Transcription
हेलो एवरीवन, वेलकम टू माय चैनल.
As you can see on the screen, the first chapter of Calculus is Integration of Functions. An application of this chapter, already covered, is Differentiation. The link for all the chapters from Calculus Differentiation will be in the description and on the I-button. A playlist is already made on the channel, strictly according to the topics, lecture-wise (1, 2, 3, etc.), so you can see the videos.
Now, in this chapter, Improper Integral, what are we going to do? We are going to see what is an improper integral, what is the difference between a proper and an improper integral. Along with that, all the different types of improper integrals, and for each type, we will solve questions. Total 7 types are there. It's quite easy, there's nothing much in it. This chapter is one, and for each type, we will make questions. All the questions we will do will include all the previous year question papers, including 2023. Questions from improper integrals also came in the last year's exam. So, we will do that too in this particular lecture. Watch till the end, so that all the topics are covered in one single lecture. Okay, so let's start quickly.
Also, one more thing I would like to mention is that in this chapter, all the integrals you will see are of the type of definite integral. There is no indefinite integral because in all of them, you will get limits – a lower limit and an upper limit. This is the general form. You will get numbers like 1, 2, 3, 4, 5, 6, or something like that.
So, let's see. Whatever integration we have done so far in Class 12, all integrals are proper integrals. We can say that this is the name of a famous mathematician, and that is an integral. We can say. So, this is what we have done. This is the integral, this is the interval A to B, and in between, that function was f(x). This is some function f(x), and we used to integrate it between the interval, something like this. The same concept is here. Now, there is going to be some change in this, and what is that? First, let me tell you the conditions. When will an integral be called a proper integral or a Riemann integral? When all these three conditions are satisfied. And what are those conditions?
Condition number one: The interval we are taking, A to B, in which we are integrating the function, this interval should be closed. The interval AB should be closed, meaning it should not be an open interval. That's the first point.
Second point: This interval, the integrating interval, should be bounded. What does that mean, I will explain later.
Number three condition: The function we are integrating, the function should also be bounded in the interval. The function is bounded in the interval.
Out of these three conditions, the first condition is a bit considerable, and we will not consider it. And in our syllabus, mainly these two conditions are there: condition number two and condition number three. If one or more of these three conditions are not satisfied, then that integral is called an improper integral.
So, in our syllabus, what is there? First, this number two condition. If it is not satisfied, condition of proper integral. So, what will be the opposite statement? This is also not bounded. Or, I am writing it in short.
Total three types. Type number one. And what is it? You will find that the upper limit is unbounded, like this. It is going to infinity. Meaning, the integration I explained here, this A to B, both A and B are finite values. Suppose, let's say, 2 to 5. Even like 3 to 1000. As many as like large numbers, but they are still finite numbers. But if we are getting something like this, that the upper limit is going towards infinity. Meaning, we have to integrate from, suppose, you have starting from 1, and we have to integrate up to infinity, plus infinity. So, this will be an example of an improper integral. This will become an improper integral. When, if in the limit, we see infinity, then this is it. Yes, why not?
Second type. Or, both the lower limit and upper limit are infinite. We are integrating over the real number line, from minus infinity to plus infinity. It is said that you have to integrate the function over this. So, all these three types are examples. As soon as you see infinity in the limit, either the upper limit or the lower limit, either plus infinity or minus infinity, immediately identify that it is an improper integral. And we will solve it from the perspective of an improper integral. Because now the question arises, sir, if an integral is improper, can we not solve it? Absolutely, we can solve it. And we will see how to solve it. That is, its... What does this mean? Tell me if you didn't understand.
In this, what will happen? In this, you will not see infinity. In both the upper and lower limits, there will be no infinity. Because in this, the limits are finite. But the value of the function will go to infinity. How? Let me give an example. Suppose, suppose in this also, we will see a total of four types. Total four types we will see. A, B, C, and D. So, what is that, I am telling you. So, the first type will be something like this. Suppose I am integrating from, say, 1 to 4, and I have 1 by x - 1 dx. Let's integrate this. Sir, here the function is, whatever is multiplied with dx, that is my function f(x). Something like this. So, this means here, with dx, I see 1 divided by x - 1 multiplied. So, is this my f(x)? Yes, sir, absolutely. See, this lower limit, x = 1. Here, is something happening to it? Something is happening. At x = 1, the function's value is 1 by 0, and 1 by 0 is infinity. This means the value of the function is going to infinity? Yes. So, this means in this category, meaning this, when the function is unbounded, this is type three. Now, what is the problem here, sir? The problem here is that either at the lower limit, at the lower limit, the value of the function is going to infinity, either plus infinity or minus infinity. That's the problem. It will go to infinity. The same can happen at the upper limit. At the upper limit, the value of the function is going to infinity. Then, both upper limit, both can happen. Like, suppose I told you to integrate 2 to 5, 1 by x - 2 dx. See, for this factor 2, there is a problem. As soon as I put 2 here, due to this factor, the value of the function will be 1 by 0, and that is infinity. Meaning, the function is becoming unbounded. So, there is a problem at 2. And due to this factor, there is no problem at the lower limit. So, where is the problem? At some point C between the upper limit and the lower limit, if the value of the function goes to infinity. Okay. So, total here, 3 + 4, total 7 types we will see from improper integral. Okay.
So, let's start one by one. So, type number one. Meaning, in this, you will see infinity either at the upper limit or at the lower limit. Plus infinity and minus infinity, or both upper and lower limits. And its type one I have taken here. You can see, type one means we will see infinity only in the upper limit. And this question is from 2023. As you can see, it's a question from last year's previous year question paper. Okay. And you can see the limit is 0 to infinity e to the power minus 6x. So, what is the problem with this? So, the problem with this, in short, the problem is at... Let me take some other link. Let's say its problem is at this plus infinity. Meaning, this question, let me tell you on the number line here. Suppose this is 0, and this is plus infinity. So, we have to integrate from 0 to infinity. Now, we cannot integrate at infinity. So, we will use some technique. We will do its integration up to some finite value M. Meaning, from 0 to this M, we will do its integration. So, what will we do? Instead of infinity, we will integrate from 0 to M. But sir, this is not right. What is the value of M? Suppose M is 1000, or 1 lakh, whatever it is. We can only do it up to the value of M. Not up to infinity. But sir, this is not right. Because we had to do it up to infinity. So, after this, we will put a limit in front of it, and it will become infinity. So, there is no problem. Because sir, the question had infinity. I have made it. But like this, by twisting it, you have to form it. So, work is done here. The problem was up to plus infinity. That's why we took some finite value, 1 lakh, 10 lakh, 1 crore, whatever. 1 crore is also finite. No problem. We will keep it. You will not write any particular value. We will write M. And then cleverly, we will put a limit in front of it. Limit tends to infinity. As soon as the limit tends to infinity, indirectly the upper limit will become infinity. What will happen? Minus of minus. It will become plus. There is no space. So, I will put the limit here in place of M. So, I will write minus of minus infinity. What happened? In place of M, I will put the limit. And anything to the power zero is one. We will know that. Now, let's understand what this is. Minus infinity. See, what is 1 by x? It is an increasing function. So, as the value of x increases and goes to infinity, similarly, the value of e to the power x also goes to infinity. The value of infinity is simple. This means what happened, sir? This became minus 0 plus 1. The final answer is obtained. Clear, everyone? Absolutely like this, you have to do it. Problem.
Next, type two. What is the problem here? No problem. We cannot integrate. The same concept will be applied. Automatically, it will become. So, many people have not subscribed yet. We have to reach 1000 subscribers. Please do it. It will be a great help. So, like, share with friends as much as possible. And subscribe. 2018.
Okay, so one thing I would like to mention here is that as soon as you get a problem like this, where both the lower limit and the upper limit have some problem, then what you have to do is, we have to break the interval. Break the main interval, minus infinity to infinity, in between some point. Meaning, break the interval at some point. You can take any particular value. Suppose some student did it. Sir, I will break it from minus infinity to 0, and then 0 to infinity. Okay, no problem. And you can do minus infinity to, say, 1, 1 to infinity. It's fine. Minus infinity to minus 1, minus 1 to infinity. It's fine. No problem. But I will suggest keeping it in a general form. Meaning, without taking any particular value, use some C. What example? Okay. I did not take 0, 1, nothing. I took C. Now, C can be anything. Any 0, 1, 2, 3, 4, 5, 6. Any value. No problem. This breaking is important. You have to break it in the very first step. And be absolutely sure. Because do not worry about this C. Because in the end, this C term will cancel out. And that's why I took it without any worry. Because I know it will cancel out in the end. No problem. So, you can take 1, 2, 3, 4, whatever you take. Ultimately, the answer will be the same because that term will cancel out. And some okay. No.
Okay. So, this became minus 2x. C to infinity. Now, what did I say? Here, you can take any number. No problem. Now, the first part, this has become. Now, we have done type two. Because its lower limit is minus infinity. Type one. Because its upper limit has infinity. Right? So, now how to do this? Just the way we did type one and type two. Absolutely like that, you have to solve it. Also, one thing to do here is to integrate this x squared minus x squared separately. So, it will help. Directly putting the value you get after integrating it here will be right. Because otherwise, what will happen is, you have to integrate twice. You have to write the same thing here and here. So, because of this, you can do it separately on the side. No problem. But it should be kept with the main question, slightly on the right side or left side, somewhere in the copy. It is not a graph. It is part of the solution of the question. Okay.
Also, how to integrate this? Simple. x squared. So, this will become 2x. It is a substitution. Basic integration of Class 12. And its integration will be 1 by 2 minus x. And a minus sign will come in front. M to C. x into dx. Plus right. C to M. C to capital. Okay. So, as soon as you put this here, you will get. I am standing on the next page. Okay. Taking out such a large minus. I am writing it outside. Okay. Because both have minus half. So, I can take it common and write it outside. This minus half is in both. So, I will get a limit. Limit tends to smallest. Minus infinity tends to plus infinity. To the power minus capital M square. Getting confused? If you can do M1, M2 as well. And x, y, whatever you feel comfortable with. I mean, my convention is convention. And I am okay with it. Small M, capital M. And whatever else you feel comfortable with. Okay.
So, this is done. Now, pay attention. N. Here, these two terms with C. Meaning, this one and this one. There is no M in this. So, there is no function of limit here. So, what can I write? I will put my limit here. So, this will become. It to the power minus 2. Minus of e to the power. Now, see what is here. M square. And here, the value of M will be put. Minus infinity. So, minus infinity whole square will be infinity. And infinity is a very large number. So, its square will also be a larger number. Meaning, infinity. Simple. Plus. Now, what is here? Same logic for this also. Okay. Plus minus infinity. Yes. So, this will become. And here. Okay. No problem. Because this will cancel out. This will cancel out. Something like this. And this term, this is zero anyway. Because as I told you, minus infinity. Once minus infinity means 1 by infinity, which is actually zero. 1 by infinity means zero. Okay.
[Music]
Okay. So, what did we do? Once, let me quickly recharge. Because we have to integrate this from infinity. So, what did we do? We broke it at some point C in between. And then automatically, it became type two and type one, which we already did. The same concept I used. M10 to minus infinity and capital infinity to plus infinity. Okay. Then, after that, we did the same upper limit minus lower limit. Then, using the limit, this C term. There is no problem with the C term. Because the C term will ultimately cancel out. The C term. Meaning, instead of C, you can take any particular value. Like I told you, 10, 2, 3, 4, or minus 1, 2, 1, whatever. Ultimately, it will cancel out. So, no problem. The final result will be something like zero. Okay.
Also, now we can start. Integral of the second kind. And what will happen in this? As I told you, in this, the function will be unbounded. Second kind. In this, the function is unbounded. Meaning, the value of the function will go up to infinity. And one more thing I would like to mention is that the point at which the value of the function goes to infinity, that point is also called the point of infinite discontinuity or singularity. That point is also called singularity, point of infinite discontinuity, or singularity. Okay. So, this is a good question. Test the convergence of the improper integral. This and hence evaluate if possible. Okay. Meaning, the question here is integration 2 to 4 dx by square root of x - 2. At 2, that is, at the lower limit, the value, the value of the function will be unbounded. And x = 2, we call it a singular point. Point. You will only see this at the lower limit. The point of continuity and singularity is only at the upper limit. Problem. Okay. So, sir, how to solve this? Yes, there is a solution for this too. And what is it? Let me show you a bit. So, we have to integrate this. I will tell you. Suppose this 2 is from 2 to 4, we have to integrate this. Here is plus infinity. Here is minus infinity. 0. 4. Suppose from 2 to 4. We have to integrate. Now, there is no problem at 4. The function at 4 is well and good. But its problem is at 2. And we cannot start integration from 2. Because at 2, we have singularity. The function will be unbounded at x = 2. So, how to do this? So, what will we do? We will again use some technique. And we will start this from some point. Let this point be some 2 minus a very small quantity. Opposite epsilon. The value of epsilon is 0.01 or something like that. A small quantity, but positive. 0.01, suppose. If I have to put a particular value. Okay. Now, from this, what will happen? Now, see. Now, this is... What am I saying? We will start integrating from 2 minus epsilon. Now, what happened? This is not 2 anymore. The lower limit where we had a problem, at 2, it is no longer 2. It has become 2 minus. Meaning, it has become something like 1.99. But it is not exactly 2. So, now we can integrate it from 1.99. And we did that. And after that, we will again use some technique in front of it. And we will put a limit epsilon tends to 0 plus. And as soon as epsilon tends to 0 plus, 2 minus epsilon will automatically become 2. Same concept. Twisting the nose. Not directly. Twisting it. Okay. So, again, we will use the same limit concept. And I am telling you again and again, if you are saying that this is an improper integral, you cannot solve it normally without using limits. As soon as you know that. Okay. Now, as we do this, we have to do this. After this, the integration of dx by x - 2. Everything else is the same. Then, after that, we have to integrate. And after integrating, you have to do upper limit minus. Same concept. I hope this concept I explained here is understood. We had a problem at 2. That's why we did 2 minus. Oh, one thing I realized is that if I do 2 minus, I will go out of this interval. Which we cannot do. We have to stay within this interval. From 2 to 4, the interval is. We have to stay within this. We cannot go outside. We cannot do 2 minus here. Okay. Good that I made a mistake. So, this is known. Doing 2 minus, we go out of this interval. To something like 1.99, which is not allowed. Because we have to stay within the interval 2 and 4. To stay inside the interval 2 and 4. Okay. And because of this, what we did is, we did 2 plus epsilon. 2 plus epsilon equals to 0 plus. Important limit. Right-hand limit. So, this plus minus is for the direction. A bit. Okay. So, there is no important. So, much deep going is not necessary. Just write plus there. And nothing else. And then its integration will be 2 into 1 of x - 2. After integrating. Then, the same upper limit. Write it. Where will I write it? I will write it here. So, 2 to 2 will cancel. So, what am I getting here? Limit epsilon to root 2 minus value will be zero. Okay. Are you understanding? Please like, subscribe, and share. And let's go. Now, this question. [Music]
Now, sir, what is the problem here? Because at x = 1, the value of the function becomes singularity and infinity. Discontinuity. Right? Getting my point? Okay. So, this also has the same technique. So, what do you see here? We have to integrate from 0 to 1. Okay. This is plus infinity. This is minus infinity. 0. But we cannot integrate directly at 1. Because at 1, the function... We have to stay between the function. You have to stay inside the interval. We cannot go outside the interval 0 and 1. So, what will we do? Very small quantity. Suppose 0.1. Even smaller. Even smaller. But positive. But positive. We cannot integrate. So, we will stop a bit ahead of 1. That is, up to some 1 minus epsilon. We will integrate it. And again, cleverly, we will use a limit in front of it. Limit epsilon tends to 0 plus. As soon as epsilon tends to 0 plus, on 1 minus. It will automatically become 1. Same concept. What is the concept of limit? It seems to be 2. Meaning, it is. So, this will become sin inverse. So, sin inverse 0 is 0. And sin inverse 1 is pi by 2. This means my final answer became, sir, pi by 2. Very good. Very good. Clear. Okay. [Music]
Right? Absolutely. One by one. So, the value of the function is becoming 1 by 0. That is, it is going to infinity. So, both upper limit and lower limit. Both points are the only points of infinity. See here. There is no problem at any other point. This denominator. At x = 0 and x = 2, there is only a problem. Otherwise, at any other point, there is no problem. So, this means 0 and x = 2. And 2 only. Points of continuity. Problem is found. So, I told you at the beginning. The previous kind, its type three. It also had minus infinity to plus infinity. And there I told you that if there is a problem in both limits, then learn to break it. Break the interval at any point in between. You could have broken it at 1. 0 to 1 and 1 to 2. I kept it general. Because I already told you, whatever you take in the middle, it will cancel out in the end. It will cancel out. No problem. Okay.
Now, we have to integrate this. What is the technique for its integration? You have to integrate it yourself. Solve it. I will give you a hint. This is absolutely basic integration of Class 12. Here, I will show you its integration a bit. How to do it. x - 9 + 2 - 6. I will write it on top. What will come on top? [Music] This x cancels with x. Only 2 will remain, which was not there. So, to adjust, just 1 by 2 outside. Now, we have to break it. It will come in log. The answer will come in log. One will be dx by 2 - x. And one will be 1 by 2 into dx by. We will do this. Here, it will become 1 by 2. Thanks by 2 minus x. Okay.
It has become. Now, we just did type two. And this has become type one. The technique is the same. It's a very easy chapter. I am telling you again and again. So, now, what is it? This factor has a problem at 0. So, we will not integrate it from 0. Meaning, see what happened. We have to integrate this. From 0 to 2. Okay. We broke it at some point C in between. The interval. So, now this has become 0 to C and C to 2. Now, we cannot integrate from 0. Because at 0, it has infinite discontinuity. Point of infinity. So, we have to start from 0 plus. 0 plus epsilon. But positive. We cannot integrate. So, up to some ahead of 2. We have to stop. That is, up to some 2 minus epsilon. We will integrate it. Some 2 minus by x minus x into 2 minus x. And put a limit in front. Limit epsilon tends to 0 plus. Okay. Now, after integrating this, upper limit minus lower limit. Integration is already shown here. Now, after integrating, you have to do upper limit minus. And as soon as you do it, we will get. I am writing it directly here. Okay. So, this will become. We will get something like this. Limit. A half will remain. Same common will come. Half will come outside, common. And here also, it will become limit epsilon. Log of C by 2 minus log of C by 2 minus C. And third bracket. This log C term. Here, this log C term. There is no function of limit on this. So, I will put it in another bracket. So, that will understand even better. This minus. Now, here, limit epsilon. I am explaining things. That's why the video will be a bit lengthy. Because explaining all the terms in very detail. And nothing else. So, be patient. Nothing. Okay. It will become defined. Infinity. And log infinity's value. And here also, same problem. What will be its final answer? And the answer will be limit does not exist. Not exist. Will be my final answer. Answer. Because here, the question is like this. The examiner. And this has come in the exam in 2012. In the convergence of the improper integral. Now, what does convergence mean? When we calculate. When we solve the improper integral and find any finite value. Suppose, after solving, we get 1, or 0, minus 1, 5 by 2. All these finite values. And when you are getting a finite value, then we say that the improper integral is convergent. It is converging. It is converging to that value. In the later chapters, we will do this. So, in short, convergence means when we are getting a finite value after solving. Otherwise, it is not convergent. Divergent. And those questions where it does not converge. In such questions, we have to write that the limit does not exist. And this will be my final answer. Limit does not exist. It is not wrong. It is absolutely correct. Okay. So, this will be your final answer. That the limit does not exist.
Let's do the question. This question. [Music] And this is a type four question. Why? Because see here, the lower limit is -1. At -1, there is no problem. Meaning, absolutely okay. At x = -1, absolutely okay. At x = +1, no problem. No problem. No problem. No problem. So, sir, where is its problem? Problem means singular point. Point of infinite discontinuity. Where is it? How to find it? Make the denominator zero. If it falls between this interval, -1 to 2. That's why this will become. Let's start. Same concept. Equal to. Break the interval at the point at which we have the point of discontinuity. Which is 0 here. There is a problem at 0. So, you have to break it like this. Something like this. Okay.
It has become. Now, we just did. And the second kind. Type one and type two has become. Because now this part. It has a problem at the upper limit. So, we cannot integrate it up to the upper limit. [Music] dx by x square. [Music] Plus limit epsilon tends to 0 plus. Minus 1. It will cancel. I am telling you every time. This middle point, at which we break it, it cancels out. Yes. Take any particular value. Or take it. Okay. Equals to. Now, understand. 1 by epsilon. Here, as soon as I put 0. Or here, as soon as I put 0, it will become infinity. What will be its answer? The final answer will be limit does not exist. Finally, it will be the final answer. Because you are not getting any finite value. Right? Below zero means the value is becoming infinity. Right? So, the final answer for me will be limit does not exist. Finally, it will be the final answer for this question. Once again. Okay.
Also, now this is a slightly important part. Let me tell you about this. Cauchy Principal Value. Cauchy Principal Value. See, generally, when they are not asking for it, and Cauchy Principal Value, we will not do it. We will do it generally. As we have been doing. But if they are asking us to find the general Cauchy Principal Value. 2018. What do I write? I write anything. Anything. Which general form? For the general, it will exist. How? See, same concept. Meaning, the initial concept is the same. Because here, the point of infinite discontinuity is 0. So, we will break it. -1 to 0. dx by. Same concept. Then, here we will limit. Meaning, we cannot integrate up to 0. So, -1 to 0. Now, I will do something. 0 minus. Because we cannot do that. So, 0 minus. And we cannot start it. Up to 1. The फंडा of Cauchy Principal Value is that you have to take two different epsilons. Epsilon, epsilon. There is no need to take two. You will take the same quantity for both. You will take it in one epsilon. This is the फंडा of Cauchy Principal Value. So, it will go from both. So, limit epsilon tends to 0 plus. Limit epsilon tends to 0 plus. Okay.
Now, we will integrate. What is integration? Minus 3. Meaning, minus 3 plus 1. Below also minus 3 plus 1. Meaning, minus 2. Log. Okay. So, its integration will be minus 1 by 2 x square. Okay. So, I will write it once. Outside limit. Epsilon 1. Something like this. Right? So, this will become minus 1 by 2. It will become plus. Done. Cancelled. And these two will also cancel. Zero. This means in the sense of principal value, this is convergence. And the value. If I don't get the final answer, then I will say that the limit does not exist. Finally, I got the value of the limit. In the general sense, when we are keeping both separately, that is the general sense. A little bit of subscription. I will cover the entire syllabus. For that, subscribe. Press the bell icon. Because otherwise, you will not get the notification. And share with your friends. And like it if you like it. So, that's all for this video, everyone. Yes, see you in the next video. Thank you.