Find the area of the region, (x,y): y² ≤ 4x, 4x² + 4y² ≤ 9 using method of integration.

The point of intersection of both the curves are (1/2, √2) and (1/2, -√2)
The required area is given by OABCO. It can be observed that area OABCO is symmetrical about x-axis. Therefore, Area OABCO = 2 × Area of OBC
Area OBCO = Area OMC + Area MBC = ∫₀¹⁄₂ 2√x dx + ∫¹⁄₂³/₂ √(9/4 - x²) dx
= ∫₀¹⁄₂ 2√x dx + ∫¹⁄₂³/₂ √(9/4 - x²) dx = 2[2x³/² / 3]₀¹⁄₂ + (1/2)[(9/4)sin⁻¹(2x/3) + (2x/3)√(9/4 - x²)]³/₂¹⁄₂ + c
= (4/3)(1/2)³/² + (1/2)[(9/4)sin⁻¹(1) + (1/2)√(9/4 - 1/4) - (9/4)sin⁻¹(1/3) - (1/3)√(9/4 - 1/4)]
= (4√2)/12 + (1/2)[(9π/8) + (√2/2) - (9/4)sin⁻¹(1/3) - (2√2)/6]
= (√2)/3 + (9π/16) + (√2/4) - (9/8)sin⁻¹(1/3) - (√2)/6
= (√2)/3 + (9π/16) + (√2)/4 - (9/8)sin⁻¹(1/3) - (√2)/6
= (2√2 + 3√2 - √2)/6 + (9π/16) - (9/8)sin⁻¹(1/3)
= (4√2)/6 + (9π/16) - (9/8)sin⁻¹(1/3)
= (2√2)/3 + (9π/16) - (9/8)sin⁻¹(1/3)