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Calculus: Fundamental Theorem of Calculus, Part II

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About this Lesson

  • Type: Video Tutorial
  • Length: 16:28
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 177 MB
  • Posted: 11/18/2008

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus Review (48 lessons, $95.04)
Calculus: Final Exam Test Prep and Review (45 lessons, $64.35)

The Second part of the Fundamental Theorem of Calculus provides the link between velocity and area. It states that the sum of the area under the curve between two points (A and B) is equal to the difference of the antiderivatives of A and B. Thus, to find the area under a curve between two points, you will take the difference of the derivatives calculated at the end points, A and B. This theorem enables you to evaluate definite integrals by finding the area between the function described and the X axis. The lesson will also cover proper notation that should be used to denote what you're evaluating over which interval. You will also work problems that involve trigonometric functions (like finding the area under a portion of the sine curve or cosine curve)

This lesson explains the second half of the Fundamental Theorem of Calculus. To see the fist half of the explanation, check out: http://www.mindbites.com/lesson/843

This lesson is perfect for review for a CLEP test, mid-term, final, summer school, or personal growth!

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. 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'Hôpital'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.

About this Author

Thinkwell
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Founded in 1997, Thinkwell has succeeded in creating "next-generation" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technology-based textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.

Thinkwell lessons feature a star-studded cast of outstanding university professors: Edward Burger (Pre-Algebra through...

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Recent Reviews

Nopic_tan
Great!
07/21/2010
~ Susan38

A clear, enjoyable presentation of examples illustrating The Fundamental Theorem of Calculus. "Painless!"

Nachan_homepage
Great Explanation!
01/19/2009
~ nachan

Professor Burger explains the idea of derivatives and the idea of trying to find areas. He shows many graphs and explains how to find the area of the sum a to b under f of x. He also explain anti-derivitaves of F and F of x. This a great lesson to help solve such a difficult problem.

Jason_reneau_pic2_homepage
Wish I had Dr. Burger when I was taking Calculus
11/19/2008
~ Jason

Calculus was one of those subjects that I was able to learn in school, but I always had problems intuitively understanding what it was all about. In this lesson, Dr. Burger does a great job of explaining one of the fundamental purposes of calculus, which is to define the area under a curve. He is a gifted lecturer and has a way of explaining things in an easy to understand way. Especially for a topic such as calculus, having a second professor such as Dr. Burger explaining core concepts will greatly benefit a lot of students.

Nopic_tan
Great!
07/21/2010
~ Susan38

A clear, enjoyable presentation of examples illustrating The Fundamental Theorem of Calculus. "Painless!"

Nachan_homepage
Great Explanation!
01/19/2009
~ nachan

Professor Burger explains the idea of derivatives and the idea of trying to find areas. He shows many graphs and explains how to find the area of the sum a to b under f of x. He also explain anti-derivitaves of F and F of x. This a great lesson to help solve such a difficult problem.

Jason_reneau_pic2_homepage
Wish I had Dr. Burger when I was taking Calculus
11/19/2008
~ Jason

Calculus was one of those subjects that I was able to learn in school, but I always had problems intuitively understanding what it was all about. In this lesson, Dr. Burger does a great job of explaining one of the fundamental purposes of calculus, which is to define the area under a curve. He is a gifted lecturer and has a way of explaining things in an easy to understand way. Especially for a topic such as calculus, having a second professor such as Dr. Burger explaining core concepts will greatly benefit a lot of students.

The Basics of Integration
The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus Page [1 of 3]
All right. Well now it’s time for me to tell you about the fundamental theorem of calculus, the theorem that really links together the 2 ideas, the idea of the derivative and the idea of trying to find areas. And basically, to think about this, let’s begin with this picture. We want to find the area under the curve and above the floor, above the x-axis between a and b. And here is the answer. The answer is, if you want to find the area, remember how we write this, I write that the area is the sort of the sum from a all the way up to b of the area under f(x)dx. And remember how to think about this. The way you think about this is I’m summing up a lot of small rectangles. They have height f(x) and they have a very, very tiny base, change in x. So this is base x height, and I’m summing them up from a to b. That’s the area under the curve. And what does it equal? Well it turns out that this equals nothing more than capital F(b)-capital F(a). Now that you have capital Fs, well where—capital F is the antiderivative of little f. So where capital F is the function whose derivative is f(x). So to find the area under a curve from a to b, the first thing you do is find the antiderivative and then take that answer, take the antiderivative, plug in b, the right hand point. And then subtract off what you get when you plug in a. This is the fundamental theorem of calculus. That areas under curves are nothing more than the antiderivative evaluated at one end point, evaluated at the other end point, and then subtracted.
It’s utterly amazing that somehow antidifferentiation, undoing the derivative, gives rise to areas under curves. And if you think back, or if you remember, or I now invite you to look at the very beginning of our discussions together, I mentioned that the two basic questions, finding instantaneous velocity and finding areas under curves, at the very end of that discussion, I said to you, “It turns out that the answers to both those questions are related.” And now we’ve finally come to the point where we can see the relationship. The relationship is that, while taking a derivative gives me the instantaneous velocity, taking an antiderivative allows me to find area, absolutely surprising, wonderful and amazing.
Okay well so much for surprising, wonderful and amazing, let’s try some examples. First of all, to see if this even is believable. Is this even believable? So let’s try and see if it’s going to be believable or not. So let’s consider the following example. How do you begin understanding something that you’re just immediately told? Well begin with simple examples where you already know the answer. That will at least give us some sense of whether things are going okay or not. So let’s consider the following simple function, f(x) = x. If you look at f(x) = x, that’s just a line. And that’s slope 1. It goes right through the origin. So this is f(x) = x. So it’s a very simple, simple curve.
And now let’s find the area under that curve from 0, so right here, to let’s say 3. From 0 to 3. I want us to find the area under that curve. Now on the one hand, this is not very hard, because this curve is so simple it’s straight. In fact, we know the area of that. This is just a triangle. In fact, it’s a right triangle. So we could find the area of this. There’s no need for calculus here. Why bother with calculus? Let’s just do it. This length is three and this height is three, because the function at x = 3. So this is 3 by 3. It’s a triangle. You might remember that the area of a triangle is ½ base x height. Or you may think of the entire square and realize this is half of the square. In any case, we see that the area is 9/2, because it’s 3 x 3, divided by 2. Okay so we know the area. Let’s now try to redo this problem using the fundamental theorem of calculus that we just talked about and learned and see if we get the same answer. How do I set this up? What I have to do, looking at the fundamental theorem, I have to now integrate from the left hand endpoint to the right hand endpoint. So I write the smaller number here and the larger number here. Those give me my endpoints, my points of integration. The starting point and then the ending point. And then I write the function, the top part of this shape, which is the function xdx. Remember for thinking purposes, think of this as the area of a little rectangle. And this is base x height, and I’m summing them up from 0 to 3. So imagine putting a lot of rectangles there. Each rectangle has the area base, delta x, dx times height function. Okay now how do we figure out what this equals? Well I need to find the antiderivative of this function. And the antiderivative of this is actually going to be x2/2. And you might be saying, “Oh yeah, x2/2 + C.” Well okay I’m going to put in the +C just this time and you’ll see that, in fact, that’s no longer going to be necessary when I have these endpoints here, I’m not going need the +C. But for now let’s put it in. Now what I want to do is—the fundamental theorem says that I have to evaluate this at the upper endpoint. So plug in 3 for x and then subtract off what I get when I plug in 0 for x. The way I’m going to write that here, by the way, and the way everyone uses is this notation of drawing a little vertical line, and then write 0 to 3. And what this means is take this thing and evaluate it. This symbol just means evaluate this first at 3 and then subtract off 0. So this symbol is just a symbol to say take this antiderivative and now evaluate it at 3 and subtract evaluate from 0. So this could equal—but when I plug in 3 into here, I would see 9/2+C. You’ll see the constant. You don’t plug anything in for that. This is just x you’re plugging in for. And then subtract what you get when you plug in 0. So 02, which is 0/2+C. And now what do you notice? Well I have a (9/2+C)-(0+ C). So just look at the C for a second. Forget about everything else. Notice that +C here? Well it’s here and it’s also here and I’m subtracting. So the Cs drop out. And you can see that, in fact, the Cs will always drop out. So in reality, when we actually do what’s called a definite integral, where we actually have these endpoints, and we’re going to subtract like this, there’s no need to write the +C. We only write the +C when we have an indefinite integral, an integral where there are no endpoints. Then we have to have the +C. Anyway, what does this give us? Well this is 9/2-0, so it’s 9/2. And look, what do we see? The same answer as what we already knew. So doing this calculus method actually produces the same answer as what we know the answer to be. So this at least is an illustration that this amazing method actually works for this example.
Let’s try a couple of other examples. That was an example where we actually could find the area, just because it was such a simple shape. It was just a little teeny triangle. Let’s try one that’s a little more exotic. How about part of a parabola? Now I hope you realize the power that can have in your disposal. You have just gone from not knowing how to find—yeah, we didn’t know how to find areas at all, to finding out areas under very interesting curves. So let’s suppose we take part of a parabola here. Let’s say f(x) = x2. And how about we look at the area under the curve from, again, 0, let’s just go to 1 this time. So let’s look between the area from 0 out to 1. Let’s look at the area under the curve, but above the floor. So now it sort of looks like a triangle, doesn’t it? But it’s not quite a triangle, because that hypotenuse sort of is curvy. So it’s sort of a curvy triangle, so the area is not obvious. You just can’t use a formula, because it’s a little curvy. Well calculus allows us to deal with that curvy part to find the area of the yellow. How do we set this up? Well the area would be an integral. We’re going to sum up infinitely small rectangles from 0 to 1. I put the left hand starting point here, the right hand point here. And I’m going to now sum up the areas of tiny rectangles. I draw in sort of a generic rectangle in here for thinking purposes, just to help us to understand there are a lot of rectangles, they are all stacked here. And I’m just pointing one out. Well then what’s the area of that? It’s base, which is a tiny change in x, so dx, x height, which is the value of the function x2. So first I have to find the antiderivative and then plug in these two points. So the antiderivative is just x3/3. I can put the + C in, but I don’t have to if I’m doing a definite integral, because that constant will subtract out when I put the C-C. And then I’m going to evaluate this. This means I have to evaluate this first at 1, and then at 0 and subtract. So this equals, plugging in a 1 first—you always plug in the big one first, the right hand endpoint—and then subtract off what you get when you plug in 0. Well what do we get? Well we get 13, which is 1/3-0. So we just get a third. So the area of this entire thing here, this thing which just moments ago was a complete mystery to us, we’ve now computed that the area turns out to be a very simple number. It’s just 1/3 units2. Isn’t that amazing. We now know what the area is under this extremely subtle curve. It’s curving very gently and lightly, and we see the exact area is 1/3. Look at the power we have at our disposal now. It’s amazing.
Let’s try one last example before I close this little discussion, and I hope this one will knock your socks off if those didn’t, or if that last one didn’t. If that last one didn’t, this one definitely will. Let’s take a look at the area under part of the sin function. The sin function, now what would that look like here? I’m just drawing part of the sin function. This is f(x) = sin(x). And how about we look at the area under sort of this first little hump here. So that would be from 0, and do you remember where this is? Oh boy, I hope you remember where that is. That’s actually at pi. It makes a complete cycle in 2pi and that’s half of it. So the area I want to look at is the area under that first hump and above the floor. So it’s this very pretty, very nicely curved region right here. What is the area of that? Well that seems like a complete and utter mystery. And it seems like what—I wish you’d guess what the answer would be. It seems very complicated. The sin function is so complicated. Maybe there’ll be square roots in there. Maybe there’ll be pi’s, maybe square roots of pi, pi2, because it’s the area2. It’ll be very complicated. There’ll be roots and things… it’s very subtle. Let’s see what the exact value of the area is.
So I set it up as the integral or the sum from 0, my left hand point all the way out to my right hand point, pi. And remember how I’m thinking about this. I pretend in my mind to stack this with a whole bunch of thin rectangles. I’ll draw just one in for your perusal, but understand that, in fact, there are a lot of them. They’re all there stacked up like little soldiers. And so what’s the area of that? Well it’s base, which is a tiny change in x, x height, which is the value of the function of x, which is sin(x). So the first thing I have to do is evaluate that integral, and then plug in pi and plug in 0 and subtract. So what’s the integral, what’s the antiderivative of sin(x)? Well the antiderivative of sin(x), we’ve already seen, is –cos(x). And remember you can always check your answer by taking the derivative and making sure you get sin. And if you do –cos(x), the derivative is –(-sin), which is sin. And then I draw this symbol which means evaluate it first at pi, and then at 0, and subtract those things. So what this means is first plug in pi. So I get –cos at pi, and then subtract off what I get when I plug in zero. Don’t forget there’s a minus sign there, cos(0). Please note that this minus sign is just the rule. We always take this value inside and subtract off this value inside. So that negative sign is really that one. And this is what we subtract, we always subtract the 2 things. Well what’s cos(pi)? Well I hope you remember your trig functions. Cos(pi), if you think about the function, I’ll draw it for you right now. It starts way up here. Here’s the axis. It goes down and then comes up. So where is it down? It’s down at –1. So in fact, this is –1, there’s a negative in front of it. So it’s –(-1) and then minus. And then what’s the cos(0)? Well it starts up here at 1 before it goes down, so I have a negative sign in front of that, -1. And what do I see, -1, -(-1) is 1. The area is 2. So the area under this extremely delicate and subtle curve turns out to be the extremely simple answer 2.
Isn’t that amazing? There’s real power to these ideas of calculus. They allow us now to investigate these very interesting objects, these very interesting shapes. Up next we’ll take a look at some more examples and start to build even more exotic shapes where we remove the floor and look at other things. Congratulations and I’ll see you in a bit.

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