Module 7 Practice Problems: Applications of Integration
Instructions: Work through each problem on paper first, then reveal the solution to check your work.
Problem 1
Find the area between y = x² and y = 2x.
Solution
Intersections: x² = 2x, x(x − 2) = 0, x = 0, 2. Top: 2x. A = ∫02 (2x − x²) dx = [x² − x³/3]02 = 4 − 8/3 = 4/3.
Problem 2
Find the area enclosed by y = √x and y = x/2.
Solution
√x = x/2, so x = x²/4, 4x = x², x = 0, 4. A = ∫04 (√x − x/2) dx = [(2/3)x3/2 − x²/4]04 = (16/3 − 4) = 4/3.
Problem 3
Find the volume when y = x² + 1 on [0, 2] is revolved about the x-axis (disk method).
Solution
V = π ∫02 (x² + 1)² dx = π ∫02 (x4 + 2x² + 1) dx = π[x5/5 + 2x³/3 + x]02 = π(32/5 + 16/3 + 2) = 206π/15.
Problem 4
Find the volume when the region between y = x and y = x² on [0, 1] is revolved about the x-axis (washer).
Solution
V = π ∫01 (x² − x4) dx = π[x³/3 − x5/5]01 = π(1/3 − 1/5) = 2π/15.
Problem 5
Use the shell method to find the volume when y = 1 − x² (y ≥ 0) is revolved about the y-axis.
Solution
y = 0 when x = 1. V = 2π ∫01 x(1 − x²) dx = 2π ∫01 (x − x³) dx = 2π[x²/2 − x4/4]01 = 2π(1/4) = π/2.
Problem 6
Use the shell method to find the volume when y = x(3 − x) on [0, 3] is revolved about the y-axis.
Solution
V = 2π ∫03 x · x(3 − x) dx = 2π ∫03 (3x² − x³) dx = 2π[x³ − x4/4]03 = 2π(27 − 81/4) = 2π(27/4) = 27π/2.
Problem 7
Find the volume when y = √x on [0, 4] is revolved about y = −2 (washer).
Solution
Outer R = √x + 2, inner r = 2. V = π ∫04 [(√x + 2)² − 4] dx = π ∫04 (x + 4√x) dx = π[x²/2 + (8/3)x3/2]04 = π(8 + 64/3) = 88π/3.
Problem 8
Find the average value of f(x) = x³ on [0, 2].
Solution
favg = (1/2) ∫02 x³ dx = (1/2)[x4/4]02 = (1/2)(4) = 2.
Problem 9
Find the average value of f(x) = 1/x on [1, e²].
Solution
favg = (1/(e² − 1)) ∫1e² (1/x) dx = (1/(e² − 1))[ln x]1e² = (1/(e² − 1))(2) = 2/(e² − 1) ≈ 0.313.
Problem 10
Find the value c guaranteed by MVT-I for f(x) = x² + 1 on [0, 3].
Solution
favg = (1/3) ∫03 (x² + 1) dx = (1/3)[x³/3 + x]03 = (1/3)(12) = 4. Then c² + 1 = 4, c² = 3, c = √3 ≈ 1.732.