1 00:00:00,040 --> 00:00:02,400 The following content is provided under a Creative 2 00:00:02,400 --> 00:00:03,690 Commons license. 3 00:00:03,690 --> 00:00:06,630 Your support will help MIT OpenCourseWare continue to 4 00:00:06,630 --> 00:00:09,990 offer high-quality educational resources for free. 5 00:00:09,990 --> 00:00:12,830 To make a donation, or to view additional materials from 6 00:00:12,830 --> 00:00:16,760 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:16,760 --> 00:00:18,010 ocw.mit.edu. 8 00:00:31,070 --> 00:00:32,040 PROFESSOR: Hi. 9 00:00:32,040 --> 00:00:35,950 Today we begin our study of integral calculus, which had 10 00:00:35,950 --> 00:00:42,240 its roots back in ancient Greece, roughly 600 BC, and 11 00:00:42,240 --> 00:00:45,560 began with an investigation of the study of area. 12 00:00:45,560 --> 00:00:47,520 In a certain manner of speaking, today's lecture 13 00:00:47,520 --> 00:00:51,000 could be called 'Calculus Revisited Revisited', in the 14 00:00:51,000 --> 00:00:54,580 sense that integral calculus can be studied quite apart 15 00:00:54,580 --> 00:00:56,140 from the fact that differential 16 00:00:56,140 --> 00:00:57,900 calculus was ever invented. 17 00:00:57,900 --> 00:01:01,260 We'll see later in this block of material that there is a 18 00:01:01,260 --> 00:01:04,129 wondrous relationship between integral and differential 19 00:01:04,129 --> 00:01:07,480 calculus, but perhaps the best proof that integral calculus 20 00:01:07,480 --> 00:01:11,160 can be studied independently of differential calculus lies 21 00:01:11,160 --> 00:01:14,700 in the fact that the ancient Greek was doing integral 22 00:01:14,700 --> 00:01:19,010 calculus in 600 BC, whereas differential calculus did not 23 00:01:19,010 --> 00:01:23,690 begin until about 1680 AD with Sir Isaac Newton. 24 00:01:23,690 --> 00:01:26,980 Well, at any rate then, what we'll call today's lecture is 25 00:01:26,980 --> 00:01:29,420 simply 'Two-dimensional Area'. 26 00:01:29,420 --> 00:01:31,900 In other words, this is how the subject began, with 27 00:01:31,900 --> 00:01:36,240 studying the amount of space in plane regions, 28 00:01:36,240 --> 00:01:39,370 two-dimensional area, as opposed to, say, as we'll talk 29 00:01:39,370 --> 00:01:43,070 about later, three-dimensional area, which is really a fancy 30 00:01:43,070 --> 00:01:44,700 word for volume, et cetera. 31 00:01:44,700 --> 00:01:46,650 But enough about that for the time being. 32 00:01:46,650 --> 00:01:50,350 Two-dimensional area, or Calculus Revisited Revisited. 33 00:01:50,350 --> 00:01:53,780 And to see what's happening over here, the study of area 34 00:01:53,780 --> 00:01:57,790 was the forerunner of integral calculus as we now know it. 35 00:01:57,790 --> 00:02:01,630 It began in ancient Greece, roughly 600 BC. 36 00:02:01,630 --> 00:02:04,840 The branch of calculus that we've been studying up until 37 00:02:04,840 --> 00:02:08,229 now in our course, differential calculus, did not 38 00:02:08,229 --> 00:02:10,979 begin until 1680 AD. 39 00:02:10,979 --> 00:02:14,730 Notice, then, the interesting juxtaposition in time. 40 00:02:14,730 --> 00:02:17,510 In other words, pedagogically, one seems to study 41 00:02:17,510 --> 00:02:20,790 differential calculus before integral calculus. 42 00:02:20,790 --> 00:02:24,450 Chronologically, integral calculus preceded differential 43 00:02:24,450 --> 00:02:27,670 calculus by more than 2,000 years. 44 00:02:27,670 --> 00:02:30,110 Now this is a rather beautiful study. 45 00:02:30,110 --> 00:02:33,260 It's one of the most aesthetic parts of elementary 46 00:02:33,260 --> 00:02:36,800 mathematics, namely the simplicity with which the 47 00:02:36,800 --> 00:02:41,220 Greek was able to tackle the sophisticated problem of 48 00:02:41,220 --> 00:02:43,160 finding areas in general. 49 00:02:43,160 --> 00:02:46,500 He started with three basically simple properties of 50 00:02:46,500 --> 00:02:50,430 area, so simple that most of us are willing to accept them 51 00:02:50,430 --> 00:02:52,070 almost intuitively. 52 00:02:52,070 --> 00:02:54,740 We'll call these the 'Axioms for Area'. 53 00:02:54,740 --> 00:02:56,830 And simply stated, they were the following. 54 00:02:56,830 --> 00:03:00,250 One, that the area of a rectangle is the base times 55 00:03:00,250 --> 00:03:01,150 the height. 56 00:03:01,150 --> 00:03:06,060 Two, if one region is contained within another, the 57 00:03:06,060 --> 00:03:09,830 area of the contained region is less than or equal to-- in 58 00:03:09,830 --> 00:03:11,740 other words, can be no greater than-- that of 59 00:03:11,740 --> 00:03:12,990 the containing region. 60 00:03:12,990 --> 00:03:15,390 Roughly speaking, the smaller the region, 61 00:03:15,390 --> 00:03:17,160 the smaller its area. 62 00:03:17,160 --> 00:03:21,220 And finally, written slightly more formally here, the old 63 00:03:21,220 --> 00:03:24,670 high-school axiom for plane geometry, that the area of the 64 00:03:24,670 --> 00:03:27,700 whole equals the sum of the areas of the parts, that if a 65 00:03:27,700 --> 00:03:32,580 region 'R' is subdivided into a union of mutually exclusive 66 00:03:32,580 --> 00:03:36,670 pieces, then the area of the entire region is equal to the 67 00:03:36,670 --> 00:03:39,640 sum of the areas of the constituent parts. 68 00:03:39,640 --> 00:03:43,290 And with just these three axioms, the ancient Greek 69 00:03:43,290 --> 00:03:47,320 invented a technique for finding areas that today is 70 00:03:47,320 --> 00:03:50,250 still known by its original name, the 'Method of 71 00:03:50,250 --> 00:03:51,980 Exhaustion'. 72 00:03:51,980 --> 00:03:55,850 And here, exhaustion does not refer to the fact that we get 73 00:03:55,850 --> 00:03:59,210 tired using the system, even though, as we shall soon see, 74 00:03:59,210 --> 00:04:00,380 it's quite intricate. 75 00:04:00,380 --> 00:04:03,530 It refers to the fact that we take the region whose area we 76 00:04:03,530 --> 00:04:09,540 want to find and exhaust the space by squeezing it between 77 00:04:09,540 --> 00:04:12,690 regions which are made up of rectangles. 78 00:04:12,690 --> 00:04:16,140 Stated more precisely, given a region 'R'-- 79 00:04:16,140 --> 00:04:19,220 and I'll define this more precisely later-- we squeeze 80 00:04:19,220 --> 00:04:22,089 it between two networks of rectangles, 81 00:04:22,089 --> 00:04:23,570 the idea being what? 82 00:04:23,570 --> 00:04:26,550 That we know what the area of a rectangle is, and 83 00:04:26,550 --> 00:04:30,090 consequently, by the fact that the area of the whole equals 84 00:04:30,090 --> 00:04:32,630 the sum of the areas of the parts, if we have a 85 00:04:32,630 --> 00:04:35,940 rectangular network, knowing how to find the area of each 86 00:04:35,940 --> 00:04:38,460 rectangle, we know how to find the area of 87 00:04:38,460 --> 00:04:40,020 the rectangular network. 88 00:04:40,020 --> 00:04:43,470 Now, rather than to wax on like this philosophically, 89 00:04:43,470 --> 00:04:45,550 let's tackle a specific problem. 90 00:04:45,550 --> 00:04:49,540 In fact, give or take a little bit, this is specifically the 91 00:04:49,540 --> 00:04:53,470 problem that Archimedes dealt with in finding the area of 92 00:04:53,470 --> 00:04:56,800 parabolic segments way back, as I say, in the 93 00:04:56,800 --> 00:05:00,050 300 to 600 BC era. 94 00:05:00,050 --> 00:05:01,930 As an example, consider the following. 95 00:05:01,930 --> 00:05:06,160 We wish to determine the area of the region 'R', 'A sub R', 96 00:05:06,160 --> 00:05:10,330 where the region 'R' is that region which is bounded above 97 00:05:10,330 --> 00:05:15,930 by the curve 'y' equals 'x squared', below by the x-axis, 98 00:05:15,930 --> 00:05:19,300 on the left by the y-axis, and on the right by the 99 00:05:19,300 --> 00:05:20,860 line 'x' equals 1. 100 00:05:20,860 --> 00:05:23,960 In other words, this region in here. 101 00:05:23,960 --> 00:05:26,950 Now here's the way we put the squeeze on it, and we'll start 102 00:05:26,950 --> 00:05:29,410 this quite gradually. 103 00:05:29,410 --> 00:05:32,900 The first thing that we observe is that if we draw a 104 00:05:32,900 --> 00:05:37,170 line parallel to the x-axis through the point (1 , 1), the 105 00:05:37,170 --> 00:05:42,000 highest point of our region here, we construct a rectangle 106 00:05:42,000 --> 00:05:44,390 which contains our region 'R'. 107 00:05:44,390 --> 00:05:48,000 And since 'R' is contained in the rectangle, the area of the 108 00:05:48,000 --> 00:05:51,410 region 'R' must be less than the area of the rectangle. 109 00:05:51,410 --> 00:05:53,650 But notice that this particular rectangle that 110 00:05:53,650 --> 00:05:57,020 we've just constructed has its base equal to 1, its height 111 00:05:57,020 --> 00:05:58,000 equal to 1. 112 00:05:58,000 --> 00:06:03,160 Therefore by our first axiom, its area is 1 times 1, or 1. 113 00:06:03,160 --> 00:06:07,140 Also, since intuitively, area is a positive thing, we see 114 00:06:07,140 --> 00:06:09,950 just from this quick diagram that whatever the area of the 115 00:06:09,950 --> 00:06:14,370 region is, we now have it bounded between 0 and 1. 116 00:06:14,370 --> 00:06:15,410 We have an upper bound. 117 00:06:15,410 --> 00:06:17,180 We have a lower bound. 118 00:06:17,180 --> 00:06:19,430 We have this thing caught, OK? 119 00:06:19,430 --> 00:06:21,760 Now, there's still a lot of space between 0 and 1. 120 00:06:21,760 --> 00:06:25,110 The method of exhaustion refines this idea. 121 00:06:25,110 --> 00:06:27,460 Namely, what we do next is we say, look at-- 122 00:06:27,460 --> 00:06:31,020 instead of just drawing one big rectangle like this, why 123 00:06:31,020 --> 00:06:34,010 don't we partition the base of our region 'R' 124 00:06:34,010 --> 00:06:35,230 into two equal parts? 125 00:06:35,230 --> 00:06:38,510 In other words, let's locate the point 1/2, which of course 126 00:06:38,510 --> 00:06:41,090 is midway between 0 and 1. 127 00:06:41,090 --> 00:06:46,230 Now what I can do is I can circumscribe two rectangles, 128 00:06:46,230 --> 00:06:50,400 one of which is the rectangle whose height corresponds to 129 00:06:50,400 --> 00:06:52,890 the x-coordinate equaling 1/2-- 130 00:06:52,890 --> 00:06:55,450 that means the y-coordinate is 1/4-- 131 00:06:55,450 --> 00:06:59,160 and the second rectangle, the one whose x-coordinate-- 132 00:06:59,160 --> 00:07:02,350 the x-coordinate of the height is 1, and since 'y' equals 'x 133 00:07:02,350 --> 00:07:04,600 squared', the height is also 1. 134 00:07:04,600 --> 00:07:07,960 You see, notice that by picking a region which is a 135 00:07:07,960 --> 00:07:11,000 curve which is always rising, the lowest point of each 136 00:07:11,000 --> 00:07:13,780 region is the point that's furthest to the left. 137 00:07:13,780 --> 00:07:16,380 And the highest point of each region is the point that's 138 00:07:16,380 --> 00:07:17,510 furthest to the right. 139 00:07:17,510 --> 00:07:20,980 And keeping this in mind, you see by making rectangles 140 00:07:20,980 --> 00:07:24,110 corresponding to the points furthest to the right in each 141 00:07:24,110 --> 00:07:28,170 interval, I get a rectangle which contains the portion of 142 00:07:28,170 --> 00:07:30,020 the region 'R' that I'm talking about. 143 00:07:30,020 --> 00:07:33,270 In other words, if I look at this rectangular network, my 144 00:07:33,270 --> 00:07:35,770 region 'R' is contained inside that. 145 00:07:35,770 --> 00:07:40,550 Consequently, its area is less than the area of the 146 00:07:40,550 --> 00:07:41,860 rectangular network. 147 00:07:41,860 --> 00:07:44,350 Now, what is the area of this rectangular network? 148 00:07:44,350 --> 00:07:48,740 Well, the big rectangle has its base equal to 1/2 and its 149 00:07:48,740 --> 00:07:50,030 height equal to 1. 150 00:07:50,030 --> 00:07:51,970 So its area is 1/2. 151 00:07:51,970 --> 00:07:55,370 The small rectangle has its base equal to 1/2 and its 152 00:07:55,370 --> 00:07:56,720 height equal to 1/4. 153 00:07:56,720 --> 00:07:58,390 Since its area is the base times the 154 00:07:58,390 --> 00:08:00,390 height, its area is 1/8. 155 00:08:00,390 --> 00:08:02,960 1/8 plus 1/2 is 5/8. 156 00:08:02,960 --> 00:08:06,420 In other words, the area of the rectangular network which 157 00:08:06,420 --> 00:08:08,260 contains 'R' is 5/8. 158 00:08:08,260 --> 00:08:10,450 Consequently, the area of the region 'R' must 159 00:08:10,450 --> 00:08:12,050 be less than 5/8. 160 00:08:12,050 --> 00:08:16,040 On the other hand, notice that the lowest point in the second 161 00:08:16,040 --> 00:08:18,900 interval corresponds to 'x' equaling 1/2. 162 00:08:18,900 --> 00:08:22,310 In other words, notice that this particular rectangle, 163 00:08:22,310 --> 00:08:25,710 whose base is 1/2 and whose height is 1/4, that this 164 00:08:25,710 --> 00:08:29,440 rectangle here is inscribed in the region 'R'. 165 00:08:29,440 --> 00:08:31,170 Its area is 1/8. 166 00:08:31,170 --> 00:08:34,380 And since it's contained within the region 'R', its 167 00:08:34,380 --> 00:08:36,760 area must be less than the area of the region 'R'. 168 00:08:36,760 --> 00:08:38,880 And so we're now certain that whatever the area of the 169 00:08:38,880 --> 00:08:42,549 region 'R' is, it's between 1/8 and 5/8. 170 00:08:42,549 --> 00:08:45,510 And by the way, notice in terms of these shaded regions, 171 00:08:45,510 --> 00:08:48,680 that even though this is an approximation which is too 172 00:08:48,680 --> 00:08:51,700 large to be the right answer, it is closer to being the 173 00:08:51,700 --> 00:08:54,420 right answer than this approximation here. 174 00:08:54,420 --> 00:08:56,210 In other words, notice that the difference between the 175 00:08:56,210 --> 00:08:59,480 overestimate here and the overestimate here is this 176 00:08:59,480 --> 00:09:03,220 amount in here. 177 00:09:03,220 --> 00:09:05,120 You see, we've chopped off part of the error. 178 00:09:05,120 --> 00:09:07,370 But we'll talk about that, as I say, in more 179 00:09:07,370 --> 00:09:08,470 detail in the notes. 180 00:09:08,470 --> 00:09:11,220 What I'd like to do is to give you an overall view of what's 181 00:09:11,220 --> 00:09:12,440 happening over here. 182 00:09:12,440 --> 00:09:15,280 You see, to generalize the method of exhaustion, we do 183 00:09:15,280 --> 00:09:17,010 what the mathematician usually does. 184 00:09:17,010 --> 00:09:19,980 Instead of saying, let's divide the base of the region 185 00:09:19,980 --> 00:09:23,250 into two equal parts, or three equal parts, or four equal 186 00:09:23,250 --> 00:09:27,440 parts, we say, let's divide it into 'n' equal parts. 187 00:09:27,440 --> 00:09:30,870 Now if we divide this base into 'n' equal parts, since 188 00:09:30,870 --> 00:09:32,370 the whole region-- 189 00:09:32,370 --> 00:09:35,720 the whole length is 0 to 1, dividing into 'n' equal parts 190 00:09:35,720 --> 00:09:39,550 means that my points of division will be '1/n', '2/n', 191 00:09:39,550 --> 00:09:44,520 '3/n', et cetera, right down to 'n/n', which is 1. 192 00:09:44,520 --> 00:09:50,460 What I do now is I pick the right endpoint of each point 193 00:09:50,460 --> 00:09:53,720 in that partition to draw my rectangle. 194 00:09:53,720 --> 00:09:56,370 You see, each of these rectangles that I draw this 195 00:09:56,370 --> 00:10:00,130 way contains the corresponding part of my region 'R'. 196 00:10:00,130 --> 00:10:05,530 Consequently, the area of this rectangular network must be an 197 00:10:05,530 --> 00:10:08,660 upper bound for the area of my region 'R'. 198 00:10:08,660 --> 00:10:11,370 Now, because it's going to be an upper bound, and because it 199 00:10:11,370 --> 00:10:14,830 depends on what the value of 'n' is, I will denote that 200 00:10:14,830 --> 00:10:17,670 upper bound by 'U sub n'. 201 00:10:17,670 --> 00:10:19,350 And what will 'U sub n' be? 202 00:10:19,350 --> 00:10:21,980 It's the sum of the areas of these circumscribed 203 00:10:21,980 --> 00:10:23,010 rectangles. 204 00:10:23,010 --> 00:10:25,320 And it's not difficult to see from this picture. 205 00:10:25,320 --> 00:10:26,820 Let's take a look. 206 00:10:26,820 --> 00:10:30,280 Since this rectangle has its height corresponding to the 207 00:10:30,280 --> 00:10:33,980 x-coordinate equaling '1/n', and since the y-coordinate is 208 00:10:33,980 --> 00:10:36,090 the square of the x-coordinate, the height of 209 00:10:36,090 --> 00:10:38,780 this rectangle will be ''1/n' squared'. 210 00:10:38,780 --> 00:10:40,320 The base is '1/n'. 211 00:10:40,320 --> 00:10:43,260 So the area of that circumscribed rectangle is 212 00:10:43,260 --> 00:10:46,180 ''1/n' squared' times '1/n'. 213 00:10:46,180 --> 00:10:49,600 Similarly, the next rectangle here has its height 214 00:10:49,600 --> 00:10:52,690 corresponding to the x-coordinate '2/n'. 215 00:10:52,690 --> 00:10:55,550 So its height is ''2/n' squared'. 216 00:10:55,550 --> 00:10:57,530 Keep in mind the y-coordinate is the square of the 217 00:10:57,530 --> 00:10:58,880 x-coordinate here. 218 00:10:58,880 --> 00:11:01,730 The base of all of these rectangles is '1/n'. 219 00:11:01,730 --> 00:11:04,930 And proceeding in this way, I finally come down to the last 220 00:11:04,930 --> 00:11:07,520 rectangle, whose height is what? 221 00:11:07,520 --> 00:11:11,080 Well, the x-coordinate is 1, so the y-coordinate is 1. 222 00:11:11,080 --> 00:11:14,400 To keep my form here, I'll write that as 'n/n'. 223 00:11:14,400 --> 00:11:18,320 Its height is therefore ''n/n' squared'. 224 00:11:18,320 --> 00:11:21,810 The base of that rectangle is '1/n', so the area of that 225 00:11:21,810 --> 00:11:26,760 last rectangle is ''n/n' squared' times '1/n'. 226 00:11:26,760 --> 00:11:28,330 Now you see that's what? 227 00:11:28,330 --> 00:11:32,170 That's an area which is too large to be 'A sub R'. 228 00:11:32,170 --> 00:11:34,740 Well, let's take a look now and see what's happening here. 229 00:11:34,740 --> 00:11:38,510 Notice that each of these terms has an 'n cubed' in the 230 00:11:38,510 --> 00:11:39,450 denominator. 231 00:11:39,450 --> 00:11:42,060 So I can factor out the 'n cubed' term. 232 00:11:42,060 --> 00:11:44,050 And what I'm left with is what? 233 00:11:44,050 --> 00:11:46,370 The sum of the first 'n' squares. 234 00:11:46,370 --> 00:11:49,995 1 squared plus 2 squared plus 3 squared, et cetera, plus et 235 00:11:49,995 --> 00:11:51,180 cetera, 'n squared'. 236 00:11:51,180 --> 00:11:54,340 In other words, this tells me how to find 'U sub n' for any 237 00:11:54,340 --> 00:11:55,370 value of 'n'. 238 00:11:55,370 --> 00:11:58,040 And we'll talk about that some more in a little while. 239 00:11:58,040 --> 00:11:59,690 That's an upper bound. 240 00:11:59,690 --> 00:12:03,810 In a corresponding way, I can also find a lower bound. 241 00:12:03,810 --> 00:12:08,350 Namely, what I'll do now is pick the smallest rectangle in 242 00:12:08,350 --> 00:12:08,910 each region. 243 00:12:08,910 --> 00:12:11,860 In other words, I'll now inscribe rectangles inside my 244 00:12:11,860 --> 00:12:16,140 region 'R', and therefore find a rectangular region whose 245 00:12:16,140 --> 00:12:19,150 area is less than that of the region 'R'. 246 00:12:19,150 --> 00:12:22,080 And to do that, without going through the details, notice 247 00:12:22,080 --> 00:12:24,970 that essentially all I have to do is shift each of these 248 00:12:24,970 --> 00:12:28,010 rectangles over by one partition. 249 00:12:28,010 --> 00:12:33,180 Namely, notice, for example, that the smallest height in 250 00:12:33,180 --> 00:12:36,280 the second partition here is the one which corresponds to 251 00:12:36,280 --> 00:12:37,530 the height '1/n'. 252 00:12:37,530 --> 00:12:40,010 In other words, what I do now is, to inscribe the 253 00:12:40,010 --> 00:12:43,430 rectangles, I just shift everything over like this. 254 00:12:43,430 --> 00:12:46,660 And leaving the details out, and letting you verify these 255 00:12:46,660 --> 00:12:48,510 for yourselves, I can again mimic 256 00:12:48,510 --> 00:12:50,160 exactly what I did before. 257 00:12:50,160 --> 00:12:54,030 The only difference being now, that instead of the height of 258 00:12:54,030 --> 00:12:59,690 the last rectangle being 'n/n', notice that it's 'n - 259 00:12:59,690 --> 00:13:01,160 1' over 'n', squared. 260 00:13:01,160 --> 00:13:02,650 In other words, the x-coordinate is 261 00:13:02,650 --> 00:13:04,140 'n - 1' over 'n'. 262 00:13:04,140 --> 00:13:06,620 The y-coordinate is the square of the x-coordinate. 263 00:13:06,620 --> 00:13:09,340 In other words, notice that this point here gives rise to 264 00:13:09,340 --> 00:13:12,620 the height of the lowest rectangle that can be 265 00:13:12,620 --> 00:13:14,490 inscribed in my last portion here. 266 00:13:14,490 --> 00:13:17,970 So without further ado, it turns out that 'L sub n', the 267 00:13:17,970 --> 00:13:22,060 lower estimate, is '1 over 'n cubed'' times the sum of the 268 00:13:22,060 --> 00:13:24,820 first 'n - 1' squares. 269 00:13:24,820 --> 00:13:25,710 OK. 270 00:13:25,710 --> 00:13:28,540 Now, let's keep track of just what it is that we've done 271 00:13:28,540 --> 00:13:30,380 over here so far. 272 00:13:30,380 --> 00:13:31,410 What we've done is what? 273 00:13:31,410 --> 00:13:37,890 For each 'n', we have squeezed 'A sub R' between two numbers, 274 00:13:37,890 --> 00:13:40,800 one number being an upper approximation, one number 275 00:13:40,800 --> 00:13:42,720 being a lower approximation. 276 00:13:42,720 --> 00:13:45,370 Now notice that the 'U sub n' and the 'L sub n' are 277 00:13:45,370 --> 00:13:46,880 functions of 'n'. 278 00:13:46,880 --> 00:13:48,380 'A sub R' is a constant. 279 00:13:48,380 --> 00:13:50,070 That's the area of the region 'R'. 280 00:13:50,070 --> 00:13:52,670 What the method of exhaustion means is simply this. 281 00:13:52,670 --> 00:13:53,710 We say, look. 282 00:13:53,710 --> 00:13:56,860 Let's take the limit of the 'L sub n's as 'n' approaches 283 00:13:56,860 --> 00:14:00,630 infinity, and let's take the limit of the 'U sub n's as 'n' 284 00:14:00,630 --> 00:14:04,095 approaches infinity, observing that for each 'n', 'A sub R' 285 00:14:04,095 --> 00:14:06,010 is squeezed between these two. 286 00:14:06,010 --> 00:14:08,960 Consequently, since 'A sub R' is squeezed between these two 287 00:14:08,960 --> 00:14:12,640 for each 'n', it must be squeezed between these two 288 00:14:12,640 --> 00:14:13,770 when we go to the limit. 289 00:14:13,770 --> 00:14:18,100 In other words, whatever 'A sub R' is, it must be between 290 00:14:18,100 --> 00:14:19,460 these two limits. 291 00:14:19,460 --> 00:14:21,170 Now here's the key step. 292 00:14:21,170 --> 00:14:25,420 It turns out, at least in this particular problem, that the 293 00:14:25,420 --> 00:14:29,700 limit of 'L sub n' as 'n' goes to infinity is the same as the 294 00:14:29,700 --> 00:14:32,180 limit of 'U sub n' as 'n' goes to infinity. 295 00:14:32,180 --> 00:14:34,740 And the best way to see that without becoming too fancy at 296 00:14:34,740 --> 00:14:38,880 this stage of the game is to observe that for a given 'n', 297 00:14:38,880 --> 00:14:41,120 the difference between 'U sub n' and 'L 298 00:14:41,120 --> 00:14:43,880 sub n' is just '1/n'. 299 00:14:43,880 --> 00:14:47,560 And again, I'll just indicate to you from this diagram how 300 00:14:47,560 --> 00:14:48,630 we can see that. 301 00:14:48,630 --> 00:14:51,820 Notice that back here, the way we went from 'U sub n' to 'L 302 00:14:51,820 --> 00:14:54,580 sub n' was we just pushed over this whole 303 00:14:54,580 --> 00:14:56,370 network by one unit. 304 00:14:56,370 --> 00:14:59,770 In other words, the rectangle that we squeezed out in going 305 00:14:59,770 --> 00:15:03,050 from the upper sum to the lower sum was just this 306 00:15:03,050 --> 00:15:05,940 particular rectangle whose height was 1 and 307 00:15:05,940 --> 00:15:07,600 whose base was '1/n'. 308 00:15:07,600 --> 00:15:10,720 In other words, the area that was kicked out in going from 309 00:15:10,720 --> 00:15:13,870 'L sub n' to 'U sub n' is '1/n'. 310 00:15:13,870 --> 00:15:16,400 This is what this thing here says here. 311 00:15:16,400 --> 00:15:18,180 This is worked out in detail in the notes. 312 00:15:18,180 --> 00:15:20,440 But again, I just want to go through this thing quickly so 313 00:15:20,440 --> 00:15:22,500 we get the idea of what's happening over here. 314 00:15:22,500 --> 00:15:26,490 In other words, since 'U sub n' minus 'L sub n' is '1/n', 315 00:15:26,490 --> 00:15:28,170 that's just another way of saying what? 316 00:15:28,170 --> 00:15:30,610 That the limit of this difference is just the limit 317 00:15:30,610 --> 00:15:32,720 of '1/n' as 'n' goes to infinity. 318 00:15:32,720 --> 00:15:34,260 But that's clearly 0. 319 00:15:34,260 --> 00:15:38,140 In other words, these two limits are the same because 320 00:15:38,140 --> 00:15:40,840 their difference goes to 0 in the limit. 321 00:15:40,840 --> 00:15:43,960 Consequently, since these two things are the same, and 'A 322 00:15:43,960 --> 00:15:48,170 sub R' is caught between these two, it must be that 'A sub R' 323 00:15:48,170 --> 00:15:51,320 is equal to this common limit. 324 00:15:51,320 --> 00:15:54,390 And that's precisely what the method of exhaustion was. 325 00:15:54,390 --> 00:15:57,880 We squeeze what we were looking for between two 326 00:15:57,880 --> 00:16:00,070 estimates which converge towards each 327 00:16:00,070 --> 00:16:01,450 other as 'n' got large. 328 00:16:01,450 --> 00:16:04,130 Now, by the way, I'm going to point out the fact that it's 329 00:16:04,130 --> 00:16:06,510 very, very difficult in general to find 330 00:16:06,510 --> 00:16:07,480 what this limit is. 331 00:16:07,480 --> 00:16:09,220 We're going to see that in working a 332 00:16:09,220 --> 00:16:10,440 few specific problems. 333 00:16:10,440 --> 00:16:12,590 What I thought might be informative-- 334 00:16:12,590 --> 00:16:15,000 without going through the details right now, I have 335 00:16:15,000 --> 00:16:18,800 taken the liberty of computing 'L sub n' and 'U sub n' in 336 00:16:18,800 --> 00:16:21,970 this problem for 'n' equals 1,000. 337 00:16:21,970 --> 00:16:24,980 In other words, if we divided this region into 1,000 equal 338 00:16:24,980 --> 00:16:28,350 parts and took the network of circumscribed rectangles and 339 00:16:28,350 --> 00:16:32,710 inscribed rectangles, it would turn out that the area of the 340 00:16:32,710 --> 00:16:40,690 inscribed rectangles would be 0.3328335, and for the 341 00:16:40,690 --> 00:16:46,950 circumscribed rectangles, 0.3338335. 342 00:16:46,950 --> 00:16:49,300 And by the way, notice how this is borne out. 343 00:16:49,300 --> 00:16:52,440 Notice that the difference between these two is precisely 344 00:16:52,440 --> 00:16:57,270 1/1000, and that is '1/n' with 'n' equal to 1,000. 345 00:16:57,270 --> 00:16:59,310 But that's not the important point here. 346 00:16:59,310 --> 00:17:01,980 The important point is that just looking at this decimal 347 00:17:01,980 --> 00:17:05,359 expansion, if I didn't know anything else, I know that 'A 348 00:17:05,359 --> 00:17:08,720 sub R' is caught between these two. 349 00:17:08,720 --> 00:17:12,200 Consequently, to two decimal places, I can be sure that 'A 350 00:17:12,200 --> 00:17:15,020 sub R' is 0.33. 351 00:17:15,020 --> 00:17:17,470 And notice that if I want more and more decimal place 352 00:17:17,470 --> 00:17:21,579 accuracy, especially if I have access to a desk calculator, I 353 00:17:21,579 --> 00:17:24,040 don't really have to compute the limit exactly. 354 00:17:24,040 --> 00:17:27,490 I can feed the formula into the machine and put the 355 00:17:27,490 --> 00:17:29,740 squeeze on and get as many decimal place 356 00:17:29,740 --> 00:17:31,940 accuracy as I want. 357 00:17:31,940 --> 00:17:34,720 By the way, as an aside, this is exactly what we did in high 358 00:17:34,720 --> 00:17:39,470 school, when we said things like let pi equal 22/7. 359 00:17:39,470 --> 00:17:42,090 Pi is not equal to 22/7. 360 00:17:42,090 --> 00:17:45,450 Among other things, 22/7 is a rational number. 361 00:17:45,450 --> 00:17:47,030 It's the ratio of two whole numbers. 362 00:17:47,030 --> 00:17:48,610 Pi is an irrational number. 363 00:17:48,610 --> 00:17:51,260 What people really meant was that you can't tell the 364 00:17:51,260 --> 00:17:55,470 difference between pi and 22/7 to two decimal places. 365 00:17:55,470 --> 00:17:58,050 They both begin 3.14. 366 00:17:58,050 --> 00:18:01,200 And consequently, if all you wanted to measure was to two 367 00:18:01,200 --> 00:18:04,200 decimal digits, you could use 22/7 as the 368 00:18:04,200 --> 00:18:05,610 approximation for pi. 369 00:18:05,610 --> 00:18:08,150 But if you wanted to squeeze out more places, you'd have to 370 00:18:08,150 --> 00:18:10,210 use a more refined approach. 371 00:18:10,210 --> 00:18:13,200 By the way, let me point out that one of the reasons that I 372 00:18:13,200 --> 00:18:17,130 chose the curve 'y' equals 'x squared' to work with was-- 373 00:18:17,130 --> 00:18:22,010 you may recall that back in our discussion of mathematical 374 00:18:22,010 --> 00:18:25,890 induction, one of the problems that we worked on was to show 375 00:18:25,890 --> 00:18:29,830 how we find by induction the recipe for the sum of the 376 00:18:29,830 --> 00:18:31,740 first 'n' squares. 377 00:18:31,740 --> 00:18:36,260 And by way of review, let me recall for you the fact that 378 00:18:36,260 --> 00:18:39,830 the sum of the first 'n' squares was given by ''n' 379 00:18:39,830 --> 00:18:45,220 times 'n + 1' times '2n + 1'' over 6. 380 00:18:45,220 --> 00:18:45,860 OK? 381 00:18:45,860 --> 00:18:49,250 In other words, going back to our recipe for 'U sub n', I 382 00:18:49,250 --> 00:18:52,570 can now replace 1 squared plus 2 squared plus et cetera 'n 383 00:18:52,570 --> 00:18:57,580 squared' by ''n' times 'n + 1' times '2n + 1'' over 6. 384 00:18:57,580 --> 00:19:01,570 And if I now divide through by 'n cubed' judiciously, namely 385 00:19:01,570 --> 00:19:05,330 canceling out one 'n' with this 'n', dividing 'n + 1' by 386 00:19:05,330 --> 00:19:10,800 'n' and '2n + 1' by 'n', I get that 'U sub n' is '1/6 'n + 1' 387 00:19:10,800 --> 00:19:13,440 over 'n'' times ''2n + 1' over 'n''. 388 00:19:13,440 --> 00:19:17,510 And that can be written even more suggestively as 1/6 times 389 00:19:17,510 --> 00:19:20,680 '1 + '1/n'' times '2 + '1/n''. 390 00:19:20,680 --> 00:19:23,100 And I can now put a very nice mathematical 391 00:19:23,100 --> 00:19:24,610 interpretation on this. 392 00:19:24,610 --> 00:19:29,680 Mainly, notice that no matter how big 'n' is, '1 + '1/n'' is 393 00:19:29,680 --> 00:19:32,880 bigger than 1, you see, because '1/n' is positive. 394 00:19:32,880 --> 00:19:35,350 And '2 + '1/n'' is bigger than 2. 395 00:19:35,350 --> 00:19:40,450 Consequently, whatever this is, it's bigger than 1/6 times 396 00:19:40,450 --> 00:19:43,930 1 times 2, which is 1/3. 397 00:19:43,930 --> 00:19:48,790 In other words, for each 'n', 'U sub n' is greater than 1/3. 398 00:19:48,790 --> 00:19:52,340 It also happens that as 'n' gets larger, 399 00:19:52,340 --> 00:19:54,300 'U sub n' gets smaller. 400 00:19:54,300 --> 00:19:58,060 Again, looking at this recipe, the bigger 'n' is, the bigger 401 00:19:58,060 --> 00:20:00,890 is our denominator, and the bigger the denominator, the 402 00:20:00,890 --> 00:20:02,950 smaller the fraction. 403 00:20:02,950 --> 00:20:07,560 And finally, notice that as 'n' goes to infinity, 'U sub 404 00:20:07,560 --> 00:20:12,140 n' gets arbitrarily close to 1/3 in value, because '1/n' 405 00:20:12,140 --> 00:20:13,850 approaches 0 in the limit. 406 00:20:13,850 --> 00:20:15,580 In fact, it is 0 in the limit. 407 00:20:15,580 --> 00:20:19,380 So summarizing then, each 'U sub n' is bigger than 1/3. 408 00:20:19,380 --> 00:20:23,280 As 'n' increases, 'U sub n' decreases, and the limit of 'U 409 00:20:23,280 --> 00:20:26,490 sub n' as 'n' approaches infinity is 1/3. 410 00:20:26,490 --> 00:20:29,790 Pictorially, what this means is that if we locate 1/3 on 411 00:20:29,790 --> 00:20:35,580 the number line, the 'U sub n's converge uniformly in the 412 00:20:35,580 --> 00:20:40,130 sense of moving steadily towards the left, towards 1/3 413 00:20:40,130 --> 00:20:41,950 as the limit. 414 00:20:41,950 --> 00:20:43,770 Well, that's an upper squeeze. 415 00:20:43,770 --> 00:20:46,960 The lower squeeze comes from the fact that in a similar 416 00:20:46,960 --> 00:20:51,000 way, we can show that 'L sub n' is 1/6-- 417 00:20:51,000 --> 00:20:53,590 and look at how close this comes to parallelling the 418 00:20:53,590 --> 00:20:59,460 structure of 'U sub n'-- '1 - '1/n'' times '2 - '1/n''. 419 00:20:59,460 --> 00:21:02,310 Minuses here instead of pluses as we had above. 420 00:21:02,310 --> 00:21:05,000 Now, mimicking what we did before, notice now 421 00:21:05,000 --> 00:21:06,390 that for each 'n'-- 422 00:21:06,390 --> 00:21:08,325 since we're subtracting over here-- 423 00:21:08,325 --> 00:21:12,220 for each 'n', 'L sub n' is less than 1/3. 424 00:21:12,220 --> 00:21:14,940 But now since the fractions get smaller as 'n' gets 425 00:21:14,940 --> 00:21:17,130 bigger, and you're subtracting them, the 426 00:21:17,130 --> 00:21:19,120 difference becomes larger. 427 00:21:19,120 --> 00:21:21,770 In other words, now notice that each 'L sub n' 428 00:21:21,770 --> 00:21:23,170 is less than 1/3. 429 00:21:23,170 --> 00:21:26,470 As 'n' increases, 'L sub n' increases. 430 00:21:26,470 --> 00:21:28,540 And the limit of 'L sub n' as 'n' approaches 431 00:21:28,540 --> 00:21:30,750 infinity is also 1/3. 432 00:21:30,750 --> 00:21:33,920 In other words, if we draw this in back here, look what 433 00:21:33,920 --> 00:21:35,290 the 'L sub n's are doing. 434 00:21:35,290 --> 00:21:38,750 They're all less than 1/3, but they move steadily towards the 435 00:21:38,750 --> 00:21:42,020 right as 'n' increases, pushing in on 1/3. 436 00:21:42,020 --> 00:21:46,040 And since 'A sub R' is always caught between these two, and 437 00:21:46,040 --> 00:21:49,460 these two converge relentlessly upon 1/3, it must 438 00:21:49,460 --> 00:21:52,690 be, by definition, if there is an area at all, that the area 439 00:21:52,690 --> 00:21:55,290 of the region 'R' must be exactly 1/3. 440 00:21:55,290 --> 00:21:58,250 And by the way, just as a quick aside, notice how 441 00:21:58,250 --> 00:22:01,840 important it is that we not only have upper and lower 442 00:22:01,840 --> 00:22:03,790 bounds which converge. 443 00:22:03,790 --> 00:22:06,570 They must converge to the same value. 444 00:22:06,570 --> 00:22:09,640 In other words, what I'm saying is suppose all the 'U 445 00:22:09,640 --> 00:22:13,290 sub n's get arbitrarily close to what I call script 'L sub 446 00:22:13,290 --> 00:22:16,120 1', and all the 'L sub n's get arbitrarily close 447 00:22:16,120 --> 00:22:17,750 to 'L2' over here. 448 00:22:17,750 --> 00:22:22,160 Then all I'm saying is, all we would know is that the area 449 00:22:22,160 --> 00:22:24,380 was someplace between 'L2' and 'L1'. 450 00:22:24,380 --> 00:22:27,470 We couldn't conclude that it was exactly equal to 1/3. 451 00:22:27,470 --> 00:22:30,460 You see, the method of exhaustion implies that all 452 00:22:30,460 --> 00:22:34,560 the error, all the room for doubt, is squeezed out. 453 00:22:34,560 --> 00:22:37,100 By the way, there was nothing sacred about choosing 'y' 454 00:22:37,100 --> 00:22:38,020 equals 'x squared'. 455 00:22:38,020 --> 00:22:39,630 We could generalize this. 456 00:22:39,630 --> 00:22:42,190 And we'll do this kind of rapidly because I just want 457 00:22:42,190 --> 00:22:43,880 you to hear what I'm saying. 458 00:22:43,880 --> 00:22:46,280 This is all written out in great detail in our 459 00:22:46,280 --> 00:22:47,280 supplementary notes. 460 00:22:47,280 --> 00:22:48,530 The idea is this. 461 00:22:48,530 --> 00:22:51,775 Let 'f' be any positive, continuous function on 'a', 462 00:22:51,775 --> 00:22:54,100 'b' and non-decreasing. 463 00:22:54,100 --> 00:22:57,000 The non-decreasing part is just for the convenience of 464 00:22:57,000 --> 00:23:00,210 being able to locate the high and low points of each 465 00:23:00,210 --> 00:23:02,930 partition point conveniently. 466 00:23:02,930 --> 00:23:06,850 Partition 'a', 'b' into 'n' equal parts, calling the first 467 00:23:06,850 --> 00:23:11,460 partition point 'x sub 0', the last partition point 'x sub 468 00:23:11,460 --> 00:23:17,470 n', and the points in between 'x1' up through ''x 'n - 1''. 469 00:23:17,470 --> 00:23:18,760 In other words, we've partitioned this closed 470 00:23:18,760 --> 00:23:21,230 interval into 'n' equal parts. 471 00:23:21,230 --> 00:23:22,980 And what we can now do is what? 472 00:23:22,980 --> 00:23:26,630 Pick the lowest point in each region, the highest point in 473 00:23:26,630 --> 00:23:28,360 each region, to form a rectangle. 474 00:23:28,360 --> 00:23:32,280 We can form 'U sub n' and 'L sub n'. 475 00:23:32,280 --> 00:23:35,490 And you see it's just what? 476 00:23:35,490 --> 00:23:39,830 For 'L sub n', you just pick this height, which is 'f of 'x 477 00:23:39,830 --> 00:23:42,310 sub 0'', 'f of a'. 478 00:23:42,310 --> 00:23:44,270 The base is 'delta x'. 479 00:23:44,270 --> 00:23:47,620 You just add these all up until you get to your last 480 00:23:47,620 --> 00:23:49,340 inscribed partition point. 481 00:23:49,340 --> 00:23:51,810 That's 'x sub 'n - 1''. 482 00:23:51,810 --> 00:23:52,570 OK. 483 00:23:52,570 --> 00:23:56,950 'U sub n' is the same thing, only now your first rectangle 484 00:23:56,950 --> 00:24:00,220 has its height corresponding to 'x' equals 'x1'. 485 00:24:00,220 --> 00:24:02,670 Consequently, the height is 'f of x1'. 486 00:24:02,670 --> 00:24:04,460 The base is 'delta x'. 487 00:24:04,460 --> 00:24:06,940 Where in each of these, 'delta x' is just what? 488 00:24:06,940 --> 00:24:09,730 This total length, which is 'b - a', divided 489 00:24:09,730 --> 00:24:11,680 into 'n' equal parts. 490 00:24:11,680 --> 00:24:16,630 And to review the so-called sigma notation that we have in 491 00:24:16,630 --> 00:24:17,760 our course-- 492 00:24:17,760 --> 00:24:20,110 in this section in the textbook, and we'll have 493 00:24:20,110 --> 00:24:23,130 exercises on this to make sure that you are familiar with 494 00:24:23,130 --> 00:24:27,580 this notation, the shortcut notation for writing this sum 495 00:24:27,580 --> 00:24:28,890 is given by this. 496 00:24:28,890 --> 00:24:32,260 We add up 'f of 'x sub k'' times 'delta x'. 497 00:24:32,260 --> 00:24:37,680 As a subscript, 'k' is allowed to vary from 1 to 'n'. 498 00:24:37,680 --> 00:24:42,220 And here we form the same sum, only now the subscript 'k' 499 00:24:42,220 --> 00:24:45,960 varies from 0 to 'n - 1'. 500 00:24:45,960 --> 00:24:49,510 At any rate, the observations are this. 501 00:24:49,510 --> 00:24:53,230 That 'U sub n' and 'L sub n' converge to the same limit. 502 00:24:53,230 --> 00:24:55,150 In fact, this isn't too hard to show. 503 00:24:55,150 --> 00:24:58,780 If you just algebraically subtract 'L sub n' from 'U sub 504 00:24:58,780 --> 00:25:03,280 n', notice that the only terms that won't cancel are 505 00:25:03,280 --> 00:25:04,890 the last term here. 506 00:25:04,890 --> 00:25:08,400 See, 'f of 'x sub n'' times 'delta x' has no counterpart 507 00:25:08,400 --> 00:25:12,310 here, because this ends with the subscript 'n - 1'. 508 00:25:12,310 --> 00:25:16,590 Similarly, there is no 0 subscript in 'U sub n', so 509 00:25:16,590 --> 00:25:18,350 this term won't cancel. 510 00:25:18,350 --> 00:25:21,630 To make a long story short, if we just subtract 'L sub n' 511 00:25:21,630 --> 00:25:25,420 from 'U sub n', that difference will be 'f of 'x 512 00:25:25,420 --> 00:25:29,430 sub n'' times 'delta x' minus 'f of 'x sub 0'' 513 00:25:29,430 --> 00:25:30,910 times 'delta x'. 514 00:25:30,910 --> 00:25:33,740 See, that's just this difference over here. 515 00:25:33,740 --> 00:25:36,610 'Delta x' is 'b - a' over 'n'. 516 00:25:36,610 --> 00:25:38,780 'x sub n' was called 'b'. 517 00:25:38,780 --> 00:25:41,220 'x sub 0' was called 'a'. 518 00:25:41,220 --> 00:25:44,480 In other words, the difference between 'U sub n' and 'L sub 519 00:25:44,480 --> 00:25:48,020 n' is just 'f of b' minus 'f of a' times 520 00:25:48,020 --> 00:25:49,760 ''b - a' over 'n''. 521 00:25:49,760 --> 00:25:54,000 And the important thing to observe is that 'a', 'b', 'f 522 00:25:54,000 --> 00:25:56,960 of a', and 'f of b' are fixed constants. 523 00:25:56,960 --> 00:26:00,460 The only thing that depends on 'n' is this denominator. 524 00:26:00,460 --> 00:26:03,620 Consequently, as 'n' goes to infinity, since the rest of 525 00:26:03,620 --> 00:26:07,990 this thing is a constant, the whole term goes to 0. 526 00:26:07,990 --> 00:26:11,050 In other words, the difference between 'Un' and 'Ln' goes to 527 00:26:11,050 --> 00:26:12,200 0 in the limit. 528 00:26:12,200 --> 00:26:15,060 That means that the limit of 'U sub n', as 'n' approaches 529 00:26:15,060 --> 00:26:18,190 infinity, equals the limit of 'L sub n' as 530 00:26:18,190 --> 00:26:19,740 'n' approaches infinity. 531 00:26:19,740 --> 00:26:23,900 'A sub R' is always caught between 'L sub n' and 'U sub 532 00:26:23,900 --> 00:26:26,480 n' by this very construction. 533 00:26:26,480 --> 00:26:30,170 Consequently, what this means is that the area of the region 534 00:26:30,170 --> 00:26:34,830 'R' must equal this common limit. 535 00:26:34,830 --> 00:26:37,360 Now by the way, this looks very hard. 536 00:26:37,360 --> 00:26:38,360 And it is difficult. 537 00:26:38,360 --> 00:26:41,510 In fact, in general, these limits are very hard to find. 538 00:26:41,510 --> 00:26:43,780 The point that I thought would be interesting to mention at 539 00:26:43,780 --> 00:26:47,210 this stage of the game is that if all we want is an estimate 540 00:26:47,210 --> 00:26:50,120 for the area under the curve, there are faster and better 541 00:26:50,120 --> 00:26:52,020 ways of doing this. 542 00:26:52,020 --> 00:26:54,790 You see, the beauty of this technique here is that it's a 543 00:26:54,790 --> 00:26:57,240 technique for exhausting the space completely. 544 00:26:57,240 --> 00:26:59,850 We find the exact areas this way. 545 00:26:59,850 --> 00:27:02,690 Let me give you a for instance here. 546 00:27:02,690 --> 00:27:05,650 Namely, let me explain to you what we mean by trapezoidal 547 00:27:05,650 --> 00:27:06,760 approximations. 548 00:27:06,760 --> 00:27:10,510 Let's take the same region, 'y' equals 'x squared'. 549 00:27:10,510 --> 00:27:13,370 And now, we'll divide it into two parts here. 550 00:27:13,370 --> 00:27:17,130 And what we'll do is we'll replace the arc of the curve 551 00:27:17,130 --> 00:27:20,360 by the straight line segment, the chord, that joins two 552 00:27:20,360 --> 00:27:21,640 points here. 553 00:27:21,640 --> 00:27:24,540 In other words, using the accented chalk here to 554 00:27:24,540 --> 00:27:25,410 illustrate this. 555 00:27:25,410 --> 00:27:28,310 What I'm going to do is instead of finding the area of 556 00:27:28,310 --> 00:27:31,400 the region 'R', I'm going to find the area of the region 557 00:27:31,400 --> 00:27:34,640 which has the top of 'R' replaced by 558 00:27:34,640 --> 00:27:36,520 these two line segments. 559 00:27:36,520 --> 00:27:40,050 All I want you to observe is that this first region here is 560 00:27:40,050 --> 00:27:45,080 a triangle whose height is 1/4 and whose base is 1/2. 561 00:27:45,080 --> 00:27:47,700 Consequently, its area, being one half the base times the 562 00:27:47,700 --> 00:27:50,380 height, is 1/16. 563 00:27:50,380 --> 00:27:55,370 The second region here is a trapezoid whose bases are 1 564 00:27:55,370 --> 00:27:58,280 and 1/4 and whose height is 1/2. 565 00:27:58,280 --> 00:28:01,150 And since the area of a trapezoid is half the sum of 566 00:28:01,150 --> 00:28:06,870 the bases times the height, we get that the area of this 567 00:28:06,870 --> 00:28:09,310 trapezoid is 5/16. 568 00:28:09,310 --> 00:28:12,950 Therefore, the area of the triangle plus the trapezoid is 569 00:28:12,950 --> 00:28:17,400 6/16 or 3/8, which is 0.375. 570 00:28:17,400 --> 00:28:21,110 Recall that 1/3, we've just seen, was the exact answer. 571 00:28:21,110 --> 00:28:24,450 And notice how close this approximation is to the exact 572 00:28:24,450 --> 00:28:28,360 answer, even with just two subdivisions over here. 573 00:28:28,360 --> 00:28:32,770 In fact, sparing you the details, if you now divide the 574 00:28:32,770 --> 00:28:36,820 base here into four equal parts, thus forming your 575 00:28:36,820 --> 00:28:39,790 trapezoidal approximations corresponding to what? 576 00:28:39,790 --> 00:28:46,430 Altitudes of 1/16, 1/4, 9/16, and 1, and adding up the areas 577 00:28:46,430 --> 00:28:48,400 of all these trapezoids-- 578 00:28:48,400 --> 00:28:50,660 by the way, this is a bit unfortunate. 579 00:28:50,660 --> 00:28:52,110 This is a triangle. 580 00:28:52,110 --> 00:28:56,110 But we can view a triangle as being a degenerate trapezoid, 581 00:28:56,110 --> 00:28:58,990 meaning the height here just happens to be 0 at this 582 00:28:58,990 --> 00:29:00,030 particular point. 583 00:29:00,030 --> 00:29:01,790 But let's not worry about that. 584 00:29:01,790 --> 00:29:03,560 Let's just get the idea of what the trapezoidal 585 00:29:03,560 --> 00:29:05,080 approximation idea means. 586 00:29:05,080 --> 00:29:10,280 We find the area of these four trapezoids, and we use that as 587 00:29:10,280 --> 00:29:13,060 an approximation for the area under the curve. 588 00:29:13,060 --> 00:29:16,070 By the way, just coming back here for a moment, notice that 589 00:29:16,070 --> 00:29:19,850 because this curve is always holding water, the chord will 590 00:29:19,850 --> 00:29:22,620 always lie above the arc. 591 00:29:22,620 --> 00:29:25,390 And consequently, not only do we get an approximation using 592 00:29:25,390 --> 00:29:28,710 trapezoids this way that's reasonably close, but we also 593 00:29:28,710 --> 00:29:32,210 know by the geometry here that our approximation must be too 594 00:29:32,210 --> 00:29:34,170 large to be the right answer. 595 00:29:34,170 --> 00:29:36,730 In other words, we're going to get an over-approximation. 596 00:29:36,730 --> 00:29:38,380 But watch how close we come. 597 00:29:38,380 --> 00:29:43,350 Leaving these details for you to verify, all I show here is 598 00:29:43,350 --> 00:29:46,690 that if you add up the areas of these four regions, we wind 599 00:29:46,690 --> 00:29:53,010 up with 44/128, which is 11/32. 600 00:29:53,010 --> 00:29:57,590 And 11/32 is mighty close to 1/3, being what? 601 00:29:57,590 --> 00:29:59,120 11/33. 602 00:29:59,120 --> 00:30:01,420 In other words, look at how close, with just four 603 00:30:01,420 --> 00:30:04,900 subdivisions, we get to an approximation that's good for 604 00:30:04,900 --> 00:30:07,360 the area under the curve, without having to put the 605 00:30:07,360 --> 00:30:08,460 squeeze on. 606 00:30:08,460 --> 00:30:11,410 But to get the exact area, we need the squeeze. 607 00:30:11,410 --> 00:30:14,740 We have to push the thing from above, from below, and show 608 00:30:14,740 --> 00:30:17,630 that these two things that we're squeezing it between, as 609 00:30:17,630 --> 00:30:20,700 gruesome as it sounds, converge towards each other. 610 00:30:20,700 --> 00:30:23,970 The thing that we're looking for is caught between them, 611 00:30:23,970 --> 00:30:26,560 and consequently must be the common limit. 612 00:30:26,560 --> 00:30:29,690 That was the method of exhaustion as known by the 613 00:30:29,690 --> 00:30:30,370 ancient Greeks. 614 00:30:30,370 --> 00:30:32,530 And if this seems tough to you, think of it from two 615 00:30:32,530 --> 00:30:33,290 points of view. 616 00:30:33,290 --> 00:30:35,040 One, it is tough. 617 00:30:35,040 --> 00:30:39,510 And secondly, if people of some 2,500 years ago were able 618 00:30:39,510 --> 00:30:42,420 to do this, it should be at least plausible that with a 619 00:30:42,420 --> 00:30:44,800 little bit of effort, we can get a good feeling for what's 620 00:30:44,800 --> 00:30:45,840 going on here. 621 00:30:45,840 --> 00:30:48,880 In fact, hopefully, the exercises to this particular 622 00:30:48,880 --> 00:30:51,440 unit will make this a little bit easier for 623 00:30:51,440 --> 00:30:53,440 you to see in action. 624 00:30:53,440 --> 00:30:55,100 But at any rate-- 625 00:30:55,100 --> 00:30:57,790 let me just make a few asides over here. 626 00:30:57,790 --> 00:31:01,200 The deeper asides will be made in greater detail in our 627 00:31:01,200 --> 00:31:02,750 supplementary notes. 628 00:31:02,750 --> 00:31:05,830 Notice that when we dealt with differential calculus, it was 629 00:31:05,830 --> 00:31:08,960 very, very important in differential calculus to have 630 00:31:08,960 --> 00:31:10,380 smooth curves. 631 00:31:10,380 --> 00:31:14,750 For finding areas, all you need are continuous curves. 632 00:31:14,750 --> 00:31:16,490 In other words, for example, we can find 633 00:31:16,490 --> 00:31:18,280 the area of a square. 634 00:31:18,280 --> 00:31:21,270 But certainly, a square has sharp corners. 635 00:31:21,270 --> 00:31:23,960 In other words, those corners are continuous, but they're 636 00:31:23,960 --> 00:31:25,820 not differentiable in terms of the language of 637 00:31:25,820 --> 00:31:27,240 differential calculus. 638 00:31:27,240 --> 00:31:29,930 The point is that even continuity can be weakened. 639 00:31:29,930 --> 00:31:31,630 Let me give you a definition. 640 00:31:31,630 --> 00:31:34,820 'f' is called piecewise continuous on the interval 641 00:31:34,820 --> 00:31:39,910 from 'a' to 'b' if and only if 'f' is continuous except at a 642 00:31:39,910 --> 00:31:43,700 finite number of points where it has jump discontinuities. 643 00:31:43,700 --> 00:31:47,210 For example, in terms of this particular diagram, notice 644 00:31:47,210 --> 00:31:51,270 that my curve, 'y' equals 'f of x', is discontinuous at a 645 00:31:51,270 --> 00:31:53,320 finite number of points, namely at 'c1' 646 00:31:53,320 --> 00:31:55,290 and 'c2', two points. 647 00:31:55,290 --> 00:31:57,810 Notice that what happens at those two points is you have 648 00:31:57,810 --> 00:32:00,300 just a finite jump discontinuity. 649 00:32:00,300 --> 00:32:03,520 The point is that whereas this curve is not continuous, 650 00:32:03,520 --> 00:32:06,550 notice that since a straight line having no thickness has 651 00:32:06,550 --> 00:32:12,060 no area, if we replace the given curve by this one-- 652 00:32:12,060 --> 00:32:14,010 you see, putting in these vertical lines-- 653 00:32:14,010 --> 00:32:17,200 notice that this does form a closed region. 654 00:32:17,200 --> 00:32:20,890 And to find the area of this closed region, I can pretend 655 00:32:20,890 --> 00:32:24,880 it was made up of these three particular regular regions. 656 00:32:24,880 --> 00:32:28,370 In other words, that even if I just have jumps, since the 657 00:32:28,370 --> 00:32:31,250 jump does not contribute towards the area, there is no 658 00:32:31,250 --> 00:32:34,920 harm done when one talks about piecewise continuous in 659 00:32:34,920 --> 00:32:38,500 finding areas rather than continuous. 660 00:32:38,500 --> 00:32:39,800 A second aside-- 661 00:32:39,800 --> 00:32:42,090 well, I've actually called this one aside number one, 662 00:32:42,090 --> 00:32:43,870 because this definition I didn't call an aside-- 663 00:32:43,870 --> 00:32:45,230 but the idea is this. 664 00:32:45,230 --> 00:32:49,010 Notice that up until now, we were assuming that we had to 665 00:32:49,010 --> 00:32:52,680 form our 'U sub n's and our 'L sub n's by choosing either the 666 00:32:52,680 --> 00:32:56,350 left endpoint of a partition or the right endpoint. 667 00:32:56,350 --> 00:32:59,910 What I'd like you to see also as a generalization is that if 668 00:32:59,910 --> 00:33:05,340 I pick any point between these two extremes and form-- 669 00:33:05,340 --> 00:33:07,720 look, I'll call it 'c sub k' between 670 00:33:07,720 --> 00:33:09,000 these two points here-- 671 00:33:09,000 --> 00:33:13,510 and form this particular sum, where 'c sub k' replaces 672 00:33:13,510 --> 00:33:18,530 either 'x sub k' or 'x sub 'k - 1'', that this sum here also 673 00:33:18,530 --> 00:33:20,990 gives me the area of the region 'R'. 674 00:33:20,990 --> 00:33:23,540 In other words, that whatever this sum is, it's caught 675 00:33:23,540 --> 00:33:25,970 between 'U sub n' and 'L sub n'. 676 00:33:25,970 --> 00:33:28,900 And since 'U sub n' and 'L sub n' have a common 677 00:33:28,900 --> 00:33:30,620 limit of 'A sub R'-- 678 00:33:30,620 --> 00:33:32,855 in other words, since both of these squeeze in 679 00:33:32,855 --> 00:33:34,220 towards 'A sub R'-- 680 00:33:34,220 --> 00:33:37,860 this being caught between them must also be 'A sub R'. 681 00:33:37,860 --> 00:33:39,790 That's what I've said over here. 682 00:33:39,790 --> 00:33:42,650 And to see this thing in terms of a picture, all we're saying 683 00:33:42,650 --> 00:33:48,460 is that when you pick the lowest point in the interval, 684 00:33:48,460 --> 00:33:52,870 you get the rectangle that contributes to 'L sub n'. 685 00:33:52,870 --> 00:33:55,440 When you pick the highest point in the interval, you get 686 00:33:55,440 --> 00:33:58,320 the rectangle that contributes to 'U sub n'. 687 00:33:58,320 --> 00:34:01,990 Consequently, for any point between these two extremes-- 688 00:34:01,990 --> 00:34:04,630 call that--that's what 'c sub k' is, it's any point between 689 00:34:04,630 --> 00:34:05,525 these two-- 690 00:34:05,525 --> 00:34:07,640 pick the height that corresponds to that. 691 00:34:07,640 --> 00:34:11,409 And whatever rectangle you form this way, that rectangle 692 00:34:11,409 --> 00:34:15,040 must have a greater area than the rectangle that went into 693 00:34:15,040 --> 00:34:18,800 forming 'L sub n', but a lesser area than the rectangle 694 00:34:18,800 --> 00:34:21,100 that went into forming 'U sub n'. 695 00:34:21,100 --> 00:34:23,909 And consequently, that's where this particular 696 00:34:23,909 --> 00:34:25,260 result comes from. 697 00:34:25,260 --> 00:34:28,389 Again, this is done in more detail in the notes, but I 698 00:34:28,389 --> 00:34:31,050 think some of these things you should hear me say out loud 699 00:34:31,050 --> 00:34:33,300 rather than to rely on your reading it. 700 00:34:33,300 --> 00:34:36,620 And finally, one more important point. 701 00:34:36,620 --> 00:34:39,429 Up until now, it's been clear that since we're talking about 702 00:34:39,429 --> 00:34:43,790 area, our region has to lie above the x-axis. 703 00:34:43,790 --> 00:34:48,570 We can remove the restriction that 'f' be non-negative if we 704 00:34:48,570 --> 00:34:51,210 replace area by net area. 705 00:34:51,210 --> 00:34:54,159 In other words, all I want you to see here is that if my 706 00:34:54,159 --> 00:34:57,390 region happens to look something like this, notice 707 00:34:57,390 --> 00:35:00,800 that between 'c' and 'b', 'f of x' is negative. 708 00:35:00,800 --> 00:35:04,010 Consequently, when I form things of the form 'f of x' 709 00:35:04,010 --> 00:35:07,990 times 'delta x', 'delta x' being positive, 'f of x' being 710 00:35:07,990 --> 00:35:10,960 negative, is going to give me a negative result. 711 00:35:10,960 --> 00:35:15,270 In other words, algebraically, if I form my summation, this 712 00:35:15,270 --> 00:35:17,840 portion in here will give me a positive result. 713 00:35:17,840 --> 00:35:20,350 This portion in here will give me a negative result. 714 00:35:20,350 --> 00:35:23,230 Consequently, working algebraically, what I will 715 00:35:23,230 --> 00:35:27,280 find is not the true area but the positive 716 00:35:27,280 --> 00:35:29,670 minus this amount here. 717 00:35:29,670 --> 00:35:32,680 In other words, what I'll find is the net area. 718 00:35:32,680 --> 00:35:36,860 And again, this will all be worked out in the notes. 719 00:35:36,860 --> 00:35:39,670 In addition, there'll be other refinements made in the notes, 720 00:35:39,670 --> 00:35:43,130 such as that these partitions don't have to be into equal 721 00:35:43,130 --> 00:35:44,950 parts or things like this. 722 00:35:44,950 --> 00:35:48,570 The important point from today's lecture is this. 723 00:35:48,570 --> 00:35:55,000 Observe that this entire study of area is done independently 724 00:35:55,000 --> 00:35:56,880 of differential calculus. 725 00:35:56,880 --> 00:36:00,730 In our next lecture, we are going to show a truly 726 00:36:00,730 --> 00:36:05,020 remarkable, a wonderful relationship between the 727 00:36:05,020 --> 00:36:08,620 differential calculus of before and the so-called area 728 00:36:08,620 --> 00:36:10,980 or integral calculus that we did today. 729 00:36:10,980 --> 00:36:13,030 At any rate, until next time, goodbye. 730 00:36:15,910 --> 00:36:19,110 Funding for the publication of this video was provided by the 731 00:36:19,110 --> 00:36:23,160 Gabriella and Paul Rosenbaum Foundation. 732 00:36:23,160 --> 00:36:27,330 Help OCW continue to provide free and open access to MIT 733 00:36:27,330 --> 00:36:31,540 courses by making a donation at ocw.mit.edu/donate.