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Question on Mean Value Theorem & Intermediate Value Theorem

  1. Aug 18, 2015 #1

    Titan97

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    Gold Member

    1. The problem statement, all variables and given/known data
    for ##0<\alpha,\beta<2##, prove that ##\int_0^4f(t)dt=2[\alpha f(\alpha)+\beta f(\beta)]##

    2. Relevant equations
    Mean value theorem: ##f'(c)=\frac{f(b)-f(a)}{b-a}##

    3. The attempt at a solution
    I got the answer for the question but I have made an assumption but I don't know if it's correct.

    Let ##g(x)=\int_0^{x^2}f(t)dt## and let ##h(x)=g'(x)=2xf(x^2)##

    now, applying intermediate value theorem for ##h(x)## in ##x\in (\alpha,\beta)##, there exists a point ##x=k## such that,
    $$h(k)=\frac{h(\alpha)+h(\beta)}{2}=\alpha f(\alpha)+\beta f(\beta)$$
    $$g'(k)=\alpha f(\alpha)+\beta f(\beta)$$
    assumption: Now, for g(x), if ##x=k## also satisfies Mean value theorem and since ##0<k<2##,

    $$g'(k)=\frac{f(2)-f(0)}{2}=\frac{1}{2}\int_0^{4}f(t)dt=\alpha f(\alpha)+\beta f(\beta)$$

    Hence, $$\int_0^{4}f(t)dt=2[\alpha f(\alpha)+\beta f(\beta)]$$
     
    Last edited by a moderator: Aug 19, 2015
  2. jcsd
  3. Aug 19, 2015 #2

    HallsofIvy

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    What, exactly, are you to prove? You say "for [itex]0<α,β< 2[/itex] [itex] \int_0^4 f(t)dt=2[\alpha f(\alpha)+\beta f(\beta)][/itex]"
    That reads "for all [itex]\alpha[/itex] and [itex]\beta[/itex] strictly between 0 and 2, and for any integrable function, f" this equality is true.

    But that is clearly NOT true. In your "proof" you appear to determine that there exist numbers [itex]\alpha[/itex] and [itex]\beta[/itex] that work.
     
  4. Aug 19, 2015 #3

    pasmith

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    Homework Helper

    What exactly is the question? To prove that there exist [itex]0 < \alpha < 2[/itex] and [itex]0 < \beta < 2[/itex] such that [tex]
    \int_0^4 f(t)\,dt = 2(\alpha f(\alpha) + \beta f(\beta))?[/tex] Are there any conditions on [itex]f[/itex]?

    Unfortunately
    [tex]\frac{h(\alpha) + h(\beta)}{2} = \alpha f(\alpha^2) + \beta f(\beta^2)[/tex] which doesn't help you.

    EDIT: It occurs to me that [itex]\alpha f(\alpha) + \beta f(\beta)[/itex] doesn't involve the value of [itex]f[/itex] on [itex](2, 4][/itex] at all, which seems suspicious.
     
    Last edited: Aug 19, 2015
  5. Aug 21, 2015 #4

    Titan97

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    Gold Member

    I am very sorry. My teacher gave me a wrong question.
     
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