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Integral mean value theorem

  1. Jun 14, 2009 #1
    http://en.wikipedia.org/wiki/Mean_value_theorem#First_mean_value_theorem_for_integration"

    Take a look at the Wikipedia proof. Now, wouldn't it be easier to prove it like this:
    The ordinary mean value theorem says that
    [tex]
    G(b)-G(a)=(b-a)G'(\xi)
    [/tex]

    And the fundamental theorem of calculus says that
    [tex]
    G(b)-G(a)=\int_a^b G'(x)dx
    [/tex]

    So the conclusion is
    [tex]
    \int_a^b G'(x)dx=(b-a)G'(\xi)
    [/tex]
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jun 15, 2009 #2

    Office_Shredder

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    You proved the case where [tex] \phi = 1[/tex] identically. Notice the wikipedia proof covers a far broader case than you did.

    Also, you probably used the mean value theorem to prove the fundamental theorem of calculus, so this is circular
     
  4. Jun 15, 2009 #3
    I see.
     
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