Preview
You Might Also Like

Calculus: Integrals of Trig and Exponent Functions 
Calculus: Derivatives of Vector Functions 
Calculus: Solving Separable Differential Equations 
Calculus: Trig Substitution  Definite Integral 2 
Calculus: Manipulation, Integration by Parts 
Calculus: Derivatives of Parametric Equations 
Calculus: Calculus I in 20 Minutes 
Calculus: Cusp Points and the Derivative 
Calculus: A Shortcut for Finding Derivatives 
Calculus: Derivatives of Inverse Trig Functions 
College Algebra: Solving for x in Log Equations 
College Algebra: Finding Log Function Values 
College Algebra: Exponential to Log Functions 
College Algebra: Using Exponent Properties 
College Algebra: Finding the Inverse of a Function 
College Algebra: Graphing Polynomial Functions 
College Algebra: Polynomial Zeros & Multiplicities 
College Algebra: PiecewiseDefined Functions 
College Algebra: Decoding the Circle Formula 
College Algebra: Rationalizing Denominators

Calculus: Derivatives of Inverse Trig Functions 
Calculus: A Shortcut for Finding Derivatives 
Calculus: Cusp Points and the Derivative 
Calculus: Calculus I in 20 Minutes 
Calculus: Derivatives of Parametric Equations 
Calculus: Manipulation, Integration by Parts 
Calculus: Trig Substitution  Definite Integral 2 
Calculus: Solving Separable Differential Equations 
Calculus: Derivatives of Vector Functions 
Calculus: Integrals of Trig and Exponent Functions
About this Lesson
 Type: Video Tutorial
 Length: 14:00
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 150 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: Basics of Integration (14 lessons, $23.76)
Calculus: Antiderivatives (3 lessons, $5.94)
This lesson is an introduction to integration and antiderivatives. Where calculus begins with differential calculus (studying about derivatives), this is the beginning of the next portion of calculus in which we learn how to undo derivatives or antidifferentiate. Antidifferentiation is the process by which we look at a function and determine what function it was before it was differentiated. To this end, Professor Burger will show you how to use the derivative formulas in reverse.
This lesson is perfect for review for a CLEP test, midterm, 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 UTAustin 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, padic 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
 2174 lessons
 Joined:
11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" 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 technologybased 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 starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
 You have reached me!
 09/03/2010

excellent  getting an A in calculus (because I watch these videos) I may be speaking for myself but I almost literally never go to class and yet feel so motivated by these videos that I often spend time outside of class and outside of these videos, thinking about about calculus! May sound far fetched but watch Prof Burgers videos and I bet you find yourself doing the same thing! Thanks Professor for the excellent instruction!
The Basics of Integration
Antiderivatives
Antidifferentiation Page [1 of 2]
First of all, I really want to extend my sincerest congratulations on how far you’ve come. You know, calculus really is, I think, a tremendous and dramatic intellectual challenge, and part of it is to conquer and understand some really deep important ideas and you have just journeyed through that process. And we just covered that material, that body of work, that’s known as differential calculus, studying about derivatives. And really, that’s all we’re going to say about derivatives. And so, in some sense, we’ve come to a closing act here of our show, but now, what I want to do is regroup and start it fresh. So what better to start than with the games? I love games; games are great. So in fact, it’s time for one of my favorite games; it’s time for Math Jeopardy. That’s right, Math Jeopardy, where it gives me the answers and all you have to do is come up with the questions. In fact, wouldn’t you love Math Jeopardy on your next exam, to give you the answers? Okay, let’s try here and see if we can make some progress in Math Jeopardy and, in fact, we really want you to play along right now. And so, in fact, we’re going to give you the answer and we are going to ask you to produce the question, and then you actually enter it in. Please make sure that your response is in the form of a question. Here we go. One hundred dollars. Its derivative is 2x.
Okay, so now we know the answer. Let’s see if we know what the answer is. Okay, what we want to do here is think to ourselves, what question has its answer that’s derivative is 2x? Well, let’s think about that, that theory comes from a function and what function is it? So, we have to ask ourselves what is F of X equals—now what function has the property that if you take the derivative of it, you get 2x? Well, this is the first time you’ve ever been given an answer like this. In particular, usually they ask us the question, find the derivative, and then we say 2x. Now, we want to work backwards. So what would the answer be? Well, let’s see, I guess since we subtract one when we take the derivative, I guess we have to put an extra one up here to compensate for the fact that if we subtract one to get the derivative, so maybe we should put a two there. So then the answer should be x2 let’s think about that. If F of X = x2, what is its derivative? Well, its derivative would be, well, 2x. So instead, I think that is a good question, because its derivative is, in fact, 2x. Great, there you have $100—there’s a hundreddollar category, hope you got that one right. We are entering the $200 category; it’s going to get harder. Here we go. Its derivative is 3x2.
All right, let’s see how we make out on this one. This one’s going to be a question, so what is F of X equals? Okay, well now I see a power two there, so let’s see, how am I going to compensate for that before hand in order for my derivative to actually have a two there. When I take the derivative, I subtract one from the exponent, which means I probably should add one in now, so maybe I mean x3. Now if I put x3 here, what would the derivative of that be? Well, that would be 3x2, oh look! That’s exactly this, so in fact, x3 seems like a great question to ask. Terrific, how are you making out? You could be as rich as $300. Let’s continue. Now, we come to the $400 category. These are getting harder. Let’s see how we do with this answer. Its derivative is sign X.
Well, let’s see, there’s no powers to add in here because there’s no exponents really. So, let’s think about what the question’s going to be. What is F of X equals? Well, it’s a trig function, let’s think about the trig functions we’ve studied. Maybe the answer should be cosign. Let’s think about that, the derivative of cosign, you remember, is actually minus the sign of X, and that’s not what’s written here. So, what should we do? Well, in fact, what I want here is a plus sign of X, so maybe I need an extra negative sign. An extra negative sign, what will that do? Well, if you take the derivative of minus cosign of X, then what would I see? Well, the minus sign I put a whole new pattern as a constant, the derivative of cosign is minus, sign and that minus and this minus combined to give me a plus. So, in fact, the derivative of minus cosign X is indeed sign X. It required me to sort of mess up a little bit the coefficient here to make that work out just right. So that question, by the way, I need a question mark, otherwise I will get no credit. That question was a bit harder. Okay, that was a hard answer, but it was worth $400. Okay, $800, this should be twice as hard as the last one, get ready. Here’s your chance to make some money. Math pays. Its derivative is 1/ x2.
Well, let’s take a look and see if we can come out with the right question. So, what is—we’ll have to start this way—F of X equals… Okay now, what should we do here? Well, I’ve got a denominator here, maybe there’s a need for the quotient rule somehow to untangle, I don’t know, but I could write this, of course, as x2. Now, if I write that as x2, then I’m back to these other problems where the exponents are there. So, if I wanted to have the derivative be x2, maybe I should add one since I take the derivative, I subtract one. So, if I add one to the x2 in the exponent, I would see x1. So a good guess seems to be that we should guess x1. Let me write that down as a good guessing, this is the stuff that you never see on TV, the people making a good guess. This would be x1 as our guess. Well, let’s take the derivative of that and see what that is going to give us this as an answer. So, F prime of X would be, now what do I do, I bring down the negative one in front and then I have X to the—and I subtract one and I get –2, which equals –1/ x2. Again, so close to the actual answer, but notice this time again, I’m off by a negative sign. How can I fix that? Well, if I just put in an extra negative sign right in front there? If I put in a negative sign there, then it would actually appear right in here and that negative and this negative would collapse and give me a positive. Great. So, in fact, the answer I do want is not x1, but it is in fact, x1, which I could write very fancily as –1/x because that negative exponent means down below. I could write this as –1/x. So, that’s the question to ask for this one. Terrific, how are you making out? How much money have you won? Well, we’ll see what you are going to do in the final Math Jeopardy.
So, now, what you have to do is you have to wager as much of your winnings as you want on this last answer. Okay, I’ll give you a second to think about how much want to actually risk. Okay, all right, now, the question that you have to deal with is the following answer and please make sure you phrase your final question as a question, here we go, get ready. Its derivative is the square root of X.
All right, well this one, I think, is pretty difficult. Let’s take a look at this together and see if we can figure out what function has that as its derivative. Well, how would you possibly even attack this? Again, let me get a little scratch work here. What are we are going to think about first? Well, first of all, the square root of X is X½. Now, if I want that to be the answer, to take the derivative of this, well, that means that I should probably add one to this to figure out what the original function was because remember, when you take a derivative, you subtract, and if I subtract one, I end up with a half. I probably have to start off with X½ + 1 which is X3/2. So, my first guess is going to be that F of X equals X3/2, let’s see if that is a good guess or not. Let’s see that’s a good guess, so F of X equals X3/2, what would the derivative of that be? Well, the derivative, I would bring the 3/2’s in front, 3/2 X 3/2 –1 which is ½. Well, that’s actually pretty close because that’s 3/2 times the square root of X, that’s really close, but I don’t want those 3/2’s in front of there, that’s not so good. I want to have them not occur so maybe I should put some coefficient in here so that this will cancel whatever coefficient I write. Well, what coefficient could I write in there so that I get that cancellation? Well, let’s put the reciprocal of that, maybe 2/3 up here, because when I put that in a holding pattern and I have a 2/3 out in front and I have a 2/3 times 3/2, then they are canceling. So, I come up with x½. So, in fact, I think the final answer, or the final question, is what is 2/3X3/2?
All right, well, how did you make out? I bet you got a lot of money on that last one and you know, I think that last few were actually pretty challenging because, in fact, you had to think about how to modify the function from the original guess. You remember, the first answer here was 2x, the next answer was 3x2, the next answer was sign of X, and then we had 1/x2, and then here we had the square root of x. So, all of these represent the derivatives of these functions individually. Well, this whole exercise, I think, is a wonderful illustration of the kind of process we now want to undertake. In particular, the process of moving backwards. Up to this point people have handed us functions complicated functions or easy functions and that asked us to find the derivative. Now, what I invite us to think about is the following interesting question: what happens if someone hands us the derivative and we want to find the original function back? Well, it turns out that this issue is known as antidifferentiation, antiderivative. It’s undoing a derivative. How do you proceed in undoing derivatives? Well, actually we are getting inspiration of how to proceed right here with these examples. In fact, let’s try to look at a very general case; let’s look at a general case here, for example.
Let’s see if we can figure out how to proceed. Let’s suppose that I take a function, let’s say F of X, and it equals X to a power. So, let’s say it equals X to a power; let’s use an example, like let’s say seven. Well, I want to do—I want to find a function, which has the property that, if you take the derivative of the function, you get back to here. So, what would I do, what’s my thinking? Well, my thinking is to write something down and then take the derivative of it and have it return to this. So, the process we’ve been seeing is we take X and add one, so we have an eight, and if we take the derivative of that just as it is, I bring the eight in front. Do you remember how to do it? We bring the eight in front, that means I would have an 8x7, well, I have no eight here. So, I just immediately divide through by that constant, so that when I bring the eight down, it would be annihilated by this division by the eight over here and it goes away and I’m left a clean x7. So, it’s my guess would be this function G of X, and notice, that if you take the derivative of G of X, we get just X7 and this is a notion of antidifferentiation, the notion of given a derivative, now find the first function.
Okay, up next, we’ll take a look at a formula that will give us the antiderivative for any function of this sort. And not surprisingly, we’ve already did all the work, it’s just a matter of thinking backward, so it’s going to be sort of exciting for us to sort of engage our minds. We’ve always been going this direction, and now we are going to put ourselves in reverse and try to intellectually think and untangle all of the stuff we’ve done. Okay, stay tuned and I hope you won a lot of money. Math always pays. I’ll see you in a bit.
Get it Now and Start Learning
Embed this video on your site
Copy and paste the following snippet:
Link to this page
Copy and paste the following snippet:
excellent  getting an A in calculus (because I watch these videos) I may be speaking for myself but I almost literally never go to class and yet feel so motivated by these videos that I often spend time outside of class and outside of these videos, thinking about about calculus! May sound far fetched but watch Prof Burgers videos and I bet you find yourself doing the same thing! Thanks Professor for the excellent instruction!