_ - / 2 ^ / 2 x d x

Step-by-step solution

Let I = _ - / 2 ^ / 2 x , dx . The integrand f(x) = x is an odd function because f(-x) = (-x) = - x = -f(x) . For any odd function f(x) integrated over a symmetric interval [-a, a] , the integral is zero: _ -a ^ a f(x) , dx = 0 . Therefore, _ - / 2 ^ / 2 x , dx = 0 . Answer: 0