Prove that $ f(\xi)=f'(\xi)\int_{0}^{\xi}{f(x)dx} $

Question

Let $f:[0,1]\rightarrow\mathbb{R}$ be a differentiable function with continuous derivative such that $$ \int_{0}^{1}{f(x)dx}=\int_{0}^{1}{xf(x)dx} $$ How can we prove that there exists $\xi\in(0,1)$ such that $$ f(\xi)=f'(\xi)\int_{0}^{\xi}{f(x)dx} $$

I tried to use $$ F(x)=\int_{0}^{x}{f(t)dt} $$ then the condition gives $$ \int_{0}^{1}{F(x)dx}=0 $$ and I have to show there exists $\xi\in(0,1)$ such that $$ F'(\xi)=F''(\xi)F(\xi) $$ I was stuck here.

Can you explain the "then the condition gives $ \int_{0}^{1}{F(x)dx}=0 $" please? — P.., Mar 11 '13 at 13:37

Yimin · Accepted Answer · 2013-03-11T14:32:31.600

5

Note that $F(0) = 0$.

Consider function $G(x) = e^{-F'(x)} F(x)$, $G'(x) = e^{-F'}(F'-F''F)$, thus $x=0$ is a zero for $G$, then

if $F(x)=0$ for all $x$, then it is done.
Otherwise, there are $x_1,x_2$ such that $F(x_1)>0$, $F(x_2)<0$. Then $G(x_1)>0$, $G(x_2)<0$, by Roll's theorem (Since we have another zero at $x=0$) we have there is a $\xi$ such that $G'(\xi) = 0$.

Done.

edited Mar 11 '13 at 14:32

answered Mar 11 '13 at 14:19

Yimin

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How does that prove it? If $G'=0$ it's proved, yes but not if $G=0$. – Ishan Banerjee Mar 11 '13 at 14:26
@IshanBanerjee I missed the prime... apologize for the typo. :/ – Yimin Mar 11 '13 at 14:28
How do you conclude $G'(x)=0$? $G(x_1)=G(x_2)$ isn't implied. – Ishan Banerjee Mar 11 '13 at 14:30
Now we have two zeros, one is at $x=0$, the other is at somewhere else.@IshanBanerjee – Yimin Mar 11 '13 at 14:31
Ah, OK, seems fine now. – Ishan Banerjee Mar 11 '13 at 14:32

Prove that $ f(\xi)=f'(\xi)\int_{0}^{\xi}{f(x)dx} $

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