Integrating a differential

  • Thread starter fluidistic
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  • #1
fluidistic
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Homework Statement


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.

Homework Equations





The Attempt at a Solution

Clueless.
 

Answers and Replies

  • #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:
 
  • #3
fluidistic
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Thanks a lot. Crystal clear to me now. I can proceed further. :smile:
 

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