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 $f\equiv 0$.

Hagen von Eitzen above has indeed given an elegant solution valid when the given g is strictly increasing but no assumption that g be non-negative is needed or used. However it is not necessary assume the integral (1) I[a,b]f(x)(g(x))^n dx =0 for all non negative integers n= 0.1,2,3...   . It suffices to assume this for integers 
n >= M where M is any positive integer and by increasing M we  assume M is odd .We use the Stone Weierstrass theorem instead of the Weierstrass Theorem .Note that (1) implies that I[a,b]f(x) (g(x)^M (p(g(x))dx =0  for all polynomials If g(x) is never 0 in [a,b] then g(x)^M is strictly increasing hence the algebra of functions of the form (2)q(x)= (g(x)*M (p(g(x)) separates the points of [a,b] and don't all vanish at any one point and so  dense in the continuous functions for the mac norm .Using such a sequence converging to f(x) uniformly we see that I(f(x)^2 ) =0 so f vanishes identically as before
 What happens if g(c) = 0 for some c in [a,b] (exactly one ,g is strictly increasing . Then  the functions of the form C + q (q as in (2),C a real constant ) are a dense set . But if h(c)=0 H continuous then a sequence of functions c+q oonverges uniformly to h and all q(c)=0  hence the sequence of constants c tends to 0 so the the sequence of q 's =c+q -c tends to h uniformly . Take h = f T  where T(x) is between 0 and 1 ,T(c) =0 .and T=1 off and interval of length e containing c , T continuous  then see that I (f^2T) =0 Let e tend to 0 to get I(f*2)=0 and hence f=0 .  Stuart M.N.  
