1 00:00:00,040 --> 00:00:01,940 ANNOUNCER: The following content is provided under a 2 00:00:01,940 --> 00:00:03,690 Creative 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 00:00:33,290 --> 00:00:36,330 HERBERT GROSS: Hi, our topic today sounds a little bit 9 00:00:36,330 --> 00:00:37,870 naughty at first glance. 10 00:00:37,870 --> 00:00:40,860 It's called improper integrals. 11 00:00:40,860 --> 00:00:43,780 And rather than give a long speech about why they're 12 00:00:43,780 --> 00:00:48,015 called improper, let me try and start our lecture a little 13 00:00:48,015 --> 00:00:51,370 bit different from our usual approach, by giving you a 14 00:00:51,370 --> 00:00:54,200 little bit of a mathematical calculus riddle. 15 00:00:54,200 --> 00:00:56,010 Namely, let me just say this. 16 00:00:56,010 --> 00:00:59,110 We'll call our lecture today improper integrals, and we'll 17 00:00:59,110 --> 00:01:02,150 start off by saying find the flaw. 18 00:01:02,150 --> 00:01:04,714 I would like to integrate 'dx' over 'x squared' 19 00:01:04,714 --> 00:01:06,760 from minus 1 to 1. 20 00:01:06,760 --> 00:01:09,880 Using the first fundamental theorem of integral calculus, 21 00:01:09,880 --> 00:01:14,800 I say this is 'G of 1' minus 'G of minus 1' where 'G prime' 22 00:01:14,800 --> 00:01:16,910 is any function whose derivative 23 00:01:16,910 --> 00:01:18,880 is '1 over 'x squared''. 24 00:01:18,880 --> 00:01:23,840 Well, in particular, do I know a function whose derivative is 25 00:01:23,840 --> 00:01:25,230 '1 over 'x squared''? 26 00:01:25,230 --> 00:01:31,550 Well, I guess minus '1 over x'. 27 00:01:31,550 --> 00:01:39,070 In other words, the derivative of 'x' to the minus 1 is minus 28 00:01:39,070 --> 00:01:40,360 'x' to the minus 2. 29 00:01:40,360 --> 00:01:42,560 With the minus in front that makes it plus. 30 00:01:42,560 --> 00:01:43,880 'x' to the minus 2. 31 00:01:43,880 --> 00:01:45,900 That's plus '1 over 'x squared''. 32 00:01:45,900 --> 00:01:47,950 In any event, I can then compute that 'G 33 00:01:47,950 --> 00:01:49,660 of 1' is minus 1. 34 00:01:49,660 --> 00:01:51,720 'G of minus 1' is 1. 35 00:01:51,720 --> 00:01:56,390 Therefore, 'G of 1' minus 'G of minus 1' is minus 1 minus 36 00:01:56,390 --> 00:01:58,190 1, is minus 2. 37 00:01:58,190 --> 00:02:03,260 Therefore, the integral from minus 1 to 1, 'dx' over 'x 38 00:02:03,260 --> 00:02:06,030 squared', is minus 2. 39 00:02:06,030 --> 00:02:09,550 And the question is, find the flaw. 40 00:02:09,550 --> 00:02:11,090 Find the flaw. 41 00:02:11,090 --> 00:02:14,660 And since again, we're on limited time, let me, if you 42 00:02:14,660 --> 00:02:18,420 haven't already discovered the flaw, at least point out why 43 00:02:18,420 --> 00:02:20,550 you should be suspicious of a flaw. 44 00:02:20,550 --> 00:02:24,950 Remember, we visualized this as an area under a curve. 45 00:02:24,950 --> 00:02:27,510 In other words, in a way, the base of our region is the 46 00:02:27,510 --> 00:02:30,480 closed interval from minus 1 to 1, and the height of our 47 00:02:30,480 --> 00:02:33,490 region at any point is given by '1 over 'x squared''. 48 00:02:33,490 --> 00:02:35,870 Well, look at '1 over 'x squared''. 49 00:02:35,870 --> 00:02:38,390 Certainly, '1 over 'x squared'' can't be negative 50 00:02:38,390 --> 00:02:40,820 because for any real number 'x', its square is 51 00:02:40,820 --> 00:02:41,710 non-negative. 52 00:02:41,710 --> 00:02:44,250 And one over a non-negative number is certainly 53 00:02:44,250 --> 00:02:45,240 non-negative. 54 00:02:45,240 --> 00:02:48,870 In particular then, since my region never goes below the 55 00:02:48,870 --> 00:02:52,700 x-axis, I would expect that the area represented by this 56 00:02:52,700 --> 00:02:54,200 should be positive. 57 00:02:54,200 --> 00:02:56,960 In other words, this is the sum of positive numbers taken 58 00:02:56,960 --> 00:02:57,830 in the limit. 59 00:02:57,830 --> 00:03:01,110 In other words, the fact that the integrand is non-negative 60 00:03:01,110 --> 00:03:04,100 tells me that whenever the integral turns out to be, it 61 00:03:04,100 --> 00:03:09,470 must be at least as big as 0, and minus 2 does not fulfill 62 00:03:09,470 --> 00:03:11,230 that criteria. 63 00:03:11,230 --> 00:03:14,600 In other words, something must be wrong here because we get a 64 00:03:14,600 --> 00:03:18,280 negative answer in a situation where a negative answer is 65 00:03:18,280 --> 00:03:19,580 preposterous. 66 00:03:19,580 --> 00:03:22,590 And the reason that I wanted to pick this approach is I 67 00:03:22,590 --> 00:03:25,400 think that the best way to motivate that something is 68 00:03:25,400 --> 00:03:29,400 wrong is precisely by winding up with a preposterous answer 69 00:03:29,400 --> 00:03:32,250 in a situation where we know the answer is preposterous. 70 00:03:32,250 --> 00:03:36,020 Then we know we must be on our guard to see what went wrong. 71 00:03:36,020 --> 00:03:37,180 And what did go wrong? 72 00:03:37,180 --> 00:03:39,080 Well, the key point is this. 73 00:03:39,080 --> 00:03:41,590 And again, it's something that we've said many, many times. 74 00:03:41,590 --> 00:03:44,290 But I think it takes a counter example before it 75 00:03:44,290 --> 00:03:45,560 really sinks in. 76 00:03:45,560 --> 00:03:47,900 People are always quoting the first 77 00:03:47,900 --> 00:03:49,780 fundamental theorem in general. 78 00:03:49,780 --> 00:03:52,200 It's easy to remember that the integral from 'a' to 'b', ''f 79 00:03:52,200 --> 00:03:55,530 of x' dx' is 'G of b' minus 'G of a', where 'G 80 00:03:55,530 --> 00:03:56,860 prime' equals 'f'. 81 00:03:56,860 --> 00:04:00,600 But the thing that we must be extremely careful of is that 82 00:04:00,600 --> 00:04:04,970 this definition requires that 'f' be at least piecewise 83 00:04:04,970 --> 00:04:07,500 continuous on the interval from 'a' to 'b'. 84 00:04:07,500 --> 00:04:09,810 Remember when we took the limit and put the squeeze on, 85 00:04:09,810 --> 00:04:13,150 we required that there be continuity, so we could pass 86 00:04:13,150 --> 00:04:14,040 to the limit. 87 00:04:14,040 --> 00:04:16,730 The important point is in this particular example that we're 88 00:04:16,730 --> 00:04:21,070 dealing with, observe that '1 over 'x squared'' is infinite. 89 00:04:21,070 --> 00:04:23,970 I write this in quotation marks to point out you have a 90 00:04:23,970 --> 00:04:27,060 1/0 form, which we usually abbreviate as infinity. 91 00:04:27,060 --> 00:04:29,120 In other words, '1 over 'x squared'' increases without 92 00:04:29,120 --> 00:04:31,380 bound as 'x' approaches 0. 93 00:04:31,380 --> 00:04:35,600 And notice that 0 was within the limits of our integration 94 00:04:35,600 --> 00:04:37,260 from minus 1 to 1. 95 00:04:37,260 --> 00:04:40,490 At any rate, without further ado, this leads to a 96 00:04:40,490 --> 00:04:41,670 fundamental definition. 97 00:04:41,670 --> 00:04:43,900 Which in terms of today's lesson, I call 98 00:04:43,900 --> 00:04:45,260 definition number 1. 99 00:04:45,260 --> 00:04:49,460 And that is the integral from 'a' to 'b', 'f of x dx is 100 00:04:49,460 --> 00:04:51,280 called 'improper'. 101 00:04:51,280 --> 00:04:53,930 And there are going to be two types of improper integrals as 102 00:04:53,930 --> 00:04:55,450 we'll find out very shortly. 103 00:04:55,450 --> 00:05:00,050 But it's called improper of the first kind, if and only if 104 00:05:00,050 --> 00:05:04,730 'f' is infinite for at least one 'c' in the interval from 105 00:05:04,730 --> 00:05:05,700 'a' to 'b'. 106 00:05:05,700 --> 00:05:09,190 And again, let me caution you about this and I'll summarize 107 00:05:09,190 --> 00:05:11,260 at the end of the lecture again to remind you. 108 00:05:11,260 --> 00:05:14,340 If you look at this particular integral, it doesn't look at 109 00:05:14,340 --> 00:05:16,040 all suspicious. 110 00:05:16,040 --> 00:05:19,620 In fact, that's where the trap came in in this 111 00:05:19,620 --> 00:05:20,790 problem over here. 112 00:05:20,790 --> 00:05:23,630 The reason that we wound up with minus 2 is that we just 113 00:05:23,630 --> 00:05:26,430 integrated '1 over 'x squared'', forgetting that 114 00:05:26,430 --> 00:05:30,100 there was a point at which the integrand was infinite in the 115 00:05:30,100 --> 00:05:32,050 given interval. 116 00:05:32,050 --> 00:05:34,640 And so the question that comes up is, how shall we handle 117 00:05:34,640 --> 00:05:37,100 this if we have an infinity in here? 118 00:05:37,100 --> 00:05:40,120 And the answer is, let's work around the point at which the 119 00:05:40,120 --> 00:05:43,700 integral or the integrand is infinite. 120 00:05:43,700 --> 00:05:45,720 Let's see what that means. 121 00:05:45,720 --> 00:05:47,320 And this again, is a continuation of the 122 00:05:47,320 --> 00:05:48,950 definition, I guess. 123 00:05:48,950 --> 00:05:51,650 If 'c' is the only point in the closed interval from 'a' 124 00:05:51,650 --> 00:05:54,930 to 'b' at which 'f' is infinite, we defined the 125 00:05:54,930 --> 00:05:57,780 integral from 'a' to 'b', ''f of x' dx' to be 126 00:05:57,780 --> 00:05:59,130 the following limit. 127 00:05:59,130 --> 00:06:01,330 It's the limit as 'h' approaches 0 128 00:06:01,330 --> 00:06:03,190 through positive values. 129 00:06:03,190 --> 00:06:06,650 The definite integral from 'a' to 'c' minus 'h'. 130 00:06:06,650 --> 00:06:09,700 In other words, you're stopping somewhere before you 131 00:06:09,700 --> 00:06:13,270 get to this bad point, evaluating ''f of x' dx'. 132 00:06:13,270 --> 00:06:16,200 And there's no harm done in here because you see, 'f of x' 133 00:06:16,200 --> 00:06:19,540 is infinite only at 'x' equals 'c' in this problem. 134 00:06:19,540 --> 00:06:23,130 Consequently, if you stop prior to 'c', 'f of x' is at 135 00:06:23,130 --> 00:06:25,720 least piecewise continuous every place in here. 136 00:06:25,720 --> 00:06:28,860 Then you jump over the bad point, pick up again at 'c 137 00:06:28,860 --> 00:06:33,230 plus h' and say plus the integral from 'c plus h' to 138 00:06:33,230 --> 00:06:34,910 'b', ''f of x' dx'. 139 00:06:34,910 --> 00:06:38,630 This limit as 'h' approaches 0 through positive values. 140 00:06:38,630 --> 00:06:41,250 Now again, this may sound very abstract. 141 00:06:41,250 --> 00:06:44,050 So to give this a more realistic meaning to you, 142 00:06:44,050 --> 00:06:47,680 let's interpret this again in terms of an area. 143 00:06:47,680 --> 00:06:49,730 What we're saying is this pictorially. 144 00:06:49,730 --> 00:06:53,500 We have a function which is piecewise continuous at least 145 00:06:53,500 --> 00:06:56,370 on the entire interval from 'a' to 'b', except at the 146 00:06:56,370 --> 00:06:59,370 point 'c' where we have an infinite discontinuity, an 147 00:06:59,370 --> 00:07:00,420 infinite jump. 148 00:07:00,420 --> 00:07:04,160 What we do is, is we pick a positive number 'h' and knock 149 00:07:04,160 --> 00:07:08,580 off the interval 'c minus h' to 'c plus h' surrounding 'c'. 150 00:07:08,580 --> 00:07:11,460 We then compute the area. 151 00:07:11,460 --> 00:07:14,450 We then compute the area of this shaded region. 152 00:07:14,450 --> 00:07:17,470 That's exactly what's inside the brackets over here. 153 00:07:17,470 --> 00:07:19,180 You see, notice that this sum is what? 154 00:07:19,180 --> 00:07:25,290 It's the sum of two integrals, one of which names this area, 155 00:07:25,290 --> 00:07:28,020 and the other of which names this area. 156 00:07:28,020 --> 00:07:32,510 And what we say is, let's put the squeeze on and see what 157 00:07:32,510 --> 00:07:36,730 limiting value this area approaches as we squeeze all 158 00:07:36,730 --> 00:07:39,650 the space out around 'c'. 159 00:07:39,650 --> 00:07:42,960 In other words, what's really happening over here? 160 00:07:42,960 --> 00:07:46,640 What we're really saying is, why can't we treat an infinite 161 00:07:46,640 --> 00:07:50,140 region the same as if it were a finite region taking the 162 00:07:50,140 --> 00:07:54,570 limit as the size of the strip in here goes to 0? 163 00:07:54,570 --> 00:07:55,840 And here's the key point. 164 00:07:55,840 --> 00:07:59,430 You see, as 'h' approaches 'c' from both sides, notice that 165 00:07:59,430 --> 00:08:03,450 the width of this region approaches 0. 166 00:08:03,450 --> 00:08:06,640 By the same token, the height of the region 167 00:08:06,640 --> 00:08:08,800 is approaching infinity. 168 00:08:08,800 --> 00:08:12,960 In essence, what we're running into again, is the problem of 169 00:08:12,960 --> 00:08:16,610 determining the area of an infinite region involves the 170 00:08:16,610 --> 00:08:19,150 study of infinity times 0. 171 00:08:19,150 --> 00:08:24,370 In fact, the question sort of is, does 'h' go to 0 faster 172 00:08:24,370 --> 00:08:28,450 than 'f of 'c plus or minus h'' goes to infinity? 173 00:08:28,450 --> 00:08:31,070 In fact, I think that's an important enough observation, 174 00:08:31,070 --> 00:08:33,070 so I've written it down here to make sure 175 00:08:33,070 --> 00:08:33,980 that you notice it. 176 00:08:33,980 --> 00:08:36,870 The question centers about the question, the problem of 177 00:08:36,870 --> 00:08:40,190 whether 'h' approaches 0 faster than-- 178 00:08:40,190 --> 00:08:41,380 whatever this means. 179 00:08:41,380 --> 00:08:45,710 Faster than 'f of 'c plus or minus h'' approaches infinity. 180 00:08:45,710 --> 00:08:48,930 In our particular example, this didn't happen. 181 00:08:48,930 --> 00:08:52,820 Now again, if you're afraid of the symbol infinity, let's 182 00:08:52,820 --> 00:08:56,010 deal with the finite case and take the limit later. 183 00:08:56,010 --> 00:08:59,110 In other words, instead of looking at the integral from 0 184 00:08:59,110 --> 00:09:04,340 to 1, 'dx' over 'x squared', let's look at the integral 185 00:09:04,340 --> 00:09:06,190 from 'h' to 1. 186 00:09:06,190 --> 00:09:09,580 Integral from 'h' to 1, 'dx' over 'x squared', and then 187 00:09:09,580 --> 00:09:12,520 later we'll take the limit as 'h' approaches 0. 188 00:09:12,520 --> 00:09:15,820 In particular, if this is my region 'R', namely bounded 189 00:09:15,820 --> 00:09:19,250 above by the curve 'y' equals '1 over 'x squared'', below by 190 00:09:19,250 --> 00:09:20,290 the x-axis. 191 00:09:20,290 --> 00:09:23,590 On the left by the line 'x' equals 'h' and on the right by 192 00:09:23,590 --> 00:09:26,910 the line 'x' equals 1, the area of the region 'R' is 193 00:09:26,910 --> 00:09:27,490 written what? 194 00:09:27,490 --> 00:09:30,790 Integral 'h' to 1, 'dx' over 'x squared'. 195 00:09:30,790 --> 00:09:33,880 Notice that as long as 'h' is not 0, our integrand is 196 00:09:33,880 --> 00:09:34,900 continuous. 197 00:09:34,900 --> 00:09:37,600 Therefore, we can use the first fundamental theorem to 198 00:09:37,600 --> 00:09:39,590 evaluate it. 199 00:09:39,590 --> 00:09:40,140 It's what? 200 00:09:40,140 --> 00:09:44,110 It's minus '1 over x' evaluated between 'h' and 1. 201 00:09:44,110 --> 00:09:47,460 Putting in the upper limit gives me a minus 1. 202 00:09:47,460 --> 00:09:50,580 Putting in the lower limit gives me minus '1/h'. 203 00:09:50,580 --> 00:09:54,000 But I'm subtracting that, so it becomes plus '1/h'. 204 00:09:54,000 --> 00:09:58,870 In other words, the area of my region 'R' is 1/h minus 1. 205 00:09:58,870 --> 00:10:03,850 Notice then that 'A sub R' as 'h' approaches 0 from the 206 00:10:03,850 --> 00:10:07,280 right becomes the limit of '1/h' as 'h' approaches 0 from 207 00:10:07,280 --> 00:10:09,330 the right minus 1. 208 00:10:09,330 --> 00:10:12,670 And the limit of '1/h' as 'h' approaches 0 from the right is 209 00:10:12,670 --> 00:10:13,560 clearly infinity. 210 00:10:13,560 --> 00:10:16,510 In other words, as 'h' gets arbitrarily close to 0 through 211 00:10:16,510 --> 00:10:20,450 positive values, '1/h' increases without any bound. 212 00:10:20,450 --> 00:10:24,230 In other words, what this means in plain English is that 213 00:10:24,230 --> 00:10:27,730 we can make the area of the region 'R' as arbitrarily as 214 00:10:27,730 --> 00:10:31,110 large as we wish just by choosing 'h' to be 215 00:10:31,110 --> 00:10:34,300 sufficiently close to 0 here. 216 00:10:34,300 --> 00:10:36,690 You see, in other words, as 'h' approaches 0, what's 217 00:10:36,690 --> 00:10:40,110 happening here is that this curve is not approaching the 218 00:10:40,110 --> 00:10:44,160 y-axis fast enough to allow us to get a finite area. 219 00:10:44,160 --> 00:10:47,550 And by the way, let me make one little aside comparing 220 00:10:47,550 --> 00:10:49,740 this with our previous answer. 221 00:10:49,740 --> 00:10:52,350 You may recall, and let me just pull this board down here 222 00:10:52,350 --> 00:10:53,730 to remind you of that. 223 00:10:53,730 --> 00:10:56,580 You may recall that when we worked this problem before, we 224 00:10:56,580 --> 00:10:59,520 got the wrong answer, negative 2. 225 00:10:59,520 --> 00:11:00,540 OK, now look it. 226 00:11:00,540 --> 00:11:05,370 In terms of my problem over here, we went from minus 1 to 227 00:11:05,370 --> 00:11:08,120 1 so this is only 1/2 of the region that 228 00:11:08,120 --> 00:11:09,170 we're talking about. 229 00:11:09,170 --> 00:11:12,200 Notice that the minus 2 comes from this. 230 00:11:12,200 --> 00:11:16,390 See twice minus 1 by symmetry. 231 00:11:16,390 --> 00:11:19,030 The interesting thing is notice here that when you take 232 00:11:19,030 --> 00:11:20,010 the limit-- 233 00:11:20,010 --> 00:11:22,540 well, forgetting that there's an infinite discontinuity 234 00:11:22,540 --> 00:11:25,590 here, when you let 'h' approach 0, you tend to say 235 00:11:25,590 --> 00:11:28,430 let's forget about the function of 'h' in here. 236 00:11:28,430 --> 00:11:30,710 Notice here as 'h' approaches 0, this 237 00:11:30,710 --> 00:11:32,110 thing went to infinity. 238 00:11:32,110 --> 00:11:35,820 By ignoring it, that's where you got the answer minus 2. 239 00:11:35,820 --> 00:11:38,540 In other words, by assuming that as 'h' got small, the 240 00:11:38,540 --> 00:11:42,460 function of 'h' got small, we threw out an infinite part of 241 00:11:42,460 --> 00:11:43,400 our answer. 242 00:11:43,400 --> 00:11:44,420 In other words, this was the thing that we 243 00:11:44,420 --> 00:11:45,350 had to worry about. 244 00:11:45,350 --> 00:11:48,340 In this particular case, we slurred over something that 245 00:11:48,340 --> 00:11:50,470 wasn't going to 0 fast enough. 246 00:11:50,470 --> 00:11:54,300 By the way, let me caution you and point out that in this 247 00:11:54,300 --> 00:11:57,920 particular example it happened that infinity 248 00:11:57,920 --> 00:12:00,370 times 0 became infinite. 249 00:12:00,370 --> 00:12:06,120 It is possible, however, that as 'h' approaches 0, it may 250 00:12:06,120 --> 00:12:09,620 happen that 'h' approaches 0 fast enough, so that we do get 251 00:12:09,620 --> 00:12:10,560 a finite area. 252 00:12:10,560 --> 00:12:13,880 To show you this, let's just change the problem slightly. 253 00:12:13,880 --> 00:12:17,360 Instead of taking 'y' equals '1 over 'x squared'', let's 254 00:12:17,360 --> 00:12:20,180 take 'y' equals '1 over the 'square root of x''. 255 00:12:20,180 --> 00:12:22,700 The same problem, and by the way, here's the subtlety of 256 00:12:22,700 --> 00:12:23,410 this again. 257 00:12:23,410 --> 00:12:26,200 If you were to sketch 'y' equals '1 over the 'square 258 00:12:26,200 --> 00:12:32,760 root of x'' alongside of 'y' equals '1 over 'x squared''-- 259 00:12:32,760 --> 00:12:36,160 in fact, fast enough so that even though we get an infinite 260 00:12:36,160 --> 00:12:38,620 region, it has only a finite area. 261 00:12:38,620 --> 00:12:40,680 In particular, here's what I'm saying now. 262 00:12:40,680 --> 00:12:44,840 Let 'R' be the region bounded above by 'y' equals '1 over 263 00:12:44,840 --> 00:12:47,900 the 'square root of x', below by the x-axis. 264 00:12:47,900 --> 00:12:50,360 On the left by 'x' equals 'h' and on the right 265 00:12:50,360 --> 00:12:51,850 by 'x' equals 1. 266 00:12:51,850 --> 00:12:55,450 Again, since the only place that '1 over the 'square root 267 00:12:55,450 --> 00:12:59,140 of x'' has an infinite discontinuity is when 'x' is 268 00:12:59,140 --> 00:13:02,430 0, and since 'h' is greater than 0 here, we have a 269 00:13:02,430 --> 00:13:03,900 continuous integrand. 270 00:13:03,900 --> 00:13:07,490 The area of the region 'R' therefore is given by what? 271 00:13:07,490 --> 00:13:10,170 Any function whose derivative is '1 over the 'square root of 272 00:13:10,170 --> 00:13:12,710 x'' evaluated between 'h' and 1. 273 00:13:12,710 --> 00:13:14,600 Again, sparing you the details. 274 00:13:14,600 --> 00:13:17,830 Observe that if you differentiate twice 'x to the 275 00:13:17,830 --> 00:13:21,070 1/2', the 1/2 and the 2 cancel and you get '1 over the 276 00:13:21,070 --> 00:13:22,020 'square root of x''. 277 00:13:22,020 --> 00:13:25,130 In other words, 2 square root of 'x' is the function whose 278 00:13:25,130 --> 00:13:27,250 derivative is '1 over the 'square root of x''. 279 00:13:27,250 --> 00:13:32,120 If I now evaluate that between 1 and 'h', I get 2 minus 2 280 00:13:32,120 --> 00:13:33,530 'square root of h'. 281 00:13:33,530 --> 00:13:37,280 If I now take the limit of 'A sub R' as 'h' approaches 0, 282 00:13:37,280 --> 00:13:40,590 notice that what happens here is that as 'h' approaches 0, 283 00:13:40,590 --> 00:13:42,160 so does the square root of 'h'. 284 00:13:42,160 --> 00:13:45,040 And the area turns out to be 2 in the limit. 285 00:13:45,040 --> 00:13:47,310 In other words, the area of the region 'R' in this 286 00:13:47,310 --> 00:13:49,980 particular case, is also a function of 'h'. 287 00:13:49,980 --> 00:13:53,360 But the part that depends on 'h' goes to 0 288 00:13:53,360 --> 00:13:55,480 as 'h' goes to 0. 289 00:13:55,480 --> 00:13:58,530 In fact, notice in this particular case, and why I 290 00:13:58,530 --> 00:14:01,290 couldn't have started with this example is that if I took 291 00:14:01,290 --> 00:14:04,170 this example and you did it the incorrect way, in other 292 00:14:04,170 --> 00:14:08,230 words, if you failed to notice that '1 over the 'square root 293 00:14:08,230 --> 00:14:12,260 of x'' was discontinuous at 'x' equals 0, and you just did 294 00:14:12,260 --> 00:14:15,970 this problem mechanically, and evaluated this between 0 and 295 00:14:15,970 --> 00:14:18,030 1, you would have got the right answer. 296 00:14:18,030 --> 00:14:20,130 And the reason that you would have got the right answer, is 297 00:14:20,130 --> 00:14:23,320 that in the limit process, notice here that the part of 298 00:14:23,320 --> 00:14:26,310 the answer that depends on 'h' does actually 299 00:14:26,310 --> 00:14:28,250 go to 0 in the limit. 300 00:14:28,250 --> 00:14:32,590 At any rate, using all of this as motivation, we come to a 301 00:14:32,590 --> 00:14:36,000 new piece of terminology that we should define. 302 00:14:36,000 --> 00:14:40,390 Namely, if 'f' is infinite at 'x' equals 'c', where 'c' is 303 00:14:40,390 --> 00:14:44,120 someplace in the interval from 'a' to 'b', then the improper 304 00:14:44,120 --> 00:14:49,710 integral 'a' to 'b', ''f of x' dx' is called convergent if 305 00:14:49,710 --> 00:14:52,850 the particular limit that we talked about previously in the 306 00:14:52,850 --> 00:14:55,550 definition of how you compute integral from 'a' to 'b', ''f 307 00:14:55,550 --> 00:14:59,140 of x' dx', if that particular limit exists, meaning it's a 308 00:14:59,140 --> 00:15:02,980 finite number, then we call the integral a convergent 309 00:15:02,980 --> 00:15:04,610 improper integral. 310 00:15:04,610 --> 00:15:06,290 On the other hand, meaning what? 311 00:15:06,290 --> 00:15:09,440 If this limit doesn't exist, if for example, the limit is 312 00:15:09,440 --> 00:15:12,670 infinite, then we call the integral divergent. 313 00:15:12,670 --> 00:15:14,680 In other words, in our first example, the 314 00:15:14,680 --> 00:15:16,590 integral was divergent. 315 00:15:16,590 --> 00:15:20,130 In the second example, the integral was convergent. 316 00:15:20,130 --> 00:15:23,720 By the way, it's rather interesting to point out too-- 317 00:15:23,720 --> 00:15:24,990 I shouldn't say it's rather interesting. 318 00:15:24,990 --> 00:15:27,980 It's just my excuse of I don't know of any other way of 319 00:15:27,980 --> 00:15:31,230 motivating improper integrals of the second kind. 320 00:15:31,230 --> 00:15:33,350 But the thing that's worth observing is this. 321 00:15:37,250 --> 00:15:43,650 When we evaluated the area under the curve here from 0 to 322 00:15:43,650 --> 00:15:47,380 1, we elected to pick strips like this. 323 00:15:47,380 --> 00:15:48,550 Remember this is why we got into that infinite 324 00:15:48,550 --> 00:15:51,580 discontinuity; we picked out strips to look like this. 325 00:15:51,580 --> 00:15:54,290 The question that comes up is, why couldn't we have picked 326 00:15:54,290 --> 00:15:55,790 our strips to be horizontal? 327 00:15:55,790 --> 00:15:58,390 In other words, if we had in terms of say, inverse 328 00:15:58,390 --> 00:16:01,860 functions, inverted the role of 'x' and 'y', why couldn't 329 00:16:01,860 --> 00:16:05,940 we have computed our area by picking strips like this and 330 00:16:05,940 --> 00:16:09,670 adding them all up from 1 to 'b', and then taking the limit 331 00:16:09,670 --> 00:16:11,360 as 'b' went to infinity? 332 00:16:11,360 --> 00:16:13,820 In other words, if I wanted to compute the area of this 333 00:16:13,820 --> 00:16:17,960 region using horizontal strips, I would have had what? 334 00:16:17,960 --> 00:16:20,730 I would have had the area of the region 'R' is the integral 335 00:16:20,730 --> 00:16:22,660 from 1 to 'b'. 336 00:16:22,660 --> 00:16:26,090 And remember, if 'y' equals '1 over 'x squared'', 'x' is '1 337 00:16:26,090 --> 00:16:27,580 over the 'square root of y''. 338 00:16:27,580 --> 00:16:30,020 Therefore, the area of the region 'R' is the interval 339 00:16:30,020 --> 00:16:34,210 from 1 to 'b', 'dy' over the square root of 'y'. 340 00:16:34,210 --> 00:16:37,680 Notice that the square root of 'y' is certainly continuous. 341 00:16:37,680 --> 00:16:39,190 '1 over the 'square root of y'' is 342 00:16:39,190 --> 00:16:40,610 continuous in this interval. 343 00:16:40,610 --> 00:16:44,240 At any rate, I could then have evaluated this integrand here 344 00:16:44,240 --> 00:16:47,150 and taken the limit as 'b' approached infinity. 345 00:16:47,150 --> 00:16:49,860 And by the way, the abbreviation for writing the 346 00:16:49,860 --> 00:16:54,010 limit of the integral 1 to 'b' as 'b' approaches infinity, is 347 00:16:54,010 --> 00:16:57,720 to simply write integral from 1 to infinity 'dy' over the 348 00:16:57,720 --> 00:16:58,940 square root of 'y'. 349 00:16:58,940 --> 00:17:00,370 And this is all this thing means. 350 00:17:00,370 --> 00:17:03,130 For example, in this particular case, notice that 351 00:17:03,130 --> 00:17:07,050 as 'b' goes to infinity, so does the square root of 'b'. 352 00:17:07,050 --> 00:17:11,089 Therefore, 2 'square root of b' minus 2 goes to infinity as 353 00:17:11,089 --> 00:17:12,440 'b' approaches infinity. 354 00:17:12,440 --> 00:17:16,040 And we find by a second method that the area of 355 00:17:16,040 --> 00:17:17,460 our region is infinite. 356 00:17:17,460 --> 00:17:20,450 And again, we have another kind of improper integral. 357 00:17:20,450 --> 00:17:23,849 But structurally, this improper integral is 358 00:17:23,849 --> 00:17:25,880 completely different from the improper 359 00:17:25,880 --> 00:17:27,630 integral that we had before. 360 00:17:27,630 --> 00:17:31,300 Namely, before we had an improper integral where the 361 00:17:31,300 --> 00:17:34,900 integrand became infinite, even though the limits of 362 00:17:34,900 --> 00:17:36,590 integration stayed finite. 363 00:17:36,590 --> 00:17:39,180 Notice what happens in this case. 364 00:17:39,180 --> 00:17:42,160 In this particular case, notice that the integrand 365 00:17:42,160 --> 00:17:43,240 stays finite. 366 00:17:43,240 --> 00:17:46,660 It's the limits of integration that become infinite. 367 00:17:46,660 --> 00:17:50,450 In other words, this leads to our next definition of a new 368 00:17:50,450 --> 00:17:52,810 type of improper integral. 369 00:17:52,810 --> 00:17:57,210 Namely, the integral say, from 'a' to infinity ''f of x' dx'. 370 00:17:57,210 --> 00:18:01,200 Whereby this we mean the abbreviation limit as 'b' 371 00:18:01,200 --> 00:18:04,410 approaches infinity, integral from 'a' to 'b', ''f of x' 372 00:18:04,410 --> 00:18:08,560 dx', where 'f' is continuous in this entire range. 373 00:18:08,560 --> 00:18:11,050 In other words, for 'x' greater than or equal to 'a'. 374 00:18:11,050 --> 00:18:15,030 That type of an integral is called an improper integral of 375 00:18:15,030 --> 00:18:16,630 the second kind. 376 00:18:16,630 --> 00:18:19,350 In other words, in the first kind, the integrand becomes 377 00:18:19,350 --> 00:18:22,170 infinite, but the limits of integration are finite. 378 00:18:22,170 --> 00:18:25,970 In the second kind, the integrand stays finite, but 379 00:18:25,970 --> 00:18:29,330 the limits of integration take on an infinite range. 380 00:18:29,330 --> 00:18:32,120 And the important point is that these two are very 381 00:18:32,120 --> 00:18:33,030 closely connected. 382 00:18:33,030 --> 00:18:35,520 In other words, in terms of the example that I just showed 383 00:18:35,520 --> 00:18:38,700 you, I wanted you to see how you can get from an integral 384 00:18:38,700 --> 00:18:41,130 of the first kind to an improper integral of the 385 00:18:41,130 --> 00:18:44,410 second kind just by replacing your vertical strips by 386 00:18:44,410 --> 00:18:45,720 horizontal strips. 387 00:18:45,720 --> 00:18:49,770 In fact, in terms of my first example over here notice that 388 00:18:49,770 --> 00:18:53,240 indeed, '1 over 'x squared'' and '1 over the 'square root 389 00:18:53,240 --> 00:18:56,960 of x'', where 'x' is at least as big as 0, are inverses. 390 00:18:56,960 --> 00:18:59,850 By the way, the reason I say 'x' is, in fact, bigger than 391 00:18:59,850 --> 00:19:02,400 0, otherwise I'll have a 0 denominator here. 392 00:19:02,400 --> 00:19:05,295 The reason I point this out is that if you don't presuppose 393 00:19:05,295 --> 00:19:09,320 that x is positive, remember, the square root function or 394 00:19:09,320 --> 00:19:12,470 the squaring function is not 1:1, so I have to be very 395 00:19:12,470 --> 00:19:14,400 careful when I talk about inverse functions. 396 00:19:14,400 --> 00:19:17,270 The thing I want to see however is, the graph of 'y' 397 00:19:17,270 --> 00:19:21,410 equals '1 over 'x squared'' is the inverse of the graph 'y' 398 00:19:21,410 --> 00:19:23,370 equals '1 over the 'square root of x''. 399 00:19:23,370 --> 00:19:25,740 In other words, essentially, by interchanging the role of 400 00:19:25,740 --> 00:19:29,270 the x- and y-axes, you go from one graph to the other. 401 00:19:29,270 --> 00:19:34,750 You see, in particular, the integral 1 to infinity, 'dx' 402 00:19:34,750 --> 00:19:36,430 over 'x squared'. 403 00:19:36,430 --> 00:19:39,280 Namely, finding the area of this particular region is the 404 00:19:39,280 --> 00:19:43,130 inverse of finding the area of this particular region. 405 00:19:43,130 --> 00:19:45,850 In other words, notice that 'y' equals '1 over 'x 406 00:19:45,850 --> 00:19:50,590 squared'' behaves for large values of 'x' the same as 'y' 407 00:19:50,590 --> 00:19:53,880 equals '1 over the 'square root of x'' behaves for small 408 00:19:53,880 --> 00:19:55,430 parts of the values of 'x'. 409 00:19:55,430 --> 00:19:59,520 Again, let me point out that much of this will be done much 410 00:19:59,520 --> 00:20:04,090 more hard core in our learning exercises and in the reading 411 00:20:04,090 --> 00:20:05,220 material and the like. 412 00:20:05,220 --> 00:20:08,480 And so I pass over some of this in a fairly quick way. 413 00:20:08,480 --> 00:20:10,800 But what I want to make sure is clear before we begin the 414 00:20:10,800 --> 00:20:13,700 exercises and the reading material is that we have a 415 00:20:13,700 --> 00:20:16,530 good idea as to what improper integrals mean. 416 00:20:16,530 --> 00:20:19,480 At any rate, continuing on this way, what do I mean by 417 00:20:19,480 --> 00:20:21,170 the integral from 1 to infinity, 418 00:20:21,170 --> 00:20:22,590 'dx' over 'x squared'? 419 00:20:22,590 --> 00:20:25,290 Well, it's an improper integral of a second type 420 00:20:25,290 --> 00:20:29,680 because the integrand is finite between 1 and infinity. 421 00:20:29,680 --> 00:20:33,050 The limits of integration span an infinite space over here. 422 00:20:33,050 --> 00:20:33,840 At any rate, it's what? 423 00:20:33,840 --> 00:20:36,710 The limit as 'b' approaches infinity, integral from 1 to 424 00:20:36,710 --> 00:20:38,910 'b', 'dx' over 'x squared'. 425 00:20:42,130 --> 00:20:44,140 At any rate, what is the integral 426 00:20:44,140 --> 00:20:44,926 of '1 over 'x squared''. 427 00:20:44,926 --> 00:20:47,520 It's minus '1/x'. 428 00:20:47,520 --> 00:20:51,840 If I evaluate that that's minus ''1 over b' plus 1'. 429 00:20:51,840 --> 00:20:54,820 And if I now take the limit as 'b' approaches infinity that 430 00:20:54,820 --> 00:20:56,190 answer is 1. 431 00:20:56,190 --> 00:20:57,390 In other words, what this means 432 00:20:57,390 --> 00:20:59,250 pictorially is the following. 433 00:20:59,250 --> 00:21:03,570 If I look at the area of this region 'R' for any value of 434 00:21:03,570 --> 00:21:09,080 'b' whatsoever, that area is equal to 1 minus '1/b'. 435 00:21:09,080 --> 00:21:12,400 In other words, no matter how big 'b' becomes, the area of 436 00:21:12,400 --> 00:21:16,240 this region 'R' can never be as big as one unit. 437 00:21:16,240 --> 00:21:19,320 And in the limit, as 'b' goes out as far as we wish, what 438 00:21:19,320 --> 00:21:20,850 we're really saying is what? 439 00:21:20,850 --> 00:21:24,040 That the area of this region 'R' can be made arbitrarily 440 00:21:24,040 --> 00:21:28,360 close to 1 just by choosing 'b' to be arbitrarily large. 441 00:21:28,360 --> 00:21:30,700 And by the way, this you see leads to an interesting 442 00:21:30,700 --> 00:21:31,710 observation. 443 00:21:31,710 --> 00:21:35,050 Namely, at this end of the spectrum, the curve was not 444 00:21:35,050 --> 00:21:37,580 approaching the y-axis fast enough to make 445 00:21:37,580 --> 00:21:39,250 this integral finite. 446 00:21:39,250 --> 00:21:43,640 Remember, the integral from 0 to 1, 'dx' over 'x squared' 447 00:21:43,640 --> 00:21:44,830 was infinite. 448 00:21:44,830 --> 00:21:48,570 However, for large values of 'x', the curve does approach 449 00:21:48,570 --> 00:21:51,910 the x-axis fast enough, so that not only is the enclosed 450 00:21:51,910 --> 00:21:55,700 area finite, it can't even be bigger than 1. 451 00:21:55,700 --> 00:21:57,750 I thought that might be an interesting example for you to 452 00:21:57,750 --> 00:21:59,440 think about for a while. 453 00:21:59,440 --> 00:22:01,960 Another type of interesting example worth thinking about 454 00:22:01,960 --> 00:22:04,530 for a while is this second example. 455 00:22:04,530 --> 00:22:08,250 Let's take the region 'R' to be bounded above by 'y' equals 456 00:22:08,250 --> 00:22:12,950 '1/x', on the left by 'x' equals 1, on the right by 'x' 457 00:22:12,950 --> 00:22:15,680 equals 'b', and below by the x-axis. 458 00:22:15,680 --> 00:22:18,960 And let's compute the area of the region 'R'. 459 00:22:18,960 --> 00:22:21,780 Again, the area of the region 'R' is a definite integral 460 00:22:21,780 --> 00:22:24,210 from 1 to 'b', 'dx' over 'x'. 461 00:22:24,210 --> 00:22:27,840 And that's equal to the natural log of 'b'. 462 00:22:27,840 --> 00:22:32,390 Namely, this integral is log absolute value of 'x'. 463 00:22:32,390 --> 00:22:34,920 Log absolute value of 'b' is just 'log b'. 464 00:22:34,920 --> 00:22:36,200 Log 1 is 0. 465 00:22:36,200 --> 00:22:39,020 So the area of the region 'R' is 'log b'. 466 00:22:39,020 --> 00:22:42,140 As 'b' increases without bound, its log increases 467 00:22:42,140 --> 00:22:44,390 without bound, we find that the area of the 468 00:22:44,390 --> 00:22:46,230 region 'R' is infinite. 469 00:22:46,230 --> 00:22:49,380 In fact, another way of saying this is what? 470 00:22:49,380 --> 00:22:54,730 This is the integral 1 to infinity 'dx' over 'x'. 471 00:22:54,730 --> 00:22:56,820 And this is, therefore, what? 472 00:22:56,820 --> 00:23:02,170 A divergent improper integral of the second type. 473 00:23:02,170 --> 00:23:04,300 All right, so far so good at any rate. 474 00:23:04,300 --> 00:23:08,540 Let me review our volumes of revolution and rotation. 475 00:23:08,540 --> 00:23:11,710 Let's take the region 'R' and rotate it about the x-axis. 476 00:23:15,000 --> 00:23:17,820 If we rotate 'R' about the x-axis, remember a volume of 477 00:23:17,820 --> 00:23:23,430 revolution is just pi 'y squared' 'dx' from 1 to 'b'. 478 00:23:23,430 --> 00:23:26,530 Well, my 'y' is '1/x'. 479 00:23:26,530 --> 00:23:28,430 That's '1 over x squared'. 480 00:23:28,430 --> 00:23:32,110 The integral of '1 over x squared' is minus '1/x'. 481 00:23:32,110 --> 00:23:35,580 Notice then as before, the volume when I rotate 'R' about 482 00:23:35,580 --> 00:23:40,440 the x-axis is just pi times '1 minus '1/b''. 483 00:23:40,440 --> 00:23:43,710 In other words, what this say is as 'b' gets very, very 484 00:23:43,710 --> 00:23:45,810 large, so does 'R'. 485 00:23:45,810 --> 00:23:50,230 Yet, if I rotate 'R' about the x-axis, no matter how big 'b' 486 00:23:50,230 --> 00:23:56,520 is the volume is just pi times '1 minus '1/b''. 487 00:23:56,520 --> 00:24:01,090 As 'b' approaches infinity, this goes to 0 and my volume 488 00:24:01,090 --> 00:24:02,270 is just pi. 489 00:24:02,270 --> 00:24:05,170 In other words, this area gets infinitely large. 490 00:24:05,170 --> 00:24:08,700 Yet, the volume generated by rotating that 491 00:24:08,700 --> 00:24:10,840 area remains finite. 492 00:24:10,840 --> 00:24:14,860 Again, another idiosyncrasy of working with improper 493 00:24:14,860 --> 00:24:16,730 integrals and infinite regions. 494 00:24:16,730 --> 00:24:19,820 You've got to be very, very careful about how rapidly 495 00:24:19,820 --> 00:24:23,550 something approaches 0 or infinity because infinity 496 00:24:23,550 --> 00:24:25,080 times 0 is indeterminate. 497 00:24:25,080 --> 00:24:28,350 In other words, here is a very important observation, an 498 00:24:28,350 --> 00:24:33,610 infinite area can enclose or can generate a finite volume. 499 00:24:33,610 --> 00:24:35,530 Be very, very careful in dealing 500 00:24:35,530 --> 00:24:37,050 with improper integrals. 501 00:24:37,050 --> 00:24:41,890 And by the way, let me make one computational aside. 502 00:24:41,890 --> 00:24:46,070 In trying to determine whether a particular improper integral 503 00:24:46,070 --> 00:24:50,400 is convergent or not, we do not have to be able to compute 504 00:24:50,400 --> 00:24:52,870 the limit itself. 505 00:24:52,870 --> 00:24:54,490 I mention this because it's very important. 506 00:24:54,490 --> 00:24:57,860 Sometime you'll be given an integral which you don't know 507 00:24:57,860 --> 00:24:59,220 how to integrate directly. 508 00:24:59,220 --> 00:25:02,010 A typical example that usually comes up, integral ''e to the 509 00:25:02,010 --> 00:25:06,520 minus 'x squared' dx', say between 1 and infinity. 510 00:25:06,520 --> 00:25:08,430 And you want to know whether this converges or not. 511 00:25:08,430 --> 00:25:10,570 You don't care what the limit is because once you know it 512 00:25:10,570 --> 00:25:13,200 converges, you can approximate it to as great a degree of 513 00:25:13,200 --> 00:25:14,550 accuracy as you wish. 514 00:25:14,550 --> 00:25:17,140 So the question is, does this converge or doesn't it? 515 00:25:17,140 --> 00:25:20,120 And again, the answer is, let's see what happens on this 516 00:25:20,120 --> 00:25:21,980 particular range, for example. 517 00:25:21,980 --> 00:25:24,060 In other words, we know, for example, that if 'x' is 518 00:25:24,060 --> 00:25:29,210 greater than or equal to 1 for numbers in magnitude at least 519 00:25:29,210 --> 00:25:33,180 as big as 1, the square exceeds the number. 520 00:25:33,180 --> 00:25:36,930 In particular by the way, if 'x squared' exceeds 'x', minus 521 00:25:36,930 --> 00:25:40,270 'x squared' cannot exceed minus 'x'. 522 00:25:40,270 --> 00:25:42,170 In other words, on this particular interval that we're 523 00:25:42,170 --> 00:25:44,890 dealing with, say from 1 to 'b', as 'b' approaches 524 00:25:44,890 --> 00:25:47,620 infinity, 'e to the minus 'x squared'' is 525 00:25:47,620 --> 00:25:49,465 less than minus 'x'. 526 00:25:52,280 --> 00:25:57,670 Therefore, because the exponential function is 1:1, 527 00:25:57,670 --> 00:26:01,100 ''e to the minus 'x squared' dx' from 1 to 'b' cannot 528 00:26:01,100 --> 00:26:04,850 exceed the integral from 1 to 'b', ''e to the minus x' dx'. 529 00:26:04,850 --> 00:26:07,290 In other words, since this exponent is less than this 530 00:26:07,290 --> 00:26:10,440 exponent, this integral is less than this one. 531 00:26:10,440 --> 00:26:13,650 However, it turns out that we know how to evaluate this one. 532 00:26:13,650 --> 00:26:15,630 How do we integrate 'e to the minus x'? 533 00:26:15,630 --> 00:26:17,810 It's just minus 'e to the minus x'. 534 00:26:17,810 --> 00:26:21,260 And if we evaluate that between 1 and 'b', we find 535 00:26:21,260 --> 00:26:25,260 that we get '1/e' minus '1 over 'e to the b'' power. 536 00:26:25,260 --> 00:26:27,040 'b' is some positive number. 537 00:26:27,040 --> 00:26:29,340 This thing here is less than infinity. 538 00:26:29,340 --> 00:26:33,390 In fact, as 'b' approaches infinity, this approaches 0, 539 00:26:33,390 --> 00:26:35,790 and the limiting value is '1/e'. 540 00:26:35,790 --> 00:26:39,360 In other words, what we now know is, is that whatever this 541 00:26:39,360 --> 00:26:43,690 improper integral is, it cannot exceed '1/e' even 542 00:26:43,690 --> 00:26:46,600 though we don't know what the exact answer is. 543 00:26:46,600 --> 00:26:50,110 Pictorially, you see what happens over here is that the 544 00:26:50,110 --> 00:26:53,870 curve which I've drawn in the accentuated chalk, this curve 545 00:26:53,870 --> 00:26:56,260 is 'y' equals 'e to the minus 'x squared''. 546 00:26:56,260 --> 00:26:58,960 The light curve is 'y' equals 'e to the minus x'. 547 00:26:58,960 --> 00:27:02,560 Notice that these two curves crisscross when 'x' equals 1. 548 00:27:02,560 --> 00:27:07,280 And all we're saying is that since the white curve chops 549 00:27:07,280 --> 00:27:11,050 off a finite area and the accentuated curve, the dark 550 00:27:11,050 --> 00:27:15,260 curve is inside the white one, certainly the area enclosed by 551 00:27:15,260 --> 00:27:18,610 the black curve cannot exceed that of the white curve. 552 00:27:18,610 --> 00:27:21,970 In other words, if this region here approaches a finite value 553 00:27:21,970 --> 00:27:25,240 as 'x' goes to infinity, so must the other region. 554 00:27:25,240 --> 00:27:27,170 And this is essentially the theory 555 00:27:27,170 --> 00:27:29,490 behind improper integrals. 556 00:27:29,490 --> 00:27:32,700 In closing, there is one thing I would like 557 00:27:32,700 --> 00:27:33,570 to say about this. 558 00:27:33,570 --> 00:27:37,310 It's a short summary, but here's the whole idea. 559 00:27:37,310 --> 00:27:40,700 It may be difficult to evaluate improper integrals of 560 00:27:40,700 --> 00:27:41,550 a second kind. 561 00:27:41,550 --> 00:27:42,830 But in the certain manner of 562 00:27:42,830 --> 00:27:45,470 speaking, they're not dangerous. 563 00:27:45,470 --> 00:27:46,660 In what manner of speaking? 564 00:27:46,660 --> 00:27:50,560 Well, when you see the limits on the integral sign like a to 565 00:27:50,560 --> 00:27:54,490 infinity, or minus infinity to 'b', or minus infinity to 566 00:27:54,490 --> 00:27:57,135 infinity, you're on the lookout. 567 00:27:57,135 --> 00:27:59,970 You're on the lookout to be careful about 568 00:27:59,970 --> 00:28:01,170 something going wrong. 569 00:28:01,170 --> 00:28:03,480 There's something about the symbol infinity, which 570 00:28:03,480 --> 00:28:07,860 hopefully strikes a certain fear or respect in your minds 571 00:28:07,860 --> 00:28:09,870 when you work mathematically. 572 00:28:09,870 --> 00:28:13,560 See, we have a warning to be ware in this case. 573 00:28:13,560 --> 00:28:17,120 However, the dangerous part is that when you're given an 574 00:28:17,120 --> 00:28:21,660 improper integral of the first type, and I don't know the 575 00:28:21,660 --> 00:28:24,210 best way of saying this other than to say perhaps, that 576 00:28:24,210 --> 00:28:26,930 mathematics has no built-in burglar alarms. 577 00:28:26,930 --> 00:28:29,990 What I mean by that is if you're supposed to multiply 5 578 00:28:29,990 --> 00:28:32,930 and 3 to get the right answer to a problem, and instead you 579 00:28:32,930 --> 00:28:36,170 add 5 and 3, you're going to get 8 for the answer even 580 00:28:36,170 --> 00:28:37,160 though it's incorrect. 581 00:28:37,160 --> 00:28:39,160 There's not going to be a buzzer that goes off and says, 582 00:28:39,160 --> 00:28:41,180 please do not add these two numbers. 583 00:28:41,180 --> 00:28:44,520 In other words, if you go ahead over here and you forget 584 00:28:44,520 --> 00:28:47,660 that 'f of x' may be infinite and you go ahead and integrate 585 00:28:47,660 --> 00:28:49,670 this using the first fundamental theorem, you're 586 00:28:49,670 --> 00:28:54,650 going to get an answer just like we did in our beginning 587 00:28:54,650 --> 00:28:56,280 exercise at the beginning of the lecture. 588 00:28:56,280 --> 00:28:58,130 It happened to be the wrong answer. 589 00:28:58,130 --> 00:29:00,830 So in other words, the word of warning here is whenever you 590 00:29:00,830 --> 00:29:04,310 see an integral with finite limits of integration, check 591 00:29:04,310 --> 00:29:07,500 the integrand and make sure it doesn't blow up, it doesn't 592 00:29:07,500 --> 00:29:08,850 become infinite. 593 00:29:08,850 --> 00:29:12,170 At any rate, with these words we finish our block on 594 00:29:12,170 --> 00:29:13,240 integration. 595 00:29:13,240 --> 00:29:15,190 And until next time, good bye. 596 00:29:17,990 --> 00:29:20,520 ANNOUNCER: Funding for the publication of this video was 597 00:29:20,520 --> 00:29:25,240 provided by the Gabriella and Paul Rosenbaum Foundation. 598 00:29:25,240 --> 00:29:29,410 Help OCW continue to provide free and open access to MIT 599 00:29:29,410 --> 00:29:33,610 courses by making a donation at ocw.mit.edu/donate.