[Script Info] Title: [Events] Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text Dialogue: 0,0:00:02.29,0:00:04.27,Default,,0000,0000,0000,,2 videos in this series. Dialogue: 0,0:00:05.67,0:00:10.52,Default,,0000,0000,0000,,Areas as summation and\Nintegration as the reverse of Dialogue: 0,0:00:10.52,0:00:15.37,Default,,0000,0000,0000,,differentiation have shown that\Nthe area under a curve. Dialogue: 0,0:00:16.32,0:00:20.94,Default,,0000,0000,0000,,Above the X axis and between two\Nordinates. That's two values of Dialogue: 0,0:00:20.94,0:00:25.56,Default,,0000,0000,0000,,X can be calculated by using\Nintegration. And what we want to Dialogue: 0,0:00:25.56,0:00:30.95,Default,,0000,0000,0000,,do is develop that idea and look\Nat some more examples of that in Dialogue: 0,0:00:30.95,0:00:32.10,Default,,0000,0000,0000,,this particular video. Dialogue: 0,0:00:33.00,0:00:39.39,Default,,0000,0000,0000,,So. Let's begin by exploring\Nthe question. We've got. The Dialogue: 0,0:00:39.39,0:00:46.21,Default,,0000,0000,0000,,function Y equals X times X\Nminus one times X minus 2. Dialogue: 0,0:00:46.96,0:00:51.23,Default,,0000,0000,0000,,What's the area contained\Nbetween this curve? Dialogue: 0,0:00:51.78,0:00:56.34,Default,,0000,0000,0000,,On the X axis, well, the very\Nnature of the question suggests Dialogue: 0,0:00:56.34,0:01:01.28,Default,,0000,0000,0000,,hang on a minute. There might be\Nsomething a bit odd here. Let's Dialogue: 0,0:01:01.28,0:01:05.08,Default,,0000,0000,0000,,draw a picture. So let's start\Nby sketching this curve. Dialogue: 0,0:01:06.80,0:01:12.20,Default,,0000,0000,0000,,Well, if we put Y equals 0 then\Nwe can see that each one of Dialogue: 0,0:01:12.20,0:01:16.88,Default,,0000,0000,0000,,these brackets could be 0 and\Nthat will give us a series of Dialogue: 0,0:01:16.88,0:01:22.28,Default,,0000,0000,0000,,points on the curve. So why is 0\Nthen? This could be 0 EX could Dialogue: 0,0:01:22.28,0:01:28.01,Default,,0000,0000,0000,,be 0. Or X minus one could be 0.\NIn other words, X could be equal Dialogue: 0,0:01:28.01,0:01:34.49,Default,,0000,0000,0000,,to 1. Or X minus two could be 0.\NIn other words, X could be equal Dialogue: 0,0:01:34.49,0:01:40.13,Default,,0000,0000,0000,,to two, so there on the X axis.\NI know that this curve is going Dialogue: 0,0:01:40.13,0:01:43.52,Default,,0000,0000,0000,,to go through their through\Nthere and through there. Dialogue: 0,0:01:44.67,0:01:49.49,Default,,0000,0000,0000,,What about when X is very large?\NWhat's going to happen to Y when Dialogue: 0,0:01:49.49,0:01:53.96,Default,,0000,0000,0000,,X is very large, X minus One X\Nminus two are really very Dialogue: 0,0:01:53.96,0:01:58.43,Default,,0000,0000,0000,,similar to X, so effectively\Nwe've got X times by X times by Dialogue: 0,0:01:58.43,0:02:03.25,Default,,0000,0000,0000,,X. You know as if X is very\Nlarge and positive, so most why Dialogue: 0,0:02:03.25,0:02:05.100,Default,,0000,0000,0000,,be so? We've got a bit of curve Dialogue: 0,0:02:05.100,0:02:10.50,Default,,0000,0000,0000,,there. Similarly, if X is large\Nbut negative, again taking one Dialogue: 0,0:02:10.50,0:02:15.18,Default,,0000,0000,0000,,off and taking two off isn't\Ngoing to make a great deal of Dialogue: 0,0:02:15.18,0:02:19.50,Default,,0000,0000,0000,,difference, so we've got a large\Nnegative number times by a large Dialogue: 0,0:02:19.50,0:02:23.46,Default,,0000,0000,0000,,negative number times by a large\Nnegative number. The answer is Dialogue: 0,0:02:23.46,0:02:27.06,Default,,0000,0000,0000,,going to be very large, but\Nnegative 'cause we three Dialogue: 0,0:02:27.06,0:02:30.66,Default,,0000,0000,0000,,negative numbers multiplied\Ntogether. So there's a bit of a Dialogue: 0,0:02:30.66,0:02:31.74,Default,,0000,0000,0000,,curve down here. Dialogue: 0,0:02:32.38,0:02:36.54,Default,,0000,0000,0000,,Now how can we join that up?\NWell, assuming the curve is Dialogue: 0,0:02:36.54,0:02:38.63,Default,,0000,0000,0000,,continuous, it's going to go up. Dialogue: 0,0:02:39.80,0:02:43.28,Default,,0000,0000,0000,,Over. Down I'm back up again. Dialogue: 0,0:02:44.10,0:02:48.67,Default,,0000,0000,0000,,So that's the picture of our\Ncurve and straight away we can Dialogue: 0,0:02:48.67,0:02:53.24,Default,,0000,0000,0000,,see what the question is asking\Nus. Find the area contained by Dialogue: 0,0:02:53.24,0:02:57.44,Default,,0000,0000,0000,,the curve and the X axis it\Nwants these two areas. Dialogue: 0,0:02:58.20,0:03:02.65,Default,,0000,0000,0000,,But they're in different\Npositions. One area a is above Dialogue: 0,0:03:02.65,0:03:08.88,Default,,0000,0000,0000,,the X axis and the other one\Narea B is below the X axis. Dialogue: 0,0:03:09.84,0:03:11.94,Default,,0000,0000,0000,,So perhaps we better be Dialogue: 0,0:03:11.94,0:03:15.70,Default,,0000,0000,0000,,cautious. And work out these two Dialogue: 0,0:03:15.70,0:03:21.54,Default,,0000,0000,0000,,areas separately. Now what we've\Nbeen told, what we know is that Dialogue: 0,0:03:21.54,0:03:22.73,Default,,0000,0000,0000,,if we integrate. Dialogue: 0,0:03:23.44,0:03:29.36,Default,,0000,0000,0000,,Between the limits. So in this\Ncase that's not an one hour Dialogue: 0,0:03:29.36,0:03:35.76,Default,,0000,0000,0000,,function X times by X minus one\Ntimes by X minus two with Dialogue: 0,0:03:35.76,0:03:43.16,Default,,0000,0000,0000,,respect to X. We will get this\Narea a, so the area a is equal Dialogue: 0,0:03:43.16,0:03:47.60,Default,,0000,0000,0000,,to the result of carrying out\Nthis integration and Dialogue: 0,0:03:47.60,0:03:52.53,Default,,0000,0000,0000,,substituting in the limits of\Nintegration. So let's do that Dialogue: 0,0:03:52.53,0:03:54.01,Default,,0000,0000,0000,,first of all. Dialogue: 0,0:03:54.01,0:03:56.93,Default,,0000,0000,0000,,We will need to multiply out Dialogue: 0,0:03:56.93,0:04:04.11,Default,,0000,0000,0000,,these brackets. Well, X times by\NX is X squared and times by that Dialogue: 0,0:04:04.11,0:04:06.38,Default,,0000,0000,0000,,ex again is X cubed. Dialogue: 0,0:04:07.37,0:04:11.87,Default,,0000,0000,0000,,I've got X times by minus two\Ngives me minus 2X. Dialogue: 0,0:04:12.54,0:04:17.92,Default,,0000,0000,0000,,And minus one times by X gives\Nme another minus X. So Dialogue: 0,0:04:17.92,0:04:22.84,Default,,0000,0000,0000,,altogether I've got minus 3X\Ncoming out of these brackets and Dialogue: 0,0:04:22.84,0:04:28.22,Default,,0000,0000,0000,,I need times it by X, so that's\Nminus three X squared. Dialogue: 0,0:04:29.15,0:04:36.51,Default,,0000,0000,0000,,Then I've got minus one times by\Nminus two. That's +2 and I need Dialogue: 0,0:04:36.51,0:04:40.20,Default,,0000,0000,0000,,to times that by X. So that's Dialogue: 0,0:04:40.20,0:04:46.41,Default,,0000,0000,0000,,plus 2X. To be integrated with\Nrespect to X so it's carry out Dialogue: 0,0:04:46.41,0:04:52.04,Default,,0000,0000,0000,,the integration. Remember we had\None to the index and divide by Dialogue: 0,0:04:52.04,0:04:58.60,Default,,0000,0000,0000,,the new index, so that's one to\Nthe index. Makes that X to the Dialogue: 0,0:04:58.60,0:05:01.42,Default,,0000,0000,0000,,power 4 / 4 - 3. Dialogue: 0,0:05:02.02,0:05:07.09,Default,,0000,0000,0000,,To integrate X squared, we had\None to the index that's X cubed Dialogue: 0,0:05:07.09,0:05:13.33,Default,,0000,0000,0000,,divided by the new index. That's\N3 + 2. XX is X to the power one. Dialogue: 0,0:05:13.33,0:05:18.40,Default,,0000,0000,0000,,We don't always right the one\Nthere, but we know that it means Dialogue: 0,0:05:18.40,0:05:24.25,Default,,0000,0000,0000,,X to the power one. So we add 1\Nto the index that's X squared Dialogue: 0,0:05:24.25,0:05:26.98,Default,,0000,0000,0000,,and we divide by the new index. Dialogue: 0,0:05:27.76,0:05:31.94,Default,,0000,0000,0000,,These are our limits of\Nintegration nought and one. Dialogue: 0,0:05:32.58,0:05:36.91,Default,,0000,0000,0000,,We can do a little bit of simply\Nsimplifying in here. We can Dialogue: 0,0:05:36.91,0:05:39.24,Default,,0000,0000,0000,,cancel three with a 3 under, two Dialogue: 0,0:05:39.24,0:05:45.35,Default,,0000,0000,0000,,with a 2. And now we can put\Nin our limit of integration. The Dialogue: 0,0:05:45.35,0:05:46.99,Default,,0000,0000,0000,,top 1 one in. Dialogue: 0,0:05:48.26,0:05:54.84,Default,,0000,0000,0000,,So that's. X to the\Npower four. When X is one is Dialogue: 0,0:05:54.84,0:06:00.69,Default,,0000,0000,0000,,just one over 4 minus X cubed.\NWhen X is one. So that's just Dialogue: 0,0:06:00.69,0:06:06.13,Default,,0000,0000,0000,,one plus X squared. When X is\None. So that's just one again. Dialogue: 0,0:06:06.65,0:06:11.25,Default,,0000,0000,0000,,Minus and now we substitute in\Nour lower limit. That's zero, Dialogue: 0,0:06:11.25,0:06:16.68,Default,,0000,0000,0000,,but when we put zero in there\Nthat's zero and that's zero as Dialogue: 0,0:06:16.68,0:06:22.53,Default,,0000,0000,0000,,well, and that's zero as well,\Nbecause X is equal to 0 in each Dialogue: 0,0:06:22.53,0:06:27.55,Default,,0000,0000,0000,,case. So the whole of the SEC\Nbracket here is just zero. Dialogue: 0,0:06:28.07,0:06:31.84,Default,,0000,0000,0000,,So we simplify that we get a Dialogue: 0,0:06:31.84,0:06:38.53,Default,,0000,0000,0000,,quarter. So the area a is\N1/4 of a unit of area. Dialogue: 0,0:06:39.06,0:06:44.66,Default,,0000,0000,0000,,What about area B? We can find\Nthat by doing the same Dialogue: 0,0:06:44.66,0:06:49.33,Default,,0000,0000,0000,,integration, this time taking\Nthe limits between one and two. Dialogue: 0,0:06:49.33,0:06:52.14,Default,,0000,0000,0000,,So let's look at that one. Dialogue: 0,0:06:52.66,0:06:55.40,Default,,0000,0000,0000,,B. Dialogue: 0,0:06:56.75,0:07:03.79,Default,,0000,0000,0000,,Will be the integral between one\Nand two of X cubed. Dialogue: 0,0:07:04.34,0:07:11.60,Default,,0000,0000,0000,,Minus three X squared\Nplus 2X with respect Dialogue: 0,0:07:11.60,0:07:18.65,Default,,0000,0000,0000,,to X. So let's have\Na look what we're going to get Dialogue: 0,0:07:18.65,0:07:24.63,Default,,0000,0000,0000,,the integral of. This should not\Nbe any different to what we had Dialogue: 0,0:07:24.63,0:07:31.07,Default,,0000,0000,0000,,last time. So if we integrate\Nthe X cubed, we have X to the Dialogue: 0,0:07:31.07,0:07:32.45,Default,,0000,0000,0000,,4th over 4. Dialogue: 0,0:07:32.97,0:07:36.27,Default,,0000,0000,0000,,Minus three X cubed Dialogue: 0,0:07:36.27,0:07:43.41,Default,,0000,0000,0000,,over 3. +2 X\Nsquared over 2 and Dialogue: 0,0:07:43.41,0:07:46.71,Default,,0000,0000,0000,,this is to be Dialogue: 0,0:07:46.71,0:07:52.73,Default,,0000,0000,0000,,evaluated. The limits one and\Ntwo, and again we can cancel the Dialogue: 0,0:07:52.73,0:07:55.56,Default,,0000,0000,0000,,threes and we can cancel the Dialogue: 0,0:07:55.56,0:08:00.86,Default,,0000,0000,0000,,tools. So now let's substitute\Nin our upper limit. Dialogue: 0,0:08:02.09,0:08:09.62,Default,,0000,0000,0000,,X to the power four when X\Nis 2, so that's 2 multiplied by Dialogue: 0,0:08:09.62,0:08:15.00,Default,,0000,0000,0000,,itself four times and that gives\Nus 16 over 4. Dialogue: 0,0:08:15.52,0:08:21.19,Default,,0000,0000,0000,,Minus X cubed which is 2\Nmultiplied by itself three Dialogue: 0,0:08:21.19,0:08:23.46,Default,,0000,0000,0000,,times, so that's eight. Dialogue: 0,0:08:24.72,0:08:30.88,Default,,0000,0000,0000,,Plus X squared which is 2\Nmultiplied by itself twice, so Dialogue: 0,0:08:30.88,0:08:38.22,Default,,0000,0000,0000,,that's four. Minus now we put\Nthe one in, so that's a quarter. Dialogue: 0,0:08:38.97,0:08:44.62,Default,,0000,0000,0000,,Because X to the power four is\Njust 1 - 1 because X cubed is Dialogue: 0,0:08:44.62,0:08:46.89,Default,,0000,0000,0000,,just one when X is one. Dialogue: 0,0:08:47.50,0:08:53.97,Default,,0000,0000,0000,,Plus one. Now\Nlet's simplify each of these Dialogue: 0,0:08:53.97,0:08:59.59,Default,,0000,0000,0000,,brackets separately. 4 into 16\Ngoes four times, so we four Dialogue: 0,0:08:59.59,0:09:04.70,Default,,0000,0000,0000,,takeaway 8 + 4. Will that\Nbracket is now 0. Dialogue: 0,0:09:05.41,0:09:10.82,Default,,0000,0000,0000,,Minus or we have 1/4 takeaway\None add one so this bracket is Dialogue: 0,0:09:10.82,0:09:12.07,Default,,0000,0000,0000,,just a quarter. Dialogue: 0,0:09:12.67,0:09:18.50,Default,,0000,0000,0000,,So I'll answer\Nis minus 1/4. Dialogue: 0,0:09:19.62,0:09:22.78,Default,,0000,0000,0000,,Let's just have a look where B Dialogue: 0,0:09:22.78,0:09:29.75,Default,,0000,0000,0000,,was. Bees here\Nit's underneath the X Dialogue: 0,0:09:29.75,0:09:30.68,Default,,0000,0000,0000,,axis. Dialogue: 0,0:09:32.48,0:09:37.26,Default,,0000,0000,0000,,Now, if you remember, or if you\Nlook back at the video on Dialogue: 0,0:09:37.26,0:09:41.31,Default,,0000,0000,0000,,integration as summation, one of\Nthe things that we were looking Dialogue: 0,0:09:41.31,0:09:44.26,Default,,0000,0000,0000,,at when we had a piece of curve. Dialogue: 0,0:09:44.95,0:09:48.55,Default,,0000,0000,0000,,And we were looking at the area\Nunderneath that curve. Dialogue: 0,0:09:49.58,0:09:53.01,Default,,0000,0000,0000,,We took small strips. Dialogue: 0,0:09:53.55,0:09:59.67,Default,,0000,0000,0000,,And these small strips were of\Nheight Y and thickness Delta X, Dialogue: 0,0:09:59.67,0:10:05.79,Default,,0000,0000,0000,,and so the little bit of area\Nthat they represented. Delta a Dialogue: 0,0:10:05.79,0:10:12.42,Default,,0000,0000,0000,,was why times by Delta X. Now\Nthat's fine for what we've done. Dialogue: 0,0:10:12.42,0:10:15.48,Default,,0000,0000,0000,,But notice this Y is positive. Dialogue: 0,0:10:16.31,0:10:20.44,Default,,0000,0000,0000,,And so when we add up all\Nthese strips that are above Dialogue: 0,0:10:20.44,0:10:24.91,Default,,0000,0000,0000,,the X axis, we get a positive\Nanswer, IE for a the answer Dialogue: 0,0:10:24.91,0:10:26.63,Default,,0000,0000,0000,,we got was a quarter. Dialogue: 0,0:10:27.90,0:10:29.05,Default,,0000,0000,0000,,But for B. Dialogue: 0,0:10:29.64,0:10:36.48,Default,,0000,0000,0000,,These why values are actually\Nbelow the X axis and so this Dialogue: 0,0:10:36.48,0:10:43.32,Default,,0000,0000,0000,,area, in a sense is a negative\Narea because the wise are Dialogue: 0,0:10:43.32,0:10:49.59,Default,,0000,0000,0000,,negative. Hence our answer is\Nminus 1/4. But the question said Dialogue: 0,0:10:49.59,0:10:55.86,Default,,0000,0000,0000,,what is the area and clearly\Nphysically there are two blocks Dialogue: 0,0:10:55.86,0:10:59.85,Default,,0000,0000,0000,,of area and so the total area. Dialogue: 0,0:10:59.86,0:11:06.14,Default,,0000,0000,0000,,Must be 1/4. That's the area of\Nthat and a quarter because 1/4 Dialogue: 0,0:11:06.14,0:11:12.42,Default,,0000,0000,0000,,is the area of that. The minus\Nsign merely tells us it's below Dialogue: 0,0:11:12.42,0:11:18.21,Default,,0000,0000,0000,,the X axis. So the answer to our\Noriginal question, what's the Dialogue: 0,0:11:18.21,0:11:23.53,Default,,0000,0000,0000,,area that the that's contained\Nbetween the curve and the X Dialogue: 0,0:11:23.53,0:11:30.41,Default,,0000,0000,0000,,axis? Is given by\Nthe area is 1/4 + Dialogue: 0,0:11:30.41,0:11:33.25,Default,,0000,0000,0000,,1/4 which is 1/2. Dialogue: 0,0:11:36.04,0:11:39.52,Default,,0000,0000,0000,,Let's just check on something\Nwhat would happen. Dialogue: 0,0:11:40.12,0:11:46.17,Default,,0000,0000,0000,,If we took the integral of our\Nfunction between North and two. Dialogue: 0,0:11:48.07,0:11:55.87,Default,,0000,0000,0000,,Remember what our function is.\NIt's X cubed minus three Dialogue: 0,0:11:55.87,0:11:58.99,Default,,0000,0000,0000,,X squared plus 2X. Dialogue: 0,0:11:59.82,0:12:03.43,Default,,0000,0000,0000,,Integrated with respect to X. Dialogue: 0,0:12:04.77,0:12:10.20,Default,,0000,0000,0000,,So let's do that integration and\Nagain the answer shouldn't be Dialogue: 0,0:12:10.20,0:12:16.63,Default,,0000,0000,0000,,any different from what we've\Nhad before X to the 4th over 4 Dialogue: 0,0:12:16.63,0:12:24.04,Default,,0000,0000,0000,,- 3 X cubed over 3 + 2\NX squared over 2 to be evaluated Dialogue: 0,0:12:24.04,0:12:29.47,Default,,0000,0000,0000,,between North two. Again, you\Ncan do the canceling the threes Dialogue: 0,0:12:29.47,0:12:31.94,Default,,0000,0000,0000,,there and the tools there. Dialogue: 0,0:12:33.15,0:12:36.27,Default,,0000,0000,0000,,Equals that substituting the Dialogue: 0,0:12:36.27,0:12:43.26,Default,,0000,0000,0000,,upper limit. So that's X to\Nthe power four. When X is 2, two Dialogue: 0,0:12:43.26,0:12:48.59,Default,,0000,0000,0000,,multiplied by itself four times\Nis 16, and to be divided by 4. Dialogue: 0,0:12:50.30,0:12:56.03,Default,,0000,0000,0000,,Minus X cubed minus 2 multiplied\Nby itself three times, that's Dialogue: 0,0:12:56.03,0:13:03.28,Default,,0000,0000,0000,,eight. Plus X squared that's\N2 multiplied by itself twice, so Dialogue: 0,0:13:03.28,0:13:05.11,Default,,0000,0000,0000,,that's plus 4. Dialogue: 0,0:13:05.12,0:13:11.16,Default,,0000,0000,0000,,Minus. And when we put zero\Nin that zero, that's zero and Dialogue: 0,0:13:11.16,0:13:15.57,Default,,0000,0000,0000,,that one zero. So the whole of\Nthat bracket is 0. Dialogue: 0,0:13:16.39,0:13:22.68,Default,,0000,0000,0000,,Counseling here. This leaves us\Nwith a four, and so we Dialogue: 0,0:13:22.68,0:13:26.11,Default,,0000,0000,0000,,have 0 - 0 is 0. Dialogue: 0,0:13:27.61,0:13:32.29,Default,,0000,0000,0000,,Which is what we would expect,\Nbecause in a sense, what this Dialogue: 0,0:13:32.29,0:13:34.24,Default,,0000,0000,0000,,process is done is added Dialogue: 0,0:13:34.24,0:13:38.05,Default,,0000,0000,0000,,together. The two\Nareas that we got. Dialogue: 0,0:13:39.26,0:13:45.17,Default,,0000,0000,0000,,With respect to their signs, 1/4\Nplus minus 1/4 is Nord. Dialogue: 0,0:13:45.70,0:13:51.94,Default,,0000,0000,0000,,What that means is that we have\Nto be very, very careful when Dialogue: 0,0:13:51.94,0:13:53.38,Default,,0000,0000,0000,,we're calculating areas. Dialogue: 0,0:13:54.08,0:13:58.62,Default,,0000,0000,0000,,Clearly if we just integrate it\Nbetween the given limits, what Dialogue: 0,0:13:58.62,0:14:04.82,Default,,0000,0000,0000,,we might do is end up with an\Nanswer that is the value of the Dialogue: 0,0:14:04.82,0:14:09.77,Default,,0000,0000,0000,,integral and not the area and\Nwhat this example shows is is Dialogue: 0,0:14:09.77,0:14:13.49,Default,,0000,0000,0000,,that the area that is contained\Nby our function. Dialogue: 0,0:14:14.38,0:14:20.52,Default,,0000,0000,0000,,The X axis and specific values\Nof the ordinance I values of X Dialogue: 0,0:14:20.52,0:14:27.12,Default,,0000,0000,0000,,may not be the same as the value\Nof the integral, so bearing that Dialogue: 0,0:14:27.12,0:14:31.84,Default,,0000,0000,0000,,in mind, let's have a look at\Nsome more examples. Dialogue: 0,0:14:32.46,0:14:35.88,Default,,0000,0000,0000,,This time. Let's take. Dialogue: 0,0:14:36.72,0:14:43.87,Default,,0000,0000,0000,,Function of the curve Y equals X\Ntimes by X minus three and let's Dialogue: 0,0:14:43.87,0:14:48.98,Default,,0000,0000,0000,,investigate what's the area\Ncontained by the curve X access Dialogue: 0,0:14:48.98,0:14:55.63,Default,,0000,0000,0000,,and the audience X equals 0 and\NX equals 5. Well, from what Dialogue: 0,0:14:55.63,0:15:02.78,Default,,0000,0000,0000,,we've learned so far, one of the\Nthings we've got to do is a Dialogue: 0,0:15:02.78,0:15:05.85,Default,,0000,0000,0000,,sketch. So here's our X&Y axes. Dialogue: 0,0:15:07.28,0:15:14.22,Default,,0000,0000,0000,,If we set Y equals 0, then we\Ncan see that either X is 0 or X Dialogue: 0,0:15:14.22,0:15:19.93,Default,,0000,0000,0000,,can be equal to three. So that\Ngives us two points on the curve Dialogue: 0,0:15:19.93,0:15:25.23,Default,,0000,0000,0000,,there at the origin and there\Nwhere X equals 3. We also know Dialogue: 0,0:15:25.23,0:15:30.54,Default,,0000,0000,0000,,that we're going up to X equals\N5, so let's Mark that there. Dialogue: 0,0:15:31.48,0:15:35.53,Default,,0000,0000,0000,,When we multiply out this\Nbracket, we can see we've got Dialogue: 0,0:15:35.53,0:15:40.31,Default,,0000,0000,0000,,X times by. X gives us X\Nsquared and it's a positive X Dialogue: 0,0:15:40.31,0:15:44.36,Default,,0000,0000,0000,,squared, which means that\Nit's going to be a U shape Dialogue: 0,0:15:44.36,0:15:45.83,Default,,0000,0000,0000,,coming down like that. Dialogue: 0,0:15:49.09,0:15:55.07,Default,,0000,0000,0000,,Now it wants the area between\Nthe X axis, the curve and these Dialogue: 0,0:15:55.07,0:16:00.59,Default,,0000,0000,0000,,ordinance. So let's draw that in\Nthere. And again we can see Dialogue: 0,0:16:00.59,0:16:07.49,Default,,0000,0000,0000,,we've got two lots of area and\Narea a which is below the X axis Dialogue: 0,0:16:07.49,0:16:13.93,Default,,0000,0000,0000,,in an area B which is above it.\NThis means we've got to be Dialogue: 0,0:16:13.93,0:16:18.99,Default,,0000,0000,0000,,careful, workout each area\Nseparately so the area a is the Dialogue: 0,0:16:18.99,0:16:20.59,Default,,0000,0000,0000,,integral. Between North Dialogue: 0,0:16:21.13,0:16:27.02,Default,,0000,0000,0000,,And three of X times by X minus\Nthree with respect to X. Dialogue: 0,0:16:27.69,0:16:30.77,Default,,0000,0000,0000,,Let's multiply out the brackets Dialogue: 0,0:16:30.77,0:16:37.40,Default,,0000,0000,0000,,first. So X times by X gives\Nus X squared X times Y minus Dialogue: 0,0:16:37.40,0:16:39.39,Default,,0000,0000,0000,,three gives us minus 3X. Dialogue: 0,0:16:39.92,0:16:43.36,Default,,0000,0000,0000,,We're integrating that with\Nrespect to X. Dialogue: 0,0:16:44.80,0:16:46.61,Default,,0000,0000,0000,,So let's do the integration. Dialogue: 0,0:16:47.27,0:16:52.96,Default,,0000,0000,0000,,X squared integrates 2X cubed\Nover three and one to the index Dialogue: 0,0:16:52.96,0:16:59.59,Default,,0000,0000,0000,,and divide by that new index,\Nminus three X. This is X to the Dialogue: 0,0:16:59.59,0:17:05.76,Default,,0000,0000,0000,,power one, add 1 to the index\Nand divide by the new index. Dialogue: 0,0:17:06.40,0:17:09.24,Default,,0000,0000,0000,,That's to be evaluated between Dialogue: 0,0:17:09.24,0:17:11.18,Default,,0000,0000,0000,,North. And three. Dialogue: 0,0:17:12.44,0:17:15.87,Default,,0000,0000,0000,,Equals we put the upper limiting Dialogue: 0,0:17:15.87,0:17:23.08,Default,,0000,0000,0000,,first. X cubed X is equal\Nto three, so this is 3 cubed. Dialogue: 0,0:17:23.08,0:17:25.07,Default,,0000,0000,0000,,That's 27 over 3. Dialogue: 0,0:17:25.99,0:17:33.23,Default,,0000,0000,0000,,Minus three times by X squared\Nand X is 3, so we Dialogue: 0,0:17:33.23,0:17:39.26,Default,,0000,0000,0000,,have minus three times by 9\Nis 27 over 2. Dialogue: 0,0:17:40.22,0:17:46.04,Default,,0000,0000,0000,,Minus now we put the lower limit\Nin X cubed is zero when X is Dialogue: 0,0:17:46.04,0:17:51.47,Default,,0000,0000,0000,,zero and X squared is zero when\NX is zero. That is minus 0. Dialogue: 0,0:17:52.21,0:17:59.71,Default,,0000,0000,0000,,Equals so now we've got 27 over\N3 - 27 over 2. We can cancel Dialogue: 0,0:17:59.71,0:18:06.71,Default,,0000,0000,0000,,by the three here. That's nine,\Nso we have 9 minus and twos into Dialogue: 0,0:18:06.71,0:18:13.71,Default,,0000,0000,0000,,27 go 13 1/2. That will simplify\Nto that and so we have minus Dialogue: 0,0:18:13.71,0:18:21.21,Default,,0000,0000,0000,,four and a half. So that's the\Narea a. Let's now have a look at Dialogue: 0,0:18:21.21,0:18:22.71,Default,,0000,0000,0000,,the area be. Dialogue: 0,0:18:23.39,0:18:29.04,Default,,0000,0000,0000,,So that's the integral of X\Nsquared minus 3X with Dialogue: 0,0:18:29.04,0:18:34.12,Default,,0000,0000,0000,,respect to X. This time\Nbetween 3:00 and 5:00. Dialogue: 0,0:18:35.16,0:18:41.13,Default,,0000,0000,0000,,Will get exactly the same as\Nwe've got here, 'cause we're Dialogue: 0,0:18:41.13,0:18:43.30,Default,,0000,0000,0000,,integrating exactly the same Dialogue: 0,0:18:43.30,0:18:50.19,Default,,0000,0000,0000,,function. But the limits\Nare different there Dialogue: 0,0:18:50.19,0:18:53.35,Default,,0000,0000,0000,,between 3:00 and Dialogue: 0,0:18:53.35,0:18:57.15,Default,,0000,0000,0000,,5:00. So let's turn over the Dialogue: 0,0:18:57.15,0:19:04.77,Default,,0000,0000,0000,,page. Area B is equal\Nto. Remember that it's this that Dialogue: 0,0:19:04.77,0:19:06.68,Default,,0000,0000,0000,,we are evaluating. Dialogue: 0,0:19:09.34,0:19:16.97,Default,,0000,0000,0000,,Between 3:00 and 5:00, so we\Nput in our upper limit first. Dialogue: 0,0:19:17.55,0:19:24.49,Default,,0000,0000,0000,,X cubed when X is equal\Nto five, is 5 multiplied by Dialogue: 0,0:19:24.49,0:19:29.11,Default,,0000,0000,0000,,itself 3 times and that's\N125 over 3. Dialogue: 0,0:19:30.18,0:19:31.68,Default,,0000,0000,0000,,Minus. Dialogue: 0,0:19:32.75,0:19:37.43,Default,,0000,0000,0000,,X squared when X is 5, that's\Nfive times by 5 is 25. Dialogue: 0,0:19:38.00,0:19:43.14,Default,,0000,0000,0000,,25 times by three is 75\Nover 2. Dialogue: 0,0:19:44.15,0:19:45.84,Default,,0000,0000,0000,,That's our first. Dialogue: 0,0:19:46.48,0:19:50.52,Default,,0000,0000,0000,,Bracket minus now we put\Nin the three. Dialogue: 0,0:19:51.73,0:19:54.84,Default,,0000,0000,0000,,So that's 27 over 3. Dialogue: 0,0:19:55.37,0:19:59.47,Default,,0000,0000,0000,,Because if you think about it,\Nwe've actually worked this out Dialogue: 0,0:19:59.47,0:20:01.34,Default,,0000,0000,0000,,before. Minus X is 3. Dialogue: 0,0:20:02.19,0:20:07.23,Default,,0000,0000,0000,,3 squared is 9 times by that is\N27 over 2. Dialogue: 0,0:20:07.94,0:20:15.27,Default,,0000,0000,0000,,OK. Let's\Ntidy up these bits first. 125 Dialogue: 0,0:20:15.27,0:20:22.72,Default,,0000,0000,0000,,over 3 - 27 over three.\NWell, that's 98 over 3 - Dialogue: 0,0:20:22.72,0:20:30.17,Default,,0000,0000,0000,,5 - 75 over 2 minus\Nminus 27 over 2. So I've Dialogue: 0,0:20:30.17,0:20:37.62,Default,,0000,0000,0000,,got minus 75 + 27 and\Nthat is going to give me Dialogue: 0,0:20:37.62,0:20:41.37,Default,,0000,0000,0000,,48. Over. 2. Dialogue: 0,0:20:43.38,0:20:45.16,Default,,0000,0000,0000,,So if you divide. Dialogue: 0,0:20:45.72,0:20:51.76,Default,,0000,0000,0000,,By three there threes into nine\Ngoes three and threes into eight Dialogue: 0,0:20:51.76,0:20:58.30,Default,,0000,0000,0000,,goes two, and there are two\Novers that's 32 and 2/3 - 2 Dialogue: 0,0:20:58.30,0:21:05.34,Default,,0000,0000,0000,,into 48 goes 24. So the area\NB is 32 and 2/3 - 24. Dialogue: 0,0:21:05.34,0:21:10.87,Default,,0000,0000,0000,,That's eight and 2/3, so the\Nactual area that I've got. Dialogue: 0,0:21:11.54,0:21:18.21,Default,,0000,0000,0000,,Is the area of a the actual area\Nof a which is four and a half. Dialogue: 0,0:21:18.21,0:21:22.80,Default,,0000,0000,0000,,The minus sign tells us that\Nit's below the X axis. Dialogue: 0,0:21:24.50,0:21:32.05,Default,,0000,0000,0000,,4 1/2 plus the area of B which\Nis 8 and 2/3 if I want to Dialogue: 0,0:21:32.05,0:21:38.19,Default,,0000,0000,0000,,add a half and 2/3 together then\Nthey've really got to be in Dialogue: 0,0:21:38.19,0:21:43.85,Default,,0000,0000,0000,,terms of the same denominator,\Nso the four and the eight gives Dialogue: 0,0:21:43.85,0:21:50.74,Default,,0000,0000,0000,,me 12. Plus 1/2 now the\Ndenominator is going to be 6, so Dialogue: 0,0:21:50.74,0:21:56.20,Default,,0000,0000,0000,,that's three sixths plus four\Nsixths, which is what the 2/3 is Dialogue: 0,0:21:56.20,0:22:02.58,Default,,0000,0000,0000,,equal to. I've got 7 sixth here,\Nwhich is one and a six. That's Dialogue: 0,0:22:02.58,0:22:07.12,Default,,0000,0000,0000,,13 and 1/6 is the actual value\Nof my area. Dialogue: 0,0:22:07.65,0:22:13.46,Default,,0000,0000,0000,,Let's take.\NAnother example, this time. Dialogue: 0,0:22:13.46,0:22:17.79,Default,,0000,0000,0000,,Let's have a look at the\Nfunction Y is equal to. Dialogue: 0,0:22:18.39,0:22:21.52,Default,,0000,0000,0000,,X squared plus Dialogue: 0,0:22:21.52,0:22:24.07,Default,,0000,0000,0000,,X. +4. Dialogue: 0,0:22:26.76,0:22:33.25,Default,,0000,0000,0000,,Now if we set Y equals 0,\Nthen we're trying to solve Dialogue: 0,0:22:33.25,0:22:38.12,Default,,0000,0000,0000,,this equation X squared plus\NX +4 equals not. Dialogue: 0,0:22:40.00,0:22:43.39,Default,,0000,0000,0000,,To find where the curve crosses\Nthe X axis. Dialogue: 0,0:22:44.04,0:22:48.68,Default,,0000,0000,0000,,Doesn't look very promising.\NEverything here looks sort of Dialogue: 0,0:22:48.68,0:22:54.86,Default,,0000,0000,0000,,positive got plus terms in it.\NLet's check B squared minus four Dialogue: 0,0:22:54.86,0:22:59.92,Default,,0000,0000,0000,,AC. Remember the formula for\Nsolving a quadratic equation is Dialogue: 0,0:22:59.92,0:23:05.47,Default,,0000,0000,0000,,X equals minus B plus or minus\Nsquare root of be squared minus Dialogue: 0,0:23:05.47,0:23:10.60,Default,,0000,0000,0000,,four AC all over 2A, but it's\Nthis bit that's the important Dialogue: 0,0:23:10.60,0:23:15.30,Default,,0000,0000,0000,,bit. the B squared minus four\NAC, 'cause If that's negative. Dialogue: 0,0:23:16.51,0:23:20.72,Default,,0000,0000,0000,,IE, when we take a square root,\Nwe can't have the square root of Dialogue: 0,0:23:20.72,0:23:24.04,Default,,0000,0000,0000,,a negative number. What that\Ntells us is that this equation Dialogue: 0,0:23:24.04,0:23:31.22,Default,,0000,0000,0000,,has no roots. So B squared well,\Nthere's 1X, so B is equal to 1, Dialogue: 0,0:23:31.22,0:23:37.41,Default,,0000,0000,0000,,so that's one squared minus four\Ntimes a while a is one 'cause Dialogue: 0,0:23:37.41,0:23:40.74,Default,,0000,0000,0000,,there's One X squared and C is Dialogue: 0,0:23:40.74,0:23:46.85,Default,,0000,0000,0000,,4. Well, that's 1 - 16 is minus\N15. It's less than zero. There Dialogue: 0,0:23:46.85,0:23:52.44,Default,,0000,0000,0000,,are no roots for this equation,\Nso it doesn't meet the X axis. Dialogue: 0,0:23:53.57,0:23:57.99,Default,,0000,0000,0000,,Let's have a look at a picture\Nfor that. There's The X Axis. Dialogue: 0,0:23:58.52,0:24:03.100,Default,,0000,0000,0000,,Why access I'd like to be able\Nto fix a point on the curve and Dialogue: 0,0:24:03.100,0:24:09.10,Default,,0000,0000,0000,,I can buy taking X is zero.\NThose two terms would be 0, so Dialogue: 0,0:24:09.10,0:24:11.30,Default,,0000,0000,0000,,that gives me Y equals 4. Dialogue: 0,0:24:12.04,0:24:18.50,Default,,0000,0000,0000,,So that's a point on the curve,\Nbut it's always going to be Dialogue: 0,0:24:18.50,0:24:23.97,Default,,0000,0000,0000,,above the X axis, and when\Nbecause X squared is positive. Dialogue: 0,0:24:25.62,0:24:29.32,Default,,0000,0000,0000,,With The X is positive or\Nnegative, X squared is always Dialogue: 0,0:24:29.32,0:24:33.68,Default,,0000,0000,0000,,going to be positive, so it's\Nalways going to be that way up Dialogue: 0,0:24:33.68,0:24:35.70,Default,,0000,0000,0000,,and it's going to look something Dialogue: 0,0:24:35.70,0:24:42.80,Default,,0000,0000,0000,,like that. So we want the\Narea between this curve, the X Dialogue: 0,0:24:42.80,0:24:49.04,Default,,0000,0000,0000,,Axis and the ordinates X equals\N1 and X equals 3. Dialogue: 0,0:24:49.83,0:24:55.78,Default,,0000,0000,0000,,X equals 1, let's say is there,\NX equals 3, let's say is there. Dialogue: 0,0:24:57.83,0:25:03.09,Default,,0000,0000,0000,,That's the area that we're\Nafter. No real problems with Dialogue: 0,0:25:03.09,0:25:08.88,Default,,0000,0000,0000,,that area, so let's do the\Ncalculation. The area is equal Dialogue: 0,0:25:08.88,0:25:15.71,Default,,0000,0000,0000,,to the integral between one and\Nthree of X squared plus X +4 Dialogue: 0,0:25:15.71,0:25:17.82,Default,,0000,0000,0000,,with respect to X. Dialogue: 0,0:25:18.74,0:25:23.74,Default,,0000,0000,0000,,Integrating each of these at one\Nto the index. Dialogue: 0,0:25:24.25,0:25:26.77,Default,,0000,0000,0000,,And divide by the new index. Dialogue: 0,0:25:27.57,0:25:34.60,Default,,0000,0000,0000,,Same again there and then four\Nis 4X. When we integrate it Dialogue: 0,0:25:34.60,0:25:40.46,Default,,0000,0000,0000,,and this is to be evaluated\Nbetween one and three. Dialogue: 0,0:25:40.99,0:25:45.47,Default,,0000,0000,0000,,So let's just turn this over.\NRemember, the area that we're Dialogue: 0,0:25:45.47,0:25:52.76,Default,,0000,0000,0000,,wanting. Is equal to. This\Nis what we have to work Dialogue: 0,0:25:52.76,0:25:59.12,Default,,0000,0000,0000,,out X cubed over 3 plus\NX squared over 2. Dialogue: 0,0:25:59.13,0:26:05.64,Default,,0000,0000,0000,,Plus 4X that's to be evaluated\Nbetween one and three. Dialogue: 0,0:26:06.76,0:26:13.69,Default,,0000,0000,0000,,So we put the upper limiting\Nfirst, so when X is equal to 3, Dialogue: 0,0:26:13.69,0:26:17.16,Default,,0000,0000,0000,,three cubed is equal to 27 over Dialogue: 0,0:26:17.16,0:26:24.04,Default,,0000,0000,0000,,3. Plus X squared when X\Nis equal to three, that's nine Dialogue: 0,0:26:24.04,0:26:30.71,Default,,0000,0000,0000,,over 2 + 4 X when X\Nis equal to three, that's 12. Dialogue: 0,0:26:30.71,0:26:33.79,Default,,0000,0000,0000,,Now we put the lower limiting. Dialogue: 0,0:26:34.33,0:26:41.84,Default,,0000,0000,0000,,So when X is One X cubed is\None 1 / 3 + 1, X is one Dialogue: 0,0:26:41.84,0:26:47.59,Default,,0000,0000,0000,,that's X squared is one and\Ndivided by two, and when X is Dialogue: 0,0:26:47.59,0:26:49.36,Default,,0000,0000,0000,,one that's plus 4. Dialogue: 0,0:26:51.27,0:26:56.85,Default,,0000,0000,0000,,Now, some of these we can\Nworkout quite easily, but let's Dialogue: 0,0:26:56.85,0:27:03.94,Default,,0000,0000,0000,,do the thirds bit first with 27\Nover 3 takeaway. A third is 26 Dialogue: 0,0:27:03.94,0:27:10.54,Default,,0000,0000,0000,,over three we have 9 over 2\Ntakeaway one over 2. So that's Dialogue: 0,0:27:10.54,0:27:16.62,Default,,0000,0000,0000,,plus eight over 2, and we've 12\Ntakeaway 4, so that's +8. Dialogue: 0,0:27:16.63,0:27:18.38,Default,,0000,0000,0000,,Will cancel down to their. Dialogue: 0,0:27:19.01,0:27:26.57,Default,,0000,0000,0000,,If I do freeze into 26, eight\Nthrees are 24 and two over slots Dialogue: 0,0:27:26.57,0:27:33.59,Default,,0000,0000,0000,,2/3 + 4 and 80s twelve so\Naltogether that's an area of 20 Dialogue: 0,0:27:33.59,0:27:36.48,Default,,0000,0000,0000,,and two. Thirds Dialogue: 0,0:27:38.47,0:27:45.76,Default,,0000,0000,0000,,Now.\NWe can workout the area between Dialogue: 0,0:27:45.76,0:27:51.93,Default,,0000,0000,0000,,a curve, the X axis and some\Ngiven values of X the ordinance. Dialogue: 0,0:27:52.44,0:27:57.68,Default,,0000,0000,0000,,We can extend this technique\Nthough to working out the area Dialogue: 0,0:27:57.68,0:28:02.44,Default,,0000,0000,0000,,contained between two curves.\NLet's have a look at that Dialogue: 0,0:28:02.44,0:28:09.79,Default,,0000,0000,0000,,supposing. You've got the curve\NY equals X times 3 minus X, Dialogue: 0,0:28:09.79,0:28:14.63,Default,,0000,0000,0000,,and we've got the straight line\NY equals X. Dialogue: 0,0:28:15.28,0:28:20.07,Default,,0000,0000,0000,,We say what's the area that's\Ncontained between these two Dialogue: 0,0:28:20.07,0:28:23.99,Default,,0000,0000,0000,,curves? Well, let's have a\Nlook at a picture just to Dialogue: 0,0:28:23.99,0:28:25.24,Default,,0000,0000,0000,,see what that looks like. Dialogue: 0,0:28:28.44,0:28:35.48,Default,,0000,0000,0000,,If we set Y equals 0 again\Nand that will give us the values Dialogue: 0,0:28:35.48,0:28:42.02,Default,,0000,0000,0000,,of X where this curve crosses\Nthe X axis clearly does so there Dialogue: 0,0:28:42.02,0:28:48.06,Default,,0000,0000,0000,,when X is zero and also when\NX is equal to 3. Dialogue: 0,0:28:49.68,0:28:54.85,Default,,0000,0000,0000,,We've also got X times by minus.\NX gives us minus X squared, so Dialogue: 0,0:28:54.85,0:29:00.01,Default,,0000,0000,0000,,for a quadratic it's going to be\Nan upside down U. It's going to Dialogue: 0,0:29:00.01,0:29:04.81,Default,,0000,0000,0000,,do that, just missed that point.\NLet's make it bigger so it goes Dialogue: 0,0:29:04.81,0:29:09.24,Default,,0000,0000,0000,,through it. The line Y equals\NXY. That's a straight line going Dialogue: 0,0:29:09.24,0:29:14.80,Default,,0000,0000,0000,,through there. And so the line\Ncrosses the curve at these two Dialogue: 0,0:29:14.80,0:29:20.15,Default,,0000,0000,0000,,points. Let me call that one P\Nand this one is oh the origin. Dialogue: 0,0:29:21.25,0:29:25.63,Default,,0000,0000,0000,,And the error that I'm\Ninterested in is this one. In Dialogue: 0,0:29:25.63,0:29:28.02,Default,,0000,0000,0000,,here the area core between the Dialogue: 0,0:29:28.02,0:29:34.37,Default,,0000,0000,0000,,two curves. Well before I can\Nfind that area, I really need to Dialogue: 0,0:29:34.37,0:29:39.64,Default,,0000,0000,0000,,know what's this point P. Where\Ndo these two curves cross? Well, Dialogue: 0,0:29:39.64,0:29:44.90,Default,,0000,0000,0000,,they will cross when their Y\Nvalues are equal, so they will Dialogue: 0,0:29:44.90,0:29:51.05,Default,,0000,0000,0000,,intersect or meet when X times\Nby 3 minus X is equal to X. Dialogue: 0,0:29:52.27,0:29:57.66,Default,,0000,0000,0000,,Multiply out the bracket\N3X minus X squared is Dialogue: 0,0:29:57.66,0:29:59.46,Default,,0000,0000,0000,,equal to X. Dialogue: 0,0:30:00.53,0:30:07.22,Default,,0000,0000,0000,,Take X away from both sides. 2X\Nminus X squared is then zero. Dialogue: 0,0:30:07.22,0:30:12.38,Default,,0000,0000,0000,,This is a quadratic, and it\Nfactorizes. There's a common Dialogue: 0,0:30:12.38,0:30:18.56,Default,,0000,0000,0000,,factor of X in each term, so\Nwe can take that out. Dialogue: 0,0:30:18.56,0:30:23.44,Default,,0000,0000,0000,,And we're left with\Ntwo minus X equals 0. Dialogue: 0,0:30:25.40,0:30:32.10,Default,,0000,0000,0000,,Either the X can be 0 or the two\Nminus X com 0, so therefore X is Dialogue: 0,0:30:32.10,0:30:38.40,Default,,0000,0000,0000,,0 or X equals 2, so this is\Nwhere X is equal to 0. Here at Dialogue: 0,0:30:38.40,0:30:44.31,Default,,0000,0000,0000,,the origin, when X is zero, Y is\N00. In there we have not times Dialogue: 0,0:30:44.31,0:30:49.43,Default,,0000,0000,0000,,by three is nothing. If we put\Nzero in there, we get nothing. Dialogue: 0,0:30:49.43,0:30:54.56,Default,,0000,0000,0000,,Or why is equal to? Well, This\Nis why equals X, so presumably Dialogue: 0,0:30:54.56,0:30:56.92,Default,,0000,0000,0000,,why must be equal to two? Dialogue: 0,0:30:57.11,0:31:02.18,Default,,0000,0000,0000,,And let's just check it over\Nhere. If we put X equals 2 with Dialogue: 0,0:31:02.18,0:31:08.33,Default,,0000,0000,0000,,2 * 3 - 2, three minus two is\None times by two is 2. So it's Dialogue: 0,0:31:08.33,0:31:10.50,Default,,0000,0000,0000,,this here. That's the point P. Dialogue: 0,0:31:12.29,0:31:17.58,Default,,0000,0000,0000,,So let's just put that into\Ntheir. Now this is the area that Dialogue: 0,0:31:17.58,0:31:24.16,Default,,0000,0000,0000,,we want. So we can look at it\Nas finding the area under the Dialogue: 0,0:31:24.16,0:31:28.94,Default,,0000,0000,0000,,curve between North and two and\Nfinding the area under the Dialogue: 0,0:31:28.94,0:31:33.58,Default,,0000,0000,0000,,straight line. And then\Nsubtracting them in order to end Dialogue: 0,0:31:33.58,0:31:34.75,Default,,0000,0000,0000,,up with that. Dialogue: 0,0:31:35.51,0:31:39.37,Default,,0000,0000,0000,,So let's do that. Let's\Nfirst of all, find the Dialogue: 0,0:31:39.37,0:31:40.91,Default,,0000,0000,0000,,area under the curve. Dialogue: 0,0:31:43.72,0:31:51.11,Default,,0000,0000,0000,,Equals now this will\Nbe the integral between Dialogue: 0,0:31:51.11,0:31:58.50,Default,,0000,0000,0000,,North and two of\Nour curve X times Dialogue: 0,0:31:58.50,0:32:02.20,Default,,0000,0000,0000,,by 3 minus X. Dialogue: 0,0:32:02.48,0:32:07.25,Default,,0000,0000,0000,,With respect to\NX. Dialogue: 0,0:32:08.27,0:32:14.67,Default,,0000,0000,0000,,So let's first of all multiply\Nout the bracket X times by three Dialogue: 0,0:32:14.67,0:32:20.57,Default,,0000,0000,0000,,is 3 XX times Y minus X is\Nminus X squared X. Dialogue: 0,0:32:21.86,0:32:28.80,Default,,0000,0000,0000,,Do the integration integral of X\Nis X squared over 2? Dialogue: 0,0:32:30.36,0:32:33.80,Default,,0000,0000,0000,,And we need to multiply it by\Nthe three that we've got there. Dialogue: 0,0:32:34.37,0:32:41.13,Default,,0000,0000,0000,,And then the integral of X\Nsquared is going to be X cubed Dialogue: 0,0:32:41.13,0:32:46.33,Default,,0000,0000,0000,,over three that to be evaluated\Nbetween North and two. Dialogue: 0,0:32:46.98,0:32:48.74,Default,,0000,0000,0000,,So I put the two in first. Dialogue: 0,0:32:49.31,0:32:52.08,Default,,0000,0000,0000,,So we have 3. Dialogue: 0,0:32:52.79,0:32:56.64,Default,,0000,0000,0000,,Times 2 to 4. Dialogue: 0,0:32:56.64,0:32:58.15,Default,,0000,0000,0000,,Divided by two. Dialogue: 0,0:32:58.76,0:33:00.95,Default,,0000,0000,0000,,Minus X cubed over 3. Dialogue: 0,0:33:01.69,0:33:06.22,Default,,0000,0000,0000,,X is 2, so that's eight\Nover 3. Dialogue: 0,0:33:07.78,0:33:13.59,Default,,0000,0000,0000,,Minus and when we put the zero\Nin, X squared is zero and X Dialogue: 0,0:33:13.59,0:33:17.32,Default,,0000,0000,0000,,cubed is 0, so that second\Nbracket is 0. Dialogue: 0,0:33:18.34,0:33:22.16,Default,,0000,0000,0000,,Equals what we can cancel it Dialogue: 0,0:33:22.16,0:33:28.65,Default,,0000,0000,0000,,to there. 326 minus threes into\Neight goals two and there's two Dialogue: 0,0:33:28.65,0:33:35.96,Default,,0000,0000,0000,,over, so that's 2/3, so the area\Nthat we've got is 6 - 2 Dialogue: 0,0:33:35.96,0:33:39.09,Default,,0000,0000,0000,,and 2/3, which is 3 1/3. Dialogue: 0,0:33:40.14,0:33:47.30,Default,,0000,0000,0000,,So that's the area under the\Ncurve. What now we need is the Dialogue: 0,0:33:47.30,0:33:50.06,Default,,0000,0000,0000,,area under Y equals X. Dialogue: 0,0:33:51.13,0:33:57.11,Default,,0000,0000,0000,,So this will be equal to the\Nintegral between Norton two of X Dialogue: 0,0:33:57.11,0:34:00.37,Default,,0000,0000,0000,,DX. Equals. Dialogue: 0,0:34:01.68,0:34:05.19,Default,,0000,0000,0000,,Integral of X is just X squared Dialogue: 0,0:34:05.19,0:34:12.33,Default,,0000,0000,0000,,over 2. Between North and\Ntwo, substituting the two 2 Dialogue: 0,0:34:12.33,0:34:15.80,Default,,0000,0000,0000,,two to four over 2. Dialogue: 0,0:34:15.96,0:34:23.08,Default,,0000,0000,0000,,Minus zero in and the X squared\Nover 2 gives gives us 0. Dialogue: 0,0:34:23.09,0:34:28.26,Default,,0000,0000,0000,,And so we've just got two\Nchoosing toward those two, Dialogue: 0,0:34:28.26,0:34:34.98,Default,,0000,0000,0000,,nothing there. So here we've got\Nthe area under the curve 3 1/3 Dialogue: 0,0:34:34.98,0:34:38.60,Default,,0000,0000,0000,,here. We've got the area under Y Dialogue: 0,0:34:38.60,0:34:43.65,Default,,0000,0000,0000,,equals X2. And so the area\Nbetween them. Dialogue: 0,0:34:46.31,0:34:50.04,Default,,0000,0000,0000,,Let's look at that Dialogue: 0,0:34:50.04,0:34:54.97,Default,,0000,0000,0000,,positions. We've calculated this\Narea here under the curve. Dialogue: 0,0:34:54.97,0:34:59.94,Default,,0000,0000,0000,,That's the one that is free and\Nthe 3rd, and we've calculated Dialogue: 0,0:34:59.94,0:35:02.42,Default,,0000,0000,0000,,the area of this bit of. Dialogue: 0,0:35:03.04,0:35:07.48,Default,,0000,0000,0000,,Triangle that's here\Nunderneath Y equals X, and the Dialogue: 0,0:35:07.48,0:35:13.39,Default,,0000,0000,0000,,answer that is 2 and So what\Nwe're looking for is the Dialogue: 0,0:35:13.39,0:35:17.83,Default,,0000,0000,0000,,difference between the big\Narea and the triangle. So Dialogue: 0,0:35:17.83,0:35:22.27,Default,,0000,0000,0000,,that's 3 1/3 - 2 gives us 1\N1/3. Dialogue: 0,0:35:24.52,0:35:28.43,Default,,0000,0000,0000,,So we've seen how we can extend\Nthe technique of finding the Dialogue: 0,0:35:28.43,0:35:29.74,Default,,0000,0000,0000,,area under a curve. Dialogue: 0,0:35:30.57,0:35:36.29,Default,,0000,0000,0000,,Between the X axis on between\Ntwo given ordinance to finding Dialogue: 0,0:35:36.29,0:35:40.97,Default,,0000,0000,0000,,the area between two given\Ncurves. Let's take another Dialogue: 0,0:35:40.97,0:35:47.21,Default,,0000,0000,0000,,example. This time, let's use\Nthe function Y equals sign X. We Dialogue: 0,0:35:47.21,0:35:53.97,Default,,0000,0000,0000,,better define a range of values\Nof X, so I'm going to take Dialogue: 0,0:35:53.97,0:35:57.61,Default,,0000,0000,0000,,X to be between North and pie. Dialogue: 0,0:35:58.26,0:36:05.46,Default,,0000,0000,0000,,Let's say I want the area that's\Ncut off from that by the line Dialogue: 0,0:36:05.46,0:36:11.62,Default,,0000,0000,0000,,Y equals one over Route 2. So\Na little sketch as always. Dialogue: 0,0:36:12.63,0:36:16.04,Default,,0000,0000,0000,,We're finding these areas.\NIt's important to know Dialogue: 0,0:36:16.04,0:36:17.32,Default,,0000,0000,0000,,where they are. Dialogue: 0,0:36:18.47,0:36:22.74,Default,,0000,0000,0000,,So why equals sign X\Nbetween North and pie? Dialogue: 0,0:36:22.74,0:36:25.11,Default,,0000,0000,0000,,So it's going to look. Dialogue: 0,0:36:26.82,0:36:34.02,Default,,0000,0000,0000,,Like that? Naspi Nazira Why is\Nequal to one over Route 2? Well, Dialogue: 0,0:36:34.02,0:36:35.92,Default,,0000,0000,0000,,that's a straight line. Dialogue: 0,0:36:36.90,0:36:39.04,Default,,0000,0000,0000,,Somewhere, Let's say across. Dialogue: 0,0:36:39.79,0:36:42.45,Default,,0000,0000,0000,,There, so this is the area that Dialogue: 0,0:36:42.45,0:36:47.30,Default,,0000,0000,0000,,we're after. Area between the\Nline and that curve. What we Dialogue: 0,0:36:47.30,0:36:50.45,Default,,0000,0000,0000,,need to know. What are these\Nvalues of X? Dialogue: 0,0:36:51.54,0:36:58.18,Default,,0000,0000,0000,,Well, why is equal to one over\NRoute 2 and Y is equal to sign Dialogue: 0,0:36:58.18,0:37:04.39,Default,,0000,0000,0000,,X, so one over Route 2 is equal\Nto sign X where they meet? Dialogue: 0,0:37:05.18,0:37:08.82,Default,,0000,0000,0000,,So the values of X will be the\Nsolutions to this equation. Dialogue: 0,0:37:09.39,0:37:14.71,Default,,0000,0000,0000,,Well, one of the values we do\Nknow is that when sign X equals Dialogue: 0,0:37:14.71,0:37:18.89,Default,,0000,0000,0000,,one over route 2X must be equal\Nto π over 4. Dialogue: 0,0:37:19.51,0:37:23.49,Default,,0000,0000,0000,,I working in radians, so I've\Ngot to give the answer in Dialogue: 0,0:37:23.49,0:37:26.48,Default,,0000,0000,0000,,radians, but the equivalent\Nanswer in degrees is 45. Dialogue: 0,0:37:27.85,0:37:34.29,Default,,0000,0000,0000,,So that's that one there and the\None there is 3 Pi over 4. Dialogue: 0,0:37:35.41,0:37:41.73,Default,,0000,0000,0000,,So now I can workout what it is\NI've got to do I can find the Dialogue: 0,0:37:41.73,0:37:46.08,Default,,0000,0000,0000,,area under the sign curve\Nbetween these two values. I can Dialogue: 0,0:37:46.08,0:37:50.82,Default,,0000,0000,0000,,find the area of this rectangle\Nand take it away. That will Dialogue: 0,0:37:50.82,0:37:55.56,Default,,0000,0000,0000,,leave me that area there. Just\Nlet me put these answers in. Dialogue: 0,0:37:56.18,0:38:02.24,Default,,0000,0000,0000,,I found their and there, so it's\Nfirst of all find the area under Dialogue: 0,0:38:02.24,0:38:05.09,Default,,0000,0000,0000,,this curve. So the area. Dialogue: 0,0:38:06.77,0:38:13.84,Default,,0000,0000,0000,,Under The\Ncurve is equal to Dialogue: 0,0:38:13.84,0:38:20.35,Default,,0000,0000,0000,,the integral between pie\Nby four and three Dialogue: 0,0:38:20.35,0:38:27.28,Default,,0000,0000,0000,,Pi 4. Of\Nsign XDX equal so Dialogue: 0,0:38:27.28,0:38:30.78,Default,,0000,0000,0000,,I have to integrate Dialogue: 0,0:38:30.78,0:38:38.04,Default,,0000,0000,0000,,sign X. Well, the\Nintegral of sine X is minus Cos Dialogue: 0,0:38:38.04,0:38:45.91,Default,,0000,0000,0000,,X. To evaluate that\Nbetween pie by four and Dialogue: 0,0:38:45.91,0:38:48.31,Default,,0000,0000,0000,,three Pi 4. Dialogue: 0,0:38:49.02,0:38:56.37,Default,,0000,0000,0000,,So we put in the upper limit\Nfirst minus the cause of three π Dialogue: 0,0:38:56.37,0:39:00.36,Default,,0000,0000,0000,,by 4. That's the first bracket. Dialogue: 0,0:39:01.11,0:39:07.48,Default,,0000,0000,0000,,Minus minus the cause of\NΠ by 4. Dialogue: 0,0:39:08.25,0:39:15.50,Default,,0000,0000,0000,,Now cause of three π by 4 is\Nminus one over Route 2 and that Dialogue: 0,0:39:15.50,0:39:22.26,Default,,0000,0000,0000,,minus sign gives me plus one\Nover Route 2. The cause of Π by Dialogue: 0,0:39:22.26,0:39:28.54,Default,,0000,0000,0000,,4 is one over Route 2, minus\Nminus gives me plus one over Dialogue: 0,0:39:28.54,0:39:35.30,Default,,0000,0000,0000,,Route 2, so that is 2 over Route\N2, which is just Route 2. Dialogue: 0,0:39:36.72,0:39:40.29,Default,,0000,0000,0000,,So I now know this area here. Dialogue: 0,0:39:40.84,0:39:44.37,Default,,0000,0000,0000,,I need to do is calculate the\Narea of this rectangle. Dialogue: 0,0:39:45.56,0:39:50.19,Default,,0000,0000,0000,,Then I can take it away from\Nthat. Well, it is a rectangle. Dialogue: 0,0:39:50.19,0:39:52.68,Default,,0000,0000,0000,,Its height is one over Route 2. Dialogue: 0,0:39:53.26,0:39:55.77,Default,,0000,0000,0000,,So area. Dialogue: 0,0:39:57.02,0:40:02.97,Default,,0000,0000,0000,,Equals one over Route 2. That's\Nthe height of that rectangle. Dialogue: 0,0:40:02.97,0:40:05.68,Default,,0000,0000,0000,,Times by this base it's. Dialogue: 0,0:40:06.24,0:40:12.93,Default,,0000,0000,0000,,Width, well, it runs from Π by\N4, three π by 4, so the Dialogue: 0,0:40:12.93,0:40:19.62,Default,,0000,0000,0000,,difference is 2π by 4. In other\Nwords, pie by two, so the area Dialogue: 0,0:40:19.62,0:40:24.40,Default,,0000,0000,0000,,of the rectangle is π over 2\Ntimes Route 2. Dialogue: 0,0:40:25.39,0:40:31.43,Default,,0000,0000,0000,,So the total area that I want\Nis the difference between these Dialogue: 0,0:40:31.43,0:40:34.95,Default,,0000,0000,0000,,two π over 2 Route 2 and Dialogue: 0,0:40:34.95,0:40:38.29,Default,,0000,0000,0000,,Route 2. In the area. Dialogue: 0,0:40:40.14,0:40:48.14,Default,,0000,0000,0000,,Equals Route 2 -\Nπ over 2 Route Dialogue: 0,0:40:48.14,0:40:54.24,Default,,0000,0000,0000,,2. Put all that over a common\Ndenominator of two Route 2, Dialogue: 0,0:40:54.24,0:40:56.10,Default,,0000,0000,0000,,which makes this 4. Dialogue: 0,0:40:56.77,0:41:04.54,Default,,0000,0000,0000,,Minus pie.\NLeave that as that. That's the Dialogue: 0,0:41:04.54,0:41:06.45,Default,,0000,0000,0000,,area that's contained. Dialogue: 0,0:41:06.97,0:41:12.15,Default,,0000,0000,0000,,In there, between the straight\Nline and the curve. Dialogue: 0,0:41:13.03,0:41:20.12,Default,,0000,0000,0000,,I want now just to\Ngo back to the example Dialogue: 0,0:41:20.12,0:41:27.21,Default,,0000,0000,0000,,where we were looking at\Nthe area that was between Dialogue: 0,0:41:27.21,0:41:29.34,Default,,0000,0000,0000,,these two curves. Dialogue: 0,0:41:29.35,0:41:32.82,Default,,0000,0000,0000,,If you remember we Dialogue: 0,0:41:32.82,0:41:36.56,Default,,0000,0000,0000,,had. This sketch. Dialogue: 0,0:41:37.28,0:41:39.49,Default,,0000,0000,0000,,The curve. Dialogue: 0,0:41:40.67,0:41:46.16,Default,,0000,0000,0000,,So that\Nwas the Dialogue: 0,0:41:46.16,0:41:48.90,Default,,0000,0000,0000,,picture that Dialogue: 0,0:41:48.90,0:41:54.85,Default,,0000,0000,0000,,we had.\NAnd we calculated the area Dialogue: 0,0:41:54.85,0:41:57.08,Default,,0000,0000,0000,,between the two curves here. Dialogue: 0,0:41:58.04,0:42:00.47,Default,,0000,0000,0000,,By finding the area underneath. Dialogue: 0,0:42:01.08,0:42:03.33,Default,,0000,0000,0000,,X times by 3 minus X. Dialogue: 0,0:42:03.89,0:42:07.55,Default,,0000,0000,0000,,Then the area underneath\Nthe line Y equals X and Dialogue: 0,0:42:07.55,0:42:08.65,Default,,0000,0000,0000,,taking them away. Dialogue: 0,0:42:11.02,0:42:16.13,Default,,0000,0000,0000,,Another way of looking at this\Nmight be to go back to 1st Dialogue: 0,0:42:16.13,0:42:20.45,Default,,0000,0000,0000,,principles if you remember when\Nwe did that, we took little Dialogue: 0,0:42:20.45,0:42:23.99,Default,,0000,0000,0000,,rectangular strips which were\Nroughly of height Y and Dialogue: 0,0:42:23.99,0:42:27.92,Default,,0000,0000,0000,,thickness Delta X. Well,\Ncouldn't we do the same here, Dialogue: 0,0:42:27.92,0:42:31.46,Default,,0000,0000,0000,,have little rectangular strips\Nfor that. Not quite rectangles, Dialogue: 0,0:42:31.46,0:42:36.56,Default,,0000,0000,0000,,but they are thin strips and\Ntheir why bit, so to speak would Dialogue: 0,0:42:36.56,0:42:40.50,Default,,0000,0000,0000,,be the difference between these\Ntwo values of Y here. Dialogue: 0,0:42:41.39,0:42:47.50,Default,,0000,0000,0000,,So if we were to call this, why\None, let's say, and this one Y Dialogue: 0,0:42:47.50,0:42:52.38,Default,,0000,0000,0000,,2, then might not the area that\Nwe're looking for the area Dialogue: 0,0:42:52.38,0:42:54.82,Default,,0000,0000,0000,,between these two curves Y1 and Dialogue: 0,0:42:54.82,0:43:00.15,Default,,0000,0000,0000,,Y2? Between these ordinates Here\NLet's call them A&B. Dialogue: 0,0:43:00.78,0:43:03.29,Default,,0000,0000,0000,,Might not we be able to work it Dialogue: 0,0:43:03.29,0:43:05.89,Default,,0000,0000,0000,,out? By doing that. Dialogue: 0,0:43:06.56,0:43:11.72,Default,,0000,0000,0000,,In other words, take the upper\Ncurve and call it why one type Dialogue: 0,0:43:11.72,0:43:14.90,Default,,0000,0000,0000,,the lower curve and call it Y 2. Dialogue: 0,0:43:15.58,0:43:19.38,Default,,0000,0000,0000,,Subtract them and then integrate\Nbetween the required limits. Dialogue: 0,0:43:19.38,0:43:24.86,Default,,0000,0000,0000,,Well, let's see if this will\Ngive us the same answer as we Dialogue: 0,0:43:24.86,0:43:26.97,Default,,0000,0000,0000,,had in the previous case. Dialogue: 0,0:43:27.76,0:43:32.13,Default,,0000,0000,0000,,So our limits\Nhere are not 2. Dialogue: 0,0:43:33.25,0:43:35.33,Default,,0000,0000,0000,,And why one is this one? Dialogue: 0,0:43:36.04,0:43:41.23,Default,,0000,0000,0000,,Let's multiply it out first, X\Ntimes by three, is 3X and X Dialogue: 0,0:43:41.23,0:43:44.02,Default,,0000,0000,0000,,times Y minus X is minus X Dialogue: 0,0:43:44.02,0:43:50.91,Default,,0000,0000,0000,,squared. Take away why two? So\Nwe're taking away X and Dialogue: 0,0:43:50.91,0:43:57.33,Default,,0000,0000,0000,,integrating with respect to X.\NSo if 2X minus X, sorry, Dialogue: 0,0:43:57.33,0:44:04.34,Default,,0000,0000,0000,,3X minus X. That gives us\N2X minus X squared to be Dialogue: 0,0:44:04.34,0:44:10.18,Default,,0000,0000,0000,,evaluated between Norton two\Nwith respect to X. So let's Dialogue: 0,0:44:10.18,0:44:12.52,Default,,0000,0000,0000,,carry out this integration. Dialogue: 0,0:44:13.79,0:44:21.32,Default,,0000,0000,0000,,The integral of X is X squared\Nover 2, so that's two times X Dialogue: 0,0:44:21.32,0:44:27.78,Default,,0000,0000,0000,,squared over 2. The integral\Nof X squared is X cubed over Dialogue: 0,0:44:27.78,0:44:34.23,Default,,0000,0000,0000,,three between North and two. I\Ncan cancel it to their and Dialogue: 0,0:44:34.23,0:44:35.31,Default,,0000,0000,0000,,their substituting. Dialogue: 0,0:44:36.53,0:44:40.07,Default,,0000,0000,0000,,2 Twos are 4. Dialogue: 0,0:44:40.82,0:44:41.85,Default,,0000,0000,0000,,Minus. Dialogue: 0,0:44:43.03,0:44:50.71,Default,,0000,0000,0000,,X cube when X is 2, that's eight\Nover 3 minus and when I put zero Dialogue: 0,0:44:50.71,0:44:54.07,Default,,0000,0000,0000,,in that second bracket is just 0 Dialogue: 0,0:44:54.07,0:44:56.84,Default,,0000,0000,0000,,equals. 4 Dialogue: 0,0:44:56.84,0:45:03.87,Default,,0000,0000,0000,,minus. 2 and 2/3 which\Njust gives me one and third Dialogue: 0,0:45:03.87,0:45:10.67,Default,,0000,0000,0000,,units of area, which is exactly\Nwhat we had last time. So this. Dialogue: 0,0:45:12.00,0:45:15.30,Default,,0000,0000,0000,,Gives us a convenient formula. Dialogue: 0,0:45:15.95,0:45:21.27,Default,,0000,0000,0000,,Being able to workout the area\Nthat is caught between two Dialogue: 0,0:45:21.27,0:45:27.57,Default,,0000,0000,0000,,curves, we call the upper curve.\NWhy won the lower curve Y two Dialogue: 0,0:45:27.57,0:45:32.41,Default,,0000,0000,0000,,and we subtract them and then\Nintegrate between the ordinance Dialogue: 0,0:45:32.41,0:45:34.83,Default,,0000,0000,0000,,of the points of intersection?