Given $f : \Bbb{R} \to \Bbb{R}$ is a continuous function. To prove that $\int_{0}^{1}f(x)x^2 dx = \frac{f(c)}{3}$ for some $c \in [0,1]$

Question

My approach was to start by integration by parts. $$\int_{0}^{1}f(x)x^2 dx = \frac{1}{3}(f(1) - \int_{0}^{1} f'(x)x^3 dx)$$ Now if I can bound $(f(1) - \int_{0}^{1} f'(x)x^3 dx)$ by $f(0)$ and $f(1)$ then we can use the intermediate value theorem and write it as some $f(c)$.

But this doesn't work. Can someone provide some hints?

Answer 1

\begin{align*} \dfrac{1}{3}m=m\int_{0}^{1}x^{2}dx\leq\int_{0}^{1}f(x)x^{2}dx\leq M\int_{0}^{1}x^{2}dx=\dfrac{1}{3}M, \end{align*} where $m=\min_{x\in[0,1]}f(x)$ and $M=\max_{x\in[0,1]}f(x)$, so by Intermediate Value Theorem we have some $c\in[0,1]$ such that \begin{align*} f(c)=3\int_{0}^{1}f(x)x^{2}dx. \end{align*}