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From: h holzer on 28 Mar 2010 17:41
(I'm using tex because I don't know how else I should express
partial derivatives via this medium)

I'm given the following:
If

\frac{\partial^2 F}{\partial x \partial y} = f(x,y)

then

\int \int_R f(x,y) dA = F(b,d) - F(a, d) - F(b, c) + F(a, c)

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

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

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

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

\frac{\partial^2 F}{\partial x \partial y} = F_xy


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?

If something was unclear, then let me know and I'll elaborate.

--
hholzer
From: Ray Vickson on 28 Mar 2010 18:09
On Mar 28, 2:41 pm, h holzer <h.holzer.m...(a)gmail.com> wrote:
> (I'm using tex because I don't know how else I should express
> partial derivatives via this medium)
>
> I'm given the following:
> If
>
> \frac{\partial^2 F}{\partial x \partial y} = f(x,y)

Let's use Maple-like notation. Assuming that diff(diff(F,x),y) = F_xy
and diff(diff(F,y),x) = F_yx are continuous, they are equal
(Clairaut). Thus, int(f(x,y),x=a..b) = F_y(b,y) - F_y(a,y), so
int[int(f, x=a..b), y=c..d] = int(F_y(b,y)-F_y(a,y),y=c..d) =
int(F_y(b,y), y = c..d) - int(F_y(a,y), y=c..d) = F(b,d) - F(b,c) -
F(a,d) + F(a,c).

R.G. Vickson

>
> then
>
> \int \int_R f(x,y) dA = F(b,d) - F(a, d) - F(b, c) + F(a, c)
>
> Where R = [a,b] x [c,d]
>
> My question: by integrating the inner integral, we get:
>
> \int_a^b \frac{\partial^2 F}{\partial x \partial y} dx =
> \left[ \frac{\partial F}{\partial y}\right]_a^b
>
> But this result is F_y.  I was expecting F_x. Why? Because:
>
> \frac{\partial^2 F}{\partial x \partial y} = F_xy
>
> 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?
>
> If something was unclear, then let me know and I'll elaborate.
>
> --
> hholzer

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