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Proving an integration

  1. Jan 24, 2005 #1
    Can someone help me with this question? gladly appreciate any help on this :smile:

    Suppose f is continuous. Prove that

    [tex]\int_0^{x} f(u)(x-u) du = [/tex][tex]\int_0^{x}[/tex] ( [tex]\int_0^{u} f(t) dt[/tex]) du.
     
    Last edited: Jan 24, 2005
  2. jcsd
  3. Jan 25, 2005 #2

    matt grime

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    differentiate both sides wrt x.
     
  4. Jan 25, 2005 #3
    I dont ever understand how you write these out

    do you mean [tex]\int_0^{x} (x-u) dx[/tex]

    to me for some reason it always seems like you write in weird form, but maybe its just me.

    also for that what I wrote above for the solution wouldn't you just substitute, but is the u susposed to be a constant or what? im confused sorry.
     
    Last edited: Jan 25, 2005
  5. Jan 25, 2005 #4

    shmoe

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    You can also integrate the left hand side by parts.
     
  6. Jan 25, 2005 #5
    oh ok i see what he was asking now... so x is a constant?
     
  7. Jan 25, 2005 #6

    shmoe

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    x is independant of u, that's what's important here. You can think of both sides as a function of x if you like.
     
  8. Jan 25, 2005 #7
    thx ppl for the help, i'm trying to solve it now
     
  9. Jan 25, 2005 #8
    for the left side, this is how i did,

    [tex]\int_0^{x} f(u)(x-u) du [/tex]
    = [tex]\int_0^{x} f(u)(x) - f(u)u du [/tex]
    = [tex]\int_0^{x} f(u)(x) du -[/tex] [tex]\int_0^{x} f(u)(u) du [/tex]
    = x [tex]\int_0^{x} f(u) du -[/tex] [tex]\int_0^{x} f(u)(u) du [/tex]

    and then derive it..

    = x f(x) - xf(0) - f(x)x
    =-xf(0)

    is what i'm doing right now correct?
    if so, can someone help me on how to make this equal to the right side?
     
    Last edited: Jan 25, 2005
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