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Integrating a differential

  1. Apr 1, 2012 #1

    fluidistic

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    Gold Member

    1. The problem statement, all variables and given/known data
    There's a step I don't really understand in some "proof".
    [itex]d \left ( \frac{\mu }{T} \right )=-\frac{3R du}{2u}-\frac{Rdv}{v}[/itex]. Now he integrates both sides to get [itex]\frac{\mu}{T}- \left ( \frac{\mu}{T} \right ) _0=-\frac{3R}{2} \ln \frac{u}{u_0}-R \ln \frac{v}{v_0}[/itex].
    I don't understand how an integration leads to such an expression.

    2. Relevant equations



    3. The attempt at a solution Clueless.
     
  2. jcsd
  3. Apr 1, 2012 #2

    tiny-tim

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    hi fluidistic! :smile:

    this is standard …

    if we integrate dx from xo to x1, we get x1 - xo

    we then put x = x1, and it becomes ∫ dx = x - xo

    they've done it for x = µ/T, dx = du/u, and dx = dv/v :wink:
     
  4. Apr 1, 2012 #3

    fluidistic

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    Thanks a lot. Crystal clear to me now. I can proceed further. :smile:
     
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