Find all the continuous functions such that $\int_{-1}^{1}f(x)x^ndx=0$. Find all the continuous functions on $[-1,1]$ such that $\int_{-1}^{1}f(x)x^ndx=0$ fof all the even integers $n$.
Clearly, if $f$ is an odd function, then it satisfies this condition. What else?

Rewrite the integral as
$$\int_0^1 [f(x)+f(-x)]x^n dx = 0,$$
which holds for all even $n$, and do a change of variables $y=x^2$ so that for all $m$, even or odd, we have
$$ \int_0^1 \left[\frac{f(\sqrt{y})+f(-\sqrt{y})}{2 \sqrt{y}}\right] y^{m} dy = 0. $$
By the Stone Weierstrass Theorem, polynomials are uniformly dense in $C([0,1])$, and therefore, if the above is satisfied, we must have
$$\frac{f(\sqrt{y})+f(-\sqrt{y})}{2 \sqrt{y}}=0$$
provided it is in $C([0,1])$. [EDIT: I just realized the singularity can be a problem for continuity--so perhaps jump to the density of polynomials in $L^1([0,1])$.] Substituting $x=\sqrt{y}$ and doing algebra, it follows $f$ is an odd function.
