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Cauchy Mean Value Theorem

  1. Jul 29, 2003 #1
    Hi, I really need some help in sovling this proof!!!

    Prove the Cauchy Mean Value Theorem:
    If f,g : [a,b]->R satisfy f continuous, g integrable and
    g(x)>=0 for all x then there exists element c is a member of set [a,b] so that
    int(x=b,a)f(x)g(x)dx=f(c)int(x=b,a)g(x)dx.

    Thanks for your help :D
     
  2. jcsd
  3. Jul 29, 2003 #2
    Hi iceman,
    imagine I'm a complete ignorant in mathematics. Then I can still type your key words into google, and get for instance this:

    http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node42.html



    Edit:
    Sorry, couldn't help it. But if I was you, I'd ask my prof what's the good in posing problems the answer to which is in the literature. Can't he come up with something more creative?
     
    Last edited: Jul 29, 2003
  4. Jul 30, 2003 #3

    HallsofIvy

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    Arcnets: You also have to be able to UNDERSTAND the results of a google search.
    The "Cauchy Mean Value Theorem" your reference gives is clearly NOT the same as the "Cauchy Mean Value Theorem" in the original post. For one thing, the Cauchy Mean Value Theorem the OP asked about is an integral mean value theorem.

    (I would use the phrase "extended mean value theorem" for the result given in Arcnet's link.)

    Iceman: You can, however, USE the (extended) mean value theorem.

    Let F(x)= int(t=a to x) f(t)g(t)dt and let G(x)= int(t= a to x)g(t)dt. Use those in the extended mean value theorem:

    (F(b)- F(a))/(G(b)- G(a))= F'(c)/G'(c) for some c in [a,b].

    That's not the complete answer- you will still need to do some work.
     
  5. Jul 30, 2003 #4
    OK,OK. Back to being helpful not provocative.
     
  6. Jul 31, 2003 #5

    HallsofIvy

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    Oh, darn! Provocative is so MUCH more fun!

    (And yes, I have complained about doing peoples "google" work for them myself.)
     
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