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
Therefore
So the result follows.
Next post: we’ll give a couple of applications of this
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