Let $F(x)= \int_{0}^{x} f(t) \, dt$. Show that if f is even, F is odd.

Question

I have been working on this question. I know that if a function is even then f(x)=f(-x). Take derivative of both sides and you get f'(x)=-f'(-x), hence f' is odd. However, I am not sure how to go about this one with the integration and different independent variable $t$. We are also given a hint: apply the u-substitution u=-t. Any help or hints is greatly appreciated thanks!

Answer 1

Consider $F(-x)=\int^{-x}_{0}f(t)dt$. From there you can use your substitution that $u=-t \Rightarrow du=-dt$ which then gives

$$F(-x)=\int^{-x}_{0}f(t) dt=-\int^{x}_{0}f(-u)du=-\int^{x}_{0}f(u)du=-F(x)$$

Which is exactly your definition of 'oddness': $F(-x)=-F(x)$.