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These are the user uploaded subtitles that are being translated: 1 00:00:00,540 --> 00:00:03,830 Today we're going to be talking about how to evaluate a definite integral. 2 00:00:04,020 --> 00:00:09,390 And in this particular problem we've been given the interval from negative 1 to 1 of either the negative 3 00:00:09,420 --> 00:00:11,370 2 x x. 4 00:00:11,370 --> 00:00:17,070 Now what makes this a definite integral which is visibly obvious is these upper and lower limits of 5 00:00:17,070 --> 00:00:20,620 integration that are attached to this integral notation here. 6 00:00:20,790 --> 00:00:22,810 The negative one and the positive one. 7 00:00:22,980 --> 00:00:27,660 If you don't have those little numbers right there and you just have your integral notation and then 8 00:00:27,660 --> 00:00:33,540 let's say either the negative 2 x x without these numbers then we call it an indefinite integral where 9 00:00:33,540 --> 00:00:36,240 there are no upper and lower limits of integration. 10 00:00:36,250 --> 00:00:39,920 Remember that an integral is all about finding area under the curve. 11 00:00:40,050 --> 00:00:46,530 So without these limits we're saying find the area under the curve everywhere under the curve and above 12 00:00:46,530 --> 00:00:52,960 the x axis from negative infinity to positive infinity or from wherever the graph intersects the x axis. 13 00:00:52,980 --> 00:00:58,080 But in this particular case where we added these limits of integration the negative one and the positive 14 00:00:58,080 --> 00:01:04,740 one were saying give me the area under this curve and above the x axis between X equals negative 1 and 15 00:01:04,770 --> 00:01:06,930 X equals ones on that interval. 16 00:01:07,080 --> 00:01:08,920 That's essentially what we're finding here. 17 00:01:09,210 --> 00:01:14,430 And definite integrals are just like indefinite integrals in the sense that we're going to evaluate 18 00:01:14,430 --> 00:01:19,110 the integral but then we have to plug in our upper and lower limits of integration. 19 00:01:19,110 --> 00:01:21,470 So we'll look at that when we get to it. 20 00:01:21,510 --> 00:01:26,370 But the first thing we want to do is just evaluate this integral and you'll remember that the integral 21 00:01:26,490 --> 00:01:35,100 of each of the X is just to the X but when we have this coefficient on the exponent here this negative 22 00:01:35,280 --> 00:01:44,010 two value we have to divide by that coefficients are integral will be easy to the negative 2 x. 23 00:01:44,070 --> 00:01:51,510 But then we need to divide by that negative twos will say 1 over negative 2 will be our new coefficient. 24 00:01:51,510 --> 00:01:57,750 This is opposite of when we take the derivative of E to the negative 2 x we get negative 2 to the negative 25 00:01:57,750 --> 00:02:03,540 2 x instead of just multiplying by that negative 2 coefficient where dividing by the negative two coefficient 26 00:02:03,540 --> 00:02:06,260 because we're going in the opposite direction taking the integral. 27 00:02:06,270 --> 00:02:11,940 So that's our integral and two things to know before we evaluate this integral on the interval negative 28 00:02:11,940 --> 00:02:12,800 1 to 1. 29 00:02:13,080 --> 00:02:16,500 First we're not going to add the constant of integration. 30 00:02:16,510 --> 00:02:22,230 See Remember normally when we take an integral we add to this plus C we're not going to do that because 31 00:02:22,230 --> 00:02:27,200 when we have a definite integral we don't need to add some arbitrary constant. 32 00:02:27,200 --> 00:02:32,370 The fact that it's a definite integral removes the ambiguity that would normally exist with an indefinite 33 00:02:32,430 --> 00:02:35,980 integral where we have to add that constant integration. 34 00:02:36,030 --> 00:02:40,570 But because we have a definite integral we never have to add that plus c so we won't be adding that. 35 00:02:40,830 --> 00:02:45,720 And then the way that we didn't know that this is a definite integral for which we're going to evaluate 36 00:02:45,720 --> 00:02:51,720 on a specific interval is we draw this line here on the right hand side and we say negative 1 to 1. 37 00:02:51,720 --> 00:02:57,630 So these upper and lower limits of integration here transfer from the integral to this kind of notation 38 00:02:57,630 --> 00:03:03,260 here and what this tells us is that we're evaluating on the interval negative one positive one. 39 00:03:03,300 --> 00:03:05,000 So how do we evaluate now. 40 00:03:05,010 --> 00:03:10,410 Well when you're evaluating a definite integral you always plug in the upper limit of integration first. 41 00:03:10,410 --> 00:03:12,330 In our case that's one here. 42 00:03:12,370 --> 00:03:18,480 We plug that in first then we subtract whatever we get when we plug in our lower limit of integration. 43 00:03:18,510 --> 00:03:19,520 Negative 1. 44 00:03:19,770 --> 00:03:23,200 So first of all plug in 1 will get 1 over negative. 45 00:03:23,210 --> 00:03:29,130 To each to the negative two times positive one our upper limit of integration. 46 00:03:29,400 --> 00:03:32,820 Then we're going to subtract whatever we get when plug our lower limit. 47 00:03:32,820 --> 00:03:37,440 Keep in mind that sometimes because you're subtracting here you can end up with a double negative so 48 00:03:37,470 --> 00:03:42,810 it's really important to have parentheses here so that you make sure you distribute this negative sign 49 00:03:43,140 --> 00:03:44,380 across multiple terms. 50 00:03:44,430 --> 00:03:45,660 If you have them. 51 00:03:45,810 --> 00:03:53,310 So we're going to say minus and then 1 over negative to E to the negative two times are lower limit 52 00:03:53,310 --> 00:03:54,690 of integration here. 53 00:03:54,690 --> 00:03:55,460 Negative 1. 54 00:03:55,500 --> 00:03:57,660 So negative 1 like this. 55 00:03:57,900 --> 00:04:01,910 And now it's just a matter of simplifying whatever we have here. 56 00:04:01,920 --> 00:04:08,060 So we've got negative 1 1/2 E to the negative to the to the negative 2. 57 00:04:08,070 --> 00:04:15,150 Now we have minus a negative one half which is going to give us plus one half B to the negative two 58 00:04:15,150 --> 00:04:19,790 times negative one gives us positive 2 and we could leave our answer like this. 59 00:04:19,800 --> 00:04:24,120 But it's nice to always lead with a positive term instead of a negative term. 60 00:04:24,120 --> 00:04:25,020 If we can. 61 00:04:25,170 --> 00:04:32,270 So we're just going to flip the order of the terms we're going to say one half the squared minus 1 1/2 62 00:04:32,720 --> 00:04:34,590 E to the negative 2. 63 00:04:34,650 --> 00:04:43,180 Then if we want to go ahead and factor out a 1 1/2 we can still say 1 1/2 times the squared minus E 64 00:04:43,200 --> 00:04:49,470 to the negative 2 it would be nice not to have this negative xponent here's what we'll do is we'll move 65 00:04:49,470 --> 00:04:54,480 that to the denominator which will make the xponent positive instead of negative. 66 00:04:54,480 --> 00:05:01,450 So we'll get e squared minus 1 over e squared and that negative becomes a positive tool and move it 67 00:05:01,450 --> 00:05:04,140 to the denominator and that's it that's our final answer. 68 00:05:04,150 --> 00:05:06,910 That's how you evaluate a definite integral. 7891