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 )

I think i proved it myself .
using the $  "a-x" property $ which states $\int_0^a f(x)dx=\int_0^a f(a-x)dx $
$$\int_0^{nT} f(x)dx =\int_0^{nT} f(nT-x)dx$$
we know from properties of periodic function that if $f$ is periodic with period T then  $f(nT+x)=f(x)$ where $n$ is integer.
replacing x with -x we have
$f(-x+nT)=f(-x)$
Hence $$\int_0^{nT} f(x)dx =\int_0^{nT} f(nT-x)dx=\int_0^{nT} f(-x)dx $$
Hence proved .
