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Proof of Integral Property

  1. Dec 2, 2007 #1
    Prove that if the functions g:[a,b] --> R and h:[a,b] --> R are continuous, with h(x)[tex]\geq[/tex]0 for all x in [a,b] then there is a point c in (a,b) such that

    [tex]\int h(x)g(x)dx = g(c) \int h(x)dx [/tex]
    when the integrals go from a to b


    ---
    My immediate thought was Integral MVT, so I said

    [tex] \int h(x)dx = (b-a)*h(c)[/tex]

    I then multiply by some g(c') with c' in (a,b), to obtain


    [tex]g(c') \int h(x)dx = (b-a)*h(c)*g(c')[/tex]

    I'd then like to conclude
    [tex](b-a)*h(c)*g(c') = \int h(x)*g(x) [/tex]
    But I'm almost 100% positive that's not a valid step for c'=c, and then I'm not really using the integral MVT if I use c'=/= c.

    My thoughts on this are that I'd like to show that because h and g are continuous there is some other number c'' in (a,b) such that h(c)*g(c')=h(c'')*g(c''), and that is equal to the integral i'm looking for. I also have so far completely ignored h(x) being greater than or equal to 0, and I'm pretty sure that should factor in in some way.

    Also, this problem appears in a section about the Cauchy Integral Remainder Theorem, although this book frequently gives problems where the necessary theorems do not lie in the chapter you are on to keep you on your toes.

    Thanks in advance.
     
  2. jcsd
  3. Dec 2, 2007 #2

    Dick

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    Let M=max(g) and m=min(g) on [a,b]. Write bounds on the integral of g*h.
     
  4. Dec 2, 2007 #3
    Ah, good call, even simpler than I thought, thanks!
     
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