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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?

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?

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