0:56

The fundamental theorem gives an equivalent between definite and

indefinite integrals. Specifically for f, a continuous function

on the interval from a to b. The definite integral of f of (x), dx as

x goes from a to b. Is equal to the indefinite integral of f

evaluated from a to b. Now again, these are ostensibly different

objects. On the left and the right hand side, the

definite integral is a number. The indefinite integral is a class of

antiderivatives. The equivalence comes from the fact that

we evaluate those anti-derivatives of a and b.

Another way to write this is more illustrative.

The definite integral is x goes from a to b of dF, where capital F is some function

is that anti-derivative f evaluated from x equals a to b.

Now this is a more compact way to write the result.

In practice, you're going to want to think of it expanded out a little bit.

Though the chain rule dF is really dFdxdx, and that definite integral from a

to b, is simply the anti-derivative f evaluated at b minus f evaluated at a.

Now, this is not a new idea to you. You've certainly used this result before.

We've certainly observed it if nothing else.

When we did, the definite integral of x d x, as x goes from a to b.

By computing the Riemann sum we found that the answer was 1 half, quantity, b

squared, minus a squared. If we apply the fundamental theorem of

integral Calculus, what it says is that we can compute the anti-derivative of x.

Which is, of course, 1 half x squared. And then evaluate that.

Add a to b. First, we plug in b and obtain, 1 half b

squared. Then we subtract what we get, when we

plug in a. Namely, 1 half a squared.

And that of course is the same answer that we obtained earlier through more

difficult means. This is the value of the fundamental

theorem. It makes computations simple.

3:49

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.

11:07

But we're integrating with respect to u. Be careful with your limits so you know

which variable you're talking about. Well let's proceed.

The integral of u plus 1, times u to the n du is expanding the integral of u to

the n plus 1, plus u to the n du. That's a simple integral, that gives us u

to the n plus 2, over n plus 2, plus u to the n plus 1, over n plus 1.

And we need to evaluate. That anti-derivative as x goes from 0 to

1, but that's in terms of x. So to compute the answer we could

substitute back in x minus 1 for you. This gives us x, minus 1, to the n plus

2, over n plus 2. Plus, x minus 1 to the n plus 1 over n

plus 1. Evaluating as x goes from 0 to 1, gives

minus negative 1 to the n plus 2, over n plus 2, minus negative 1 to the n plus 1

over n plus 1. With a little bit of simplification,

factoring out on negative 1 to the n plus 2, and then simplifying that to negative

1 to the n, we get a final answer of negative 1 to the n, over n plus 1 times

quantity n plus 2. Now, you can see how you could get into

trouble. If you weren't careful labeling the x

limits versus the u limits. Now another way to do this, would be to

change from x limits to u limits. When x is 0, u, x minus 1, is negative 1.

When x is 1, u is equal to 0. By changing the limits to u limits

directly, we can obtain the same answer very simply.

And in some cases, without the opportunity for confusion.

Now, that's not the only way to solve this integral.

13:28

We could have used the integration by parts formula.

If, in this case, we let u be x, and dv be x minus 1 to the n.

Then, setting du equal to dx, and v equal to x minus 1 to the n plus 1, over n plus

1. What do we obtain?

Well, we get u times v, that is x times quantity x minus 1 to the n plus 1 over n

plus 1 minus the integral of v. Max minus 1 to the n plus 1 over n plus 1

times du, that is dx. And that's a simple enough integral to

do, however, we must be careful with the limits.

The integration by parts formula for definite integrals follows the pattern

you would expect but you have to evaluate the uv term from a to b.

So let's do so in this case. Evaluating as x goes from 0 to 1.

What does this give? Well, when we evaluate the u times v

quantity from 0 to 1, we get at 1, 0. At 0, 0.

And, that's simple enough. Fortunately, it goes away.

And we're left with the integral of x minus 1 to the n plus 1, over n plus 1.

That is of course x minus 1 to the n plus 2, over n plus 2 times that negative 1

over n plus 1 that was hanging around. Evaluating that from 0 to 1 gives us our

answer very simply. Negative 1 to the n plus 2 over quantity,

n plus 1, times n plus 2. Pulling out a negative 1 squared, gives

us the same answer that we saw before. The fundamental theorem, of the integral

of calculus is fantastic. With it, your going to be able to compute

definite integrals galore. With it, we're going to fuel all of the

applications that we'll see in chapter four.

But in our next lesson, we'll see that when pushed to the limit, this theorem

can run into problems.