1 00:00:02,290 --> 00:00:04,270 2 videos in this series. 2 00:00:05,670 --> 00:00:10,521 Areas as summation and integration as the reverse of 3 00:00:10,521 --> 00:00:15,372 differentiation have shown that the area under a curve. 4 00:00:16,320 --> 00:00:20,940 Above the X axis and between two ordinates. That's two values of 5 00:00:20,940 --> 00:00:25,560 X can be calculated by using integration. And what we want to 6 00:00:25,560 --> 00:00:30,950 do is develop that idea and look at some more examples of that in 7 00:00:30,950 --> 00:00:32,105 this particular video. 8 00:00:33,000 --> 00:00:39,392 So. Let's begin by exploring the question. We've got. The 9 00:00:39,392 --> 00:00:46,208 function Y equals X times X minus one times X minus 2. 10 00:00:46,960 --> 00:00:51,230 What's the area contained between this curve? 11 00:00:51,780 --> 00:00:56,340 On the X axis, well, the very nature of the question suggests 12 00:00:56,340 --> 00:01:01,280 hang on a minute. There might be something a bit odd here. Let's 13 00:01:01,280 --> 00:01:05,080 draw a picture. So let's start by sketching this curve. 14 00:01:06,800 --> 00:01:12,200 Well, if we put Y equals 0 then we can see that each one of 15 00:01:12,200 --> 00:01:16,880 these brackets could be 0 and that will give us a series of 16 00:01:16,880 --> 00:01:22,280 points on the curve. So why is 0 then? This could be 0 EX could 17 00:01:22,280 --> 00:01:28,012 be 0. Or X minus one could be 0. In other words, X could be equal 18 00:01:28,012 --> 00:01:34,494 to 1. Or X minus two could be 0. In other words, X could be equal 19 00:01:34,494 --> 00:01:40,134 to two, so there on the X axis. I know that this curve is going 20 00:01:40,134 --> 00:01:43,518 to go through their through there and through there. 21 00:01:44,670 --> 00:01:49,486 What about when X is very large? What's going to happen to Y when 22 00:01:49,486 --> 00:01:53,958 X is very large, X minus One X minus two are really very 23 00:01:53,958 --> 00:01:58,430 similar to X, so effectively we've got X times by X times by 24 00:01:58,430 --> 00:02:03,246 X. You know as if X is very large and positive, so most why 25 00:02:03,246 --> 00:02:05,998 be so? We've got a bit of curve 26 00:02:05,998 --> 00:02:10,500 there. Similarly, if X is large but negative, again taking one 27 00:02:10,500 --> 00:02:15,180 off and taking two off isn't going to make a great deal of 28 00:02:15,180 --> 00:02:19,500 difference, so we've got a large negative number times by a large 29 00:02:19,500 --> 00:02:23,460 negative number times by a large negative number. The answer is 30 00:02:23,460 --> 00:02:27,060 going to be very large, but negative 'cause we three 31 00:02:27,060 --> 00:02:30,660 negative numbers multiplied together. So there's a bit of a 32 00:02:30,660 --> 00:02:31,740 curve down here. 33 00:02:32,380 --> 00:02:36,544 Now how can we join that up? Well, assuming the curve is 34 00:02:36,544 --> 00:02:38,626 continuous, it's going to go up. 35 00:02:39,800 --> 00:02:43,280 Over. Down I'm back up again. 36 00:02:44,100 --> 00:02:48,672 So that's the picture of our curve and straight away we can 37 00:02:48,672 --> 00:02:53,244 see what the question is asking us. Find the area contained by 38 00:02:53,244 --> 00:02:57,435 the curve and the X axis it wants these two areas. 39 00:02:58,200 --> 00:03:02,650 But they're in different positions. One area a is above 40 00:03:02,650 --> 00:03:08,880 the X axis and the other one area B is below the X axis. 41 00:03:09,840 --> 00:03:11,945 So perhaps we better be 42 00:03:11,945 --> 00:03:15,705 cautious. And work out these two 43 00:03:15,705 --> 00:03:21,540 areas separately. Now what we've been told, what we know is that 44 00:03:21,540 --> 00:03:22,728 if we integrate. 45 00:03:23,440 --> 00:03:29,356 Between the limits. So in this case that's not an one hour 46 00:03:29,356 --> 00:03:35,765 function X times by X minus one times by X minus two with 47 00:03:35,765 --> 00:03:43,160 respect to X. We will get this area a, so the area a is equal 48 00:03:43,160 --> 00:03:47,597 to the result of carrying out this integration and 49 00:03:47,597 --> 00:03:52,527 substituting in the limits of integration. So let's do that 50 00:03:52,527 --> 00:03:54,006 first of all. 51 00:03:54,010 --> 00:03:56,926 We will need to multiply out 52 00:03:56,926 --> 00:04:04,110 these brackets. Well, X times by X is X squared and times by that 53 00:04:04,110 --> 00:04:06,385 ex again is X cubed. 54 00:04:07,370 --> 00:04:11,869 I've got X times by minus two gives me minus 2X. 55 00:04:12,540 --> 00:04:17,916 And minus one times by X gives me another minus X. So 56 00:04:17,916 --> 00:04:22,844 altogether I've got minus 3X coming out of these brackets and 57 00:04:22,844 --> 00:04:28,220 I need times it by X, so that's minus three X squared. 58 00:04:29,150 --> 00:04:36,514 Then I've got minus one times by minus two. That's +2 and I need 59 00:04:36,514 --> 00:04:40,196 to times that by X. So that's 60 00:04:40,196 --> 00:04:46,409 plus 2X. To be integrated with respect to X so it's carry out 61 00:04:46,409 --> 00:04:52,037 the integration. Remember we had one to the index and divide by 62 00:04:52,037 --> 00:04:58,603 the new index, so that's one to the index. Makes that X to the 63 00:04:58,603 --> 00:05:01,417 power 4 / 4 - 3. 64 00:05:02,020 --> 00:05:07,090 To integrate X squared, we had one to the index that's X cubed 65 00:05:07,090 --> 00:05:13,330 divided by the new index. That's 3 + 2. XX is X to the power one. 66 00:05:13,330 --> 00:05:18,400 We don't always right the one there, but we know that it means 67 00:05:18,400 --> 00:05:24,250 X to the power one. So we add 1 to the index that's X squared 68 00:05:24,250 --> 00:05:26,980 and we divide by the new index. 69 00:05:27,760 --> 00:05:31,945 These are our limits of integration nought and one. 70 00:05:32,580 --> 00:05:36,909 We can do a little bit of simply simplifying in here. We can 71 00:05:36,909 --> 00:05:39,240 cancel three with a 3 under, two 72 00:05:39,240 --> 00:05:45,350 with a 2. And now we can put in our limit of integration. The 73 00:05:45,350 --> 00:05:46,990 top 1 one in. 74 00:05:48,260 --> 00:05:54,840 So that's. X to the power four. When X is one is 75 00:05:54,840 --> 00:06:00,692 just one over 4 minus X cubed. When X is one. So that's just 76 00:06:00,692 --> 00:06:06,126 one plus X squared. When X is one. So that's just one again. 77 00:06:06,650 --> 00:06:11,248 Minus and now we substitute in our lower limit. That's zero, 78 00:06:11,248 --> 00:06:16,682 but when we put zero in there that's zero and that's zero as 79 00:06:16,682 --> 00:06:22,534 well, and that's zero as well, because X is equal to 0 in each 80 00:06:22,534 --> 00:06:27,550 case. So the whole of the SEC bracket here is just zero. 81 00:06:28,070 --> 00:06:31,836 So we simplify that we get a 82 00:06:31,836 --> 00:06:38,530 quarter. So the area a is 1/4 of a unit of area. 83 00:06:39,060 --> 00:06:44,664 What about area B? We can find that by doing the same 84 00:06:44,664 --> 00:06:49,334 integration, this time taking the limits between one and two. 85 00:06:49,334 --> 00:06:52,136 So let's look at that one. 86 00:06:52,660 --> 00:06:55,400 B. 87 00:06:56,750 --> 00:07:03,790 Will be the integral between one and two of X cubed. 88 00:07:04,340 --> 00:07:11,604 Minus three X squared plus 2X with respect 89 00:07:11,604 --> 00:07:18,650 to X. So let's have a look what we're going to get 90 00:07:18,650 --> 00:07:24,630 the integral of. This should not be any different to what we had 91 00:07:24,630 --> 00:07:31,070 last time. So if we integrate the X cubed, we have X to the 92 00:07:31,070 --> 00:07:32,450 4th over 4. 93 00:07:32,970 --> 00:07:36,270 Minus three X cubed 94 00:07:36,270 --> 00:07:43,406 over 3. +2 X squared over 2 and 95 00:07:43,406 --> 00:07:46,710 this is to be 96 00:07:46,710 --> 00:07:52,732 evaluated. The limits one and two, and again we can cancel the 97 00:07:52,732 --> 00:07:55,564 threes and we can cancel the 98 00:07:55,564 --> 00:08:00,858 tools. So now let's substitute in our upper limit. 99 00:08:02,090 --> 00:08:09,622 X to the power four when X is 2, so that's 2 multiplied by 100 00:08:09,622 --> 00:08:15,002 itself four times and that gives us 16 over 4. 101 00:08:15,520 --> 00:08:21,190 Minus X cubed which is 2 multiplied by itself three 102 00:08:21,190 --> 00:08:23,458 times, so that's eight. 103 00:08:24,720 --> 00:08:30,880 Plus X squared which is 2 multiplied by itself twice, so 104 00:08:30,880 --> 00:08:38,225 that's four. Minus now we put the one in, so that's a quarter. 105 00:08:38,970 --> 00:08:44,625 Because X to the power four is just 1 - 1 because X cubed is 106 00:08:44,625 --> 00:08:46,887 just one when X is one. 107 00:08:47,500 --> 00:08:53,966 Plus one. Now let's simplify each of these 108 00:08:53,966 --> 00:08:59,587 brackets separately. 4 into 16 goes four times, so we four 109 00:08:59,587 --> 00:09:04,697 takeaway 8 + 4. Will that bracket is now 0. 110 00:09:05,410 --> 00:09:10,818 Minus or we have 1/4 takeaway one add one so this bracket is 111 00:09:10,818 --> 00:09:12,066 just a quarter. 112 00:09:12,670 --> 00:09:18,496 So I'll answer is minus 1/4. 113 00:09:19,620 --> 00:09:22,777 Let's just have a look where B 114 00:09:22,777 --> 00:09:29,748 was. Bees here it's underneath the X 115 00:09:29,748 --> 00:09:30,676 axis. 116 00:09:32,480 --> 00:09:37,264 Now, if you remember, or if you look back at the video on 117 00:09:37,264 --> 00:09:41,312 integration as summation, one of the things that we were looking 118 00:09:41,312 --> 00:09:44,256 at when we had a piece of curve. 119 00:09:44,950 --> 00:09:48,550 And we were looking at the area underneath that curve. 120 00:09:49,580 --> 00:09:53,008 We took small strips. 121 00:09:53,550 --> 00:09:59,670 And these small strips were of height Y and thickness Delta X, 122 00:09:59,670 --> 00:10:05,790 and so the little bit of area that they represented. Delta a 123 00:10:05,790 --> 00:10:12,420 was why times by Delta X. Now that's fine for what we've done. 124 00:10:12,420 --> 00:10:15,480 But notice this Y is positive. 125 00:10:16,310 --> 00:10:20,438 And so when we add up all these strips that are above 126 00:10:20,438 --> 00:10:24,910 the X axis, we get a positive answer, IE for a the answer 127 00:10:24,910 --> 00:10:26,630 we got was a quarter. 128 00:10:27,900 --> 00:10:29,049 But for B. 129 00:10:29,640 --> 00:10:36,480 These why values are actually below the X axis and so this 130 00:10:36,480 --> 00:10:43,320 area, in a sense is a negative area because the wise are 131 00:10:43,320 --> 00:10:49,590 negative. Hence our answer is minus 1/4. But the question said 132 00:10:49,590 --> 00:10:55,860 what is the area and clearly physically there are two blocks 133 00:10:55,860 --> 00:10:59,850 of area and so the total area. 134 00:10:59,860 --> 00:11:06,139 Must be 1/4. That's the area of that and a quarter because 1/4 135 00:11:06,139 --> 00:11:12,418 is the area of that. The minus sign merely tells us it's below 136 00:11:12,418 --> 00:11:18,214 the X axis. So the answer to our original question, what's the 137 00:11:18,214 --> 00:11:23,527 area that the that's contained between the curve and the X 138 00:11:23,527 --> 00:11:30,412 axis? Is given by the area is 1/4 + 139 00:11:30,412 --> 00:11:33,248 1/4 which is 1/2. 140 00:11:36,040 --> 00:11:39,520 Let's just check on something what would happen. 141 00:11:40,120 --> 00:11:46,168 If we took the integral of our function between North and two. 142 00:11:48,070 --> 00:11:55,870 Remember what our function is. It's X cubed minus three 143 00:11:55,870 --> 00:11:58,990 X squared plus 2X. 144 00:11:59,820 --> 00:12:03,430 Integrated with respect to X. 145 00:12:04,770 --> 00:12:10,204 So let's do that integration and again the answer shouldn't be 146 00:12:10,204 --> 00:12:16,626 any different from what we've had before X to the 4th over 4 147 00:12:16,626 --> 00:12:24,036 - 3 X cubed over 3 + 2 X squared over 2 to be evaluated 148 00:12:24,036 --> 00:12:29,470 between North two. Again, you can do the canceling the threes 149 00:12:29,470 --> 00:12:31,940 there and the tools there. 150 00:12:33,150 --> 00:12:36,274 Equals that substituting the 151 00:12:36,274 --> 00:12:43,260 upper limit. So that's X to the power four. When X is 2, two 152 00:12:43,260 --> 00:12:48,590 multiplied by itself four times is 16, and to be divided by 4. 153 00:12:50,300 --> 00:12:56,031 Minus X cubed minus 2 multiplied by itself three times, that's 154 00:12:56,031 --> 00:13:03,280 eight. Plus X squared that's 2 multiplied by itself twice, so 155 00:13:03,280 --> 00:13:05,113 that's plus 4. 156 00:13:05,120 --> 00:13:11,161 Minus. And when we put zero in that zero, that's zero and 157 00:13:11,161 --> 00:13:15,572 that one zero. So the whole of that bracket is 0. 158 00:13:16,390 --> 00:13:22,682 Counseling here. This leaves us with a four, and so we 159 00:13:22,682 --> 00:13:26,114 have 0 - 0 is 0. 160 00:13:27,610 --> 00:13:32,290 Which is what we would expect, because in a sense, what this 161 00:13:32,290 --> 00:13:34,240 process is done is added 162 00:13:34,240 --> 00:13:38,050 together. The two areas that we got. 163 00:13:39,260 --> 00:13:45,167 With respect to their signs, 1/4 plus minus 1/4 is Nord. 164 00:13:45,700 --> 00:13:51,940 What that means is that we have to be very, very careful when 165 00:13:51,940 --> 00:13:53,380 we're calculating areas. 166 00:13:54,080 --> 00:13:58,623 Clearly if we just integrate it between the given limits, what 167 00:13:58,623 --> 00:14:04,818 we might do is end up with an answer that is the value of the 168 00:14:04,818 --> 00:14:09,774 integral and not the area and what this example shows is is 169 00:14:09,774 --> 00:14:13,491 that the area that is contained by our function. 170 00:14:14,380 --> 00:14:20,516 The X axis and specific values of the ordinance I values of X 171 00:14:20,516 --> 00:14:27,124 may not be the same as the value of the integral, so bearing that 172 00:14:27,124 --> 00:14:31,844 in mind, let's have a look at some more examples. 173 00:14:32,460 --> 00:14:35,880 This time. Let's take. 174 00:14:36,720 --> 00:14:43,874 Function of the curve Y equals X times by X minus three and let's 175 00:14:43,874 --> 00:14:48,984 investigate what's the area contained by the curve X access 176 00:14:48,984 --> 00:14:55,627 and the audience X equals 0 and X equals 5. Well, from what 177 00:14:55,627 --> 00:15:02,781 we've learned so far, one of the things we've got to do is a 178 00:15:02,781 --> 00:15:05,847 sketch. So here's our X&Y axes. 179 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 180 00:15:14,216 --> 00:15:19,928 can be equal to three. So that gives us two points on the curve 181 00:15:19,928 --> 00:15:25,232 there at the origin and there where X equals 3. We also know 182 00:15:25,232 --> 00:15:30,536 that we're going up to X equals 5, so let's Mark that there. 183 00:15:31,480 --> 00:15:35,528 When we multiply out this bracket, we can see we've got 184 00:15:35,528 --> 00:15:40,312 X times by. X gives us X squared and it's a positive X 185 00:15:40,312 --> 00:15:44,360 squared, which means that it's going to be a U shape 186 00:15:44,360 --> 00:15:45,832 coming down like that. 187 00:15:49,090 --> 00:15:55,070 Now it wants the area between the X axis, the curve and these 188 00:15:55,070 --> 00:16:00,590 ordinance. So let's draw that in there. And again we can see 189 00:16:00,590 --> 00:16:07,490 we've got two lots of area and area a which is below the X axis 190 00:16:07,490 --> 00:16:13,930 in an area B which is above it. This means we've got to be 191 00:16:13,930 --> 00:16:18,990 careful, workout each area separately so the area a is the 192 00:16:18,990 --> 00:16:20,590 integral. Between North 193 00:16:21,130 --> 00:16:27,019 And three of X times by X minus three with respect to X. 194 00:16:27,690 --> 00:16:30,770 Let's multiply out the brackets 195 00:16:30,770 --> 00:16:37,401 first. So X times by X gives us X squared X times Y minus 196 00:16:37,401 --> 00:16:39,386 three gives us minus 3X. 197 00:16:39,920 --> 00:16:43,364 We're integrating that with respect to X. 198 00:16:44,800 --> 00:16:46,610 So let's do the integration. 199 00:16:47,270 --> 00:16:52,958 X squared integrates 2X cubed over three and one to the index 200 00:16:52,958 --> 00:16:59,594 and divide by that new index, minus three X. This is X to the 201 00:16:59,594 --> 00:17:05,756 power one, add 1 to the index and divide by the new index. 202 00:17:06,400 --> 00:17:09,240 That's to be evaluated between 203 00:17:09,240 --> 00:17:11,180 North. And three. 204 00:17:12,440 --> 00:17:15,866 Equals we put the upper limiting 205 00:17:15,866 --> 00:17:23,076 first. X cubed X is equal to three, so this is 3 cubed. 206 00:17:23,076 --> 00:17:25,068 That's 27 over 3. 207 00:17:25,990 --> 00:17:33,226 Minus three times by X squared and X is 3, so we 208 00:17:33,226 --> 00:17:39,256 have minus three times by 9 is 27 over 2. 209 00:17:40,220 --> 00:17:46,040 Minus now we put the lower limit in X cubed is zero when X is 210 00:17:46,040 --> 00:17:51,472 zero and X squared is zero when X is zero. That is minus 0. 211 00:17:52,210 --> 00:17:59,710 Equals so now we've got 27 over 3 - 27 over 2. We can cancel 212 00:17:59,710 --> 00:18:06,710 by the three here. That's nine, so we have 9 minus and twos into 213 00:18:06,710 --> 00:18:13,710 27 go 13 1/2. That will simplify to that and so we have minus 214 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 215 00:18:21,210 --> 00:18:22,710 the area be. 216 00:18:23,390 --> 00:18:29,040 So that's the integral of X squared minus 3X with 217 00:18:29,040 --> 00:18:34,125 respect to X. This time between 3:00 and 5:00. 218 00:18:35,160 --> 00:18:41,133 Will get exactly the same as we've got here, 'cause we're 219 00:18:41,133 --> 00:18:43,305 integrating exactly the same 220 00:18:43,305 --> 00:18:50,186 function. But the limits are different there 221 00:18:50,186 --> 00:18:53,354 between 3:00 and 222 00:18:53,354 --> 00:18:57,150 5:00. So let's turn over the 223 00:18:57,150 --> 00:19:04,770 page. Area B is equal to. Remember that it's this that 224 00:19:04,770 --> 00:19:06,678 we are evaluating. 225 00:19:09,340 --> 00:19:16,972 Between 3:00 and 5:00, so we put in our upper limit first. 226 00:19:17,550 --> 00:19:24,486 X cubed when X is equal to five, is 5 multiplied by 227 00:19:24,486 --> 00:19:29,110 itself 3 times and that's 125 over 3. 228 00:19:30,180 --> 00:19:31,680 Minus. 229 00:19:32,750 --> 00:19:37,430 X squared when X is 5, that's five times by 5 is 25. 230 00:19:38,000 --> 00:19:43,136 25 times by three is 75 over 2. 231 00:19:44,150 --> 00:19:45,839 That's our first. 232 00:19:46,480 --> 00:19:50,520 Bracket minus now we put in the three. 233 00:19:51,730 --> 00:19:54,840 So that's 27 over 3. 234 00:19:55,370 --> 00:19:59,473 Because if you think about it, we've actually worked this out 235 00:19:59,473 --> 00:20:01,338 before. Minus X is 3. 236 00:20:02,190 --> 00:20:07,228 3 squared is 9 times by that is 27 over 2. 237 00:20:07,940 --> 00:20:15,267 OK. Let's tidy up these bits first. 125 238 00:20:15,267 --> 00:20:22,719 over 3 - 27 over three. Well, that's 98 over 3 - 239 00:20:22,719 --> 00:20:30,171 5 - 75 over 2 minus minus 27 over 2. So I've 240 00:20:30,171 --> 00:20:37,623 got minus 75 + 27 and that is going to give me 241 00:20:37,623 --> 00:20:41,370 48. Over. 2. 242 00:20:43,380 --> 00:20:45,160 So if you divide. 243 00:20:45,720 --> 00:20:51,756 By three there threes into nine goes three and threes into eight 244 00:20:51,756 --> 00:20:58,295 goes two, and there are two overs that's 32 and 2/3 - 2 245 00:20:58,295 --> 00:21:05,337 into 48 goes 24. So the area B is 32 and 2/3 - 24. 246 00:21:05,337 --> 00:21:10,870 That's eight and 2/3, so the actual area that I've got. 247 00:21:11,540 --> 00:21:18,212 Is the area of a the actual area of a which is four and a half. 248 00:21:18,212 --> 00:21:22,799 The minus sign tells us that it's below the X axis. 249 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 250 00:21:32,052 --> 00:21:38,188 add a half and 2/3 together then they've really got to be in 251 00:21:38,188 --> 00:21:43,852 terms of the same denominator, so the four and the eight gives 252 00:21:43,852 --> 00:21:50,745 me 12. Plus 1/2 now the denominator is going to be 6, so 253 00:21:50,745 --> 00:21:56,205 that's three sixths plus four sixths, which is what the 2/3 is 254 00:21:56,205 --> 00:22:02,575 equal to. I've got 7 sixth here, which is one and a six. That's 255 00:22:02,575 --> 00:22:07,125 13 and 1/6 is the actual value of my area. 256 00:22:07,650 --> 00:22:13,456 Let's take. Another example, this time. 257 00:22:13,456 --> 00:22:17,790 Let's have a look at the function Y is equal to. 258 00:22:18,390 --> 00:22:21,525 X squared plus 259 00:22:21,525 --> 00:22:24,070 X. +4. 260 00:22:26,760 --> 00:22:33,252 Now if we set Y equals 0, then we're trying to solve 261 00:22:33,252 --> 00:22:38,121 this equation X squared plus X +4 equals not. 262 00:22:40,000 --> 00:22:43,393 To find where the curve crosses the X axis. 263 00:22:44,040 --> 00:22:48,675 Doesn't look very promising. Everything here looks sort of 264 00:22:48,675 --> 00:22:54,855 positive got plus terms in it. Let's check B squared minus four 265 00:22:54,855 --> 00:22:59,923 AC. Remember the formula for solving a quadratic equation is 266 00:22:59,923 --> 00:23:05,474 X equals minus B plus or minus square root of be squared minus 267 00:23:05,474 --> 00:23:10,598 four AC all over 2A, but it's this bit that's the important 268 00:23:10,598 --> 00:23:15,295 bit. the B squared minus four AC, 'cause If that's negative. 269 00:23:16,510 --> 00:23:20,724 IE, when we take a square root, we can't have the square root of 270 00:23:20,724 --> 00:23:24,035 a negative number. What that tells us is that this equation 271 00:23:24,035 --> 00:23:31,222 has no roots. So B squared well, there's 1X, so B is equal to 1, 272 00:23:31,222 --> 00:23:37,410 so that's one squared minus four times a while a is one 'cause 273 00:23:37,410 --> 00:23:40,742 there's One X squared and C is 274 00:23:40,742 --> 00:23:46,850 4. Well, that's 1 - 16 is minus 15. It's less than zero. There 275 00:23:46,850 --> 00:23:52,440 are no roots for this equation, so it doesn't meet the X axis. 276 00:23:53,570 --> 00:23:57,990 Let's have a look at a picture for that. There's The X Axis. 277 00:23:58,520 --> 00:24:03,995 Why access I'd like to be able to fix a point on the curve and 278 00:24:03,995 --> 00:24:09,105 I can buy taking X is zero. Those two terms would be 0, so 279 00:24:09,105 --> 00:24:11,295 that gives me Y equals 4. 280 00:24:12,040 --> 00:24:18,501 So that's a point on the curve, but it's always going to be 281 00:24:18,501 --> 00:24:23,968 above the X axis, and when because X squared is positive. 282 00:24:25,620 --> 00:24:29,316 With The X is positive or negative, X squared is always 283 00:24:29,316 --> 00:24:33,684 going to be positive, so it's always going to be that way up 284 00:24:33,684 --> 00:24:35,700 and it's going to look something 285 00:24:35,700 --> 00:24:42,800 like that. So we want the area between this curve, the X 286 00:24:42,800 --> 00:24:49,037 Axis and the ordinates X equals 1 and X equals 3. 287 00:24:49,830 --> 00:24:55,780 X equals 1, let's say is there, X equals 3, let's say is there. 288 00:24:57,830 --> 00:25:03,090 That's the area that we're after. No real problems with 289 00:25:03,090 --> 00:25:08,876 that area, so let's do the calculation. The area is equal 290 00:25:08,876 --> 00:25:15,714 to the integral between one and three of X squared plus X +4 291 00:25:15,714 --> 00:25:17,818 with respect to X. 292 00:25:18,740 --> 00:25:23,735 Integrating each of these at one to the index. 293 00:25:24,250 --> 00:25:26,770 And divide by the new index. 294 00:25:27,570 --> 00:25:34,602 Same again there and then four is 4X. When we integrate it 295 00:25:34,602 --> 00:25:40,462 and this is to be evaluated between one and three. 296 00:25:40,990 --> 00:25:45,467 So let's just turn this over. Remember, the area that we're 297 00:25:45,467 --> 00:25:52,760 wanting. Is equal to. This is what we have to work 298 00:25:52,760 --> 00:25:59,120 out X cubed over 3 plus X squared over 2. 299 00:25:59,130 --> 00:26:05,640 Plus 4X that's to be evaluated between one and three. 300 00:26:06,760 --> 00:26:13,690 So we put the upper limiting first, so when X is equal to 3, 301 00:26:13,690 --> 00:26:17,155 three cubed is equal to 27 over 302 00:26:17,155 --> 00:26:24,043 3. Plus X squared when X is equal to three, that's nine 303 00:26:24,043 --> 00:26:30,712 over 2 + 4 X when X is equal to three, that's 12. 304 00:26:30,712 --> 00:26:33,790 Now we put the lower limiting. 305 00:26:34,330 --> 00:26:41,844 So when X is One X cubed is one 1 / 3 + 1, X is one 306 00:26:41,844 --> 00:26:47,590 that's X squared is one and divided by two, and when X is 307 00:26:47,590 --> 00:26:49,358 one that's plus 4. 308 00:26:51,270 --> 00:26:56,847 Now, some of these we can workout quite easily, but let's 309 00:26:56,847 --> 00:27:03,945 do the thirds bit first with 27 over 3 takeaway. A third is 26 310 00:27:03,945 --> 00:27:10,536 over three we have 9 over 2 takeaway one over 2. So that's 311 00:27:10,536 --> 00:27:16,620 plus eight over 2, and we've 12 takeaway 4, so that's +8. 312 00:27:16,630 --> 00:27:18,380 Will cancel down to their. 313 00:27:19,010 --> 00:27:26,570 If I do freeze into 26, eight threes are 24 and two over slots 314 00:27:26,570 --> 00:27:33,590 2/3 + 4 and 80s twelve so altogether that's an area of 20 315 00:27:33,590 --> 00:27:36,480 and two. Thirds 316 00:27:38,470 --> 00:27:45,764 Now. We can workout the area between 317 00:27:45,764 --> 00:27:51,926 a curve, the X axis and some given values of X the ordinance. 318 00:27:52,440 --> 00:27:57,676 We can extend this technique though to working out the area 319 00:27:57,676 --> 00:28:02,436 contained between two curves. Let's have a look at that 320 00:28:02,436 --> 00:28:09,788 supposing. You've got the curve Y equals X times 3 minus X, 321 00:28:09,788 --> 00:28:14,630 and we've got the straight line Y equals X. 322 00:28:15,280 --> 00:28:20,070 We say what's the area that's contained between these two 323 00:28:20,070 --> 00:28:23,990 curves? Well, let's have a look at a picture just to 324 00:28:23,990 --> 00:28:25,235 see what that looks like. 325 00:28:28,440 --> 00:28:35,482 If we set Y equals 0 again and that will give us the values 326 00:28:35,482 --> 00:28:42,021 of X where this curve crosses the X axis clearly does so there 327 00:28:42,021 --> 00:28:48,057 when X is zero and also when X is equal to 3. 328 00:28:49,680 --> 00:28:54,846 We've also got X times by minus. X gives us minus X squared, so 329 00:28:54,846 --> 00:29:00,012 for a quadratic it's going to be an upside down U. It's going to 330 00:29:00,012 --> 00:29:04,809 do that, just missed that point. Let's make it bigger so it goes 331 00:29:04,809 --> 00:29:09,237 through it. The line Y equals XY. That's a straight line going 332 00:29:09,237 --> 00:29:14,800 through there. And so the line crosses the curve at these two 333 00:29:14,800 --> 00:29:20,148 points. Let me call that one P and this one is oh the origin. 334 00:29:21,250 --> 00:29:25,628 And the error that I'm interested in is this one. In 335 00:29:25,628 --> 00:29:28,016 here the area core between the 336 00:29:28,016 --> 00:29:34,369 two curves. Well before I can find that area, I really need to 337 00:29:34,369 --> 00:29:39,637 know what's this point P. Where do these two curves cross? Well, 338 00:29:39,637 --> 00:29:44,905 they will cross when their Y values are equal, so they will 339 00:29:44,905 --> 00:29:51,051 intersect or meet when X times by 3 minus X is equal to X. 340 00:29:52,270 --> 00:29:57,661 Multiply out the bracket 3X minus X squared is 341 00:29:57,661 --> 00:29:59,458 equal to X. 342 00:30:00,530 --> 00:30:07,225 Take X away from both sides. 2X minus X squared is then zero. 343 00:30:07,225 --> 00:30:12,375 This is a quadratic, and it factorizes. There's a common 344 00:30:12,375 --> 00:30:18,555 factor of X in each term, so we can take that out. 345 00:30:18,560 --> 00:30:23,438 And we're left with two minus X equals 0. 346 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 347 00:30:32,098 --> 00:30:38,402 0 or X equals 2, so this is where X is equal to 0. Here at 348 00:30:38,402 --> 00:30:44,312 the origin, when X is zero, Y is 00. In there we have not times 349 00:30:44,312 --> 00:30:49,434 by three is nothing. If we put zero in there, we get nothing. 350 00:30:49,434 --> 00:30:54,556 Or why is equal to? Well, This is why equals X, so presumably 351 00:30:54,556 --> 00:30:56,920 why must be equal to two? 352 00:30:57,110 --> 00:31:02,178 And let's just check it over here. If we put X equals 2 with 353 00:31:02,178 --> 00:31:08,332 2 * 3 - 2, three minus two is one times by two is 2. So it's 354 00:31:08,332 --> 00:31:10,504 this here. That's the point P. 355 00:31:12,290 --> 00:31:17,581 So let's just put that into their. Now this is the area that 356 00:31:17,581 --> 00:31:24,160 we want. So we can look at it as finding the area under the 357 00:31:24,160 --> 00:31:28,945 curve between North and two and finding the area under the 358 00:31:28,945 --> 00:31:33,580 straight line. And then subtracting them in order to end 359 00:31:33,580 --> 00:31:34,750 up with that. 360 00:31:35,510 --> 00:31:39,370 So let's do that. Let's first of all, find the 361 00:31:39,370 --> 00:31:40,914 area under the curve. 362 00:31:43,720 --> 00:31:51,112 Equals now this will be the integral between 363 00:31:51,112 --> 00:31:58,504 North and two of our curve X times 364 00:31:58,504 --> 00:32:02,200 by 3 minus X. 365 00:32:02,480 --> 00:32:07,248 With respect to X. 366 00:32:08,270 --> 00:32:14,666 So let's first of all multiply out the bracket X times by three 367 00:32:14,666 --> 00:32:20,570 is 3 XX times Y minus X is minus X squared X. 368 00:32:21,860 --> 00:32:28,801 Do the integration integral of X is X squared over 2? 369 00:32:30,360 --> 00:32:33,805 And we need to multiply it by the three that we've got there. 370 00:32:34,370 --> 00:32:41,130 And then the integral of X squared is going to be X cubed 371 00:32:41,130 --> 00:32:46,330 over three that to be evaluated between North and two. 372 00:32:46,980 --> 00:32:48,744 So I put the two in first. 373 00:32:49,310 --> 00:32:52,078 So we have 3. 374 00:32:52,790 --> 00:32:56,638 Times 2 to 4. 375 00:32:56,640 --> 00:32:58,149 Divided by two. 376 00:32:58,760 --> 00:33:00,950 Minus X cubed over 3. 377 00:33:01,690 --> 00:33:06,218 X is 2, so that's eight over 3. 378 00:33:07,780 --> 00:33:13,590 Minus and when we put the zero in, X squared is zero and X 379 00:33:13,590 --> 00:33:17,325 cubed is 0, so that second bracket is 0. 380 00:33:18,340 --> 00:33:22,156 Equals what we can cancel it 381 00:33:22,156 --> 00:33:28,650 to there. 326 minus threes into eight goals two and there's two 382 00:33:28,650 --> 00:33:35,958 over, so that's 2/3, so the area that we've got is 6 - 2 383 00:33:35,958 --> 00:33:39,090 and 2/3, which is 3 1/3. 384 00:33:40,140 --> 00:33:47,303 So that's the area under the curve. What now we need is the 385 00:33:47,303 --> 00:33:50,058 area under Y equals X. 386 00:33:51,130 --> 00:33:57,110 So this will be equal to the integral between Norton two of X 387 00:33:57,110 --> 00:34:00,370 DX. Equals. 388 00:34:01,680 --> 00:34:05,194 Integral of X is just X squared 389 00:34:05,194 --> 00:34:12,334 over 2. Between North and two, substituting the two 2 390 00:34:12,334 --> 00:34:15,799 two to four over 2. 391 00:34:15,960 --> 00:34:23,084 Minus zero in and the X squared over 2 gives gives us 0. 392 00:34:23,090 --> 00:34:28,260 And so we've just got two choosing toward those two, 393 00:34:28,260 --> 00:34:34,981 nothing there. So here we've got the area under the curve 3 1/3 394 00:34:34,981 --> 00:34:38,600 here. We've got the area under Y 395 00:34:38,600 --> 00:34:43,650 equals X2. And so the area between them. 396 00:34:46,310 --> 00:34:50,038 Let's look at that 397 00:34:50,038 --> 00:34:54,972 positions. We've calculated this area here under the curve. 398 00:34:54,972 --> 00:34:59,940 That's the one that is free and the 3rd, and we've calculated 399 00:34:59,940 --> 00:35:02,424 the area of this bit of. 400 00:35:03,040 --> 00:35:07,477 Triangle that's here underneath Y equals X, and the 401 00:35:07,477 --> 00:35:13,393 answer that is 2 and So what we're looking for is the 402 00:35:13,393 --> 00:35:17,830 difference between the big area and the triangle. So 403 00:35:17,830 --> 00:35:22,267 that's 3 1/3 - 2 gives us 1 1/3. 404 00:35:24,520 --> 00:35:28,432 So we've seen how we can extend the technique of finding the 405 00:35:28,432 --> 00:35:29,736 area under a curve. 406 00:35:30,570 --> 00:35:36,290 Between the X axis on between two given ordinance to finding 407 00:35:36,290 --> 00:35:40,970 the area between two given curves. Let's take another 408 00:35:40,970 --> 00:35:47,210 example. This time, let's use the function Y equals sign X. We 409 00:35:47,210 --> 00:35:53,970 better define a range of values of X, so I'm going to take 410 00:35:53,970 --> 00:35:57,610 X to be between North and pie. 411 00:35:58,260 --> 00:36:05,456 Let's say I want the area that's cut off from that by the line 412 00:36:05,456 --> 00:36:11,624 Y equals one over Route 2. So a little sketch as always. 413 00:36:12,630 --> 00:36:16,038 We're finding these areas. It's important to know 414 00:36:16,038 --> 00:36:17,316 where they are. 415 00:36:18,470 --> 00:36:22,736 So why equals sign X between North and pie? 416 00:36:22,736 --> 00:36:25,106 So it's going to look. 417 00:36:26,820 --> 00:36:34,024 Like that? Naspi Nazira Why is equal to one over Route 2? Well, 418 00:36:34,024 --> 00:36:35,920 that's a straight line. 419 00:36:36,900 --> 00:36:39,040 Somewhere, Let's say across. 420 00:36:39,790 --> 00:36:42,450 There, so this is the area that 421 00:36:42,450 --> 00:36:47,300 we're after. Area between the line and that curve. What we 422 00:36:47,300 --> 00:36:50,450 need to know. What are these values of X? 423 00:36:51,540 --> 00:36:58,185 Well, why is equal to one over Route 2 and Y is equal to sign 424 00:36:58,185 --> 00:37:04,387 X, so one over Route 2 is equal to sign X where they meet? 425 00:37:05,180 --> 00:37:08,816 So the values of X will be the solutions to this equation. 426 00:37:09,390 --> 00:37:14,710 Well, one of the values we do know is that when sign X equals 427 00:37:14,710 --> 00:37:18,890 one over route 2X must be equal to π over 4. 428 00:37:19,510 --> 00:37:23,494 I working in radians, so I've got to give the answer in 429 00:37:23,494 --> 00:37:26,482 radians, but the equivalent answer in degrees is 45. 430 00:37:27,850 --> 00:37:34,290 So that's that one there and the one there is 3 Pi over 4. 431 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 432 00:37:41,730 --> 00:37:46,075 area under the sign curve between these two values. I can 433 00:37:46,075 --> 00:37:50,815 find the area of this rectangle and take it away. That will 434 00:37:50,815 --> 00:37:55,555 leave me that area there. Just let me put these answers in. 435 00:37:56,180 --> 00:38:02,242 I found their and there, so it's first of all find the area under 436 00:38:02,242 --> 00:38:05,090 this curve. So the area. 437 00:38:06,770 --> 00:38:13,840 Under The curve is equal to 438 00:38:13,840 --> 00:38:20,352 the integral between pie by four and three 439 00:38:20,352 --> 00:38:27,280 Pi 4. Of sign XDX equal so 440 00:38:27,280 --> 00:38:30,784 I have to integrate 441 00:38:30,784 --> 00:38:38,038 sign X. Well, the integral of sine X is minus Cos 442 00:38:38,038 --> 00:38:45,912 X. To evaluate that between pie by four and 443 00:38:45,912 --> 00:38:48,309 three Pi 4. 444 00:38:49,020 --> 00:38:56,370 So we put in the upper limit first minus the cause of three π 445 00:38:56,370 --> 00:39:00,360 by 4. That's the first bracket. 446 00:39:01,110 --> 00:39:07,478 Minus minus the cause of Π by 4. 447 00:39:08,250 --> 00:39:15,495 Now cause of three π by 4 is minus one over Route 2 and that 448 00:39:15,495 --> 00:39:22,257 minus sign gives me plus one over Route 2. The cause of Π by 449 00:39:22,257 --> 00:39:28,536 4 is one over Route 2, minus minus gives me plus one over 450 00:39:28,536 --> 00:39:35,298 Route 2, so that is 2 over Route 2, which is just Route 2. 451 00:39:36,720 --> 00:39:40,290 So I now know this area here. 452 00:39:40,840 --> 00:39:44,371 I need to do is calculate the area of this rectangle. 453 00:39:45,560 --> 00:39:50,188 Then I can take it away from that. Well, it is a rectangle. 454 00:39:50,188 --> 00:39:52,680 Its height is one over Route 2. 455 00:39:53,260 --> 00:39:55,770 So area. 456 00:39:57,020 --> 00:40:02,971 Equals one over Route 2. That's the height of that rectangle. 457 00:40:02,971 --> 00:40:05,676 Times by this base it's. 458 00:40:06,240 --> 00:40:12,932 Width, well, it runs from Π by 4, three π by 4, so the 459 00:40:12,932 --> 00:40:19,624 difference is 2π by 4. In other words, pie by two, so the area 460 00:40:19,624 --> 00:40:24,404 of the rectangle is π over 2 times Route 2. 461 00:40:25,390 --> 00:40:31,426 So the total area that I want is the difference between these 462 00:40:31,426 --> 00:40:34,947 two π over 2 Route 2 and 463 00:40:34,947 --> 00:40:38,290 Route 2. In the area. 464 00:40:40,140 --> 00:40:48,140 Equals Route 2 - π over 2 Route 465 00:40:48,140 --> 00:40:54,244 2. Put all that over a common denominator of two Route 2, 466 00:40:54,244 --> 00:40:56,100 which makes this 4. 467 00:40:56,770 --> 00:41:04,538 Minus pie. Leave that as that. That's the 468 00:41:04,538 --> 00:41:06,452 area that's contained. 469 00:41:06,970 --> 00:41:12,154 In there, between the straight line and the curve. 470 00:41:13,030 --> 00:41:20,120 I want now just to go back to the example 471 00:41:20,120 --> 00:41:27,210 where we were looking at the area that was between 472 00:41:27,210 --> 00:41:29,337 these two curves. 473 00:41:29,350 --> 00:41:32,822 If you remember we 474 00:41:32,822 --> 00:41:36,560 had. This sketch. 475 00:41:37,280 --> 00:41:39,490 The curve. 476 00:41:40,670 --> 00:41:46,158 So that was the 477 00:41:46,158 --> 00:41:48,902 picture that 478 00:41:48,902 --> 00:41:54,850 we had. And we calculated the area 479 00:41:54,850 --> 00:41:57,080 between the two curves here. 480 00:41:58,040 --> 00:42:00,470 By finding the area underneath. 481 00:42:01,080 --> 00:42:03,330 X times by 3 minus X. 482 00:42:03,890 --> 00:42:07,550 Then the area underneath the line Y equals X and 483 00:42:07,550 --> 00:42:08,648 taking them away. 484 00:42:11,020 --> 00:42:16,129 Another way of looking at this might be to go back to 1st 485 00:42:16,129 --> 00:42:20,452 principles if you remember when we did that, we took little 486 00:42:20,452 --> 00:42:23,989 rectangular strips which were roughly of height Y and 487 00:42:23,989 --> 00:42:27,919 thickness Delta X. Well, couldn't we do the same here, 488 00:42:27,919 --> 00:42:31,456 have little rectangular strips for that. Not quite rectangles, 489 00:42:31,456 --> 00:42:36,565 but they are thin strips and their why bit, so to speak would 490 00:42:36,565 --> 00:42:40,495 be the difference between these two values of Y here. 491 00:42:41,390 --> 00:42:47,495 So if we were to call this, why one, let's say, and this one Y 492 00:42:47,495 --> 00:42:52,379 2, then might not the area that we're looking for the area 493 00:42:52,379 --> 00:42:54,821 between these two curves Y1 and 494 00:42:54,821 --> 00:43:00,154 Y2? Between these ordinates Here Let's call them A&B. 495 00:43:00,780 --> 00:43:03,292 Might not we be able to work it 496 00:43:03,292 --> 00:43:05,889 out? By doing that. 497 00:43:06,560 --> 00:43:11,721 In other words, take the upper curve and call it why one type 498 00:43:11,721 --> 00:43:14,897 the lower curve and call it Y 2. 499 00:43:15,580 --> 00:43:19,378 Subtract them and then integrate between the required limits. 500 00:43:19,378 --> 00:43:24,864 Well, let's see if this will give us the same answer as we 501 00:43:24,864 --> 00:43:26,974 had in the previous case. 502 00:43:27,760 --> 00:43:32,128 So our limits here are not 2. 503 00:43:33,250 --> 00:43:35,326 And why one is this one? 504 00:43:36,040 --> 00:43:41,227 Let's multiply it out first, X times by three, is 3X and X 505 00:43:41,227 --> 00:43:44,020 times Y minus X is minus X 506 00:43:44,020 --> 00:43:50,910 squared. Take away why two? So we're taking away X and 507 00:43:50,910 --> 00:43:57,334 integrating with respect to X. So if 2X minus X, sorry, 508 00:43:57,334 --> 00:44:04,342 3X minus X. That gives us 2X minus X squared to be 509 00:44:04,342 --> 00:44:10,182 evaluated between Norton two with respect to X. So let's 510 00:44:10,182 --> 00:44:12,518 carry out this integration. 511 00:44:13,790 --> 00:44:21,322 The integral of X is X squared over 2, so that's two times X 512 00:44:21,322 --> 00:44:27,778 squared over 2. The integral of X squared is X cubed over 513 00:44:27,778 --> 00:44:34,234 three between North and two. I can cancel it to their and 514 00:44:34,234 --> 00:44:35,310 their substituting. 515 00:44:36,530 --> 00:44:40,070 2 Twos are 4. 516 00:44:40,820 --> 00:44:41,850 Minus. 517 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 518 00:44:50,710 --> 00:44:54,070 in that second bracket is just 0 519 00:44:54,070 --> 00:44:56,845 equals. 4 520 00:44:56,845 --> 00:45:03,870 minus. 2 and 2/3 which just gives me one and third 521 00:45:03,870 --> 00:45:10,669 units of area, which is exactly what we had last time. So this. 522 00:45:12,000 --> 00:45:15,300 Gives us a convenient formula. 523 00:45:15,950 --> 00:45:21,274 Being able to workout the area that is caught between two 524 00:45:21,274 --> 00:45:27,566 curves, we call the upper curve. Why won the lower curve Y two 525 00:45:27,566 --> 00:45:32,406 and we subtract them and then integrate between the ordinance 526 00:45:32,406 --> 00:45:34,826 of the points of intersection?