# Thread:

1. ## if f is continuous on [a,b]....

Prove that if f is continuous on [a,b] with $f(x) \geq 0 \ , \forall x \in [a,b]$ then there is $c \in [a,b] \ such \ that \ f(c)=[\frac{1}{b-a} \int_a^bf^2]$

2. Originally Posted by flower3
Prove that if f is continuous on [a,b] with $f(x) \geq 0 \ , \forall x \in [a,b]$ then there is $c \in [a,b] \ such \ that \ f(c)=[\frac{1}{b-a} \int_a^bf^2]$

This is false: take $f(x)=10\,,\,\,[a,b]=[0,1]\Longrightarrow \frac{1}{1-0}\int\limits_0^1 10^2\,dx=100\neq 10=f(c)\,\,\forall\,c\in [0,1]$ .

The claim though is true if instead $f^2$ we put $f$ in the integral: is just Roll's theorem.

Tonio

3. Originally Posted by flower3
Prove that if f is continuous on [a,b] with $f(x) \geq 0 \ , \forall x \in [a,b]$ then there is $c \in [a,b] \ such \ that \ f(c)=[\frac{1}{b-a} \int_a^bf^2]$
I agree with tonio, except I think he meant MVT. Let $F(x)=\int_a^x f(x)\text{ }dx$. By the FTC this is continuous on $[a,b]$ and diff. on $(a,b)$ so there exists some $c\in(a,b)$ such that $f(c)=F'(c)=\frac{1}{b-a}\left\{\int_a^b \text{ }f(x)dx-\int_a^a\text{ }f(x)dx\right\}=\frac{1}{b-a}\int_a^b\text{ }f(x)\text{ }dx$
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