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Proof involving FTC to two variables

  1. Mar 28, 2010 #1
    I'm given the following:
    If [tex] \frac{\partial^2 F}{\partial x \partial y} = f(x,y)[/tex]
    then
    [tex]
    \int \int_R f(x,y) dA = F(b,d) - F(a, d) - F(b, c) + F(a, c)
    [/tex]

    Where R = [a,b] x [c,d]

    My question: by integrating the inner integral, we get:

    [tex]
    \int_a^b \frac{\partial^2 F}{\partial x \partial y} dx = \left[ \frac{\partial F}{\partial y}\right]_a^b
    [/tex]

    But this result is F_y. I was expecting F_x. Why? Because:

    [tex]
    \frac{\partial^2 F}{\partial x \partial y} = F_xy
    [/tex]

    Which means: we differentiate with respect to x, then with respect to y.
    The result above seemingly implies that we differentiated with
    respect to y, then with respect to x. Hence, integrating F_yx,
    gives F_y, is this not correct? Or do they apply fubini's theorem
    and Clairaut's theorem above?
     
    Last edited: Mar 28, 2010
  2. jcsd
  3. Mar 29, 2010 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    As long as the derivatives are continuous, the to ways of differentiating a mixed second derivative are equal:
    [tex]\frac{\partial }{\partial x}\frac{\partial f}{\partial y}= \frac{\partial}{\partial y}\frac{\partial f}{\partial x}[/tex].

    By integrating [tex]\frac{\partial^2 f}{\partial x\partial y}[/tex] with respect to x, you've "undone" the differentiation with respect to x leaving [tex]\frac{\partial f}{\partial y}[/tex].
     
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