Let A ₁ be the bounded area enclosed by the curves y=x²+2, x+y=8 and y -axis that lies in the first quadrant. Let A ₂ be the bounded area enclosed by the curves y=x²+2, y²=x, x=2 , and y -axis that lies in the first quadrant. Then A ₁- A ₂ is equal to

Step-by-step solution

For A₁ : intersection of y = x^2 + 2 and x + y = 8 at x = 2, y = 6 A₁ = ₀^2 (8-x-(x^2+2)) dx = ₀^2 (6-x-x^2) dx = [6x - x^2 2 - x^3 3 ]₀^2 = 12 - 2 - 8 3 = 22 3 For A₂ : bounded by y = x^2 + 2 , y^2 = x (i.e., x = y^2 ), x = 2 , and y -axis. In first quadrant, y^2 = x goes from (0,0) to (2, 2 ). The parabola y = x^2 + 2 passes through (0,2) and (2,6). A₂ = ₀^ 2 (2 - y^2) dy = [2y - y^3 3 ]₀^ 2 = 2 2 - 2 2 3 = 4 2 3 A₁ - A₂ = 22 3 - 4 2 3 = 2 3 (2 2 +1)