Proof involving FTC to two variables

  • Thread starter hholzer
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  • #1
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Main Question or Discussion Point

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:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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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