Problem:
Prove that if $f:\left [ 0,1 \right ]\rightarrow \mathbb{R}$ is a continuous function such that: $\int_{0}^{1}f(x)e^{nx}dx=0$ for all $n=0,1,2,...$. Prove that $f(x)=0$ for all $0\leqslant x\leqslant 1$ using two methods: 1/ change of variables and then apply Weiestrasss Theorem. 2/ Apply Stone Weierstrass Theorem .
I already know how to prove that $f(x)=0$ for all $0\leqslant x\leqslant 1$ if $\int_{0}^{1}f(x)x^{n}dx=0$. I did the following change of variable: $e^{x}=y$ and then I got $\int_{1}^{e}f(y)y^{n-1}dy$. since $\int_{0}^{1}f(y)x^{n-1}dy=0$, we are left with $\int_{1}^{e}f(y)x^{n-1}dy$ which I want to prove equal to zero. Any help with this?
For the second part, I don't have any idea how to use the Stone Weierstrass Theorem to prove it. I have never used this theorem before in solving problems, so I appreciate if someone helps with details for this part.