If f(x) is a periodic function, with period T, then

$\int\limits_a^b f(x) d x=\int\limits_{a+T}^{b+T} f(x) d x$

We have,

$\int\limits_{a+T}^{b+T} f(x) d x=\int\limits_a^b f(y+T) d y$, where $x=y+T$

$\Rightarrow \int\limits_{a+T}^{b+T} f(x) d x=\int\limits_a^b f(y) d y$             $\left[\begin{array}{cc} ∵ f \text { is periodic with period } T \\ ∴ f(y+T)=f(y)\end{array}\right]$

$\Rightarrow \int\limits_{a+T}^{b+T} f(x) d x=\int\limits_a^b f(x) d x$