If $f$ is continuous and $\int^{a}_{-a} f(x) \,dx=0$, is $f$ necessarily odd? I understand that if $f$ is an odd function then
$$\int^{a}_{-a} f(x) \,dx =0$$
Can I say that $f$ is odd function if $\int^{a}_{-a} f(x)=0$ and $f$ is continuous on $[-a,a]$? 

Unless you put some additional hypotheses on $f$ and/or demand that the equality holds for all $a$, you can not.
Consider the function $$f=\begin{cases}2x,&0\le x\le1,\\- 3x^2,&-1\le x\le 0.\end{cases}$$ For obvious reasons, $\int_{-1}^1f(x)dx=0$, yet $f$ is not odd.
However, if you ask that the identity holds for all $a>0$ and that $f$ is, say, continuous, you  can write $$F(a) = \int_{-a}^af(x)dx \equiv 0,$$
hence $F'(a)=0\quad\forall a$. Yet $F'(a)=f(a)+f(-a)$, hence $f(a)=-f(-a)$. Impose now that $f(0)=0$ and $f$ becomes odd.
