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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.

rick
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    Are you taking this exam? – Adam Smith Sep 14 '11 at 02:35
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    @Adam: Not again – The Chaz 2.0 Sep 14 '11 at 02:37
  • 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. – The Chaz 2.0 Sep 14 '11 at 02:46
  • I'm sorry, I'm new here. – rick Sep 14 '11 at 02:53
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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.

     – Andrew Christianson Sep 14 '11 at 02:56
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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. – rick Sep 14 '11 at 03:03
  • rick, (+1) for an ideal question (not in the algebraic sense...). Thanks for the edit. – The Chaz 2.0 Sep 14 '11 at 03:12
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    If $f(x)=1, f''(x)=0$ and there is no $c$ that fits the requirement. – Ross Millikan Sep 14 '11 at 03:18
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    @Ross: That's a much better counter example than the one I was going to propose: $f(x) = 4(x-\frac{1}{2})^2$. – Jason DeVito - on hiatus Sep 14 '11 at 03:20
  • I'm really sorry, should have been $f(0)=f(1)=0$. – rick Sep 14 '11 at 03:28
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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?

     – rick Sep 14 '11 at 03:30
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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 $f''(x) \equiv 2$. – JavaMan Sep 14 '11 at 03:32
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    well, the typos prove that i didn't merely copy down the question. :P – rick Sep 14 '11 at 03:33
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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 $\int_{0}^{1} f(x) , dx = f(c)$. – Zhen Lin Sep 14 '11 at 03:47
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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.

     – rick Sep 14 '11 at 03:54

3 Answers3

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Hint: $$ \int_0^1x(1-x)f''(x)\mathrm dx=-2\int_0^1f(x)\mathrm dx. $$

Andrew
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  • Finally, a hint that really helps! Thanks! I can solve it now.

    Only one question, how did you come up with that equation?

     – rick Sep 14 '11 at 07:58
  • Nice answer (the above equation can be obtained using integration by parts twice). I'm also interested in its origin. My hunch was that the coefficient $1/12$ could be related to the trapezoidal rule, but I wasn't able to complete the proof along that line. – pharmine Sep 14 '11 at 09:15
  • @rick the idea is to use some kernel in order to apply the mean value theorem. So a hint could look like this: find function $g$ s.t. $$ \int_0^1 g(x)f''(x)\mathrm dx=\int_0^1f(x)\mathrm dx. $$ – Andrew Sep 14 '11 at 09:21
  • Okay, I start to see what's going on here. But how exactly did you find $g(x)$? Integration by parts doesn't lead us directly to $g(x)=\frac{x(x-1)}2$. – rick Sep 14 '11 at 09:47
  • Integrating twice by parts leads to $g''(x)\equiv1$ so it's a polynomial of the second order. Plus there will be terms $g(0)f'(0)$ and $g(1)f'(1)$. To annihilate them we can request that $g(0)=g(1)=0;$ etc. – Andrew Sep 14 '11 at 14:45
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Here is how I continue from Andrew's hint. Please check it:

Firstly we prove that $\int_0^1x(1-x)f''(x)dx=-2\int_0^1f(x)dx$: $$\begin{align*}-2\int_0^1f(x)dx&=-2\left(xf(x)\Big|^1_0-\int_0^1xf'(x)dx\right)\\&=2\left(\frac12x^2f'(x)\Big|^1_0-\int_0^1\frac12x^2f''(x)dx\right)\\&=f'(1)-\int_0^1x^2f''(x)dx\\&=xf'(x)\Big|^1_0-\int_0^1x^2f''(x)dx\\&=xf'(x)\Big|^1_0 - \int_0^1f'(x)dx -\int_0^1x^2f''(x)dx\\&=\int_0^1xf''(x)dx-\int_0^1x^2f''(x)dx\\&=\int_0^1x(1-x)f''(x)dx\end{align*}$$ So we will prove that there exists $c\in[0,1]$ such that $\int_0^1x(1-x)f''(x)dx=\frac16f''(c)$. Note that $x(1-x)\ge0$ for $x\in[0,1]$. Since $f''$ is continuous, by EVT it has a maximum $M$ and minimum $m$ on $[0,1]$. Hence $$\frac16m=\int_0^1x(1-x)m dx\le\int_0^1x(1-x)f''(x)dx\le \int_0^1x(1-x)M dx=\frac16M.$$ Since $f''$ is continuous, by IVT we know that there exists $c$ such that $\frac16f''(c)=\int_0^1x(1-x)f''(x)dx$, as desired.

QED

rick
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0

Let be $$f(x)= x^2-x$$ Clearly, $f(0)=f(1)=0$. And $$\int_0^1 f(x) \ dx= \frac{1}{3}-\frac{1}{2}= \frac{-1}{6}$$ In the other hand, $$f''(x)= 0$$for all $x \in \mathbb{R}$. Therefore, on this case, do not exists $c \in \mathbb{R}$ such that $ \int_0^1 f(x) \ dx = f''(c)$.

Mario De León
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