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Integrating [itex]\frac{\partial f(x,y)}{\partial y(x)}[/itex] with respect to x

  1. Sep 14, 2012 #1
    I don't know how to apply Leibniz's rule for this integrand. I believe there is a substitution to allow me to express this integral differently, but the partial being a function of what we're integrating by is confusing me.

    [itex]\int^{b}_{a} \frac{\partial f(x,y)}{\partial y(x)} dx[/itex]

    Knowns are that [itex][a,b][/itex] is a subset of the domain of [itex]x[/itex] and that [itex]y[/itex] is a function of [itex]x[/itex].

    Can someone please suggest something to get me started, or ask me a question that may get my brain going in the right direction?

    I attempted working with [itex]\int^{b}_{a} \frac{\partial f(x,y)}{\partial x} \frac{dx}{dy}dx[/itex] by expanding the partial according to the chain rule, but I didn't think it helped me.

    Edit: I didn't think the above would help because what exactly does [itex][f(b,y) - f(a,y)]\frac{dx}{dy}[/itex] mean? I don't know how to interpret the factor [itex]\frac{dx}{dy}[/itex].
     
    Last edited: Sep 14, 2012
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  3. Sep 14, 2012 #2

    HallsofIvy

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    I don't understand your notation. [itex]\partial f/\partial y[/itex] is defined as the derivative of y, treating x as a constant. With that definition, [itex]\partial f/\partial y(x)[/itex] is meaningless. If y is a function of x and x is a constant, then y is a constant, not a variable and we cannot differentiate with respect to it. And the problem you are working on, [itex]\int_a^b\frac{\partial f}{\partial x}\frac{dx}{dy}dx[/itex] does not imply that. You are, I think, using the chain rule incorrectly. You want, of course, "df" so that you can integrate. But the chain rule gives [itex]df= \left(\frac{\partial f}{\partial x}\frac{dx}{dy}+ \frac{\partial f}{\partial y}\right)dy[/itex], not at all what you have- in addition to an additional term, you end with "dx", not "dy".

    The fundamental problem is that your integral, [itex]\int_a^b\frac{\partial f}{\partial x}\frac{dx}{dy}dx[/itex] has an "extra" [itex]\frac{1}{dy}[/itex] that just doesn't make sense in an integral.
     
  4. Sep 14, 2012 #3
    You can apply it but I don't think it will make much difference. Note:

    [tex]\frac{\partial}{\partial y}\int_a^b f(x,y)dx=\int_a^b \frac{\partial f}{\partial y} dx[/tex]
     
    Last edited: Sep 14, 2012
  5. Sep 14, 2012 #4
    My original equation was taken from something with messy notation, and I lost some subscripts while working part of the larger question out. So I gave you a poorly framed question to answer -- I'm sorry.

    In my original question, the function [itex]f(x,y)[/itex] is vector valued, and the integrand is with respect to [itex]dx_{1}[[/itex].

    Since I just noticed how sloppy my notation is getting, I'll go back through my problem with more precision before I ask another question.
     
  6. Sep 14, 2012 #5
    . . . then I want my money back.
     
  7. Sep 14, 2012 #6
    Don't worry, it's... uh... already in the mail! :uhh:
     
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