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 $[-a,a]$? 

No. Let $a = 1$ and consider the function 
$$ f(x) = \begin{cases}\frac{3}{2}x^2&\text{if $x\geq 0$} \\ x&\text{if $x<0$} \end{cases} $$
$f$ is continuous at $0$ since $\lim_{x\to 0^+} f(x) = \lim_{x \to 0^-}$. Furthermore, we have 
$$ \int_{0}^1 f(x) = \int_{-1}^0 \frac{3}{2} x^2 = \frac{1}{2} $$
and
$$ \int_{-1}^0 f(x) = \int_{-1}^0 x = -\frac{1}{2} $$
so 
$$ \int_{-1}^1 f(x) = 0 .$$
But $f$ is obviously not odd.
