Calculus II: Integration & Series

Calculus I built the derivative to measure instantaneous rates. Calculus II begins with the opposite problem: finding the total accumulation, the area under a curve. Suppose f(x) >= 0 on [a, b] and we want the area between the graph and the x-axis. Curved regions have no simple area formula, so we approximate with rectangles and then let the rectangles get infinitely thin.

Building a Riemann sum

Divide [a, b] into n equal subintervals, each of width dx = (b - a)/n. On each subinterval pick a sample point and build a rectangle whose height is f at that point. Adding the rectangle areas gives a Riemann sum:

S = f(x_1) dx + f(x_2) dx + ... + f(x_n) dx

Choosing the left endpoint of each subinterval gives a left sum, the right endpoint a right sum, and the middle a midpoint sum. For an increasing function the left sum underestimates and the right sum overestimates the true area.

Worked example, four rectangles

Estimate the area under f(x) = x^2 on [0, 2] using a right Riemann sum with n = 4.

1. Width: dx = (2 - 0)/4 = 0.5. Right endpoints: 0.5, 1.0, 1.5, 2.0.
2. Heights: f(0.5) = 0.25, f(1) = 1, f(1.5) = 2.25, f(2) = 4.
3. Sum the areas: (0.25 + 1 + 2.25 + 4)(0.5) = (7.5)(0.5) = 3.75.

The estimate is 3.75. The exact area, as we will soon confirm, is 8/3 approximately 2.667; the right sum overestimates because x^2 is increasing. More rectangles would sharpen the estimate.

From sum to integral

Letting n to infinity so the rectangles become infinitely thin gives the exact area. That limit is the definite integral:

integral from a to b of f(x) dx = lim (n to infinity) sum of f(x_i) dx

The elongated S is a stretched sum, and dx is the width of an infinitely thin strip. The numbers a and b are the limits of integration. When f can be negative, the integral computes signed area: area above the axis counts positive, area below counts negative.

Basic properties

• integral from a to a of f dx = 0 (no width, no area).
• integral from a to b of f dx = - integral from b to a of f dx (reversing the limits flips the sign).
• integral from a to b of [f + g] dx = integral of f + integral of g (integrals add).
• integral from a to c of f dx + integral from c to b of f dx = integral from a to b of f dx (splitting the interval).

These rules let you break complicated regions into manageable pieces, and they mirror the limit laws you already trust from Calculus I.

Computing integrals as limits of sums is exhausting. The Fundamental Theorem of Calculus (FTC) reveals that integration and differentiation are inverse operations, which makes definite integrals easy to evaluate. It comes in two parts.

Part 1: derivative of an accumulation function

Define an accumulation function g(x) = integral from a to x of f(t) dt, the running area from a fixed a out to a moving upper limit x. Part 1 says that differentiating this area recovers the original function:

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

In words, the rate at which accumulated area grows equals the height of the curve at the leading edge, which makes intuitive sense: a taller curve piles on area faster. With the chain rule, if the upper limit is u(x), then d/dx integral from a to u(x) of f(t) dt = f(u(x)) u'(x). For example, d/dx integral from 0 to x^2 of cos(t) dt = cos(x^2) (2x).

Part 2: evaluating definite integrals

Part 2 is the workhorse. If F is any antiderivative of f (so F' = f), then:

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

You find an antiderivative, plug in the top limit, plug in the bottom limit, and subtract. The constant of integration cancels in the subtraction, so it is dropped for definite integrals. The bracket notation [F(x)] from a to b means F(b) - F(a).

Worked example one

Evaluate integral from 0 to 2 of x^2 dx.

1. An antiderivative of x^2 is F(x) = x^3/3.
2. Apply Part 2: F(2) - F(0) = 8/3 - 0 = 8/3.

So the exact area under x^2 on [0, 2] is 8/3, confirming that the right sum of 3.75 was an overestimate.

Worked example two

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

1. An antiderivative of sin(x) is -cos(x).
2. Evaluate: [-cos(x)] from 0 to pi = -cos(pi) - (-cos(0)) = -(-1) + 1 = 2.

The area under one hump of the sine curve is exactly 2, a strikingly clean result. The Fundamental Theorem is the bridge between the two halves of calculus, and nearly every integral you compute from here on rests on it.

The Fundamental Theorem has a powerful interpretation. Since integral from a to b of F'(x) dx = F(b) - F(a), integrating a rate of change gives the net change in the quantity. This is the Net Change Theorem, and it turns integrals into a tool for accumulation across science and economics.

Displacement versus total distance

If v(t) is the velocity of an object, then integrating velocity gives displacement, the net change in position:

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

Because velocity can be negative (moving backward), displacement can be smaller than the actual path length. To get total distance traveled, integrate the speed, which is the absolute value of velocity:

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

Worked example. An object has velocity v(t) = t - 2 (in m/s) for t in [0, 3]. Find its displacement and total distance.

1. Displacement: integral from 0 to 3 of (t - 2) dt = [t^2/2 - 2t] from 0 to 3 = (4.5 - 6) - 0 = -1.5 m. The object ends up 1.5 m behind its start.
2. Total distance: the velocity is negative on [0, 2) and positive on (2, 3], so split at t = 2.
3. From 0 to 2: integral of (t - 2) dt = [t^2/2 - 2t] from 0 to 2 = (2 - 4) = -2, contributing distance 2.
4. From 2 to 3: [t^2/2 - 2t] from 2 to 3 = (4.5 - 6) - (2 - 4) = -1.5 - (-2) = 0.5, contributing distance 0.5.
5. Total distance: 2 + 0.5 = 2.5 m.

Displacement is -1.5 m but total distance is 2.5 m, because the object first backed up, then moved forward.

The average value of a function

The average value of f on [a, b] spreads its total accumulation evenly across the interval:

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

Worked example. Find the average value of f(x) = x^2 on [0, 3].

1. integral from 0 to 3 of x^2 dx = [x^3/3] from 0 to 3 = 9.
2. Divide by the width: f_avg = (1/3)(9) = 3.

The average value is 3. The Mean Value Theorem for Integrals guarantees a point c in [0, 3] where f(c) = 3, here c = sqrt(3), so the curve actually attains its own average somewhere on the interval.

The most important integration technique is substitution, which reverses the chain rule. When an integrand contains a function and (up to a constant) its own derivative, we substitute a new variable u for the inner function.

The method

To integrate integral of f(g(x)) g'(x) dx, let u = g(x), so du = g'(x) dx. The integral becomes integral of f(u) du, which is usually easy. The whole point is to make du match the leftover part of the integrand.

Worked example one. Evaluate integral of 2x cos(x^2) dx.

1. Let u = x^2, so du = 2x dx. The 2x dx is exactly present.
2. Substitute: integral of cos(u) du = sin(u) + C.
3. Back-substitute: sin(x^2) + C.

Adjusting for a missing constant

Often the constant is not a perfect match, and we fix it with a factor.

Worked example two. Evaluate integral of x (x^2 + 1)^5 dx.

1. Let u = x^2 + 1, so du = 2x dx, which means x dx = du/2.
2. Substitute: integral of u^5 (du/2) = (1/2) integral of u^5 du = (1/2)(u^6/6) = u^6/12.
3. Back-substitute: (x^2 + 1)^6 / 12 + C.

Definite integrals: change the limits

For a definite integral you can convert the x-limits into u-limits and skip back-substituting.

Worked example three. Evaluate integral from 0 to 2 of x (x^2 + 1)^3 dx.

1. Let u = x^2 + 1, du = 2x dx, so x dx = du/2.
2. Change limits: when x = 0, u = 1; when x = 2, u = 5.
3. Rewrite: (1/2) integral from 1 to 5 of u^3 du = (1/2)[u^4/4] from 1 to 5 = (1/8)(625 - 1) = 624/8 = 78.

The answer is 78. A common special case is integral of g'(x)/g(x) dx = ln|g(x)| + C, because u = g(x) turns it into integral of du/u. For instance, integral of (2x)/(x^2 + 1) dx = ln(x^2 + 1) + C. Choosing the inner function as u and checking that its derivative appears is the entire strategy.

Integration by parts reverses the product rule, and it handles integrals of products that substitution cannot, such as x e^x or ln(x). The formula is:

integral of u dv = u v - integral of v du

You split the integrand into a part u to differentiate and a part dv to integrate. The goal is that the new integral integral of v du is simpler than the one you started with.

Choosing u with LIATE

Pick u to be whichever factor comes first in this priority list, because that factor simplifies when differentiated:

• L - Logarithmic functions (ln x)
• I - Inverse trig functions (arctan x)
• A - Algebraic functions (x, x^2)
• T - Trigonometric functions (sin x, cos x)
• E - Exponential functions (e^x)

Whatever is left, together with dx, becomes dv.

Worked example one

Evaluate integral of x e^x dx.

1. By LIATE, the algebraic x beats the exponential, so u = x and dv = e^x dx.
2. Then du = dx and v = e^x.
3. Apply the formula: u v - integral of v du = x e^x - integral of e^x dx = x e^x - e^x + C.

So integral of x e^x dx = e^x(x - 1) + C. Check by differentiating: the product rule returns x e^x.

Worked example two, a logarithm

Evaluate integral of ln(x) dx. There is no obvious antiderivative, so use parts with the whole thing as u.

1. Let u = ln(x) and dv = dx. Then du = (1/x) dx and v = x.
2. Apply: x ln(x) - integral of x (1/x) dx = x ln(x) - integral of 1 dx = x ln(x) - x + C.

Applying parts twice

Some integrals need parts repeated. For integral of x^2 e^x dx, set u = x^2, dv = e^x dx to get x^2 e^x - 2 integral of x e^x dx; the leftover integral is the one solved above, giving x^2 e^x - 2(x e^x - e^x) + C = e^x(x^2 - 2x + 2) + C. Occasionally the original integral reappears (the boomerang), as with integral of e^x sin(x) dx; you then solve for it algebraically by moving it to the left side.

Integrals involving trig functions and integrals with square roots like sqrt(a^2 - x^2) yield to identity-based methods. Two families cover most cases.

Powers of sine and cosine

For integral of sin^m(x) cos^n(x) dx, use parity:

• If the power of sine is odd, peel off one sin(x) for du, convert the rest with sin^2 = 1 - cos^2, and let u = cos(x).
• If the power of cosine is odd, peel off one cos(x), convert with cos^2 = 1 - sin^2, and let u = sin(x).
• If both are even, use the half-angle identities sin^2(x) = (1 - cos 2x)/2 and cos^2(x) = (1 + cos 2x)/2.

Worked example. Evaluate integral of sin^3(x) cos^2(x) dx.

1. Sine is odd. Write sin^3 = sin^2 (sin) = (1 - cos^2)(sin).
2. The integral is integral of (1 - cos^2 x) cos^2 x sin(x) dx. Let u = cos(x), so du = -sin(x) dx.
3. This becomes -integral of (1 - u^2) u^2 du = -integral of (u^2 - u^4) du = -(u^3/3 - u^5/5).
4. Back-substitute: -cos^3(x)/3 + cos^5(x)/5 + C.

Trigonometric substitution

When an integrand contains a square root of a quadratic, a clever trig substitution turns it into a plain trig integral using a Pythagorean identity. The three patterns are:

Expression	Substitute	Identity used
sqrt(a^2 - x^2)	x = a sin(theta)	1 - sin^2 = cos^2
sqrt(a^2 + x^2)	x = a tan(theta)	1 + tan^2 = sec^2
sqrt(x^2 - a^2)	x = a sec(theta)	sec^2 - 1 = tan^2

Worked example. Evaluate integral of dx / sqrt(4 - x^2).

1. Here a = 2, so let x = 2 sin(theta), giving dx = 2 cos(theta) d(theta).
2. The root becomes sqrt(4 - 4 sin^2 theta) = 2 cos(theta).
3. The integral is integral of (2 cos theta)/(2 cos theta) d(theta) = integral of 1 d(theta) = theta + C.
4. Since x = 2 sin(theta), we have theta = arcsin(x/2), so the answer is arcsin(x/2) + C.

This confirms the standard result integral of dx/sqrt(a^2 - x^2) = arcsin(x/a) + C. Trig substitution converts algebra with roots into trig integrals you already know how to handle.

Rational functions, ratios of polynomials, can be integrated by breaking them into a sum of simpler fractions. This method of partial fractions reverses the process of adding fractions over a common denominator.

The setup

First the fraction must be proper (numerator degree less than denominator degree); if not, do polynomial long division first. Then factor the denominator and write one partial fraction for each factor. For distinct linear factors, each contributes a term with an unknown constant on top:

(numerator)/((x - r_1)(x - r_2)) = A/(x - r_1) + B/(x - r_2)

Worked example, distinct linear factors

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

1. Factor the denominator: x^2 - x - 2 = (x - 2)(x + 1).
2. Set up: (5x - 4)/((x - 2)(x + 1)) = A/(x - 2) + B/(x + 1).
3. Clear denominators: 5x - 4 = A(x + 1) + B(x - 2).
4. Let x = 2: 10 - 4 = A(3), so A = 2. Let x = -1: -5 - 4 = B(-3), so B = 3.
5. Integrate each piece: integral of [2/(x - 2) + 3/(x + 1)] dx = 2 ln|x - 2| + 3 ln|x + 1| + C.

The key trick in step 4, substituting the roots one at a time, isolates each unknown instantly.

Repeated and irreducible quadratic factors

Two more cases round out the method:

• Repeated linear factor (x - r)^2: include both A/(x - r) and B/(x - r)^2, one term per power.
• Irreducible quadratic factor x^2 + bx + c (no real roots): use a linear numerator, (Ax + B)/(x^2 + bx + c). These often integrate to a logarithm plus an arctangent.

For example, a term like 1/(x^2 + 1) integrates to arctan(x) + C, and a term like x/(x^2 + 1) integrates by substitution to (1/2) ln(x^2 + 1) + C. Partial fractions turns any rational function into a sum of logarithms and arctangents, completing our toolkit of integration techniques. With substitution, parts, trig methods, and partial fractions, you can now integrate a very wide range of functions.

The definite integral was defined for a bounded function on a bounded interval. An improper integral relaxes those conditions in one of two ways: the interval runs to infinity, or the integrand becomes infinite somewhere. Each is defined as a limit of ordinary integrals. If the limit is a finite number, the integral converges; otherwise it diverges.

Type 1: infinite interval

We define integral from a to infinity of f(x) dx = lim (t to infinity) integral from a to t of f(x) dx. Compute the ordinary integral to t, then take the limit.

Worked example, convergent. Evaluate integral from 1 to infinity of 1/x^2 dx.

1. Integrate to t: integral from 1 to t of x^(-2) dx = [-1/x] from 1 to t = -1/t + 1.
2. Take the limit: lim (t to infinity) (1 - 1/t) = 1 - 0 = 1.

The integral converges to 1. The area under 1/x^2 stretching forever to the right is finite.

Worked example, divergent. Evaluate integral from 1 to infinity of 1/x dx.

1. Integrate to t: [ln x] from 1 to t = ln t - 0 = ln t.
2. Take the limit: lim (t to infinity) ln t = infinity.

This integral diverges. The graphs of 1/x and 1/x^2 look similar, yet one bounds finite area and the other does not; the power matters. In general the p-integral integral from 1 to infinity of 1/x^p dx converges exactly when p > 1.

Type 2: unbounded integrand

If f blows up at an endpoint, say at x = 0, define integral from 0 to 1 of f dx = lim (t to 0^+) integral from t to 1 of f dx. For example, integral from 0 to 1 of 1/sqrt(x) dx = lim (t to 0^+) [2 sqrt(x)] from t to 1 = 2 - 0 = 2, which converges, even though the integrand is infinite at 0.

The comparison test

Sometimes we cannot integrate f but can still decide convergence by comparing it with a simpler function. If 0 <= f(x) <= g(x) and the larger integral of g converges, then the smaller integral of f converges too. If the smaller integral of f diverges, the larger integral of g diverges. For instance, integral from 1 to infinity of 1/(x^2 + 1) dx converges because 1/(x^2 + 1) <= 1/x^2 and the 1/x^2 integral converges. Comparison lets you judge convergence without an explicit antiderivative.

The definite integral finds area under one curve. Subtracting two integrals finds the area between two curves, a basic tool for comparing accumulated quantities.

The formula

If f(x) >= g(x) on [a, b], the area of the region between them is the integral of the gap, top minus bottom:

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

The limits a and b are usually the x-values where the curves cross, found by setting f(x) = g(x). Always subtract the lower curve from the upper curve so the integrand stays non-negative across the region.

Worked example

Find the area enclosed between y = x + 2 and y = x^2.

1. Find intersections: x + 2 = x^2 gives x^2 - x - 2 = 0 = (x - 2)(x + 1), so x = -1 and x = 2.
2. Decide which is on top. Test x = 0: the line gives 2, the parabola gives 0, so the line x + 2 is above.
3. Set up: A = integral from -1 to 2 of [(x + 2) - x^2] dx.
4. Integrate: [x^2/2 + 2x - x^3/3] from -1 to 2.
5. At x = 2: 2 + 4 - 8/3 = 6 - 8/3 = 10/3. At x = -1: 1/2 - 2 + 1/3 = -7/6.
6. Subtract: 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2.

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

Integrating with respect to y

When curves are easier to describe as x in terms of y (for instance, sideways parabolas), integrate horizontally instead: A = integral from c to d of [x_right(y) - x_left(y)] dy. The principle is identical, right minus left, top minus bottom; choose the direction that keeps the boundaries as simple functions.

Rotating a region around an axis sweeps out a three-dimensional solid of revolution. We find its volume by slicing it into thin pieces, computing each piece's volume, and integrating.

The disk method

Rotate the region under y = f(x) from a to b about the x-axis. Each thin vertical strip sweeps out a disk (a coin) of radius f(x) and thickness dx. A disk's volume is pi (radius)^2 (thickness), so:

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

Worked example. Rotate y = sqrt(x) on [0, 4] about the x-axis.

1. Radius is f(x) = sqrt(x), so [f(x)]^2 = x.
2. Set up: V = integral from 0 to 4 of pi x dx = pi [x^2/2] from 0 to 4 = pi (16/2) = 8 pi.

The volume is 8 pi cubic units. Squaring the radius before integrating is the step students most often skip.

The washer method

If the region is rotated about an axis it does not touch, each slice is a washer, a disk with a hole. With outer radius R (to the far curve) and inner radius r (to the near curve), the slice area is pi(R^2 - r^2), so:

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

Worked example. Rotate the region between y = x and y = x^2 about the x-axis. On [0, 1] the line y = x is the outer boundary and the parabola y = x^2 is the inner.

1. Outer radius R = x, inner radius r = x^2.
2. Set up: V = integral from 0 to 1 of pi (x^2 - x^4) dx.
3. Integrate: pi [x^3/3 - x^5/5] from 0 to 1 = pi (1/3 - 1/5) = pi (5/15 - 3/15) = 2 pi/15.

The volume is 2 pi/15. Subtract the squares, never subtract then square: R^2 - r^2 is not (R - r)^2. The key to every slicing problem is drawing one representative slice and finding its radius from the axis of rotation.

The shell method is a second way to find volumes of revolution, especially handy when the region is rotated about a vertical axis and describing it with horizontal slices would be awkward.

The idea

Instead of slicing the solid into flat disks, imagine it built from thin concentric cylindrical shells, like the rings of a tree or nested tin cans. A shell at radius x with height f(x) and thickness dx, when unrolled, is a thin rectangular sheet of dimensions circumference by height by thickness: 2 pi x times f(x) times dx. Integrating over all shells gives, for rotation about the y-axis:

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

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

Worked example

Rotate the region under y = x^2 on [0, 2] about the y-axis.

1. Shell radius x, shell height f(x) = x^2.
2. Set up: V = integral from 0 to 2 of 2 pi x (x^2) dx = 2 pi integral from 0 to 2 of x^3 dx.
3. Integrate: 2 pi [x^4/4] from 0 to 2 = 2 pi (16/4) = 2 pi (4) = 8 pi.

The volume is 8 pi. Solving this with washers would require rewriting y = x^2 as x = sqrt(y) and handling the region carefully; shells keep it in terms of x.

Choosing a method

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

Both methods always give the same volume; the choice is about which integral is simpler to set up. A reliable rule: if a representative strip runs parallel to the axis of rotation, use shells; if it runs perpendicular, use disks or washers. Sketching one strip and noting its orientation tells you immediately which method to reach for.

Integration also measures the length of a curve. Approximate the curve by many short straight segments; each has length given by the Pythagorean theorem, and summing them in the limit produces an integral.

The formula

A tiny piece of curve has horizontal run dx and vertical rise dy, so its length is sqrt(dx^2 + dy^2). Factoring out dx and using dy/dx = f'(x) gives the arc length of y = f(x) from a to b:

L = integral from a to b of sqrt(1 + [f'(x)]^2) dx

The 1 comes from the horizontal run and the [f'(x)]^2 from the slope. The whole method is the Pythagorean theorem applied to infinitely many infinitesimal segments.

Worked example

Find the length of y = (2/3) x^(3/2) from x = 0 to x = 3. This function is chosen because it produces a perfect square under the root.

1. Differentiate: f'(x) = (2/3)(3/2) x^(1/2) = x^(1/2) = sqrt(x).
2. Square it: [f'(x)]^2 = x.
3. Form the integrand: sqrt(1 + x).
4. Integrate: L = integral from 0 to 3 of sqrt(1 + x) dx. Let u = 1 + x: integral of u^(1/2) du = (2/3) u^(3/2).
5. Evaluate: (2/3)[(1 + x)^(3/2)] from 0 to 3 = (2/3)(4^(3/2) - 1^(3/2)) = (2/3)(8 - 1) = 14/3.

The arc length is 14/3. Notice we chose 4^(3/2) = (sqrt 4)^3 = 2^3 = 8.

When the integral resists

Arc length integrands often contain a square root of a messy expression that has no elementary antiderivative. For example, the length of the parabola y = x^2 leads to integral of sqrt(1 + 4x^2) dx, which requires trig substitution or numerical methods. This is why textbook arc length problems are carefully rigged so that 1 + [f'(x)]^2 becomes a perfect square. In practice, most arc lengths are computed numerically, but the integral setup remains exactly the same, and understanding the setup is the real skill.

The second half of Calculus II studies the infinite. We begin with sequences, the raw material of infinite series. A sequence is an ordered list of numbers, a_1, a_2, a_3, ..., defined by a rule for the nth term a_n. For example a_n = 1/n gives 1, 1/2, 1/3, 1/4, ....

The limit of a sequence

A sequence converges to a limit L if the terms get arbitrarily close to L as n to infinity, written lim (n to infinity) a_n = L. If no finite limit exists, the sequence diverges. Because n marches to infinity, evaluating lim (n to infinity) a_n uses the same techniques as limits at infinity from Calculus I.

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

Worked example two. Does a_n = (3n^2 + 1)/(2n^2 - n) converge? Divide top and bottom by n^2: (3 + 1/n^2)/(2 - 1/n). As n to infinity the small terms vanish, leaving 3/2. It converges to 3/2.

Geometric and factorial behavior

Two patterns appear constantly:

• Powers r^n: if |r| < 1, then r^n to 0 (for example (1/2)^n to 0); if |r| > 1, the terms blow up; if r = 1, they stay at 1.
• Factorial growth: n! grows faster than any power or exponential, so terms like n^{10}/n! and 2^n/n! tend to 0.

Worked example three. Does a_n = (-1)^n converge? The terms alternate -1, 1, -1, 1, ... forever, never settling, so the sequence diverges by oscillation.

A useful bridge

If a_n = f(n) for a function f and lim (x to infinity) f(x) = L, then lim (n to infinity) a_n = L as well. This lets you apply calculus tools, including L'Hopital's rule, to sequences. For instance, lim (n to infinity) (ln n)/n = 0 because the function version (ln x)/x to 0. Understanding when the terms of a sequence die out is the foundation for the central question of the next lessons: when does adding infinitely many terms give a finite total?

An infinite series is the sum of all terms of a sequence: a_1 + a_2 + a_3 + ..., written sum from n=1 to infinity of a_n. Adding infinitely many numbers sounds impossible, but we make sense of it with partial sums.

Partial sums define the total

The nth partial sum S_n = a_1 + a_2 + ... + a_n adds just the first n terms. The infinite series converges to a sum S if the sequence of partial sums converges: lim (n to infinity) S_n = S. If the partial sums have no finite limit, the series diverges. So a series is really a limit of finite sums in disguise.

The geometric series

The most important series is the geometric series, where each term is a fixed ratio r times the one before: a + a r + a r^2 + a r^3 + .... It converges precisely when |r| < 1, and then its sum has a clean closed form:

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

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

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

1. First term a = 1, ratio r = 1/2, and |r| < 1, so it converges.
2. Apply the formula: a/(1 - r) = 1/(1 - 1/2) = 1/(1/2) = 2.

The sum is exactly 2. Adding infinitely many positive numbers gave a finite total because the terms shrink fast enough. Another example: sum from n=1 to infinity of (1/3)^n = (1/3)/(1 - 1/3) = (1/3)/(2/3) = 1/2 (note this starts at n = 1, so the first term is 1/3).

The nth term test for divergence

Here is a quick necessary condition: if a series converges, its terms must shrink to 0. Contrapositively, the nth term test says that if lim (n to infinity) a_n is not 0 (or does not exist), the series diverges. For example sum of n/(n + 1) diverges because a_n to 1, not 0. Warning: the test only detects divergence. Terms going to 0 does NOT guarantee convergence, as the next lesson's harmonic series will show. Always run the nth term test first; it is the cheapest check available.

Most series are not geometric, so we need tests that decide convergence without computing the sum. For series with positive terms, several powerful tests apply. Start every problem with the nth term test, then choose from the following.

The integral test and p-series

If a_n = f(n) for a positive, decreasing, continuous function f, then the series sum of a_n and the improper integral integral from 1 to infinity of f(x) dx either both converge or both diverge. This links series back to the improper integrals of Module 3.

The most important consequence is the p-series result:

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

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

The comparison test

Compare a mysterious series to a known one, term by term. If 0 <= a_n <= b_n and the larger sum of b_n converges, then the smaller sum of a_n converges. If the smaller sum of a_n diverges, the larger sum of b_n diverges.

Worked example. Does sum of 1/(n^2 + 1) converge? Since 1/(n^2 + 1) < 1/n^2 and sum of 1/n^2 converges (p-series, p = 2), the smaller series converges by comparison.

The limit comparison test

Direct comparison can be finicky, so the limit comparison test is often smoother. If a_n and b_n are positive and lim (n to infinity) a_n/b_n = c where c is a finite positive number, then both series do the same thing, converge together or diverge together.

Worked example. Does sum of (2n + 1)/(n^3 + 5) converge? For large n it behaves like 2n/n^3 = 2/n^2. Compare with b_n = 1/n^2: the ratio a_n/b_n = (2n + 1)n^2/(n^3 + 5) to 2, a finite positive number. Since sum of 1/n^2 converges, the original series converges too. The strategy is to keep only the dominant powers to guess a comparison series, then confirm with the ratio.

Series with mixed signs and series with factorials or exponentials call for two more tests, plus a finer notion of convergence.

The alternating series test

An alternating series flips sign every term, like sum of (-1)^n b_n with b_n > 0. The alternating series test says it converges if two conditions hold: the terms b_n are eventually decreasing, and lim (n to infinity) b_n = 0.

Worked example. The alternating harmonic series 1 - 1/2 + 1/3 - 1/4 + ... has b_n = 1/n, which decreases to 0, so it converges. Remarkably, its sum is ln 2. Contrast this with the ordinary harmonic series, which diverges: alternating signs can rescue convergence.

The ratio test

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

L = lim (n to infinity) |a_(n+1)/a_n|

• If L < 1, the series converges (absolutely).
• If L > 1, the series diverges.
• If L = 1, the test is inconclusive; try another test.

Worked example. Does sum of 2^n/n! converge? Form the ratio: a_(n+1)/a_n = [2^(n+1)/(n+1)!] * [n!/2^n] = 2/(n + 1). As n to infinity this ratio goes to 0, which is less than 1, so the series converges. Factorials in the denominator almost always win, driving the ratio to 0.

Absolute versus conditional convergence

A series sum of a_n converges absolutely if the series of absolute values sum of |a_n| converges. Absolute convergence is the stronger, better-behaved kind, and it guarantees ordinary convergence. If a series converges but the absolute-value series diverges, we call it conditionally convergent. The alternating harmonic series is the star example: sum of (-1)^n/n converges, but sum of 1/n diverges, so it is conditionally convergent. This distinction matters because rearranging a conditionally convergent series can change its sum, while absolutely convergent series can be safely rearranged. With these tests, geometric, nth term, integral, comparison, limit comparison, alternating series, and ratio, you can classify almost any series you meet.

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

sum from n=0 to infinity of a_n (x - c)^n = a_0 + a_1(x - c) + a_2(x - c)^2 + ...

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

Radius and interval of convergence

A power series converges on an interval centered at c. The radius of convergence R is half the length of that interval: the series converges for |x - c| < R and diverges for |x - c| > R. The endpoints must be checked separately. We usually find R with the ratio test.

Worked example. Find the interval of convergence of sum from n=0 to infinity of x^n/n! (centered at 0).

1. Ratio test: |a_(n+1)/a_n| = |x^(n+1)/(n+1)! * n!/x^n| = |x|/(n + 1).
2. Take the limit: lim (n to infinity) |x|/(n + 1) = 0 for every x.
3. Since the limit is 0 < 1 always, the series converges for all real x, so R = infinity.

This particular series turns out to equal e^x everywhere, as the next lesson shows.

A finite radius

Worked example. Find the radius of convergence of sum from n=0 to infinity of x^n.

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

So 1/(1 - x) = 1 + x + x^2 + x^3 + ... for |x| < 1, our first example of a function written as a power series. Inside the radius of convergence, power series behave like polynomials: you can add, differentiate, and integrate them term by term, which makes them an extraordinarily flexible tool for representing and approximating functions.

The previous lesson wrote a couple of functions as power series by luck. Taylor series give a systematic recipe to expand almost any smooth function as an infinite polynomial, using its derivatives at a single point.

The Taylor series formula

The Taylor series of f centered at c is:

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

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

Worked example, building e^x

Find the Maclaurin series of f(x) = e^x.

1. Every derivative of e^x is e^x, and at 0 each equals 1: f^(n)(0) = 1.
2. Plug into the formula: sum of [1/n!] x^n = 1 + x + x^2/2! + x^3/3! + ....

So e^x = sum from n=0 to infinity of x^n/n!, matching the series we found earlier. This series converges for all x.

The essential Maclaurin series

Three expansions are worth memorizing; they recur throughout mathematics and physics:

Function	Maclaurin series
e^x	1 + x + x^2/2! + x^3/3! + ...
sin(x)	x - x^3/3! + x^5/5! - x^7/7! + ...
cos(x)	1 - x^2/2! + x^4/4! - x^6/6! + ...

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

Taylor polynomials as approximations

Truncating the series after a few terms gives a Taylor polynomial that approximates f near the center. The more terms, the better the fit.

Worked example. Approximate cos(0.1) with two terms. Using cos(x) approximately 1 - x^2/2: 1 - (0.1)^2/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995. The true value is 0.995004..., so two terms already give four correct digits. This is exactly how calculators and computers evaluate transcendental functions: they sum enough Taylor terms to reach the required precision. Taylor series turn hard functions into polynomials we can actually compute.

Not every curve is the graph of a function; a circle fails the vertical line test, for instance. Parametric equations describe a curve by giving both coordinates as functions of a third variable, the parameter t (often time):

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

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

Slope of a parametric curve

To find the slope dy/dx without eliminating the parameter, use the chain rule. Since dy/dt = (dy/dx)(dx/dt), solve for the slope:

dy/dx = (dy/dt)/(dx/dt), provided dx/dt is not 0

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

1. Differentiate: dx/dt = 2t and dy/dt = 3t^2 - 3.
2. Form the ratio: dy/dx = (3t^2 - 3)/(2t).
3. Evaluate at t = 2: (3(4) - 3)/(2(2)) = (12 - 3)/4 = 9/4.

The slope at that point is 9/4. A horizontal tangent occurs where dy/dt = 0 (here t = 1, giving t^2 = 1) and a vertical tangent where dx/dt = 0 (here t = 0).

Arc length in parametric form

The arc length formula adapts naturally. A tiny step has length sqrt(dx^2 + dy^2); factoring out dt gives the parametric arc length from t = a to t = b:

L = integral from a to b of sqrt((dx/dt)^2 + (dy/dt)^2) dt

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

1. dx/dt = -sin(t), dy/dt = cos(t).
2. Sum of squares: sin^2 t + cos^2 t = 1, so the integrand is sqrt(1) = 1.
3. Integrate: L = integral from 0 to 2 pi of 1 dt = 2 pi.

The circumference is 2 pi, exactly matching the known formula 2 pi r with r = 1. Parametric calculus lets us handle loops, spirals, and any path a moving point can trace.

Polar coordinates locate a point by its distance from the origin and its angle, rather than by horizontal and vertical position. A point is written (r, theta), where r is the distance from the origin (the pole) and theta is the angle measured counterclockwise from the positive x-axis. Polar coordinates make circles and spirals far simpler to describe than rectangular ones do.

Converting between systems

Basic trigonometry links the two coordinate systems:

• Polar to rectangular: x = r cos(theta) and y = r sin(theta).
• Rectangular to polar: r^2 = x^2 + y^2 and tan(theta) = y/x.

For example, the point (r, theta) = (2, pi/3) becomes x = 2 cos(pi/3) = 2(1/2) = 1 and y = 2 sin(pi/3) = 2(sqrt(3)/2) = sqrt(3), so it is (1, sqrt(3)) in rectangular form.

Polar curves

A polar curve is usually given as r = f(theta). A few standard shapes:

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

Area in polar coordinates

Area enclosed by a polar curve is built from thin circular sectors, not rectangles. A sector of radius r and tiny angle d(theta) has area (1/2) r^2 d(theta). Integrating over the swept angle gives:

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

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

1. Set up: A = integral from -pi/2 to pi/2 of (1/2)(2 cos theta)^2 d(theta) = (1/2) integral from -pi/2 to pi/2 of 4 cos^2 theta d(theta).
2. Use the half-angle identity cos^2 theta = (1 + cos 2theta)/2: the integrand becomes 2 (1 + cos 2theta)/2 = 1 + cos 2theta after the constants combine, giving A = integral from -pi/2 to pi/2 of (1 + cos 2theta) d(theta).
3. Integrate: [theta + (1/2) sin 2theta] from -pi/2 to pi/2 = (pi/2 + 0) - (-pi/2 + 0) = pi.

The area is pi. As a check, r = 2 cos theta is a circle of radius 1 (diameter 2), and a radius-1 circle has area pi r^2 = pi. Polar coordinates turn awkward rectangular integrals for round regions into simple ones, completing your calculus toolkit for curves of every kind.