Given: $f:[0, 27] \to \mathbb R$ such that, $f(0)=0$ , $f(10)=1$ , $f(27)=1$ , where $f(x)$ is differentiable.

Prove that , for some $\alpha$, $\beta$ $\in(0,3)$ , the relation

$$2\int_0^{27} f(x)\, dx = 9[\alpha^{2}f(\alpha^{3})+\beta ^{2} f(\beta^{3})]$$ holds.

I think this question is a question on the lagrange's mean value theorem. By the form of the right hand side, I think I should use another function $g(x)=\int_0^{x^{3}} f(x) dx$

so that $g'(x)= 3 x^{2} f(x^{3})$

but I cannot figure out the limits to apply, or how to proceed next

  • $\begingroup$ Is there a multiple of 2 in the integrand on the left hand side?Just need to be clear. $\endgroup$ Mar 4, 2015 at 8:09
  • $\begingroup$ @Mathemagician1234 Yes, there is. $\endgroup$
    – Saraswat
    Mar 4, 2015 at 8:12
  • $\begingroup$ That's a tricky problem,it's not obvious how to proceed and I'm too tired to help. I think you have a good observation about the right hand side, but I have no idea how to go from there either. Life without caffeine is hard. I'll have to sit this one out,hopefully one of the analysts in here can help you out. : ) $\endgroup$ Mar 4, 2015 at 8:20

1 Answer 1


Put $x=y^{3}.$ You have $$2\int_{0}^{27}f\left(x\right)dx=\int_{0}^{27}f\left(x\right)dx+\int_{0}^{27}f\left(x\right)dx=3\int_{0}^{3}y^{2}f\left(y^{3}\right)dy+3\int_{0}^{3}y^{2}f\left(y^{3}\right)dy.$$ Now for the first mean value theorem for integration we have that exists some $\gamma\in\left(0,3\right)$ such that $$3\int_{0}^{3}y^{2}f\left(y^{3}\right)dy=3\gamma^{2}f\left(\gamma^{3}\right)\int_{0}^{3}1dy=9\gamma^{2}f\left(\gamma^{3}\right)$$ so $$2\int_{0}^{27}f\left(x\right)dx=9\left(\gamma^{2}f\left(\gamma^{3}\right)+\gamma^{2}f\left(\gamma^{3}\right)\right)$$ and so your equality holds with the choice $\alpha=\beta=\gamma.$

  • 1
    $\begingroup$ Damn,almost had it. I'd have had it if I could still have coffee........lol $\endgroup$ Mar 4, 2015 at 8:29
  • $\begingroup$ @Mathemagician1234 LOL! $\endgroup$ Mar 4, 2015 at 8:30
  • $\begingroup$ There may exist more than $1$ such $\gamma$, thus $\alpha$ and $\beta$ are not necessarily the same. $\endgroup$
    – mastrok
    Mar 4, 2015 at 8:36
  • $\begingroup$ @mastrok We have only to prove that exists one right choice. $\endgroup$ Mar 4, 2015 at 8:42
  • $\begingroup$ Ur right. It is just a complementary comment to your answer. $\endgroup$
    – mastrok
    Mar 4, 2015 at 8:46

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