Unit 8 Practice Test

Solution:

1) Find intersection points: x³ = x → x(x² - 1) = 0 → x = -1, 0, 1

2) On [-1,0]: x³ ≥ x, On [0,1]: x ≥ x³

3) A = ∫_-1⁰ (x³ - x) dx + ∫₀¹ (x - x³) dx

4) = [x⁴/4 - x²/2]_-1⁰ + [x²/2 - x⁴/4]₀¹

5) = (0 - 0) - (1/4 - 1/2) + (1/2 - 1/4) - (0 - 0) = 1/4 + 1/4 = 1/2

Answer: 1/2

Solution:

1) Find intersection points: x² = 2x → x = 0, 2

2) On [0,2]: 2x ≥ x², so R(x) = 2x, r(x) = x²

3) V = π ∫₀² [(2x)² - (x²)²] dx = π ∫₀² (4x² - x⁴) dx

4) = π[4x³/3 - x⁵/5]₀² = π(32/3 - 32/5)

5) = π(160/15 - 96/15) = 64π/15

Answer: 64π/15

Solution:

1) Find spring constant: F = kx → 8 = k(0.04) → k = 200 N/m

2) W = ∫_0.04^0.08 200x dx = 200[x²/2]_0.04^0.08

3) = 100[(0.08)² - (0.04)²] = 100(0.0064 - 0.0016)

4) = 100(0.0048) = 0.48 J

Answer: 0.48 J

Solution:

1) Find intersection: x² = 4 → x = ±2

2) A(x) = (4 - x²)² = 16 - 8x² + x⁴

3) V = ∫_-2² (16 - 8x² + x⁴) dx = 2∫₀² (16 - 8x² + x⁴) dx

4) = 2[16x - 8x³/3 + x⁵/5]₀²

5) = 2(32 - 64/3 + 32/5) = 2(480/15 - 320/15 + 96/15) = 512/15

Answer: 512/15

Solution:

1) f_avg = (1/(3-0)) ∫₀³ x² dx

2) = (1/3) ∫₀³ x² dx = (1/3)[x³/3]₀³

3) = (1/3)(27/3) = (1/3)(9) = 3

Answer: 3

Solution:

1) Using shell method: V = 2π ∫₀² x(x²) dx

2) = 2π ∫₀² x³ dx = 2π[x⁴/4]₀²

3) = 2π(16/4) = 2π(4) = 8π

Answer: 8π