AP Calc AB Applications of Integration — Questions and Answers
Question 1: What is the average value of f(x) = sin x on the interval [0, π]?
- 1/π
- 2
- 2/π (Correct answer)
- π/2
Correct answer: 2/π
The average value is (1/(π−0))∫₀^π sin x dx = (1/π)(2) = 2/π.
The average value of a continuous function f on [a, b] is (1/(b−a))∫ₐᵇ f(x) dx. Here ∫₀^π sin x dx = [−cos x]₀^π = 1 − (−1) = 2. The interval has length π, so the average value is 2/π. Choice 2 is the integral itself, which is the area under the curve. It is not the average height.
Question 2: A particle moves along a line with velocity v(t) = t² − 4t + 3 for 0 ≤ t ≤ 3. What is the total distance the particle travels on this interval?
- 0
- 8/3 (Correct answer)
- 4/3
- 3
Correct answer: 8/3
v(t) = (t−1)(t−3) is positive on (0,1) and negative on (1,3). The distance is ∫₀¹ v dt − ∫₁³ v dt = 4/3 + 4/3 = 8/3.
Displacement is ∫v dt, but total distance is ∫|v| dt. Here the displacement over [0,3] is 0, because the forward and backward motion cancel. Factoring v(t) = (t−1)(t−3) shows the particle moves forward on [0,1] and backward on [1,3]. The antiderivative is t³/3 − 2t² + 3t. It gives ∫₀¹ v dt = 4/3 and ∫₁³ v dt = −4/3. Adding the absolute values gives a distance of 8/3.
Question 3: What is the area of the region enclosed by the graphs of y = x and y = x²?
- 1/6 (Correct answer)
- 1/3
- 1/2
- 1/12
Correct answer: 1/6
The graphs intersect at x = 0 and x = 1, and y = x lies above y = x² between them. The area is ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6.
Setting x = x² gives x = 0 or x = 1, so these are the limits of integration. On (0,1), x > x², so the line is the upper curve. The area is ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6. Choice 1/2 comes from integrating only the line. Choice 1/3 comes from integrating only the parabola.
Question 4: The region bounded by y = √x, the x-axis, and the line x = 4 is revolved about the x-axis. What is the volume of the resulting solid?
- 4π
- 16π
- 32π/5
- 8π (Correct answer)
Correct answer: 8π
By the disk method, V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π(16/2) = 8π.
Revolving about the x-axis gives disks of radius √x. The disk method is V = π∫ₐᵇ [R(x)]² dx, so V = π∫₀⁴ x dx = π[x²/2]₀⁴ = 8π. Choice 32π/5 comes from integrating (√x)⁴. Choice 16π comes from forgetting the 1/2 factor in the antiderivative.
Question 5: The region enclosed by y = x and y = x² is revolved about the x-axis. What is the volume of the solid generated?
- π/6
- 2π/15 (Correct answer)
- π/15
- 4π/15
Correct answer: 2π/15
Use washers with outer radius x and inner radius x²: V = π∫₀¹ (x² − x⁴) dx = π(1/3 − 1/5) = 2π/15.
The solid has a hole, so use the washer method: V = π∫(R² − r²) dx. On [0,1] the outer radius is R = x and the inner radius is r = x². This gives V = π∫₀¹ (x² − x⁴) dx = π[x³/3 − x⁵/5]₀¹ = π(1/3 − 1/5) = 2π/15. Squaring the difference, (x − x²)², is a common error. It does not account for the hole correctly.
Question 6: The base of a solid is the region in the first quadrant bounded by y = 1 − x² and the coordinate axes. Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?
- 2/3
- 4/15
- 8/15 (Correct answer)
- 1/5
Correct answer: 8/15
Each square has side 1 − x², so V = ∫₀¹ (1 − x²)² dx = ∫₀¹ (1 − 2x² + x⁴) dx = 1 − 2/3 + 1/5 = 8/15.
For known cross-sections, the volume is V = ∫ A(x) dx, where A(x) is the cross-sectional area. The side of each square runs from the x-axis up to the curve, so s = 1 − x² and A(x) = (1 − x²)². Expanding gives 1 − 2x² + x⁴, and the antiderivative is x − 2x³/3 + x⁵/5. Evaluating from 0 to 1 gives 1 − 2/3 + 1/5 = 8/15. Choice 2/3 is the area of the base region, which is not the volume.
Question 7: A particle moves along the x-axis with velocity v(t) = 6t for t ≥ 0. If its position at t = 0 is x = 2, what is its position at t = 2?
- 14 (Correct answer)
- 12
- 24
- 10
Correct answer: 14
x(2) = x(0) + ∫₀² 6t dt = 2 + 3t²|₀² = 2 + 12 = 14.
The Fundamental Theorem of Calculus gives x(b) = x(a) + ∫ₐᵇ v(t) dt. Here ∫₀² 6t dt = [3t²]₀² = 12, which is the displacement. Adding the initial position of 2 gives 14. Choice 12 forgets the initial position. It gives the displacement, not the final location.
Question 8: Water is in a tank at the rate of change V′(t) = 6 − 2t gallons per minute for 0 ≤ t ≤ 6, where V(t) is the volume of water in gallons. If the tank holds 20 gallons at t = 0, what is the maximum volume of water in the tank on this interval?
- 20
- 18
- 38
- 29 (Correct answer)
Correct answer: 29
V′ changes from positive to negative at t = 3, so the maximum is at t = 3. V(3) = 20 + ∫₀³ (6 − 2t) dt = 20 + 9 = 29.
V′(t) = 6 − 2t is positive for t < 3 and negative for t > 3, so V increases and then decreases. The maximum is at t = 3. The net change is ∫₀³ (6 − 2t) dt = [6t − t²]₀³ = 18 − 9 = 9. Adding the initial 20 gallons gives V(3) = 29. Using the whole interval would give 20 + 0 = 20 gallons, because the net change over [0,6] is zero.
Question 9: What is the area of the region bounded by the curves x = y² and x = y + 2?
- 9/2 (Correct answer)
- 9
- 3/2
- 27/2
Correct answer: 9/2
The curves meet at y = −1 and y = 2, and x = y + 2 is to the right of x = y². The area is ∫₋₁² (y + 2 − y²) dy = 9/2.
Setting y² = y + 2 gives (y − 2)(y + 1) = 0, so y = −1 or y = 2. Because both curves are given as x in terms of y, integrate with respect to y. The right curve is x = y + 2. The area is ∫₋₁² (y + 2 − y²) dy = [y²/2 + 2y − y³/3]₋₁². This equals (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 + 7/6 = 27/6 = 9/2.
Question 10: Let f(x) = x² on the interval [0, 3]. At what value of c in (0, 3) does f(c) equal the average value of f on [0, 3]?
- 3
- √3 (Correct answer)
- 3/2
- 3√3
Correct answer: √3
The average value is (1/3)∫₀³ x² dx = (1/3)(9) = 3. Setting c² = 3 gives c = √3, which lies in (0, 3).
The Mean Value Theorem for Integrals says a continuous f on [a, b] takes its average value at some c in (a, b). First, ∫₀³ x² dx = 9, so the average value is 9/3 = 3. Next solve c² = 3 to get c = √3 ≈ 1.73, which is in the interval. The value 3 is the average value of f, not the input c where it occurs.
