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These are the user uploaded subtitles that are being translated: 1 00:00:00,520 --> 00:00:04,910 In this video we're talking about how to use the integral test to say whether or not a series converges 2 00:00:05,240 --> 00:00:08,220 and the integral test is one of our many convergence tests. 3 00:00:08,240 --> 00:00:13,670 In this particular problem we've been given this series here one divided by n square plus one we're 4 00:00:13,670 --> 00:00:16,610 taking the sum of that series from end equals one to infinity. 5 00:00:16,610 --> 00:00:22,040 And what we want to do is use the integral test to say whether or not this series converges or diverges 6 00:00:22,370 --> 00:00:29,170 so the integral test tells us that if a sub that's our series here one over and square plus one is positive 7 00:00:29,270 --> 00:00:30,960 decreasing and continuous. 8 00:00:30,960 --> 00:00:38,330 So these three points here are extremely important if our series isn't positive and or is not decreasing 9 00:00:38,420 --> 00:00:44,750 and or is not continuous we can't use the integral test at all to determine convergence or divergence. 10 00:00:44,750 --> 00:00:48,950 We'll have to use a different test or maybe we won't be able to determine convergence at all. 11 00:00:48,950 --> 00:00:54,170 So we have to first make sure that this series is positive decreasing and continuous. 12 00:00:54,170 --> 00:00:59,420 If it is all three of these things then we can move on with the integral test and then the integral 13 00:00:59,420 --> 00:01:06,200 test tells us basically that the behavior of this series and the behavior of this integral the integral 14 00:01:06,200 --> 00:01:12,530 of f of x from 1 to infinity is the same meaning that either both of these things converge or they both 15 00:01:12,620 --> 00:01:13,540 diverge. 16 00:01:13,550 --> 00:01:17,690 So in other words it gives us something else we can evaluate to determine convergence. 17 00:01:17,690 --> 00:01:24,380 So we can find the integral from one to infinity of this function f of x and we just set a 7 and f of 18 00:01:24,380 --> 00:01:29,380 x equal to one another so we use this value here one over end squared plus one. 19 00:01:29,390 --> 00:01:34,100 We just switch out the variable and for the variable x and then we would take this integral integral 20 00:01:34,100 --> 00:01:38,580 from one to infinity of 1 divided by x squared plus 1. 21 00:01:38,680 --> 00:01:43,940 We'd evaluate that integral and then if that integral converges we know that the original series converges 22 00:01:44,360 --> 00:01:45,860 if that integral diverges. 23 00:01:45,860 --> 00:01:50,570 We know that the original series diverges because the integral test tells us that these two things will 24 00:01:50,570 --> 00:01:52,290 always have the same behavior. 25 00:01:52,460 --> 00:01:58,220 As long as the series meets these first three criteria and therefore allows us to use the interval test 26 00:01:58,520 --> 00:01:59,440 in the first place. 27 00:01:59,540 --> 00:02:01,480 So let's tackle these first three criteria. 28 00:02:01,490 --> 00:02:05,300 First we have to show that the series is positive decreasing and continuous. 29 00:02:05,300 --> 00:02:11,330 So we'll start with positive and oftentimes the easiest way to figure these three out is just to find 30 00:02:11,330 --> 00:02:14,960 the first couple of terms and look at the trend that you see between them. 31 00:02:14,960 --> 00:02:18,600 So what this tells you here this and equals one on the sun. 32 00:02:18,620 --> 00:02:21,470 This tells you that the series starts at end equals 1. 33 00:02:21,470 --> 00:02:26,200 So I'm going to plug one in for n then I'm going to plug into them 3 and 4 then 5. 34 00:02:26,300 --> 00:02:29,540 And those are the first few terms of this series. 35 00:02:29,570 --> 00:02:33,540 So if I plug in and equals one I'm going to get one squared which is one. 36 00:02:33,710 --> 00:02:40,280 One plus one is two one divided by two is one half so the first term in the series is one half the next 37 00:02:40,280 --> 00:02:45,080 term in the series is going to be the termit and equal to Pelion n equals 2 I get two square which is 38 00:02:45,080 --> 00:02:50,620 4 4 plus 1 is 5 1 divided by 5 is one fifth. 39 00:02:50,840 --> 00:02:52,460 Then I'm going to plug in and equals three. 40 00:02:52,460 --> 00:02:54,340 I'm going to get nine plus one is 10. 41 00:02:54,350 --> 00:02:58,130 So I'm going to have one tenth then I'll play again and equals four. 42 00:02:58,130 --> 00:03:00,560 I'm going to get 16 plus 1 or 17. 43 00:03:00,650 --> 00:03:05,400 So I'm going to get one over 17 so you can see the pattern here these are the first four terms of the 44 00:03:05,400 --> 00:03:07,880 series and I couldn't keep going. 45 00:03:07,880 --> 00:03:12,800 Now I want to look at these terms and say Can I use these terms to say whether or not the series is 46 00:03:12,800 --> 00:03:13,460 positive. 47 00:03:13,490 --> 00:03:16,840 And in fact I can I can see that the terms are getting smaller. 48 00:03:16,840 --> 00:03:18,780 One fifth is smaller than one half. 49 00:03:18,830 --> 00:03:22,450 One tenth is smaller than one fifth 117 is smaller than one tenth. 50 00:03:22,460 --> 00:03:26,590 But no matter how small they get the terms are always going to be positive. 51 00:03:26,690 --> 00:03:27,940 There's no value of. 52 00:03:27,950 --> 00:03:31,130 And I can plug in because my end values are always going to be positive. 53 00:03:31,130 --> 00:03:36,730 They're going to start at positive 1 and count up forever positive two positive three positive four. 54 00:03:36,890 --> 00:03:41,630 So I'm always going to plug in a positive value and there's no value I can plug in for n that's going 55 00:03:41,630 --> 00:03:47,060 to make this term negative therefore I can say that the terms in my series are always going to be positive 56 00:03:47,390 --> 00:03:52,400 and therefore that the series itself will always be positive then decreasing. 57 00:03:52,400 --> 00:03:58,700 So if I look at my series here I can see that one half is greater than one fifth one fifth is greater 58 00:03:58,700 --> 00:04:01,640 than one tenth one tenth is greater than 1 17th. 59 00:04:01,790 --> 00:04:04,590 The numerator is always going to stay one. 60 00:04:04,820 --> 00:04:09,380 The denominator is going to get larger and larger and larger which means the fraction as a whole gets 61 00:04:09,380 --> 00:04:10,550 smaller and smaller and smaller. 62 00:04:10,550 --> 00:04:16,100 So every term is always going to be smaller than or less than the previous term. 63 00:04:16,130 --> 00:04:22,070 And when each term is less than the previous term then I can say that my series is decreasing so that 64 00:04:22,070 --> 00:04:23,720 one's taking care of as well. 65 00:04:23,720 --> 00:04:25,580 Now what about continuous. 66 00:04:25,760 --> 00:04:31,640 Well basically I just want to show that there are no discontinuities in the domain of this series and 67 00:04:31,640 --> 00:04:37,510 I know that's true because since my series starts it equals one and I count up and equals positive 1 68 00:04:37,750 --> 00:04:41,120 to posit three positive for on into infinity. 69 00:04:41,120 --> 00:04:44,890 I'm always just going to have a positive number in the denominator of each term. 70 00:04:44,900 --> 00:04:51,050 I'm never going to be able to plug in a value for n that's going to make this denominator zero and therefore 71 00:04:51,050 --> 00:04:56,600 result in a discontinuity of the function there's no way I can make this function discontinuous so I 72 00:04:56,600 --> 00:05:03,510 can say that the function is in fact continuous therefore I can say A7 the series is positive decreasing 73 00:05:03,510 --> 00:05:06,920 and continuous so I know I can now use the integral test. 74 00:05:07,050 --> 00:05:13,380 So use the integral test what I want to do is I want to say a Subban is equal to f of X.. 75 00:05:13,380 --> 00:05:19,690 So in other words this value here my series that's a seben that's this value right here. 76 00:05:19,730 --> 00:05:26,730 So then I want to set a sub an equal to f of x and I just want to put it in here into this integral 77 00:05:26,790 --> 00:05:27,910 for f objects. 78 00:05:27,930 --> 00:05:31,760 The only thing I need to do because my integral here is in terms of x variables. 79 00:05:31,830 --> 00:05:36,870 I just want to go ahead and change the end variable to an x variable and then evaluate this integral 80 00:05:36,870 --> 00:05:42,230 so I'm going to have the integral from 1 to infinity and then ever X is going to be dysfunction here. 81 00:05:42,270 --> 00:05:43,440 One over. 82 00:05:43,590 --> 00:05:49,420 But now instead of and I want to say X so I'm going to say X squared plus 1 D x. 83 00:05:49,530 --> 00:05:54,780 Now we need to evaluate this integral and figure out whether or not this integral converges or diverges 84 00:05:54,990 --> 00:06:00,360 if it converges in the original series will also converge if it diverges than the original series will 85 00:06:00,370 --> 00:06:01,870 also diverged. 86 00:06:01,890 --> 00:06:08,010 So to take the integral what we need to realize is that this function here one over x squared plus 1 87 00:06:08,250 --> 00:06:12,220 is a special type of function and this just happens to be this particular problem. 88 00:06:12,240 --> 00:06:20,070 But the integral of this quotient is arctan or Tanev the negative one of x and this function actually 89 00:06:20,070 --> 00:06:25,770 comes up a lot in calculus so it's helpful to know that the specific value 1 divided by x squared plus 90 00:06:25,770 --> 00:06:29,780 1 the integral of that function is arctan of x. 91 00:06:29,820 --> 00:06:34,950 So we're going to have arctan of X evaluated over the interval 1 to infinity. 92 00:06:34,950 --> 00:06:40,530 Now I'm going to be technical about this what I want to do is change this to the limit as B goes to 93 00:06:40,530 --> 00:06:47,280 infinity arctan or 10 and negative 1 of X evaluated over the interval one to be. 94 00:06:47,280 --> 00:06:53,820 And so then I would say the limit as B goes to infinity and I actually could have done that before I 95 00:06:53,820 --> 00:06:58,560 took the integral but I've got the limit as Bigos to infinity evaluating over the interval. 96 00:06:58,560 --> 00:07:01,560 I plug in my upper limit of integration first which is B. 97 00:07:01,590 --> 00:07:08,670 So I'm going to get arctan of B and then minus whatever I get when I plug in my lower limit of integration 98 00:07:08,670 --> 00:07:13,320 once I'm going to get minus arctan of one. 99 00:07:13,320 --> 00:07:19,680 Now you can plug these values arctan of infinity and arctan of one into your calculator or you can do 100 00:07:19,680 --> 00:07:25,380 the math and use the unit circle and then we'll just take arctan of one as an example here. 101 00:07:25,530 --> 00:07:34,110 If we say 10 to the negative one of 1 is equal to x and then we take a tangent of both sides. 102 00:07:34,150 --> 00:07:43,540 We're going to take tangent of the left side here and then tangent of x we get tangent and arctan to 103 00:07:43,540 --> 00:07:50,230 cancel and we get one is equal to tangent of x we remember that tangent is sign over cosign. 104 00:07:50,230 --> 00:07:57,970 So what we get is sign of X over cosign of x is equal to 1 and the only place where this happens on 105 00:07:57,970 --> 00:08:03,730 the unit circle member on a unit circle sign represents the y value of each coordinate point cosen represents 106 00:08:03,730 --> 00:08:09,550 the x value each coordinate point so really what we're looking for is why over x is equal to one or 107 00:08:09,820 --> 00:08:11,760 the point where y is equal to x. 108 00:08:11,890 --> 00:08:16,790 The only point where that happens on the unit circle is the angle PI over 4. 109 00:08:16,810 --> 00:08:21,410 So there's arctan of 1 is PI over 4. 110 00:08:21,430 --> 00:08:26,350 You could also do that with arctan of infinity or taking the limit is Bigos to infinity so we're kind 111 00:08:26,350 --> 00:08:29,470 of plugging infinity in for be here. 112 00:08:29,620 --> 00:08:35,590 And if we did the same Matthew what we get is y over x is equal to infinity instead of Y over x is equal 113 00:08:35,590 --> 00:08:36,850 to 1. 114 00:08:36,850 --> 00:08:43,220 The closest we ever get on the unit circle to infinity the largest value we ever obtain is that PI over 115 00:08:43,240 --> 00:08:43,950 two. 116 00:08:43,960 --> 00:08:48,640 So we're going to say is that arctan of B is PI over 2. 117 00:08:48,640 --> 00:08:52,050 So we end up with PI over to minus PI over 4. 118 00:08:52,120 --> 00:08:58,060 We can multiply both the numerator and denominator of this first fraction by two to get a common denominator 119 00:08:58,060 --> 00:09:05,250 so we end up with two pi over four minus PI over four to PI minus pi is just pi. 120 00:09:05,380 --> 00:09:07,300 And then we have four in the denominator. 121 00:09:07,300 --> 00:09:09,670 So our answer then is PI over 4. 122 00:09:09,670 --> 00:09:13,110 What's important here is not the exact value of the result. 123 00:09:13,120 --> 00:09:17,790 What's important is whether or not you get a real number answer here or an infinite answer. 124 00:09:17,830 --> 00:09:22,020 So the fact that we got a real number pi over 4 is a real number. 125 00:09:22,060 --> 00:09:23,110 It's a constant. 126 00:09:23,110 --> 00:09:27,930 The fact that we got a real number means that this original integral here converges. 127 00:09:27,970 --> 00:09:30,690 So a real number answer indicates that this converges. 128 00:09:30,790 --> 00:09:36,820 If we had found instead a result of infinity or negative infinity that would indicate that this original 129 00:09:36,880 --> 00:09:42,460 integral was divergent but because the value of this integral was a real number that means the integral 130 00:09:42,460 --> 00:09:49,180 converges and the integral test tells us that when this integral converges the original series also 131 00:09:49,180 --> 00:09:52,750 converges so we can say that this original series is convergent. 132 00:09:52,900 --> 00:09:57,670 Again if we had found an infinite value here that would mean that the integral was divergent and we 133 00:09:57,670 --> 00:10:00,780 would say that the original integral was also divergent. 134 00:10:00,880 --> 00:10:06,040 So the last thing I'd say is the key to knowing whether or not you should use the integral test is to 135 00:10:06,040 --> 00:10:11,050 look at this original series and see whether or not it's easy to integrate because remember you're just 136 00:10:11,050 --> 00:10:13,890 going to take this function and plug it into an integral. 137 00:10:13,900 --> 00:10:19,810 Yes we change the variable from end to X but we just evaluated this series as an integral. 138 00:10:19,960 --> 00:10:24,700 So if this original series is easy to integrate and you know you can integrate it then you might be 139 00:10:24,700 --> 00:10:26,800 able to find the value of this integral. 140 00:10:26,890 --> 00:10:29,040 So the integral test might be a good choice. 141 00:10:29,110 --> 00:10:33,700 Sometimes you're going to get a series that's impossible to integrate if you imagine the series in an 142 00:10:33,700 --> 00:10:39,340 integral and you think to yourself how would I ever take the integral of this series the integral test 143 00:10:39,430 --> 00:10:41,560 might not be the test you're looking for. 144 00:10:41,560 --> 00:10:45,610 But if this looks easy to integrate then you might want to try the integral test and then you should 145 00:10:45,610 --> 00:10:52,450 go immediately to this positive decreasing continuous checklist because if this series is positive decreasing 146 00:10:52,450 --> 00:10:55,970 and continuous and that's when you know you'll be able to use the integral test. 147 00:10:56,110 --> 00:10:58,810 And again the result of the integral. 148 00:10:58,870 --> 00:11:03,400 All that matters is whether or not it's a real number answer in which case this integral converges and 149 00:11:03,400 --> 00:11:09,480 so does the series or it's an infinite answer in which case the integral diverges and therefore so does 150 00:11:09,490 --> 00:11:10,360 the series. 17626