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.