Solutions — Applications of Integration

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1. Show Solution
   Displacement = ∫₀⁴ (4 − 2t) dt = [4t − t²]₀⁴
   = (16 − 16) − 0 = 0 m
   
   The particle returns to its starting position. Note: v(t) = 0 at t = 2; the particle moves forward for 2 s then back for 2 s.
2. Show Solution
   Volume = ∫₀⁴ (6 + t) dt = [6t + t²/2]₀⁴
   = (24 + 8) − 0 = 32 litres
3. Show Solution
   f̄ = (1/(3−0)) ∫₀³ (x²+1) dx = (1/3)[x³/3 + x]₀³
   = (1/3)(9 + 3) = (1/3)(12) = 4
4. Show Solution
   v(t) = 3t² − 6t = 3t(t − 2) = 0 at t = 0, 2
   Sign check: v(1) = 3 − 6 = −3 < 0; v(2.5) = 3(6.25) − 15 = 3.75 > 0
   
   D₁ = |∫₀² (3t²−6t) dt| = |[t³−3t²]₀²| = |8−12| = 4
   D₂ = ∫₂³ (3t²−6t) dt = [t³−3t²]₂³ = (27−27)−(8−12) = 0+4 = 4
   Total distance = 4 + 4 = 8 m
5. Show Solution
   v(t) = t² − 4t + 3 = (t−1)(t−3) = 0 at t = 1, 3
   Sign check: v(0.5) = 0.25−2+3 = 1.25 > 0; v(2) = 4−8+3 = −1 < 0; v(3.5) = 12.25−14+3 = 1.25 > 0
   
   (a) Displacement = ∫₀⁴ (t²−4t+3) dt = [t³/3−2t²+3t]₀⁴
   = (64/3−32+12) − 0 = 64/3−20 = (64−60)/3 = 4/3 m
   
   (b) Total distance:
   I₁ = [t³/3−2t²+3t]₀¹ = 1/3−2+3 = 4/3
   I₂ = [t³/3−2t²+3t]₁³ = (9−18+9)−(4/3) = 0−4/3 = −4/3, so |I₂| = 4/3
   I₃ = [t³/3−2t²+3t]₃⁴ = (64/3−32+12)−0 = 4/3
   Total = 4/3 + 4/3 + 4/3 = 4 m
6. Show Solution
   P(5) = P(0) + ∫₀⁵ 200e^0.1t dt
   = 3000 + [200e^0.1t/0.1]₀⁵
   = 3000 + 2000[e^0.5 − e⁰]
   = 3000 + 2000(e^0.5 − 1)
   = 3000 + 2000(1.6487 − 1)
   = 3000 + 1297.4 ≈ 4297 cells
7. Show Solution
   v(t) = v(0) + ∫₀^t (6s−2) ds = 1 + [3s²−2s]₀^t = 1 + 3t² − 2t
   So v(t) = 3t² − 2t + 1
   
   Check if v(t) changes sign: discriminant of 3t²−2t+1 is 4−12 = −8 < 0, so v(t) > 0 for all t (no direction change).
   
   Distance = ∫₀² (3t²−2t+1) dt = [t³−t²+t]₀² = 8−4+2 = 6 m
8. Show Solution
   Average value = (1/π) ∫₀^π sin t dt = (1/π)[−cos t]₀^π
   = (1/π)[−cos(π)+cos(0)] = (1/π)[1+1] = 2/π ≈ 0.637
   
   If v(t) is a velocity function, the average value 2/π m/s is the average velocity. Since sin t ≥ 0 on [0,π], the average velocity equals the average speed here, and displacement = 2 m = average velocity × time interval (π s).
9. Show Solution
   Trapezoidal rule with w = 2, values: 0, 8, 14, 18, 20
   Distance ≈ (2/2)[0 + 2(8) + 2(14) + 2(18) + 20]
   = 1[0 + 16 + 28 + 36 + 20] = 100 m
   
   Average speed = total distance / time = 100/8 = 12.5 m/s
10. Show Solution
    Total energy = ∫₀¹² 500 sin(πt/12) dt
    = 500 [−12/π cos(πt/12)]₀¹²
    = −6000/π [cos(π) − cos(0)]
    = −6000/π [−1 − 1]
    = −6000/π × (−2) = 12000/π ≈ 3820 Wh
    
    Average power = (1/12) ∫₀¹² 500 sin(πt/12) dt = (1/12) × 12000/π = 1000/π ≈ 318 W