Topic Review — Introduction to Integration — Solutions

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This review covers all three lessons in Introduction to Integration: anti-differentiation and indefinite integrals, integration rules for composite functions, and displacement and velocity applications.

Review Questions

1. Find ∫ (5x⁴ − 3x² + 2) dx. ▶ Show Answer
   = 5 · x⁵/5 − 3 · x³/3 + 2x + c = x⁵ − x³ + 2x + c
2. Find ∫ (4e^x − 6 cos x + 5/x) dx. ▶ Show Answer
   = 4e^x − 6 sin x + 5 ln|x| + c
3. Find ∫ (3x + 2)⁴ dx. ▶ Show Answer
   a = 3, n = 4: = (3x+2)⁵/[3×5] + c = (3x+2)⁵/15 + c
4. Find ∫ e^5−2x dx. ▶ Show Answer
   a = −2: = (1/(−2)) e^5−2x + c = −(1/2)e^5−2x + c
5. Find ∫ sin(3x − 1) dx. ▶ Show Answer
   a = 3: = −(1/3) cos(3x−1) + c = −(1/3)cos(3x − 1) + c
6. Find ∫ 4/(1 − 2x) dx. ▶ Show Answer
   = 4 × (1/(−2)) ln|1−2x| + c = −2 ln|1 − 2x| + c
7. Given f′(x) = 6x² − 4 sin x and f(0) = 1, find f(x). ▶ Show Answer
   f(x) = 2x³ + 4 cos x + c
   f(0) = 0 + 4 + c = 1 ⇒ c = −3
   f(x) = 2x³ + 4 cos x − 3
8. A particle has velocity v(t) = 6t − 12 m/s and initial position s(0) = 2 m.
   (a) Find s(t).   (b) Find the position when the particle is at rest. ▶ Show Answer
   (a) s(t) = 3t² − 12t + c; s(0) = c = 2. s(t) = 3t² − 12t + 2
   (b) v = 0: t = 2. s(2) = 12 − 24 + 2 = −10 m
9. Find ∫ [cos(x/3) − 2e^−x] dx. ▶ Show Answer
   ∫ cos(x/3) dx = (1/(1/3)) sin(x/3) = 3 sin(x/3)
   ∫ −2e^−x dx = −2 × (−1) e^−x = 2e^−x
   Answer: 3 sin(x/3) + 2e^−x + c
10. The gradient of a curve is dy/dx = (2x − 1)³ and it passes through (1, 2). Find y. ▶ Show Answer
    y = (2x−1)⁴/[2×4] + c = (2x−1)⁴/8 + c
    y(1) = (1)⁴/8 + c = 2 ⇒ c = 15/8
    y = (2x−1)⁴/8 + 15/8
11. A particle has velocity v(t) = t² − 6t + 8 m/s, starting at s = 0 when t = 0.
    (a) When is the particle at rest?   (b) Find total distance from t = 0 to t = 5. ▶ Show Answer
    (a) v = 0: (t−2)(t−4) = 0 ⇒ t = 2 s and t = 4 s
    (b) s(t) = t³/3 − 3t² + 8t
    s(2) = 8/3−12+16 = 8/3+4 = 20/3
    s(4) = 64/3−48+32 = 64/3−16 = 16/3
    s(5) = 125/3−75+40 = 125/3−35 = 20/3
    Dist = 20/3 + |16/3−20/3| + |20/3−16/3| = 20/3+4/3+4/3 = 28/3 ≈ 9.33 m
12. Explain why ∫ x⁻¹ dx ≠ x⁰/0 + c, and state the correct result. ▶ Show Answer
    The power rule formula requires dividing by (n+1). When n = −1, this means dividing by 0, which is undefined. The power rule cannot be applied to this case.
    The correct result is ∫ 1/x dx = ln|x| + c, which follows from d/dx(ln|x|) = 1/x.
13. An object moves with acceleration a(t) = 12t − 6 m/s². At t = 0, velocity is 4 m/s and position is −1 m. Find s(t) and the position when v = 0 (for t > 0). ▶ Show Answer
    v(t) = ∫(12t−6)dt = 6t²−6t + c; v(0)=4 ⇒ c=4. v(t) = 6t²−6t+4
    s(t) = ∫(6t²−6t+4)dt = 2t³−3t²+4t+k; s(0)=−1 ⇒ k=−1
    s(t) = 2t³ − 3t² + 4t − 1
    v = 0: 6t²−6t+4=0 ⇒ 3t²−3t+2=0
    Discriminant = 9−24 < 0 ⇒ v is never zero (always positive). Particle never at rest.
14. A ball is launched upward with velocity 20 m/s from a height of 1 m above ground. Taking upward as positive, a = −10 m/s².
    (a) Find v(t) and s(t). (b) Maximum height. (c) Time when it hits the ground. ▶ Show Answer
    (a) v(t) = −10t + 20; s(t) = −5t² + 20t + 1
    (b) v = 0 at t = 2: s(2) = −20+40+1 = 21 m
    (c) s = 0: −5t²+20t+1 = 0 ⇒ t = (−20−√(400+20))/(−10) = (20+√420)/10 = 2 + √(4.2) ≈ 4.05 s
15. Given f′(x) = 3/(2x + 1) for x > −1/2 and f(0) = 5, find f(3). Give an exact answer. ▶ Show Answer
    f(x) = ∫ 3/(2x+1) dx = 3 × (1/2) ln(2x+1) + c = (3/2)ln(2x+1) + c
    f(0) = 0 + c = 5 ⇒ c = 5
    f(3) = (3/2)ln(7) + 5 = (3/2)ln 7 + 5