AP Calculus BC, AP exams, 6–9%

Unit 8: Applications of Integration

Unit 8 uses definite integrals to compute average value, accumulated change, area between curves, volumes by cross sections and revolution, and, in BC, arc length. As in AB, most of the credit lives in the setup, so a correct integrand and correct limits matter more than the final arithmetic.

Position: 8 of 10 units
Published weighting: 6–9% in the official framework
Topics covered: 6 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 BC. 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 applied contexts.
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 either variable, including multi-piece regions.
Volumes with known cross sections: Integrating cross-sectional area for square, triangular, and semicircular slices.
Volumes of revolution: Disk and washer setups about horizontal and vertical lines, including non-axis lines.
Arc length: Computing the length of a curve given in function form, including setting up the integral only.

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 the integral, every time.
Separate displacement from total distance explicitly; the distinction is a standard trap.
Practice problems that ask only for the setup, which is common and rewards careful notation.

Where this fits

Unit 8: Applications of Integration is 8 of 10 units in the AP Calculus BC guide, and it carries a published weighting of 6–9% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus BC 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) = sin(x) on the interval [0, π]?
1. A2/π
2. B1/π
3. Cπ/2
4. D0
Show answer
Answer: A. 2/π
The average value is (1/π) times ∫ from 0 to π of sin(x) dx. The integral equals -cos(π) + cos(0) = 2, so the average value is 2/π ≈ 0.64. Answering 0 uses the average over a full period, but on [0, π] the sine is never negative.
Key insight: Average value divides the accumulated area by the width of the interval.
Question 2Foundational
What is the area of the region bounded by y = e^x, the x-axis, x = 0, and x = 1?
1. Ae - 1
2. Be
3. Ce + 1
4. D1
Show answer
Answer: A. e - 1
The area is ∫ from 0 to 1 of e^x dx = e^1 - e^0 = e - 1 ≈ 1.718. Answering e forgets to subtract the value at the lower limit, since e^0 = 1 rather than 0.
Key insight: e^0 = 1, so the lower limit of an exponential integral almost never contributes zero.
Question 3Intermediate
Which integral gives the length of the curve y = x^2 from x = 0 to x = 3?
1. A∫ from 0 to 3 of √(1 + 4x^2) dx
2. B∫ from 0 to 3 of √(1 + x^4) dx
3. C∫ from 0 to 3 of √(1 + 2x) dx
4. D∫ from 0 to 3 of (1 + 4x^2) dx
Show answer
Answer: A. ∫ from 0 to 3 of √(1 + 4x^2) dx
Arc length is ∫ √(1 + (dy/dx)^2) dx. Here dy/dx = 2x, so (dy/dx)^2 = 4x^2 and the integrand is √(1 + 4x^2). Squaring the function instead of its derivative produces √(1 + x^4), and forgetting to square 2x produces √(1 + 2x).
Key insight: Arc length squares the derivative, not the function.
Question 4Intermediate
The region enclosed by y = √x and y = x is revolved about the x-axis. What is the volume of the resulting solid?
1. Aπ/6
2. Bπ/3
3. Cπ/2
4. D2π/15
Show answer
Answer: A. π/6
The curves meet at x = 0 and x = 1, and √x lies above x on that interval, so the washer method gives π∫ from 0 to 1 of (x - x^2) dx = π(1/2 - 1/3) = π/6. Squaring √x gives x, which is why the integrand is a difference of a linear and a quadratic term.
Key insight: Identify which curve is farther from the axis of revolution before assigning outer and inner radii.
Question 5Advanced
What is the length of the curve y = (2/3)x^(3/2) from x = 0 to x = 3?
1. A14/3
2. B7/3
3. C8/3
4. D16/3
Show answer
Answer: A. 14/3
Here dy/dx = x^(1/2), so 1 + (dy/dx)^2 = 1 + x and the length is ∫ from 0 to 3 of √(1 + x) dx = (2/3)(1 + x)^(3/2) evaluated from 0 to 3 = (2/3)(8 - 1) = 14/3. Forgetting the lower-limit term 1 gives 16/3.
Key insight: This curve is chosen so that 1 + (y')^2 is a perfect setup for the power rule; check for that structure first.

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Unit 8: Applications of Integration