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Integration by parts involving partial derivatives

  1. Jul 27, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]
    \int x \frac {\partial f} {\partial x} dx

    [/tex]

    where

    [tex]f=f(x,t)[/tex]
    2. Relevant equations
    [tex]
    \int u \, dv = uv - \int v \, du
    [/tex]

    3. The attempt at a solution
    [tex]
    u = x [/tex] so [tex]du = dx [/tex]

    and

    [tex]
    dv = \frac {\partial f} {\partial x} dx[/tex] so [tex]v = \int \frac {\partial f} {\partial x} dx = f?
    [/tex]

    Are these the correct substitutions? I am especially hesitant about the last term, so can anyone explain why it's right/wrong?
     
  2. jcsd
  3. Jul 27, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    you'd get v=f(x,t) + h(t)


    EDIT: Sorry, there'd be no h(t)...
     
    Last edited: Jul 27, 2009
  4. Jul 27, 2009 #3
    The term you are hesitant about is correct. You are integrating, with respect to x, the derivative of f with respect to x, so by the fundamental theorem of calculus that integral is just f.

    So, you're right. Just finish the integration by parts and you're done!
     
  5. Jul 27, 2009 #4

    Just be sure to remember that the "constant" of integration in this case isn't constant--it's an arbitrary function of t.
     
  6. Jul 27, 2009 #5
    That's correct. Thanks.
     
  7. Jul 28, 2009 #6
    What if the integral was a definite integral? Would the final result be some function of t only? And would you disregard the arbitrary function of t?
     
  8. Jul 28, 2009 #7

    Cyosis

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    Homework Helper

    You wouldn't disregard it nor do you just disregard it for a normal integral. What happens when you evaluate a definite integral is that the constants get subtracted from each other.

    [tex]
    f(x,t)+h(t)]_{x_1}^{x_2}=f(x_1,t)+h(t)-(f(x_2,t)+h(t))=f(x_1,t)-f(x_2,t)
    [/tex]
     
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