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Suppose that $f$ is a continuous real-valued function on the interval $[0,1]$. Show that $$\int_0^1x^2f(x)\,dx=\frac{1}{3}f(\xi)$$ for some $\xi\in[0,1]$.

I guess that MVT is involved, but I don't know how to apply it.

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  • $\begingroup$ Extreme Value Theorem and Intermediate Value Theorem. $\endgroup$ – Sungjin Kim Apr 24 '17 at 20:58
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$f $ is continuous at $[0,1]$ $$ \implies $$

$$\exists m,M \in \mathbb R \;: \;\;f ([0,1])=[m,M] $$ $$\implies $$ $$\forall x\in [0,1]\;\;mx^2\leq x^2f (x)\leq Mx^2$$ $$\implies $$

$$\frac {m}{3}\leq \int_0^1x^2f (x)dx\leq \frac {M}{3} $$ (because $\int_0^1x^2dx=\frac {1}{3} $.)

$$\implies $$

$$m\leq 3\int_0^1x^2f (x)dx\leq M $$

$$\implies $$

$$\exists \xi \in [0,1]\;\;:\; 3\int_0^1x^2f (x)dx=f (\xi) $$ you can finish.

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  • $\begingroup$ For this to work, $m$ and $M$ cannot be arbitrary bounds for $f$, but its infimum and supremum. $\endgroup$ – Michał Miśkiewicz Apr 24 '17 at 21:06
  • $\begingroup$ Any upper bound that is not the supremum is larger by definition (and similar argument for infimum with 'smaller'). The intermediate value theorem can be applied just as well in that case, it just changes the interval over which there is a solution (which is, of course, still on [0,1]) $\endgroup$ – infinitylord Apr 24 '17 at 22:03
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The second mean value theorem for integrals states that

$$\int_a^b f(x)g(x)dx = f(c)\int_a^b g(x)dx $$ For some $c \in (a,b)$

Apply this theorem with $f(x) = f(x)$ and $g(x) = x^2$

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  • $\begingroup$ There is no need to clarify that $f(x)=f(x)$. =) $\endgroup$ – Pedro Tamaroff Apr 24 '17 at 21:08
  • $\begingroup$ @PedroTamaroff and even less need to clarify that there was no need to clarify $\endgroup$ – infinitylord Apr 24 '17 at 21:28

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