Given an odd function $f$, defined everywhere, periodic with period $2$, and integrable on every interval. Let $g(x) = \displaystyle\int_{0}^{x}f(t)dt$.
I know that $\displaystyle\int_{-b}^{b}f(t)dt=0$ for $b\in\mathbb{R}$ if it is odd function, and $f(t)= f(t+2n)$ where $n$ is integer if it is periodic with period $2$.
Prove that $g(2n)= 0$ for every integer n.
And I do the next:
$g(2n)= \displaystyle\int_{0}^{2n}f(t)dt= \displaystyle\int_{-n}^{n}f(t-n)dt$
if $n=2k$ where $k$ is integer, so $\displaystyle\int_{-n}^{n}f(t)dt=0$
But if $n = 2k+1$ then $f(t+n)=f(t+1)$ then I can be sure that $\displaystyle\int_{-n}^{n}f(t+1)dt=0$, some hint to solve that.
If I take $h(t) = f(t+1)$, we can see that $h(t)=h(t+2)$ (because $f$ is 2-periodic function too) and $h(-t)=-h(t)$ (because $f$ odd function).
$h(t+2)= f((t+2)+1)=f(t+1)= h(t)$
$h(-t)= f(-t+1)=f(-(t-1))=-f(t-1)=-f(t+1)=-h(t)$
Then $\displaystyle\int_{-n}^{n}f(t+1)dt=\displaystyle\int_{-n}^{n}h(t)dt=0$
Best Answer
You don't need to divide into the cases when $n$ is even or odd: $$ g(2n)=\int_0^{2n}f(t)\,dt= \int_{-n}^n f(u)\,du=0 $$ with the substitution $t=u+n$.