WEBVTT 00:00:01.000 --> 00:00:10.608 Let's take a look at finding area. If we want to find the area of a geometric figure that's a regular 00:00:10.610 --> 00:00:16.502 geometric figure, we have formulas for that. What we're going to look at now is what happens if we want 00:00:16.504 --> 00:00:29.003 the area of a figure that is not regular. So let's take a look at how we would find the area below this line 00:00:29.005 --> 00:00:45.100 and above the x-axis and bounded on the sides by vertical lines that I'll label a and b. Well, we have 00:00:45.102 --> 00:00:54.602 no formula for this, so let's see what we could do. What has been done is we've divided this area into 00:00:54.604 --> 00:01:04.200 some familiar figures and then tried to work from there. So let's try that. Let's divide this into rectangles. 00:01:04.202 --> 00:01:19.102 When we do, you can see that what we get is not going to be exactly correct. So if I divide this into four 00:01:19.104 --> 00:01:34.102 intervals and I draw a rectangle that closely, but not exactly, approximates each of the four areas, 00:01:34.104 --> 00:01:40.604 I can get something that's close to the correct area, but we would like something a little more accurate 00:01:40.606 --> 00:01:54.302 than that. So here's what we do. The distance from a to b is given by b minus a. We're going to divide this 00:01:54.304 --> 00:02:00.003 by the number of rectangles in this. The number of, well they're rectangles, but the number of breaks in the 00:02:00.005 --> 00:02:09.701 x-axis. Here I have four, but I'm going to use n. Then what we're going to do is we're going to let this 00:02:09.703 --> 00:02:15.601 move over and every time you do that you get a more accurate rectangle. So the area is going to be more 00:02:15.603 --> 00:02:25.304 accurate. You will have more rectangles, and n will be larger. So that means we're letting n go to infinity, the 00:02:25.306 --> 00:02:35.605 number of divisions which is also the number of rectangles. Now how do we represent the area of one rectangle? 00:02:35.607 --> 00:02:47.700 Alright, the area of one rectangle would be this length, which is b minus a over n. We're going to call that delta x. 00:02:47.702 --> 00:02:57.000 Now because these divisions do not have to be exactly the same, it's easier if they are, but they don't 00:02:57.002 --> 00:03:05.202 have to be, we're going to call this delta x i to indicate that there are several of these and 00:03:05.204 --> 00:03:16.598 as n goes to infinity, more and more and more. This endpoint here at this side of this rectangle is f of a. 00:03:16.600 --> 00:03:26.502 We're going to call it f of x 1, and then we'll call this one f of x 2, and we'll continue 00:03:26.504 --> 00:03:36.805 in that fashion, so what we're actually going to do is call these f of x-sub-i. So what we have here is 00:03:36.807 --> 00:03:54.204 this is the width of the rectangle, and this is the length of each rectangle represented by delta x-sub-i 00:03:54.206 --> 00:04:03.201 and f of x-sub-i. As n goes to infinity, this representation that we're going to make becomes more 00:04:03.203 --> 00:04:12.901 and more accurate. So let's write it down. The area of this figure is going to be found by taking the 00:04:12.903 --> 00:04:23.101 limit as the number of rectangles goes to infinity, and then we're going to take the sum of the areas of 00:04:23.103 --> 00:04:35.404 the individual rectangles. The sum, and we're going to take it as i, this index, goes from 1 to n, 00:04:35.406 --> 00:04:46.502 which is the number of rectangles. Now how do we represent the area? It's the length, f of x-sub-i, times 00:04:46.504 --> 00:05:01.301 the width, delta x-sub-i. Now this is amazing, but it's also a lot of work, and it's very difficult to do. 00:05:01.303 --> 00:05:09.501 Over time mathematicians studied this and discovered something which is really amazing. They discovered 00:05:09.503 --> 00:05:15.202 that the value you get when you do this and you let n get larger and larger and larger, the number 00:05:15.202 --> 00:05:23.204 of rectangles larger and larger and larger, is exactly the same thing as.. 00:05:23.206 --> 00:05:33.203 Ok, this is the function f. You will get exactly the same value, the area, if you take the antiderivative 00:05:33.205 --> 00:05:49.105 of this function, evaluate it at b, take the antiderivative of the same function, and evaluate it at a. 00:05:49.107 --> 00:05:57.901 Now does this look like anything we recognize? Well, I hope so. I hope that you recognize that this is 00:05:57.903 --> 00:06:10.105 simply the evaluation of the definite integral. So we have defined this term to be the integral 00:06:10.107 --> 00:06:26.804 from a to b of f of x dx. So the area is this limit, and the area is also found by doing this. 00:06:26.806 --> 00:06:32.600 So what can we do? Well, we put this all together, and we discover that if we want to find this area 00:06:32.602 --> 00:06:43.703 all we actually have to do is evaluate this definite integral, which is this process. Let's write this down, 00:06:43.705 --> 00:06:57.700 and then let's see how we can apply this to other situations. We can find this area, capital A, by 00:06:57.702 --> 00:07:12.403 evaluating the integral from a to b of f of x dx. Well, that's great, but not every figure is bounded 00:07:12.405 --> 00:07:21.800 below by the x-axis or on the sides by two vertical lines. So what we do here is, it takes some work, but 00:07:21.802 --> 00:07:30.002 I think intuitively you can see what's going to happen. Suppose we had something where we had two functions, 00:07:30.004 --> 00:07:41.404 and we want to take this arrangement and generalize it to work in a situation where we have say, maybe, 00:07:41.406 --> 00:07:58.600 f of x and then maybe g of x, bounded by a and b. Can you figure how we would do this? Well we could find 00:07:58.602 --> 00:08:09.007 the area between f of x and the x-axis. Get that value. Then we could find the area between g of x and the x-axis. 00:08:09.009 --> 00:08:17.107 Get that value, and then subtract. So let's put that all together and see if we can come up with something 00:08:17.109 --> 00:08:29.404 that works here. Well, let's try this. What about if we integrate from a to b and we would have f of x dx. 00:08:29.406 --> 00:08:40.701 Then we'd have minus. So that would be the whole area here. Then from that we would subtract the [integral] 00:08:40.703 --> 00:08:52.203 from a to b of g of x dx because that would be the area down in here. So we're going to do the whole area here, 00:08:52.205 --> 00:08:58.599 and then just this part here. And so what is the difference? The difference is the part we're looking for. 00:08:58.601 --> 00:09:04.605 Now think about the rules for integration that you know. Can you combine this into one term? 00:09:04.607 --> 00:09:12.601 Yes we can, as long as we are careful to remember that this function has to be what we call the upper, and 00:09:12.603 --> 00:09:20.604 throughout the domain from a to b both functions have to be continuous and f of x has to be above 00:09:20.606 --> 00:09:28.602 or equal to g of x in this interval. Which means, intuitively, these two functions can't cross within 00:09:28.604 --> 00:09:34.506 the interval, and this function has to always have y-values that are greater than or equal to this function. 00:09:34.508 --> 00:09:42.102 Alright, that one's called the upper, so we might as well call that one the lower. We put it all together, 00:09:42.104 --> 00:09:52.100 and we end up with something that looks like this. The area, capital A, that we are looking for here 00:09:52.102 --> 00:10:09.000 is the integral of f of x minus g of x dx. We need some brackets here. Let's take a look at this on a slide. 00:10:09.002 --> 00:10:21.903 The area between two curves. If f and g are continuous and f of x is greater than or equal to g of x 00:10:21.905 --> 00:10:28.904 over the interval from a to b, so both have to be continuous and f of x has to be greater than or equal to 00:10:28.906 --> 00:10:37.704 g of x over the interval. Then the area bounded by the two curves, and of course x has to fall between the 00:10:37.706 --> 00:10:46.303 closed interval from a to b, is given by this formula, which is exactly what we just developed here. 00:10:46.305 --> 00:11:04.001 Example one. We want to find the area bounded by y equals 2x plus 3, y equals negative x squared plus 1, 00:11:04.003 --> 00:11:15.302 x equals negative 1, and x equals 1. Now a picture here, at least a rough sketch, is essential because 00:11:15.304 --> 00:11:24.203 we have to be certain that we meet the criteria. Both functions have to be continuous from negative 1 to 1, 00:11:24.205 --> 00:11:34.903 and one has to be above the other, and of course we have to determine which one that is. So let's get a sketch. 00:11:34.905 --> 00:11:45.103 Ok, this is a line. It has an intercept, y-intercept at (0,3) and a slope of 2. Here's the y-intercept. 00:11:45.105 --> 00:11:53.603 Slope of 2 means you can find another point on the line by going up 2 and over 1. We'll try this and see 00:11:53.605 --> 00:12:01.602 if this is accurate enough to determine upper and lower, and we only want the part from negative 1 to 1. We'll 00:12:01.604 --> 00:12:08.007 draw it all, and then we'll go back and put in side boundaries. This is a parabola. Opens down, 00:12:08.009 --> 00:12:24.605 shifted up 1. So there are some key points on the parabola, and there it is. We want the area between 00:12:24.607 --> 00:12:34.101 these two functions bounded by x equals negative 1, and I usually make that line just a little squiggly 00:12:34.103 --> 00:12:43.701 because you end up with a lot of lines on here, and so that's the area that we want. Now at this point you 00:12:43.703 --> 00:12:51.702 estimate because arithmetic on this is so tedious sometimes, you want to have a realistic answer. So let's 00:12:51.704 --> 00:13:05.201 just guess. How many square units do you think we have here? Well we have one, two, three whole squares, 00:13:05.203 --> 00:13:14.101 and then maybe that makes four, and then maybe that makes something like, oh let's say four and three fourths 00:13:14.103 --> 00:13:27.207 or maybe five. So our estimate here is that this would be five square units. Let's see if we're right. 00:13:27.209 --> 00:13:38.200 The area is found by integrating from negative 1 to 1. The upper function is the line throughout the 00:13:38.202 --> 00:13:51.105 entire interval from negative 1 to 1. So there's upper minus lower. Alright, now all we have to do is 00:13:51.107 --> 00:13:57.001 clean this up, get the antiderivative, and evaluate using the fundamental theorem. So here is where you 00:13:57.003 --> 00:14:05.704 actually use all of this wonderful knowledge that you have. This is the integral from negative 1 to 1 of 00:14:05.706 --> 00:14:24.195 positive x squared, then 2x, and then 3 minus 1, which is 2. It's a good thing to check here. If you 00:14:24.197 --> 00:14:30.200 make a mistake on your simple subtraction here, obviously your area is going to be wrong. Another way you 00:14:30.202 --> 00:14:35.099 can tell the area is wrong is if the answer comes out negative because obviously the area is going to be 00:14:35.101 --> 00:14:44.905 positive. So let's check. This is x squared, 2x, and 3 minus 1 is 2. Ok, so the antiderivative would be 00:14:44.907 --> 00:14:55.508 x cubed over 3 plus, and let's save ourselves some work here. The antiderivative would be 2 x squared over 2, 00:14:55.510 --> 00:15:14.601 which is x squared, and then plus 2x all evaluated from negative 1 to 1. Let's sub in the 1, and again, watch 00:15:14.603 --> 00:15:36.604 your arithmetic on this. This would be 1/3 plus 1 plus 2. That's F of b. Minus F of a. Negative 1/3 cubed, 00:15:36.606 --> 00:15:49.105 well negative 1 cubed is negative 1, over 3. Negative 1 here, squared is 1. Negative 1 here is negative 2. 00:15:49.107 --> 00:16:03.002 Alright what do we have? Well here we have 3 and a third, and here we have negative 1 and negative 1/3. 00:16:03.004 --> 00:16:13.301 So that's negative 4/3, but it's negative, so we end up with positive 4/3. This is 1 and a third and 00:16:13.303 --> 00:16:24.901 that's 3 and a third. Together we get 4 and 2/3. Now this is area. Our estimate was 5. Our actual answer 00:16:24.903 --> 00:16:38.301 is 4 and 2/3 square units. Now let's stop and appreciate what we did here. You used the Fundamental Theorem 00:16:38.303 --> 00:16:48.600 of Calculus to find the area of an irregular figure bounded by a line, a parabola, and two vertical lines. 00:16:48.602 --> 00:17:10.602 Example two. Let's find the area bounded by y equals 5 minus x squared and y equals x squared minus 3. 00:17:10.604 --> 00:17:19.603 Ok, what's wrong here? You're right. We have no side boundaries, so what this indicates is that the graphs of 00:17:19.605 --> 00:17:27.204 these two functions must cross somewhere and form some boundary. We're going to have to take a look at this. 00:17:27.206 --> 00:17:43.500 So what do we do? Well, we need to sketch. This is a parabola, opens downward, shifted up 5, so it's going 00:17:43.502 --> 00:17:49.600 to look something like this. I'm not sure exactly where it goes without plotting a lot of points. 00:17:49.602 --> 00:17:57.100 So let's see. This is a parabola that opens upward and is shifted down 3. So it looks something like this. 00:17:57.102 --> 00:18:06.899 What we want to know is yes, they're going to cross. Where? Let's see. They cross where 5 minus x squared 00:18:06.901 --> 00:18:22.700 equals x squared minus 3. So where is that? Well, this would be 2 x squared equals 8, so x squared is 4. 00:18:22.702 --> 00:18:31.601 We get two values here, plus or minus 2, and that's not a surprise because it's clear that they cross twice. 00:18:31.603 --> 00:18:41.208 This is going to help us. Now, where exactly is really not going to be that essential as long as we know 00:18:41.210 --> 00:18:56.500 that the x-value is 2. We don't need to know what the y-values are here. We have enough now to find the area. 00:18:56.502 --> 00:19:18.502 Let's estimate. Now we may not have this exactly right. You see how this looks off right here. It looks 00:19:18.504 --> 00:19:24.900 like this is 1, this is 2, and so that really needs to be back here a little more. That's ok though. 00:19:24.902 --> 00:19:31.105 We're really looking for upper and lower, and we're looking for an estimate. So one, two, three, four, five, 00:19:31.107 --> 00:19:39.003 six, seven, eight, nine, ten, eleven, twelve, maybe thirteen, fourteen, maybe fifteen, sixteen, seventeen, 00:19:39.005 --> 00:19:52.402 eighteen, nineteen, twenty. Rough. Rough estimate. Twenty square units. Why do we do this? Well because 00:19:52.404 --> 00:20:07.101 if you get 135, you know something is wrong. Ok, let's find the area. The area is the integral from 00:20:07.103 --> 00:20:15.406 negative 2 to 2 of the upper function which is which one? Carefully. It is the one that opens down. 00:20:15.408 --> 00:20:26.502 So it's 5 minus x squared. That's the upper. Minus the lower which is x squared minus 3, and then don't 00:20:26.504 --> 00:20:36.701 forget your dx. You actually need brackets here too. Alright, let's clean it up. So the area is the integral 00:20:36.703 --> 00:20:55.104 from negative 2 to 2 of negative 2 x squared plus 5 plus 3 is 8 dx. Now we need the antiderivative. 00:20:55.106 --> 00:21:13.410 The antiderivative would be negative 2 x cubed over 3 plus 8x. Let's check that. Negative 2, then x cubed 00:21:13.412 --> 00:21:29.503 over 3, plus 8x, evaluated from negative 2 to 2. Let's find F of 2. 2 cubed is 8 times 2 is 16. So that's 00:21:29.505 --> 00:21:38.701 negative 16 over 3 plus 16. Then F of negative 2, carefully. Negative 2 cubed is negative 8 times 00:21:38.703 --> 00:21:51.504 [negative] 2 is positive 16, over 3, minus 16. Now see the setup here. This happens a lot, so let's be 00:21:51.506 --> 00:22:06.206 careful with our signs. We have negative 16/3 plus 16 minus 16/3 plus 16. So 16 and 16 is 32, but 00:22:06.208 --> 00:22:16.701 negative 16/3 and negative 16/3 is negative 32/3. You get a really good fraction review on these problems. 00:22:16.703 --> 00:22:31.200 Now we want a common denominator of 3, so this is going to be 96/3 minus 32/3 is about 64/3, which is 00:22:31.202 --> 00:22:46.699 about 21 and a third. Actually it's exactly 21 and a third square units. What was our estimate? Twenty. 00:22:46.701 --> 00:22:54.504 That's not bad because the graph is certainly not perfect. So the area bounded by this irregular figure, 00:22:54.506 --> 00:23:04.903 using calculus and integration, is found to be exactly 21 and 1/3 square units. Before we do the example, 00:23:04.905 --> 00:23:17.900 let's read this again. It says f and g have to be continuous over the interval from a to b. Well, we've been 00:23:17.902 --> 00:23:23.802 using lines and parabolas, so we've had no problem with that. But also one of the functions has to be 00:23:23.804 --> 00:23:28.604 either greater than or equal to the other function over the entire interval. What happens if that's 00:23:28.606 --> 00:23:45.099 not the case? So here's example three. Find the area bounded by y equals x squared minus 4, the x-axis. 00:23:45.101 --> 00:23:59.904 What's the equation of the x-axis? Just think about it for a minute. x equals 0, and x equals 3. 00:23:59.907 --> 00:24:11.800 Let's draw it, and let's see what we're going to do. Did you remember this? This is y equals 0. This is a 00:24:11.802 --> 00:24:26.801 parabola. It's shifted down 4, and it opens up. I'm only going to draw part of it because I only want the 00:24:26.803 --> 00:24:40.006 part from 0 to 3. So here's the parabola. I'll just do like this, and you can see it goes and continues here. 00:24:40.008 --> 00:24:49.300 Now, the second line that I want to look at is the x-axis. I'm going to use maybe a different color for this. 00:24:49.302 --> 00:24:59.800 Here's the x-axis, y equals 0, and then I want to bound it from 0 to 3. Well, I think we need to know what 00:24:59.802 --> 00:25:08.603 this is right here. Now if you remember much about the function x squared minus 4, you can tell that this 00:25:08.605 --> 00:25:15.703 is going to be x equals 2, right here at this intercept, but you can always check it. You can take this 00:25:15.705 --> 00:25:29.304 function, set it equal to 0, and solve for x. You get plus or minus 2, and those are your two x-intercepts. 00:25:29.306 --> 00:25:37.302 We're not interested in this one. We are very interested in this one though because that's 2 and we want the area 00:25:37.304 --> 00:25:48.905 from 0 to 3. When we get ready to draw the area, we're going to have to break this up into two areas 00:25:48.907 --> 00:25:58.603 and therefore two integrals. Alright, here's the 3, and here's the 0. So what are we looking for? 00:25:58.605 --> 00:26:10.103 Well, we have area one, and then we have area two. Now estimating is going to be difficult here. We didn't 00:26:10.105 --> 00:26:19.301 actually calculate this, but we know that the answer is somewhere, certainly somewhere between let's say, 00:26:19.303 --> 00:26:26.705 maybe, say, let's just count these. One, two, three, four, five, six, so the answer's probably somewhere 00:26:26.707 --> 00:26:32.301 between 0 and 20. Let's see what we get when we get when we try to calculate it. First we're going to do 00:26:32.303 --> 00:26:49.000 this area. Alright, that one we'll call A-sub-one. It is the integral from 0 to 2. Which one is the upper? 00:26:49.002 --> 00:27:07.500 It's the line y equals 0, so we put 0. That's the f of x, minus x squared minus 4. Now be careful. 00:27:07.502 --> 00:27:21.506 It's f of x minus g of x, so what I really have is the integral from 0 to 2 of minus x squared plus 4 dx. 00:27:21.508 --> 00:27:42.704 The antiderivative is negative x cubed over 3 plus 4x, evaluated from 0 to 2. F of 2 would be negative 8/3 00:27:42.706 --> 00:28:00.005 plus 8, and F of 0 is 0. Our common denominator here would make this 24/3, so that area is 16/3. 00:28:00.007 --> 00:28:08.404 Let's come over here and say that area one is 16/3, which is 5 and a third, and that looks pretty reasonable, 00:28:08.406 --> 00:28:38.802 square units of course. Let's go for A-two. A-two is the integral from 2 to 3. Now which one is the upper? 00:28:38.804 --> 00:28:48.604 Well, it's the parabola, so it's the x squared minus 4, and the lower is the line y equals 0. So this is 00:28:48.606 --> 00:29:01.101 x squared minus 4 minus 0. So we're ready. The antiderivative is x cubed over 3, and the antiderivative 00:29:01.103 --> 00:29:27.206 of this is 4x, and we evaluate from 2 to 3. F of 3 would be 27/3 minus 12. F of 2 would be 2 cubed is 8, 00:29:27.208 --> 00:29:42.602 so that would be 8/3 minus 8. Alright, what do we have? Let's check it. 27 over 3 minus 12. 8/3 minus 8. 00:29:42.604 --> 00:30:00.902 Let's check. Alright, 27 over 3 is 9. So we have 9 minus 12 minus 8/3 plus 8. 8 and 9 is 17, minus 12 is 5. 00:30:00.904 --> 00:30:22.203 5 is 15/3, so we have 15/3 minus 8/3 is 7/3. That's area two. So let's see what we have for total area. 00:30:22.204 --> 00:30:43.904 What do we have to do? We simply add A-one and A-two. A-one plus A-two is 16/3 plus 7/3, 23/3, which is 00:30:43.906 --> 00:31:05.101 7 and 2/3. Now let's check this. Let's see, that would be 21 and 2, thirds, so we get 7 and 2/3 square units. 00:31:05.103 --> 00:31:17.805 That is the total area bounded by these boundaries. You now have a powerful tool at your disposal. 00:31:17.807 --> 00:31:26.301 If you can describe what the function, the curve, any curve, you can find the area between two curves, whether 00:31:26.303 --> 00:31:32.702 the curves intersect, whether one is greater than the other at any point, or whether you have side boundaries. 00:31:32.704 --> 00:31:40.000 We now turn our attention to an interesting business application that uses this process.