Calculus question about integration. Consider  $ f$ from $[0,1]$to $[0,1] $ be continuous and non constant .
Then, is there $c\in[0,1]$ such that $f(c) =\int^1 _0 f^2(t) dt $ ? 

Counterexample
Consider $f(x) = \frac{1}{2} (1-x) + \frac{1}{4} x = \frac{1}{4} \left(2-x\right)$. Then 
$$
  \int_0^1 f^2(x) \mathrm{d} x = \frac{1}{16} \int_0^1 \left(2-x\right)^2 \mathrm{d} x = \frac{1}{16} \int_1^2 y^2 \mathrm{d} y = \frac{1}{16} \cdot \frac{2^3-1^3}{3} = \frac{7}{48}
$$
But the equation
$$
   \frac{1}{4} \left(2-x\right) = \frac{7}{48} \quad x = 2 - \frac{7}{12} = \frac{24-7}{12} = \frac{17}{12}
$$
but the solution $ x= \frac{17}{12}$ lies outside $[0,1]$ interval.
