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$f$ is twice differentiable, $f(0)=f(1)=0$ and $f''$ is continuous. Prove that there exists $c\in[0,1]$ such that $$\int_0^1f(x)dx=-\frac1{12}f''(c).$$

I haven't progressed much on this problem. A lot of ideas came up to my mind but none seems to work. Obviously, this has something to do with mean value theorem. In fact, we only need to show that there exist $a,b\in[0,1]$ such that $\int_0^1f(x)dx=\frac{f'(a)-f'(b)}{a-b}$. By mean value theorem, there exists $c\in[a,b]$ such that $f''(c)=\frac{f'(a)-f'(b)}{a-b}$. But this does not seem to be the correct path, because we haven't used that $f''$ is continuous.

This leads to the second idea to show that $f''(x)$ attains some values below and above $\int_0^1f(x)dx$. So I think we need to work out some inequalities, which I don't have any idea.

Anyway, I just started learning calculus for a few weeks. This question is from the previous exam paper, it's the only question I can't solve. Any help is appreciated, thanks.

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    Are you taking this exam?2011-09-14
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    @Adam: Not *again*2011-09-14
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    rick, it is culturally unacceptable to merely copy your problem on this site. "*This question does not show any research effort*" is the description of why a question would receive a downvote. And hence, I am voting down.2011-09-14
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    I'm sorry, I'm new here.2011-09-14
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    @rick, any indication that you've worked on this problem yourself would most likely generate a much greater response from the community. No one here wants to do your homework for you, but most people would be willing to teach you how to do it yourself. That being said, you may be so lost you have no idea of where to begin. That's ok too, just communicate that.2011-09-14
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    I just added some details on what I've done. Anyway, it's not a homework. I'm a self learner.2011-09-14
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    rick, (+1) for an ideal question (not in the algebraic sense...). Thanks for the edit.2011-09-14
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    If $f(x)=1, f''(x)=0$ and there is no $c$ that fits the requirement.2011-09-14
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    @Ross: That's a much better counter example than the one I was going to propose: $f(x) = 4(x-\frac{1}{2})^2$.2011-09-14
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    I'm really sorry, should have been $f(0)=f(1)=0$.2011-09-14
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    now it should be correct, sorry for the typo. what's interesting here is that $-\frac1{12}$ is the only constant that works. take $f(x)=x^2-x$ to convince yourself. but that's also what makes it difficult. where on earth does that strange $-\frac1{12}$ come from?2011-09-14
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    @rick: You might need additional assumptions. Consider $f(x) = x^2 - x$. Then $f(0) = f(1) = 0$ and $\int_0^1 f(x)dx = - \frac{1}{6}$, while $f''(x) \equiv 2$.2011-09-14
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    well, the typos prove that i didn't merely copy down the question. :P2011-09-14
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    Here's a heuristic that might help point in the right direction. Let's replace the interval $[0, 1]$ with $[0, L]$. Then the LHS of your equation scales linearly with $L$, but the RHS scales inverse-quadratically. So there's a factor of $L^3$ missing. You may also want to consider a baby version of this problem: given $f : [0, 1] \to \mathbb{R}$ continuous, show that there is a $c$ in $[0, 1]$ such that $\int_{0}^{1} f(x) \, dx = f(c)$.2011-09-14
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    the baby version is MVT for integral, which i already know. in fact, i tried using that for the proof. unfortunately, it's not enough, because there are cases when $f''(x)$ cannot attain some values of $f(c)$.. so $c$ cannot be any number, we need to be more specific if we're using this approach. possibly we can somehow adapt the proof of this mvt for integral. i'll try again.2011-09-14

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