0:00:02.290,0:00:04.270 2 videos in this series. 0:00:05.670,0:00:10.521 Areas as summation and[br]integration as the reverse of 0:00:10.521,0:00:15.372 differentiation have shown that[br]the area under a curve. 0:00:16.320,0:00:20.940 Above the X axis and between two[br]ordinates. That's two values of 0:00:20.940,0:00:25.560 X can be calculated by using[br]integration. And what we want to 0:00:25.560,0:00:30.950 do is develop that idea and look[br]at some more examples of that in 0:00:30.950,0:00:32.105 this particular video. 0:00:33.000,0:00:39.392 So. Let's begin by exploring[br]the question. We've got. The 0:00:39.392,0:00:46.208 function Y equals X times X[br]minus one times X minus 2. 0:00:46.960,0:00:51.230 What's the area contained[br]between this curve? 0:00:51.780,0:00:56.340 On the X axis, well, the very[br]nature of the question suggests 0:00:56.340,0:01:01.280 hang on a minute. There might be[br]something a bit odd here. Let's 0:01:01.280,0:01:05.080 draw a picture. So let's start[br]by sketching this curve. 0:01:06.800,0:01:12.200 Well, if we put Y equals 0 then[br]we can see that each one of 0:01:12.200,0:01:16.880 these brackets could be 0 and[br]that will give us a series of 0:01:16.880,0:01:22.280 points on the curve. So why is 0[br]then? This could be 0 EX could 0:01:22.280,0:01:28.012 be 0. Or X minus one could be 0.[br]In other words, X could be equal 0:01:28.012,0:01:34.494 to 1. Or X minus two could be 0.[br]In other words, X could be equal 0:01:34.494,0:01:40.134 to two, so there on the X axis.[br]I know that this curve is going 0:01:40.134,0:01:43.518 to go through their through[br]there and through there. 0:01:44.670,0:01:49.486 What about when X is very large?[br]What's going to happen to Y when 0:01:49.486,0:01:53.958 X is very large, X minus One X[br]minus two are really very 0:01:53.958,0:01:58.430 similar to X, so effectively[br]we've got X times by X times by 0:01:58.430,0:02:03.246 X. You know as if X is very[br]large and positive, so most why 0:02:03.246,0:02:05.998 be so? We've got a bit of curve 0:02:05.998,0:02:10.500 there. Similarly, if X is large[br]but negative, again taking one 0:02:10.500,0:02:15.180 off and taking two off isn't[br]going to make a great deal of 0:02:15.180,0:02:19.500 difference, so we've got a large[br]negative number times by a large 0:02:19.500,0:02:23.460 negative number times by a large[br]negative number. The answer is 0:02:23.460,0:02:27.060 going to be very large, but[br]negative 'cause we three 0:02:27.060,0:02:30.660 negative numbers multiplied[br]together. So there's a bit of a 0:02:30.660,0:02:31.740 curve down here. 0:02:32.380,0:02:36.544 Now how can we join that up?[br]Well, assuming the curve is 0:02:36.544,0:02:38.626 continuous, it's going to go up. 0:02:39.800,0:02:43.280 Over. Down I'm back up again. 0:02:44.100,0:02:48.672 So that's the picture of our[br]curve and straight away we can 0:02:48.672,0:02:53.244 see what the question is asking[br]us. Find the area contained by 0:02:53.244,0:02:57.435 the curve and the X axis it[br]wants these two areas. 0:02:58.200,0:03:02.650 But they're in different[br]positions. One area a is above 0:03:02.650,0:03:08.880 the X axis and the other one[br]area B is below the X axis. 0:03:09.840,0:03:11.945 So perhaps we better be 0:03:11.945,0:03:15.705 cautious. And work out these two 0:03:15.705,0:03:21.540 areas separately. Now what we've[br]been told, what we know is that 0:03:21.540,0:03:22.728 if we integrate. 0:03:23.440,0:03:29.356 Between the limits. So in this[br]case that's not an one hour 0:03:29.356,0:03:35.765 function X times by X minus one[br]times by X minus two with 0:03:35.765,0:03:43.160 respect to X. We will get this[br]area a, so the area a is equal 0:03:43.160,0:03:47.597 to the result of carrying out[br]this integration and 0:03:47.597,0:03:52.527 substituting in the limits of[br]integration. So let's do that 0:03:52.527,0:03:54.006 first of all. 0:03:54.010,0:03:56.926 We will need to multiply out 0:03:56.926,0:04:04.110 these brackets. Well, X times by[br]X is X squared and times by that 0:04:04.110,0:04:06.385 ex again is X cubed. 0:04:07.370,0:04:11.869 I've got X times by minus two[br]gives me minus 2X. 0:04:12.540,0:04:17.916 And minus one times by X gives[br]me another minus X. So 0:04:17.916,0:04:22.844 altogether I've got minus 3X[br]coming out of these brackets and 0:04:22.844,0:04:28.220 I need times it by X, so that's[br]minus three X squared. 0:04:29.150,0:04:36.514 Then I've got minus one times by[br]minus two. That's +2 and I need 0:04:36.514,0:04:40.196 to times that by X. So that's 0:04:40.196,0:04:46.409 plus 2X. To be integrated with[br]respect to X so it's carry out 0:04:46.409,0:04:52.037 the integration. Remember we had[br]one to the index and divide by 0:04:52.037,0:04:58.603 the new index, so that's one to[br]the index. Makes that X to the 0:04:58.603,0:05:01.417 power 4 / 4 - 3. 0:05:02.020,0:05:07.090 To integrate X squared, we had[br]one to the index that's X cubed 0:05:07.090,0:05:13.330 divided by the new index. That's[br]3 + 2. XX is X to the power one. 0:05:13.330,0:05:18.400 We don't always right the one[br]there, but we know that it means 0:05:18.400,0:05:24.250 X to the power one. So we add 1[br]to the index that's X squared 0:05:24.250,0:05:26.980 and we divide by the new index. 0:05:27.760,0:05:31.945 These are our limits of[br]integration nought and one. 0:05:32.580,0:05:36.909 We can do a little bit of simply[br]simplifying in here. We can 0:05:36.909,0:05:39.240 cancel three with a 3 under, two 0:05:39.240,0:05:45.350 with a 2. And now we can put[br]in our limit of integration. The 0:05:45.350,0:05:46.990 top 1 one in. 0:05:48.260,0:05:54.840 So that's. X to the[br]power four. When X is one is 0:05:54.840,0:06:00.692 just one over 4 minus X cubed.[br]When X is one. So that's just 0:06:00.692,0:06:06.126 one plus X squared. When X is[br]one. So that's just one again. 0:06:06.650,0:06:11.248 Minus and now we substitute in[br]our lower limit. That's zero, 0:06:11.248,0:06:16.682 but when we put zero in there[br]that's zero and that's zero as 0:06:16.682,0:06:22.534 well, and that's zero as well,[br]because X is equal to 0 in each 0:06:22.534,0:06:27.550 case. So the whole of the SEC[br]bracket here is just zero. 0:06:28.070,0:06:31.836 So we simplify that we get a 0:06:31.836,0:06:38.530 quarter. So the area a is[br]1/4 of a unit of area. 0:06:39.060,0:06:44.664 What about area B? We can find[br]that by doing the same 0:06:44.664,0:06:49.334 integration, this time taking[br]the limits between one and two. 0:06:49.334,0:06:52.136 So let's look at that one. 0:06:52.660,0:06:55.400 B. 0:06:56.750,0:07:03.790 Will be the integral between one[br]and two of X cubed. 0:07:04.340,0:07:11.604 Minus three X squared[br]plus 2X with respect 0:07:11.604,0:07:18.650 to X. So let's have[br]a look what we're going to get 0:07:18.650,0:07:24.630 the integral of. This should not[br]be any different to what we had 0:07:24.630,0:07:31.070 last time. So if we integrate[br]the X cubed, we have X to the 0:07:31.070,0:07:32.450 4th over 4. 0:07:32.970,0:07:36.270 Minus three X cubed 0:07:36.270,0:07:43.406 over 3. +2 X[br]squared over 2 and 0:07:43.406,0:07:46.710 this is to be 0:07:46.710,0:07:52.732 evaluated. The limits one and[br]two, and again we can cancel the 0:07:52.732,0:07:55.564 threes and we can cancel the 0:07:55.564,0:08:00.858 tools. So now let's substitute[br]in our upper limit. 0:08:02.090,0:08:09.622 X to the power four when X[br]is 2, so that's 2 multiplied by 0:08:09.622,0:08:15.002 itself four times and that gives[br]us 16 over 4. 0:08:15.520,0:08:21.190 Minus X cubed which is 2[br]multiplied by itself three 0:08:21.190,0:08:23.458 times, so that's eight. 0:08:24.720,0:08:30.880 Plus X squared which is 2[br]multiplied by itself twice, so 0:08:30.880,0:08:38.225 that's four. Minus now we put[br]the one in, so that's a quarter. 0:08:38.970,0:08:44.625 Because X to the power four is[br]just 1 - 1 because X cubed is 0:08:44.625,0:08:46.887 just one when X is one. 0:08:47.500,0:08:53.966 Plus one. Now[br]let's simplify each of these 0:08:53.966,0:08:59.587 brackets separately. 4 into 16[br]goes four times, so we four 0:08:59.587,0:09:04.697 takeaway 8 + 4. Will that[br]bracket is now 0. 0:09:05.410,0:09:10.818 Minus or we have 1/4 takeaway[br]one add one so this bracket is 0:09:10.818,0:09:12.066 just a quarter. 0:09:12.670,0:09:18.496 So I'll answer[br]is minus 1/4. 0:09:19.620,0:09:22.777 Let's just have a look where B 0:09:22.777,0:09:29.748 was. Bees here[br]it's underneath the X 0:09:29.748,0:09:30.676 axis. 0:09:32.480,0:09:37.264 Now, if you remember, or if you[br]look back at the video on 0:09:37.264,0:09:41.312 integration as summation, one of[br]the things that we were looking 0:09:41.312,0:09:44.256 at when we had a piece of curve. 0:09:44.950,0:09:48.550 And we were looking at the area[br]underneath that curve. 0:09:49.580,0:09:53.008 We took small strips. 0:09:53.550,0:09:59.670 And these small strips were of[br]height Y and thickness Delta X, 0:09:59.670,0:10:05.790 and so the little bit of area[br]that they represented. Delta a 0:10:05.790,0:10:12.420 was why times by Delta X. Now[br]that's fine for what we've done. 0:10:12.420,0:10:15.480 But notice this Y is positive. 0:10:16.310,0:10:20.438 And so when we add up all[br]these strips that are above 0:10:20.438,0:10:24.910 the X axis, we get a positive[br]answer, IE for a the answer 0:10:24.910,0:10:26.630 we got was a quarter. 0:10:27.900,0:10:29.049 But for B. 0:10:29.640,0:10:36.480 These why values are actually[br]below the X axis and so this 0:10:36.480,0:10:43.320 area, in a sense is a negative[br]area because the wise are 0:10:43.320,0:10:49.590 negative. Hence our answer is[br]minus 1/4. But the question said 0:10:49.590,0:10:55.860 what is the area and clearly[br]physically there are two blocks 0:10:55.860,0:10:59.850 of area and so the total area. 0:10:59.860,0:11:06.139 Must be 1/4. That's the area of[br]that and a quarter because 1/4 0:11:06.139,0:11:12.418 is the area of that. The minus[br]sign merely tells us it's below 0:11:12.418,0:11:18.214 the X axis. So the answer to our[br]original question, what's the 0:11:18.214,0:11:23.527 area that the that's contained[br]between the curve and the X 0:11:23.527,0:11:30.412 axis? Is given by[br]the area is 1/4 + 0:11:30.412,0:11:33.248 1/4 which is 1/2. 0:11:36.040,0:11:39.520 Let's just check on something[br]what would happen. 0:11:40.120,0:11:46.168 If we took the integral of our[br]function between North and two. 0:11:48.070,0:11:55.870 Remember what our function is.[br]It's X cubed minus three 0:11:55.870,0:11:58.990 X squared plus 2X. 0:11:59.820,0:12:03.430 Integrated with respect to X. 0:12:04.770,0:12:10.204 So let's do that integration and[br]again the answer shouldn't be 0:12:10.204,0:12:16.626 any different from what we've[br]had before X to the 4th over 4 0:12:16.626,0:12:24.036 - 3 X cubed over 3 + 2[br]X squared over 2 to be evaluated 0:12:24.036,0:12:29.470 between North two. Again, you[br]can do the canceling the threes 0:12:29.470,0:12:31.940 there and the tools there. 0:12:33.150,0:12:36.274 Equals that substituting the 0:12:36.274,0:12:43.260 upper limit. So that's X to[br]the power four. When X is 2, two 0:12:43.260,0:12:48.590 multiplied by itself four times[br]is 16, and to be divided by 4. 0:12:50.300,0:12:56.031 Minus X cubed minus 2 multiplied[br]by itself three times, that's 0:12:56.031,0:13:03.280 eight. Plus X squared that's[br]2 multiplied by itself twice, so 0:13:03.280,0:13:05.113 that's plus 4. 0:13:05.120,0:13:11.161 Minus. And when we put zero[br]in that zero, that's zero and 0:13:11.161,0:13:15.572 that one zero. So the whole of[br]that bracket is 0. 0:13:16.390,0:13:22.682 Counseling here. This leaves us[br]with a four, and so we 0:13:22.682,0:13:26.114 have 0 - 0 is 0. 0:13:27.610,0:13:32.290 Which is what we would expect,[br]because in a sense, what this 0:13:32.290,0:13:34.240 process is done is added 0:13:34.240,0:13:38.050 together. The two[br]areas that we got. 0:13:39.260,0:13:45.167 With respect to their signs, 1/4[br]plus minus 1/4 is Nord. 0:13:45.700,0:13:51.940 What that means is that we have[br]to be very, very careful when 0:13:51.940,0:13:53.380 we're calculating areas. 0:13:54.080,0:13:58.623 Clearly if we just integrate it[br]between the given limits, what 0:13:58.623,0:14:04.818 we might do is end up with an[br]answer that is the value of the 0:14:04.818,0:14:09.774 integral and not the area and[br]what this example shows is is 0:14:09.774,0:14:13.491 that the area that is contained[br]by our function. 0:14:14.380,0:14:20.516 The X axis and specific values[br]of the ordinance I values of X 0:14:20.516,0:14:27.124 may not be the same as the value[br]of the integral, so bearing that 0:14:27.124,0:14:31.844 in mind, let's have a look at[br]some more examples. 0:14:32.460,0:14:35.880 This time. Let's take. 0:14:36.720,0:14:43.874 Function of the curve Y equals X[br]times by X minus three and let's 0:14:43.874,0:14:48.984 investigate what's the area[br]contained by the curve X access 0:14:48.984,0:14:55.627 and the audience X equals 0 and[br]X equals 5. Well, from what 0:14:55.627,0:15:02.781 we've learned so far, one of the[br]things we've got to do is a 0:15:02.781,0:15:05.847 sketch. So here's our X&Y axes. 0:15:07.280,0:15:14.216 If we set Y equals 0, then we[br]can see that either X is 0 or X 0:15:14.216,0:15:19.928 can be equal to three. So that[br]gives us two points on the curve 0:15:19.928,0:15:25.232 there at the origin and there[br]where X equals 3. We also know 0:15:25.232,0:15:30.536 that we're going up to X equals[br]5, so let's Mark that there. 0:15:31.480,0:15:35.528 When we multiply out this[br]bracket, we can see we've got 0:15:35.528,0:15:40.312 X times by. X gives us X[br]squared and it's a positive X 0:15:40.312,0:15:44.360 squared, which means that[br]it's going to be a U shape 0:15:44.360,0:15:45.832 coming down like that. 0:15:49.090,0:15:55.070 Now it wants the area between[br]the X axis, the curve and these 0:15:55.070,0:16:00.590 ordinance. So let's draw that in[br]there. And again we can see 0:16:00.590,0:16:07.490 we've got two lots of area and[br]area a which is below the X axis 0:16:07.490,0:16:13.930 in an area B which is above it.[br]This means we've got to be 0:16:13.930,0:16:18.990 careful, workout each area[br]separately so the area a is the 0:16:18.990,0:16:20.590 integral. Between North 0:16:21.130,0:16:27.019 And three of X times by X minus[br]three with respect to X. 0:16:27.690,0:16:30.770 Let's multiply out the brackets 0:16:30.770,0:16:37.401 first. So X times by X gives[br]us X squared X times Y minus 0:16:37.401,0:16:39.386 three gives us minus 3X. 0:16:39.920,0:16:43.364 We're integrating that with[br]respect to X. 0:16:44.800,0:16:46.610 So let's do the integration. 0:16:47.270,0:16:52.958 X squared integrates 2X cubed[br]over three and one to the index 0:16:52.958,0:16:59.594 and divide by that new index,[br]minus three X. This is X to the 0:16:59.594,0:17:05.756 power one, add 1 to the index[br]and divide by the new index. 0:17:06.400,0:17:09.240 That's to be evaluated between 0:17:09.240,0:17:11.180 North. And three. 0:17:12.440,0:17:15.866 Equals we put the upper limiting 0:17:15.866,0:17:23.076 first. X cubed X is equal[br]to three, so this is 3 cubed. 0:17:23.076,0:17:25.068 That's 27 over 3. 0:17:25.990,0:17:33.226 Minus three times by X squared[br]and X is 3, so we 0:17:33.226,0:17:39.256 have minus three times by 9[br]is 27 over 2. 0:17:40.220,0:17:46.040 Minus now we put the lower limit[br]in X cubed is zero when X is 0:17:46.040,0:17:51.472 zero and X squared is zero when[br]X is zero. That is minus 0. 0:17:52.210,0:17:59.710 Equals so now we've got 27 over[br]3 - 27 over 2. We can cancel 0:17:59.710,0:18:06.710 by the three here. That's nine,[br]so we have 9 minus and twos into 0:18:06.710,0:18:13.710 27 go 13 1/2. That will simplify[br]to that and so we have minus 0:18:13.710,0:18:21.210 four and a half. So that's the[br]area a. Let's now have a look at 0:18:21.210,0:18:22.710 the area be. 0:18:23.390,0:18:29.040 So that's the integral of X[br]squared minus 3X with 0:18:29.040,0:18:34.125 respect to X. This time[br]between 3:00 and 5:00. 0:18:35.160,0:18:41.133 Will get exactly the same as[br]we've got here, 'cause we're 0:18:41.133,0:18:43.305 integrating exactly the same 0:18:43.305,0:18:50.186 function. But the limits[br]are different there 0:18:50.186,0:18:53.354 between 3:00 and 0:18:53.354,0:18:57.150 5:00. So let's turn over the 0:18:57.150,0:19:04.770 page. Area B is equal[br]to. Remember that it's this that 0:19:04.770,0:19:06.678 we are evaluating. 0:19:09.340,0:19:16.972 Between 3:00 and 5:00, so we[br]put in our upper limit first. 0:19:17.550,0:19:24.486 X cubed when X is equal[br]to five, is 5 multiplied by 0:19:24.486,0:19:29.110 itself 3 times and that's[br]125 over 3. 0:19:30.180,0:19:31.680 Minus. 0:19:32.750,0:19:37.430 X squared when X is 5, that's[br]five times by 5 is 25. 0:19:38.000,0:19:43.136 25 times by three is 75[br]over 2. 0:19:44.150,0:19:45.839 That's our first. 0:19:46.480,0:19:50.520 Bracket minus now we put[br]in the three. 0:19:51.730,0:19:54.840 So that's 27 over 3. 0:19:55.370,0:19:59.473 Because if you think about it,[br]we've actually worked this out 0:19:59.473,0:20:01.338 before. Minus X is 3. 0:20:02.190,0:20:07.228 3 squared is 9 times by that is[br]27 over 2. 0:20:07.940,0:20:15.267 OK. Let's[br]tidy up these bits first. 125 0:20:15.267,0:20:22.719 over 3 - 27 over three.[br]Well, that's 98 over 3 - 0:20:22.719,0:20:30.171 5 - 75 over 2 minus[br]minus 27 over 2. So I've 0:20:30.171,0:20:37.623 got minus 75 + 27 and[br]that is going to give me 0:20:37.623,0:20:41.370 48. Over. 2. 0:20:43.380,0:20:45.160 So if you divide. 0:20:45.720,0:20:51.756 By three there threes into nine[br]goes three and threes into eight 0:20:51.756,0:20:58.295 goes two, and there are two[br]overs that's 32 and 2/3 - 2 0:20:58.295,0:21:05.337 into 48 goes 24. So the area[br]B is 32 and 2/3 - 24. 0:21:05.337,0:21:10.870 That's eight and 2/3, so the[br]actual area that I've got. 0:21:11.540,0:21:18.212 Is the area of a the actual area[br]of a which is four and a half. 0:21:18.212,0:21:22.799 The minus sign tells us that[br]it's below the X axis. 0:21:24.500,0:21:32.052 4 1/2 plus the area of B which[br]is 8 and 2/3 if I want to 0:21:32.052,0:21:38.188 add a half and 2/3 together then[br]they've really got to be in 0:21:38.188,0:21:43.852 terms of the same denominator,[br]so the four and the eight gives 0:21:43.852,0:21:50.745 me 12. Plus 1/2 now the[br]denominator is going to be 6, so 0:21:50.745,0:21:56.205 that's three sixths plus four[br]sixths, which is what the 2/3 is 0:21:56.205,0:22:02.575 equal to. I've got 7 sixth here,[br]which is one and a six. That's 0:22:02.575,0:22:07.125 13 and 1/6 is the actual value[br]of my area. 0:22:07.650,0:22:13.456 Let's take.[br]Another example, this time. 0:22:13.456,0:22:17.790 Let's have a look at the[br]function Y is equal to. 0:22:18.390,0:22:21.525 X squared plus 0:22:21.525,0:22:24.070 X. +4. 0:22:26.760,0:22:33.252 Now if we set Y equals 0,[br]then we're trying to solve 0:22:33.252,0:22:38.121 this equation X squared plus[br]X +4 equals not. 0:22:40.000,0:22:43.393 To find where the curve crosses[br]the X axis. 0:22:44.040,0:22:48.675 Doesn't look very promising.[br]Everything here looks sort of 0:22:48.675,0:22:54.855 positive got plus terms in it.[br]Let's check B squared minus four 0:22:54.855,0:22:59.923 AC. Remember the formula for[br]solving a quadratic equation is 0:22:59.923,0:23:05.474 X equals minus B plus or minus[br]square root of be squared minus 0:23:05.474,0:23:10.598 four AC all over 2A, but it's[br]this bit that's the important 0:23:10.598,0:23:15.295 bit. the B squared minus four[br]AC, 'cause If that's negative. 0:23:16.510,0:23:20.724 IE, when we take a square root,[br]we can't have the square root of 0:23:20.724,0:23:24.035 a negative number. What that[br]tells us is that this equation 0:23:24.035,0:23:31.222 has no roots. So B squared well,[br]there's 1X, so B is equal to 1, 0:23:31.222,0:23:37.410 so that's one squared minus four[br]times a while a is one 'cause 0:23:37.410,0:23:40.742 there's One X squared and C is 0:23:40.742,0:23:46.850 4. Well, that's 1 - 16 is minus[br]15. It's less than zero. There 0:23:46.850,0:23:52.440 are no roots for this equation,[br]so it doesn't meet the X axis. 0:23:53.570,0:23:57.990 Let's have a look at a picture[br]for that. There's The X Axis. 0:23:58.520,0:24:03.995 Why access I'd like to be able[br]to fix a point on the curve and 0:24:03.995,0:24:09.105 I can buy taking X is zero.[br]Those two terms would be 0, so 0:24:09.105,0:24:11.295 that gives me Y equals 4. 0:24:12.040,0:24:18.501 So that's a point on the curve,[br]but it's always going to be 0:24:18.501,0:24:23.968 above the X axis, and when[br]because X squared is positive. 0:24:25.620,0:24:29.316 With The X is positive or[br]negative, X squared is always 0:24:29.316,0:24:33.684 going to be positive, so it's[br]always going to be that way up 0:24:33.684,0:24:35.700 and it's going to look something 0:24:35.700,0:24:42.800 like that. So we want the[br]area between this curve, the X 0:24:42.800,0:24:49.037 Axis and the ordinates X equals[br]1 and X equals 3. 0:24:49.830,0:24:55.780 X equals 1, let's say is there,[br]X equals 3, let's say is there. 0:24:57.830,0:25:03.090 That's the area that we're[br]after. No real problems with 0:25:03.090,0:25:08.876 that area, so let's do the[br]calculation. The area is equal 0:25:08.876,0:25:15.714 to the integral between one and[br]three of X squared plus X +4 0:25:15.714,0:25:17.818 with respect to X. 0:25:18.740,0:25:23.735 Integrating each of these at one[br]to the index. 0:25:24.250,0:25:26.770 And divide by the new index. 0:25:27.570,0:25:34.602 Same again there and then four[br]is 4X. When we integrate it 0:25:34.602,0:25:40.462 and this is to be evaluated[br]between one and three. 0:25:40.990,0:25:45.467 So let's just turn this over.[br]Remember, the area that we're 0:25:45.467,0:25:52.760 wanting. Is equal to. This[br]is what we have to work 0:25:52.760,0:25:59.120 out X cubed over 3 plus[br]X squared over 2. 0:25:59.130,0:26:05.640 Plus 4X that's to be evaluated[br]between one and three. 0:26:06.760,0:26:13.690 So we put the upper limiting[br]first, so when X is equal to 3, 0:26:13.690,0:26:17.155 three cubed is equal to 27 over 0:26:17.155,0:26:24.043 3. Plus X squared when X[br]is equal to three, that's nine 0:26:24.043,0:26:30.712 over 2 + 4 X when X[br]is equal to three, that's 12. 0:26:30.712,0:26:33.790 Now we put the lower limiting. 0:26:34.330,0:26:41.844 So when X is One X cubed is[br]one 1 / 3 + 1, X is one 0:26:41.844,0:26:47.590 that's X squared is one and[br]divided by two, and when X is 0:26:47.590,0:26:49.358 one that's plus 4. 0:26:51.270,0:26:56.847 Now, some of these we can[br]workout quite easily, but let's 0:26:56.847,0:27:03.945 do the thirds bit first with 27[br]over 3 takeaway. A third is 26 0:27:03.945,0:27:10.536 over three we have 9 over 2[br]takeaway one over 2. So that's 0:27:10.536,0:27:16.620 plus eight over 2, and we've 12[br]takeaway 4, so that's +8. 0:27:16.630,0:27:18.380 Will cancel down to their. 0:27:19.010,0:27:26.570 If I do freeze into 26, eight[br]threes are 24 and two over slots 0:27:26.570,0:27:33.590 2/3 + 4 and 80s twelve so[br]altogether that's an area of 20 0:27:33.590,0:27:36.480 and two. Thirds 0:27:38.470,0:27:45.764 Now.[br]We can workout the area between 0:27:45.764,0:27:51.926 a curve, the X axis and some[br]given values of X the ordinance. 0:27:52.440,0:27:57.676 We can extend this technique[br]though to working out the area 0:27:57.676,0:28:02.436 contained between two curves.[br]Let's have a look at that 0:28:02.436,0:28:09.788 supposing. You've got the curve[br]Y equals X times 3 minus X, 0:28:09.788,0:28:14.630 and we've got the straight line[br]Y equals X. 0:28:15.280,0:28:20.070 We say what's the area that's[br]contained between these two 0:28:20.070,0:28:23.990 curves? Well, let's have a[br]look at a picture just to 0:28:23.990,0:28:25.235 see what that looks like. 0:28:28.440,0:28:35.482 If we set Y equals 0 again[br]and that will give us the values 0:28:35.482,0:28:42.021 of X where this curve crosses[br]the X axis clearly does so there 0:28:42.021,0:28:48.057 when X is zero and also when[br]X is equal to 3. 0:28:49.680,0:28:54.846 We've also got X times by minus.[br]X gives us minus X squared, so 0:28:54.846,0:29:00.012 for a quadratic it's going to be[br]an upside down U. It's going to 0:29:00.012,0:29:04.809 do that, just missed that point.[br]Let's make it bigger so it goes 0:29:04.809,0:29:09.237 through it. The line Y equals[br]XY. That's a straight line going 0:29:09.237,0:29:14.800 through there. And so the line[br]crosses the curve at these two 0:29:14.800,0:29:20.148 points. Let me call that one P[br]and this one is oh the origin. 0:29:21.250,0:29:25.628 And the error that I'm[br]interested in is this one. In 0:29:25.628,0:29:28.016 here the area core between the 0:29:28.016,0:29:34.369 two curves. Well before I can[br]find that area, I really need to 0:29:34.369,0:29:39.637 know what's this point P. Where[br]do these two curves cross? Well, 0:29:39.637,0:29:44.905 they will cross when their Y[br]values are equal, so they will 0:29:44.905,0:29:51.051 intersect or meet when X times[br]by 3 minus X is equal to X. 0:29:52.270,0:29:57.661 Multiply out the bracket[br]3X minus X squared is 0:29:57.661,0:29:59.458 equal to X. 0:30:00.530,0:30:07.225 Take X away from both sides. 2X[br]minus X squared is then zero. 0:30:07.225,0:30:12.375 This is a quadratic, and it[br]factorizes. There's a common 0:30:12.375,0:30:18.555 factor of X in each term, so[br]we can take that out. 0:30:18.560,0:30:23.438 And we're left with[br]two minus X equals 0. 0:30:25.400,0:30:32.098 Either the X can be 0 or the two[br]minus X com 0, so therefore X is 0:30:32.098,0:30:38.402 0 or X equals 2, so this is[br]where X is equal to 0. Here at 0:30:38.402,0:30:44.312 the origin, when X is zero, Y is[br]00. In there we have not times 0:30:44.312,0:30:49.434 by three is nothing. If we put[br]zero in there, we get nothing. 0:30:49.434,0:30:54.556 Or why is equal to? Well, This[br]is why equals X, so presumably 0:30:54.556,0:30:56.920 why must be equal to two? 0:30:57.110,0:31:02.178 And let's just check it over[br]here. If we put X equals 2 with 0:31:02.178,0:31:08.332 2 * 3 - 2, three minus two is[br]one times by two is 2. So it's 0:31:08.332,0:31:10.504 this here. That's the point P. 0:31:12.290,0:31:17.581 So let's just put that into[br]their. Now this is the area that 0:31:17.581,0:31:24.160 we want. So we can look at it[br]as finding the area under the 0:31:24.160,0:31:28.945 curve between North and two and[br]finding the area under the 0:31:28.945,0:31:33.580 straight line. And then[br]subtracting them in order to end 0:31:33.580,0:31:34.750 up with that. 0:31:35.510,0:31:39.370 So let's do that. Let's[br]first of all, find the 0:31:39.370,0:31:40.914 area under the curve. 0:31:43.720,0:31:51.112 Equals now this will[br]be the integral between 0:31:51.112,0:31:58.504 North and two of[br]our curve X times 0:31:58.504,0:32:02.200 by 3 minus X. 0:32:02.480,0:32:07.248 With respect to[br]X. 0:32:08.270,0:32:14.666 So let's first of all multiply[br]out the bracket X times by three 0:32:14.666,0:32:20.570 is 3 XX times Y minus X is[br]minus X squared X. 0:32:21.860,0:32:28.801 Do the integration integral of X[br]is X squared over 2? 0:32:30.360,0:32:33.805 And we need to multiply it by[br]the three that we've got there. 0:32:34.370,0:32:41.130 And then the integral of X[br]squared is going to be X cubed 0:32:41.130,0:32:46.330 over three that to be evaluated[br]between North and two. 0:32:46.980,0:32:48.744 So I put the two in first. 0:32:49.310,0:32:52.078 So we have 3. 0:32:52.790,0:32:56.638 Times 2 to 4. 0:32:56.640,0:32:58.149 Divided by two. 0:32:58.760,0:33:00.950 Minus X cubed over 3. 0:33:01.690,0:33:06.218 X is 2, so that's eight[br]over 3. 0:33:07.780,0:33:13.590 Minus and when we put the zero[br]in, X squared is zero and X 0:33:13.590,0:33:17.325 cubed is 0, so that second[br]bracket is 0. 0:33:18.340,0:33:22.156 Equals what we can cancel it 0:33:22.156,0:33:28.650 to there. 326 minus threes into[br]eight goals two and there's two 0:33:28.650,0:33:35.958 over, so that's 2/3, so the area[br]that we've got is 6 - 2 0:33:35.958,0:33:39.090 and 2/3, which is 3 1/3. 0:33:40.140,0:33:47.303 So that's the area under the[br]curve. What now we need is the 0:33:47.303,0:33:50.058 area under Y equals X. 0:33:51.130,0:33:57.110 So this will be equal to the[br]integral between Norton two of X 0:33:57.110,0:34:00.370 DX. Equals. 0:34:01.680,0:34:05.194 Integral of X is just X squared 0:34:05.194,0:34:12.334 over 2. Between North and[br]two, substituting the two 2 0:34:12.334,0:34:15.799 two to four over 2. 0:34:15.960,0:34:23.084 Minus zero in and the X squared[br]over 2 gives gives us 0. 0:34:23.090,0:34:28.260 And so we've just got two[br]choosing toward those two, 0:34:28.260,0:34:34.981 nothing there. So here we've got[br]the area under the curve 3 1/3 0:34:34.981,0:34:38.600 here. We've got the area under Y 0:34:38.600,0:34:43.650 equals X2. And so the area[br]between them. 0:34:46.310,0:34:50.038 Let's look at that 0:34:50.038,0:34:54.972 positions. We've calculated this[br]area here under the curve. 0:34:54.972,0:34:59.940 That's the one that is free and[br]the 3rd, and we've calculated 0:34:59.940,0:35:02.424 the area of this bit of. 0:35:03.040,0:35:07.477 Triangle that's here[br]underneath Y equals X, and the 0:35:07.477,0:35:13.393 answer that is 2 and So what[br]we're looking for is the 0:35:13.393,0:35:17.830 difference between the big[br]area and the triangle. So 0:35:17.830,0:35:22.267 that's 3 1/3 - 2 gives us 1[br]1/3. 0:35:24.520,0:35:28.432 So we've seen how we can extend[br]the technique of finding the 0:35:28.432,0:35:29.736 area under a curve. 0:35:30.570,0:35:36.290 Between the X axis on between[br]two given ordinance to finding 0:35:36.290,0:35:40.970 the area between two given[br]curves. Let's take another 0:35:40.970,0:35:47.210 example. This time, let's use[br]the function Y equals sign X. We 0:35:47.210,0:35:53.970 better define a range of values[br]of X, so I'm going to take 0:35:53.970,0:35:57.610 X to be between North and pie. 0:35:58.260,0:36:05.456 Let's say I want the area that's[br]cut off from that by the line 0:36:05.456,0:36:11.624 Y equals one over Route 2. So[br]a little sketch as always. 0:36:12.630,0:36:16.038 We're finding these areas.[br]It's important to know 0:36:16.038,0:36:17.316 where they are. 0:36:18.470,0:36:22.736 So why equals sign X[br]between North and pie? 0:36:22.736,0:36:25.106 So it's going to look. 0:36:26.820,0:36:34.024 Like that? Naspi Nazira Why is[br]equal to one over Route 2? Well, 0:36:34.024,0:36:35.920 that's a straight line. 0:36:36.900,0:36:39.040 Somewhere, Let's say across. 0:36:39.790,0:36:42.450 There, so this is the area that 0:36:42.450,0:36:47.300 we're after. Area between the[br]line and that curve. What we 0:36:47.300,0:36:50.450 need to know. What are these[br]values of X? 0:36:51.540,0:36:58.185 Well, why is equal to one over[br]Route 2 and Y is equal to sign 0:36:58.185,0:37:04.387 X, so one over Route 2 is equal[br]to sign X where they meet? 0:37:05.180,0:37:08.816 So the values of X will be the[br]solutions to this equation. 0:37:09.390,0:37:14.710 Well, one of the values we do[br]know is that when sign X equals 0:37:14.710,0:37:18.890 one over route 2X must be equal[br]to π over 4. 0:37:19.510,0:37:23.494 I working in radians, so I've[br]got to give the answer in 0:37:23.494,0:37:26.482 radians, but the equivalent[br]answer in degrees is 45. 0:37:27.850,0:37:34.290 So that's that one there and the[br]one there is 3 Pi over 4. 0:37:35.410,0:37:41.730 So now I can workout what it is[br]I've got to do I can find the 0:37:41.730,0:37:46.075 area under the sign curve[br]between these two values. I can 0:37:46.075,0:37:50.815 find the area of this rectangle[br]and take it away. That will 0:37:50.815,0:37:55.555 leave me that area there. Just[br]let me put these answers in. 0:37:56.180,0:38:02.242 I found their and there, so it's[br]first of all find the area under 0:38:02.242,0:38:05.090 this curve. So the area. 0:38:06.770,0:38:13.840 Under The[br]curve is equal to 0:38:13.840,0:38:20.352 the integral between pie[br]by four and three 0:38:20.352,0:38:27.280 Pi 4. Of[br]sign XDX equal so 0:38:27.280,0:38:30.784 I have to integrate 0:38:30.784,0:38:38.038 sign X. Well, the[br]integral of sine X is minus Cos 0:38:38.038,0:38:45.912 X. To evaluate that[br]between pie by four and 0:38:45.912,0:38:48.309 three Pi 4. 0:38:49.020,0:38:56.370 So we put in the upper limit[br]first minus the cause of three π 0:38:56.370,0:39:00.360 by 4. That's the first bracket. 0:39:01.110,0:39:07.478 Minus minus the cause of[br]Π by 4. 0:39:08.250,0:39:15.495 Now cause of three π by 4 is[br]minus one over Route 2 and that 0:39:15.495,0:39:22.257 minus sign gives me plus one[br]over Route 2. The cause of Π by 0:39:22.257,0:39:28.536 4 is one over Route 2, minus[br]minus gives me plus one over 0:39:28.536,0:39:35.298 Route 2, so that is 2 over Route[br]2, which is just Route 2. 0:39:36.720,0:39:40.290 So I now know this area here. 0:39:40.840,0:39:44.371 I need to do is calculate the[br]area of this rectangle. 0:39:45.560,0:39:50.188 Then I can take it away from[br]that. Well, it is a rectangle. 0:39:50.188,0:39:52.680 Its height is one over Route 2. 0:39:53.260,0:39:55.770 So area. 0:39:57.020,0:40:02.971 Equals one over Route 2. That's[br]the height of that rectangle. 0:40:02.971,0:40:05.676 Times by this base it's. 0:40:06.240,0:40:12.932 Width, well, it runs from Π by[br]4, three π by 4, so the 0:40:12.932,0:40:19.624 difference is 2π by 4. In other[br]words, pie by two, so the area 0:40:19.624,0:40:24.404 of the rectangle is π over 2[br]times Route 2. 0:40:25.390,0:40:31.426 So the total area that I want[br]is the difference between these 0:40:31.426,0:40:34.947 two π over 2 Route 2 and 0:40:34.947,0:40:38.290 Route 2. In the area. 0:40:40.140,0:40:48.140 Equals Route 2 -[br]π over 2 Route 0:40:48.140,0:40:54.244 2. Put all that over a common[br]denominator of two Route 2, 0:40:54.244,0:40:56.100 which makes this 4. 0:40:56.770,0:41:04.538 Minus pie.[br]Leave that as that. That's the 0:41:04.538,0:41:06.452 area that's contained. 0:41:06.970,0:41:12.154 In there, between the straight[br]line and the curve. 0:41:13.030,0:41:20.120 I want now just to[br]go back to the example 0:41:20.120,0:41:27.210 where we were looking at[br]the area that was between 0:41:27.210,0:41:29.337 these two curves. 0:41:29.350,0:41:32.822 If you remember we 0:41:32.822,0:41:36.560 had. This sketch. 0:41:37.280,0:41:39.490 The curve. 0:41:40.670,0:41:46.158 So that[br]was the 0:41:46.158,0:41:48.902 picture that 0:41:48.902,0:41:54.850 we had.[br]And we calculated the area 0:41:54.850,0:41:57.080 between the two curves here. 0:41:58.040,0:42:00.470 By finding the area underneath. 0:42:01.080,0:42:03.330 X times by 3 minus X. 0:42:03.890,0:42:07.550 Then the area underneath[br]the line Y equals X and 0:42:07.550,0:42:08.648 taking them away. 0:42:11.020,0:42:16.129 Another way of looking at this[br]might be to go back to 1st 0:42:16.129,0:42:20.452 principles if you remember when[br]we did that, we took little 0:42:20.452,0:42:23.989 rectangular strips which were[br]roughly of height Y and 0:42:23.989,0:42:27.919 thickness Delta X. Well,[br]couldn't we do the same here, 0:42:27.919,0:42:31.456 have little rectangular strips[br]for that. Not quite rectangles, 0:42:31.456,0:42:36.565 but they are thin strips and[br]their why bit, so to speak would 0:42:36.565,0:42:40.495 be the difference between these[br]two values of Y here. 0:42:41.390,0:42:47.495 So if we were to call this, why[br]one, let's say, and this one Y 0:42:47.495,0:42:52.379 2, then might not the area that[br]we're looking for the area 0:42:52.379,0:42:54.821 between these two curves Y1 and 0:42:54.821,0:43:00.154 Y2? Between these ordinates Here[br]Let's call them A&B. 0:43:00.780,0:43:03.292 Might not we be able to work it 0:43:03.292,0:43:05.889 out? By doing that. 0:43:06.560,0:43:11.721 In other words, take the upper[br]curve and call it why one type 0:43:11.721,0:43:14.897 the lower curve and call it Y 2. 0:43:15.580,0:43:19.378 Subtract them and then integrate[br]between the required limits. 0:43:19.378,0:43:24.864 Well, let's see if this will[br]give us the same answer as we 0:43:24.864,0:43:26.974 had in the previous case. 0:43:27.760,0:43:32.128 So our limits[br]here are not 2. 0:43:33.250,0:43:35.326 And why one is this one? 0:43:36.040,0:43:41.227 Let's multiply it out first, X[br]times by three, is 3X and X 0:43:41.227,0:43:44.020 times Y minus X is minus X 0:43:44.020,0:43:50.910 squared. Take away why two? So[br]we're taking away X and 0:43:50.910,0:43:57.334 integrating with respect to X.[br]So if 2X minus X, sorry, 0:43:57.334,0:44:04.342 3X minus X. That gives us[br]2X minus X squared to be 0:44:04.342,0:44:10.182 evaluated between Norton two[br]with respect to X. So let's 0:44:10.182,0:44:12.518 carry out this integration. 0:44:13.790,0:44:21.322 The integral of X is X squared[br]over 2, so that's two times X 0:44:21.322,0:44:27.778 squared over 2. The integral[br]of X squared is X cubed over 0:44:27.778,0:44:34.234 three between North and two. I[br]can cancel it to their and 0:44:34.234,0:44:35.310 their substituting. 0:44:36.530,0:44:40.070 2 Twos are 4. 0:44:40.820,0:44:41.850 Minus. 0:44:43.030,0:44:50.710 X cube when X is 2, that's eight[br]over 3 minus and when I put zero 0:44:50.710,0:44:54.070 in that second bracket is just 0 0:44:54.070,0:44:56.845 equals. 4 0:44:56.845,0:45:03.870 minus. 2 and 2/3 which[br]just gives me one and third 0:45:03.870,0:45:10.669 units of area, which is exactly[br]what we had last time. So this. 0:45:12.000,0:45:15.300 Gives us a convenient formula. 0:45:15.950,0:45:21.274 Being able to workout the area[br]that is caught between two 0:45:21.274,0:45:27.566 curves, we call the upper curve.[br]Why won the lower curve Y two 0:45:27.566,0:45:32.406 and we subtract them and then[br]integrate between the ordinance 0:45:32.406,0:45:34.826 of the points of intersection?