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The fundamental theorem gives an equivalent between definite and

Â indefinite integrals. Specifically for f, a continuous function

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

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

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

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

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

Â antiderivatives. The equivalence comes from the fact that

Â we evaluate those anti-derivatives of a and b.

Â Another way to write this is more illustrative.

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

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

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

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

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

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

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

Â We've certainly observed it if nothing else.

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

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

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

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

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

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

Â squared. Then we subtract what we get, when we

Â plug in a. Namely, 1 half a squared.

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

Â difficult means. This is the value of the fundamental

Â theorem. It makes computations simple.

Â 3:49

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

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

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

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

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

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

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

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

Â already known. Mainly that it is the area under the

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â cetera. Lets do an example of marginal

Â quantities. Lets assume a publisher is printing

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

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

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

Â 2000. Then, what change in profit would result

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

Â First, we're going to need some variables.

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

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

Â The rate of change of the number of books.

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

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

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

Â times dx. What is this marginal revenue function?

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

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

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

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

Â element, dP. P is for profit.

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

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

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

Â what do we do? We integrate the profit element.

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

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

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

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

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

Â integral. We can to that anti-derivative easily.

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

Â 4,000. Subtract and evaluate as x goes from

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

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

Â 11:07

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Â trouble. If you weren't careful labeling the x

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

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

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

Â directly, we can obtain the same answer very simply.

Â And in some cases, without the opportunity for confusion.

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

Â 13:28

We could have used the integration by parts formula.

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

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

Â 1. What do we obtain?

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

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

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

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

Â The integration by parts formula for definite integrals follows the pattern

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

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

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

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

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

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

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

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

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

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

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

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

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

Â applications that we'll see in chapter four.

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

Â can run into problems.

Â