Module 6 Practice Problems: Integration
Instructions: Work through each problem on paper first, then reveal the solution to check your work.
Problem 1
Find the antiderivative: ∫ (3x² − 4x + 5) dx.
Solution
∫ (3x² − 4x + 5) dx = x³ − 2x² + 5x + C.
Problem 2
Evaluate: ∫ (x1/2 + x−1/2) dx.
Solution
= (2/3)x3/2 + 2x1/2 + C = (2/3)x√x + 2√x + C.
Problem 3
Approximate ∫04 x² dx using a right Riemann sum with n = 4.
Solution
Δx = 1. Right endpoints: 1, 2, 3, 4. R4 = 1(1 + 4 + 9 + 16) = 30. (Exact value = 64/3 ≈ 21.33, so the right sum overestimates for this increasing function.)
Problem 4
Evaluate: ∫14 (2x − 3/√x) dx.
Solution
= [x² − 6√x]14 = (16 − 12) − (1 − 6) = 4 − (−5) = 9.
Problem 5
Evaluate: ∫0π (sin x + cos x) dx.
Solution
= [−cos x + sin x]0π = (−(−1) + 0) − (−1 + 0) = 1 + 1 = 2.
Problem 6
Find d/dx [∫3x ln(t² + 1) dt].
Solution
By FTC Part 2: ln(x² + 1).
Problem 7
Find d/dx [∫1x² e−t dt].
Solution
FTC Part 2 + chain rule: e−x² · 2x = 2x e−x².
Problem 8
A particle has velocity v(t) = t − 3 on [0, 5]. Find the displacement.
Solution
∫05 (t − 3) dt = [t²/2 − 3t]05 = (25/2 − 15) − 0 = −5/2. Displacement = −5/2 (2.5 units in the negative direction).
Problem 9
For the same particle in Problem 8, find the total distance traveled on [0, 5].
Solution
v = 0 at t = 3. ∫03 |t − 3| dt + ∫35 (t − 3) dt = ∫03 (3 − t) dt + ∫35 (t − 3) dt = [3t − t²/2]03 + [t²/2 − 3t]35 = 9/2 + 2 = 13/2.
Problem 10
Given ∫08 f(x) dx = 12 and ∫03 f(x) dx = 5, find ∫38 [2f(x) + 3] dx.
Solution
∫38 f(x) dx = 12 − 5 = 7. Then ∫38 [2f(x) + 3] dx = 2(7) + 3(5) = 14 + 15 = 29.