Transcription
Visually, the idea of integration is very intuitive. To find the area, we divide the region into small strips. Then we approximate the area of a strip with a rectangle like this. Then we patch the area of all these rectangles to get the approximate area under this curve. Now, as we increase the number of strips, our approximation gets better and better. And in the limit, the width of the strips tends to zero. The approximate area approaches the area under the curve. This is the idea of integration that we've seen so far. To find the area, we break it into smaller parts and then add the areas of the small parts to get the area of the whole.
But does this idea actually work? In this video, we will see how to exactly apply this idea to find the area under the graph or curve of a function. Consider this function. Let's say we want to find the area under the graph between "X equal to 2 and 6." Can you tell me what is this shape? It's a trapezoid. Even without integration, we know how to find its area. It's equal to the half of its altitude times the sum of its bases. Here, the base of this trapezoid will be these two lines, and its altitude will be this line. Solving this will give us the area of the trapezoid as 16. But how can we use integration to find this area? Will we get the same answer? Let's see that in the next part of the lesson.
How do you think we can find this area using integration? What is the first thing we should do? Correct... we divide this interval into 'n' equal subintervals. Let's denote these subintervals like this. Now, notice that the width of this interval is "6 minus 2," that's "4". So the width of any subinterval will be "equal to 4" over "n". Let's denote it by Delta X. This enables us to measure these subintervals in terms of Delta X. This means "X not is equal to 2," "X1 is equal to 2 plus Delta X." "X2 will be 2 plus 2 times Delta X," and so on. In the end, "6 which is equal to Xn will be equal to 2 plus n times Delta X."
Now, what should we do next? In each subinterval, we find the approximate area under the graph using rectangles. But now observe any one subinterval. We can draw many rectangles here depending on the height we take. So which rectangle should we choose? Notice that the rectangles which cross the graph will have areas closer to the area under the graph. So we should take any rectangle whose height is equal to the value of 'y' corresponding to a point on the graph. For simplicity, let's take the minimum 'y' value in each subinterval as the height of the respective rectangles. We can see that in each subinterval, the value of 'y' corresponding to the left-hand value of X will be the minimum 'y' value. So in the first subinterval, the minimum "Y value will be equal to f of 2." In the second, it will be "f of 2 plus Delta X," and so on. And in the end interval, it will be "f of 2 plus n minus 1 times Delta X."
Now we draw the rectangles choosing these as the heights. The base of each rectangle has width equal to Delta X. So the sum of the areas of these rectangles will be equal to this. On putting the values of the function, we will get this. Now, if we simplify it, we will get this expression. Notice that the series here in the bracket is an arithmetic progression. We know that the sum of an arithmetic progression series is equal to this. Here 'n' is the number of terms. In this series, we have "n minus 1 terms." So we will get the sum of this series to be "n times n minus 1 over 2." So the sum of the areas of these rectangles will be equal to this.
We see that the sum of the areas of these rectangles depends on the number of the subintervals "n" and the width of the subintervals Delta X. But, we know according to this relation that as we increase the number of subintervals, their width decreases. So let's substitute "Delta X as 4 over n" here. Simplifying it will give us this. What should we do next? Let's observe one particular subinterval. If we divide this subinterval further into two parts and draw the rectangle considering the minimum 'y' value in each, we see that the sum of these areas will be closer to the area under the graph than this area. So this means that to get a better approximation, we have to increase the number of subintervals. Notice that the sum of the area of rectangles now only depends on the number of intervals "n". So if we increase the number of intervals, the value of the term "8 over n" will decrease. This means that in the limit 'n' tends to infinity this term "8 over n" will approach "0". So the sum of the areas of rectangles will approach the number 16. This is the area under the graph we found out earlier.
So we see that in the limit 'n' tends to infinity the area of these rectangles approaches the area under the graph. Here, in each subinterval, we took the minimum 'y' value. In this case, we can see that the sum of the areas of the rectangles will always be less than the area under the graph. So this sum of the areas of the rectangles is called the lower sum. For this reason, we denote the sum by putting a lower bar. But now what if instead of the minimum 'y' value in each subinterval we had taken any other 'y' value? Will that also work? What do you think? Let's continue this in the next part.