Prove that the integral of an even function is odd I'm given the function$$F(x)=\int \limits _0^xf(t)\,dt$$and that $f(t)$ is an even function. The assignment asks me to prove that $F(x)$ is an odd function.
I've tried doing something like$$F(-x)=\int \limits _0^{-x}f(-t)\,dt=\int \limits _0^{-x}f(t)\,dt=-\int \limits _{-x}^0f(t)\,dt$$and now since $f(t)$ is even the area in the intervals $[-x,0]$ and $[0,x]$ should be the same,but I'm not sure whether this is a rigorous enough proof. Any tips?
 A: We have
$$
F(x) = \int_{0}^{x}f(t)dt,
$$
then
$$F(-x) =\int_{0}^{-x}f(t)dt=\overbrace{ \int_{0}^{x}f(-u)(-du)}^{\large u\:=-t}=\int_{0}^{x}f(u)(-du)=-F(x).$$
A: Notice that
$$\begin{align}
F(x)&=\int_0^{x}f(t)\;dt\tag{definition of $F$}\\\\
&=\int_0^{x}f(-t)\;dt\tag{$f$ is even}\\\\
&=-\int_0^{-x}f(s)\;ds\tag{substitution $s=-t$}\\\\
&=-F(-x)\tag{definition of $F$}
\end{align}$$
and thus
$$F(-x)=-F(x)$$
A: I would go the other way!  The DERIVATIVE of an even function is odd and the derivative of an odd function is an even function.  If f is even then f(-x)= f(x).  Differentiating, -f'(-x)=  F'(x) so f'(-x)= -f'(x)  and f'(x) is an odd function.  If f is odd then f(-x)= -f(x).  Differetiating, -f'(-x)= -f'(x) so f'(-x)= f(x) and f' s an even function.  That can be rephrased as "if' is odd then f is even and if f' is even then f is odd".  Since integration is the inverse operation to differentiation, replacing f' with f and r with $\int f dx$"  we  have "if f is odd the $\int f dx$ is even and if f is even $\int f dx$ is odd'.
