Integral mean value theorem

  • #1
302
0
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]
 
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Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
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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
 
  • #3
302
0
I see.
 

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