For $f$ Riemann integrable prove $\lim_{n\to\infty} \int_0^1x^nf(x)dx=0.$ 
Suppose $f$ is a Riemann integrable function on $[0,1]$. Prove that $\lim_{n\to\infty} \int_0^1x^nf(x)dx=0.$

This is what I am thinking: Fix $n$. Then by Jensen's Inequality we have $$0\leq\left(\int_0^1x^nf(x)dx\right)^2 \leq \left(\int_0^1x^{2n}dx\right)\left(\int_0^1f^2(x)dx\right)=\left(\frac{1}{2n+1}\right)\left(\int_0^1f^2(x)dx\right).$$Thus, if $n\to\infty$ then $$0\leq \lim_{n\to \infty}\left(\int_0^1x^nf(x)dx\right)^2 \leq 0$$ and hence we get what we want. How correct (or incorrect) is this?
 A: If $f$ is Riemann integrable on $I$ then it is bounded in $I$, i.e. $m \leq f \leq M$. Thus one has
$$\eqalign{
  & m\int\limits_0^1 {{x^n}dx}  < \int\limits_0^1 {{x^n}f\left( x \right)dx}  < M\int\limits_0^1 {{x^n}dx}   \cr 
  & \frac{m}{{n + 1}} < \int\limits_0^1 {{x^n}f\left( x \right)dx}  < \frac{M}{{n + 1}} \cr} $$
This, with the squeeze theorem proves the assertion.
A: Given $\epsilon$, choose $\delta$ and $n$ such that $x^n$ is small for $0\le x\le1-\delta$ and $\int_{1-\delta}^1f$ is small. 
A: That looks great. If someone doesn't know Jensen's inequality, this is still seen just with Cauchy-Schwarz. Another quick method is the dominated convergence theorem. Gerry's and Peters answers are both far simpler though.
A: Just so people can agree : Wikipedia states that a function is Riemann integrable if and only if it is bounded and continuous almost everywhere (just type Riemann integral on wiki). Since the function "squaring" is continuous and that composition of continuous function at a point preserves continuity, $f^2$ is continuous almost everywhere as well, and an obvious bound for $f^2$ is the bound for $f$, squared. The rest is taken care of by $OP$'s proof.
Hope that helps,
