0:00:01.670,0:00:06.990
Going to have a look in this[br]video and how we might be able

0:00:06.990,0:00:09.650
to find the area under a curve.

0:00:10.310,0:00:15.640
But the really important bit of[br]this video is to do with looking

0:00:15.640,0:00:20.970
at integration as summation, IE[br]the idea that we can add up lots

0:00:20.970,0:00:25.890
of little bits of something in[br]order to arrive at a concrete

0:00:25.890,0:00:27.120
block of something.

0:00:27.970,0:00:33.052
Let's begin by having a look at[br]something that's probably quite

0:00:33.052,0:00:39.686
familiar. It's a problem that's[br]often set GCSE, another.

0:00:40.900,0:00:42.640
Levels of Maths.

0:00:43.730,0:00:45.870
Supposing you were asked.

0:00:47.230,0:00:48.499
Find the area.

0:00:49.470,0:00:52.142
Of this shape. How might we do

0:00:52.142,0:00:56.510
it? One way that we would do it[br]perhaps is the divided up.

0:00:58.180,0:01:01.948
Into a series of shapes like

0:01:01.948,0:01:07.650
that. That's one way each of[br]these shapes is a standard area.

0:01:07.650,0:01:11.550
A triangle or square or[br]rectangle, whatever. And we can

0:01:11.550,0:01:17.010
workout what its area is. Then[br]we can get the area of the whole

0:01:17.010,0:01:21.690
shape by adding together the[br]area of each of these shapes. So

0:01:21.690,0:01:26.760
from a shape that had no defined[br]formula for its area by breaking

0:01:26.760,0:01:31.440
it up into more recognizable[br]shapes, we were able to sort out

0:01:31.440,0:01:33.000
what it's area was.

0:01:34.070,0:01:38.219
OK, can we take that[br]a little bit further?

0:01:39.430,0:01:42.730
For a circle, there is[br]a formula for the area

0:01:42.730,0:01:44.380
a equals π R squared.

0:01:45.480,0:01:49.080
But if there was no formula, how[br]might we proceed?

0:01:51.120,0:01:55.735
One way might be to place a grid[br]of squares over the circle,

0:01:55.735,0:01:59.640
workout the area of each square[br]and add them all together.

0:02:01.040,0:02:05.027
However, some of the squares[br]will not coincide completely

0:02:05.027,0:02:06.356
with the circle.

0:02:06.930,0:02:10.846
Thus, if we add together all[br]the squares that contain any

0:02:10.846,0:02:14.762
part of the circle, we will[br]arrive at an over estimate

0:02:14.762,0:02:16.898
of the area of the circle.

0:02:19.440,0:02:24.115
We could try to attempt to[br]correct this by ignoring all

0:02:24.115,0:02:27.940
those squares that do not[br]completely coincide with the

0:02:27.940,0:02:32.074
circle. This would mean[br]ignoring some of the area

0:02:32.074,0:02:34.978
of the circle and we[br]would gain an

0:02:34.978,0:02:37.519
underestimate of the area[br]of the circle.

0:02:40.120,0:02:46.035
However, we would now have two[br]limits on the area of the circle

0:02:46.035,0:02:47.400
and upper one.

0:02:47.990,0:02:48.968
Under lower one.

0:02:50.080,0:02:54.964
To trap the area of the circle[br]ever tighter between an upper

0:02:54.964,0:02:59.848
limit and a lower limit, we[br]would need to make the squares

0:02:59.848,0:03:04.732
in the grid smaller and smaller[br]so that the parts that were

0:03:04.732,0:03:08.395
included or discarded were in[br]total smaller and smaller.

0:03:12.570,0:03:18.258
Now we're going to do is[br]take that idea.

0:03:18.260,0:03:20.108
And use it.

0:03:20.680,0:03:27.628
To attempt to find the[br]area underneath a curve.

0:03:28.160,0:03:33.269
So what we've got here is a[br]curve, and we're looking to find

0:03:33.269,0:03:35.627
is can we find the area?

0:03:36.340,0:03:42.151
Underneath the curve, so can we[br]find in some way this area and

0:03:42.151,0:03:47.962
can we use the same sorts of[br]ideas that we just spoke about

0:03:47.962,0:03:49.303
with the circle?

0:03:51.360,0:03:57.205
Well, let's take a[br]very specific example.

0:03:58.320,0:03:59.936
I'm going to have a look at the

0:03:59.936,0:04:01.750
curve. Why?

0:04:02.930,0:04:05.189
Equals X squared.

0:04:08.560,0:04:12.226
Going to draw its graph between

0:04:12.226,0:04:19.539
a. And. A[br]square so this end point here

0:04:19.539,0:04:22.167
is the point a.

0:04:22.760,0:04:24.930
A squared.

0:04:26.280,0:04:30.120
Now. How can we divide

0:04:30.120,0:04:35.880
this up? Well, I just put a[br]limit to where we want to be.

0:04:36.510,0:04:42.474
One way might be to say, OK,[br]let's divide this in half, so we

0:04:42.474,0:04:44.178
take a over 2.

0:04:44.880,0:04:46.608
And let's take.

0:04:47.830,0:04:50.498
Some rectangles so there.

0:04:54.280,0:04:55.790
And there.

0:04:57.800,0:05:00.744
So there are two rectangles.[br]This one here.

0:05:02.180,0:05:03.420
And this big one.

0:05:04.030,0:05:11.246
Here. Now, at this point here[br]on the curve, X is equal to

0:05:11.246,0:05:17.850
a over 2 where dealing with the[br]curve Y equals X squared, so

0:05:17.850,0:05:24.454
that we know what this point is.[br]Here, it's a over 2A squared

0:05:24.454,0:05:30.986
over 4. So we know[br]the height of this rectangle.

0:05:30.986,0:05:37.730
It's A squared over 4, so I[br]guess an approximation at this

0:05:37.730,0:05:44.474
area underneath the curve. The[br]area a. We know that a is

0:05:44.474,0:05:51.218
in fact going to be less than[br]the area of this rectangle.

0:05:52.310,0:05:59.978
Which is a over two times[br]a squared over 4 plus the

0:05:59.978,0:06:05.729
area of this rectangle, which is[br]a over 2.

0:06:06.460,0:06:12.374
Times A. Squared because that's[br]the height of that rectangle.

0:06:13.710,0:06:19.478
So we've got this area a what[br]we've got is an upper limit for

0:06:19.478,0:06:22.774
it. We know it must be less than

0:06:22.774,0:06:28.440
that. OK, if that's the case,[br]then what about a lower

0:06:28.440,0:06:32.598
limit? One way might be to[br]take other rectangles.

0:06:34.690,0:06:38.490
Such as one down here of[br]zero height.

0:06:40.770,0:06:42.399
And this rectangle.

0:06:43.290,0:06:43.820
Here.

0:06:45.110,0:06:50.414
Where again, this height is[br]going to be a squared over 4 and

0:06:50.414,0:06:55.718
we know very clearly that that[br]means that this area a must be

0:06:55.718,0:06:59.798
greater than zero. That's the[br]area of this rectangle plus.

0:07:00.380,0:07:06.148
A over two times a[br]squared over 4.

0:07:07.400,0:07:12.008
Now what did we say about a[br]circle? To get a better

0:07:12.008,0:07:13.928
approximation, we said we take

0:07:13.928,0:07:16.060
smaller. And smaller.

0:07:16.770,0:07:21.112
Squares and put it on the[br]circle. So in this case, what we

0:07:21.112,0:07:25.120
want to do is take rectangles[br]which are smaller and smaller in

0:07:25.120,0:07:26.456
terms of their width.

0:07:27.170,0:07:34.766
So let's draw this picture again[br]and this time let's divide these

0:07:34.766,0:07:37.931
two segments into half again.

0:07:38.010,0:07:41.130
So let's draw.

0:07:41.670,0:07:42.840
Our curve.

0:07:43.910,0:07:44.640
Gain.

0:07:48.520,0:07:51.238
Take a there.

0:07:51.790,0:07:54.280
So this will be a squared.

0:07:55.060,0:08:00.044
There for the curve[br]Y equals X squared.

0:08:01.860,0:08:07.034
This area that we're trying to[br]find and we said that we would

0:08:07.034,0:08:09.180
divide. The base.

0:08:10.060,0:08:16.798
Into half And then[br]in two half again.

0:08:17.750,0:08:18.780
And.

0:08:20.800,0:08:23.470
Again there.

0:08:24.760,0:08:28.288
Now a over 2 is half

0:08:28.288,0:08:33.440
way. But I want to keep this[br]denominator of four if I can, so

0:08:33.440,0:08:37.940
I'm not going to call that a[br]over 2. I'm going to call it to

0:08:37.940,0:08:42.440
a over 4 and you'll see why I[br]want to do that in a moment.

0:08:44.210,0:08:47.380
Now, where are our rectangles?

0:08:48.920,0:08:51.790
Well, there's a rectangle there.

0:08:57.060,0:08:58.239
There's another one.

0:08:58.810,0:08:59.550
There.

0:09:03.700,0:09:05.728
There's another one there.

0:09:06.630,0:09:09.618
And the final one.

0:09:10.680,0:09:14.816
There, so here we've got four[br]rectangles, all of which contain

0:09:14.816,0:09:18.952
area that is underneath the[br]curve, but all of which contain

0:09:18.952,0:09:23.840
extra bits of area. So when we[br]write down the areas of these

0:09:23.840,0:09:28.728
rectangles and add them up, we[br]will get an area that is greater

0:09:28.728,0:09:33.240
than the area under the curve.[br]So let's just write down the

0:09:33.240,0:09:36.624
coordinates of these points,[br]because the coordinates of these

0:09:36.624,0:09:40.760
points will be the Heights of[br]the rectangles, so the Y

0:09:40.760,0:09:43.542
coordinate. There will[br]give us the height of

0:09:43.542,0:09:45.894
that rectangle, the Y[br]coordinate their the

0:09:45.894,0:09:48.246
height of that rectangle[br]and so on.

0:09:49.430,0:09:56.668
So this point here has an X[br]coordinate of X over 4 or a

0:09:56.668,0:10:03.906
over 4 in this case, so that's[br]a over 4. It will be a

0:10:03.906,0:10:05.457
squared over 16.

0:10:06.250,0:10:13.190
This one. Has an[br]X coordinate of two 8 over 4.

0:10:14.490,0:10:21.546
So it will have[br]a Y coordinate of

0:10:21.546,0:10:25.074
2 squared, A squared

0:10:25.074,0:10:32.460
over 16. This one[br]has an X coordinate of three,

0:10:32.460,0:10:38.895
a over 4, and so it[br]will have a Y coordinate.

0:10:39.730,0:10:47.110
Of three squared, A squared[br]over 60 and finally this

0:10:47.110,0:10:49.324
one will be.

0:10:51.030,0:10:58.842
It's X coordinate is A and[br]so it will be a square.

0:10:59.520,0:11:06.252
Another way of writing that to[br]keep the pattern going would be

0:11:06.252,0:11:14.106
to call it 4A over 4 so[br]it will be 16 A squared over

0:11:14.106,0:11:20.910
60. OK, let's write down[br]the area of each of these

0:11:20.910,0:11:28.190
rectangles. The width is a over[br]4, so we'll have a over four

0:11:28.190,0:11:33.230
times its height, which is a[br]squared over 16.

0:11:34.400,0:11:40.042
Plus the area of this rectangle,[br]its width is again a over 4 is

0:11:40.042,0:11:42.460
that all of the same width?

0:11:43.000,0:11:47.080
Times its height, which this[br]time is 2.

0:11:47.800,0:11:50.932
Squared A squared

0:11:50.932,0:11:55.970
over 16.[br]Plus

0:11:57.540,0:12:00.244
And now we want to move to this

0:12:00.244,0:12:03.940
one. Its width is again a

0:12:03.940,0:12:10.886
over 4. Its height is[br]3 squared, A squared over

0:12:10.886,0:12:18.470
16 and then finally the end[br]one again. Its width is a

0:12:18.470,0:12:19.734
over 4.

0:12:20.770,0:12:26.672
Times its height, which is now[br]I've written 16 and of course we

0:12:26.672,0:12:33.028
know that 16 is 4 squared, A[br]squared over 60 and that this is

0:12:33.028,0:12:37.568
bigger than a. We know that is[br]bigger than A.

0:12:38.220,0:12:42.854
Let me just have a look at this.[br]Can we see anything here that

0:12:42.854,0:12:47.157
might be of interest to us?[br]Well, first of all, we can see

0:12:47.157,0:12:49.143
there's a common factor of a

0:12:49.143,0:12:56.710
over 4. There's also a common[br]factor of a 16 and a squared, so

0:12:56.710,0:12:59.650
let's write that as a squared

0:12:59.650,0:13:06.758
over 60. And what are[br]we left with 1 +

0:13:06.758,0:13:13.768
2 squared, +3 squared +4[br]squared is bigger than a.

0:13:14.800,0:13:21.892
Now, what about our lower limit?[br]Well, there will be a rectangle

0:13:21.892,0:13:24.256
here of zero height.

0:13:24.950,0:13:27.130
There will be another rectangle

0:13:27.130,0:13:34.598
here. Another one here[br]and another one here.

0:13:35.140,0:13:39.456
So we're looking at the area of[br]that rectangle. The area of that

0:13:39.456,0:13:43.108
one, that one, and that one[br]where the upper boundaries of

0:13:43.108,0:13:44.768
the rectangles are marked in

0:13:44.768,0:13:51.556
red. What does that area come to[br]where we know that it will be

0:13:51.556,0:13:56.860
less than the area under the[br]curve and it will be 0?

0:13:57.680,0:14:05.504
Plus a over 4 times by[br]now that's its height, the Y

0:14:05.504,0:14:09.416
value there a squared over 16.

0:14:10.870,0:14:17.562
Plus a over 4 the width of that[br]one times its height, which is

0:14:17.562,0:14:25.064
this one. 2 squared A squared[br]over 16 plus the last one where

0:14:25.064,0:14:32.120
again we have a over 4 for[br]the width and for the height we

0:14:32.120,0:14:37.664
have 3 squared. A squared over[br]16. Three squared, A squared

0:14:37.664,0:14:44.216
over 16 and so a is greater[br]than like we look what we've

0:14:44.216,0:14:49.760
got. Again, we've got this[br]common factor of a over 4.

0:14:49.800,0:14:55.597
Again, we've got this common[br]factor of A squared over 16.

0:14:56.360,0:15:03.459
Bracket. And 0[br]+ 1 + 2 squared,

0:15:03.459,0:15:04.993
+3 squared.

0:15:05.680,0:15:10.598
OK. We could go on[br]and do this again.

0:15:12.100,0:15:17.492
But what we want is to try and[br]do it. In general, if we can. So

0:15:17.492,0:15:22.547
what I'm going to do is to take[br]the same piece of area, the same

0:15:22.547,0:15:26.928
piece of curve, the same piece[br]of area I'm going to divide the

0:15:26.928,0:15:31.646
X axis into N equal parts. Here[br]I had it divided into 4 equal

0:15:31.646,0:15:36.027
parts. Now I'm going to divide[br]it into N equal parts, and in

0:15:36.027,0:15:41.082
doing that I'm going to look at[br]only a part of it. But the idea

0:15:41.082,0:15:43.104
is going to be the same.

0:15:43.110,0:15:47.180
And the same patterns that we[br]picked out here, namely this

0:15:47.180,0:15:51.990
sums of squares of the integers[br]is going to be there again for

0:15:51.990,0:15:53.100
us to see.

0:15:53.990,0:15:55.720
So.

0:15:57.030,0:15:59.494
Here's a section of the X axis.

0:16:02.010,0:16:03.340
A piece of the curve.

0:16:04.180,0:16:06.328
Interested in the.

0:16:06.840,0:16:13.164
Off strip, so let's just mark[br]off a few of these strips.

0:16:15.750,0:16:22.428
So there are a few[br]of the strips now

0:16:22.428,0:16:24.654
remember each one.

0:16:26.190,0:16:28.478
Is the same width.

0:16:29.050,0:16:35.658
The total length of the X axis[br]we were looking at was A and

0:16:35.658,0:16:42.266
there are N strips, so the width[br]of each of these strips is a

0:16:42.266,0:16:47.930
over N, so each one has a width[br]of a over N.

0:16:48.800,0:16:54.080
Now that helps us count because[br]the arts trip. Let's say it's

0:16:54.080,0:17:00.240
this one here must be our times[br]a over N for its X coordinate

0:17:00.240,0:17:07.860
their. And here for its[br]X coordinate our minus one times

0:17:07.860,0:17:09.852
a over N.

0:17:11.260,0:17:17.436
Let's build up the rectangle[br]on this strip.

0:17:18.010,0:17:22.760
So there's the rectangle[br]now what's the area of that

0:17:22.760,0:17:25.610
rectangle? The area of the[br]rectangle?

0:17:27.570,0:17:32.530
Equals well its width,[br]which we know is a over

0:17:32.530,0:17:35.010
N times by its height.

0:17:36.660,0:17:40.040
Which is this? Why coordinate

0:17:40.040,0:17:46.186
here? And remember, this is[br]the curve Y equals X

0:17:46.186,0:17:50.458
squared, and so the[br]coordinates of this point

0:17:50.458,0:17:56.866
are a over N and then we[br]must square it R-squared A

0:17:56.866,0:18:03.274
squared over end square. So[br]the area is a over N times

0:18:03.274,0:18:07.546
that height R-squared, A[br]squared over N squared.

0:18:08.640,0:18:10.320
OK.

0:18:11.570,0:18:18.430
So this is the area of the Strip[br]and it's bigger than this area

0:18:18.430,0:18:22.350
underneath the curve. Let's call[br]that Delta A.

0:18:23.010,0:18:27.798
Let's just tidy this up a little[br]bit. We've gotten a cubed.

0:18:29.500,0:18:35.620
And we've gotten N cubed, and we[br]got this thing R-squared and

0:18:35.620,0:18:38.680
that is bigger than Delta A.

0:18:39.610,0:18:45.756
And what we want is the whole[br]area. So we want all of these

0:18:45.756,0:18:52.780
Delta Azan. We want to add them[br]all up. So we need to add up all

0:18:52.780,0:18:58.926
of these but that will still be[br]bigger than the area as a whole.

0:18:58.926,0:19:05.072
So let's add them all up Sigma.[br]That means some add up a cubed

0:19:05.072,0:19:09.901
over N cubed times by R-squared[br]and we'll have from our.

0:19:09.940,0:19:14.428
Equals 1 up to an. Added[br]all those up that will

0:19:14.428,0:19:18.100
still be bigger than the[br]area that we're after.

0:19:19.300,0:19:23.428
Now let's look at the[br]other rectangle. This

0:19:23.428,0:19:25.492
one the lower one.

0:19:27.890,0:19:34.390
Now we can say that the height[br]of this one depends upon this

0:19:34.390,0:19:38.890
point here and its coordinate is[br]R minus one.

0:19:39.470,0:19:46.130
Times a over N for X and so it's[br]Y coordinate is R minus 1

0:19:46.130,0:19:49.864
squared. A squared over N

0:19:49.864,0:19:57.315
squared. So what's the[br]area of this smaller

0:19:57.315,0:20:04.602
rectangle area? Equals well,[br]the width is still a over N,

0:20:04.602,0:20:10.938
but the height is different. Are[br]minus one all squared A squared

0:20:10.938,0:20:14.634
over N squared and this will be

0:20:14.634,0:20:19.458
less than? That[br]area Delta A.

0:20:20.500,0:20:26.372
We now need to Add all of these[br]up to get an idea of the lower

0:20:26.372,0:20:31.510
limit of the area. So let's do[br]that. A will be greater than the

0:20:31.510,0:20:37.015
sum of an. Let's tidy this one[br]up in the same way that we did

0:20:37.015,0:20:41.786
that we've gotten a cubed formed[br]by the A Times by a squared.

0:20:42.390,0:20:47.746
So that will give us the A[br]cubed and we've got an end

0:20:47.746,0:20:52.278
times by N squared which[br]will give us N cubed. And

0:20:52.278,0:20:56.810
then we've got this R minus[br]one all squared and will

0:20:56.810,0:21:00.518
some that from our equals 1[br]up to N.

0:21:01.630,0:21:07.696
So now let's write down[br]these two inequalities together.

0:21:07.700,0:21:11.388
So we've got Sigma.

0:21:12.120,0:21:15.580
R equals 1 to end.

0:21:16.390,0:21:22.194
Of a.[br]Cubed of

0:21:22.194,0:21:27.745
N cubed.[br]Times

0:21:27.745,0:21:34.890
by. All[br]squared is greater than

0:21:34.890,0:21:41.148
a. Is greater than[br]the sum from R equals

0:21:41.148,0:21:47.814
12 N of a cubed over[br]N cubed times by R

0:21:47.814,0:21:50.238
minus one or squared?

0:21:52.660,0:21:59.184
OK. This a cubed over NQ[br]bisa common factor in each one

0:21:59.184,0:22:01.908
so we can take that out.

0:22:23.260,0:22:27.964
Now question is can we add up[br]the sums of the squares of the

0:22:27.964,0:22:31.324
integers 1 squared +2 squared,[br]+3 squared, +4 squared, +5

0:22:31.324,0:22:35.020
squared? Remember, we had that[br]before when we were looking at

0:22:35.020,0:22:39.388
dividing it up into quarters and[br]I pointed out that we had one

0:22:39.388,0:22:43.084
squared +2 squared, +3 squared,[br]+4 squared, and which see that

0:22:43.084,0:22:47.452
again, and this is it. Here we[br]can see quite clearly 1 squared,

0:22:47.452,0:22:51.484
+2 squared, +3 squared, +4[br]squared, and so on, and the same

0:22:51.484,0:22:53.836
over here. Can we add those up?

0:22:53.910,0:22:58.837
Well, yes, there is a formula[br]that is a formula that tells us

0:22:58.837,0:23:01.490
what the sum of the first end

0:23:01.490,0:23:08.970
squares is. And the sum of[br]the first N squares from R

0:23:08.970,0:23:11.438
equals 1 to N.

0:23:12.450,0:23:14.240
Is N.

0:23:15.540,0:23:17.160
N plus one.

0:23:18.500,0:23:24.464
2 N plus one all over 6 looks a[br]bit complicated, but that's what

0:23:24.464,0:23:30.854
it is and in fact it's easy to[br]work with, so let's put that in

0:23:30.854,0:23:32.984
here instead of this one.

0:23:34.490,0:23:39.250
So we get a cubed[br]over and cubed.

0:23:40.450,0:23:47.542
Times by N over times by[br]M Plus One times by 2

0:23:47.542,0:23:50.497
N plus one over 6.

0:23:52.160,0:23:57.841
Is bigger than a is bigger[br]than now? This is if you like

0:23:57.841,0:24:03.522
R minus one. So in a sense,[br]we're always one less than N.

0:24:03.522,0:24:09.640
So we want an minus one times[br]N times 2 N minus one, which

0:24:09.640,0:24:16.632
is what we get when we replace[br]N by two N all over 6 times by

0:24:16.632,0:24:19.254
the A cubed over and cubed.

0:24:20.470,0:24:25.636
Now in each case we've got this[br]a cubed over and cubed times by

0:24:25.636,0:24:30.802
6, and in fact we've gotten end.[br]Look here on either side that we

0:24:30.802,0:24:35.968
can cancel so we can cancel that[br]we get a square and cancel that

0:24:35.968,0:24:41.503
and get a square. So I'm going[br]to take the A cubed over 6 out.

0:24:41.550,0:24:47.010
And I'm going to multiply out[br]each of these brackets and put

0:24:47.010,0:24:49.285
it all over N squared.

0:24:49.290,0:24:54.936
So have a cubed[br]over 6.

0:24:56.150,0:25:03.280
All over and squared, greater[br]than a greater than a

0:25:03.280,0:25:10.410
cubed over 6, all over[br]and squared. Let's just go

0:25:10.410,0:25:15.650
back. And look at these[br]brackets. This is N plus one

0:25:15.650,0:25:20.928
times by 2 N plus one, so it[br]will give us two and squared.

0:25:22.910,0:25:29.499
N and two N which will[br]be 3 N plus one.

0:25:29.500,0:25:36.836
So we'll have two[br]and squared plus 3N

0:25:36.836,0:25:43.890
plus one. This bracket will give[br]us a gain two and squared.

0:25:44.430,0:25:51.290
But then minus two and minus[br]sign will be minus 3 N and plus

0:25:51.290,0:25:59.051
one. So again, two and squared[br]minus 3 N plus one. Let's

0:25:59.051,0:26:06.983
divide each term by end squared[br]a over a cubed over 6.

0:26:08.210,0:26:15.050
2 + 3 over N plus one[br]over N squared that's dividing

0:26:15.050,0:26:22.460
each term in turn by the end[br]squared is greater than a is

0:26:22.460,0:26:25.880
greater than a cubed over 6.

0:26:26.560,0:26:33.570
2 - 3 over N[br]plus one over and squared.

0:26:34.700,0:26:37.989
Now it's taking us rather[br]along time to arrive at this

0:26:37.989,0:26:40.979
position, but we're here[br]where we want to be now.

0:26:42.880,0:26:49.750
Let's imagine that we let[br]an become very, very large,

0:26:49.750,0:26:56.620
so the number of strips[br]where taking is infinite there

0:26:56.620,0:27:02.034
infinitely thin. Infinitely thin[br]because N is very very large.

0:27:02.034,0:27:08.379
Now let's think what happens to[br]a number like 3 over N when N is

0:27:08.379,0:27:13.878
very very large. Well, it goes[br]to zero and one over N squared

0:27:13.878,0:27:19.800
goes to zero as those three over[br]N this side and one over N

0:27:19.800,0:27:21.492
squared at this side.

0:27:22.260,0:27:28.981
So in fact the area a[br]is trapped between a cubed

0:27:28.981,0:27:36.313
over 6 times by two and[br]a cubed over 6 times by

0:27:36.313,0:27:41.812
two again. So therefore as N[br]tends to Infinity.

0:27:43.260,0:27:44.940
In the limit.

0:27:46.200,0:27:53.484
A is equal to a cubed[br]over 6 times by two, which

0:27:53.484,0:28:00.161
is 1/3 of a cubed. It's[br]trapped between these two bits

0:28:00.161,0:28:03.803
that are the same here, because

0:28:03.803,0:28:07.499
these bits. Disappear to 0.

0:28:08.590,0:28:14.830
Now we've done that for[br]one particular curve.

0:28:15.460,0:28:21.972
To show that we can add up a lot[br]of little bits to arrive at a

0:28:21.972,0:28:26.042
finite answer. A lot of[br]infinitely small bits to arrive

0:28:26.042,0:28:32.554
at a finite answer. What we want[br]to be able to do now is to do

0:28:32.554,0:28:37.438
it in general to arrive at[br]something that will work for all

0:28:37.438,0:28:44.916
curves. So let's see if[br]we can extend what we've

0:28:44.916,0:28:48.500
done. To a more general curve.

0:28:49.490,0:28:51.470
So here's our curve, let's say.

0:28:52.060,0:28:59.860
And again, let's say that it[br]runs from X equals North to

0:28:59.860,0:29:01.810
X equals A.

0:29:02.380,0:29:04.095
And what we're going to do is

0:29:04.095,0:29:10.792
divide. The X axis up into very,[br]very thin strips, each one of

0:29:10.792,0:29:18.232
which is a width Delta X. So we[br]have X there and the next mark

0:29:18.232,0:29:25.176
on the X axis is X Plus Delta[br]X. Delta axes are small positive

0:29:25.176,0:29:26.664
amount of X.

0:29:27.910,0:29:28.969
So we've got.

0:29:29.820,0:29:33.570
Our. Ordinance up to the curve.

0:29:34.470,0:29:37.476
And we can complete a rectangle.

0:29:38.970,0:29:46.317
So. What are the coordinates[br]of that point? Well, this is the

0:29:46.317,0:29:53.428
curve Y equals F of X, and[br]so the coordinates of this point

0:29:53.428,0:29:55.069
are just XY.

0:29:56.160,0:29:59.368
That's the point, P.

0:30:00.430,0:30:03.006
What are the coordinates[br]of this point?

0:30:04.350,0:30:10.961
While the coordinates at this[br]point, let's call it the point

0:30:10.961,0:30:18.173
QRX plus Delta X&Y plus Delta[br]Y, where Delta. Why is that

0:30:18.173,0:30:19.976
small increment there?

0:30:21.780,0:30:26.993
That we've added on as a result[br]of the small increment that we

0:30:26.993,0:30:28.196
made in X.

0:30:28.710,0:30:34.606
Let's say that the area under[br]the curve is Delta A.

0:30:35.160,0:30:40.459
So this is the[br]area Delta A.

0:30:41.750,0:30:48.955
And the area Delta A is[br]contained between the areas of

0:30:48.955,0:30:53.699
two rectangles. The[br]larger rectangle.

0:30:54.730,0:30:57.964
Whose height is[br]why plus Delta Y?

0:30:59.460,0:31:03.906
And whose width is[br]Delta X.

0:31:05.280,0:31:11.064
And so that's it's area and the[br]smaller rectangle. Let me just

0:31:11.064,0:31:12.510
mark it's top.

0:31:13.250,0:31:20.198
Their whose width is Delta X[br]that whose height is just, why?

0:31:21.620,0:31:27.400
So it's area will be Y[br]times by Delta X.

0:31:28.560,0:31:34.736
So if I want the whole of this[br]area, I have to add up all of

0:31:34.736,0:31:40.526
these little bits, which means I[br]have to be able to add up all of

0:31:40.526,0:31:44.041
these bits. So let's do

0:31:44.041,0:31:51.030
that. A. The[br]total area underneath this curve

0:31:51.030,0:31:58.410
is the sum of all these[br]little bits of area from X

0:31:58.410,0:32:05.790
equals not up to X equals[br]A and so therefore that area

0:32:05.790,0:32:13.170
a is caught between the sum[br]of all of these little bits

0:32:13.170,0:32:16.860
from X equals not to X

0:32:16.860,0:32:19.650
equals. A at the bottom end.

0:32:20.180,0:32:27.161
And at the top end, the sum[br]of all of these little bits

0:32:27.161,0:32:34.142
from X equals nor two X[br]equals a or Y plus Delta Y

0:32:34.142,0:32:36.290
times by Delta X.

0:32:37.490,0:32:44.738
OK, let me multiply out this[br]bit so we have Sigma X

0:32:44.738,0:32:51.986
equals North to a of Y[br]times by Delta X Plus Delta

0:32:51.986,0:32:59.234
Y times by Delta X greater[br]than a greater than the sum

0:32:59.234,0:33:06.482
from X equals not to a[br]of Y times by Delta X.

0:33:07.340,0:33:13.836
Now we're going to let Delta X[br]tend to 0, just like we let N go

0:33:13.836,0:33:18.708
off to Infinity, which made the[br]strips under Y equals X squared

0:33:18.708,0:33:19.926
come down to.

0:33:20.550,0:33:24.788
A over N 10 down to a zero[br]thickness, we're going to let

0:33:24.788,0:33:29.026
Delta X come down to zero. Now[br]what's going to happen when we

0:33:29.026,0:33:32.612
do that? Will certainly here[br]we've got two very, very small

0:33:32.612,0:33:36.850
terms. We've got Delta. It's[br]going to 0, so the change in Y

0:33:36.850,0:33:41.414
is going to be very very small[br]as well. So this term here is

0:33:41.414,0:33:43.696
actually going to run off to 0.

0:33:44.360,0:33:49.400
But when that happens, let's[br]just cover that up. We see that

0:33:49.400,0:33:53.180
a is trapped between two things[br]which are identical.

0:33:53.710,0:33:56.790
And So what we have is that.

0:33:57.440,0:34:04.024
A is equal to[br]the limit as Delta

0:34:04.024,0:34:10.608
X tends to zero[br]of the sum from

0:34:10.608,0:34:17.192
X equals not a[br]of Y Delta X.

0:34:18.130,0:34:23.160
Now. With actually made an awful[br]lot of assumptions in what's

0:34:23.160,0:34:28.370
going on. And we've had to make[br]them because this level it's

0:34:28.370,0:34:32.759
very difficult to do anything[br]other than make these kinds of

0:34:32.759,0:34:36.172
assumptions. Let's just have a[br]look at two of them.

0:34:36.970,0:34:41.434
For instance, one of the things[br]that we have assumed is that

0:34:41.434,0:34:46.642
this curve is increasing all the[br]time for all values of X. But as

0:34:46.642,0:34:51.106
we know, curves can decrease as[br]X increases, curves can wave up

0:34:51.106,0:34:55.570
and down, they can go up and[br]down as X increases. However,

0:34:55.570,0:35:00.778
the results are still the same.[br]We need to do it a little bit

0:35:00.778,0:35:04.498
differently, but the results[br]still come out the same. The

0:35:04.498,0:35:08.590
other assumption that we're[br]making is that in fact we can

0:35:08.590,0:35:15.140
add up. In this form, lots of[br]little bits and still get a

0:35:15.140,0:35:16.700
definite finite answer.

0:35:17.420,0:35:22.865
And there's a whole raft of pure[br]mathematics, which assures us

0:35:22.865,0:35:29.300
that these things do work, so[br]here we are at this stage. Now

0:35:29.300,0:35:34.745
this is very, very cumbersome[br]language, and so in fact we

0:35:34.745,0:35:41.675
choose to write this as a equals[br]the integral from not a of YDX.

0:35:42.300,0:35:49.636
A is the integral from North to[br]a of Y with respect to X,

0:35:49.636,0:35:53.828
and it's this notation which[br]replaces that one.

0:35:54.570,0:36:00.546
OK, supposing we actually want[br]the area not from North to a,

0:36:00.546,0:36:07.518
but between, let us say two[br]limits A and B2 values of X and

0:36:07.518,0:36:14.640
the curve. So we want, let's[br]say, to find the area, let's

0:36:14.640,0:36:17.946
have a curve here that goes

0:36:17.946,0:36:20.560
between there. And.

0:36:21.590,0:36:22.440
There.

0:36:23.610,0:36:28.174
What would we want to do? Take[br]that back to the Y axis there?

0:36:28.174,0:36:30.130
What would we want to do?

0:36:30.630,0:36:35.417
Well. This is the area[br]that we're after.

0:36:36.380,0:36:41.490
If we think about what that area[br]is, it's the area from nought to

0:36:41.490,0:36:46.747
be. Minus the area from[br]North to A.

0:36:48.110,0:36:51.725
So this area that we're

0:36:51.725,0:36:59.104
wanting. Is equal to the[br]area from North to be, which is

0:36:59.104,0:37:04.456
that. Minus the[br]area from North to

0:37:04.456,0:37:07.260
a, which is that.

0:37:08.440,0:37:12.230
Now it seems quite natural to[br]write that as the.

0:37:13.050,0:37:19.023
Integral from A to B of Y[br]with respect to X.

0:37:20.570,0:37:25.180
So we would evaluate whatever[br]the result of this computation

0:37:25.180,0:37:30.712
was at B and subtract from it[br]the result of whatever the

0:37:30.712,0:37:32.556
computation was at a.

0:37:32.580,0:37:38.280
And so there we've seen how area[br]can be represented as a

0:37:38.280,0:37:43.505
summation and that that leads us[br]to the traditional way of

0:37:43.505,0:37:48.730
finding an area which is to[br]integrate the equation of the

0:37:48.730,0:37:50.630
curve and substituting limits.

0:37:51.200,0:37:57.750
To conclude, this video will[br]just look one further result.

0:37:58.420,0:38:01.910
This result will enable us to[br]actually calculate areas by

0:38:01.910,0:38:06.098
chopping them up into convenient[br]segments just the same way as we

0:38:06.098,0:38:09.937
began. If you remember that[br]shape rather like a dog were

0:38:09.937,0:38:13.776
able to chop it up in a[br]convenient segments, find the

0:38:13.776,0:38:17.266
area of the separate segments[br]added altogether. Well, this is

0:38:17.266,0:38:22.152
this result is going to show us[br]that we can do very much the

0:38:22.152,0:38:23.897
same with a curve, so.

0:38:26.290,0:38:32.230
Let's suppose that 4 hour curve.[br]We have three ordinance X equals

0:38:32.230,0:38:39.655
a X equals B&X equals C and that[br]they are in fact in order of

0:38:39.655,0:38:43.615
size a is less than B is less

0:38:43.615,0:38:50.012
than C. Let's begin by[br]asking ourselves what's the area

0:38:50.012,0:38:57.149
contained by the curve Y equals[br]F of X, and these two ordinate's

0:38:57.149,0:39:04.835
and the X axis. Well, it's from[br]A to see the integral of YDX.

0:39:05.670,0:39:12.594
And we know that that is[br]the integral from North. To see

0:39:12.594,0:39:18.941
of YDX minus the integral from[br]nought to a of YDX.

0:39:19.730,0:39:22.462
Equals, well,

0:39:22.462,0:39:28.450
rights. Introduce the[br]ordinate be now.

0:39:29.410,0:39:32.930
By taking away that.

0:39:33.800,0:39:40.688
Area if I've taken away, let's[br]add it on in order to

0:39:40.688,0:39:43.558
keep the quality the same.

0:39:44.270,0:39:50.067
Now if we look at this bit[br]natural interpretation of that

0:39:50.067,0:39:55.864
is again the integral from B to[br]CAYDX and then natural

0:39:55.864,0:40:02.188
interpretation of this bit is[br]the integral from A to BYX. So

0:40:02.188,0:40:09.566
what we've shown is that we want[br]if we want to find the area

0:40:09.566,0:40:13.255
from A to C and there's some

0:40:13.255,0:40:17.618
sort of. Inconvenience there[br]around be. Then we can work from

0:40:17.618,0:40:23.772
be to see an from A to B and add[br]the two together to give us the

0:40:23.772,0:40:25.582
area contained by the curve.