Basic question on double integrals

[mnb96]

Hello,
sorry for the trivial question: what's the correct way of computing the following double integral:

$$\int_a^b \int_c^d \frac{\partial^2 f}{\partial x \partial y} dy dx$$

 

[mathman]

Assuming things are reasonably well behaved, the answer is f(b,d)-f(a,d)-f(b,c)+f(a,c), where f is f(x,y).

 

[HallsofIvy]

Hello,
sorry for the trivial question: what's the correct way of computing the following double integral:

$$\int_a^b \int_c^d \frac{\partial^2 f}{\partial x \partial y} dy dx$$

Since the differentials are ordered "dy dx", this means:
$$\int_a^b\left(\int_c^d \frac{\partial^2 f}{\partial x\partial y} dy\right)dx$$

(I would consider it better to write
$$\int_{x= a}^b\int_{y= c}^d \frac{\partial^2 f}{\partial x \partial y} dy dx$$)

By the fundamental theorem of of calculus,
$$\int_c^d \frac{\partial^2 f}{partial y\partial x} dy= \frac{df(x,d)}{dx}- \frac{df(x,c)}{dx}$$
so that
$$\int_{x=a}^b\int_{y= c}^d \frac{\partial^2 f}{\partial x\partial y}dy dx= \int_a^b \left(\frac{df(x,d)}{dx}- \frac{df(x,c)}{dx}\right)dx$$

Applying the fundamental theorem of calculus again gives mathman's solution.

 

[mnb96]

thank you both.
very clear answers.