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 $[-a,a]$? 
 A: If you mean that the identity $\int_{-a}^a f(x) \,dx = 0$ holds for some fixed $a$ then no---consider, for example, $$f(x) = \cos x, a = \pi .$$
If you mean that the identity holds for all real $a$, then the answer is yes: Differentiating both sides of the identity with respect to $a$ gives $0 = f(a) - (-f(a)) = f(a) + f(-a)$, and rearranging then gives $$f(-a) = -f(a) .$$
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.
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.
