The value of the integral _ - /2 ^ /2 ( x^2 x 1+e^x )dx is equal to ( ^2 A ) - B . Then ( A B ) =

Step-by-step solution

Let I = _ - /2 ^ /2 ( x^2 x 1+e^x ) dx Using the property _ -a ^ a f(x) dx = ₀^ a (f(x) + f(-x)) dx , we get: I = ₀^ /2 ( x^2 x 1+e^x + (-x)^2 (-x) 1+e^ -x ) dx I = ₀^ /2 x^2 x ( 1 1+e^x + e^x e^x+1 ) dx I = ₀^ /2 x^2 x dx Applying integration by parts: I = [x^2 x]₀^ /2 - ₀^ /2 2x x dx I = ^2 4 - ( [-2x x]₀^ /2 - ₀^ /2 (-2 x) dx ) I = ^2 4 - ( 0 + [2 x]₀^ /2 ) I = ^2 4 - 2 Comparing this with ^2 A - B , we get A = 4 and B = 2 . Therefore, A B = 4 2 = 2 . Answer: 2