I am stuck on a question that involves the intergral of a periodic function. The question is phrased as follows:
Definition. A function is periodic with period $a$ if $f(x)=f(x+a)$ for all $x$.
Question. If $f$ is continuous and periodic with period $a$, then show that $\int_{0}^{a}f(t)dt=\int_{b}^{b+a}f(t)dt$ for all $b\in \mathbb{R}$.
I understand the equality, but I am having trouble showing that it is true for all $b$. I've tried writing it in different forms such as $F(a)=F(b+a)-F(b)$. This led me to the following, though I am not sure how this shows the equality is true for all $b$,
$\int_{0}^{a}f(t)dt-\int_{b}^{b+a}f(t)dt=0$ $=F(a)-F(0)-F(b+a)-F(b)$ $=(F(b+a)-F(a))-F(b)$ $=\int_{a}^{b+a}f(t)dt-\int_{0}^{b+a}f(t)dt=0$
So, this leaves me with
$\int_{a}^{b+a}f(t)dt-\int_{0}^{b+a}f(t)dt=\int_{0}^{a}f(t)dt-\int_{b}^{b+a}f(t)dt$
I feel I am close, and I've made myself a diagram of a sine function to visualize what each of the above integrals might describe, but the power to explain the above equality evades me.