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!
Best Answer
$$ \begin{align} \int_a^{a+T}f(x)\,\mathrm{d}x-\int_0^{T}f(x)\,\mathrm{d}x &=\left(\color{red}{\int_a^{T}f(x)\,\mathrm{d}x}+\int_T^{a+T}f(x)\,\mathrm{d}x\right)\\ &-\left(\int_0^{a}f(x)\,\mathrm{d}x+\color{red}{\int_a^{T}f(x)\,\mathrm{d}x}\right)\\ &=\int_T^{a+T}f(x)\,\mathrm{d}x-\int_0^{a}f(x)\,\mathrm{d}x\\ &=\int_0^{a}f(x+T)\,\mathrm{d}x-\int_0^{a}f(x)\,\mathrm{d}x\\ &=\int_0^{a}(f(x+T)-f(x))\,\mathrm{d}x\\ &=\int_0^{a}0\,\mathrm{d}x\\ &=0 \end{align} $$