Application of Definite Integration

2∫₀² (4 - x²) dx, ∫₋₂² (4 - x²) dx

The curves intersect at x = ±2. The line y = 4 is above the parabola y = x² in this interval. By symmetry, we can use 2∫₀² (4 - x²) dx or directly ∫₋₂² (4 - x²) dx.

12π square units

For ellipse x²/a² + y²/b² = 1, area = πab. Here a² = 9 so a = 3, and b² = 16 so b = 4. Area = π × 3 × 4 = 12π square units.

4 square units

Area = ∫₀² x³ dx = [x⁴/4]₀² = 16/4 - 0 = 4 square units. The antiderivative of x³ is x⁴/4.

∫₁⁴ 2√y dy

From x² = 4y, we get x = 2√y (taking positive root in first quadrant). Area = ∫₁⁴ x dy = ∫₁⁴ 2√y dy.