Difficulty: Intermediate to Advanced | Prerequisites: Limits, Derivatives, Curve Sketching, and FTC (Docs 1 and 2 of this set).
Integration is the reverse of differentiation, and you will need a toolkit of techniques to handle the variety of integrands the exam throws at you. Beyond computing integrals, this section covers what you can do with them: finding areas, volumes, and solving motion problems. The volume methods (disk, washer, shell, cross-section) are among the most heavily tested topics on the AP exam.
Know your common integral formulas cold. Use u-substitution when you spot a composite function, and integration by parts when you have a product of two different types of function. For applications, velocity is the integral of acceleration, position is the integral of velocity, and volumes of revolution come from disk, washer, or shell methods depending on the axis and the shape.
Indefinite integral
The general antiderivative of a function, written without bounds. Always includes + C (the constant of integration).
Definite integral
The integral evaluated between two bounds a and b. Gives a numerical value representing signed area under the curve.
U-substitution
A technique for simplifying integrals by substituting u for part of the integrand. Think of it as the chain rule in reverse.
Integration by parts
A technique based on the product rule in reverse. The formula is: integral of u dv = uv - integral of v du. Use it when you have a product of two different types of function (e.g. x * e^x).
Riemann sum
An approximation of a definite integral by dividing the area into rectangles. Can use left endpoints, right endpoints, or midpoints.
Disk method
Computes the volume of a solid of revolution by integrating cross-sectional disks: V = pi * integral [R(x)]^2 dx.
Washer method
Like the disk method but with a hole in the middle: V = pi * integral ([R(x)]^2 - [r(x)]^2) dx.
Shell method
Computes volume using cylindrical shells: V = 2pi * integral r(x) * h(x) dx. Useful when the axis of revolution is parallel to the variable of integration.
integral of k dx = kx + C
integral of x^n dx = x^(n+1) / (n+1) + C (n ≠ -1)
integral of 1/x dx = ln|x| + C
integral of e^x dx = e^x + C
integral of e^(ax) dx = e^(ax) / a + C
integral of a^x dx = a^x / ln(a) + C
integral of cos(x) dx = sin(x) + C
integral of sin(x) dx = -cos(x) + C
integral of sec^2(x) dx = tan(x) + C
integral of csc^2(x) dx = -cot(x) + C
integral of sec(x)tan(x) dx = sec(x) + C
integral of csc(x)cot(x) dx = -csc(x) + C
integral of 1/(1+x^2) dx = arctan(x) + C
integral of 1/sqrt(1-x^2) dx = arcsin(x) + C
integral of tan(x) dx = ln|sec(x)| + C
integral of cot(x) dx = ln|sin(x)| + C
Commit these to memory. They appear frequently and recognising them quickly saves time on the exam.
When the integrand contains a composite function, let u equal the inner function. Then du = (du/dx) dx. Rewrite the entire integral in terms of u and du, integrate, then substitute back.
Look for a function and its derivative both present in the integrand. That is the signal to use u-sub.
integral of u dv = uv - integral of v du
Choose u and dv using LIATE (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential): let u be whichever type comes first in that list, and dv be the rest.
Sometimes you need to apply integration by parts more than once. If after two rounds you get back to the original integral, set it equal to a variable, collect terms, and solve.
A Riemann sum approximates the definite integral by breaking the interval [a, b] into n rectangles.
Delta x = (b - a) / n
The height of each rectangle is determined by the endpoint you choose: left, right, or midpoint. Sum all the heights, multiply by delta x.
Left Riemann sums underestimate the integral for increasing functions and overestimate for decreasing functions. Right Riemann sums do the opposite. The trapezoidal rule averages the left and right sums.
Velocity = integral of acceleration
Position = integral of velocity
Displacement over [a, b] = integral from a to b of v(t) dt
Total distance travelled = integral from a to b of |v(t)| dt
The distinction between displacement and total distance is tested regularly.
Disk method: V = pi * integral from a to b of [R(x)]^2 dx. Use when the region has no gap between it and the axis of revolution.
Washer method: V = pi * integral from a to b of ([R(x)]^2 - [r(x)]^2) dx. Use when there is a gap (an inner radius and an outer radius).
Shell method: V = 2pi * integral from a to b of r(x) * h(x) dx. Use when slicing parallel to the axis of revolution is simpler.
Perpendicular to the x-axis: V = integral from a to b of A(x) dx, where A(x) is the cross-sectional area.
Perpendicular to the y-axis: V = integral from c to d of A(y) dy.
Common cross-section shapes: squares, semicircles, equilateral triangles, isosceles right triangles. The exam tells you the shape; you supply the area formula.
Students forget + C on indefinite integrals. This costs points every single exam.
Confusing displacement with total distance is extremely common. Displacement uses v(t); total distance uses |v(t)|.
When setting up volume integrals, students sometimes square the wrong function or forget to subtract the inner radius in the washer method.
With u-substitution, students often forget to change the bounds when computing a definite integral. Either convert the bounds to u-values or substitute back to x before evaluating.
⚠️ Volume problems (disk/washer/shell) appear on the free-response section almost every year.
⚠️ The integral table entries for arctan and arcsin pop up in multiple-choice regularly. Recognise 1/(1+x^2) and 1/sqrt(1-x^2) on sight.
⚠️ Integration by parts: if the exam gives you integral of x * e^x dx or integral of x * ln(x) dx, this is the intended technique.
Fill in the blank: integral of sin(x) dx = ______
True or false: The shell method formula is V = pi * integral r(x) * h(x) dx.
Fill in the blank: Total distance travelled = integral of ______ dt.
True or false: You can always use the same bounds for a definite integral after u-substitution without converting them.
Answers: 1. -cos(x) + C. 2. False (it is 2pi, not pi). 3. |v(t)|. 4. False (you must either convert the bounds to u-values or substitute back to x before plugging in).
Q: Evaluate integral of x * e^(x^2) dx.
A: Let u = x^2, du = 2x dx, so x dx = du/2. The integral becomes (1/2) * integral of e^u du = (1/2) e^(x^2) + C.
Q: Find the volume of the solid formed by revolving y = sqrt(x) from x = 0 to x = 4 about the x-axis.
A: Use the disk method. V = pi * integral from 0 to 4 of [sqrt(x)]^2 dx = pi * integral from 0 to 4 of x dx = pi * [x^2/2] from 0 to 4 = pi * 8 = 8pi.
Q: Evaluate integral of x * ln(x) dx.
A: Integration by parts. Let u = ln(x), dv = x dx. Then du = 1/x dx, v = x^2/2. Result: (x^2/2) ln(x) - integral of x/2 dx = (x^2/2) ln(x) - x^2/4 + C.
Q: A particle moves along a line with velocity v(t) = t^2 - 4. Find the total distance travelled from t = 0 to t = 3.
A: v(t) = 0 at t = 2. Total distance = integral from 0 to 2 of |t^2 - 4| dt + integral from 2 to 3 of |t^2 - 4| dt = integral from 0 to 2 of (4 - t^2) dt + integral from 2 to 3 of (t^2 - 4) dt = [4t - t^3/3] from 0 to 2 + [t^3/3 - 4t] from 2 to 3 = 16/3 + 4/3 = 20/3.
Integration techniques are essential for the series and sequences material in Doc 4, where the integral test uses improper integrals to determine convergence.
Volume methods connect back to curve sketching: you need to know where curves intersect to set up the correct bounds.
U-substitution and integration by parts reappear in differential equations and in computing Taylor series coefficients.
integration, antiderivative, indefinite integral, definite integral, u-substitution, u-sub, integration by parts, LIATE, Riemann sum, left endpoint, right endpoint, midpoint, trapezoidal rule, disk method, washer method, shell method, volume of revolution, cross-section, velocity, acceleration, displacement, total distance, trig integrals, arctan integral, arcsin integral, common integrals, AP Calculus BC