Integral of periodic function over the length of the period is the same everywhere 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.
 A: You have made various false steps in your four line block and should have ended up with $$\int_{a}^{b+a}f(t)dt-\int_{0}^{b}f(t)dt=0$$ but this does not take you much further forward.  
Instead note that somewhere in the interval $[b, b+a]$ is an integer multiple of $a$, say $na$.  Then using $f(t)=f(t+a)=f(t+na)$: $$\int_{b}^{b+a}f(t)dt = \int_{b}^{na}f(t)dt+\int_{na}^{b+a}f(t)dt = \int_{b+a}^{(n+1)a}f(t)dt+\int_{an}^{b+a}f(t)dt = \int_{na}^{(n+1)a}f(t)dt = \int_{0}^{a}f(t)dt.$$ 
A: No differentiation is needed: 
Pick the unique integer $n$ such that $b\leqslant na\lt b+a$, decompose the integral of $f(t)$ over $t$ from $b$ to $b+a$ into the sum of the integrals from $b$ to $na$ and from $na$ to $b+a$, apply the changes of variable $t=x+(n-1)a$ in the former and $t=x+na$ in the latter, then the periodicity of $f$ implies that $f(x)=f(t)$, hence the result is the sum of the integrals of $f(x)$ over $x$ from $b-(n-1)a$ to $a$ and from $0$ to $b-(n-1)a$...
...Et voilà !
A: $$\begin{align}
\int_{b}^{a+b}f(x)\ dx&= \int_{a}^{a+b}f(x)\ dx +\int_{b}^{a}f(x)\ dx\\&\overset{y=x-a}{=} \color{red}{\int_{0}^{a}f(y+a)\ dx} +\int_{b}^{a}f(x)\ dx\\&\overset{periodic}{=} \color{red}{\int_{0}^{b}f(y)\ dx} +\int_{b}^{a}f(x)\ dx\\&=\int_0^af(x)\ dx.
\end{align}$$
A: Let $H(x)=\int_x^{x+a}f(t)\,dt$. Then
$$\frac{dH}{dx}=f(x+a)-f(x)=0.$$ 
It follows that $H(x)$ is constant. In particular, $H(b)=H(0)$.
A: We have
$$
\int_{0}^{a}f(t)\ dt+\int_{a}^{a+b}f(x)\ dx=\int_{0}^{b}f(y)\ dy+\int_{b}^{a+b}f(t)\ dt,
$$ 
and setting $x=y-a$ turns the second integral into the third one.
