Prove that if f is continuous on [a,b] with $\displaystyle f(x) \geq 0 \ , \forall x \in [a,b]$ then there is $\displaystyle c \in [a,b] \ such \ that \ f(c)=[\frac{1}{b-a} \int_a^bf^2]$
This is false: take $\displaystyle f(x)=10\,,\,\,[a,b]=[0,1]\Longrightarrow \frac{1}{1-0}\int\limits_0^1 10^2\,dx=100\neq 10=f(c)\,\,\forall\,c\in [0,1]$ .
The claim though is true if instead $\displaystyle f^2$ we put $\displaystyle f$ in the integral: is just Roll's theorem.
Tonio
I agree with tonio, except I think he meant MVT. Let $\displaystyle F(x)=\int_a^x f(x)\text{ }dx$. By the FTC this is continuous on $\displaystyle [a,b]$ and diff. on $\displaystyle (a,b)$ so there exists some $\displaystyle c\in(a,b)$ such that $\displaystyle f(c)=F'(c)=\frac{1}{b-a}\left\{\int_a^b \text{ }f(x)dx-\int_a^a\text{ }f(x)dx\right\}=\frac{1}{b-a}\int_a^b\text{ }f(x)\text{ }dx$