AP Calculus AB, AP exams, 10–15%

Unit 8: Applications of Integration

Unit 8 uses the definite integral to compute quantities: average value, area between curves, accumulated change in motion problems, and volumes by cross sections and by revolution. Setting up the correct integral, with the right integrand and limits, carries most of the credit even before the arithmetic.

Position: 8 of 8 units
Published weighting: 10–15% in the official framework
Topics covered: 5 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Calculus AB. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Average value of a function: The integral definition of average value and its interpretation in context.
Accumulated change and motion: Position from velocity, total distance from speed, and net change from a rate function.
Area between curves: Setting up integrals with respect to x or y, including regions requiring more than one integral.
Volumes with known cross sections: Integrating the area of square, rectangular, triangular, or semicircular cross sections.
Volumes of revolution: disk and washer: Revolving a region about a horizontal or vertical line, including axes other than the coordinate axes.

Study tips

How to practice Applications of Integration.

Section-specific guidance that follows from how this material is usually tested.

Sketch the region and a representative slice before writing any volume or area integral.
Distinguish displacement from total distance every time; the difference is a standard free-response trap.
Practice revolution about lines that are not the axes, since those setups appear regularly.

Where this fits

Unit 8: Applications of Integration is 8 of 8 units in the AP Calculus AB guide, and it carries a published weighting of 10–15% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus AB guide Official exam information

Sample Applications of Integration questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
What is the average value of f(x) = x^2 on the interval [0, 3]?
1. A3
2. B4.5
3. C9
4. D27
Show answer
Answer: A. 3
The average value is (1/(b - a)) times ∫ from a to b of f(x) dx. Here ∫ from 0 to 3 of x^2 dx = 9, and dividing by the interval length 3 gives 3. Reporting 9 gives the integral without dividing by the length of the interval.
Key insight: Average value is the integral divided by the width of the interval, not the integral itself.
Question 2Foundational
What is the area of the region enclosed by the graphs of y = x and y = x^2?
1. A1/6
2. B1/3
3. C1/2
4. D1
Show answer
Answer: A. 1/6
The curves meet where x = x^2, that is at x = 0 and x = 1, and on that interval the line lies above the parabola. The area is ∫ from 0 to 1 of (x - x^2) dx = 1/2 - 1/3 = 1/6. Answering 1/2 integrates only the line and forgets to subtract the lower curve.
Key insight: Area between curves integrates top minus bottom, and the intersection points set the limits.
Question 3Intermediate
A particle moves along a line with velocity v(t) = t - 3 meters per second for 0 ≤ t ≤ 5. What is the total distance the particle travels?
1. A2.5 meters
2. B5 meters
3. C6.5 meters
4. D12.5 meters
Show answer
Answer: C. 6.5 meters
Total distance is ∫ from 0 to 5 of |v(t)| dt. The velocity is negative on [0, 3] and positive on [3, 5], giving ∫ from 0 to 3 of (3 - t) dt = 4.5 and ∫ from 3 to 5 of (t - 3) dt = 2, for a total of 6.5 meters. The value 2.5 is the magnitude of the displacement, which cancels the two directions.
Key insight: Displacement integrates velocity; total distance integrates speed, so split at every sign change.
Question 4Intermediate
The base of a solid is the triangular region bounded by y = 2 - x, the x-axis, and the y-axis. Every cross section perpendicular to the x-axis is a square. What is the volume of the solid?
1. A4/3
2. B8/3
3. C16/3
4. D8
Show answer
Answer: B. 8/3
Each square has side length equal to the height of the region, 2 - x, so its area is (2 - x)^2 and the volume is ∫ from 0 to 2 of (2 - x)^2 dx = 8/3. Using (2 - x)^2/2, the area of a triangular cross section rather than a square one, gives the tempting 4/3.
Key insight: For known cross sections, integrate the area formula of the named shape; no π appears unless the shape is round.
Question 5Advanced
The region enclosed by y = 2x and y = x^2 is revolved about the x-axis. What is the volume of the resulting solid?
1. A8π/3
2. B16π/15
3. C64π/15
4. D128π/15
Show answer
Answer: C. 64π/15
The curves meet at x = 0 and x = 2, and the line is farther from the axis, so the washer method gives π∫ from 0 to 2 of ((2x)^2 - (x^2)^2) dx = π(32/3 - 32/5) = 64π/15. Subtracting the radii before squaring gives π∫ from 0 to 2 of (2x - x^2)^2 dx = 16π/15, the classic error here.
Key insight: In the washer method, square each radius separately: (R^2 - r^2), never (R - r)^2.

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Unit 8: Applications of Integration