I'm trying to integrate f(x) = (2/3)(4-x^2)^(3/2) with respect to x from -2 to 2. So far I've used trig sub with x/2 = cos u to get f(u) = (-32/3)(sin(u)) ^ 4. I think this is correct but I'm not sure what bounds to use when I integrate f(u). When I tried going from -pi to 0 I get - 4pi rather than 4pi which I know to be the correct answer. Can someone explain to me why I can't go from -pi to 0? Or if I've made a mistake elseware can you explain what I did? Thanks a lot!