# Proof involving FTC to two variables

1. Mar 28, 2010

### hholzer

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?

Last edited: Mar 28, 2010
2. Mar 29, 2010

### HallsofIvy

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

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