1): at x = 1 it is e · 0 = 0; at x = 0 it is 1 · (−1) = −1. The definite integral is 0 − (−1) = 1.

Common misconceptions

• "You need a +C when you find v." You do not. Carrying v + C1 through the formula adds C1u and subtracts ∫ C1 du = C1u, so it cancels. What you must not drop is the + C at the very end of an indefinite integral, which is worth a point on the exam.
• "LIATE is a rule you must obey." It is a reliable habit. The real test is whether ∫ v du is easier than what you started with. If it is not, swap and rerun.
• "Parts always terminates." It terminates when u is a polynomial. For ex sin x it cycles, and you finish with algebra instead.
• "The minus sign belongs to the second term." The minus sign belongs to the whole second integral, including any sign that comes out of it. −∫ sin x dx is +cos x.

Pulling it together

Integration by parts is the product rule integrated and rearranged into ∫ u dv = uv − ∫ v du, and every use of it is a trade you should check before committing: the new integral has to be easier. LIATE picks u correctly almost every time, and when it does not, the harder second integral tells you immediately to swap. A lone ln x or arctan x is a product with 1. A polynomial factor means repeated parts, which the tabular layout compresses into three columns and one diagonal. An exponential against a sine or cosine cycles back to the original integral, and you solve for it as an unknown, arriving at ex(sin x − cos x)/2 + C. For a definite integral, bracket the uv term and keep the limits on the integral that remains. Every answer here is checkable by differentiating it, and on a technique with this many sign traps, checking is not optional.

Sources

1. OpenStax. (2016). 3.1 Integration by parts. In Calculus Volume 2. openstax.org
2. Dawkins, P. (n.d.). Integration by parts. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Massachusetts Institute of Technology. (2010). 18.01SC Single Variable Calculus, Unit 4: Techniques of integration. MIT OpenCourseWare. ocw.mit.edu
4. College Board. (2020). AP Calculus AB and AP Calculus BC course and exam description. Topic 6.11, Integration by parts, is BC only.

A fraction you cannot integrate and a sum you can

Evaluate ∫ dx/(x² − 1). Nothing in the antiderivative table matches it. Substitution needs the derivative of the denominator, 2x, somewhere in the numerator, and there is no x at all. Parts needs a product worth splitting, and this is a single quotient.

Now watch what happens if the same expression is written differently:

1/(x² − 1) = 1/2 · 1/(x − 1) − 1/2 · 1/(x + 1)

Both pieces on the right are logarithms you already know. The integral falls out in one line:

∫ dx/(x² − 1) = (1/2)ln|x − 1| − (1/2)ln|x + 1| + C = (1/2)ln|(x − 1)/(x + 1)| + C

Differentiate to confirm: (1/2)[1/(x − 1) − 1/(x + 1)] = (1/2) · 2/((x − 1)(x + 1)) = 1/(x² − 1). Correct.

The only difficulty was finding that split. Partial fraction decomposition is the algebra that finds it, for any rational function whose denominator you can factor.

The point: partial fractions is not an integration technique at all. It is algebra done before the integration, and once it is done the integrals are ones you have known since AB.

Two things to check before you start

First, the degree condition. The numerator's degree must be strictly less than the denominator's. If it is not, divide first. For x³/(x² − 1), polynomial long division gives

x³/(x² − 1) = x + x/(x² − 1)

and only the leftover fraction gets decomposed. Skip this check and the algebra produces an inconsistent system with no solution, which is a confusing way to discover a step you should have done first.

Second, factor the denominator completely over the real numbers. Every real polynomial factors into linear pieces and quadratic pieces that have no real roots. Which pieces you get determines the shape of the decomposition:

Two rules generate that whole table. A repeated factor gets one term for every power up to its multiplicity. A quadratic factor gets a linear numerator, not a constant one, because a constant does not have enough freedom to match a general numerator of degree 1.

Distinct linear factors, and the fastest way to find the constants

Decompose 1/(x² − 1) properly. Factor: (x − 1)(x + 1). Write

1/((x − 1)(x + 1)) = A/(x − 1) + B/(x + 1)

Multiply both sides by the full denominator to clear fractions:

1 = A(x + 1) + B(x − 1)

This is an identity, true for every x, so you may substitute any convenient value. Choose the values that kill a term. At x = 1: 1 = A(2) + B(0), so A = 1/2. At x = −1: 1 = A(0) + B(−2), so B = −1/2. That matches the split used above.

Choosing roots of the factors is exactly the Heaviside cover-up method: to get the constant over (x − a), cover that factor in the original fraction and evaluate everything else at x = a. For distinct linear factors it gives every constant in seconds with no system to solve. It does not reach the higher-power terms of a repeated factor, which is why the next section needs a second technique alongside it.

A repeated factor

Evaluate ∫ (3x + 2)/(x(x + 1)²) dx. The denominator has a distinct linear factor and a repeated one, so the decomposition needs three terms:

(3x + 2)/(x(x + 1)²) = A/x + B/(x + 1) + C/(x + 1)²

Clear denominators:

3x + 2 = A(x + 1)² + Bx(x + 1) + Cx

Substituting roots gets two of the three. At x = 0: 2 = A(1), so A = 2. At x = −1: −1 = C(−1), so C = 1. There is no value of x that isolates B, so compare coefficients instead. Expanding the right side, the x² terms are A x² + B x², and the left side has no x² at all, so A + B = 0 and B = −2.

Check with the x coefficient before integrating: 2A + B + C = 4 − 2 + 1 = 3, which matches the 3 on the left. Now integrate term by term:

∫ (3x + 2)/(x(x + 1)²) dx = 2ln|x| − 2ln|x + 1| − 1/(x + 1) + C

Note the third term. ∫ dx/(x + 1)² is a power rule integral, not a logarithm: it is −(x + 1)−1. Only the first power of a linear factor produces a log.

An irreducible quadratic, where an arctangent appears

Evaluate ∫ (2x² − x + 4)/(x³ + 4x) dx. Factor the denominator: x(x² + 4). The quadratic has no real roots, so it stays whole and takes a linear numerator:

(2x² − x + 4)/(x(x² + 4)) = A/x + (Bx + C)/(x² + 4)

Clear denominators and collect:

2x² − x + 4 = A(x² + 4) + (Bx + C)x = (A + B)x² + Cx + 4A

Matching coefficients: the constant term gives 4A = 4, so A = 1; the x term gives C = −1; the x² term gives A + B = 2, so B = 1. The integral splits into three standard pieces:

∫ dx/x + ∫ x dx/(x² + 4) − ∫ dx/(x² + 4)

The first is ln|x|. The second is a substitution with u = x² + 4, giving (1/2)ln(x² + 4), with no absolute value needed because x² + 4 is always positive. The third is the arctangent form ∫ dx/(x² + a²) = (1/a)arctan(x/a) with a = 2. Altogether:

ln|x| + (1/2)ln(x² + 4) − (1/2)arctan(x/2) + C

That pattern is worth naming, because it recurs: from a quadratic factor, the Bx part of the numerator becomes a logarithm by substitution and the C part becomes an arctangent. Split the fraction into those two pieces on purpose rather than hoping one integral will handle both.

The split the logistic equation runs on

One decomposition earns its own section because Module 2 cannot proceed without it. A population P growing toward a ceiling K satisfies a differential equation whose separated form contains ∫ dP/(P(K − P)). Decompose it:

1/(P(K − P)) = A/P + B/(K − P), so 1 = A(K − P) + BP.

At P = 0: 1 = AK, so A = 1/K. At P = K: 1 = BK, so B = 1/K. Both constants are the same, which is the useful surprise:

1/(P(K − P)) = (1/K)[1/P + 1/(K − P)]

Integrating, and remembering that the chain rule puts a minus sign on the second logarithm because the inside is K − P:

∫ dP/(P(K − P)) = (1/K)[ln|P| − ln|K − P|] + C = (1/K)ln|P/(K − P)| + C

Take K = 1000 and check the arithmetic at a number: the decomposition claims 1/(P(1000 − P)) = (1/1000)(1/P + 1/(1000 − P)). At P = 200, the left side is 1/(200 · 800) = 1/160000 = 6.25 × 10−6. The right side is (1/1000)(1/200 + 1/800) = (1/1000)(0.005 + 0.00125) = 6.25 × 10−6. They agree.

Why this matters: the closed-form logistic curve every biology textbook prints is nothing more than this one decomposition, integrated and then solved for P. You will do exactly that two lessons from now.

Common misconceptions

• "Every denominator factors into linear pieces." Over the reals it does not. x² + 4 has no real roots and stays intact, carrying a numerator Bx + C.
• "A squared factor needs only one term." (x + 1)² needs both B/(x + 1) and C/(x + 1)². With only the squared term the system is usually unsolvable, and when it is solvable it is wrong.
• "Decompose first, then worry about degree." The degree condition is a precondition. If the numerator's degree is greater than or equal to the denominator's, divide first and decompose the remainder.
• "Every piece integrates to a logarithm." Only the first power of a linear factor does. Higher powers give the power rule, and the constant part over an irreducible quadratic gives an arctangent.

What to carry forward

Partial fractions turns a rational function you cannot integrate into a sum of pieces you can. Check the degree first and divide if you must, factor the denominator completely, then write one term per distinct linear factor, one term per power of a repeated factor, and a linear numerator over each irreducible quadratic. Find the constants by substituting the roots, which is the cover-up method and takes seconds, and fall back on matching coefficients for whatever the roots cannot reach. Then integrate: first powers of linear factors give logarithms with absolute values, higher powers give the power rule, the Bx piece of a quadratic gives a logarithm by substitution, and the constant piece gives an arctangent. The one decomposition to memorise outright is 1/(P(K − P)) = (1/K)[1/P + 1/(K − P)], because the entire logistic model rests on it.

Sources

1. OpenStax. (2016). 3.4 Partial fractions. In Calculus Volume 2. openstax.org
2. Dawkins, P. (n.d.). Partial fractions. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Wikipedia. (n.d.). Partial fraction decomposition. en.wikipedia.org
4. College Board. (n.d.). AP Calculus BC course framework. Unit 6, Integration and Accumulation of Change, carries 15 to 20 percent of the multiple-choice section. apcentral.collegeboard.org

Two regions that look identical and are not

Draw the region under y = 1/x² from x = 1 rightward, forever. Its area is exactly 1. Draw the region under y = 1/x from x = 1 rightward, forever. Its area is infinite.

Past x = 10 the two curves are nearly indistinguishable on any graph you could print: one sits at 0.01 and the other at 0.1, and both are pressed against the axis. Yet one region can be painted with a finite amount of paint and the other cannot. Nothing in the picture tells you which is which. Only the arithmetic does.

Here is the arithmetic. Neither region is an integral in the sense AB defined, because the definite integral was built on a closed bounded interval. So define it as a limit and see what the limit does:

∫(1 to b) dx/x² = [−1/x](1 to b) = 1 − 1/b, and as b → ∞ this tends to 1.

∫(1 to b) dx/x = [ln x](1 to b) = ln b, and as b → ∞ this grows without bound, so the integral diverges.

What matters here: an improper integral is not a new object. It is a limit of ordinary integrals, and the word convergent means that limit exists and is finite.

The two kinds, and the notation that keeps them honest

Type 1: an infinite limit of integration. Replace the infinity by a letter and take a limit.

∫(a to ∞) f(x) dx = lim(b → ∞) ∫(a to b) f(x) dx

Type 2: an integrand that blows up at an endpoint. Replace the bad endpoint by a letter and take a one-sided limit toward it.

∫(0 to 1) dx/√x = lim(a → 0+) ∫(a to 1) dx/√x

Write the limit down every time. Examiners award the setup, and more importantly the limit notation is what stops you evaluating an antiderivative at a point where it does not exist. Once the limit is written, the rest is AB.

Work the Type 2 example: ∫(a to 1) x−1/2 dx = [2√x](a to 1) = 2 − 2√a. As a → 0+, √a → 0, so the integral converges to 2. The region is infinitely tall and has area 2.

The p-integrals, derived once

Both families are worth deriving, because once you have them you will recognise half the improper integrals you meet without computing anything.

Out at infinity. For p ≠ 1,

∫(1 to b) x−p dx = [x1−p/(1 − p)](1 to b) = (b1−p − 1)/(1 − p)

If p > 1 then 1 − p is negative, so b1−p → 0 and the limit is −1/(1 − p) = 1/(p − 1). If p < 1 then 1 − p is positive and b1−p → ∞, so it diverges. The excluded case p = 1 gives ln b, which also diverges.

Near zero. The same antiderivative, different end:

∫(a to 1) x−p dx = (1 − a1−p)/(1 − p)

Now it is a → 0+ that matters. If p < 1 then 1 − p > 0 and a1−p → 0, so the value is 1/(1 − p). If p > 1 the exponent is negative and a1−p → ∞, so it diverges. Again p = 1 diverges, since −ln a → ∞.

The inequality flips, and the reason is not symmetry for its own sake. Out at infinity the danger is a tail that decays too slowly, so you need a large p to force the function down fast. Near zero the danger is a spike that grows too fast, so you need a small p to keep the spike shallow. p = 1, the harmonic exponent, fails at both ends. Hold on to that number: it returns in Module 4 as the boundary between convergent and divergent p-series, and for the same reason.

Reading the decisions off

A discontinuity hiding inside the interval

Evaluate ∫(−1 to 1) dx/x². Apply the fundamental theorem carelessly:

[−1/x](−1 to 1) = (−1) − (1) = −2

That answer is impossible. The integrand 1/x² is positive everywhere it is defined, so any area it encloses is positive, and a negative number cannot be right. The error is that −1/x is not an antiderivative of 1/x² on the whole interval, because neither function exists at x = 0, which sits in the middle.

The correct treatment splits at the bad point and requires both halves to converge:

∫(−1 to 0) dx/x² + ∫(0 to 1) dx/x²

The right half is lim(a → 0+) [−1/x](a to 1) = lim (−1 + 1/a) = ∞. It diverges, so the whole integral diverges, and you stop there. One divergent piece is enough. Before writing any antiderivative, look for points inside the interval where the denominator vanishes or a logarithm is taken of zero. A silently infinite integrand is the most expensive mistake in this lesson.

The same splitting applies when both ends are infinite. ∫(−∞ to ∞) dx/(1 + x²) must be cut at any convenient point, say 0, and each half done separately: each is arctan approaching π/2, so the value is π/2 + π/2 = π = 3.1416.

Deciding convergence without evaluating

Some integrands have no elementary antiderivative, so the limit cannot be computed directly. Comparison settles them. If 0 ≤ f(x) ≤ g(x) for all x past some point, then a convergent ∫ g forces ∫ f to converge, and a divergent ∫ f forces ∫ g to diverge. Smaller than finite is finite; bigger than infinite is infinite.

Does ∫(1 to ∞) e−x² dx converge? There is no elementary antiderivative for e−x². But for x ≥ 1 we have x² ≥ x, so −x² ≤ −x, so e−x² ≤ e−x. And ∫(1 to ∞) e−x dx = e−1 = 0.368 converges. So the original converges, to some number at most 0.368. Comparison gives the verdict and a bound, never the exact value.

A solid with finite volume and infinite surface

Rotate the curve y = 1/x for x ≥ 1 about the x-axis. The volume by the disc method is π∫(1 to ∞) dx/x² = π(1) = π, a perfectly ordinary number. The surface area integral is bounded below by 2π∫(1 to ∞) dx/x, which diverges. The shape, called Gabriel's horn, holds π cubic units of paint and has a wall you could never finish painting. It is not a paradox, only a reminder that a one-dimensional decay rate of 1/x is enough to make an area diverge while 1/x² is not.

Common misconceptions

• "An unbounded region must have infinite area." It need not. ∫(1 to ∞) dx/x² = 1 and ∫(0 to 1) dx/√x = 2 are both finite, and both regions are unbounded.
• "Just plug the endpoints into the antiderivative." That is what produces −2 for a positive integrand. Write the limit, and check for interior discontinuities first.
• "The p rule is the same at both ends." It reverses. Large p converges at infinity and diverges at zero; p = 1 fails at both.
• "If one half of a split integral converges, the integral converges." Both halves must converge. A finite piece plus an infinite piece is infinite.

The short version

An improper integral is a limit of proper ones, and writing that limit is the first line of the answer, not an optional formality. Type 1 has an infinite limit of integration; Type 2 has an integrand that blows up, which can happen at an endpoint or, more dangerously, inside the interval, where you must split and demand convergence of every piece. The two p-integrals settle most cases on sight: ∫(1 to ∞) dx/xp converges exactly when p > 1, to 1/(p − 1), and ∫(0 to 1) dx/xp converges exactly when p < 1, to 1/(1 − p), with p = 1 divergent at both ends. When there is no antiderivative, comparison still decides: bound the integrand above by something convergent or below by something divergent. And an unbounded region can perfectly well have a finite area, which is the whole point of the lesson and the reason the horn holds paint.

Sources

1. OpenStax. (2016). 3.7 Improper integrals. In Calculus Volume 2. openstax.org
2. Dawkins, P. (n.d.). Improper integrals. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Wikipedia. (n.d.). Improper integral. en.wikipedia.org
4. Wikipedia. (n.d.). Gabriel's horn. en.wikipedia.org

Measuring a curve with a straight ruler

Take the parabola y = x² from (0, 0) to (1, 1). The straight distance between those two points is √2 = 1.41421. The curve itself is longer, because it bulges below the chord and has to come back up. How much longer? The answer, which this lesson will produce, is 1.47894. Only four percent longer, which is not a number you could have guessed from the picture.

The obstacle is that length is defined for straight things. A ruler measures a segment. There is no primitive operation that measures a bend. So do what calculus always does with a quantity defined only for straight objects: break the curve into pieces short enough to be nearly straight, measure each piece with the tool you have, add them, and take a limit as the pieces shrink.

The derivation, one triangle at a time

Cut the interval [a, b] into n subintervals of width Δx. Over one of them, the curve runs from (xi, f(xi)) to (xi + Δx, f(xi + Δx)). Replace that arc by the straight chord joining its ends. The chord is the hypotenuse of a right triangle with horizontal leg Δx and vertical leg Δy, so the Pythagorean theorem gives its length exactly:

Δs = √((Δx)² + (Δy)²)

That is the whole idea. Everything after this is algebra whose only purpose is to turn that expression into something with a single Δx in it, so it can become an integral. Factor (Δx)² out of the radical:

Δs = √((Δx)²[1 + (Δy/Δx)²]) = √(1 + (Δy/Δx)²) · Δx

The step is legitimate for Δx > 0, which it is, since we are moving left to right. Now the quantity Δy/Δx is a difference quotient, and by the mean value theorem it equals f'(xi*) for some point xi* inside the subinterval, provided f is differentiable there. So

Δs = √(1 + [f'(xi*)]²) · Δx

Add up all n pieces and you have a Riemann sum. Let n → ∞ and it becomes an integral:

L = ∫(a to b) √(1 + [f'(x)]²) dx

The core of it: the arc length formula is the distance formula summed over infinitely many infinitesimal triangles. If you can rebuild it from √((Δx)² + (Δy)²) you never have to memorise it.

When a curve is easier to describe with x as a function of y, the same derivation factoring out Δy instead gives

L = ∫(c to d) √(1 + [g'(y)]²) dy

Use whichever version makes the derivative simpler. For x = y³ from y = 0 to y = 1, integrating in y avoids a fractional power that integrating in x would force on you.

Two checks worth running on every answer

The lower bound. The radicand is 1 + [f'(x)]², and a square is never negative, so the integrand is at least 1 everywhere. Therefore L ≥ b − a always. An arc length shorter than the horizontal span of the interval is arithmetically impossible, and spotting that saves you from reporting a sign error.

The straight-line case. For a line f(x) = mx + c, the derivative is the constant m and the formula gives √(1 + m²)(b − a). Compare that with the distance formula on the endpoints: the horizontal run is b − a and the rise is m(b − a), so the distance is √((b − a)² + m²(b − a)²) = √(1 + m²)(b − a). Identical. A formula that reproduces the answer you already know on the easy case is a formula you can trust on the hard one.

An exact one, where the radicand collapses

Find the length of y = (2/3)x3/2 from x = 0 to x = 3.

1. Differentiate: f'(x) = x1/2.
2. Square and add one: 1 + [f'(x)]² = 1 + x.
3. Integrate: L = ∫(0 to 3) √(1 + x) dx = [(2/3)(1 + x)3/2](0 to 3).
4. Evaluate: (2/3)(43/2) − (2/3)(13/2) = (2/3)(8) − (2/3)(1) = 14/3 = 4.6667.

Check the bound: the interval has width 3 and the answer is 4.667, comfortably larger. Check the chord: the endpoints are (0, 0) and (3, 2√3) = (3, 3.464), a straight distance of √(9 + 12) = 4.583. The curve is 4.667, slightly longer than the chord, exactly as it must be.

A second exact one, engineered on purpose

Find the length of y = x²/8 − ln x from x = 1 to x = 2.

f'(x) = x/4 − 1/x, so [f'(x)]² = x²/16 − 1/2 + 1/x², and

1 + [f'(x)]² = x²/16 + 1/2 + 1/x² = (x/4 + 1/x)²

The cross term in that last square is 2 · (x/4) · (1/x) = 1/2, which is precisely the −1/2 from the squared derivative with its sign flipped by the added 1. The square root now comes out cleanly, and since both terms are positive on [1, 2] no absolute value is needed:

L = ∫(1 to 2) (x/4 + 1/x) dx = [x²/8 + ln x](1 to 2) = (0.5 + 0.6931) − (0.125 + 0) = 1.0681

Notice what made both examples work: the +1 completed a perfect square. That is not an accident of these functions, it is why textbook arc length problems look artificial. Functions of the form g(x) + 1/(16g(x)) are constructed backwards from the answer, because almost nothing else integrates in closed form.

The honest case: the parabola from the opening

For y = x² on [0, 1], f'(x) = 2x and

L = ∫(0 to 1) √(1 + 4x²) dx

No substitution touches this. The antiderivative exists but needs a trigonometric substitution well past the AP syllabus, and it is

[x√(1 + 4x²)/2 + (1/4)ln(2x + √(1 + 4x²))](0 to 1)

Evaluate it: √5/2 = 1.11803, plus (1/4)ln(2 + 2.23607) = (1/4)(1.44363) = 0.36091. The lower limit contributes 0. So L = 1.47894, the number promised at the start.

On the exam you would never produce that antiderivative. You would write the integral, state that it is to be evaluated numerically, and report 1.479 to three decimal places. That is the expected answer, and the setup carries most of the credit. Arc length questions on BC are calculator-active for exactly this reason: the integrand √(1 + [f']²) almost never has an elementary antiderivative.

Length of a curve against distance travelled

One distinction to fix now, because Module 3 leans on it. Arc length measures the geometric curve. If a particle runs along that curve and doubles back, the curve does not get longer, but the distance the particle travels does. For a graph y = f(x) traversed once from left to right the two agree, which is why they can look like the same idea here. The moment a path is given parametrically and can retrace itself, they separate, and the exam tests the difference.

Common misconceptions

• "Arc length is the integral of f." That is area. Length integrates √(1 + [f']²), which involves the derivative and never the function value.
• "The square root of a sum splits." √(1 + [f']²) is not 1 + f'. Test it at f' = 1: the left side is 1.414 and the right side is 2.
• "Most arc lengths can be computed by hand." Almost none can. The textbook cases are reverse-engineered so the radicand is a perfect square; everything else goes to a numerical integrator.
• "A shorter answer than the interval width is fine if the curve dips." It never is. The integrand is at least 1, so L ≥ b − a with equality only for a horizontal line.

What to remember

Cut the curve into chords, measure each with the Pythagorean theorem, factor Δx out of the radical, and the sum becomes L = ∫(a to b) √(1 + [f'(x)]²) dx. Swap the roles of the variables when the curve is nicer as x = g(y). Two checks cost nothing: the answer must exceed the chord joining the endpoints, and it must be at least the width of the interval. When the radicand collapses into a perfect square, as it does for y = (2/3)x3/2 with its clean answer of 14/3, finish by hand. When it does not, as for y = x² with its √(1 + 4x²), set up the integral, say clearly that it is evaluated numerically, and report 1.479. Getting the setup right is the skill; the arithmetic is the calculator's job.

Sources

1. OpenStax. (2016). 2.4 Arc length of a curve and surface area. In Calculus Volume 2. openstax.org
2. Dawkins, P. (n.d.). Arc length. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Wikipedia. (n.d.). Arc length. en.wikipedia.org
4. College Board. (n.d.). AP Calculus BC course framework. Unit 8, Applications of Integration, carries 5 to 10 percent of the multiple-choice section. apcentral.collegeboard.org

Five numbers and the number they were aiming at

Here is a table of five values: 1, 1.5, 2.5, 4.25, 7.125. They come from stepping along the differential equation dy/dx = x + y from the point (0, 1) out to x = 2, in four jumps of width 0.5.

The true value of y(2) is 11.778. The estimate missed it by 4.653, which is nearly forty percent. That is not a rounding error, it is the method being crude on purpose, and the lesson is about knowing exactly how crude and in which direction.

Euler's method exists because most differential equations have no closed-form solution. AB taught you to separate variables, which works for dy/dx = ky and a handful of others. Everything else needs a numerical method, and Euler's is the simplest one there is.

The idea in one sentence

A differential equation hands you the slope at every point. A point plus a slope is a tangent line. So: stand at the known point, read the slope off the equation, walk along that tangent line a short distance h, and call where you land the new point. Repeat.

In symbols, with dy/dx = f(x, y) and a starting point (x0, y0):

xn+1 = xn + h and yn+1 = yn + h · f(xn, yn)

The second formula is just the point-slope equation of a line, Δy = (slope)(Δx), applied over and over. If you picture a slope field, Euler's method is a path that follows one little dash, then re-reads the field at wherever it ends up, then follows the next.

The error enters immediately. The slope is correct at the start of each step and wrong everywhere else in it, because the true solution curves away while the tangent line keeps going straight. Then the next step starts from a point that is already off the true curve, so the next slope is the right slope for the wrong curve. Errors compound.

The table, filled in

Solve dy/dx = x + y with y(0) = 1, approximating y(2) with h = 0.5. Lay out one row per step and compute left to right:

So Euler's method gives y(2) ≈ 7.125. Two habits make this reliable under time pressure. Keep the slope column separate from the Δy column, because the commonest arithmetic slip is adding the slope instead of adding h times the slope. And carry more decimal places than you intend to report, since the error accumulates.

How wrong, exactly

This equation happens to be solvable, which is what makes it a good teaching example. The solution through (0, 1) is

y = 2ex − x − 1

Verify it rather than trusting it: differentiating gives y' = 2ex − 1, while x + y = x + 2ex − x − 1 = 2ex − 1. The two agree, and y(0) = 2 − 0 − 1 = 1 matches the initial condition. At x = 2, y = 2e² − 3 = 14.7781 − 3 = 11.7781.

Now run the same method with smaller steps and watch the error:

Read the last column downward: 4.653, 2.857, 1.646, 0.885. Each halving of h cuts the error by a factor tending toward 2, not toward 4. That makes Euler a first-order method: the global error is roughly proportional to h. To get one extra decimal place of accuracy you need about ten times as many steps, which is why nobody uses plain Euler for real computation and why every numerical analysis course moves on to Runge-Kutta methods within a week.

In short: Euler's method is exact in the limit and mediocre in practice. Its value on the AP exam is that it is transparent: every number in the table can be checked by hand.

Which way does the error point?

Every Euler estimate above is too small. That is not luck, and the exam asks you to justify it in words.

Differentiate the equation itself: from y' = x + y, y'' = 1 + y' = 1 + x + y. Along this solution, x and y are both positive on [0, 2], so y'' > 0 and the solution curve is concave up throughout.

A tangent line to a concave-up curve lies below the curve, everywhere except the point of tangency. Euler steps along tangent lines. So every step lands below the true solution, and the shortfalls accumulate. The rule is short enough to remember:

• Solution concave up on the interval: Euler underestimates.
• Solution concave down on the interval: Euler overestimates.

The honest caveat is that the rule needs the concavity to keep one sign across the whole interval. If the solution has an inflection point in the middle, tangent lines lie below on one side and above on the other, the two effects fight, and the sign of the final error is not determined by concavity alone. Say so if a question hands you such a curve.

A second run, with a differential equation you cannot separate

Approximate y(1.2) for dy/dx = x² − y with y(1) = 2, using two steps of h = 0.1.

1. At (1, 2): slope = 1 − 2 = −1. Then Δy = 0.1(−1) = −0.1, so y(1.1) ≈ 1.9.
2. At (1.1, 1.9): slope = 1.21 − 1.9 = −0.69. Then Δy = 0.1(−0.69) = −0.069, so y(1.2) ≈ 1.831.

Which direction is this one off? Differentiate the equation: y'' = 2x − y'= 2x − (x² − y) = 2x − x² + y. At the starting point that is 2 − 1 + 2 = 3, positive, and it stays positive over this short interval, so the solution is concave up and 1.831 is an underestimate. Notice that a negative slope does not mean a negative error direction. Concavity, not the sign of the slope, decides.

Common misconceptions

• "Add the slope to y." Add h times the slope. With h = 0.5 in the first table, adding 1.000 instead of 0.500 at the first step throws every later row off.
• "Re-use the slope from the previous step." The slope must be recomputed at the new point, using the new x and the new y. That recomputation is the method.
• "Smaller steps make it exact." They make it better, roughly in proportion to h. Four times as many steps buys about four times less error, not perfection.
• "Euler is too high when the slope is positive." The direction of the error follows the concavity of the solution, not the sign of its slope.

Where this leaves us

Euler's method turns a differential equation into arithmetic: read the slope at the point you are standing on, move h to the right, move h · f(x, y) up, and repeat, one row of a table per step. Applied to dy/dx = x + y from (0, 1) with h = 0.5, it produces 7.125 for a true value of 11.778, and shrinking h to 0.0625 narrows that gap to 0.885, roughly halving the error each time h halves. The direction of the error is not a guess: compute y'' from the equation, and concave up means the tangent lines run below the curve so the estimate is low. Keep the slope and the Δy in separate columns, recompute the slope at every new point, and state your step size in the answer, because a table without its h is not a solution.

Sources

1. OpenStax. (2016). 4.2 Direction fields and numerical methods. In Calculus Volume 2. openstax.org
2. Dawkins, P. (n.d.). Euler's method. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Wikipedia. (n.d.). Euler method. en.wikipedia.org
4. College Board. (n.d.). AP Calculus BC course framework. Unit 7, Differential Equations, carries 5 to 10 percent of the multiple-choice section. apcentral.collegeboard.org

Verhulst changes one assumption

In 1838 P. F. Verhulst published a note of a few pages that altered a single assumption in Malthus and left everything else standing. Malthus had a population growing in proportion to its own size, dP/dt = kP, which doubles and doubles and never stops. Verhulst kept that behaviour for a small population and multiplied the rate by a factor that shuts growth down as the population nears a ceiling. What he wrote is the equation the BC exam asks about:

dP/dt = kP(1 − P/K)

k is the growth constant, the proportional rate the population would grow at with nothing in its way, and K is the carrying capacity, the number the environment can support. The bracket carries the whole idea. When P is small next to K the bracket sits near 1 and the equation is Malthus. When P climbs toward K the bracket falls toward 0 and growth stalls. Notice that it multiplies the rate rather than subtracting from the population: nothing is being removed, the engine is being throttled.

The logistic function was found again twice. A. G. McKendrick fitted it to bacteria growing in broth in 1911, and Raymond Pearl and Lowell Reed at Johns Hopkins arrived at it independently in 1920, which is why ecology texts sometimes call it the Verhulst-Pearl equation. Three people writing down the same equation from three different data sets is a reason to take it seriously as a first model, and not a reason to believe any population obeys it exactly.

Three answers you can give without solving anything

Most logistic questions on the exam are answered from the differential equation alone. Take a concrete one: a pond stocked with 100 fish, where

dP/dt = 0.25P(1 − P/1000)

with P in fish and t in years.

What is the long-run population? Set dP/dt = 0. The right side is zero when P = 0 or P = 1000. Those two constants are the equilibrium solutions. Between them the product 0.25P(1 − P/1000) is positive, so any population that starts between 0 and 1000 increases; above 1000 the bracket turns negative, so the population falls. Both arrows point at 1000. The limit is 1000 fish, and the answer to the standard question is lim(t → ∞) P(t) = K = 1000.

When is the population growing fastest? Read dP/dt as a function of P rather than of t. Expanded, it is 0.25P − 0.00025P², a downward parabola in P with roots at 0 and 1000. A parabola peaks halfway between its roots, at P = 500. So growth is fastest when the pond is half full, whatever the starting number and whatever the year. In general the fastest growth is at P = K/2.

How fast is that? Put P = K/2 back in: k(K/2)(1 − 1/2) = kK/4. Here 0.25(1000)/4 = 62.5 fish per year. That is the ceiling on the growth rate; the pond never adds fish faster than that.

Worth holding on to: carrying capacity, fastest-growth population and maximum rate are K, K/2 and kK/4, and all three come out of the equation without an integral in sight.

Solving it, with partial fractions doing the work

Now the closed form. The equation is separable, so move every P to one side:

dP / [P(1 − P/K)] = k dt

Clear the fraction inside the bracket first, because 1 − P/K = (K − P)/K:

K dP / [P(K − P)] = k dt

The left side is exactly the sort of rational function Module 1 handled. Split it over the two distinct linear factors:

K / [P(K − P)] = A/P + B/(K − P)

Multiply through by P(K − P): K = A(K − P) + BP. Set P = 0 and you get K = AK, so A = 1. Set P = K and you get K = BK, so B = 1. Both coefficients are 1, which is worth remembering because it makes the integration immediate:

∫ (1/P + 1/(K − P)) dP = ∫ k dt

ln|P| − ln|K − P| = kt + C

The minus sign on the second logarithm is the one place this derivation goes wrong. The inner derivative of K − P is −1, so ∫ dP/(K − P) = −ln|K − P|. Combine the logarithms and exponentiate:

P/(K − P) = ekt + C = C1ekt

Solve for P. Cross multiply: P = C1ekt(K − P), gather the P terms as P(1 + C1ekt) = C1Kekt, and divide:

P = C1Kekt / (1 + C1ekt)

Divide top and bottom by C1ekt, which is never zero, and write A = 1/C1:

P(t) = K / (1 + Ae−kt)

Finally pin A to the initial condition. At t = 0, P(0) = K/(1 + A), so

A = (K − P0)/P0

That is the entire derivation, and it is worth doing once by hand rather than memorising the result, because the exam has asked for the separation and the partial fraction split as the credited steps.

The pond, in numbers

For the fish: K = 1000, P0 = 100, k = 0.25. So A = (1000 − 100)/100 = 9 and

P(t) = 1000 / (1 + 9e−0.25t)

Check it at t = 0: 1000/(1 + 9) = 100. Correct. Now put the throttled model beside the unthrottled one, 100e0.25t, which is what the same pond would do with no ceiling.

Read the first four years: the two columns are within fifteen percent of each other, which is the sense in which early logistic growth is exponential growth. By year twenty they differ by a factor of fifteen. The last column shows why: the bracket has collapsed from 0.900 to 0.057, so the pond is running at under six percent of its unthrottled rate.

Where the curve bends

The fastest-growth answer above came from a parabola in P. The same fact is visible in t as an inflection point, and it is worth getting there by differentiating, because the exam sometimes asks for the year rather than the population.

Differentiate dP/dt = kP − kP²/K with respect to t, using the chain rule on both terms:

d²P/dt² = kP' − (2kP/K)P' = kP'(1 − 2P/K)

Since P' > 0 for a population between 0 and K, the second derivative is zero only when 1 − 2P/K = 0, that is P = K/2, and it changes sign there: concave up below K/2, concave down above. The solution curve is the familiar S shape, and the inflection point is where the growth rate peaks.

For the pond, solve 1000/(1 + 9e−0.25t) = 500. That needs 1 + 9e−0.25t = 2, so e−0.25t = 1/9, so −0.25t = −ln 9 and

t = ln 9 / 0.25 = 2.1972/0.25 = 8.789 years

At that instant the pond holds 500 fish and is gaining kK/4 = 62.5 fish per year, faster than at any other moment in its history. Notice how little work the closed form did here: it converted a population into a date, and nothing else.

The disguised version, which is how it usually appears

An exam is more likely to hand you

dP/dt = 0.8P − 0.0004P²

and ask for the carrying capacity. There is no K in sight. Factor the leading coefficient out:

0.8P − 0.0004P² = 0.8P(1 − 0.0005P) = 0.8P(1 − P/2000)

because 0.0004/0.8 = 0.0005 = 1/2000. So k = 0.8 and K = 2000, growth is fastest at 1000, and the maximum rate is 0.8(2000)/4 = 400. The general rule: for dP/dt = aP − bP², the capacity is K = a/b and the growth constant is k = a. Derive it rather than trusting it: aP − bP² = aP(1 − (b/a)P), and matching that with kP(1 − P/K) gives k = a and 1/K = b/a.

Bottom line: the second-degree form and the bracket form are the same equation, and the fastest question on any logistic problem is which form you were handed.

Starting above the ceiling

Nothing in the derivation assumed P0 < K. Stock the same pond with 1200 fish. Then A = (1000 − 1200)/1200 = −1/6, and

P(t) = 1000/(1 − (1/6)e−0.25t)

At t = 0 the denominator is 5/6 and P = 1200, as it should be. As t grows the exponential dies and P falls to 1000 from above, never dipping below it. The bracket 1 − P/K was negative the whole time, so dP/dt was negative the whole time. A logistic population above capacity decays toward it; it does not crash to zero, and it does not oscillate. Real fish populations do sometimes crash, which is a fact about real ponds and not about this equation.

Common misconceptions

• "K is the starting population." K is the limit, P0 is the start, and the constant A = (K − P0)/P0 is what connects them. In the pond, K is 1000 and the start is 100.
• "Growth is fastest at the beginning." The rate is kP(1 − P/K), a product of a small number and a large one at the start. It peaks at the halfway point, 500 fish in year 8.789, not in year 0.
• "The population eventually reaches K." It approaches K. The term Ae−kt is never exactly zero for finite t, so P(t) < K forever when the population starts below. The answer to a limit question is K; the answer to "does it get there" is no.
• "The inflection point is where growth stops." It is where growth is fastest. Growth slows after it, which is what concave down means, but the population is still rising.
• "You can skip the partial fractions and integrate 1/[P(1 − P/K)] directly." There is no elementary antiderivative you can write down by inspection for that product; splitting it into 1/P + 1/(K − P) is the step that makes it two logarithms.

Putting it together

The logistic equation is Malthus multiplied by a brake: dP/dt = kP(1 − P/K). From the equation alone you get the equilibria 0 and K, the limit K, the fastest-growth population K/2, and the maximum rate kK/4. To get the function, separate, rewrite 1 − P/K as (K − P)/K, split K/[P(K − P)] into 1/P + 1/(K − P), integrate to ln|P| − ln|K − P| = kt + C, and unwind to P(t) = K/(1 + Ae−kt) with A = (K − P0)/P0. For the pond that is P(t) = 1000/(1 + 9e−0.25t), which passes 500 fish in year 8.789 at 62.5 fish per year and stands at 942.8 after twenty years. When the equation arrives as aP − bP², the capacity is a/b. And when a question only asks for a limit or a fastest-growth population, do not solve the differential equation at all.

Sources

1. OpenStax. (2016). 4.4 The logistic equation. In Calculus Volume 2. openstax.org
2. Wikipedia. (n.d.). Logistic function. History section, on Verhulst's 1838 note, McKendrick's 1911 bacterial fit and the Pearl and Reed rediscovery of 1920. en.wikipedia.org
3. Wikipedia. (n.d.). Carrying capacity. en.wikipedia.org
4. College Board. (n.d.). AP Calculus BC course framework. Topic 7.9, Logistic Models with Differential Equations, is a BC-only topic within Unit 7. apcentral.collegeboard.org

A paint mark on a rolling wheel

Roll a wheel of radius 1 along a flat road through one full turn. A mark on the rim starts on the ground, rises, and comes back down, tracing one arch of a cycloid. The wheel has moved forward 2π = 6.283. The mark has travelled exactly 8. Not 2π, not some number with a π in it at all: the integer 8, and this lesson derives it in four lines.

That curve cannot be written as y = f(x) anywhere near its cusps, and no amount of algebra will make it one. What describes it is a pair of coordinates driven by a third variable:

x(t) = t − sin t, y(t) = 1 − cos t

with t the angle the wheel has turned. This is a parametric equation, and t is the parameter. The picture is a curve; the equations are instructions for walking along it. Both readings matter, because some exam questions ask about the curve and some ask about the walk, and they have different answers.

Why the slope is a ratio of two rates

Nothing new is needed to differentiate a parametric curve, only the chain rule you already have. If y is a function of x along the curve, and x is a function of t, then

dy/dt = (dy/dx)(dx/dt)

Divide both sides by dx/dt, which is legal wherever that rate is not zero:

dy/dx = (dy/dt)/(dx/dt)

Read it as a comparison of rates. If y is rising three times as fast as x is advancing, the curve rises three units for every one it runs, so its slope is 3. Nothing about that reasoning required a formula for y in terms of x, which is exactly the point.

The second derivative is where people invent a rule that does not exist. d²y/dx² means the derivative of dy/dx with respect to x, so apply the same conversion again to the function dy/dx:

d²y/dx² = [d/dt (dy/dx)] / (dx/dt)

Differentiate the slope with respect to t, then divide by dx/dt. It is emphatically not (d²y/dt²)/(d²x/dt²), and the worked example below produces two different numbers to prove it.

One curve, worked end to end

Take x = t² − 2t and y = t³ − 3t.

Step 1, the rates. dx/dt = 2t − 2 and dy/dt = 3t² − 3.

Step 2, the slope. dy/dx = (3t² − 3)/(2t − 2). Factor before you evaluate anything: the top is 3(t − 1)(t + 1) and the bottom is 2(t − 1), so for every t ≠ 1

dy/dx = 3(t + 1)/2

Step 3, a tangent line. At t = 2 the point is x = 4 − 4 = 0, y = 8 − 6 = 2, so (0, 2), and the slope is 3(3)/2 = 4.5. The tangent line is y = 2 + 4.5x. Give the point in x and y, never in t: a line in the plane has no t in it.

Step 4, concavity. Differentiate the slope with respect to t: d/dt [3(t + 1)/2] = 3/2. Divide by dx/dt:

d²y/dx² = (3/2)/(2t − 2) = 3/(4(t − 1))

At t = 2 that is 3/4 = 0.75, positive, so the curve is concave up there. Now run the invented rule for comparison: d²y/dt² = 6t and d²x/dt² = 2, whose quotient is 3t, equal to 6 at t = 2. Two numbers, 0.75 and 6, and only the first one is the concavity of the curve.

The point: both parametric derivatives are conversions, not new formulas. Differentiate with respect to t, then divide by dx/dt, every single time.

Horizontal, vertical, and the case that is neither

The slope is a fraction, so its behaviour is decided by which part of the fraction is zero.

For the worked curve, dy/dt = 3(t² − 1) = 0 at t = 1 and t = −1. At t = −1 the other rate is 2(−1) − 2 = −4, not zero, so there is a genuine horizontal tangent, at the point x = 1 + 2 = 3, y = −1 + 3 = 2, that is (3, 2).

At t = 1 both rates vanish. The tempting conclusion is a vertical tangent, because dx/dt = 0. It is wrong. The simplified slope 3(t + 1)/2 tends to 3(2)/2 = 3 as t approaches 1, so the curve passes through (−1, −2) with slope 3, as ordinary a tangent as any other. The zero in the denominator was cancelled by a zero in the numerator, which is why Step 2 factors before it evaluates.

Speed, arc length, distance travelled

Lesson 4 built arc length from Δs = √((Δx)² + (Δy)²). Nothing about that triangle cared where x and y came from. Factor out Δt this time rather than Δx:

Δs = √((Δx/Δt)² + (Δy/Δt)²) · Δt

and in the limit

L = ∫(a to b) √((dx/dt)² + (dy/dt)²) dt

The integrand has a physical name. If t is time, √((dx/dt)² + (dy/dt)²) is the speed of the particle, and the integral of speed over an interval of time is the total distance travelled. Speed is a number and is never negative; velocity is the pair of rates and carries direction.

For the worked curve at t = 2: dx/dt = 2, dy/dt = 9, so the speed is √(4 + 81) = √85 = 9.2195. Its distance travelled from t = 0 to t = 2 is

∫(0 to 2) √((2t − 2)² + (3t² − 3)²) dt = 6.3655

which no elementary antiderivative will produce. Set it up, integrate numerically, report three decimals. That is the expected exam answer, and the setup is where the credit sits.

Finishing the cycloid

Now the promised 8. With x = t − sin t and y = 1 − cos t:

dx/dt = 1 − cos t, dy/dt = sin t

Square and add, then use sin²t + cos²t = 1:

(1 − cos t)² + sin²t = 1 − 2cos t + cos²t + sin²t = 2 − 2cos t

The half-angle identity 1 − cos t = 2sin²(t/2) turns that into 4sin²(t/2), whose square root is 2|sin(t/2)|. On 0 ≤ t ≤ 2π the half-angle runs from 0 to π, where the sine is never negative, so the bars come off:

L = ∫(0 to 2π) 2 sin(t/2) dt = [−4 cos(t/2)](0 to 2π) = −4(−1) + 4(1) = 8

An arch of a cycloid traced by a wheel of radius r has length 8r, a result Wikipedia's article derives the same way. Notice where the exactness came from: the radicand collapsed to a perfect square, exactly as in the engineered examples of Lesson 4. Parametric arc length is not more tractable than the Cartesian kind, it just has more chances for a trigonometric identity to rescue it.

When distance travelled is not the length of the curve

Let x = cos t, y = sin t for 0 ≤ t ≤ 4π. The speed is √(sin²t + cos²t) = 1, so the integral of speed over that interval is 4π = 12.566. But the curve being traced is the unit circle, whose length is 2π = 6.283. The particle went around twice.

So the arc length integral computes the length of the curve only when the curve is traced once. Read the question: "how far does the particle travel" always wants the integral of speed, while "what is the length of the curve" wants the parameter interval that traces it exactly once. On the same curve, the displacement is a third quantity again: here the particle ends where it started, so its displacement is zero while its distance travelled is 12.566.

Common misconceptions

• "The second derivative is the ratio of the second derivatives." For x = t² − 2t, y = t³ − 3t at t = 2, that rule gives 6 and the correct answer is 0.75.
• "dx/dt = 0 means a vertical tangent." Only when dy/dt is not also zero. At t = 1 on the worked curve both vanish and the true slope is 3.
• "The tangent line can be written with t in it." A line in the xy plane is an equation in x and y. Convert the parameter value to a point first.
• "Speed is dy/dx." Speed is √((dx/dt)² + (dy/dt)²). The slope has no units of time in it and says nothing about how fast the particle moves.
• "Arc length and distance travelled are the same integral, so they are the same number." Same integrand, different question: retracing adds to distance travelled and not to the length of the curve.

What you now know

A parametric curve is two coordinate functions of one parameter, and every calculus question about it reduces to differentiating with respect to t and dividing by dx/dt. The slope is (dy/dt)/(dx/dt); the concavity is [d/dt(dy/dx)]/(dx/dt) and never a ratio of second derivatives. Factor the slope before evaluating, so a shared zero cancels instead of frightening you: on x = t² − 2t, y = t³ − 3t that turns an apparent catastrophe at t = 1 into a plain tangent of slope 3. Horizontal tangents need dy/dt = 0 with dx/dt ≠ 0; vertical ones need the reverse. Speed is the square root of the sum of the squared rates, its integral is distance travelled, and that integral is the length of the curve only when the parameter traces the curve once. One arch of the unit cycloid is exactly 8; a particle going twice around the unit circle travels 4π along a curve of length 2π.

Sources

1. OpenStax. (2016). 7.2 Calculus of parametric curves. In Calculus Volume 2. openstax.org
2. Dawkins, P. (n.d.). Tangents with parametric equations. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
3. Dawkins, P. (n.d.). Arc length with parametric equations. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Cycloid. Measurement section: the arc length of one arch is 8r. en.wikipedia.org
5. College Board. (n.d.). AP Calculus BC course framework. Unit 9, Parametric Equations, Polar Coordinates, and Vector-Valued Functions, is BC only and carries 11 to 12 percent of the multiple-choice section. apcentral.collegeboard.org

A different pair of numbers for the same point

The point 1.616 to the right and 0.933 up can be named another way: 1.866 units from the origin, at an angle of 30 degrees. Those two numbers, r = 1.866 and θ = π/6, are its polar coordinates, and the translation runs both ways:

x = r cos θ, y = r sin θ, and r² = x² + y², tan θ = y/x

That point lies on the curve r = 1 + cos θ, whose Cartesian equation is a fourth-degree mess nobody wants to differentiate. In polar form it is five characters long. That trade is the only reason polar coordinates are worth the trouble: the equation gets short, and every formula from AB has to be rebuilt. Here is what changes and what does not.

Every difference in that table comes from one swap: the thin vertical rectangle becomes a thin wedge with its point at the origin.

Graphing without a calculator

Build a table of angles and radii, plot the points, and join them in order of increasing θ. For r = 1 + cos θ:

Start at 2 on the positive x axis, shrink steadily to 0 at θ = π, then grow back to 2. The result is a cardioid, a heart-shaped loop with a dimple where it touches the pole. Three families cover almost everything the exam shows you.

• Circles. r = a is a circle of radius a centred at the origin. r = 2a cos θ is a circle of radius a centred at (a, 0), traced completely as θ runs from −π/2 to π/2; r = 2a sin θ is the same circle rotated onto the y axis.
• Limacons. r = a + b cos θ. When a = b it is a cardioid. When a < b, as in r = 2 + 4 cos θ, the radius goes negative for some angles and the curve makes an inner loop.
• Roses. r = a cos(nθ) has n petals when n is odd and 2n when n is even. The reason is retracing: for odd n the curve has already drawn the whole figure by θ = π and the second half of the circle repeats it, while for even n the second half draws new petals. So r = cos 2θ has four petals and r = 2 sin 3θ has three.

Negative r is not an error: r = −2 at θ = 0 plots at (−2, 0), the stated distance backwards along the ray. That convention is what draws the inner loops.

Slope, by borrowing the parametric rule

A polar curve is a parametric curve whose parameter is the angle:

x(θ) = r(θ) cos θ and y(θ) = r(θ) sin θ

Differentiate each with the product rule, which is the whole derivation:

dx/dθ = r' cos θ − r sin θ and dy/dθ = r' sin θ + r cos θ

Then dy/dx = (dy/dθ)/(dx/dθ), exactly as in Lesson 7. Try it on the cardioid at θ = π/6, where r = 1 + cos(π/6) = 1.8660 and r' = −sin(π/6) = −0.5:

dy/dθ = (−0.5)(0.5) + (1.8660)(0.8660) = −0.25 + 1.6160 = 1.3660

dx/dθ = (−0.5)(0.8660) − (1.8660)(0.5) = −0.4330 − 0.9330 = −1.3660

The quotient is exactly −1. The point itself is (1.8660 · 0.8660, 1.8660 · 0.5) = (1.6160, 0.9330), so the tangent line there is y − 0.9330 = −(x − 1.6160). As always, the answer is a line in x and y with no θ left in it.

So what?: the polar slope formula is worth deriving on the spot from x = r cos θ and y = r sin θ rather than memorising, because misremembering one sign in it is the commonest way to lose these points.

Area comes from sectors, not rectangles

A full circle of radius r has area πr². A sector spanning an angle Δθ is the fraction Δθ/(2π) of it, so its area is

(Δθ/(2π)) · πr² = (1/2)r² Δθ

Cut the region between the rays θ = α and θ = β into thin wedges, approximate each by a circular sector of that radius, add, and let the wedges shrink:

A = (1/2) ∫(α to β) r² dθ

Two things go wrong here more than anywhere else in the unit. The square is not optional, and the one half is not optional. An answer computed as ∫ r dθ is not an area at all; it has the units of a length.

The whole cardioid. The curve closes as θ runs from 0 to 2π, so

A = (1/2) ∫(0 to 2π) (1 + cos θ)² dθ = (1/2) ∫(0 to 2π) (1 + 2cos θ + cos²θ) dθ

Take the three pieces separately. The constant contributes 2π. The 2cos θ contributes 0, since sin θ returns to its starting value. For the last one use cos²θ = (1 + cos 2θ)/2, whose integral over a full turn is π. So

A = (1/2)(2π + 0 + π) = 3π/2 = 4.712

One petal of a rose. For r = cos 2θ, a petal begins and ends where r = 0, that is where cos 2θ = 0, so 2θ = −π/2 and 2θ = π/2, giving θ from −π/4 to π/4. Finding those limits is the real work; the integral is routine:

A = (1/2) ∫(−π/4 to π/4) cos²2θ dθ = (1/4) ∫ (1 + cos 4θ) dθ = (1/4)[θ + (sin 4θ)/4]

Evaluated between the limits, the sine terms are 0 at both ends and the linear term gives π/2, so A = π/8 = 0.3927. Four petals make π/2.

Two curves, and the subtraction that is not a subtraction

For the region inside r = R(θ) and outside r = r(θ), subtract the sector areas, not the radii:

A = (1/2) ∫(α to β) (R² − r²) dθ

and note that R² − r² is not (R − r)². Work the standard case: the area inside the circle r = 3 sin θ and outside the cardioid r = 1 + sin θ.

1. Find the limits. Set the two equal: 3 sin θ = 1 + sin θ, so sin θ = 1/2 and θ = π/6 or 5π/6.
2. Decide which is outer. Test the midpoint θ = π/2: the circle gives 3 and the cardioid gives 2. The circle is outside, so it is R.
3. Set up and expand. (3 sin θ)² − (1 + sin θ)² = 9sin²θ − 1 − 2sin θ − sin²θ = 8sin²θ − 2 sin θ − 1. Replace 8sin²θ with 4 − 4cos 2θ, leaving 3 − 4cos 2θ − 2 sin θ.
4. Integrate. An antiderivative is 3θ − 2 sin 2θ + 2 cos θ. At 5π/6 it is 7.8540 + 1.7321 − 1.7321 = 7.8540; at π/6 it is 1.5708 − 1.7321 + 1.7321 = 1.5708. The difference is 6.2832 = 2π.
5. Halve it. A = (1/2)(2π) = π = 3.1416.

The intersection your algebra will not find

Now the case that costs the most points. Find where r = 1 + cos θ meets r = 3 cos θ. Setting them equal gives 1 + cos θ = 3 cos θ, so cos θ = 1/2 and θ = ±π/3, where both curves have r = 1.5.

But look at the pole. The cardioid reaches r = 0 at θ = π. The circle reaches r = 0 at θ = π/2. Both pass through the origin, so the origin is an intersection point, and no value of θ satisfies both equations there. Solving equations finds only the points where the curves are in the same place at the same angle, and r = 0 names the origin for every θ.

The habit that fixes it: after solving, always check separately whether each curve passes through the pole, and sketch. A sketch also tells you when a region needs two integrals with different limits, which happens whenever the outer curve changes identity partway round.

Common misconceptions

• "Area is the integral of r." It is half the integral of r². The sector, not the rectangle, is the building block, and a sector's area carries r².
• "The area between two curves is half the integral of (R − r)²." It is half the integral of R² − r². Those differ: with R = 3 and r = 2 the first gives 1 and the second gives 5.
• "Negative r is meaningless." It plots backwards along the ray, and it is what draws the inner loop of r = 2 + 4 cos θ.
• "r = cos(nθ) has n petals." Only for odd n. For even n it has 2n, so r = cos 2θ has four.
• "Setting the two equations equal finds every intersection." It misses the pole whenever the curves arrive there at different angles, as r = 1 + cos θ and r = 3 cos θ do.

The takeaway

Polar coordinates trade a short equation for a rebuilt toolkit, and every rebuilt formula comes from one swap: the thin rectangle becomes a thin sector. Graph by tabulating radii against angles and joining them in order, and know the three families, with roses giving n petals for odd n and 2n for even n. Differentiate by writing x = r cos θ and y = r sin θ, which gave the cardioid a slope of exactly −1 at θ = π/6. Areas are (1/2)∫r²dθ, giving 3π/2 for the cardioid and π/8 for one petal of r = cos 2θ; regions between curves are (1/2)∫(R² − r²)dθ, giving exactly π inside r = 3 sin θ and outside r = 1 + sin θ. Remember: solve for the intersections, then check the pole by hand, because the algebra cannot see it.

Sources

1. OpenStax. (2016). 7.4 Area and arc length in polar coordinates. In Calculus Volume 2. openstax.org
2. OpenStax. (2016). 7.3 Polar coordinates. In Calculus Volume 2. openstax.org
3. Dawkins, P. (n.d.). Area with polar coordinates. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Polar coordinate system. en.wikipedia.org
5. College Board. (n.d.). AP Calculus BC course framework. Topics 9.7 to 9.9 cover polar curves, their derivatives and the area of polar regions. apcentral.collegeboard.org

One scenario, three questions, three different integrals

At time t = 1 a particle sits at the point (2, −1). For t ≥ 1 its velocity is

v(t) = (ln(t + 1), 2e−t)

Three questions get asked about this particle, and a reader who confuses them loses the whole question. Where is it at t = 3? How far did it travel between t = 1 and t = 3? How fast is it moving at t = 1? The answers are (4.159, −0.364), 2.298, and 1.011. The first needs two integrals and the initial point, the second needs one integral of a square root, and the third needs no integral at all.

The rest of the lesson builds the machinery, then comes back and computes all three.

A position is a pair of functions, and so is everything else

A vector-valued function assigns a point in the plane to each value of t:

r(t) = (x(t), y(t))

Some texts write it with angle brackets, <x(t), y(t)>, and some with i and j components; the exam accepts any of them. This is the same object as the parametric curve of Lesson 7, renamed. What the vector language adds is that the derivative is itself a vector, with a direction as well as a size, and that the second derivative is worth having a name for.

Differentiation is componentwise, because each coordinate is an ordinary function of t:

• Velocity v(t) = r'(t) = (x'(t), y'(t)), which points along the direction of motion.
• Acceleration a(t) = r''(t) = (x''(t), y''(t)), which need not point along the motion at all.
• Speed |v(t)| = √([x'(t)]² + [y'(t)]²), a single number, never negative.

Speed is the only scalar in that list, and it is the one the distance integral uses. Velocity is a vector and has no sign in the AB sense; a particle moving left and down has velocity (−3, −4) and speed 5.

Key idea: everything vector-valued is done one component at a time, except speed, which is where the two components come back together under a square root.

Going backwards: velocity to position

If you know the velocity and one position, the Fundamental Theorem of Calculus recovers the rest, again one component at a time:

x(b) = x(a) + ∫(a to b) x'(t) dt and y(b) = y(a) + ∫(a to b) y'(t) dt

The initial coordinate is not optional and is the single most frequently dropped term on this kind of question. Without it you have computed the displacement in each coordinate, which is the change in position, not the position.

Distance travelled is a different integral entirely:

distance = ∫(a to b) √([x'(t)]² + [y'(t)]²) dt

Displacement can be zero while distance travelled is large; the reverse cannot happen. A particle that goes out and comes back has displacement (0, 0) and a distance travelled equal to twice the trip.

Answering the opening question

The position at t = 3. For x, integrate ln(t + 1). That is Module 1 work: integrate by parts with u = ln(t + 1) and dv = dt, or substitute w = t + 1 and use the standard result, giving the antiderivative (t + 1)ln(t + 1) − (t + 1). Evaluate from 1 to 3:

[4 ln 4 − 4] − [2 ln 2 − 2] = (5.5452 − 4) − (1.3863 − 2) = 1.5452 + 0.6137 = 2.1589

So x(3) = 2 + 2.1589 = 4.1589. For y, ∫(1 to 3) 2e−t dt = [−2e−t](1 to 3) = −0.0996 + 0.7358 = 0.6362, so y(3) = −1 + 0.6362 = −0.3638. The particle is at (4.159, −0.364).

The distance travelled. Now the two components go under one square root:

∫(1 to 3) √([ln(t + 1)]² + 4e−2t) dt = 2.2983

No antiderivative exists for that in elementary terms, which is normal and expected. Write the integral, evaluate it numerically, report 2.298. Notice it is smaller than the 2.1589 the x coordinate moved plus the 0.6362 the y coordinate moved: the hypotenuse of each little step is shorter than the two legs added, and that is the whole difference between a distance and a pair of displacements.

The speed at t = 1. No integral. x'(1) = ln 2 = 0.6931 and y'(1) = 2e−1 = 0.7358, so the speed is √(0.4804 + 0.5414) = √1.0218 = 1.0108.

A particle that stops, and what happens there

Take a second particle with

x(t) = t² − 4t and y(t) = t³/3 − 4t

so v(t) = (2t − 4, t² − 4) and a(t) = (2, 2t).

When is it at rest? A particle is at rest when it is not moving in either direction, so both components of velocity must be zero at the same t. The first vanishes at t = 2. The second vanishes at t = 2 and t = −2. Only t = 2 satisfies both, so the particle is at rest exactly at t = 2, where its position is (−4, −16/3).

Its acceleration at that instant is (2, 4), which is not zero. Being at rest does not mean being unaccelerated: a ball at the top of its flight is momentarily still and is being pulled down the whole time.

How far does it travel from t = 0 to t = 3?

∫(0 to 3) √((2t − 4)² + (t² − 4)²) dt = 9.2287

The displacement over the same interval is a much smaller thing: x(3) − x(0) = −3 and y(3) − y(0) = 9 − 12 = −3, a displacement of (−3, −3) whose length is 3√2 = 4.243. The particle covered 9.229 units of ground to end up 4.243 units from where it started, because it reversed direction at t = 2.

Is the speed increasing?

This question confuses people because speed is a square root and differentiating it looks ugly. It is not, if you differentiate the square instead. Write s(t)² = [x'(t)]² + [y'(t)]² and differentiate both sides:

2s s' = 2x'x'' + 2y'y'', so s s' = x'x'' + y'y'' = v · a

Since speed s is positive whenever the particle is moving, s' has the same sign as the dot product v · a. So:

• v · a > 0: speed increasing. The acceleration has a component in the direction of travel.
• v · a < 0: speed decreasing. The acceleration is partly opposing the motion.
• v · a = 0: speed momentarily unchanged, even though the direction is turning.

Test the second particle at t = 1: v = (−2, −3) and a = (2, 2), so v · a = −4 − 6 = −10, negative, and the speed is decreasing as the particle heads toward its stop at t = 2. At t = 3: v = (2, 5) and a = (2, 6), so v · a = 4 + 30 = 34, positive, and the speed is climbing again.

The upshot: speed increasing is about the angle between velocity and acceleration, not about either one being positive. Circular motion at constant speed has a large acceleration and v · a = 0 at every instant.

Common misconceptions

• "Speed is the derivative of the position vector." That is velocity, a vector. Speed is the length of the velocity, a single non-negative number.
• "The particle is at rest when acceleration is zero." At rest means the velocity vector is (0, 0). The second particle is at rest at t = 2 with acceleration (2, 4).
• "Integrating the velocity gives the position." It gives the change in position. Add the known coordinate: x(3) = 2 + 2.1589, not 2.1589.
• "Distance travelled is the length of the displacement." Only if the particle never reverses. The second particle travelled 9.229 and ended 4.243 away.
• "Speed is increasing when the acceleration is positive." Acceleration is a vector and has no sign. The test is the sign of v · a.

Summing up

A vector-valued function is a parametric curve with the components kept in one object, and the calculus is componentwise: velocity is (x', y'), acceleration is (x'', y''), and speed is the length √(x'² + y'²). Going the other way, each coordinate is recovered as the starting value plus the integral of its own rate, which is how (2, −1) and v(t) = (ln(t + 1), 2e−t) produced (4.159, −0.364) at t = 3. Distance travelled is the single integral of speed, 2.298 over that interval, and it is never the length of the displacement unless the path never turns back. A particle is at rest only when both velocity components vanish together, and its speed is increasing exactly when v · a is positive. Read the question for which of the three it wants: position, distance, or speed. They are three different calculations on the same two functions.

Sources

1. OpenStax. (2016). 3.4 Motion in space. In Calculus Volume 3. Velocity, acceleration and speed as derivatives of a vector-valued function. openstax.org
2. OpenStax. (2016). 7.2 Calculus of parametric curves. In Calculus Volume 2. The arc length and distance integrals used here. openstax.org
3. Wikipedia. (n.d.). Vector-valued function. en.wikipedia.org
4. College Board. (n.d.). AP Calculus BC course framework. Topics 9.4 to 9.6, Vector-Valued Functions and motion, are BC only. apcentral.collegeboard.org

A decimal that is secretly an infinite sum

Write out 0.727272... and ask what it actually equals. Not approximately: exactly. The digits say

0.72 + 0.0072 + 0.000072 + ...

which is an infinite list of numbers being added. Each term is the previous one divided by 100. By the end of this lesson you will be able to say that the sum is exactly 8/11, and to say why adding infinitely many positive numbers produced a finite answer rather than an infinite one. That same machinery decides every question in the last third of BC.

Two objects that sound the same and are not

A sequence is an ordered list of numbers, a1, a2, a3, ..., which is to say a function whose inputs are the positive integers. A series is what you get when you add a sequence up. Keeping them apart is not pedantry: they converge under different conditions, and confusing them produces the single commonest wrong answer in this unit.

Look at the fourth row. The sequence 1/n converges to 0, and the series ∑ 1/n diverges. Both statements are about the same list of numbers. That is the whole difficulty of this module in one line.

Limits of sequences reuse what you already have

To find the limit of a sequence, treat n as a continuous variable and use every limit technique you know.

• an = (3n² + 2)/(5n² − n). Divide top and bottom by n²: (3 + 2/n²)/(5 − 1/n) → 3/5.
• an = n/2n. The form is ∞/∞, so L'Hopital on x/2x gives 1/(2x ln 2) → 0. Exponentials beat polynomials, always.
• an = (1 + 1/n)n → e = 2.71828. This is the definition of e, not something to derive here.
• an = (−1)n has no limit: it sits at 1 and −1 forever, approaching neither.
• an = (−1)n/n → 0, because the size shrinks to zero even though the sign keeps flipping.

One theorem is worth having by name. If a sequence is increasing and bounded above, it converges, and likewise if it is decreasing and bounded below. That is the monotone convergence theorem, and it guarantees a limit exists without telling you what it is. It is the reason a series of positive terms whose partial sums stay under a ceiling must converge, which is what every comparison test in the next lesson exploits.

What the sum of infinitely many numbers means

You cannot add infinitely many numbers. What you can do is add finitely many and watch what happens. Define the partial sum

Sn = a1 + a2 + ... + an

and then say that the series converges to S when the sequence S1, S2, S3, ... converges to S. The infinite sum is defined as a limit of finite sums, and no other meaning is claimed for it.

What matters here: every statement about a series is secretly a statement about its sequence of partial sums. When a test result confuses you, go back to the partial sums.

A case where you can see the partial sums directly: ∑(n = 1 to ∞) 1/(n(n + 1)). Partial fractions split the term into 1/n − 1/(n + 1), so

Sn = (1 − 1/2) + (1/2 − 1/3) + ... + (1/n − 1/(n + 1)) = 1 − 1/(n + 1)

Everything in the middle cancels, which is why this is called a telescoping series. Then Sn → 1, so the sum is exactly 1. Check it numerically: S1 = 0.5, S4 = 0.8, S99 = 0.99.

The geometric series, derived

A geometric series multiplies by a fixed ratio r at every step. Take the partial sum of n terms starting from a:

Sn = a + ar + ar² + ... + arn−1

Multiply the whole thing by r:

rSn = ar + ar² + ... + arn−1 + arn

Subtract the second line from the first. Every term in the middle appears in both lines and cancels, leaving Sn − rSn = a − arn, so for r ≠ 1

Sn = a(1 − rn)/(1 − r)

Now take the limit. The only part that depends on n is rn. If |r| < 1 then rn → 0 and

∑(n = 0 to ∞) arn = a/(1 − r)

If |r| > 1 then rn grows without bound and the partial sums do too. If r = 1 the sum is a + a + a + ..., which diverges for a ≠ 0. If r = −1 the partial sums bounce between a and 0 and never settle. So the condition is exactly |r| < 1, and the formula is never to be used outside it.

Numbers to hold on to: ∑ 3(1/4)n from n = 0 is 3/(1 − 1/4) = 4, and its partial sums run 3, 3.75, 3.9375, 3.984375, closing on 4 exactly as promised.

Two traps in the formula

The first term is whatever the series actually starts with. For ∑(n = 2 to ∞) 5(1/3)n, the first term is not 5. It is the n = 2 term, 5/9. So the sum is (5/9)/(1 − 1/3) = (5/9)(3/2) = 5/6 = 0.8333. Writing 5/(1 − 1/3) = 7.5 is the standard mistake and it is off by a factor of nine.

The ratio has to be constant. ∑ n/2n is not geometric: the ratio of consecutive terms is (n + 1)/(2n), which changes with n. It does converge, but not by this formula.

Now finish the opening. The decimal 0.727272... is 0.72 + 0.72(0.01) + 0.72(0.01)² + ..., geometric with a = 0.72 and r = 0.01. So

0.727272... = 0.72/(1 − 0.01) = 0.72/0.99 = 72/99 = 8/11

The same argument applied to 0.999... gives 0.9/(1 − 0.1) = 1 exactly, which is why 0.999... and 1 are the same number and not merely close.

The warning this module is built around

Terms shrinking to zero is necessary for convergence and nowhere near sufficient. The harmonic series ∑ 1/n has terms going to zero and diverges anyway. Group its terms:

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

The first bracket exceeds 1/4 + 1/4 = 1/2. The second exceeds 4(1/8) = 1/2. The next group of eight exceeds 8(1/16) = 1/2. Every group adds more than a half, and there are infinitely many groups, so the partial sums pass any number you name. They do it slowly: reaching 10 takes about 12367 terms. Slow divergence is still divergence.

Common misconceptions

• "If the terms go to zero the series converges." The harmonic series is the standing counterexample. The implication runs one way only.
• "A sequence and its series converge together." 1/n converges and ∑1/n diverges. Read which object the question names.
• "a is the coefficient in front." a is the first term the series actually contains. For ∑(n = 2) 5(1/3)n it is 5/9.
• "0.999... is slightly less than 1." It is a geometric series with sum 0.9/0.9 = 1. The partial sums are less than 1; the limit is not.
• "Divergent means the terms blow up." Divergent means the partial sums fail to approach a single number. They can grow slowly, as in ∑1/n, or oscillate, as in ∑(−1)n.

Looking back

A sequence is a list and a series is its running total, and the series converges exactly when the partial sums Sn converge. Sequence limits are ordinary limits with n as the variable, so L'Hopital and dividing by the highest power both still work, and an increasing sequence with a ceiling must converge even when you cannot name the ceiling. Two families can be summed exactly: telescoping series, where the partial sum collapses to a couple of terms and ∑1/(n(n+1)) comes out at 1, and geometric series, where Sn = a(1 − rn)/(1 − r) gives a/(1 − r) whenever |r| < 1 and gives nothing otherwise. Take a from the first term actually present, not from the coefficient. Remember: terms going to zero buys you nothing on its own, which is why the next lesson is nine tests rather than one.

Sources

1. OpenStax. (2016). 5.1 Sequences. In Calculus Volume 2. openstax.org
2. OpenStax. (2016). 5.2 Infinite series. In Calculus Volume 2. Partial sums, geometric and telescoping series. openstax.org
3. Dawkins, P. (n.d.). Special series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Geometric series. en.wikipedia.org
5. College Board. (n.d.). AP Calculus BC course framework. Unit 10, Infinite Sequences and Series, is BC only and carries 17 to 18 percent of the multiple-choice section. apcentral.collegeboard.org

Four series that look alike and behave differently

Here are four sums. They are built from the same materials and three of the four decisions go different ways.

∑ n/(2n + 3), ∑ 1/√n, ∑ 1/(n² + 1), ∑ 1/(n ln n)

The first diverges for a trivial reason, the second for a structural one, the third converges, and the fourth diverges even though its terms are smaller than the third's benchmark. This lesson is the procedure that settles each of them, in the order you should try things.

Step one: does the term even go to zero?

The nth-term test is one line long and should be the first thing you check every time.

If lim(n → ∞) an ≠ 0, the series diverges.

The proof is short enough to carry around. Suppose the series converges to S. Then Sn → S and Sn−1 → S as well, since dropping one term from the front does not change where a sequence is heading. But an = Sn − Sn−1, so an → S − S = 0. Convergence forces the terms to zero; therefore terms that do not go to zero forbid convergence.

Read what that does and does not license. It licenses one conclusion, divergence, from one observation, terms not tending to zero. Reversed, it says nothing. ∑1/n has terms tending to zero and diverges. So a series that passes this test has not been shown to converge; it has merely survived to the next test.

First series settled. ∑ n/(2n + 3): divide by n to get 1/(2 + 3/n) → 1/2. The terms approach 1/2, not 0, so the sum diverges. You are adding roughly half a unit forever. Total work: one line.

Step two: is it one of the series you already know?

Three families have known answers, and recognising one ends the problem.

• Geometric: constant ratio r. Converges exactly when |r| < 1, to a/(1 − r).
• Telescoping: the partial sum collapses. Evaluate it directly.
• p-series: ∑ 1/np. Converges exactly when p > 1. That rule is derived below, not assumed.

The integral test, and where the p-series rule comes from

Let f be positive, continuous and decreasing on [1, ∞), with f(n) = an. The integral test says the series ∑ an and the improper integral ∫(1 to ∞) f(x) dx converge together and diverge together.

The reason is a picture made of rectangles. Over the interval [n − 1, n] draw a rectangle of width 1 and height f(n). Because f is decreasing, f(n) is the smallest value f takes on that interval, so the rectangle fits under the curve:

an = f(n) ≤ ∫(n−1 to n) f(x) dx

Add those from n = 2 to N: the partial sum SN − a1 is at most ∫(1 to N) f. If the improper integral converges, the partial sums are increasing and bounded above, so by the monotone convergence theorem they converge. Now put the rectangle on the other side: with height f(n) over [n, n + 1] it sits above the curve, giving an ≥ ∫(n to n+1) f, so SN ≥ ∫(1 to N+1) f. If the integral diverges, the partial sums are pushed past every bound.

The point: the test compares a sum of rectangles with the area under a curve, and the three conditions on f are exactly what makes the rectangles line up on the right side of the curve.

The test does not say the sum equals the integral. ∑1/n² = π²/6 = 1.6449, while ∫(1 to ∞) dx/x² = 1. Same verdict, different numbers.

Deriving the p-series rule. Apply the test to f(x) = x−p, which is positive and decreasing on [1, ∞) for p > 0. Module 1 already evaluated this improper integral: for p ≠ 1,

∫(1 to ∞) x−p dx = [x1−p/(1 − p)](1 to ∞)

The exponent 1 − p is negative exactly when p > 1, and then the upper limit contributes 0 and the integral equals 1/(p − 1). For p < 1 the exponent is positive and the upper limit blows up. For p = 1 the antiderivative is ln x, which also blows up. So the series converges exactly when p > 1, and the harmonic series at p = 1 sits precisely on the boundary.

Second series settled. ∑1/√n = ∑1/n1/2 has p = 1/2 ≤ 1, so it diverges.

Fourth series settled. ∑(n = 2 to ∞) 1/(n ln n). The function 1/(x ln x) is positive, continuous and decreasing on [2, ∞). Substitute u = ln x, du = dx/x:

∫(2 to ∞) dx/(x ln x) = ∫(ln 2 to ∞) du/u = [ln u] = [ln(ln x)](2 to ∞)

which grows without bound. The series diverges. It diverges with almost unbelievable slowness, since ln(ln x) reaches 3 only near x = e20, about 485 million, but slow divergence is divergence.

Step three: compare it with something you know

Most series are none of the standard families but look like one. Two tests handle that, and the difference between them is worth getting straight.

Direct comparison. For series with non-negative terms, with 0 ≤ an ≤ bn for all large n:

• If ∑ bn converges, then ∑ an converges. A sum trapped under a finite ceiling cannot escape.
• If ∑ an diverges, then ∑ bn diverges. Something bigger than an unbounded sum is unbounded.

The two useless directions are worth naming out loud: smaller than a divergent series proves nothing, and bigger than a convergent series proves nothing.

Third series settled. ∑1/(n² + 1). Since n² + 1 > n², the terms satisfy 1/(n² + 1) < 1/n², and ∑1/n² converges as a p-series with p = 2. So the series converges. The inequality had to point the right way, and here it did, because adding to a denominator shrinks a fraction.

Limit comparison. Direct comparison fails whenever the inequality points the wrong way, which is often. Take ∑1/(n² − 1): the terms are larger than 1/n², so the convergent benchmark says nothing. Limit comparison repairs this. For positive terms, form

L = lim(n → ∞) an/bn

If 0 < L < ∞, the two series do the same thing. The intuition is that the terms are eventually within a constant factor of each other, and multiplying every term of a series by a fixed positive constant cannot change whether it converges.

Choosing bn is the skill, and the rule is mechanical: keep the highest power on top and the highest power on the bottom of an, and throw away everything else.

Work the first row so the mechanics are visible. With an = (2n² + 3)/(5n⁴ + n + 1) and bn = 1/n²,

an/bn = n²(2n² + 3)/(5n⁴ + n + 1) = (2n⁴ + 3n²)/(5n⁴ + n + 1) → 2/5

The limit is finite and positive, and ∑1/n² converges, so the original converges. Note what the answer is not: 2/5 is not the sum. It is only the evidence that the two series share a fate.

The order to try things

1. Terms not going to zero? Diverges. Stop.
2. Geometric, telescoping or a p-series? Use the known answer.
3. A rational function of n, or anything built from powers? Limit comparison against the ratio of leading powers.
4. Positive, decreasing and easy to integrate, especially with a logarithm in it? Integral test.
5. A clean inequality available in the right direction? Direct comparison, which is the fastest when it works.

Common misconceptions

• "The terms go to zero, so it converges." That is the nth-term test run backwards, and it is invalid. ∑1/n and ∑1/(n ln n) both fail this way.
• "The integral test gives the sum." It gives the verdict only. ∑1/n² = 1.6449 while the matching integral is 1.
• "My terms are smaller than a divergent series, so mine converges." No conclusion follows. 1/(2n) is smaller than 1/n and still diverges.
• "The limit in limit comparison is the sum." It is a ratio of terms, not a total. For the first table row it is 2/5 and the actual sum is a different number entirely.
• "The integral test works on any series." It needs positive, continuous and decreasing. ∑(−1)n/n fails the first condition and needs the next lesson's test.

Recap

Start every series with the nth-term test, because it is one limit and it ends the problem whenever the terms miss zero, as n/(2n + 3) → 1/2 does. It can never show convergence. Next, look for a family you already know: geometric, telescoping or p-series, the last of which comes straight out of the integral test, since ∫(1 to ∞)x−pdx is finite exactly when p > 1. That is why ∑1/√n diverges and ∑1/n² converges to π²/6. For anything left, compare: directly when a clean inequality runs the right way, as 1/(n² + 1) < 1/n² does, and by limits otherwise, keeping only the leading powers to build the benchmark and checking that the ratio tends to something finite and positive. Worth holding on to: every one of these tests except the nth-term test requires non-negative terms, which is exactly the gap the next lesson fills.

Sources

1. OpenStax. (2016). 5.3 The divergence and integral tests. In Calculus Volume 2. openstax.org
2. OpenStax. (2016). 5.4 Comparison tests. In Calculus Volume 2. openstax.org
3. Dawkins, P. (n.d.). Integral test. Paul's Online Math Notes, Lamar University. Includes a proof of the test from the rectangle comparison. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Basel problem. The sum of 1/n squared equals pi squared over six. en.wikipedia.org
5. College Board. (n.d.). AP Calculus BC course framework. Topics 10.2 to 10.6 cover the nth-term test, the integral test, p-series and the comparison tests. apcentral.collegeboard.org

Two series, one sign apart

Add the reciprocals of the integers and the total grows past every bound. Add them again with the signs alternating,

1 − 1/2 + 1/3 − 1/4 + 1/5 − ...

and the total settles down to ln 2 = 0.693147. Same numbers, one change, opposite verdicts. Everything in this lesson follows from understanding why a sign can do that, and what it costs: the convergence you buy this way is fragile in a precise and surprising sense, which the last section makes concrete.

Why the alternation rescues it

Write the partial sums of the alternating harmonic series: 1, 0.5, 0.8333, 0.5833, 0.7833, 0.6167. They step up, then down, then up, each step shorter than the last. So the even-numbered partial sums increase, the odd-numbered ones decrease, and every even one stays below every odd one. Two monotone bounded sequences closing on each other have to meet, and the gap between them is exactly the size of the next term, which goes to zero. That is the whole argument, and it is why the alternating series test reads as it does.

For a series ∑ (−1)nbn or ∑ (−1)n+1bn, the series converges if all three hold:

1. bn > 0 for all n;
2. bn is decreasing, at least eventually;
3. bn → 0.

All three are load-bearing. Drop the third and ∑(−1)n(2n)/(3n + 1) has sizes tending to 2/3; the partial sums bounce forever and the series diverges, which the nth-term test also reports. Drop the second and the nested-interval picture collapses, because a step longer than the one before can carry the partial sum back outside the bracket.

The error bound, and how many terms an accuracy costs

The same picture gives a free error estimate that almost no other test provides. Because the true sum is always trapped between consecutive partial sums,

|S − Sn| ≤ bn+1

The error after n terms is no larger than the first term you left out. Also, the error has the sign of that first omitted term, so you know which side of the answer you are on.

Check it on the numbers. For the alternating harmonic series, S10 = 0.645635 and the true sum is 0.693147. The actual error is 0.047512. The bound promises at most b11 = 1/11 = 0.090909. The bound held, and as usual it was conservative.

Run it backwards, which is the exam question. How many terms guarantee an error below 0.01? Solve bn+1 = 1/(n + 1) ≤ 0.01, giving n + 1 ≥ 100, so n ≥ 99. Ninety-nine terms for two decimal places is dreadful, and it is a fair picture of how slowly conditionally convergent series converge.

Now the same question for ∑(n = 0 to ∞) (−1)n/n!, whose sum is 1/e = 0.367879. For an error below 0.001, solve 1/(n + 1)! ≤ 0.001. Since 6! = 720 and 7! = 5040, take n + 1 = 7, so n = 6 and you need the terms through n = 6. Their sum is 0.368056, off by 0.000176. Seven terms against ninety-nine, for a tighter tolerance: factorials in a denominator change everything.

So what?: the error bound is the only place in BC where you can state a numerical guarantee about an infinite sum in one line. Expect to be asked for the number of terms, not just the bound.

The ratio test, which asks how fast the terms shrink

Comparison tests ask what a series resembles. The ratio test asks something internal instead: by what factor does each term shrink relative to the one before?

L = lim(n → ∞) |an+1/an|

• L < 1: the series converges absolutely.
• L > 1 or L infinite: the series diverges.
• L = 1: no conclusion. Use another test.

The reasoning behind the first case is a comparison with a geometric series. If the ratio eventually stays below some r < 1, then from that point the terms are dominated by a geometric series with ratio r, which converges. The test is essentially geometric comparison with the comparison done automatically.

That origin also explains what it is good at. Use the ratio test when the terms contain factorials, or n in an exponent, or both, because those are the things that simplify when you divide consecutive terms.

Worked: ∑ n²/2n. The ratio is

[(n + 1)²/2n+1] · [2n/n²] = (1/2)((n + 1)/n)² → 1/2

Since 1/2 < 1, the series converges absolutely. Notice the powers of 2 collapsed to a single factor of one half, which is the whole reason this test is fast here.

Worked: ∑ n!/3n. The ratio is [(n + 1)!/3n+1] · [3n/n!] = (n + 1)/3, which grows without bound. So L is infinite and the series diverges. Factorials beat exponentials, and the ratio test is how you say so.

Why L = 1 really is useless. For ∑1/n, the ratio is n/(n + 1) → 1. For ∑1/n², the ratio is n²/(n + 1)² → 1. Same L, and the first diverges while the second converges. So L = 1 is not weak evidence or a hint. It is no evidence, and writing "L = 1, so the series diverges" is a guaranteed loss of the point. Every p-series gives L = 1, which is why the ratio test is never the tool for one.

Absolute, conditional, divergent: three boxes, not two

Given any series, look at two questions and you get three possible answers.

The fourth logical box, absolute values converging while the series does not, is empty: absolute convergence implies convergence. That theorem is what lets you throw signs away and use the positive-term tests of Lesson 11 on any series you like. If ∑|an| converges you are finished; only if it diverges do you need the alternating series test to decide the signed version.

The strategy that follows is worth writing down: test ∑|an| first. If it converges, say "absolutely convergent" and stop. If it diverges, go back to the signed series with the alternating test. If that works too, say "conditionally convergent"; the word conditional means the convergence depends on the arrangement of signs, not on the sizes alone.

How fragile conditional convergence is

That last phrase is literal. The Riemann series theorem says that the terms of any conditionally convergent series can be rearranged so that the new series converges to any real number you nominate, or diverges. The alternating harmonic series adds to 0.693147 in the order written, and a rearrangement of the same terms adds to 5, or to −100, or to nothing at all.

It is possible because the positive terms alone and the negative terms alone each diverge. Take as many positive terms as you need to pass your target, then negatives until you drop below it, then positives again. Both supplies are inexhaustible, and the steps shrink to zero, so the partial sums close in on whatever number you chose. An absolutely convergent series cannot be manipulated this way: its sum does not depend on the order at all. That is the real content of the distinction.

Common misconceptions

• "Alternating signs make a series converge." Only with decreasing sizes going to zero. ∑(−1)n(2n)/(3n + 1) alternates and diverges.
• "The error bound is the nth term." It is bn+1, the first term omitted. After 10 terms of the alternating harmonic series the bound is 1/11, not 1/10.
• "L = 1 means the series diverges." It means the test failed. ∑1/n² gives L = 1 and converges.
• "Conditionally convergent means it might converge." It converges, definitely. The condition refers to the order of the terms, and Riemann's theorem shows how much that order matters.
• "Use the ratio test on a p-series." Every p-series gives L = 1, so you learn nothing. Use the p-rule directly.

The short version

An alternating series converges when its sizes are positive, decreasing and tending to zero, and all three conditions have to be checked and stated. When it does converge, the truncation error is at most the first omitted term, which turns an accuracy requirement into an inequality you can solve: 99 terms for the alternating harmonic series to reach 0.01, but only 7 terms of ∑(−1)n/n! to reach 0.001. The ratio test looks at |an+1/an| and is the right tool wherever factorials or n-th powers appear, giving 1/2 for ∑n²/2n and an infinite limit for ∑n!/3n; at L = 1 it reports nothing, as it does for every p-series. Finally, classify: test the absolute values first, because absolute convergence implies convergence and ends the question. Bottom line: conditionally convergent series converge only in the order they are written, and Riemann proved that rearranging one can produce any sum at all.

Sources

1. OpenStax. (2016). 5.5 Alternating series. In Calculus Volume 2. The test, the remainder estimate and absolute versus conditional convergence. openstax.org
2. OpenStax. (2016). 5.6 Ratio and root tests. In Calculus Volume 2. openstax.org
3. Dawkins, P. (n.d.). Alternating series test. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Riemann series theorem. Rearrangement of a conditionally convergent series can produce any prescribed sum. en.wikipedia.org
5. College Board. (n.d.). AP Calculus BC course framework. Topics 10.7 to 10.9 cover the alternating series test, the ratio test and the alternating series error bound. apcentral.collegeboard.org

A series with an x in it

Consider

∑(n = 1 to ∞) (x − 2)n/(n · 3n)

For which values of x does this add up to a number? Put x = 2 and every term is 0, so it converges there. Put x = 100 and the terms are enormous and growing, so it does not. Somewhere between those two the behaviour changes, and the answer turns out to be: it converges for every x with −1 ≤ x < 5, and nowhere else. Note the brackets. One endpoint is in and the other is out, and no single test decides both.

A power series is any series of the form

∑ cn(x − a)n

with a the centre and cn the coefficients. It is a function of x, whose domain is exactly the set of x where it converges. Finding that domain is what this lesson does; the next lesson is about which functions can be written this way.

Only three things can happen

This is a theorem, not a pattern, and it makes the whole problem finite. For any power series centred at a, exactly one of the following holds:

1. It converges only at x = a. The radius of convergence is R = 0.
2. It converges for every real x. Then R = ∞.
3. There is a positive number R such that it converges for |x − a| < R and diverges for |x − a| > R.

Every power series converges at its own centre, since every term but the constant one vanishes there. So an interval of convergence is never empty, and a radius of 0 means the domain is the single point a.

Key idea: convergence of a power series is symmetric about the centre, which is why the answer is always an interval and why one number, R, fixes almost all of it. The only thing R does not settle is what happens at the two endpoints.

Finding R with the ratio test

The ratio test is built for this, because consecutive terms of a power series differ by one factor of (x − a) and one step of the coefficient. Work the opening example. With an = (x − 2)n/(n 3n),

|an+1/an| = |x − 2|n+1/((n + 1)3n+1) · (n 3n)/|x − 2|n = (|x − 2|/3) · (n/(n + 1))

As n → ∞ the fraction n/(n + 1) → 1, so L = |x − 2|/3. Notice that L depends on x, which is the point: the test now reads as a condition on x. Convergence needs L < 1:

|x − 2| < 3, that is −1 < x < 5

So R = 3, centred at 2. For |x − 2| > 3 the ratio test gives L > 1 and the series diverges outright. What remains is the two points where |x − 2| = 3, namely x = −1 and x = 5, and there L = 1, where the ratio test says nothing at all. That is not an oversight in your work. It is structural: the ratio test is always inconclusive at the endpoints of a power series, every time, for every series. Which is why they must be checked by hand.

Checking the endpoints, one at a time

Substitute each endpoint into the original series and you get an ordinary numerical series. Then use Lesson 11 and Lesson 12.

At x = 5: (x − 2)n = 3n, so the term becomes 3n/(n 3n) = 1/n. The series is the harmonic series. It diverges. So 5 is not in the interval.

At x = −1: (x − 2)n = (−3)n, so 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 in the interval.

The interval of convergence is [−1, 5): closed on the left, open on the right. Writing the interval with the wrong bracket is a lost point even when R is right, and the two endpoints genuinely can differ, as they do here.

A second example where they agree: ∑ (x + 1)n/(n² 2n). The ratio gives L = |x + 1|/2, so R = 2 and the centre is −1, making the open interval (−3, 1). At x = 1 the series becomes ∑1/n², convergent. At x = −3 it becomes ∑(−1)n/n², which converges absolutely. Both endpoints are in, so the interval is [−3, 1]. The n² in the denominator is what closes the interval; the single n in the first example is what left it half open.

The two extremes

R = 0. Take ∑ n! xn. The ratio is (n + 1)!|x|n+1/(n!|x|n) = (n + 1)|x|, which grows without bound for every x ≠ 0. So the series converges only at x = 0. Factorials in the numerator destroy a power series.

R infinite. Take ∑ xn/n!. The ratio is |x|n+1/(n + 1)! · n!/|x|n = |x|/(n + 1) → 0 for every x. So L = 0 < 1 always, and the series converges for all real x. There are no endpoints to check, and the interval is (−∞, ∞). That series is ex, as the next lesson shows, and its infinite radius is why the exponential series can be used at any input.

The geometric series is the other one to keep in mind: ∑(n = 0 to ∞) xn = 1/(1 − x) for |x| < 1, with R = 1. Both endpoints fail, since the terms become 1 and (−1)n, neither tending to zero, so the interval is the open (−1, 1). This is the only power series so far whose sum you can name in closed form, and the next lesson builds a dozen others out of it.

What differentiating and integrating do to the interval

Inside its interval of convergence a power series can be differentiated and integrated term by term, and the radius R does not change. The endpoints can change, and in a predictable direction: differentiating tends to lose endpoints and integrating tends to gain them, because differentiating pulls a factor of n into the numerator and integrating pushes one into the denominator.

Watch it on the geometric series. Start with ∑xn on (−1, 1). Integrating term by term gives ∑xn+1/(n + 1), still with R = 1, but now x = −1 produces an alternating series with sizes 1/(n + 1), which converges. The interval became [−1, 1). The radius held; the endpoint switched. So check the endpoints again after any term-by-term operation rather than copying the old interval across.

Common misconceptions

• "The ratio test settles the endpoints too." It cannot: at |x − a| = R the limit is exactly 1, which is the case the ratio test has nothing to say about.
• "The interval is always open." It can be open, closed, or half open, and the two ends are independent. The examples above give [−1, 5) and [−3, 1].
• "R is the interval." R is a distance from the centre. The interval is (a − R, a + R) plus whichever endpoints survive testing.
• "A power series has to converge somewhere interesting." ∑n!xn converges at one point only. That is a legitimate answer, with R = 0.
• "Term-by-term integration keeps the same interval." It keeps the same radius. Endpoints have to be retested, and integrating ∑xn gains the left one.

What to carry forward

A power series ∑cn(x − a)n is a function whose domain is its set of convergence, and that set is always an interval centred at a. Find its radius with the ratio test, which turns into an inequality in x: for ∑(x − 2)n/(n3n) the condition |x − 2|/3 < 1 gives R = 3. Then substitute each endpoint separately and decide it with the tests of the previous two lessons, because the ratio test is guaranteed to fail there: x = 5 gives the harmonic series and is excluded, x = −1 gives the alternating harmonic series and is included, so the interval is [−1, 5). Two special cases are worth knowing on sight: a factorial in the numerator gives R = 0 and a factorial in the denominator gives R = ∞. The core of it: R comes from one limit, the brackets come from two separate decisions, and an answer missing either part is incomplete.

Sources

1. OpenStax. (2016). 6.1 Power series and functions. In Calculus Volume 2. openstax.org
2. OpenStax. (2016). 6.2 Properties of power series. In Calculus Volume 2. Term-by-term differentiation and integration on the interval of convergence. openstax.org
3. Dawkins, P. (n.d.). Power series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Radius of convergence. en.wikipedia.org
5. College Board. (n.d.). AP Calculus BC course framework. Topics 10.13 and 10.14 cover the radius and interval of convergence of a power series. apcentral.collegeboard.org

Eight numbers that nearly make e

Add these: 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720 + 1/5040. The total is 2.7182540. The number e is 2.7182818. Eight terms, and the first four decimal places are already right.

Those denominators are the factorials, and the sum is the first eight terms of a series that equals ex at x = 1. This lesson derives that series and five others, and then shows how a handful of algebraic moves turn those six into every series the exam will ask for. Nobody memorises a hundred expansions. They memorise four and manipulate.

Where the coefficients have to come from

Suppose a function equals a power series on some interval around a:

f(x) = c0 + c1(x − a) + c2(x − a)² + c3(x − a)³ + ...

Then the coefficients are forced, and finding them takes one trick used repeatedly: substitute x = a, which kills every term with a factor of (x − a) in it.

Straight away, f(a) = c0. Now differentiate the whole series term by term, which is legal inside the interval of convergence:

f'(x) = c1 + 2c2(x − a) + 3c3(x − a)² + ...

and set x = a again: f'(a) = c1. Differentiate once more:

f''(x) = 2c2 + 6c3(x − a) + 12c4(x − a)² + ...

so f''(a) = 2c2 and c2 = f''(a)/2. The third derivative leaves 6c3, so c3 = f'''(a)/6. The pattern is that differentiating n times turns cn(x − a)n into n! cn, so

cn = f(n)(a)/n!

and the Taylor series of f about a is

∑(n = 0 to ∞) f(n)(a)(x − a)n/n!

When a = 0 it is called a Maclaurin series. The factorial in the denominator is not decoration: it is exactly what cancels the factorial that repeated differentiation produces.

In short: a Taylor polynomial is the unique polynomial whose value and first n derivatives match the function at a single point. Everything else about Taylor series follows from that one property.

Building the three that matter, one derivative at a time

The exponential. Every derivative of ex is ex, so every derivative at 0 is 1. Then cn = 1/n! and

ex = 1 + x + x²/2! + x³/3! + ... = ∑ xn/n!

Lesson 13 showed this has infinite radius, so it holds for every x. At x = 1 it gives the eight numbers from the opening.

The sine. The derivatives cycle: sin x, cos x, −sin x, −cos x, then repeat. At x = 0 that is 0, 1, 0, −1, 0, 1, .... Half the coefficients vanish, and the survivors alternate:

sin x = x − x³/3! + x5/5! − x7/7! + ... = ∑(−1)nx2n+1/(2n + 1)!

Only odd powers appear, which is what you would expect from an odd function. Test it: sin(0.5) = 0.4794255, and 0.5 − 0.125/6 = 0.4791667, then + 0.03125/120 = 0.4794271. Two terms give three decimals; three terms give six.

The cosine. Same cycle starting one step earlier, giving 1, 0, −1, 0, ... at zero:

cos x = 1 − x²/2! + x⁴/4! − ... = ∑(−1)nx2n/(2n)!

Only even powers, as an even function requires. And differentiating the sine series term by term gives 1 − 3x²/3! + 5x⁴/5! − ... = 1 − x²/2! + x⁴/4! − ..., which is the cosine series. The two are consistent, which is the sort of check worth running when you doubt a sign.

The six to keep, and where each is valid

The first four are the ones to know cold. The last two are worth knowing because they are so easy to rebuild: both come from integrating a geometric series, which the next section does.

Four moves that generate everything else

1. Substitute. Anything can go in for x, as long as you carry the interval along with it. Replace x by −x² in the exponential series:

e−x² = 1 − x² + x⁴/2! − x6/3! + ... = ∑(−1)nx2n/n!

That function has no elementary antiderivative, and this series is how its integral gets computed, as Lesson 15 will do. Replace x by 3x in the cosine series and every term picks up a power of 3: cos 3x = 1 − 9x²/2! + 81x⁴/4! − ...

2. Multiply. x sin x = x² − x⁴/3! + x6/5! − ..., term by term. Multiplying a series by a power of x shifts every exponent and changes nothing else.

3. Differentiate. Term by term, inside the radius. Differentiating 1/(1 − x) = ∑xn gives

1/(1 − x)² = 1 + 2x + 3x² + 4x³ + ... = ∑ n xn−1

which you would not want to build from derivatives of the original function.

4. Integrate. This is the move that produces the logarithm and the arctangent. Start from the geometric series with −x in place of x:

1/(1 + x) = 1 − x + x² − x³ + ...

Integrate both sides from 0 to x. The left side gives ln(1 + x), and the right side integrates term by term:

ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + ...

The constant of integration is 0 because both sides vanish at x = 0. At x = 1 this gives the alternating harmonic series summing to ln 2, which is the promise Lesson 12 made and this line keeps. It is called the Mercator series.

Now substitute x² for x in the same geometric series and integrate:

1/(1 + x²) = 1 − x² + x⁴ − ... gives arctan x = x − x³/3 + x5/5 − ...

Put x = 1 and, since arctan 1 = π/4, you get the Leibniz formula: π/4 = 1 − 1/3 + 1/5 − 1/7 + .... It is a beautiful identity and a terrible way to compute π; by the alternating error bound, one more decimal place costs ten times as many terms, so five correct decimals need about 100000 terms.

Centring somewhere other than zero

When a question asks for a series about a = 2, the coefficients use derivatives at 2 and the powers are (x − 2)n. Build the Taylor series of f(x) = ln x about a = 1, where the derivatives are manageable:

• f(1) = ln 1 = 0
• f'(x) = 1/x, so f'(1) = 1
• f''(x) = −1/x², so f''(1) = −1 and c2 = −1/2
• f'''(x) = 2/x³, so f'''(1) = 2 and c3 = 2/6 = 1/3

So ln x = (x − 1) − (x − 1)²/2 + (x − 1)³/3 − ..., which is the Mercator series with x − 1 in place of x, exactly as it should be. Two routes, one answer, and the shortcut through substitution was much faster than four derivatives.

Common misconceptions

• "Every function equals its Taylor series." Not automatically. The series can converge to something else; only a bound on the remainder, which Lesson 15 supplies, proves the two agree.
• "Substituting keeps the same interval." Substituting x² into a series valid for |x| < 1 gives one valid for |x²| < 1, which is still |x| < 1, but substituting 3x gives |x| < 1/3. Translate the condition too.
• "The sine series has even powers in it." It has only odd ones, and the cosine only even. If your answer has both, a sign or a power is wrong.
• "You must differentiate to get any series." Substitution and integration are usually faster. Deriving e−x² from repeated differentiation is punishing; substituting takes one line.
• "A Taylor polynomial at a is a good approximation everywhere." It is built to match at a single point, and it degrades as you move away, quickly once you pass the radius of convergence.

Pulling it together

Matching a function to a power series forces the coefficients to be f(n)(a)/n!, because differentiating n times and setting x = a leaves n! cn and nothing else. Running that on the three cyclic functions gives ex = ∑xn/n!, sin x = ∑(−1)nx2n+1/(2n + 1)! and cos x = ∑(−1)nx2n/(2n)!, all valid for every x, alongside the geometric series on |x| < 1. Everything else comes from those four by substituting, multiplying by a power of x, differentiating, or integrating: e−x² in one substitution, ln(1 + x) and arctan x by integrating a geometric series, and π/4 = 1 − 1/3 + 1/5 − ... as a by-product. What matters here: four series memorised plus four moves beats fifty series memorised, and the moves are the part the exam actually tests.

Sources

1. OpenStax. (2016). 6.3 Taylor and Maclaurin series. In Calculus Volume 2. Derivation of the coefficient formula and the standard expansions. openstax.org
2. OpenStax. (2016). 6.4 Working with Taylor series. In Calculus Volume 2. Substitution, differentiation and integration of known series. openstax.org
3. Dawkins, P. (n.d.). Taylor series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Leibniz formula for pi. The arctangent series evaluated at 1, and its very slow convergence. en.wikipedia.org
5. College Board. (n.d.). AP Calculus BC course framework. Topics 10.11 and 10.15, Taylor polynomials and representing functions with known series. apcentral.collegeboard.org

An approximation is worthless until you can bound it

The fourth-degree Maclaurin polynomial for cosine, evaluated at 0.3, gives

1 − (0.3)²/2 + (0.3)⁴/24 = 1 − 0.045 + 0.0003375 = 0.9553375

A calculator says cos(0.3) = 0.9553365. The approximation is off by about 0.0000010. But you only know that because you had the true value to compare against, and the whole point of the polynomial is to be used when you do not. So the real question is: without knowing the answer, what can you promise about the error? This lesson's first half answers that in one inequality, and by the end of this page you will see that inequality predict 0.0000010 almost exactly.

Taylor's theorem, and what the remainder actually is

Write the function as its degree-n Taylor polynomial plus whatever is left over:

f(x) = Pn(x) + Rn(x)

Taylor's theorem says the leftover is not mysterious. There exists a number c strictly between a and x with

Rn(x) = f(n+1)(c)(x − a)n+1/(n + 1)!

Look at the shape of that: it is exactly the next term of the series, except that the derivative is evaluated at some unknown c rather than at a. This is the mean value theorem grown up; the same "there exists a point where it works out exactly" structure, and the same uselessness for finding c. You never find c. What you do instead is bound the derivative over the whole interval, which converts the equality into a usable inequality.

Let M be any number with |f(n+1)(t)| ≤ M for every t between a and x. Then

|Rn(x)| ≤ M|x − a|n+1/(n + 1)!

That is the Lagrange error bound, and it is the only tool in BC that turns an approximation into a guarantee.

The point: the bound needs three things and nothing else: the order n, the distance |x − a|, and a ceiling M on the next derivative. Getting M is the only step requiring any thought.

The cosine, bounded

Return to cos(0.3) with a = 0. The derivatives of cosine are −sin, −cos, sin, cos, repeating, so every one of them is bounded by 1 everywhere. Take M = 1, which needs no computation at all.

With n = 4 the bound is (1)(0.3)5/5! = 0.00243/120 = 0.0000203. True, but loose by a factor of twenty.

Here is the improvement worth knowing. The fifth-degree term of the cosine series is zero, because cosine has no odd powers. So P4 and P5 are the same polynomial, and you are free to call it degree 5 and use n = 5 in the bound:

|R5(0.3)| ≤ (1)(0.3)6/6! = 0.000729/720 = 0.00000101

Compare that with the actual error, 0.00000101. The bound is essentially exact here, because the first omitted term is the whole story when the following terms are far smaller. Free accuracy, obtained by noticing a missing term.

The exponential, where M costs a sentence

Approximate e0.5 with P3:

1 + 0.5 + 0.125 + 0.0208333 = 1.6458333

Now the bound. Every derivative of ex is ex, which is increasing, so its largest value on [0, 0.5] is e0.5. That is the number being approximated, so using it would be circular. Use a crude, honest over-estimate instead: e0.5 < e < 3, and since e0.5 is certainly under 2, take M = 2. Then

|R3(0.5)| ≤ 2(0.5)⁴/4! = 2(0.0625)/24 = 0.0052083

The true error is 1.6487213 − 1.6458333 = 0.0028880, comfortably inside. A loose M gives a loose but valid bound, and on the exam a justified crude bound earns the point while an unjustified tight one does not.

How many terms for a stated accuracy? This is the standard version of the question. For an error below 0.0001 at x = 0.5 with M = 2, solve

2(0.5)n+1/(n + 1)! ≤ 0.0001

Test values rather than trying to invert a factorial. At n = 4: 2(0.03125)/120 = 0.00052, too big. At n = 5: 2(0.015625)/720 = 0.0000434, small enough. So degree 5 does it. Checking against reality, P5 = 1.6486979, an error of 0.0000234, which indeed beats 0.0001.

Two bounds, and which to reach for

Read the question before choosing. If the series at that particular x alternates, the first omitted term is faster and usually better. For e0.5 the terms are all positive, so only Lagrange applies. For cos(0.3) either works, and they agree closely, which is the clearest evidence that the Lagrange bound really is the first omitted term with the derivative moved to an unknown point.

Integrals that have no antiderivative

Now the second use of series, which is the reason they earn their place in an applied course. The function e−x² has no elementary antiderivative. None exists; this is a theorem, not a failure of cleverness. Yet the integral is needed constantly, since it is the shape of the normal distribution.

Series make it routine. From Lesson 14, substituting −x² into the exponential series gives

e−x² = 1 − x² + x⁴/2! − x6/3! + x8/4! − ...

Integrate term by term from 0 to 0.5, which is legal because 0.5 is inside the interval of convergence:

∫(0 to 0.5) e−x² dx = [x − x³/3 + x5/10 − x7/42 + x9/216 − ...](0 to 0.5)

The denominators are (2n + 1)n!: the n! from the exponential series and the 2n + 1 from integrating x2n. Evaluate:

The true value is 0.4612810. Five terms give six correct decimal places, and because the series alternates you get the error bound free: after four terms the next term is (0.5)9/216 = 0.0000090, and the actual error was 0.0000087.

A second one, using the sine series. Divide sin x = x − x³/3! + x5/5! − ... by x term by term:

(sin x)/x = 1 − x²/6 + x⁴/120 − x6/5040 + ...

which also repairs the hole at x = 0, since the series is plainly 1 there. Integrate from 0 to 1:

1 − 1/18 + 1/600 − 1/35280 = 1 − 0.0555556 + 0.0016667 − 0.0000283 = 0.9460828

Four terms, and the value is right to seven decimal places.

Common misconceptions

• "You have to find c." You never do. c exists, is unknown, and is replaced by a bound M on the derivative over the whole interval.
• "M is the maximum of f." M bounds the (n + 1)st derivative, not the function. For cos both happen to be 1, which hides the distinction.
• "The Lagrange bound gives the error." It gives a ceiling. For e0.5 with P3 the ceiling was 0.0052 and the error was 0.0029.
• "A bigger n always needs recomputing everything." When a term of the series is zero, as with the odd terms of cosine, you can raise n for free and sharpen the bound without changing the polynomial.
• "Term-by-term integration needs the interval to be short." It needs the limits to lie inside the interval of convergence. For e−x² that is every real number.

What to remember

Taylor's theorem says the leftover after a degree-n polynomial is the next term of the series with its derivative evaluated at some unfindable point c, and bounding that derivative by M turns it into |Rn(x)| ≤ M|x − a|n+1/(n + 1)!. Choose M as the largest size the next derivative can have between a and x, and justify the choice in words: M = 1 for every derivative of sine or cosine, M = 2 for ex on [0, 0.5]. That gave 0.0000010 for cos(0.3) against an actual error of the same size, and a valid 0.0052 for e0.5 against an actual 0.0029. Turn the bound into an inequality and test values of n when the question asks how many terms an accuracy needs. And when a function has no elementary antiderivative, integrate its series instead: five terms put ∫(0 to 0.5)e−x²dx at 0.4612814 against a true 0.4612810. Remember: an approximation without a bound is an assertion, and the bound is what the exam is actually paying for.

Sources

1. OpenStax. (2016). 6.3 Taylor and Maclaurin series. In Calculus Volume 2. Taylor's theorem with remainder and the bound used here. openstax.org
2. OpenStax. (2016). 6.4 Working with Taylor series. In Calculus Volume 2. Term-by-term integration of a series to evaluate a definite integral. openstax.org
3. Dawkins, P. (n.d.). Estimating the value of a series. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. Wikipedia. (n.d.). Taylor's theorem. Statement of the Lagrange form of the remainder. en.wikipedia.org
5. College Board. (n.d.). AP Calculus BC course framework. Topic 10.12, the Lagrange error bound, and Topic 10.15, representing functions with Taylor series. apcentral.collegeboard.org

A solution that is wrong in one line

Here is a complete answer to a real BC question, written the way a student under time pressure writes it.

Question: determine whether ∑(n = 1 to ∞) 1/(n² + 3n) converges.

Answer: by the ratio test, |an+1/an| = (n² + 3n)/((n + 1)² + 3(n + 1)) → 1. Since L = 1, the series diverges.

The limit is correct. The arithmetic is correct. The conclusion is worth zero, and the series in fact converges. Everything in this lesson has that shape: the work is mostly right and one specific line destroys it. Finding that line in someone else's solution is the fastest way to stop writing it in your own.

Error 1: treating L = 1 as a verdict

Where it breaks. At the word "since". The ratio test has three cases, and L = 1 is the one where the test reports nothing. It is not weak evidence for divergence. It is the absence of evidence.

The proof that it means nothing takes two examples. ∑1/n gives L = 1 and diverges. ∑1/n² gives L = 1 and converges. Any p-series gives L = 1, because (n/(n + 1))p → 1 for every p. So whenever the terms are a rational function of n, the ratio test is guaranteed to be useless before you start.

The repair. Use limit comparison. With an = 1/(n² + 3n), keep the leading powers: bn = 1/n². Then

an/bn = n²/(n² + 3n) = 1/(1 + 3/n) → 1

a finite positive limit, and ∑1/n² converges as a p-series with p = 2. So the original converges. Three lines, and it was never a ratio-test question.

The habit. Before using the ratio test, look for a factorial or an n-th power. If there is neither, do not start.

Error 2: stopping at the radius

Question: find the interval of convergence of ∑(n = 1 to ∞) (x − 1)n/(n · 2n).

Answer: the ratio gives |x − 1|/2 < 1, so R = 2 and the interval is (−1, 3).

Where it breaks. At the word "so". The ratio test is silent at |x − 1| = 2, because there L = 1 exactly. Two points remain undecided and the answer treats them as decided.

The repair. Substitute each one.

• x = 3: the term is 2n/(n 2n) = 1/n. Harmonic, divergent. Excluded.
• x = −1: the term is (−2)n/(n 2n) = (−1)n/n. Alternating harmonic, convergent. Included.

The interval is [−1, 3). Note the two endpoints behaved differently, which is normal, so testing one and assuming the other matches is its own error.

The habit. Write the two substitutions down as a separate numbered step, before you write any brackets.

Error 3: losing a constant inside integration by parts

Question: find ∫ x e2x dx.

Answer: take u = x, dv = e2xdx, so du = dx and v = e2x. Then ∫ x e2xdx = xe2x − ∫e2xdx = xe2x − (1/2)e2x + C.

Where it breaks. At v = e2x. The antiderivative of e2x is (1/2)e2x, and the missing half propagates into both remaining terms.

The check that catches it in ten seconds. Differentiate the answer:

d/dx[xe2x − (1/2)e2x] = e2x + 2xe2x − e2x = 2xe2x

which is twice the integrand. Every antiderivative on this exam can be checked this way, and a check that costs ten seconds and catches a factor of two is the best trade available to you.

The repair. With v = (1/2)e2x:

∫ x e2xdx = (x/2)e2x − (1/2)∫e2xdx = (x/2)e2x − (1/4)e2x + C

Differentiate to confirm: (1/2)e2x + xe2x − (1/2)e2x = xe2x. Correct.

The same family of errors includes dropping the constant altogether in an accumulation problem. If a particle has x(1) = 2 and you compute ∫(1 to 3)x'(t)dt = 2.159, the position at t = 3 is 4.159, not 2.159. The integral is the change, and the change is not the answer.

Error 4: terms going to zero, read backwards

Question: does ∑(n = 1 to ∞) 1/(3n + 2) converge?

Answer: 1/(3n + 2) → 0, so by the nth-term test the series converges.

Where it breaks. At "so by the nth-term test". That test has exactly one conclusion, divergence, from exactly one hypothesis, terms not going to zero. Running it backwards is invalid, and the harmonic series is the permanent counterexample.

The repair. Limit comparison with 1/n: the ratio is n/(3n + 2) → 1/3, finite and positive, and the harmonic series diverges, so this series diverges too.

The habit. Say the test out loud in its correct direction every time: "if the terms do not go to zero, the series diverges." A test stated correctly cannot be used backwards.

Error 5: mistaking the polynomial for the function

Question: use the third-degree Maclaurin polynomial for sin x to estimate sin 3.

Answer: sin 3 ≈ 3 − 27/6 = −1.5.

Where it breaks. Not in the arithmetic, which is right, but in the expectation. sin 3 = 0.1411, and the estimate is not merely inaccurate, it is outside the range of the sine function entirely. A number that cannot possibly be a sine is a signal you can read without a calculator.

What the error bound was telling you. With M = 1 and the fourth-degree term of the sine series equal to zero, the Lagrange bound at n = 4 is

|R4(3)| ≤ (1)(3)5/5! = 243/120 = 2.025

The bound is 2.025 on an answer whose true size is 0.14. The bound was not violated; the actual error of 1.641 sits inside it. The bound was screaming that the approximation carried no information. Compute the bound before you trust the estimate, and if the bound is larger than the quantity, add terms or accept that you cannot answer this way.

The upshot: a Taylor polynomial is built to match a function at one point. Its accuracy decays with distance from that point, fast, and the error bound is the instrument that tells you how fast.

The one-minute checklist

Five checks, none of which needs more than a minute, and between them they catch every error above plus most of the others.

The last row is the cheapest of all and is skipped most often. An arc length below the width of its interval, a probability above 1, a speed below zero and a logistic population past its carrying capacity are all impossible before any calculation is examined, and noticing one sends you back to the right line immediately.

Common misconceptions

• "The answer is what earns the points." On free response, most points sit in the setup and the justification. A correct number with no shown integral typically scores 1 of 4 or fewer.
• "Calculator questions need no setup." They need the integral or equation written out before the numerical value. A bare number cannot be awarded, because nothing shows what was computed.
• "Three decimals is a style preference." It is a stated requirement on calculator-active questions. Truncating rather than rounding, or giving two places, loses the answer point.
• "Naming a test is padding." The name is the justification. "Converges by limit comparison with the p-series 1/n squared" is worth a point that "converges" is not.
• "A units answer is obvious from context." Rate questions ask for units and award them separately. Fish per year, metres per second, dollars per day: write them.

Where this leaves us

Five lines destroy most of the BC points that get destroyed. L = 1 in the ratio test is not a verdict, and a series of rational terms should be sent to limit comparison instead. A radius is not an interval, and both endpoints have to be substituted and decided separately, as [−1, 3) shows. An antiderivative found by parts should be differentiated before you move on, which catches the missing half in v = (1/2)e2x in seconds. The nth-term test runs one way only. And a Taylor polynomial far from its centre can return a value the function cannot even take, which the Lagrange bound of 2.025 predicted before the calculator did. Bottom line: every one of these has a check that costs under a minute, and the checks are worth more marks per minute than any new technique you could learn instead.

Sources

1. OpenStax. (2016). 5.6 Ratio and root tests. In Calculus Volume 2. The three cases of the ratio test, including the inconclusive one. openstax.org
2. OpenStax. (2016). 6.1 Power series and functions. In Calculus Volume 2. Endpoint testing as a separate step from the radius. openstax.org
3. Dawkins, P. (n.d.). Ratio test. Paul's Online Math Notes, Lamar University. tutorial.math.lamar.edu
4. College Board. (n.d.). AP Calculus BC exam information and free-response scoring. Scoring guidelines allocate points to setup and justification as well as to the final value. apcentral.collegeboard.org

What a complete answer looks like

A BC free-response question is worth nine points, and they are itemised. A typical part awards one point for the correct integral or equation, one for the numerical value, and one for a justification stated in words. You can compute the right number and score one of three. Everything below is written the way it should appear on the page: setup first, value to three decimals, reason in a sentence.

Two of the four questions here are calculator-active, meaning you are expected to write the integral and then evaluate it numerically. Two are not, and every number in them comes out exactly.

Question 1: vector motion, calculator active

A particle moves in the xy-plane. For t ≥ 0 its velocity is v(t) = (ln(t² + 1), 3 sin t). At t = 0 the particle is at (2, −1).

(a) Find the speed and the acceleration vector at t = 2. (b) Find the position at t = 3. (c) Find the total distance travelled on 0 ≤ t ≤ 3. (d) Is the speed increasing or decreasing at t = 2? Justify.

(a) Speed is the length of the velocity:

|v(2)| = √([ln 5]² + [3 sin 2]²) = √(1.609438² + 2.727892²) = 3.167

Acceleration differentiates each component: x''(t) = 2t/(t² + 1) and y''(t) = 3 cos t, so

a(2) = (4/5, 3 cos 2) = (0.800, −1.248)

(b) Each coordinate is its starting value plus the integral of its own rate. Write both integrals before evaluating anything:

x(3) = 2 + ∫(0 to 3) ln(t² + 1) dt = 2 + 3.4058 = 5.406

y(3) = −1 + ∫(0 to 3) 3 sin t dt = −1 + 5.970 = 4.970

The position is (5.406, 4.970). The second integral is elementary, [−3 cos t](0 to 3) = −3cos 3 + 3, and writing it exactly costs nothing and protects against a calculator slip.

(c) Distance is the integral of speed, a single integral with both components under one root:

∫(0 to 3) √([ln(t² + 1)]² + [3 sin t]²) dt = 7.212

(d) Speed increases exactly when v · a > 0:

v(2) · a(2) = (1.609438)(0.800) + (2.727892)(−1.248441) = −2.118

Since this is negative, the speed is decreasing at t = 2. Say that in words as well as in symbols: the acceleration has a component opposing the motion, so the particle is slowing down.

Question 2: polar area, calculator permitted but unnecessary

The curves r = 2 + 2 cos θ and r = 3 are graphed for 0 ≤ θ ≤ 2π. (a) Find the values of θ where they intersect. (b) Find the area of the region inside the first curve and outside the second. (c) Find dy/dx on the first curve at θ = π/2.

(a) Set them equal: 2 + 2cos θ = 3, so cos θ = 1/2 and θ = π/3 or θ = 5π/3, which is −π/3. Check the pole separately: the limacon reaches r = 0 at θ = π and the circle never does, so there is no extra intersection there.

(b) Between those angles the limacon is outside, since at θ = 0 it gives 4 against the circle's 3. So

A = (1/2)∫(−π/3 to π/3) [(2 + 2cos θ)² − 3²] dθ

Expand before integrating. (2 + 2cos θ)² = 4 + 8cos θ + 4cos²θ, and 4cos²θ = 2 + 2cos 2θ, so the bracket is 6 + 8cos θ + 2cos 2θ − 9 = −3 + 8cos θ + 2cos 2θ. An antiderivative is −3θ + 8 sin θ + sin 2θ. At π/3 that is −π + 8(√3/2) + √3/2 = 4.652635, and the lower limit gives the negative of it, since every term is odd. So

A = (1/2)(2 · 4.652635) = 4.653

(c) With r = 2 + 2cos θ and r' = −2 sin θ, at θ = π/2 we have r = 2 and r' = −2. Then

dy/dθ = r' sin θ + r cos θ = (−2)(1) + (2)(0) = −2

dx/dθ = r' cos θ − r sin θ = (−2)(0) − (2)(1) = −2

so dy/dx = 1. The point is (0, 2) in rectangular coordinates, and the tangent there is y = x + 2.

Question 3: a series, no calculator

Let f(x) = ln(1 + 2x). (a) Write the first four nonzero terms and the general term of the Maclaurin series for f. (b) Find the interval of convergence, showing the endpoint work. (c) Use the first three nonzero terms to approximate f(0.1) and bound the error. (d) Find f(4)(0).

(a) Do not differentiate four times. Start from the known series ln(1 + u) = u − u²/2 + u³/3 − u⁴/4 + ... and substitute u = 2x:

ln(1 + 2x) = 2x − 2x² + (8/3)x³ − 4x⁴ + ...

with general term (−1)n+12nxn/n for n ≥ 1. Check the second term: −(2x)²/2 = −2x². Correct.

(b) Ratio test on the general term:

|an+1/an| = (2n+1|x|n+1/(n + 1)) · (n/(2n|x|n)) = 2|x| · n/(n + 1) → 2|x|

Convergence needs 2|x| < 1, so R = 1/2 and the open interval is (−1/2, 1/2). Now the endpoints, each substituted into the series itself:

• x = 1/2: the terms become (−1)n+1/n, the alternating harmonic series, which converges. Included.
• x = −1/2: the terms become (−1)n+1(−1)n/n = −1/n, the negative of the harmonic series, which diverges. Excluded.

The interval of convergence is (−1/2, 1/2].

(c) f(0.1) ≈ 2(0.1) − 2(0.1)² + (8/3)(0.1)³ = 0.2 − 0.02 + 0.0026667 = 0.1826667. The series at x = 0.1 alternates with decreasing terms, so the error is at most the first omitted term:

|error| ≤ 4(0.1)⁴ = 0.0004

The true value is ln 1.2 = 0.1823216, and the actual error is 0.000345, inside the bound as promised.

(d) The coefficient of x⁴ is f(4)(0)/4!. From the series that coefficient is −4, so f(4)(0) = −4 · 24 = −96. This is the fastest way to get a high derivative at a point, and it is worth checking once directly: f'''(x) = 16(1 + 2x)−3, so f(4)(x) = −96(1 + 2x)−4 and f(4)(0) = −96.

Question 4: a differential equation, no calculator

A population satisfies dP/dt = 0.3P(1 − P/500) with P(0) = 100, P in thousands of fish and t in years. (a) State the carrying capacity, the population at fastest growth and that maximum rate, with units. (b) Use Euler's method with two steps of h = 1 to approximate P(2). (c) Is that an overestimate or an underestimate? Justify. (d) Solve the differential equation.

(a) K = 500 thousand fish. Growth is fastest at K/2 = 250 thousand fish, at a rate of kK/4 = 0.3(500)/4 = 37.5 thousand fish per year. Units are a scored item; write them.

(b) Two rows of a table, with the slope kept separate from the change:

So P(2) ≈ 151.974 thousand fish.

(c) Differentiate the equation: d²P/dt² = kP'(1 − 2P/K). On this interval P stays below 250, so 1 − 2P/500 > 0, and P' > 0, so the second derivative is positive and the solution is concave up. Tangent lines to a concave-up curve lie below it, and Euler's method steps along tangent lines, so 151.974 is an underestimate.

(d) Separate and split with partial fractions, as in Lesson 6. The solution is

P(t) = K/(1 + Ae−kt) with A = (500 − 100)/100 = 4, so P(t) = 500/(1 + 4e−0.3t)

Check part (c) against it: P(2) = 500/(1 + 4e−0.6) = 500/3.1952 = 156.482. The Euler estimate of 151.974 was indeed low, by 4.5 thousand fish, which is the price of two steps of width 1.

Common misconceptions

• "Polar area integrals run over x." They run over θ, and the limits are angles found by solving the curves against each other. Writing dx in a polar integral loses the setup point.
• "A justification can be a calculation." The justification is the sentence that connects the calculation to the claim: concave up, therefore tangent lines below, therefore an underestimate.
• "Exact answers are always safer." On calculator-active parts the instruction is three decimal places, and an unevaluated exact expression can lose the value point.
• "You cannot use a result from an earlier part." You can, and should. Part (d) of question 4 checks part (b), and stating that check costs one line.

What you now know

Four question types cover almost all of what BC adds. Vector motion asks for speed as a length, acceleration componentwise, position as a starting value plus an integral, distance as the integral of speed, and the sign of v · a for whether the speed is rising: 3.167, (0.800, −1.248), (5.406, 4.970), 7.212, and decreasing. Polar asks you to solve for the angles, decide which curve is outer, and integrate half the difference of the squares, giving 4.653. Series asks for a known expansion with a substitution, a radius from the ratio test, both endpoints checked by hand for (−1/2, 1/2], an alternating error bound of 0.0004, and a high derivative read off a coefficient as −96. Differential equations ask for the capacity and the fastest growth off the equation, an Euler table, and a concavity argument for the direction of the error. Key idea: in every one of them, the sentence explaining why is worth as much as the number, and it is the part most often missing.

Sources

1. College Board. (n.d.). AP Calculus BC course and exam description and released free-response questions. Scoring guidelines itemise setup, value, units and justification. apcentral.collegeboard.org
2. OpenStax. (2016). 7.4 Area and arc length in polar coordinates. In Calculus Volume 2. The area formula and the outer minus inner setup used in Question 2. openstax.org
3. OpenStax. (2016). 6.4 Working with Taylor series. In Calculus Volume 2. Substitution into a known expansion, as in Question 3. openstax.org
4. OpenStax. (2016). 4.4 The logistic equation. In Calculus Volume 2. The closed-form solution checked against the Euler estimate in Question 4. openstax.org