text: Suppose that f(x) is an even function, then f(-x)=f(x)\n\nLet g(x) = int^x_0 f(x) dx,  then\ng(-x) = int^(-x)_0 f(s) ds\n\nLet s=-u so ds = -du then\n\ng(-x) =  - int^x_0 f(-u) du\n       = - int^x_0 f(u) du\n       = -g(x)\n\nThe function is odd.\n\nNow suppose that f(x) is odd, f(-x)=-f(x) then\nlet g(x) = in^x_0 f(s) ds and then\ng(-x) = int^-x_0 f(s) ds\n\nLet s=-u then ds=-du so\n\ng(-x) = -int^x_0 f(-u) du\n       = -int^x_0 -f(u) du\n       = int^x_0 f(u) du\n       = g(x)\n\nso the function is even.\n\nIf you just look at a graph the result is counter-intuitive, but you have to remember that the bounds of integration matter!