Prove or Disprove: If $\forall a \in \mathbb{R}:\int_{-a}^a f(x)dx=0$,then $f(x)$ is odd. Problem
Prove or Disprove: Suppose $f(x)$ is continuous. If $\int_{-a}^a f(x)dx=0(\forall a \in \mathbb{R})$,then $f(x)$ is odd.
Comment
Obviously, its converse version holds. But how is itself?
 A: $I = \int^a_{-a} f(x)dx = \int^0_{-a} f(x)dx + \int^a_{0} f(x)dx$
Let in the first integral $x+a = u   \implies dx =du$

$I = \int^a_{0} f(a+u)du + \int^a_{0} f(x)dx$

As, $\int^b_ag(x)dx = \int^b_ag(a+b-x)dx$
$I = \int^a_{0} f( a- (a+u))du+ \int^a_{0} f(x)dx = \int^{a}_0f(-u)du + \int^a_{0} f(x)dx$
Substitute $u =x$
$I = \int^{a}_0f(-x)dx + \int^a_{0} f(x)dx = 0$
$\int^a_{0} f(-x)dx = -\int^{a}_0f(x)dx$

$$f(-x) = -f(x)$$

A: There are already many good answers but here's perhaps yet another way to think about it:
You can take away an arbitrary large "middle-part" from the integral, since that is $0$, too. And you are left with two tiny slices in the ends. Let those slices go to zero and the values tend to the values of the function (you might have to make a change of variable in one of those).
A: Let $g(x):=f(x)+f(-x)$ so $0=\int_0^a g(x)dx$ for all $a$. If $g$ is continuous, we may differentiate to get $g(a)=0$ so $f$ is odd.
A: 
A function $f(x)$ is said to be odd function if $f(-x)=-f(x)$

Now  $\forall a \in \mathbb{R}$
$$I=\int_{-a}^a f(x)dx=0\implies\int_{-a}^0 f(x)dx + \int_{0}^a f(x)dx = 0 \implies I_1 + I_2 =0$$ where $I_1 = \int_{-a}^0 f(x)dx$ and $I_2 = \int_{0}^a f(x)dx$
Let, $t = – x$ or $x = – t$, so that $dt = – dx$
Also, observe that when $x = – a$, $t = a$, and when $x = 0, t = 0$
Then $I_1= \int_{-a}^0 f(x)dx=-\int_{a}^0 f(-t)dt=\int_{0}^{a}f(-t) dt=\int_{0}^{a}f(-x) dx$
Now $$I=I_1 + I_2 =0\implies \int_{-a}^0 f(x)dx + \int_{0}^a f(x)dx = 0 \implies \int_{0}^{a}f(-x) dx + \int_{0}^a f(x)dx = 0 \implies \int_{0}^{a}(f(-x)+f(x)) dx=0$$ which is only possible if $f(-x)+f(x)=0\implies f(-x)=-f(x)$ i.e., if $f(x)$ is an odd function.
