Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$. 
Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$.

Attempt: this exercise has two parts. I did part a) already. From part a) we know that $g_{n} \geq 0$ is a sequence on integrable functions which satisfies  $\lim_{n→∞}\int_{a}^{b} g_{n}dx = 0$.
Then suppose $f$ is integrable, thus, it is bounded so let $m$ be the infimum and $M$ be the supremum of $f$ on the interval $[a,b]$.
Then by the Mean Value Theorem there is a $c \in [m, M]$ such that  $\int_{a}^{b} f(x)g(x)dx = c \int_{a}^{b} g(x)dx$.
Then let $g_{n} = x^{n} \geq 0$ $\forall x \in [0,1]$ Then $\int_{a}^{b} f(x)g_{n}(x)dx = c \int_{a}^{b} g_{n}(x)dx = c \int_{a}^{b} x^{n}dx$.
Can someone please help me continue?Any feedback/help would be appreciated. Thank you.

If you know that $f$ is bounded, so that $-M \le f \le M$, then you have $\int_0^1 x^n f(x)\,dx \le M \int_0^1 x^n\,dx$.  But you can compute $\int_0^1 x^n\,dx$ directly and show that it converges to 0.  Likewise, $\int_0^1 x^n f(x)\,dx \ge -M \int_0^1 x^n\,dx$.  Use the squeeze theorem.
Alternatively, use the fact that $$\left| \int_0^1 x^n f(x)\,dx \right| \le \int_0^1 x^n |f(x)|\,dx \le M \int_0^1 x^n\,dx.$$
(This assumes that "integrable" means "Riemann integrable".  If it means "Lebesgue integrable", then integrable functions need not be bounded, and this proof doesn't work.  But if you're working with Lebesgue integrable functions then you probably know the dominated convergence theorem and should just use that.)
(Acknowledgement of priority: I just noticed that Jon's comment contains exactly the same hint.)
