Hi! We show you're using Internet Explorer 6. Unfortunately, IE6 is an older browser and everything at MindBites may not work for you. We recommend upgrading (for free) to the latest version of Internet Explorer from Microsoft or Firefox from Mozilla.
Click here to read more about IE6 and why it makes sense to upgrade.

Calculus: Evaluating Definite Integrals

Preview

Like what you see? Buy now to watch it online or download.

You Might Also Like

About this Lesson

  • Type: Video Tutorial
  • Length: 12:54
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 139 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.

About this Author

Thinkwell
Thinkwell
2174 lessons
Joined:
11/13/2008

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...

More..

Recent Reviews

Nopic_grn
Definitely the best on definite integrals pract...
04/27/2014
~ Jennifer68

A great combination of "DIY" problems! Practice and use U-substitution, breaking terms apart and intuition! Exactly the kind of review I needed before my test!

Nopic_grn
Definitely the best on definite integrals pract...
04/27/2014
~ Jennifer68

A great combination of "DIY" problems! Practice and use U-substitution, breaking terms apart and intuition! Exactly the kind of review I needed before my test!

The Basics of Integration
The Fundamental Theory of Calculus
Evaluating Definite Integrals Page [1 of 2]
All right, so now we're taking a look at these definite integrals. Those are integrals that actually have the points of integration marked on the top and the bottom, and those definite integrals are found by first finding the antiderivative of the function, then plugging in that top value into the antiderivative, and then subtracting from that the value you get when you take the bottom number and plug that into the antiderivative. So I thought just to take a moment's lapse from thinking about areas of exotic regions, let's just concentrate on definite integrals. So I want to give you just a couple of question here to try and, in fact, let's let this be a little interactive--get a chance for you to actually contribute.
So the first question I'd like for you to think about is to compute this definite integral. The integral from zero to one of x square root of x^2 + 1dx. So, of course, the geometrical significance of this is that this would be the area under the curve, x square root of x^2 + 1 from zero to one. But let's forgo that and just ask you to compute this thing. And, in fact, I ask you right now to really sit down in your seat, because where else would you sit down? And try it for yourself right now, and then we'll be back and do it together. Please give it a try.
All right. Well, I realize this problem has to be done by a udu substitution, because I've got square root of a blop here, and the derivative of the inside, notice, is 2x, which is basically the outside here. So if I make a little substitution--and I'll make this literally a little substitution--let's let u = x^2 + 1, then du would equal 2xdx. But I don't want 2xdx, I just want xdx, so I'll divide both sides of this by 2 and see du equals xdx. So xdx will be replaced by just du--let's do that, du. , I'll pull it out in front, du. And then, what about this piece right here? Well, that piece is just the square root of u, which I'll right as u^1/2. Now, if we integrate, what do we get? Well, I have , and then I have u^3/2 and then I divide through by 3/2, which is the same thing as multiplying by 2/3. I see a little cancellation, and so I see 1/3u^3/2.
Now, you may be tempted at this point to plug in the zero and then the one. But you've got to curb that temptation, because you have to remember that this zero and that one, those are x values. This is a u. So first I suggest that you untangle that u by replacing it by what it equals. So I would now write this as 1/3 x the quantity x^2 + 1^3/2. And that is now the antiderivative of its original function, so now what we want to do is we want to evaluate that at these points. So what I see now--we pick up the action--if I call this thing--let's call this thing star. And what I would see is that star is the same thing as, well, 1/3 x^2 + 1^3/2. But now I'm going to evaluate that from zero to one. When I plug in the one, which I have to do first, I see 1/3 and I see 1^2, which is 1 + 1 is 2. So I see 2^3/2, and when I plug in zero into here, I get just a one with a third, so I see -1/3. So the answer to this one is one-third, 2^3/2 - 1/3. 2^3/2--we could have written that as 2^3, which is 8^1/2, which is the square root of 8. So we could write this as the square root of 8, if you wanted to. A lot of ways of writing this. But it turns out that's the exact value of this integral.
So I hope you got that correct, and I hope that if not that you made some progress. It's okay not to get these things correct, but at least, hopefully, you can make a little bit of progress. What's not really okay, or not great, is if you don't even try, because then you're not going to get a sense of whether you're really sort of making this material your own or not. So I hope you tried it, and in either case, I now invite you to try this next one.
So here's the next question. It's finding the definite integral from one to two of 2x - 1 all over x^3 dx. And before you actually move forward on this, let me just give you a little warning. This problem is a teeny bit sneaky. So see if you can find the sneakiness of this problem when you attempt to answer this one. Okay, good luck.
Well, you look at this and you probably are tempted to use integration by substitution. But you'll notice that if you let the bottom be equal to u, then the derivative would be 3x^2. And I have no x^2 here. So, in fact, I don't know how to use substitution on this problem in a manner to give me something that's easier to deal with. What do you do? Well, this is a sneaky problem that I wanted to share with you, because the answer is, that since there's only one thing on the bottom, and I'm subtracting two things on the top, I can break this fraction up into two fractions. I can write this as 2x/x^3 - 1/x^3, because remember, that if you have the same bottom, you just combine the top. So that would be 2x - 1. Well, that's 2x - 1. So in fact, this one fraction can be split apart into these two fractions, as long as the bottoms are the same. Well, now I can simplify this dramatically because I can cancel and see 2/x^2 - 1/x^3dx. And now you see each of these pieces individually I can actually integrate. So this is a problem that looks, upon first inspection, like a udu substitution, but in reality, you have to break it apart and then you can just do it directly. Let me just write out one more step here. So this is 2x^-2 - x^-3. Now you can really see what the integral will be. So the antiderivative will be what? Well, if I add one here, I would get a -1, so I would see a -2x^-1, and then when I add one here I get a -2, so I'm going to divide by -2 which makes this a +1/2x^-2. And you can check that by taking the derivative and seeing that if I bring this minus sign out, I get a 2x^-2-. And here, if I bring that -2 out, I see a -x^-3. You can check that.
Now, I evaluate this from 1 to 2. When I plug in the 2 in there, what do I see? Well, I'm going to see what? In fact, remember that this minus exponent means down beneath, so I see a -2 divided by that 2 + , and this, again, goes underneath as a 1/2^2, which is 4. So that's what I get when I plug in the 2.
Now, let's not get confused. All I just did was plug in the 2. I now have to subtract off what I get when I plug in the 1. So let's not lose sight of where we are here. I put a big old parenthesis there because I want to subtract everything I write down. Plug in a 1 here and I just see -2 and then plus--and I plug in a 1 in here I just see a half. So what does this equal? Well, this is just a -1. This is, I guess, +1/8. This is a minus and minus which is a +2, and this is a -1/2. So this 2 minus 1 is going to be a 1, and then an 8^th minus a half, what's that? Well, this is actually 4/8, and so I have 1/8 minus 4/8, which is minus 3/8, and 1 is actually 8/8, and so this would be 8 minus 3, I guess, which is 5/8. So this original integral, this definite integral, which represents a number--represents the number 5/8. So we're thinking about that as we look at the area under this curve, from 1 to 2, it will give you 5/8.
Okay, great. So here you can see the example which looks like a udu substitution, but really isn't. It's a good thing to watch out for, because people love to ask other people that kind of trick.
And now the last thing, which again, I think is just as tricky as the previous one, but is different, is the definite integral from zero to 1 of the square root of 1 - x^2 dx. And let me give you a hint here. First of all, a word of caution, warning. This is, I think, very, very tricky. That's hint one. Hint two, I'll bet you won't be able to integrate that. That's hint two. Hint three, think about what the significance of this may be. What does this represent, and maybe you can actually figure it out. I think this is really tricky. Don't be frustrated if you can't nail it, but please give it an earnest try. Give it a shot. We'll see how you do.
All right. This was really awful of me to give you, because if you attempt to do this problem you might try to do a substitution. But see, if you let u be the inside, then the derivative of that is -2x. So you would need an x somewhere on the outside to actually be able to integrate that using substitution. Well, we don't have an x on the outside, so this looks like a problem that's actually really difficult to find the antiderivative of problem. So what do you do? Well, the hint I gave you for this very special problem was to think about what this means. Well, this means some area under some curve, in fact, this curve. Now what curve is that? Well, like I said, this was really tricky, so don't be mad if you didn't think about this. What's the graph of that function? Well, I don't know, maybe, at the moment. But what if I square both sides to get rid of that radical? Then I'd see y^2 = 1 - x^2, because the radical lifts--and if I bring that -x^2 over to the left hand side, I would see x^2 + y^2 = 1. I'd recognize that that's actually a circle that's centered at the origin of radius 1. If I draw a picture of that it would look like this. I got new markers. Look how beautiful they look. I love new markers.
There's the circle of radius 1. Now what's the graph of this function then? Well, since this is just the positive square root, it's the upper half of the circle. So it's the upper half of the circle, so this graph is actually the upper half of the circle. Now, I know that a lot of people wouldn't have thought about this, but it turns out that this--not that there, but the one right on 1 is the graph of y = the square root of 1 - x^2. It's the upper hemisphere. The minus square root gives me the lower hemisphere. But this is the positive here--we have this one. Okay, and then where are we looking? From zero to 1. So from x = zero, which is here, out to 1. So I'm looking actually for this area right in here. Well, that's the area of this quarter of a circle. But we know the formula for the area of a circle. The area of a circle--what is that? Well, that's pi r^2, which in this case is pi x 1^2. So that's the whole circle. The yellow is a quarter of it, and so this integral would be equal to a quarter of this, which is pi/4.
Isn't that sneaky? That really was sneaky, and I apologize, but I think it's a great lesson. The lesson is that sometimes if you think about the graph inside of here, and then consider the problem as an area problem, you might actually be able to resolve it, even though finding the actual antiderivative of this still eludes us at this time. But we were able to answer the definite integral question.
Okay. Up next we'll look at more exotic regions, and in fact, put two curves together to take a look at really fancy regions. I'll see you there.

Embed this video on your site

Copy and paste the following snippet: