Chapter 8 Solutions — Application of Integrals

The ellipse is x²/16 + y²/9 = 1, giving a = 4 and b = 3.

From the equation: y = (3/4)√(16 - x²) (taking positive value in the first quadrant).

Since the ellipse is symmetrical about both axes, the total area is 4 times the area in the first quadrant:

Area = 4 ∫₀⁴ y dx = 4 ∫₀⁴ (3/4)√(16 - x²) dx = 3 ∫₀⁴ √(16 - x²) dx

Using the standard formula ∫√(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a) + C, with a = 4:

Area = 3 [x/2 · √(16 - x²) + 8 sin⁻¹(x/4)]₀⁴

At x = 4: (4/2)·√(16-16) + 8 sin⁻¹(1) = 0 + 8·(π/2) = 4π

At x = 0: (0/2)·√16 + 8 sin⁻¹(0) = 0

Area = 3 [4π - 0] = 12π

(i) y = x², bounded by x = 1, x = 2, and the x-axis.

Since y = x² ≥ 0 for all x, the curve lies above the x-axis on [1, 2].

Area = ∫₁² x² dx = [x³/3]₁²

= 8/3 - 1/3

= 7/3

(ii) y = x⁴, bounded by x = 1, x = 5, and the x-axis.

Since y = x⁴ ≥ 0 for all x, the curve lies above the x-axis on [1, 5].

Area = ∫₁⁵ x⁴ dx = [x⁵/5]₁⁵

= 5⁵/5 - 1⁵/5

= 3125/5 - 1/5

= 3124/5

The ellipse is x²/4 + y²/9 = 1, giving a = 2 and b = 3.

From the equation: y = (3/2)√(4 - x²) (taking positive value in the first quadrant).

Since the ellipse is symmetrical about both axes, the total area is 4 times the area in the first quadrant:

Area = 4 ∫₀² y dx = 4 ∫₀² (3/2)√(4 - x²) dx = 6 ∫₀² √(4 - x²) dx

Using the standard formula ∫√(a² - x²) dx = (x/2)√(a² - x²) + (a²/2) sin⁻¹(x/a) + C, with a = 2:

Area = 6 [x/2 · √(4 - x²) + 2 sin⁻¹(x/2)]₀²

At x = 2: (2/2)·√(4-4) + 2 sin⁻¹(1) = 0 + 2·(π/2) = π

At x = 0: (0/2)·√4 + 2 sin⁻¹(0) = 0

Area = 6 [π - 0] = 6π