Suppose we have and we’d like to know what is.
The answer is .
This is an important result in applied mathematics; I’ll give some applications (there are many!) in our next post. Both examples are from a first course in differential equations.
First, I should give the conditions on to make this result true: continuity of and on some rectangle in space which contains all of the points in question (including the interval of integration) is sufficient.
Why is the formula true? The proof isn’t hard at all and it makes use of the Mean Value Theorem and of some basic theorems concerning limits and integrals.
Some facts that we’ll use: if on some interval , then and the Mean Value Theorem.
Now recall from calculus:
We now employ one of the most common
tricks of mathematics; we guess at the “right answer” and then show that the right answer is what we guessed.
We will examine the integrand (the function being integrated). Does remind you of anything? Right; this is the fraction from the Mean Value Theorem; that is, there is some between and such that
Because we are assuming the continuity of the partial derivative, we can say that for sufficiently close to ,
This means that
Now realize that can be made as small as desired by letting get sufficiently close to so it follows by the definition of limit that:
which implies that
So the result follows.
Next post: we’ll give a couple of applications of this