Transcription
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, no 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 numbers 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 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 an 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 the 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 the 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 the concept of functions. We will also see in our upcoming videos, that in a way, differentiation and integration are OPPOSITE processes. To stay updated, subscribe to our YouTube channel by clicking on this somewhat of an illuminati sign!