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 functions. 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 sub intervals.
Let's denote these sub intervals like this.
Now notice that the width of this interval is "6 minus 2", that's "4".
So the width of any sub interval will be "equal to 4" over "n".
Let's denote it by Delta X.
This enables us to measure these sub intervals 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 sub interval, we find the approximate area,
under the graph using rectangles.
But now observe any one sub interval.
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 sub interval
as the height of the respective rectangles.
We can see that in each sub interval the value of 'y'
corresponding to the left-hand value of X
will be the minimum 'y' value.
So in the first sub interval, 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.
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 sub intervals "n"
and the width of the sub intervals Delta X.
But, we know according to this relation that
as we increase the number of sub intervals,
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 sub interval.
If we divide this sub interval 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 sub intervals.
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 sub interval, 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 sub interval
we had taken any other 'y' value?
Will that also work?
What do you think?
Let's continue this in the next part.
