Proving $g(2n)=0$ Given that $f$ is an odd function periodic with period $2$ and continuous for all $x$ and $g(x)=\int_0^x f(t) dt$
then the question is to prove $g(2n)=0$
$g(2n)=n\int_0^2 f(x) dx=n g(2)$ so I have to prove that $g(2)=0$.I could check that $g(x) $ is a even function.Any ideas?

$$g(2)=\int_0^2f(x)\text{d}x=\int_{-2}^0f(x)\text{d}x$$ 
Thus,
$$g(2)=\frac{1}{2}\left(\int_0^2f(x)\text{d}x+\int_{-2}^0f(x)\text{d}x\right)=\frac{1}{2}\int_{-2}^2f(x)\text{d}x=0$$
