Welcome to Calculus. 
I'm Professor Ghrist and we're about to begin. 
Lecture 26, of a fundamental theorem of integral calculus. 
Based on the definitions alone, definite integrals and indefinite integrals seem 
to have nothing in common, except the name, and that long squiggly sign that we 
use to denote them. In fact, they are closely related. 
They are two aspects of the same character. 
We will see that in today's lesson, where we introduce the fundamental theorem of 
integral calculus. The goal of this lesson is to understand 
and use the Fundamental Theorem of Integral Calculus. 
And while you've probably seen it before, a theorem of this importance is worth 
your careful attention. 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. 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. Let's consider another example, one that 
illustrates how we have to be careful with limits when applying integration 
techniques. Let's consider the integral as x goes 
from 0 to 1, on x time x minus 1. Well now, that's too easy. 
Let's say x times x minus 1 to the nth power dx. 
Where n is some positive integer, let's say. 
Well one way to solve this, would be by substitution. 
Letting u be x minus 1. Du is equal to dx. 
And so we obtain, simply, the integral. As x goes from 0 to 1 of quantity u plus 
1, that's x, times u to the n du. Now notice how I wrote in the limits x 
equals 0 to 1. 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. 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.