1 00:00:01,670 --> 00:00:06,990 Going to have a look in this video and how we might be able 2 00:00:06,990 --> 00:00:09,650 to find the area under a curve. 3 00:00:10,310 --> 00:00:15,640 But the really important bit of this video is to do with looking 4 00:00:15,640 --> 00:00:20,970 at integration as summation, IE the idea that we can add up lots 5 00:00:20,970 --> 00:00:25,890 of little bits of something in order to arrive at a concrete 6 00:00:25,890 --> 00:00:27,120 block of something. 7 00:00:27,970 --> 00:00:33,052 Let's begin by having a look at something that's probably quite 8 00:00:33,052 --> 00:00:39,686 familiar. It's a problem that's often set GCSE, another. 9 00:00:40,900 --> 00:00:42,640 Levels of Maths. 10 00:00:43,730 --> 00:00:45,870 Supposing you were asked. 11 00:00:47,230 --> 00:00:48,499 Find the area. 12 00:00:49,470 --> 00:00:52,142 Of this shape. How might we do 13 00:00:52,142 --> 00:00:56,510 it? One way that we would do it perhaps is the divided up. 14 00:00:58,180 --> 00:01:01,948 Into a series of shapes like 15 00:01:01,948 --> 00:01:07,650 that. That's one way each of these shapes is a standard area. 16 00:01:07,650 --> 00:01:11,550 A triangle or square or rectangle, whatever. And we can 17 00:01:11,550 --> 00:01:17,010 workout what its area is. Then we can get the area of the whole 18 00:01:17,010 --> 00:01:21,690 shape by adding together the area of each of these shapes. So 19 00:01:21,690 --> 00:01:26,760 from a shape that had no defined formula for its area by breaking 20 00:01:26,760 --> 00:01:31,440 it up into more recognizable shapes, we were able to sort out 21 00:01:31,440 --> 00:01:33,000 what it's area was. 22 00:01:34,070 --> 00:01:38,219 OK, can we take that a little bit further? 23 00:01:39,430 --> 00:01:42,730 For a circle, there is a formula for the area 24 00:01:42,730 --> 00:01:44,380 a equals π R squared. 25 00:01:45,480 --> 00:01:49,080 But if there was no formula, how might we proceed? 26 00:01:51,120 --> 00:01:55,735 One way might be to place a grid of squares over the circle, 27 00:01:55,735 --> 00:01:59,640 workout the area of each square and add them all together. 28 00:02:01,040 --> 00:02:05,027 However, some of the squares will not coincide completely 29 00:02:05,027 --> 00:02:06,356 with the circle. 30 00:02:06,930 --> 00:02:10,846 Thus, if we add together all the squares that contain any 31 00:02:10,846 --> 00:02:14,762 part of the circle, we will arrive at an over estimate 32 00:02:14,762 --> 00:02:16,898 of the area of the circle. 33 00:02:19,440 --> 00:02:24,115 We could try to attempt to correct this by ignoring all 34 00:02:24,115 --> 00:02:27,940 those squares that do not completely coincide with the 35 00:02:27,940 --> 00:02:32,074 circle. This would mean ignoring some of the area 36 00:02:32,074 --> 00:02:34,978 of the circle and we would gain an 37 00:02:34,978 --> 00:02:37,519 underestimate of the area of the circle. 38 00:02:40,120 --> 00:02:46,035 However, we would now have two limits on the area of the circle 39 00:02:46,035 --> 00:02:47,400 and upper one. 40 00:02:47,990 --> 00:02:48,968 Under lower one. 41 00:02:50,080 --> 00:02:54,964 To trap the area of the circle ever tighter between an upper 42 00:02:54,964 --> 00:02:59,848 limit and a lower limit, we would need to make the squares 43 00:02:59,848 --> 00:03:04,732 in the grid smaller and smaller so that the parts that were 44 00:03:04,732 --> 00:03:08,395 included or discarded were in total smaller and smaller. 45 00:03:12,570 --> 00:03:18,258 Now we're going to do is take that idea. 46 00:03:18,260 --> 00:03:20,108 And use it. 47 00:03:20,680 --> 00:03:27,628 To attempt to find the area underneath a curve. 48 00:03:28,160 --> 00:03:33,269 So what we've got here is a curve, and we're looking to find 49 00:03:33,269 --> 00:03:35,627 is can we find the area? 50 00:03:36,340 --> 00:03:42,151 Underneath the curve, so can we find in some way this area and 51 00:03:42,151 --> 00:03:47,962 can we use the same sorts of ideas that we just spoke about 52 00:03:47,962 --> 00:03:49,303 with the circle? 53 00:03:51,360 --> 00:03:57,205 Well, let's take a very specific example. 54 00:03:58,320 --> 00:03:59,936 I'm going to have a look at the 55 00:03:59,936 --> 00:04:01,750 curve. Why? 56 00:04:02,930 --> 00:04:05,189 Equals X squared. 57 00:04:08,560 --> 00:04:12,226 Going to draw its graph between 58 00:04:12,226 --> 00:04:19,539 a. And. A square so this end point here 59 00:04:19,539 --> 00:04:22,167 is the point a. 60 00:04:22,760 --> 00:04:24,930 A squared. 61 00:04:26,280 --> 00:04:30,120 Now. How can we divide 62 00:04:30,120 --> 00:04:35,880 this up? Well, I just put a limit to where we want to be. 63 00:04:36,510 --> 00:04:42,474 One way might be to say, OK, let's divide this in half, so we 64 00:04:42,474 --> 00:04:44,178 take a over 2. 65 00:04:44,880 --> 00:04:46,608 And let's take. 66 00:04:47,830 --> 00:04:50,498 Some rectangles so there. 67 00:04:54,280 --> 00:04:55,790 And there. 68 00:04:57,800 --> 00:05:00,744 So there are two rectangles. This one here. 69 00:05:02,180 --> 00:05:03,420 And this big one. 70 00:05:04,030 --> 00:05:11,246 Here. Now, at this point here on the curve, X is equal to 71 00:05:11,246 --> 00:05:17,850 a over 2 where dealing with the curve Y equals X squared, so 72 00:05:17,850 --> 00:05:24,454 that we know what this point is. Here, it's a over 2A squared 73 00:05:24,454 --> 00:05:30,986 over 4. So we know the height of this rectangle. 74 00:05:30,986 --> 00:05:37,730 It's A squared over 4, so I guess an approximation at this 75 00:05:37,730 --> 00:05:44,474 area underneath the curve. The area a. We know that a is 76 00:05:44,474 --> 00:05:51,218 in fact going to be less than the area of this rectangle. 77 00:05:52,310 --> 00:05:59,978 Which is a over two times a squared over 4 plus the 78 00:05:59,978 --> 00:06:05,729 area of this rectangle, which is a over 2. 79 00:06:06,460 --> 00:06:12,374 Times A. Squared because that's the height of that rectangle. 80 00:06:13,710 --> 00:06:19,478 So we've got this area a what we've got is an upper limit for 81 00:06:19,478 --> 00:06:22,774 it. We know it must be less than 82 00:06:22,774 --> 00:06:28,440 that. OK, if that's the case, then what about a lower 83 00:06:28,440 --> 00:06:32,598 limit? One way might be to take other rectangles. 84 00:06:34,690 --> 00:06:38,490 Such as one down here of zero height. 85 00:06:40,770 --> 00:06:42,399 And this rectangle. 86 00:06:43,290 --> 00:06:43,820 Here. 87 00:06:45,110 --> 00:06:50,414 Where again, this height is going to be a squared over 4 and 88 00:06:50,414 --> 00:06:55,718 we know very clearly that that means that this area a must be 89 00:06:55,718 --> 00:06:59,798 greater than zero. That's the area of this rectangle plus. 90 00:07:00,380 --> 00:07:06,148 A over two times a squared over 4. 91 00:07:07,400 --> 00:07:12,008 Now what did we say about a circle? To get a better 92 00:07:12,008 --> 00:07:13,928 approximation, we said we take 93 00:07:13,928 --> 00:07:16,060 smaller. And smaller. 94 00:07:16,770 --> 00:07:21,112 Squares and put it on the circle. So in this case, what we 95 00:07:21,112 --> 00:07:25,120 want to do is take rectangles which are smaller and smaller in 96 00:07:25,120 --> 00:07:26,456 terms of their width. 97 00:07:27,170 --> 00:07:34,766 So let's draw this picture again and this time let's divide these 98 00:07:34,766 --> 00:07:37,931 two segments into half again. 99 00:07:38,010 --> 00:07:41,130 So let's draw. 100 00:07:41,670 --> 00:07:42,840 Our curve. 101 00:07:43,910 --> 00:07:44,640 Gain. 102 00:07:48,520 --> 00:07:51,238 Take a there. 103 00:07:51,790 --> 00:07:54,280 So this will be a squared. 104 00:07:55,060 --> 00:08:00,044 There for the curve Y equals X squared. 105 00:08:01,860 --> 00:08:07,034 This area that we're trying to find and we said that we would 106 00:08:07,034 --> 00:08:09,180 divide. The base. 107 00:08:10,060 --> 00:08:16,798 Into half And then in two half again. 108 00:08:17,750 --> 00:08:18,780 And. 109 00:08:20,800 --> 00:08:23,470 Again there. 110 00:08:24,760 --> 00:08:28,288 Now a over 2 is half 111 00:08:28,288 --> 00:08:33,440 way. But I want to keep this denominator of four if I can, so 112 00:08:33,440 --> 00:08:37,940 I'm not going to call that a over 2. I'm going to call it to 113 00:08:37,940 --> 00:08:42,440 a over 4 and you'll see why I want to do that in a moment. 114 00:08:44,210 --> 00:08:47,380 Now, where are our rectangles? 115 00:08:48,920 --> 00:08:51,790 Well, there's a rectangle there. 116 00:08:57,060 --> 00:08:58,239 There's another one. 117 00:08:58,810 --> 00:08:59,550 There. 118 00:09:03,700 --> 00:09:05,728 There's another one there. 119 00:09:06,630 --> 00:09:09,618 And the final one. 120 00:09:10,680 --> 00:09:14,816 There, so here we've got four rectangles, all of which contain 121 00:09:14,816 --> 00:09:18,952 area that is underneath the curve, but all of which contain 122 00:09:18,952 --> 00:09:23,840 extra bits of area. So when we write down the areas of these 123 00:09:23,840 --> 00:09:28,728 rectangles and add them up, we will get an area that is greater 124 00:09:28,728 --> 00:09:33,240 than the area under the curve. So let's just write down the 125 00:09:33,240 --> 00:09:36,624 coordinates of these points, because the coordinates of these 126 00:09:36,624 --> 00:09:40,760 points will be the Heights of the rectangles, so the Y 127 00:09:40,760 --> 00:09:43,542 coordinate. There will give us the height of 128 00:09:43,542 --> 00:09:45,894 that rectangle, the Y coordinate their the 129 00:09:45,894 --> 00:09:48,246 height of that rectangle and so on. 130 00:09:49,430 --> 00:09:56,668 So this point here has an X coordinate of X over 4 or a 131 00:09:56,668 --> 00:10:03,906 over 4 in this case, so that's a over 4. It will be a 132 00:10:03,906 --> 00:10:05,457 squared over 16. 133 00:10:06,250 --> 00:10:13,190 This one. Has an X coordinate of two 8 over 4. 134 00:10:14,490 --> 00:10:21,546 So it will have a Y coordinate of 135 00:10:21,546 --> 00:10:25,074 2 squared, A squared 136 00:10:25,074 --> 00:10:32,460 over 16. This one has an X coordinate of three, 137 00:10:32,460 --> 00:10:38,895 a over 4, and so it will have a Y coordinate. 138 00:10:39,730 --> 00:10:47,110 Of three squared, A squared over 60 and finally this 139 00:10:47,110 --> 00:10:49,324 one will be. 140 00:10:51,030 --> 00:10:58,842 It's X coordinate is A and so it will be a square. 141 00:10:59,520 --> 00:11:06,252 Another way of writing that to keep the pattern going would be 142 00:11:06,252 --> 00:11:14,106 to call it 4A over 4 so it will be 16 A squared over 143 00:11:14,106 --> 00:11:20,910 60. OK, let's write down the area of each of these 144 00:11:20,910 --> 00:11:28,190 rectangles. The width is a over 4, so we'll have a over four 145 00:11:28,190 --> 00:11:33,230 times its height, which is a squared over 16. 146 00:11:34,400 --> 00:11:40,042 Plus the area of this rectangle, its width is again a over 4 is 147 00:11:40,042 --> 00:11:42,460 that all of the same width? 148 00:11:43,000 --> 00:11:47,080 Times its height, which this time is 2. 149 00:11:47,800 --> 00:11:50,932 Squared A squared 150 00:11:50,932 --> 00:11:55,970 over 16. Plus 151 00:11:57,540 --> 00:12:00,244 And now we want to move to this 152 00:12:00,244 --> 00:12:03,940 one. Its width is again a 153 00:12:03,940 --> 00:12:10,886 over 4. Its height is 3 squared, A squared over 154 00:12:10,886 --> 00:12:18,470 16 and then finally the end one again. Its width is a 155 00:12:18,470 --> 00:12:19,734 over 4. 156 00:12:20,770 --> 00:12:26,672 Times its height, which is now I've written 16 and of course we 157 00:12:26,672 --> 00:12:33,028 know that 16 is 4 squared, A squared over 60 and that this is 158 00:12:33,028 --> 00:12:37,568 bigger than a. We know that is bigger than A. 159 00:12:38,220 --> 00:12:42,854 Let me just have a look at this. Can we see anything here that 160 00:12:42,854 --> 00:12:47,157 might be of interest to us? Well, first of all, we can see 161 00:12:47,157 --> 00:12:49,143 there's a common factor of a 162 00:12:49,143 --> 00:12:56,710 over 4. There's also a common factor of a 16 and a squared, so 163 00:12:56,710 --> 00:12:59,650 let's write that as a squared 164 00:12:59,650 --> 00:13:06,758 over 60. And what are we left with 1 + 165 00:13:06,758 --> 00:13:13,768 2 squared, +3 squared +4 squared is bigger than a. 166 00:13:14,800 --> 00:13:21,892 Now, what about our lower limit? Well, there will be a rectangle 167 00:13:21,892 --> 00:13:24,256 here of zero height. 168 00:13:24,950 --> 00:13:27,130 There will be another rectangle 169 00:13:27,130 --> 00:13:34,598 here. Another one here and another one here. 170 00:13:35,140 --> 00:13:39,456 So we're looking at the area of that rectangle. The area of that 171 00:13:39,456 --> 00:13:43,108 one, that one, and that one where the upper boundaries of 172 00:13:43,108 --> 00:13:44,768 the rectangles are marked in 173 00:13:44,768 --> 00:13:51,556 red. What does that area come to where we know that it will be 174 00:13:51,556 --> 00:13:56,860 less than the area under the curve and it will be 0? 175 00:13:57,680 --> 00:14:05,504 Plus a over 4 times by now that's its height, the Y 176 00:14:05,504 --> 00:14:09,416 value there a squared over 16. 177 00:14:10,870 --> 00:14:17,562 Plus a over 4 the width of that one times its height, which is 178 00:14:17,562 --> 00:14:25,064 this one. 2 squared A squared over 16 plus the last one where 179 00:14:25,064 --> 00:14:32,120 again we have a over 4 for the width and for the height we 180 00:14:32,120 --> 00:14:37,664 have 3 squared. A squared over 16. Three squared, A squared 181 00:14:37,664 --> 00:14:44,216 over 16 and so a is greater than like we look what we've 182 00:14:44,216 --> 00:14:49,760 got. Again, we've got this common factor of a over 4. 183 00:14:49,800 --> 00:14:55,597 Again, we've got this common factor of A squared over 16. 184 00:14:56,360 --> 00:15:03,459 Bracket. And 0 + 1 + 2 squared, 185 00:15:03,459 --> 00:15:04,993 +3 squared. 186 00:15:05,680 --> 00:15:10,598 OK. We could go on and do this again. 187 00:15:12,100 --> 00:15:17,492 But what we want is to try and do it. In general, if we can. So 188 00:15:17,492 --> 00:15:22,547 what I'm going to do is to take the same piece of area, the same 189 00:15:22,547 --> 00:15:26,928 piece of curve, the same piece of area I'm going to divide the 190 00:15:26,928 --> 00:15:31,646 X axis into N equal parts. Here I had it divided into 4 equal 191 00:15:31,646 --> 00:15:36,027 parts. Now I'm going to divide it into N equal parts, and in 192 00:15:36,027 --> 00:15:41,082 doing that I'm going to look at only a part of it. But the idea 193 00:15:41,082 --> 00:15:43,104 is going to be the same. 194 00:15:43,110 --> 00:15:47,180 And the same patterns that we picked out here, namely this 195 00:15:47,180 --> 00:15:51,990 sums of squares of the integers is going to be there again for 196 00:15:51,990 --> 00:15:53,100 us to see. 197 00:15:53,990 --> 00:15:55,720 So. 198 00:15:57,030 --> 00:15:59,494 Here's a section of the X axis. 199 00:16:02,010 --> 00:16:03,340 A piece of the curve. 200 00:16:04,180 --> 00:16:06,328 Interested in the. 201 00:16:06,840 --> 00:16:13,164 Off strip, so let's just mark off a few of these strips. 202 00:16:15,750 --> 00:16:22,428 So there are a few of the strips now 203 00:16:22,428 --> 00:16:24,654 remember each one. 204 00:16:26,190 --> 00:16:28,478 Is the same width. 205 00:16:29,050 --> 00:16:35,658 The total length of the X axis we were looking at was A and 206 00:16:35,658 --> 00:16:42,266 there are N strips, so the width of each of these strips is a 207 00:16:42,266 --> 00:16:47,930 over N, so each one has a width of a over N. 208 00:16:48,800 --> 00:16:54,080 Now that helps us count because the arts trip. Let's say it's 209 00:16:54,080 --> 00:17:00,240 this one here must be our times a over N for its X coordinate 210 00:17:00,240 --> 00:17:07,860 their. And here for its X coordinate our minus one times 211 00:17:07,860 --> 00:17:09,852 a over N. 212 00:17:11,260 --> 00:17:17,436 Let's build up the rectangle on this strip. 213 00:17:18,010 --> 00:17:22,760 So there's the rectangle now what's the area of that 214 00:17:22,760 --> 00:17:25,610 rectangle? The area of the rectangle? 215 00:17:27,570 --> 00:17:32,530 Equals well its width, which we know is a over 216 00:17:32,530 --> 00:17:35,010 N times by its height. 217 00:17:36,660 --> 00:17:40,040 Which is this? Why coordinate 218 00:17:40,040 --> 00:17:46,186 here? And remember, this is the curve Y equals X 219 00:17:46,186 --> 00:17:50,458 squared, and so the coordinates of this point 220 00:17:50,458 --> 00:17:56,866 are a over N and then we must square it R-squared A 221 00:17:56,866 --> 00:18:03,274 squared over end square. So the area is a over N times 222 00:18:03,274 --> 00:18:07,546 that height R-squared, A squared over N squared. 223 00:18:08,640 --> 00:18:10,320 OK. 224 00:18:11,570 --> 00:18:18,430 So this is the area of the Strip and it's bigger than this area 225 00:18:18,430 --> 00:18:22,350 underneath the curve. Let's call that Delta A. 226 00:18:23,010 --> 00:18:27,798 Let's just tidy this up a little bit. We've gotten a cubed. 227 00:18:29,500 --> 00:18:35,620 And we've gotten N cubed, and we got this thing R-squared and 228 00:18:35,620 --> 00:18:38,680 that is bigger than Delta A. 229 00:18:39,610 --> 00:18:45,756 And what we want is the whole area. So we want all of these 230 00:18:45,756 --> 00:18:52,780 Delta Azan. We want to add them all up. So we need to add up all 231 00:18:52,780 --> 00:18:58,926 of these but that will still be bigger than the area as a whole. 232 00:18:58,926 --> 00:19:05,072 So let's add them all up Sigma. That means some add up a cubed 233 00:19:05,072 --> 00:19:09,901 over N cubed times by R-squared and we'll have from our. 234 00:19:09,940 --> 00:19:14,428 Equals 1 up to an. Added all those up that will 235 00:19:14,428 --> 00:19:18,100 still be bigger than the area that we're after. 236 00:19:19,300 --> 00:19:23,428 Now let's look at the other rectangle. This 237 00:19:23,428 --> 00:19:25,492 one the lower one. 238 00:19:27,890 --> 00:19:34,390 Now we can say that the height of this one depends upon this 239 00:19:34,390 --> 00:19:38,890 point here and its coordinate is R minus one. 240 00:19:39,470 --> 00:19:46,130 Times a over N for X and so it's Y coordinate is R minus 1 241 00:19:46,130 --> 00:19:49,864 squared. A squared over N 242 00:19:49,864 --> 00:19:57,315 squared. So what's the area of this smaller 243 00:19:57,315 --> 00:20:04,602 rectangle area? Equals well, the width is still a over N, 244 00:20:04,602 --> 00:20:10,938 but the height is different. Are minus one all squared A squared 245 00:20:10,938 --> 00:20:14,634 over N squared and this will be 246 00:20:14,634 --> 00:20:19,458 less than? That area Delta A. 247 00:20:20,500 --> 00:20:26,372 We now need to Add all of these up to get an idea of the lower 248 00:20:26,372 --> 00:20:31,510 limit of the area. So let's do that. A will be greater than the 249 00:20:31,510 --> 00:20:37,015 sum of an. Let's tidy this one up in the same way that we did 250 00:20:37,015 --> 00:20:41,786 that we've gotten a cubed formed by the A Times by a squared. 251 00:20:42,390 --> 00:20:47,746 So that will give us the A cubed and we've got an end 252 00:20:47,746 --> 00:20:52,278 times by N squared which will give us N cubed. And 253 00:20:52,278 --> 00:20:56,810 then we've got this R minus one all squared and will 254 00:20:56,810 --> 00:21:00,518 some that from our equals 1 up to N. 255 00:21:01,630 --> 00:21:07,696 So now let's write down these two inequalities together. 256 00:21:07,700 --> 00:21:11,388 So we've got Sigma. 257 00:21:12,120 --> 00:21:15,580 R equals 1 to end. 258 00:21:16,390 --> 00:21:22,194 Of a. Cubed of 259 00:21:22,194 --> 00:21:27,745 N cubed. Times 260 00:21:27,745 --> 00:21:34,890 by. All squared is greater than 261 00:21:34,890 --> 00:21:41,148 a. Is greater than the sum from R equals 262 00:21:41,148 --> 00:21:47,814 12 N of a cubed over N cubed times by R 263 00:21:47,814 --> 00:21:50,238 minus one or squared? 264 00:21:52,660 --> 00:21:59,184 OK. This a cubed over NQ bisa common factor in each one 265 00:21:59,184 --> 00:22:01,908 so we can take that out. 266 00:22:23,260 --> 00:22:27,964 Now question is can we add up the sums of the squares of the 267 00:22:27,964 --> 00:22:31,324 integers 1 squared +2 squared, +3 squared, +4 squared, +5 268 00:22:31,324 --> 00:22:35,020 squared? Remember, we had that before when we were looking at 269 00:22:35,020 --> 00:22:39,388 dividing it up into quarters and I pointed out that we had one 270 00:22:39,388 --> 00:22:43,084 squared +2 squared, +3 squared, +4 squared, and which see that 271 00:22:43,084 --> 00:22:47,452 again, and this is it. Here we can see quite clearly 1 squared, 272 00:22:47,452 --> 00:22:51,484 +2 squared, +3 squared, +4 squared, and so on, and the same 273 00:22:51,484 --> 00:22:53,836 over here. Can we add those up? 274 00:22:53,910 --> 00:22:58,837 Well, yes, there is a formula that is a formula that tells us 275 00:22:58,837 --> 00:23:01,490 what the sum of the first end 276 00:23:01,490 --> 00:23:08,970 squares is. And the sum of the first N squares from R 277 00:23:08,970 --> 00:23:11,438 equals 1 to N. 278 00:23:12,450 --> 00:23:14,240 Is N. 279 00:23:15,540 --> 00:23:17,160 N plus one. 280 00:23:18,500 --> 00:23:24,464 2 N plus one all over 6 looks a bit complicated, but that's what 281 00:23:24,464 --> 00:23:30,854 it is and in fact it's easy to work with, so let's put that in 282 00:23:30,854 --> 00:23:32,984 here instead of this one. 283 00:23:34,490 --> 00:23:39,250 So we get a cubed over and cubed. 284 00:23:40,450 --> 00:23:47,542 Times by N over times by M Plus One times by 2 285 00:23:47,542 --> 00:23:50,497 N plus one over 6. 286 00:23:52,160 --> 00:23:57,841 Is bigger than a is bigger than now? This is if you like 287 00:23:57,841 --> 00:24:03,522 R minus one. So in a sense, we're always one less than N. 288 00:24:03,522 --> 00:24:09,640 So we want an minus one times N times 2 N minus one, which 289 00:24:09,640 --> 00:24:16,632 is what we get when we replace N by two N all over 6 times by 290 00:24:16,632 --> 00:24:19,254 the A cubed over and cubed. 291 00:24:20,470 --> 00:24:25,636 Now in each case we've got this a cubed over and cubed times by 292 00:24:25,636 --> 00:24:30,802 6, and in fact we've gotten end. Look here on either side that we 293 00:24:30,802 --> 00:24:35,968 can cancel so we can cancel that we get a square and cancel that 294 00:24:35,968 --> 00:24:41,503 and get a square. So I'm going to take the A cubed over 6 out. 295 00:24:41,550 --> 00:24:47,010 And I'm going to multiply out each of these brackets and put 296 00:24:47,010 --> 00:24:49,285 it all over N squared. 297 00:24:49,290 --> 00:24:54,936 So have a cubed over 6. 298 00:24:56,150 --> 00:25:03,280 All over and squared, greater than a greater than a 299 00:25:03,280 --> 00:25:10,410 cubed over 6, all over and squared. Let's just go 300 00:25:10,410 --> 00:25:15,650 back. And look at these brackets. This is N plus one 301 00:25:15,650 --> 00:25:20,928 times by 2 N plus one, so it will give us two and squared. 302 00:25:22,910 --> 00:25:29,499 N and two N which will be 3 N plus one. 303 00:25:29,500 --> 00:25:36,836 So we'll have two and squared plus 3N 304 00:25:36,836 --> 00:25:43,890 plus one. This bracket will give us a gain two and squared. 305 00:25:44,430 --> 00:25:51,290 But then minus two and minus sign will be minus 3 N and plus 306 00:25:51,290 --> 00:25:59,051 one. So again, two and squared minus 3 N plus one. Let's 307 00:25:59,051 --> 00:26:06,983 divide each term by end squared a over a cubed over 6. 308 00:26:08,210 --> 00:26:15,050 2 + 3 over N plus one over N squared that's dividing 309 00:26:15,050 --> 00:26:22,460 each term in turn by the end squared is greater than a is 310 00:26:22,460 --> 00:26:25,880 greater than a cubed over 6. 311 00:26:26,560 --> 00:26:33,570 2 - 3 over N plus one over and squared. 312 00:26:34,700 --> 00:26:37,989 Now it's taking us rather along time to arrive at this 313 00:26:37,989 --> 00:26:40,979 position, but we're here where we want to be now. 314 00:26:42,880 --> 00:26:49,750 Let's imagine that we let an become very, very large, 315 00:26:49,750 --> 00:26:56,620 so the number of strips where taking is infinite there 316 00:26:56,620 --> 00:27:02,034 infinitely thin. Infinitely thin because N is very very large. 317 00:27:02,034 --> 00:27:08,379 Now let's think what happens to a number like 3 over N when N is 318 00:27:08,379 --> 00:27:13,878 very very large. Well, it goes to zero and one over N squared 319 00:27:13,878 --> 00:27:19,800 goes to zero as those three over N this side and one over N 320 00:27:19,800 --> 00:27:21,492 squared at this side. 321 00:27:22,260 --> 00:27:28,981 So in fact the area a is trapped between a cubed 322 00:27:28,981 --> 00:27:36,313 over 6 times by two and a cubed over 6 times by 323 00:27:36,313 --> 00:27:41,812 two again. So therefore as N tends to Infinity. 324 00:27:43,260 --> 00:27:44,940 In the limit. 325 00:27:46,200 --> 00:27:53,484 A is equal to a cubed over 6 times by two, which 326 00:27:53,484 --> 00:28:00,161 is 1/3 of a cubed. It's trapped between these two bits 327 00:28:00,161 --> 00:28:03,803 that are the same here, because 328 00:28:03,803 --> 00:28:07,499 these bits. Disappear to 0. 329 00:28:08,590 --> 00:28:14,830 Now we've done that for one particular curve. 330 00:28:15,460 --> 00:28:21,972 To show that we can add up a lot of little bits to arrive at a 331 00:28:21,972 --> 00:28:26,042 finite answer. A lot of infinitely small bits to arrive 332 00:28:26,042 --> 00:28:32,554 at a finite answer. What we want to be able to do now is to do 333 00:28:32,554 --> 00:28:37,438 it in general to arrive at something that will work for all 334 00:28:37,438 --> 00:28:44,916 curves. So let's see if we can extend what we've 335 00:28:44,916 --> 00:28:48,500 done. To a more general curve. 336 00:28:49,490 --> 00:28:51,470 So here's our curve, let's say. 337 00:28:52,060 --> 00:28:59,860 And again, let's say that it runs from X equals North to 338 00:28:59,860 --> 00:29:01,810 X equals A. 339 00:29:02,380 --> 00:29:04,095 And what we're going to do is 340 00:29:04,095 --> 00:29:10,792 divide. The X axis up into very, very thin strips, each one of 341 00:29:10,792 --> 00:29:18,232 which is a width Delta X. So we have X there and the next mark 342 00:29:18,232 --> 00:29:25,176 on the X axis is X Plus Delta X. Delta axes are small positive 343 00:29:25,176 --> 00:29:26,664 amount of X. 344 00:29:27,910 --> 00:29:28,969 So we've got. 345 00:29:29,820 --> 00:29:33,570 Our. Ordinance up to the curve. 346 00:29:34,470 --> 00:29:37,476 And we can complete a rectangle. 347 00:29:38,970 --> 00:29:46,317 So. What are the coordinates of that point? Well, this is the 348 00:29:46,317 --> 00:29:53,428 curve Y equals F of X, and so the coordinates of this point 349 00:29:53,428 --> 00:29:55,069 are just XY. 350 00:29:56,160 --> 00:29:59,368 That's the point, P. 351 00:30:00,430 --> 00:30:03,006 What are the coordinates of this point? 352 00:30:04,350 --> 00:30:10,961 While the coordinates at this point, let's call it the point 353 00:30:10,961 --> 00:30:18,173 QRX plus Delta X&Y plus Delta Y, where Delta. Why is that 354 00:30:18,173 --> 00:30:19,976 small increment there? 355 00:30:21,780 --> 00:30:26,993 That we've added on as a result of the small increment that we 356 00:30:26,993 --> 00:30:28,196 made in X. 357 00:30:28,710 --> 00:30:34,606 Let's say that the area under the curve is Delta A. 358 00:30:35,160 --> 00:30:40,459 So this is the area Delta A. 359 00:30:41,750 --> 00:30:48,955 And the area Delta A is contained between the areas of 360 00:30:48,955 --> 00:30:53,699 two rectangles. The larger rectangle. 361 00:30:54,730 --> 00:30:57,964 Whose height is why plus Delta Y? 362 00:30:59,460 --> 00:31:03,906 And whose width is Delta X. 363 00:31:05,280 --> 00:31:11,064 And so that's it's area and the smaller rectangle. Let me just 364 00:31:11,064 --> 00:31:12,510 mark it's top. 365 00:31:13,250 --> 00:31:20,198 Their whose width is Delta X that whose height is just, why? 366 00:31:21,620 --> 00:31:27,400 So it's area will be Y times by Delta X. 367 00:31:28,560 --> 00:31:34,736 So if I want the whole of this area, I have to add up all of 368 00:31:34,736 --> 00:31:40,526 these little bits, which means I have to be able to add up all of 369 00:31:40,526 --> 00:31:44,041 these bits. So let's do 370 00:31:44,041 --> 00:31:51,030 that. A. The total area underneath this curve 371 00:31:51,030 --> 00:31:58,410 is the sum of all these little bits of area from X 372 00:31:58,410 --> 00:32:05,790 equals not up to X equals A and so therefore that area 373 00:32:05,790 --> 00:32:13,170 a is caught between the sum of all of these little bits 374 00:32:13,170 --> 00:32:16,860 from X equals not to X 375 00:32:16,860 --> 00:32:19,650 equals. A at the bottom end. 376 00:32:20,180 --> 00:32:27,161 And at the top end, the sum of all of these little bits 377 00:32:27,161 --> 00:32:34,142 from X equals nor two X equals a or Y plus Delta Y 378 00:32:34,142 --> 00:32:36,290 times by Delta X. 379 00:32:37,490 --> 00:32:44,738 OK, let me multiply out this bit so we have Sigma X 380 00:32:44,738 --> 00:32:51,986 equals North to a of Y times by Delta X Plus Delta 381 00:32:51,986 --> 00:32:59,234 Y times by Delta X greater than a greater than the sum 382 00:32:59,234 --> 00:33:06,482 from X equals not to a of Y times by Delta X. 383 00:33:07,340 --> 00:33:13,836 Now we're going to let Delta X tend to 0, just like we let N go 384 00:33:13,836 --> 00:33:18,708 off to Infinity, which made the strips under Y equals X squared 385 00:33:18,708 --> 00:33:19,926 come down to. 386 00:33:20,550 --> 00:33:24,788 A over N 10 down to a zero thickness, we're going to let 387 00:33:24,788 --> 00:33:29,026 Delta X come down to zero. Now what's going to happen when we 388 00:33:29,026 --> 00:33:32,612 do that? Will certainly here we've got two very, very small 389 00:33:32,612 --> 00:33:36,850 terms. We've got Delta. It's going to 0, so the change in Y 390 00:33:36,850 --> 00:33:41,414 is going to be very very small as well. So this term here is 391 00:33:41,414 --> 00:33:43,696 actually going to run off to 0. 392 00:33:44,360 --> 00:33:49,400 But when that happens, let's just cover that up. We see that 393 00:33:49,400 --> 00:33:53,180 a is trapped between two things which are identical. 394 00:33:53,710 --> 00:33:56,790 And So what we have is that. 395 00:33:57,440 --> 00:34:04,024 A is equal to the limit as Delta 396 00:34:04,024 --> 00:34:10,608 X tends to zero of the sum from 397 00:34:10,608 --> 00:34:17,192 X equals not a of Y Delta X. 398 00:34:18,130 --> 00:34:23,160 Now. With actually made an awful lot of assumptions in what's 399 00:34:23,160 --> 00:34:28,370 going on. And we've had to make them because this level it's 400 00:34:28,370 --> 00:34:32,759 very difficult to do anything other than make these kinds of 401 00:34:32,759 --> 00:34:36,172 assumptions. Let's just have a look at two of them. 402 00:34:36,970 --> 00:34:41,434 For instance, one of the things that we have assumed is that 403 00:34:41,434 --> 00:34:46,642 this curve is increasing all the time for all values of X. But as 404 00:34:46,642 --> 00:34:51,106 we know, curves can decrease as X increases, curves can wave up 405 00:34:51,106 --> 00:34:55,570 and down, they can go up and down as X increases. However, 406 00:34:55,570 --> 00:35:00,778 the results are still the same. We need to do it a little bit 407 00:35:00,778 --> 00:35:04,498 differently, but the results still come out the same. The 408 00:35:04,498 --> 00:35:08,590 other assumption that we're making is that in fact we can 409 00:35:08,590 --> 00:35:15,140 add up. In this form, lots of little bits and still get a 410 00:35:15,140 --> 00:35:16,700 definite finite answer. 411 00:35:17,420 --> 00:35:22,865 And there's a whole raft of pure mathematics, which assures us 412 00:35:22,865 --> 00:35:29,300 that these things do work, so here we are at this stage. Now 413 00:35:29,300 --> 00:35:34,745 this is very, very cumbersome language, and so in fact we 414 00:35:34,745 --> 00:35:41,675 choose to write this as a equals the integral from not a of YDX. 415 00:35:42,300 --> 00:35:49,636 A is the integral from North to a of Y with respect to X, 416 00:35:49,636 --> 00:35:53,828 and it's this notation which replaces that one. 417 00:35:54,570 --> 00:36:00,546 OK, supposing we actually want the area not from North to a, 418 00:36:00,546 --> 00:36:07,518 but between, let us say two limits A and B2 values of X and 419 00:36:07,518 --> 00:36:14,640 the curve. So we want, let's say, to find the area, let's 420 00:36:14,640 --> 00:36:17,946 have a curve here that goes 421 00:36:17,946 --> 00:36:20,560 between there. And. 422 00:36:21,590 --> 00:36:22,440 There. 423 00:36:23,610 --> 00:36:28,174 What would we want to do? Take that back to the Y axis there? 424 00:36:28,174 --> 00:36:30,130 What would we want to do? 425 00:36:30,630 --> 00:36:35,417 Well. This is the area that we're after. 426 00:36:36,380 --> 00:36:41,490 If we think about what that area is, it's the area from nought to 427 00:36:41,490 --> 00:36:46,747 be. Minus the area from North to A. 428 00:36:48,110 --> 00:36:51,725 So this area that we're 429 00:36:51,725 --> 00:36:59,104 wanting. Is equal to the area from North to be, which is 430 00:36:59,104 --> 00:37:04,456 that. Minus the area from North to 431 00:37:04,456 --> 00:37:07,260 a, which is that. 432 00:37:08,440 --> 00:37:12,230 Now it seems quite natural to write that as the. 433 00:37:13,050 --> 00:37:19,023 Integral from A to B of Y with respect to X. 434 00:37:20,570 --> 00:37:25,180 So we would evaluate whatever the result of this computation 435 00:37:25,180 --> 00:37:30,712 was at B and subtract from it the result of whatever the 436 00:37:30,712 --> 00:37:32,556 computation was at a. 437 00:37:32,580 --> 00:37:38,280 And so there we've seen how area can be represented as a 438 00:37:38,280 --> 00:37:43,505 summation and that that leads us to the traditional way of 439 00:37:43,505 --> 00:37:48,730 finding an area which is to integrate the equation of the 440 00:37:48,730 --> 00:37:50,630 curve and substituting limits. 441 00:37:51,200 --> 00:37:57,750 To conclude, this video will just look one further result. 442 00:37:58,420 --> 00:38:01,910 This result will enable us to actually calculate areas by 443 00:38:01,910 --> 00:38:06,098 chopping them up into convenient segments just the same way as we 444 00:38:06,098 --> 00:38:09,937 began. If you remember that shape rather like a dog were 445 00:38:09,937 --> 00:38:13,776 able to chop it up in a convenient segments, find the 446 00:38:13,776 --> 00:38:17,266 area of the separate segments added altogether. Well, this is 447 00:38:17,266 --> 00:38:22,152 this result is going to show us that we can do very much the 448 00:38:22,152 --> 00:38:23,897 same with a curve, so. 449 00:38:26,290 --> 00:38:32,230 Let's suppose that 4 hour curve. We have three ordinance X equals 450 00:38:32,230 --> 00:38:39,655 a X equals B&X equals C and that they are in fact in order of 451 00:38:39,655 --> 00:38:43,615 size a is less than B is less 452 00:38:43,615 --> 00:38:50,012 than C. Let's begin by asking ourselves what's the area 453 00:38:50,012 --> 00:38:57,149 contained by the curve Y equals F of X, and these two ordinate's 454 00:38:57,149 --> 00:39:04,835 and the X axis. Well, it's from A to see the integral of YDX. 455 00:39:05,670 --> 00:39:12,594 And we know that that is the integral from North. To see 456 00:39:12,594 --> 00:39:18,941 of YDX minus the integral from nought to a of YDX. 457 00:39:19,730 --> 00:39:22,462 Equals, well, 458 00:39:22,462 --> 00:39:28,450 rights. Introduce the ordinate be now. 459 00:39:29,410 --> 00:39:32,930 By taking away that. 460 00:39:33,800 --> 00:39:40,688 Area if I've taken away, let's add it on in order to 461 00:39:40,688 --> 00:39:43,558 keep the quality the same. 462 00:39:44,270 --> 00:39:50,067 Now if we look at this bit natural interpretation of that 463 00:39:50,067 --> 00:39:55,864 is again the integral from B to CAYDX and then natural 464 00:39:55,864 --> 00:40:02,188 interpretation of this bit is the integral from A to BYX. So 465 00:40:02,188 --> 00:40:09,566 what we've shown is that we want if we want to find the area 466 00:40:09,566 --> 00:40:13,255 from A to C and there's some 467 00:40:13,255 --> 00:40:17,618 sort of. Inconvenience there around be. Then we can work from 468 00:40:17,618 --> 00:40:23,772 be to see an from A to B and add the two together to give us the 469 00:40:23,772 --> 00:40:25,582 area contained by the curve.