If a function is periodic with period T ,then is $\int_0^{nT} f(x)dx$=$\int_0^{nT} f(-x)dx$? So we need to prove that if f(x) is periodic with period T then this holds true -$$\int_0^{nT} f(x)dx=\int_0^{nT} f(-x)dx$$
First i tried to prove that  if $f(x)$ is periodic with period T then is $f(x)=f(-x)$ .Then i realized that Definitely it is not true .We need some more info to prove $f(x)=f(-x)$ or take example of $sinx$ , sinx is periodic with period $2\pi$ but $sin(-x)=-sinx$.
So it must be limited to the definite integral only . I tried to put $x=-y$ but it failed .
Any elegant proofs?
(Although i am not really sure weather this property holds or not .I just observed it .It would be great if someone can prove it because i have a really strong feeling its true )

The key is to see that for any $a\in\mathbb{R}$,
$$
\int^{a+T}_af(x)\,dx = \int^T_0 f(x)\,dx$$
To see this, consider let $n=\lfloor a/T\rfloor$, so that $nT\leq a < T(n+1)\leq a+T$
$$\begin{align}
\int^{a+T}_af(x)\,dx&=-\int^a_{nT} f(x)\,dx +\int^a_{nT} f(x)\,dx+\int_a^{(n+1)T} f(x)\,dx +\int^{a+T}_{(n+1)T} f(x)\,dx\\
&=\int^{(n+1)T}_{nT} f(x)\,dx + \int^{a+T}_{(n+1)T}f(x)\,dx-\int^a_{nT} f(x)\,dx\\
&=\int^T_0 f(x-nT)\,dx+\int^a_{nT}f(x-T)\,dx -\int^a_{nT}f(x)\,dx\\
&=\int^T_0f(x)\,dx
\end{align}$$
for the $T$-periodicity of $f$ implies $f(x-nT)=f(x)$ for any integer $n$.
From this we conclude that
$$\int^T_0 f(-x)\,dx\stackrel{u=-x}{=}\int^0_{-T} f(u)\,du=\int^T_0f(u)\,du$$
since the interval $[-T,0]$ has length $T$.
Not that that is taken care off, notice that for any integer $n$
$$\int^{nT}_0f(x)\,dx =\sum^n_{k=1}\int^{kT}_{(k-1)T}f(x)\,dx=n\int^T_0 f(x)\,dx$$
Hopefully you can finish from all this.
