Let A be the area of the region bounded by the curve y = |x|, x^2 and the line y = 2 . Then the value of (3A + 1)^2 is equal to

Step-by-step solution

The given curve is y = |x|, x^2 . First, we find the points of intersection of y = |x| and y = x^2 . |x| = x^2 |x| - |x|^2 = 0 |x|(1 - |x|) = 0 Thus, x = -1, 0, 1 . For x [-1, 1] , |x| x^2 , so |x|, x^2 = |x| . For x (- , -1] [1, ) , x^2 |x| , so |x|, x^2 = x^2 . The region is bounded above by the line y = 2 and below by the curve y = |x|, x^2 . The curve intersects y = 2 when x^2 = 2 x = 2 . Due to symmetry about the y-axis, the required area A is: A = 2 ( ₀^1 (2 - x) , dx + ₁^ 2 (2 - x^2) , dx ) Evaluating the in