If _ - /2 ^ /2 x^2 x 1+e^x dx = ^2 a - b , where a and b are positive integers, then the value of a+b is

Step-by-step solution

Let I = _ - /2 ^ /2 x^2 x 1+e^x dx (1) Using the property _ a ^ b f(x) dx = _ a ^ b f(a+b-x) dx , we replace x with -x : I = _ - /2 ^ /2 (-x)^2 (-x) 1+e^ -x dx = _ - /2 ^ /2 x^2 x 1 + 1 e^x dx I = _ - /2 ^ /2 e^x x^2 x 1+e^x dx (2) Adding equations (1) and (2): 2I = _ - /2 ^ /2 (1+e^x) x^2 x 1+e^x dx 2I = _ - /2 ^ /2 x^2 x dx Since f(x) = x^2 x is an even function, _ -L ^ L f(x) dx = 2 ₀^ L f(x) dx . 2I = 2 ₀^ /2 x^2 x dx I = ₀^ /2 x^2 x dx Applying integration by parts, taking x^2 as the first function and x as th