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Integration by parts Query

  1. Jul 19, 2012 #1
    FOlks,

    I am self studying a book and I have a question on

    1)what the author means by the following comment "Integrating the second term in the last step to transfer differentiation from v to u"

    2) Why does he perform integration by parts? I understand how but why? I can see that the last term has no derivatives of v in it.

    See attached jpg.

    THanks
     

    Attached Files:

  2. jcsd
  3. Jul 19, 2012 #2

    HallsofIvy

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    "The second term in the last equation" is [itex](\partial F/\partial u')v'dx[/itex] and, since that is a product of a function and a derivative, a rather obvious thing to do is to integrate by parts ([itex]\int U dV= UV- \int V dU[/itex]) with [itex]U= v'[/itex] and [itex]dV= (\partial F/\partial u')dx[/itex].
     
  4. Jul 19, 2012 #3
    I can see what he has done but why? Why integrate by parts?

    Thanks
     
  5. Jul 19, 2012 #4

    HallsofIvy

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    Because integration by parts, being the "inverse" of the product rule for derivatives, is a natural way to integrate a product. Especially here where the product is of a function and a derivative: [itex]\int u dv[/itex].
     
  6. Jul 19, 2012 #5
    integration by part has been done to arrive at something similar to euler lagrange eqn.it just occurs many times in physics in different form may be covariant form
     
  7. Jul 19, 2012 #6
    Ok, was just wondering what he meant by transfer differentiation from v to u?

    Anyhow, thanks folks.
     
  8. Jul 19, 2012 #7
    Actually, shouldnt it be the other way around [itex]U= (\partial F/\partial u')dx[/itex] and [itex]dV=v' [/itex]

    Then [itex]V=v [/itex] since [itex]\int dV=\int v'dx [/itex] and [itex] \displaystyle \frac{dU}{dx}= d(\frac{\partial F}{\partial u'})=\frac{d}{dx} \frac{ \partial F}{\partial u'} [/itex].......?
     
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