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Calculus: Fundamental Theorem of Calculus Applied

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  • Type: Video Tutorial
  • Length: 13:55
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 150 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus: Final Exam Test Prep and Review (45 lessons, $64.35)
Calculus: Basics of Integration (14 lessons, $23.76)
Calculus: The Fundamental Theorem of Calculus (5 lessons, $9.90)

Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/calculus. The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

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The Basics of Integration
The Fundamental Theorem of Calculus
Illustrating the Fundamental Theorem of Calculus Page [1 of 1]
All right. So now we're seeing illustrations of the fundamental theorem of calculus, which in some sense empowers us to find areas under curves. I thought we'd take a look at a couple more examples of this so you could really see the idea. Basic theme here is that if you want to find the area under a curve, from a to b, you first compute the antiderivative. Evaluate the antiderivative at the right hand endpoint and then subtract off the antiderivative evaluated at the left hand endpoint. And that number represents the area between those 2 endpoints; absolutely stunning fact.
All right. Let's take a look at the following example. Let's find the area under the curve f(x) = e^x. So the exponential function f(x) = e^x, let's find the area under that curve x going from 1 to x going to 2. Okay well actually this is a problem, this is a question that we can actually resolve without ever even knowing what the picture looks like. It's sort of amazing that you can actually now find areas under curves without even seeing a picture. But let's draw a picture just to get a sense of what this is, even though it's not necessary for the problem. I'm going to remind you what the exponential function looks like. We talked about this earlier on when we talked about exponentials and their derivatives. And we saw that the exponential function looks something like this, f(x) = e^x power. And I want to look at this curve from the point x = 1, right here, out to x = 2, which is right here. And so I want to see the area under the green curve, but above the floor, and I see this = that. All right. Now how do I set up the integral? Well the integral would look something like this. We would have that the area of this integral would = the integral from where? Well the left hand endpoint, put the 1 down there, to the right hand endpoint, 2 up there, and what are we going to integrate? Well we're going to integrate the function (e^x)dx. Okay so how do we proceed? Well the way we proceed is we integrate first and find the antiderivative. Well what's the antiderivative of e^x? Well, e^x has the wonderful feature that it's antiderivative is itself. So e^x and the antiderivative is e^x. And now I evaluate, this means evaluate this from 2 to 1. So I'm going 1 to 2. So I first plug in the 2 and then subtract off the one. And so this equals e^2, and then I subtract off what I get when I plug in 1, e^1 power. And so this = e^2-e. That's one way of writing the answer. Or another way of writing the answer if you want to is to factor out the e. And see e(e-1), either one of these, it's the same number. And that's the exact answer. That's what the area under the curve is. That's the area of the yellow. You wonder just roughly what that equals numerically so you get a sense of this. It's around 4.6707... and so on. So you can see it's a little bit more than 4 units^2 is this area here in yellow. Absolutely amazing that we can now find the areas of these really exotic regions.
Let's try a trig example now. We solved one earlier, let's try another one. Let's look under the cos function. And, in fact, let me state the problem and let's actually set up the problem without ever even visualizing it. And then I'll draw you a picture. Just to show you that you don't really need to visualize these things anymore, as long as you understand what you're doing. So in this example, I want you to find the area under the sin(x) function, but above the floor, above the x-axis, between x = pi/4 and x = pi/2. Okay let's set this up. Well that area, I'm deliberately not drawing a picture at this moment, but I promise I will, will be the integral or the sum. And where am I going to be going from? Well from my low point, which is pi/4, 45 degrees for you folks that are still on degrees rather than radians. These are all in radians here. All the up to pi/2, which is 90 degrees. And what am I integrating? I'm integrating (cos(x))dx. So that's the integral you have set up. And now visually, what's going on? Well just to prove to you that there was no need for this picture, I set it up in advance. But now let me draw you a picture so you can see what the picture really is, because of the visual thing, of course, around here. And we draw part of the cos function for you. There's part of the cos function, f(x) = cos(x). And where are we looking? Are we looking from pi/4, which is right over here, pi/4 and we're going to pi/2. And pi/2 is right here. So the area I want is sort of this triangle, it's not quite a triangle, because the hypotenuse sort of bulges out. You see it sort of bulging out? It's sort of bulging out a little bit there. So it's not quite a right triangle. It's sort of this curvy bulge here. And I want to find the area of that. And it turns out that's exactly this. Now you can see it. I'm summing up the rectangles' base x height, base x height cos, from pi/4 to pi/2. All right let's do it. First I have to find the antiderivative. The antiderivative of cos is sin. Don't believe me? Stop and check. Derivative of sin is cos. So sin is the antiderivative of cos. And now I evaluate first at pi/2 and then at pi/4. And what do we see? Well this gives us the sin(pi/2) - sin(pi/4). I plug in the pi/2 first and then I subtract when I plug in pi/4. And what is the sin pi/2? This requires a little bit of thinking, to think about what exactly is it, but the answer is 1 minus. And what's the sin of pi/4? Well you either look at the little chart that I showed you how to make. Or you can think about some other way of remembering that the sin 45 degrees is the square root of 2 divided by 2. So you can actually combine this and write this as 2/2 - this, so it's 2 - the square root of 2 all divided by 2. And so that number represents the area under this curved little triangle there. Absolutely incredible that we can calculate that exactly. This is not an estimate. This is the exact area under this curve from here to here, really amazing.
Okay how about one last one? And the question here is the following. Find the area under the curve -x^2+6x-8, that's above the x-axis. So I want you to find the area under that curve that is above the x-axis. And that's all I tell you. The first thing you have to figure out is where's the curve above the x-axis? Well let's write that curve down, write that function down, because that is going to require us to do a few extra steps before we actually find the area. So the function, I believe, was -x^2+6x-8. I want to find the area under that curve that's above the x-axis. So let' me see if I can, first of all, figure out what this thing is. This is obviously a parabola. It's a sad faced parabola. So the parabola's actually going to go down like this, because I see the negative sign there. But I don't know how it's going to go or how it's curving or anything. Well let me find out when this parabola crosses the x-axis, because notice that once I know my crosses the x-axis, I'll see where it's above, and that's the area they must want us to find. Everything else we can ignore. So how do I find out where this curve crosses the x-axis? I want to know when on this curve is y = 0. So I'll set this equal to 0 and figure out when it crosses the y-axis. So if we do that, let's see what we get. So I have -x^2+6x-8 = 0. And I want to factor that. And so how would you factor this? You know, in fact, you know how I'd factor this, by the way, just to let you know on a little secret? I would actually multiply everything through here by -1. If I multiply both sides by -1, it's not going to change, but then I've got a +x^2 here. It might make it a little easier. So I'm just going to multiply through by -1. So x^2-6x+8 = 0. You don't have to do that if you don't want to, but I find it sometimes a little easier to factor. So then I have an x and an x. Both signs will be the same, and they'll both be negatives. 2 numbers have to multiply to give 8 and combine to give 6. So it sounds like 4 and 2 look pretty good. This is -4x-2x = -6x, -2 x -4 = +8. Great, so therefore, I see x either equals 2, when that equals 0, or x = 4, if x-4 = 0. And so that tells me that, roughly speaking, this parabola must look something like this. It must cross at 2 and at 4. We know it's a sad parabola so, roughly speaking, the curve must go something like this. So the area above the x-axis and below the curve is right here. So notice that in the question that we weren't given, we weren't given the points of integration, we weren't given those endpoints. We had to first find them using this procedure to find out where those are. Now we see it's from 2 to 4.
Now we can set up the integral pretty easily. That area is the integral from 2 up to 4 of, well, the parabola, which is (-x^2+6x-8)dx. How do we integrate? Well we'll integrate term by term. And if we integrate term by term, what do we see? Well the f(-x^2), that's going to be -x^3/3, f(6x) is going to be +3x^2, right? Take the derivative, it's 6x, and then -8 is -8x. You can just check the answer by taking the derivative of this and see you get this. Now evaluate that whole thing from 4 to 2. So first I plug in 4. And I've got to plug in 4 everywhere. So this equals, plugging in 4 I see 4^3/3+3 x 4^2-8 x 4. So that's what I get when I plug 4 everywhere in. And then I subtract what I get when I plug 2 in everywhere. So I see -2^3/3+3/2^2-8 x 2. Okay now something is wrong here. Can you figure out what's wrong? Let me give you some choices and see if you can figure out what's wrong. Choice number 1 for what's wrong somewhere here is that the antiderivative is incorrect. Choice number 2 is that I should have first put in the 2 and then put in the 4 and subtracted them. So these should have been reversed. Choice number 3 is that I should be adding here and not subtracting. And choice number 4 is that some parentheses somewhere are missing. Vote right now for which one you think it is. Okay, all those guesses, I think, were good guesses, but of course it's the standard mistake that you've got to subtract everything. You've got to hit that negative sign. That negative sign's got to be distributed across all these terms. Don't be stingy with that negative sign. Share it amongst all these friends. That -8 there wants 2 negative signs. That +3 wants 2 negative signs, so you've got to distribute across the board. And when you do that, and you figure all this out, I think you actually see the number 24. And so the answer, I believe, if I did the arithmetic correctly, is 24. So the area under this curve is 24 units^2. So if they were inches, it'd be 24 inches^2.
All right. Well next up we're going to take a look at some more elaborate situations, where we're going to be looking at even some more exotic functions, and look at their areas under the curves. Stay with us.

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