TITLE: Prove that $\exists c\in(0,1)$ such that $f(c)=c^2$
QUESTION [5 upvotes]: Suppose that a function $f:[0,1] \to \mathbb{ R}$ is continuous  and $3 \int_0^1 f(x)dx = 1$. Prove there exists $c \in (0,1)$ such that $f(c) = c^2  $

REPLY [4 votes]: We have that $f$ is integrable. Let $$h(x)=\int_{0}^{x} f(t)dt-\frac {x^3}{3}$$.
Then $h$ is continuous in $[0,1]$ and differentiable at $(0,1)$ .Also $h(0)=h(1)$. So by Roll'es Theorem we have that there is a $c\in (0,1):h'(c)=0$.