www.mathcentre.ac.uk/.../9.1%20Integration%20as%20a%20summation.mp4

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.