= Solution
{c}
Write $a=2\pi m+r$, where $m$ is an <integer> and $0\leq r<2\pi$. The <change of variables> $x=y+2\pi m$ and the fact that $f$ is a <periodic function> give
$$
\int_a^{a+2\pi}f(x)\,dx=\int_r^{r+2\pi}f(y)\,dy.
$$
Split at $2\pi$, then substitute $y=z+2\pi$ in the second piece:
$$
\begin{aligned}
\int_r^{r+2\pi}f(y)\,dy
&=\int_r^{2\pi}f(y)\,dy+\int_{2\pi}^{2\pi+r}f(y)\,dy\\
&=\int_r^{2\pi}f(y)\,dy+\int_0^r f(z+2\pi)\,dz\\
&=\int_r^{2\pi}f(y)\,dy+\int_0^r f(z)\,dz.
\end{aligned}
$$
Therefore
$$
\boxed{\int_a^{a+2\pi}f(x)\,dx=\int_0^{2\pi}f(x)\,dx\quad\text{for every }a\in\mathbb R.}
$$
This proves <translation invariance of an integral over one period> without assuming that $f$ is <differentiable>.
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