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Definite Integrals

Transcript

In this video we will talk about the concept of definite integrals. First, we understand the importance of definite integrals in the J Syllabus as well as in mathematics. Next, we'll talk about the basic formula of how to compute definite integrals. We'll also see an associated example. And lastly we will understand how to compute an area between two of the curves.

So the basic concept of definite integrals talks about the area between a curve and the x axis. Here we will generalize it to see how we can compute area between any two curves. All right, so talking about the importance of definite integrals, this is one of the most important chapters in J in terms of calculus. So basically, every year in J, many questions are asked from definite integrals.

Some of them involve area calculation, others involve other applications of integrals. The takeaway message is that this is an important chapter and you should not be surprised if you find two or three questions from this chapter in J, okay? So now moving to the definition of definite integrals. We know that the area between a curve, so something like this.

So this is the curve, y = f(x). And the area between the curve and the x axis, from point x = a to x = b is given as a equals integration of f(x)dx starting from x = a to x = b, which is nothing but f(b)- f(a).

So f is basically a indefinite integral of small f. All right, let us see a simple problem on this. Consider the function y = x squared. Total the point x = 1 from the origin. So this is 0,0, this is 1,0. And this is a function y = x squared.

So we are asked to find this area. So this area will be, from the above dissert, integration from x = 0 which is A x = 1 which is b here of f(x) which is x squared dx. So this becomes x 3 / 3 from x = 0 to x = 2, 1, which is 1 / 3.

All right, so this is a simple example of how we can use definite integrals to compute area between a curve and the x axis. Now the tricky part happens when we learn to compute the area between two curves. One of them is not x axis. Okay, so let us take a simple example here. Suppose we have the following two curves.

So this is x axis and this is y axis. This is the line y = x. This is the curve y = x squared. So they intersect at this point, (1,1). So this is (1,0). This is origin.

We are asked to find this particular area. Okay, so now we have two curves. And we are supposed to find the area between these two curves. So suppose, let us first understand a generalization. Let the two curves be f(x) and g(x) in the interval a to b. Such that, for all x in the interval a to b, we have f(x) is more than or equal to to g(x).

In such a situation we can say that area bounded between f and g and the interval [a,b] is given as integration of f(x)- g(x), dx from a to b. So let us understand what we just said. We are talking about two functions, f and g, in an interval a to b.

And we are saying that if it holds that for all x and ab, f(x) is more than equal to g(x). Then the area between f and g in the interval a to b is given as integration a to b, f(x) minus g(x) times dx. So if you take a look at this example, in the interval 0 to 1, it is true that for all x, x, f(x) equal to x is more than equal to g(x) is equal to 2x squared.

So this becomes g (x) and this becomes f(x). So therefore, area between f and g. So area, between y = x and y = x squared is given as integration from a to b. So a is 0, b is 1. f(x), which is x, minus g(x), which is x squared, dx. So this becomes, x squared by 2, minus x cubed by 3, from 0 to 1.

So this becomes, 1 by 2, minus 1 by 3. So this is nothing but number 6, so this is the area of the shaded region, as desired. So the whole idea is to identify the curves f and g, which satisfied this criteria, and to appropriately integrate it. So this will help us find the area between any two curves.

Hope this video helps you in understanding the concept of definite integration. In particular, in understanding how to find the area between a curve and the x axis as well as area between two curves.

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