If $\int_{a}^{b}f(x)g^n(x)dx=0, \quad \forall n \in \mathbb{N}, \; $ then $\; f \equiv 0$ Let g be continuous, not negative, and strictly increasing in [a, b]. Prove that if $f$ is continuous and
$$\int_{a}^{b}f(x)g^n(x)dx=0, \quad \forall n \in \mathbb{N},$$
then $f\equiv 0$.
With a change of variable I have arrived here but I could not continue: 
$$\int_{g(a)}^{g(b)}f(u)u^n \frac{du}{u'}=0, \quad \forall n \in \mathbb{N},$$
Two particular cases already resolved by the community are:
$g(x)=x$ and $x \in [0,1]$: $\quad \int_{0}^{1}f(x)x^ndx=0, \quad \forall n \in \mathbb{N}, \; $
then $f\equiv 0$.
$g(x)=e^x$: $\quad \int_{a}^{b}f(x)e^{nx} dx=0, \quad \forall n \in \mathbb{N}, \; $
then $f\equiv 0$.

Okay, re-attempt. Assume $f$ is not identically zero. Then there exists at least one point $x_0 \in [a,b]$ such that $f(x_0) \neq 0$ and say w.l.o.g. that $f(x_0) > 0$ (otherwise just take the negation). Since $f$ is continuous, there is a small neighborhood around $x_0$ where $f>0$ in that interval. Since $g$ is non-negative and increasing, this implies that
$$
\int f(x)\,g^n(x)\,dx > 0
$$
Which is a contradiction to the original statement. Thus $f \equiv 0$.
