$F(x)$ is a periodic function with petiod $k$ If $f$ be a periodic function with period $k$ and $f(-x)=-f(x)$ in $\bigg[-\frac{k}{2}\;,\frac{k}{2}\bigg]$. Then prove that $\displaystyle \int^{x}_{a}f(t)dt$ is a periodic function with period $k$
Solution i tried
Let $\displaystyle F(x)=\int^{x}_{a}f(t)dt$. Then $\displaystyle F(x+k)=\int^{x+k}_{a}f(t)dt$
$\displaystyle F(x+k)=\int^{x}_{a}f(t)dt+\int^{x+k}_{x}f(t)dt$
$\displaystyle F(x+k)-F(x)=\int^{x+k}_{x}f(t)dt$
I want some help How to prove $F(x)$ is periodic function. Help me plaese

Hint: $\int_x^{x+k}f(t)dx=\int_{{-k}\over 2}^{k\over 2}f(t)dt=0$. Since $f$ is $k$-periodic, (make the change of variables $u=t-x-{k\over 2}$).
Write $u=-x$, $\int_0^{k\over 2}f(t)dt=-\int_0^{{-k}\over 2}f(-u)du=\int_0^{{-k}\over 2}f(u)du=-\int_{{-k}\over 2}^0f(u)du$. 
