If ₀^ ^2 x 1 - x x d x = k , where k is a positive integer, then the value of k is ________.

Step-by-step solution

Let I = ₀^ ^2 x 1 - x x d x . Split the integral at x = 2 : I = ₀^ 2 ^2 x 1 - x x d x + _ 2 ^ ^2 x 1 - x x d x In the second integral, substitute x = - y , so d x = - d y . The limits change from 2 0 : _ 2 ^ ^2 x 1 - x x d x = _ 2 ^0 ^2( - y) 1 - ( - y) ( - y) (- d y) = ₀^ 2 ^2 y 1 + y y d y Thus, I = I _A + I _B , where: I _A = ₀^ 2 ^2 x 1 - x x d x I _B = ₀^ 2 ^2 x 1 + x x d x Apply King's Rule ( x 2 - x ) to I _A : I _A = ₀^ 2 ^2 x 1 - x x d x Adding the two forms of I _A : 2 I _A = ₀^ 2 ^2 x + ^2 x 1 - x x d x