An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$. I want to prove:

For an integrable function $f(x)$ and periodic with period $T$, for every $a \in \mathbb{R}$, $$\int_{0}^{T}f(x)\;dx=\int_{a}^{a+T}f(x)\;dx.$$

I tried to change the values and define $y=a+x$ so that $dy=dx$ and the limits of the integrals are as we want, but I'm not sure how to use the fact that $f(x)$ is periodic.
Thanks a lot!
 A: Here is a simple way using the periodicity of the function and the antiderivative:
$$F(x+n\cdot T) = F(x); \ n\in \mathbb{Z}$$
$$
\int^{a+T}_a f(x)dx = F(a+T)-F(a)= F(a) - F(a) = 0 = F(T)-F(0)=\int^T_0 f(x)dx
$$ 
The periodicity carries over from the function to the antiderivative, which is easily seen if you take a look at the fourier series of the periodic function.
A: $$
\begin{align}
\int_a^{a+T}f(x)\,\mathrm{d}x-\int_0^{T}f(x)\,\mathrm{d}x
&=\left(\color{red}{\int_a^{T}f(x)\,\mathrm{d}x}+\int_T^{a+T}f(x)\,\mathrm{d}x\right)\\
&-\left(\int_0^{a}f(x)\,\mathrm{d}x+\color{red}{\int_a^{T}f(x)\,\mathrm{d}x}\right)\\
&=\int_T^{a+T}f(x)\,\mathrm{d}x-\int_0^{a}f(x)\,\mathrm{d}x\\
&=\int_0^{a}f(x+T)\,\mathrm{d}x-\int_0^{a}f(x)\,\mathrm{d}x\\
&=\int_0^{a}(f(x+T)-f(x))\,\mathrm{d}x\\
&=\int_0^{a}0\,\mathrm{d}x\\
&=0
\end{align}
$$
A: If $F$ is a primitive of $f$, then
$$\int_{a}^{a+T}f(x)\ dx-\int_{0}^{T}f(x)\ dx$$
$$=F(a+T)-F(a)-F(T)+F(0)$$
$$=\Big(F(a+T)-F(T)\Big)-\Big(F(a)-F(0)\Big)$$
$$=\int_T^{a+T}f(x)\ dx-\int_0^af(x)\ dx$$
$$=0.$$
One checks the last equality by making the obvious change of variable, and by using the periodicity.
EDIT 1. What I wrote above is how I remember the computation. Of course, it can be written like that:
$$
\int_{a}^{a+T}f(x)\ dx-\int_{0}^{T}f(x)\ dx=\int_T^{a+T}f(x)\ dx-\int_0^af(x)\ dx=0.
$$
EDIT 2. Formal justification of the first equality in the above display:
$$
\int_0^af(x)\ dx+\int_{a}^{a+T}f(x)\ dx=\int_{0}^{T}f(x)\ dx+\int_T^{a+T}f(x)\ dx.
$$
(This formula should appear somewhere...)
A: $$\begin{align}
\int_{a}^{a+T}f(x)\ dx&= \int_{T}^{a+T}f(x)\ dx +\int_{a}^{T}f(x)\ dx\\&\overset{y=x-T}{=} \color{red}{\int_{0}^{a}f(y+T)\ dx} +\int_{a}^{T}f(x)\ dx\\&\overset{periodic}{=} \color{red}{\int_{0}^{a}f(y)\ dx} +\int_{a}^{T}f(x)\ dx\\&=\int_0^Tf(x)\ dx.
\end{align}$$
A: $\int_a^{a+T}f(t)dt=\int_a^Tf(t)dt+\int_T^{a+T}f(t)dt,$
and in the last integral, making the substitution $x=t-T$, we get for $a\in [0,T)$, since $f$ is $T$-periodic 
$$\int_a^{a+T}f(t)dt=\int_a^Tf(t)dt+\int_0^{a}f(x+T)dx=\int_a^Tf(t)dt+\int_0^{a}f(x)dx=\int_0^T f(t)dt.$$
For $a\in\mathbb R$, take $n\in\mathbb Z$ such that $a+nT\in [0,T)$. Then 
$$\int_a^{a+T}f(t)dt=\int_{a+nT}^{a+(n+1)T}f(x-nT)dx=\int_{a+nT}^{a+(n+1)T}f(x)dx=\int_0^Tf(t)dt$$
using the previous case. 
A: Here's a picture illustrating the basic idea. (Compare the areas marked with the same color.)

$NT$ denotes the integer multiple of $T$ which belongs to the interval $\langle a,a+T \rangle$. (In this example $N=2$.)

In case someone wants to see metapost source for the picture, it is figure 5 from here.
