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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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    $\begingroup$ @Adam: Not again $\endgroup$ Sep 14 '11 at 2:37
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    $\begingroup$ @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. $\endgroup$ Sep 14 '11 at 2:56
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    $\begingroup$ I just added some details on what I've done. Anyway, it's not a homework. I'm a self learner. $\endgroup$
    – rick
    Sep 14 '11 at 3:03
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    $\begingroup$ If $f(x)=1, f''(x)=0$ and there is no $c$ that fits the requirement. $\endgroup$ Sep 14 '11 at 3:18
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    $\begingroup$ @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 $f''(x) \equiv 2$. $\endgroup$
    – JavaMan
    Sep 14 '11 at 3:32
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Hint: $$ \int_0^1x(1-x)f''(x)\mathrm dx=-2\int_0^1f(x)\mathrm dx. $$

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  • $\begingroup$ Finally, a hint that really helps! Thanks! I can solve it now. Only one question, how did you come up with that equation? $\endgroup$
    – rick
    Sep 14 '11 at 7:58
  • $\begingroup$ 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. $\endgroup$
    – pharmine
    Sep 14 '11 at 9:15
  • $\begingroup$ @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. $$ \int_0^1 g(x)f''(x)\mathrm dx=\int_0^1f(x)\mathrm dx. $$ $\endgroup$
    – Andrew
    Sep 14 '11 at 9:21
  • $\begingroup$ 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 $g(x)=\frac{x(x-1)}2$. $\endgroup$
    – rick
    Sep 14 '11 at 9:47
  • $\begingroup$ 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. $\endgroup$
    – 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

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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)$.

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  • $\begingroup$ I haven't learned much calculus, but I know this is definitely wrong. :( $\endgroup$
    – rick
    Sep 14 '11 at 3:46
  • $\begingroup$ Mario: $f"(x)=2$ $\endgroup$
    – gary
    Sep 14 '11 at 3:48
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    $\begingroup$ ahh Mario was giving counterexample.. fortunately f''(x)=2 $\endgroup$
    – rick
    Sep 14 '11 at 3:49
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    $\begingroup$ @Mario: Like gary hinted above, this is not a counterexample. $f''(x) = 2$ for all $x$, so $\frac{-1}{6} = \frac{-1}{12}f''(\frac{1}{2})$. $\endgroup$ Sep 14 '11 at 4:12
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    $\begingroup$ It's worth stating that this was a counterexample to a previously posted version of the problem. $\endgroup$ Sep 14 '11 at 19:37

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