Can you tell me how to find the area of this
regular polygon?
Let me give you a hint by drawing these dotted
lines. Did it click? The area of this polygon
can be looked at as the sum of the areas of
these triangles. The area of this regular
polygon will be the sum of the areas of these
8 triangles.
Similarly, we can find the areas of most of
the regions bounded by straight lines. We
simply have to decompose the areas into triangles
and rectangles and our job is done. But what
about the areas of regions bounded by curved
lines. We can’t fit the triangles and rectangles
there right?
As we mentioned previously, this is where
Calculus comes into picture. In this video,
we will see a general process called Integration
to calculate the area of any region, not matter
what its shape is. The beauty here is that
this process is also based on the idea of
limits, something that we’re now confident
about.In ancient times, people knew how to deal
with regions bounded by straight lines. But
finding the area of even a simple curved shape
like a circle posed a problem. In one of our
previous videos, we saw that the Greek mathematicians
had found some headway on it. Do you remember
the idea that was used here?
If the circle is placed between two squares
like this, we get a rough estimate of the
area of the circle. It’ll be between the
areas of the two squares.
And as we consider regular polygons with even
greater number of sides, our approximations
will get better. Eventually, the area of the
polygons will approach the area of the circle.
This method of approximation is called the
method of EXHAUSTION. Let’s look at another
major achievement of the Greeks in finding
such areas.
Look at this curve. You may have seen or learned
about this type of curve. Do you know what
it is called? It is called a PARABOLA.
Let’s say we have a horizontal straight
line like this and this point below it.
Now if we consider points such that their distance
from this line and this point is the same,
we get a parabola. So the distance of all
the points from the parabola to this straight
line will be the same as their respective
distance from this point.
Remember seeing a similar curve in our previous
videos? The stone thrown by Nora into the
sea follows a parabolic path. Actually, whenever
we throw something up, the object always comes
down following a parabolic path. Unless it’s
straight up and down!
A Greek mathematician named Archimedes was
successful in finding the area of the region
between a parabola and a chord like this.
He did it by the method of exhaustion. In
the case of circles, we had seen that to exhaust
the area of the circle, regular polygons were
used. But notice that the regular polygons
will not work here. Can you think of other
shapes we can use here?
Archimedes realised that TRIANGLES would work
here. How? And why?
First he chose a point ‘C’ on the curve
such that the tangent line at this point is
parallel to the this chord. Now look at his
triangle. Let’s say the area of this triangle
is ‘alpha’. Then it turns out that the
area of this region is equal to ‘four over
three’ times the area of this triangle,
which is alpha. How?
If we look at these two regions, we can draw
two more triangles like this. And it turns
out by geometry that the sum of the areas
of these two triangles is one fourth of the
area of triangle ‘A B C’. So now the area
of this region will be equal to ‘alpha plus
one fourth of alpha plus the area of this
remaining region.
Did you understand the idea here? If we keep
drawing triangles further and further we will
eventually cover the whole region. We get
that the sum of the areas of the triangles
added in each next step is equal to one fourth
of the sum of the areas of triangles in the
previous step. So we get infinite terms like
this. And it turns out that their sum approaches
‘four over three’ times ‘alpha’.
Note that I said APPROACHES here. As we saw
previously, we cannot add infinite terms manually.
As we keep adding the terms, we can only find
whether the sum approaches a number or not.
So we get the area bounded by this curved
line and this chord to be ‘four over three’
times the area of this triangle. Note that
this is the point at which the tangent is
parallel to this chord.
This result proved by Archimedes is called
the quadrature of parabola.
But did you notice a problem here? To find
the area bounded by different kinds of curved
lines, each time we have to find out which
shapes will exhaust the given area.
If we have area bounded by curved lines like
this, how will we find the area then? We need
to develop a GENERAL method to find the area
of any region.
It turns out that the answer to our problems
is RECTANGLES. In the next part we will see
the general method used to find area of ANY region.
Let’s consider this area. It is bounded
by this curve, these two lines parallel to
the Y axis, and the X axis.
We want to find the area under this curve.
But you may ask why! We will see in our upcoming
videos that most of our problems related to
finding areas and volumes can be reduced to
this problem.
For example, let's say we want to find the
volume and area of this solid, which is an
elongated sphere. Now let’s cut this solid
like this and look at its cross section. If
we know how to find the area under this curve
and above the X axis, then our job is done.
We can see that rotating this area by three
hundred and sixty degrees will give us the
‘volume of the solid’. So this area was
enough to find the volume of the solid.
Also, once we understand the method of integration,
we will see that it can be applied to find
the length of a curve. So if we know the length
of this curve, rotating it by three hundred
and sixty degrees will give us the surface
area of this solid. We will understand it
in detail in our upcoming videos. Now let’s
get back to our problem of finding the area
under this curve.
We know that according to these co-ordinate
axes, a curve can be looked at as being made
up of the different points. Each point on
the curve is specified by its ‘X’ co-ordinate
and ‘Y’ co-ordinate.
Now tell me which points on this curve will
have the minimum and maximum value for their
‘Y’ co-ordinate?
These will be the two extreme points of the
curve ‘A B’. Let’s say the minimum ‘Y’
value is ‘alpha’ and the maximum ‘Y’
value is ‘beta’.
Now by taking these two points, we draw two
rectangles like this. For better understanding
let’s draw these rectangles in two separate
figures.
So we see that the area under this curve will
be GREATER than the area of this rectangle.
The area of this rectangle will be ‘alpha’
time ‘this length’. And this length is
‘D minus C’. Hence the area of the rectangle
will be ‘alpha times D minus C’. And it
can be seen that the area under this curve
will be LESS than the area of this rectangle;
which is ‘beta times D minus C’.
Let’s denote the area under this curve by
‘A’. And we denote the area of this rectangle
as ‘S one’ with a dash below it. This
dash below, tells us that the area of this
rectangle is less than ‘A’. Similarly
we denote the area of this rectangle also
with ‘S one’ but with a dash above it.
Now focus on the figure on the left. We divide
this interval into two equal parts and draw
two rectangles like this.
Now compared to the previous case, we get
a better estimate of the area. Let’s denote
the SUM of the areas of these rectangles by
‘S two dash’. We can see that it will
be greater than ‘S one dash’ but less
than ‘A’. Similarly if we divide this
interval into more parts, let’s say ‘N
parts’, then we will get an even better
estimate. These rectangles are drawn by taking
the minimum value of ‘Y’ in each part.
So we can see that as we divide this interval
even further, we will get closer and closer
to the area ‘A’.
In a similar manner, in the figure on the
right, we divide the interval into ‘N’
parts and draw these rectangles. Note that
these rectangles are drawn by taking the MAXIMUM
‘Y’ value not the minimum. So the sum
of the areas of these rectangles, ‘S N dash’
will be less than ‘S one dash’ but greater
than ‘A’.
Now as we keep dividing this interval further
and find the sum of areas of such rectangles,
we will get closer to the area ‘A’. This
is the general approach to find the area.
It is called Integration.Integrating is
nothing but combining one thing with another
to form a whole!
Using this method we can find area under any curve.
What we did here is that we first divided
this interval into small parts. Then by taking
the minimum and maximum ‘Y’ value in each
part, we calculated the two approximate areas.
The first one will always be less than the
area under this curve and is hence called
the lower sum. And the second one will always
be greater than the area under this curve
and is hence called the upper sum. Now as
we divide this interval into smaller and smaller
parts, we see that the lower sum and upper
sum approach a number, which is the area under
this curve. This is the basic idea of integration.
We will learn about it in detail in our upcoming
videos.
So until now we have an idea of the limit
process. We saw how we can apply it to find
the instantaneous rate of change of any quantity.
The process of finding this rate is called
differentiation. In this video, we saw how
it is applied to find the area under a curve.
And this process is called Integration.
In the next video, we will look at one of
the most important ideas of mathematics called
functions. We can understand the true essence
of all these ideas once we understand concept
of functions. We will also see in our upcoming
videos, that in a way, differentiation and
integration are OPPOSITE processes. To stay
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