Prove that $F(x)=\int_0^x f$ is periodic

Question

Show that if $f$ is a periodic function of period $T$, continuous in $\mathbb{R}$ and $\displaystyle \int_0^T f=0$, then $\displaystyle F(x)=\int_0^x f,\;\forall x\in \mathbb{R}$ defines a periodic function of period $T$.

Definition

A function $f(x)$ is said to be periodic with period T if $f(x+T) = f(x)$ for all values of $x$ in the domain.

Thank you

Answer 1

HINT

Note that $$ F(x+T) - F(x) = \int_0^{x+T} f(s)ds - \int_0^x f(s) ds $$

Can you take it from here, proving that this difference is 0?

Answer 2

Consider

$$F(x+T) = \int_0^{x+T} f(t) \, dt = \int_T^{x+T} f(t) \, dt= \int_0^x f(u+T) \, du = \int_0^x f(u) \, du = F(x)$$ Hence $F$ is periodic.