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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Each solution identifies the limits, the top and bottom or the radii, and then evaluates the integral. If your answer differs, check which curve is on top and whether the radii were measured from the correct axis of revolution.
Standards: AP Calculus AB Unit 8 · Level: Grade 12 AP Calculus AB
Problem 1 Average value of a polynomial
∫25 (4x2 − 3x) dx = [4x3/3 − 3x2/2] from 2 to 5 = (500/3 − 75/2) − (32/3 − 6) = 124.5. Divide by the width 3: 83/2 = 41.5.
Problem 2 Average value of sine
∫0π sin x dx = [−cos x] from 0 to π = 1 − (−1) = 2. The average value is 2/π, or 2/π ≈ 0.64.
Problem 3 Position from velocity
s(3) = s(0) + ∫03 (4t − t2) dt = 5 + [2t2 − t3/3] from 0 to 3 = 5 + (18 − 9) = 14 meters.
Problem 4 Net change and total distance
An antiderivative is s(t) = t3/3 − 7t2/2 + 10t. Net change: s(6) − s(0) = 72 − 126 + 60 = 6 m. The velocity (t − 2)(t − 5) is zero at 2 and 5. Positions: s(2) = 26/3, s(5) = 25/6, s(6) = 6. Path length: 26/3 + (26/3 − 25/6) + (6 − 25/6) = 52/6 + 27/6 + 11/6 = 15 m.
Problem 5 Area between two parabolas
Set 8 − x2 = x2: x = ±2. On [−2, 2] the downward parabola is on top. Area = ∫−22 (8 − 2x2) dx = [8x − 2x3/3] from −2 to 2 = 2(16 − 16/3) = 64/3.
Problem 6 Area integrated with respect to y
Set y2 = 6 − y: y2 + y − 6 = 0, so y = 2 or y = −3. The line is on the right. Area = ∫−32 (6 − y − y2) dy = [6y − y2/2 − y3/3] from −3 to 2 = (22/3) − (−27/2) = 125/6.
Problem 7 Square cross sections
The curves meet at x = 0 and x = 2, with the line above. The edge is 2x − x2, so A(x) = (2x − x2)2 = 4x2 − 4x3 + x4. Then V = ∫02 (4x2 − 4x3 + x4) dx = [4x3/3 − x4 + x5/5] from 0 to 2 = 32/3 − 16 + 32/5 = 16/15.
Problem 8 Semicircular cross sections
The diameter is √x, so the radius is √x/2 and the slice area is (π/2)(x/4) = πx/8. Then V = ∫09 πx/8 dx = (π/8)(81/2) = 81π/16.
Problem 9 Disc method
Disc radius √(x + 1), so V = π∫03 (x + 1) dx = π[x2/2 + x] from 0 to 3 = π(9/2 + 3) = 15π/2.
Problem 10 Washer method about the x-axis
The curves meet at x = 0 and x = 1, with √x above, so the outer radius is √x and the inner radius is x2. V = π∫01 (x − x4) dx = π(1/2 − 1/5) = 3π/10.
Problem 11 Revolving about a horizontal line
Measure from the line y = −1: the outer radius is x2 + 1 and the inner radius is 0 + 1 = 1. V = π∫01 [(x2 + 1)2 − 12] dx = π∫01 (x4 + 2x2) dx = π(1/5 + 2/3) = 13π/15.
Problem 12 Revolving about the y-axis
Solve for x: x = √y, so the disc radius is √y. V = π∫04 y dy = π(16/2) = 8π.
Problem 13 Tank with water flowing in and out
The amount is A(t) = 120 + ∫0t (2s + 4 − 10) ds = 120 + t2 − 6t. So A(10) = 120 + 100 − 60 = 160 gallons. The net rate 2t − 6 changes from negative to positive at t = 3, so the least amount is at t = 3: A(3) = 111 gallons.
Problem 14 Area with three intersection points
Set x3 = 4x: x = −2, 0, 2. On [0, 2] the line is on top; on [−2, 0] the cubic is on top, and by symmetry the two areas are equal. Area = 2∫02 (4x − x3) dx = 2[2x2 − x4/4] from 0 to 2 = 2(8 − 4) = 8.
Problem 15 Area where curves cross
The curves cross at x = π/4. On [0, π/4], cos is above sin; on [π/4, π/2], sin is above cos. Area = ∫0π/4 (cos x − sin x) dx + ∫π/4π/2 (sin x − cos x) dx. Each piece equals √2 − 1. The total is 2√2 − 2 ≈ 0.83.
Problem 16 Revolving about a vertical line
Since y = √x means x = y2, at height y the region runs from x = y2 to x = 4, so the disc radius (the distance to the axis x = 4) is 4 − y2, and y runs from 0 to 2. V = π∫02 (4 − y2)2 dy = π∫02 (16 − 8y2 + y4) dy = π[16y − 8y3/3 + y5/5] from 0 to 2 = π(32 − 64/3 + 32/5) = 256π/15.
Problem 17 Greenhouse temperature
Mean = (1/12)∫012 (58 + 12 sin(πh/12)) dh. The integral of sin(πh/12) over [0, 12] is [−(12/π)cos(πh/12)] = 24/π. So the mean is 58 + (12/12)(24/π) = 58 + 24/π ≈ 65.6°F.
Problem 18 Volume of a tent
A leg is 9 − x2, so the triangle has area (1/2)(9 − x2)2. V = (1/2)∫−33 (81 − 18x2 + x4) dx = ∫03 (81 − 18x2 + x4) dx = [81x − 6x3 + x5/5] from 0 to 3 = 243 − 162 + 48.6 = 129.6 ft3 (648/5).
Problem 19 Area and volume from one region
Area = ∫1e (1/x) dx = ln e − ln 1 = 1. Volume = π∫1e (1/x2) dx = π[−1/x] from 1 to e = π(1 − 1/e) ≈ 1.99.
Problem 20 Mean Value Theorem for integrals
∫14 (x2 + 1) dx = [ x3/3 + x] from 1 to 4 = (64/3 + 4) − (1/3 + 1) = 24. The average value is 24/3 = 8. Solve c2 + 1 = 8: c = √7 ≈ 2.65, which lies in [1, 4].
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