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Calculus: Regions Bound by Several Curves


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About this Lesson

  • Type: Video Tutorial
  • Length: 11:14
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 121 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: Applications of Integration (10 lessons, $16.83)
Calculus: Finding Area between Two or More Curves (4 lessons, $7.92)

In this lesson, you will learn how to find the area of a region bounded by 3 or more curves. This is basically a semi-advanced application of integration and antidifferentiation. To calculate the area of bounded regions, you will have to break the integral into multiple different pieces wherever the curves bounding the regions switch. In the video, you will learn how to find points of intersection between different region boundaries and how to calculate the areas based upon the combination of different antiderivatives.

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

2174 lessons

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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Application of Integration
Finding the Area Between Two Curves
Regions Bound by Several Curves Page [1 of 1]
Well, up to this point we’ve actually seen that we can find the areas between two curves and figure that out by taking the top minus the bottom. I thought I’d just do one quick example with you to show that, in fact, you can find the area bounded by three or more curves. It’s just a matter of sort of cutting up and dissecting the region into the right number of spaces. So, let me actually just give you a very specific example. Let’s consider the following three functions: y = 3x + 4, y= x2, and finally, y = -x + 2. And the question that I want us to answer is: “How would you go about finding the area of the region bounded by these three curves?” No longer the area of regions bounded by two curves, but now by three. Well, the first step is to actually sketch a reasonably accurate picture of this so we get a sense of what the region looks like. And, by doing so, you’ll get a sense of how you have to go about attacking this problem. Well, I’ll try to graph this for you right now live on the fly here. Now, usually when you graph these things, by the way, it takes a couple of tries to actually graph it correctly. But, hopefully, I’ll try to do it well on the first try. So, y = x2. So, we all know what that looks like, that’s a happy face parabola. Do that one in blue here. Not too bad, but not perfect. And let’s look at these things. Well these are both y, so let’s just graph them, this is y = 3x + 4, and so that’s going to be a line that has slope 3, so very positively sloped and intercepts the y axis at 4, so it’s way up here. Let me draw that in green for you, draw that in green, something like this. This is the line y = 3x + 4, and then the other line is y = -x + 2. So that slope of –1, that is slope –1 it is a negative direction and it goes through the point 2, so it might look something like this. Try and see if I can capture the spirit of this, and this would by y = -x = 2. And now you can see the region that these three functions map out, it’s right in here. It’s sort of, what does that look like? What is that? It’s sort of rounded on the bottom and it’s got two openings on the side, I don’t quite know what that—you know those toys when you were a kid that would sort of, you know they would always wobble, but they won’t fall down. It’s sort of like one of those things, right, because it sort of rounds on the bottom, but then the top is sort of weird, I don’t know. It doesn’t really look like the figure to me, though, but you get the idea. Anyway, how would you find the area of that?
Okay, we see top minus bottom; it’s a little tricky right? The bottom I guess is always clear, the bottom is always the beautiful blue parabola, but the top actually varies. The top actually varies. So, what you have to do, what we have to do, in fact, is split up this region so that, into pieces, so that each piece within that piece, we have a uniform generic rectangle. You see, and I put a rectangle right in here, how is that described? Well, that would be described as orange minus the parabola, blue. And it’s always like that for a while, but then notice what happens after some point, all of a sudden now, it’s different. It’s not orange minus blue, which is way up here. That’s orange minus blue. What I want is green minus blue, so all of a sudden the very shape or the very description of the generic rectangle has changed and now I’m looking at green minus blue. So, somewhere I’ve changed course, so I need to cut this into pieces where each little piece is the same describable generic shape. Now, where would you cut it? Well, I think it’s reasonable to guess that I should cut it where these two things intersect. So, I should cut the region up right here. And from this point to this red, I see that all the rectangles are describable by orange minus blue. And after the red, I see all the rectangles in here are describable by green minus blue. So, in fact, that’s a good place to cut it up, find the area of this piece, find the area of that piece and then add it all together. Okay, well, let’s proceed by figuring out how to do that. I think the hardest part here is to find all the points of integration. So, how do I find these endpoints? Well, to find these endpoints, what I have to do is find out where the orange intersects with the blue. So, let’s go off and do that. How do I find this point right there? I’ve got to find where the orange intersects with the blue, so I’ve got to set the y value for the orange equal to the y value for the blue and find out what x value gives me the same y values. So, if I set them equal, I see x2 = -x + 2, which gives me x2 + x – 2 = 0 and let’s see if that can factor. That would be an x and an x, they’re both going to have different signs and it looks like a 2 here and a 1 here. So, it’s going to work quite nicely, and so I see x = -2 and x = 1. I seem to be getting two answers, which is a little disturbing. Well, no not disturbing when I realize that the orange line, of course, doesn’t realize the problem at hand and it just keeps on traveling and sure enough does hit the curve again. And that point is certainly negative and must be in fact, a –2 answer. Our answer from the picture is probably the one that’s a little closer in, but to the right; it’s the one answer. So, this must be at 1, so it we do get two answers, but in fact, only one of them is the one we want, by the picture, the –2 is way out here, which we aren’t interested in.
Okay, great, now how about this point? Well, here we’ve got to do, is solve the green and the blue together and if we solve the green and the blue, let’s see what we get. I want to set the y on the green equal to the y of the blue. So, x2 = 3x + 4, or in other words, x2 - 3x – 4 = 0. Can we factor this quadratic? Let’s see we have a negative sign so again, plus or minus, and looks like four and one is going to work quite well. So, I see x = -1 and x = 4. So again, I see two points, well the –1 is planted here and the other one is way off the computer screen here at four. So, we are not interested in that one, we’re interested in this one right here and that was the –1. So, that’s great. So at –1, this one is at 1 and now, where’s the point of intersection? Well, the point of intersection, how do we find that? Well, that’s going to be where the two lines, in fact, intersect. So, I have to set the two lines equal to each other to find out this point, this red. So, 3x + 4, look at all this algebra we have to do, we haven’t even done any calculus yet, it’s amazing the amount of algebra sometimes required for the multi-stepped problems. So, we set this equal to the other line minus 2x + 2 and we bring this over and I see 4x equals, and I’ll bring that over and I’ll see that it is a –2 and so I see that x will equal – ½. So, in fact that intersection point is at – ½. This is what I meant by the way the first time you draw it; you might not get it right. Maybe you had those two lines intersecting way up here in the positive side. If you do that, by the way, don’t panic, just check your work and if you’re right, then obviously, then you sketched the graph a little too roughly. But in fact, I actually sketched this in advance, made that mistake, and then resketched it. So I don’t want you to think that somehow I’m some sort of math superstar for the reason I can graph it. The reason why I think I’m okay in math, is because I don’t mind trying, making mistakes, learning and moving on. Okay, anyway, now we have all the points we want. We have all the points we want and we’ve got this one point, we’ve got this –1 point, we have all the things we want here. For example, unless we get a check, if I plug that one into here, then what do I see? I see –1 + 2, which gives me a net gain of 1 and if I plug a one into here, I see one. So, you can see this point really is a point of intersection and you can check these other ones if you want to make sure that everything is really okay.
All right, now you need to set up a whole bunch of integrals and what are those integrals? Well, now we come to the calculus, which I don’t think is that big of a deal anymore. Let’s look at this first region right here, the one before the red. What is the generic rectangle look like? I would draw one in if I were you. And what I see is first where the rectangle started, -1 and they go all the way up to – ½, and then what am I adding up? I’m adding up the areas of the rectangles, base times height. Well, what’s the height? Well, the height here, you see is green line minus the blue parabola, so it’s going to be green line, which is 3x + 4, we’ve got to be careful, we’re taking the right line, minus the parabola, minus the bottom x square, cx. And now we have to evaluate that. And then we have to add to that, the integral from minus a ½, we start at the red and go all the way out to 1. And what are we looking at here? The areas of these rectangles, which look different, because they are yellow, or orange, minus the blue parabola. So, I see – x + 2 minus the parabola which is x2 dx. And now I get to work out each of these integrals and I’ll let you actually try these integrals yourself. I’ll just tell you that this integral right here, this piece turns out to be 7/12. So this one I’ll let you even compute yourself; I won’t even tell you this one. I want to really see if you can do it, so this little area in here is 7/12, this area here you can just evaluate this integral and do the arithmetic to add those two numbers up, you get the area that you want. The point of all this, is when you have more than two functions, you need to be very careful with how the curves break up the shape you are interested in, and you must break up that shape into a little pieces, little bite size pieces where you can bite, bite, and that answer all the put all the answers together and add them up and get the final answer.
Okay, well up next what I want to do is tackle a slightly different issue. An issue that’s less complicationally involved, but I think more interesting. And the question is, what if the curve that we’re trying to find the areas bounded by are really really exotic? Notice I said curves, I didn’t say function. I’ll see you in the next session, bye.

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