Solutions — Anti-differentiation and Indefinite Integrals

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1. Show Solution
   ∫ (x³ + 2x − 7) dx
   = x⁴/4 + 2 · x²/2 − 7x + c
   = x⁴/4 + x² − 7x + c
   
   Check: d/dx(x⁴/4 + x² − 7x + c) = x³ + 2x − 7 ✓
2. Show Solution
   ∫ (4e^x + 3 sin x) dx
   = 4 ∫ e^x dx + 3 ∫ sin x dx
   = 4e^x + 3(−cos x) + c
   = 4e^x − 3 cos x + c
   
   Check: d/dx(4e^x − 3 cos x) = 4e^x + 3 sin x ✓
3. Show Solution
   ∫ (5 cos x − 2/x) dx
   = 5 ∫ cos x dx − 2 ∫ (1/x) dx
   = 5 sin x − 2 ln|x| + c
   = 5 sin x − 2 ln|x| + c
4. Show Solution
   ∫ (6x² − 4x + 1/x) dx
   = 6 · x³/3 − 4 · x²/2 + ln|x| + c
   = 2x³ − 2x² + ln|x| + c
5. Show Solution
   ∫ (x⁴ − x⁻²) dx
   ∫ x⁴ dx = x⁵/5
   ∫ x⁻² dx = x⁻¹/(−1) = −1/x
   Answer: x⁵/5 − (−1/x) + c = x⁵/5 + 1/x + c
   
   Check: d/dx(x⁵/5 + 1/x) = x⁴ − x⁻² = x⁴ − 1/x² ✓
6. Show Solution
   Step 1 — Integrate:
   f(x) = ∫(3x² − 6x + 2) dx = x³ − 3x² + 2x + c
   
   Step 2 — Apply initial condition f(1) = 4:
   (1)³ − 3(1)² + 2(1) + c = 4
   1 − 3 + 2 + c = 4
   0 + c = 4 ⇒ c = 4
   
   f(x) = x³ − 3x² + 2x + 4
7. Show Solution
   Step 1 — Integrate:
   f(x) = ∫(sin x + e^x) dx = −cos x + e^x + c
   
   Step 2 — Apply initial condition f(0) = 2:
   −cos(0) + e⁰ + c = 2
   −1 + 1 + c = 2
   c = 2
   
   f(x) = −cos x + e^x + 2
8. Show Solution
   The power rule ∫ xⁿ dx = xⁿ⁺¹/(n+1) + c requires dividing by (n + 1). When n = −1, this gives division by 0, which is undefined — so the formula cannot be used.
   
   The correct result is: ∫ x⁻¹ dx = ∫ 1/x dx = ln|x| + c
   
   Verification: d/dx(ln|x|) = 1/x = x⁻¹ ✓
9. Show Solution
   Rewrite 3/x² = 3x⁻².
   
   Step 1 — Integrate:
   f(x) = ∫(4x − 3x⁻²) dx
   = 4 · x²/2 − 3 · x⁻¹/(−1) + c
   = 2x² + 3/x + c
   
   Step 2 — Apply initial condition f(1) = 6:
   2(1)² + 3/1 + c = 6
   2 + 3 + c = 6
   c = 1
   
   f(x) = 2x² + 3/x + 1
10. Show Solution
    Finding v(t):
    v(t) = ∫ a(t) dt = ∫(6t − 4) dt = 3t² − 4t + c
    v(0) = 0 − 0 + c = −2 ⇒ c = −2
    v(t) = 3t² − 4t − 2 m/s
    
    Finding s(t):
    s(t) = ∫ v(t) dt = ∫(3t² − 4t − 2) dt = t³ − 2t² − 2t + k
    s(0) = 0 − 0 − 0 + k = 5 ⇒ k = 5
    s(t) = t³ − 2t² − 2t + 5 m