Point of intersection of y = √2x and y = 4 - x = √2x = 4 - x = 2x = 16 + x2 -8x = x2 - 10x + 16 = 0 x = 2,8 but at x = 8, y is negative as y = 4 - x = y = -4 at x = 8 so, for y ≥ 0, x = 2. at x = 2, y = 4-2 = 2 so point of intersection (2,2) Now, = ∫02√2xdx+∫24(4−x)dx = [