To prove using Rolle's or Mean Value Theorem Let $f:[0,1]\rightarrow R$ be a continuous function such that $\int_{0}^{1}f(x) dx=0$.
Prove that there exists some $c\in(0,1)$ such that $\int_{0}^{c}x f(x)dx=0$.
My Attempt:
Either Rolle's Theorem or the LMVT is to be used but can't figure out how
 A: We can argue by contradiction. Let $g(x)=\int_0^xtf(t)dt$. Suppose that $g(x)$ has no zeros other than $0$, say $g(x)>0,x\in (0,1]$. Then
\begin{align*}
0=&\int_0^1f(x)dx=\lim_{\epsilon\to 0}\int_{\epsilon}^1\frac{1}{x}g'(x)dx\\
=&\lim_{\epsilon\to 0}\left(\frac{g(x)}{x}|_{\epsilon}^1+\int_{\epsilon}^1\frac{g(x)}{x^2}dx\right)\\
=&g(1)+\lim_{\epsilon\to 0}\int_{\epsilon}^1\frac{g(x)}{x^2}dx>g(1)>0,
\end{align*}
which is impossible.
A: Following  notations in Jayden's answer:  
We can argue by contradiction. Let $g(x)=\int_0^xtf(t)dt$. Suppose that $g(x)$ has no zeros other than $0$, say $g(x)>0,x\in (0,1]$. Then
\begin{align*}
0=&\int_0^1f(x)dx \ge \liminf_{\epsilon\to 0}\int_{\epsilon}^1\frac{1}{x}g'(x)dx\\
=&\liminf_{\epsilon\to 0}\left(\frac{g(x)}{x}|_{\epsilon}^1+\int_{\epsilon}^1\frac{g(x)}{x^2}dx\right)\\
\ge&g(1)+\liminf_{\epsilon\to 0} -\frac{g(\epsilon)}{\epsilon}  +\liminf_{\epsilon\to 0}\int_{\epsilon}^1\frac{g(x)}{x^2}dx\\
&>g(1) + 0 +\liminf_{\epsilon\to 0}\int_{\epsilon}^1\frac{g(x)}{x^2}dx>g(1)>0,
\end{align*}
which is impossible.
P.S : $\liminf_{\epsilon\to 0} -\frac{g(\epsilon)}{\epsilon}=0$ Since
Note that $f$ is bounded on $[0,1]$ so for some $m ,M \in R$ we have  $m \leq f(t) \leq M$ for all $t \in [0,1]$ then
$$  m \frac{\epsilon^2}{2} \leq \int_{0}^{\epsilon}tf(t) dt \leq M \frac{\epsilon^2}{2} $$   $$-M \frac{\epsilon}{2} \leq - \frac{1}{\epsilon}\int_{0}^{\epsilon}tf(t) dt \leq -m \frac{\epsilon}{2}$$ $$ -M \frac{\epsilon}{2} \leq-\frac{g(\epsilon)}{\epsilon}\leq -m \frac{\epsilon}{2}$$
