Let's look at a different example. Compute, the definite integral as x goes

from 1 to t, of 1 over x, dx. Now, we can think of this geometrically

in terms of limits of Riemann sums, getting something that approximates the

area under the curve 1 over x. By the fundamental theorem, we can

anti-differentiate 1 over x to get, of course, log of x.

And evaluate that from 1 to t. That gives us log of t, minus log of 1.

Of course the natural log of 1 is 0. And so we obtain log of T as the answer,

which gives us a new interpretation of the natural log rhythm that you may have

already known. Mainly that it is the area under the

curve 1 over x, as x goes from 1 to t. Now this is all well and good, and you

will find the fundamental theorem to be extremely useful in computations, but you

must know what this theorem really means, and it has several interpretations.

Let's look at the compact form of the fundamental theorem, and rearrange the

terms a bit, so that on the left we have a function, f, evaluated from a to b,

that is equal to, on the right, the definite integral from a to b, of dF.

Otherwise said, the net change in some quantity, F, is equal to the integral of

its rate of change. Now you might say, that's obvious, but

it's not. And there are many different contexts in

which this applies in a non-trivial and non-obvious way.

Some are simple in saying that the position is equal to the integral of the

velocity. Or the net change in height, is equal to

the integral of the growth rate. Some are not so obvious, particularly in

economics, where one talks of marginal quantities as the derivative.

So the net change in supply is equal to the integral of the marginal supply et

cetera. Lets do an example of marginal

quantities. Lets assume a publisher is printing

12,000 books per month with an expected revenue of $60 per book, but it costs

money to publish these books, and the marginal cost is a function of x, the

number of books published per month. This function is given by 10 plus x over

2000. Then, what change in profit would result

from a 25% increase in production? Let's set this up as an integral problem.

First, we're going to need some variables.

The cost element, that is, the rate of change of cost to the publisher is given

by this marginal cost function, MC of (x) times dx.

The rate of change of the number of books.

This is, of course, 10 plus x over 2000 times dx.

What about revenue? Well the revenue element, that is the

rate of change of revenue, is given in terms of a marginal revenue function

times dx. What is this marginal revenue function?

Well if we look at the problem, we see that the revenue is at $60 dollars per

book. Since it's a per book quantity it is

marginal, so the revenue element is 60dx. Now, the problem is asking, for profit,

in particular, change in profit. And so we would look at the profit

element, dP. P is for profit.

This is the revenue minus the cost, or at the marginal level, the marginal revenue

minus the marginal cost. This is 50 minus x over 2,000 dx.

That is our profit element. And so, to obtain a net change in profit,

what do we do? We integrate the profit element.

50 minus x over 2,000 dx. With what limits?

Well, we began at x equals 12,000 books per month.

And, we need to get an upper limit, we were asked to consider a 25% increase in

production. That would be going to 15,000 books per

month. And so we see that the answer is a simple

integral. We can to that anti-derivative easily.

50 integrates to 50 x. X over 2,000 integrates to x squared over

4,000. Subtract and evaluate as x goes from

12000 to 15000. I'll leave it to you to determine the

numerical answer of almost $130,000. That is the net increase in profit.