Going to have a look in this video and how we might be able
to find the area under a curve.
But the really important bit of this video is to do with looking
at integration as summation, IE the idea that we can add up lots
of little bits of something in order to arrive at a concrete
block of something.
Let's begin by having a look at something that's probably quite
familiar. It's a problem that's often set GCSE, another.
Levels of Maths.
Supposing you were asked.
Find the area.
Of this shape. How might we do
it? One way that we would do it perhaps is the divided up.
Into a series of shapes like
that. That's one way each of these shapes is a standard area.
A triangle or square or rectangle, whatever. And we can
workout what its area is. Then we can get the area of the whole
shape by adding together the area of each of these shapes. So
from a shape that had no defined formula for its area by breaking
it up into more recognizable shapes, we were able to sort out
what it's area was.
OK, can we take that a little bit further?
For a circle, there is a formula for the area
a equals π R squared.
But if there was no formula, how might we proceed?
One way might be to place a grid of squares over the circle,
workout the area of each square and add them all together.
However, some of the squares will not coincide completely
with the circle.
Thus, if we add together all the squares that contain any
part of the circle, we will arrive at an over estimate
of the area of the circle.
We could try to attempt to correct this by ignoring all
those squares that do not completely coincide with the
circle. This would mean ignoring some of the area
of the circle and we would gain an
underestimate of the area of the circle.
However, we would now have two limits on the area of the circle
and upper one.
Under lower one.
To trap the area of the circle ever tighter between an upper
limit and a lower limit, we would need to make the squares
in the grid smaller and smaller so that the parts that were
included or discarded were in total smaller and smaller.
Now we're going to do is take that idea.
And use it.
To attempt to find the area underneath a curve.
So what we've got here is a curve, and we're looking to find
is can we find the area?
Underneath the curve, so can we find in some way this area and
can we use the same sorts of ideas that we just spoke about
with the circle?
Well, let's take a very specific example.
I'm going to have a look at the
curve. Why?
Equals X squared.
Going to draw its graph between
a. And. A square so this end point here
is the point a.
A squared.
Now. How can we divide
this up? Well, I just put a limit to where we want to be.
One way might be to say, OK, let's divide this in half, so we
take a over 2.
And let's take.
Some rectangles so there.
And there.
So there are two rectangles. This one here.
And this big one.
Here. Now, at this point here on the curve, X is equal to
a over 2 where dealing with the curve Y equals X squared, so
that we know what this point is. Here, it's a over 2A squared
over 4. So we know the height of this rectangle.
It's A squared over 4, so I guess an approximation at this
area underneath the curve. The area a. We know that a is
in fact going to be less than the area of this rectangle.
Which is a over two times a squared over 4 plus the
area of this rectangle, which is a over 2.
Times A. Squared because that's the height of that rectangle.
So we've got this area a what we've got is an upper limit for
it. We know it must be less than
that. OK, if that's the case, then what about a lower
limit? One way might be to take other rectangles.
Such as one down here of zero height.
And this rectangle.
Here.
Where again, this height is going to be a squared over 4 and
we know very clearly that that means that this area a must be
greater than zero. That's the area of this rectangle plus.
A over two times a squared over 4.
Now what did we say about a circle? To get a better
approximation, we said we take
smaller. And smaller.
Squares and put it on the circle. So in this case, what we
want to do is take rectangles which are smaller and smaller in
terms of their width.
So let's draw this picture again and this time let's divide these
two segments into half again.
So let's draw.
Our curve.
Gain.
Take a there.
So this will be a squared.
There for the curve Y equals X squared.
This area that we're trying to find and we said that we would
divide. The base.
Into half And then in two half again.
And.
Again there.
Now a over 2 is half
way. But I want to keep this denominator of four if I can, so
I'm not going to call that a over 2. I'm going to call it to
a over 4 and you'll see why I want to do that in a moment.
Now, where are our rectangles?
Well, there's a rectangle there.
There's another one.
There.
There's another one there.
And the final one.
There, so here we've got four rectangles, all of which contain
area that is underneath the curve, but all of which contain
extra bits of area. So when we write down the areas of these
rectangles and add them up, we will get an area that is greater
than the area under the curve. So let's just write down the
coordinates of these points, because the coordinates of these
points will be the Heights of the rectangles, so the Y
coordinate. There will give us the height of
that rectangle, the Y coordinate their the
height of that rectangle and so on.
So this point here has an X coordinate of X over 4 or a
over 4 in this case, so that's a over 4. It will be a
squared over 16.
This one. Has an X coordinate of two 8 over 4.
So it will have a Y coordinate of
2 squared, A squared
over 16. This one has an X coordinate of three,
a over 4, and so it will have a Y coordinate.
Of three squared, A squared over 60 and finally this
one will be.
It's X coordinate is A and so it will be a square.
Another way of writing that to keep the pattern going would be
to call it 4A over 4 so it will be 16 A squared over
60. OK, let's write down the area of each of these
rectangles. The width is a over 4, so we'll have a over four
times its height, which is a squared over 16.
Plus the area of this rectangle, its width is again a over 4 is
that all of the same width?
Times its height, which this time is 2.
Squared A squared
over 16. Plus
And now we want to move to this
one. Its width is again a
over 4. Its height is 3 squared, A squared over
16 and then finally the end one again. Its width is a
over 4.
Times its height, which is now I've written 16 and of course we
know that 16 is 4 squared, A squared over 60 and that this is
bigger than a. We know that is bigger than A.
Let me just have a look at this. Can we see anything here that
might be of interest to us? Well, first of all, we can see
there's a common factor of a
over 4. There's also a common factor of a 16 and a squared, so
let's write that as a squared
over 60. And what are we left with 1 +
2 squared, +3 squared +4 squared is bigger than a.
Now, what about our lower limit? Well, there will be a rectangle
here of zero height.
There will be another rectangle
here. Another one here and another one here.
So we're looking at the area of that rectangle. The area of that
one, that one, and that one where the upper boundaries of
the rectangles are marked in
red. What does that area come to where we know that it will be
less than the area under the curve and it will be 0?
Plus a over 4 times by now that's its height, the Y
value there a squared over 16.
Plus a over 4 the width of that one times its height, which is
this one. 2 squared A squared over 16 plus the last one where
again we have a over 4 for the width and for the height we
have 3 squared. A squared over 16. Three squared, A squared
over 16 and so a is greater than like we look what we've
got. Again, we've got this common factor of a over 4.
Again, we've got this common factor of A squared over 16.
Bracket. And 0 + 1 + 2 squared,
+3 squared.
OK. We could go on and do this again.
But what we want is to try and do it. In general, if we can. So
what I'm going to do is to take the same piece of area, the same
piece of curve, the same piece of area I'm going to divide the
X axis into N equal parts. Here I had it divided into 4 equal
parts. Now I'm going to divide it into N equal parts, and in
doing that I'm going to look at only a part of it. But the idea
is going to be the same.
And the same patterns that we picked out here, namely this
sums of squares of the integers is going to be there again for
us to see.
So.
Here's a section of the X axis.
A piece of the curve.
Interested in the.
Off strip, so let's just mark off a few of these strips.
So there are a few of the strips now
remember each one.
Is the same width.
The total length of the X axis we were looking at was A and
there are N strips, so the width of each of these strips is a
over N, so each one has a width of a over N.
Now that helps us count because the arts trip. Let's say it's
this one here must be our times a over N for its X coordinate
their. And here for its X coordinate our minus one times
a over N.
Let's build up the rectangle on this strip.
So there's the rectangle now what's the area of that
rectangle? The area of the rectangle?
Equals well its width, which we know is a over
N times by its height.
Which is this? Why coordinate
here? And remember, this is the curve Y equals X
squared, and so the coordinates of this point
are a over N and then we must square it R-squared A
squared over end square. So the area is a over N times
that height R-squared, A squared over N squared.
OK.
So this is the area of the Strip and it's bigger than this area
underneath the curve. Let's call that Delta A.
Let's just tidy this up a little bit. We've gotten a cubed.
And we've gotten N cubed, and we got this thing R-squared and
that is bigger than Delta A.
And what we want is the whole area. So we want all of these
Delta Azan. We want to add them all up. So we need to add up all
of these but that will still be bigger than the area as a whole.
So let's add them all up Sigma. That means some add up a cubed
over N cubed times by R-squared and we'll have from our.
Equals 1 up to an. Added all those up that will
still be bigger than the area that we're after.
Now let's look at the other rectangle. This
one the lower one.
Now we can say that the height of this one depends upon this
point here and its coordinate is R minus one.
Times a over N for X and so it's Y coordinate is R minus 1
squared. A squared over N
squared. So what's the area of this smaller
rectangle area? Equals well, the width is still a over N,
but the height is different. Are minus one all squared A squared
over N squared and this will be
less than? That area Delta A.
We now need to Add all of these up to get an idea of the lower
limit of the area. So let's do that. A will be greater than the
sum of an. Let's tidy this one up in the same way that we did
that we've gotten a cubed formed by the A Times by a squared.
So that will give us the A cubed and we've got an end
times by N squared which will give us N cubed. And
then we've got this R minus one all squared and will
some that from our equals 1 up to N.
So now let's write down these two inequalities together.
So we've got Sigma.
R equals 1 to end.
Of a. Cubed of
N cubed. Times
by. All squared is greater than
a. Is greater than the sum from R equals
12 N of a cubed over N cubed times by R
minus one or squared?
OK. This a cubed over NQ bisa common factor in each one
so we can take that out.
Now question is can we add up the sums of the squares of the
integers 1 squared +2 squared, +3 squared, +4 squared, +5
squared? Remember, we had that before when we were looking at
dividing it up into quarters and I pointed out that we had one
squared +2 squared, +3 squared, +4 squared, and which see that
again, and this is it. Here we can see quite clearly 1 squared,
+2 squared, +3 squared, +4 squared, and so on, and the same
over here. Can we add those up?
Well, yes, there is a formula that is a formula that tells us
what the sum of the first end
squares is. And the sum of the first N squares from R
equals 1 to N.
Is N.
N plus one.
2 N plus one all over 6 looks a bit complicated, but that's what
it is and in fact it's easy to work with, so let's put that in
here instead of this one.
So we get a cubed over and cubed.
Times by N over times by M Plus One times by 2
N plus one over 6.
Is bigger than a is bigger than now? This is if you like
R minus one. So in a sense, we're always one less than N.
So we want an minus one times N times 2 N minus one, which
is what we get when we replace N by two N all over 6 times by
the A cubed over and cubed.
Now in each case we've got this a cubed over and cubed times by
6, and in fact we've gotten end. Look here on either side that we
can cancel so we can cancel that we get a square and cancel that
and get a square. So I'm going to take the A cubed over 6 out.
And I'm going to multiply out each of these brackets and put
it all over N squared.
So have a cubed over 6.
All over and squared, greater than a greater than a
cubed over 6, all over and squared. Let's just go
back. And look at these brackets. This is N plus one
times by 2 N plus one, so it will give us two and squared.
N and two N which will be 3 N plus one.
So we'll have two and squared plus 3N
plus one. This bracket will give us a gain two and squared.
But then minus two and minus sign will be minus 3 N and plus
one. So again, two and squared minus 3 N plus one. Let's
divide each term by end squared a over a cubed over 6.
2 + 3 over N plus one over N squared that's dividing
each term in turn by the end squared is greater than a is
greater than a cubed over 6.
2 - 3 over N plus one over and squared.
Now it's taking us rather along time to arrive at this
position, but we're here where we want to be now.
Let's imagine that we let an become very, very large,
so the number of strips where taking is infinite there
infinitely thin. Infinitely thin because N is very very large.
Now let's think what happens to a number like 3 over N when N is
very very large. Well, it goes to zero and one over N squared
goes to zero as those three over N this side and one over N
squared at this side.
So in fact the area a is trapped between a cubed
over 6 times by two and a cubed over 6 times by
two again. So therefore as N tends to Infinity.
In the limit.
A is equal to a cubed over 6 times by two, which
is 1/3 of a cubed. It's trapped between these two bits
that are the same here, because
these bits. Disappear to 0.
Now we've done that for one particular curve.
To show that we can add up a lot of little bits to arrive at a
finite answer. A lot of infinitely small bits to arrive
at a finite answer. What we want to be able to do now is to do
it in general to arrive at something that will work for all
curves. So let's see if we can extend what we've
done. To a more general curve.
So here's our curve, let's say.
And again, let's say that it runs from X equals North to
X equals A.
And what we're going to do is
divide. The X axis up into very, very thin strips, each one of
which is a width Delta X. So we have X there and the next mark
on the X axis is X Plus Delta X. Delta axes are small positive
amount of X.
So we've got.
Our. Ordinance up to the curve.
And we can complete a rectangle.
So. What are the coordinates of that point? Well, this is the
curve Y equals F of X, and so the coordinates of this point
are just XY.
That's the point, P.
What are the coordinates of this point?
While the coordinates at this point, let's call it the point
QRX plus Delta X&Y plus Delta Y, where Delta. Why is that
small increment there?
That we've added on as a result of the small increment that we
made in X.
Let's say that the area under the curve is Delta A.
So this is the area Delta A.
And the area Delta A is contained between the areas of
two rectangles. The larger rectangle.
Whose height is why plus Delta Y?
And whose width is Delta X.
And so that's it's area and the smaller rectangle. Let me just
mark it's top.
Their whose width is Delta X that whose height is just, why?
So it's area will be Y times by Delta X.
So if I want the whole of this area, I have to add up all of
these little bits, which means I have to be able to add up all of
these bits. So let's do
that. A. The total area underneath this curve
is the sum of all these little bits of area from X
equals not up to X equals A and so therefore that area
a is caught between the sum of all of these little bits
from X equals not to X
equals. A at the bottom end.
And at the top end, the sum of all of these little bits
from X equals nor two X equals a or Y plus Delta Y
times by Delta X.
OK, let me multiply out this bit so we have Sigma X
equals North to a of Y times by Delta X Plus Delta
Y times by Delta X greater than a greater than the sum
from X equals not to a of Y times by Delta X.
Now we're going to let Delta X tend to 0, just like we let N go
off to Infinity, which made the strips under Y equals X squared
come down to.
A over N 10 down to a zero thickness, we're going to let
Delta X come down to zero. Now what's going to happen when we
do that? Will certainly here we've got two very, very small
terms. We've got Delta. It's going to 0, so the change in Y
is going to be very very small as well. So this term here is
actually going to run off to 0.
But when that happens, let's just cover that up. We see that
a is trapped between two things which are identical.
And So what we have is that.
A is equal to the limit as Delta
X tends to zero of the sum from
X equals not a of Y Delta X.
Now. With actually made an awful lot of assumptions in what's
going on. And we've had to make them because this level it's
very difficult to do anything other than make these kinds of
assumptions. Let's just have a look at two of them.
For instance, one of the things that we have assumed is that
this curve is increasing all the time for all values of X. But as
we know, curves can decrease as X increases, curves can wave up
and down, they can go up and down as X increases. However,
the results are still the same. We need to do it a little bit
differently, but the results still come out the same. The
other assumption that we're making is that in fact we can
add up. In this form, lots of little bits and still get a
definite finite answer.
And there's a whole raft of pure mathematics, which assures us
that these things do work, so here we are at this stage. Now
this is very, very cumbersome language, and so in fact we
choose to write this as a equals the integral from not a of YDX.
A is the integral from North to a of Y with respect to X,
and it's this notation which replaces that one.
OK, supposing we actually want the area not from North to a,
but between, let us say two limits A and B2 values of X and
the curve. So we want, let's say, to find the area, let's
have a curve here that goes
between there. And.
There.
What would we want to do? Take that back to the Y axis there?
What would we want to do?
Well. This is the area that we're after.
If we think about what that area is, it's the area from nought to
be. Minus the area from North to A.
So this area that we're
wanting. Is equal to the area from North to be, which is
that. Minus the area from North to
a, which is that.
Now it seems quite natural to write that as the.
Integral from A to B of Y with respect to X.
So we would evaluate whatever the result of this computation
was at B and subtract from it the result of whatever the
computation was at a.
And so there we've seen how area can be represented as a
summation and that that leads us to the traditional way of
finding an area which is to integrate the equation of the
curve and substituting limits.
To conclude, this video will just look one further result.
This result will enable us to actually calculate areas by
chopping them up into convenient segments just the same way as we
began. If you remember that shape rather like a dog were
able to chop it up in a convenient segments, find the
area of the separate segments added altogether. Well, this is
this result is going to show us that we can do very much the
same with a curve, so.
Let's suppose that 4 hour curve. We have three ordinance X equals
a X equals B&X equals C and that they are in fact in order of
size a is less than B is less
than C. Let's begin by asking ourselves what's the area
contained by the curve Y equals F of X, and these two ordinate's
and the X axis. Well, it's from A to see the integral of YDX.
And we know that that is the integral from North. To see
of YDX minus the integral from nought to a of YDX.
Equals, well,
rights. Introduce the ordinate be now.
By taking away that.
Area if I've taken away, let's add it on in order to
keep the quality the same.
Now if we look at this bit natural interpretation of that
is again the integral from B to CAYDX and then natural
interpretation of this bit is the integral from A to BYX. So
what we've shown is that we want if we want to find the area
from A to C and there's some
sort of. Inconvenience there around be. Then we can work from
be to see an from A to B and add the two together to give us the
area contained by the curve.