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