 All right guys, let's start, it's already time. Hope everybody can see my screen and can hear me loud and clear. So last class, we had started talking about Leveny's rule, right? So I'm going to start with some problems which are based directly on the use of Leveny's rule. Even though we had done few problems, but there are a couple of types of problems which we need to address, okay? So those who have forgotten about Leveny's rule, let me just quickly recap for you. Good afternoon, everyone. So Leveny's rule was used to find the derivative under the integral sign. So let's say you have a function, it may be a multivariate function. In this case, I'm using two variables, x and t, where I'm integrating with respect to t, whereas I'm differentiating with respect to x, correct? So this rule says that the derivative of this integral, you don't have to evaluate the integral, you can directly use Leveny's rule. This integral has been done from phi of x to psi of x, okay? So the rule says integral from phi of x to psi of x of the partial derivative of this function f of x comma t, x comma t, okay? With respect to x, and whatever result comes out, integral of that with respect to t, plus f of x comma psi x times psi dash x minus f of x comma phi x into phi dash x. When I say phi dash and psi dash, I mean derivative of those functions with respect to x. Is that fine? Hope everyone has joined in. Okay, Khyati just joined. So this was the Leveny's rule for derivative under the integral sign, okay? Just a recall of formula, I'm not taught anything right now. I'm going to begin this session with a problem on this. Fine, so let's start with the problem. The first problem that we are going to do here is, evaluate, evaluate, integral of x to the power b minus one by ln x from zero to one, from zero to one. Here b is a positive number. So all of you please give it a try. So this is the question which is basically a simple looking definite integral, right? But when you start solving it by the use of the properties, you would realize that it's going nowhere, okay? I would like all of you to take two minutes of your time and try to attempt this question now. So many times what happens is, the question is a question on Leveny's rule, but there is no clue, no indication whatsoever that it requires the use of Leveny's rule. Ramya, can you just switch on to your computer audio or if you're using your phone, then to your phone audio? Everybody can hear me? Yeah, there's an audio setting, just check the audio setting. Just switch to computer audio if you're using your laptop. Okay, now you would have tried all your methods or properties of definite integral, but I'm sure you must not have got success in this, okay? Now all of us first try to understand here, this is an integral where you are performing the integration with respect to X. And of course your limits are basically put on X, right? So the answer that you will get from here would be in terms of B, correct? So this is very clear to all of us that the answer for this will actually come out in terms of B, okay? So whatever is your B, okay? The answer will be in terms of that B only because your integration is performed on X and your limits are also put for X, correct? So what I'm going to do here, I'm going to first name this integral as a function of B, okay? I'm going to first name this integral as a function of B, fine? Now I'm going to differentiate both sides with respect to B, right? Isn't this surprising because you never expected this to happen? That's where I was saying that many a times it is, you know, unpredictable or you may not even think that there is a use of Leveny's rule in the question and it may require the use of it, okay? So I'll differentiate both sides with respect to B. Now many students ask me this question, sir, B is a constant. How are you differentiating with respect to B? See here, B is a parameter, not exactly a constant, okay? Because for different, different B, your answer will be different, isn't it? So you can treat this as if there is a function in B. In the question per se, B is acting like a constant, but overall, when you evaluate this, your answer comes out in terms of B, which is actually a function of B. And we can definitely differentiate with respect to B throughout, right? So you can treat this as if it is a multivariate function in X and B. So you can treat this as if this is a multivariate function in X and B, or you can say B and X, correct? So think as if you have been, you have been given a scenario like this where you are trying to integrate F of B comma X with respect to X. And you're differentiating both sides with respect to B. Right? So what will you do here? We'll use our Leveny's rule and we'll use this rule, which I have just discussed with you over here. So this rule will be used here, all right? So if you use this rule, it'll become dou by dou B of this function, X to the power B minus one by ln X, okay? The whole thing integral with respect to X from zero to one. And remember the rest of the terms, that is this term and this term you need not write. Why? Because your upper and the lower limit are constants. And they will anyways disappear when you differentiate the constants here, okay? When you do Psi dash X and Phi dash X, those constants anyways will disappear, okay? So there is no need to write the other terms, okay? So far so good. So the rest of the terms, I'm just writing it as zero minus zero for your reference. Is this clear so far? Just write CLR on your chat box if it is clear. Awesome, awesome, good, good, good, good. Fine. So now a question for all of you over here, how do you differentiate this with respect to B? That means you're treating X as a constant, partial derivative with respect to B means X is a constant. So can I say ln X will also be a constant, so it'll be just one by ln X? Now X to the power B derivative with respect to B is the same way as you differentiate A to the power X with respect to X, which is actually A to the power X ln A. So in this case, it will be X to the power B ln X, correct? One derivative anyways, it's a constant, so whether you are differentiating with respect to B or X, it becomes zero, doesn't matter. And the entire integral boils down to this. ln X, ln X goes for a toss, and this is something which anyone can integrate X to the power B, so which is X to the power B plus one by B plus one, okay? Integral from zero to one, that's going to give you one by B plus one, okay? So right now what we have done, we have found out, we have found out, I should not call this as F because F has already been used. I will use a different name over here. Let me name, let me use a name here, let's say G of something, okay? So this is your derivative of the function with respect to B. So what we have found out here was the derivative of my answer with respect to B. Remember I have to find F of B, not F dash B. But the root to F of B will go via F dash B. So here we have F dash B result. Now in order to find F of B, you will say simple, integrate both sides with respect to B. That's what we are going to do next. So F of B, F of B could be obtained by integrating one by B plus one, both sides with respect to B, which is nothing but ln of B plus one, plus a constant of integration. Guys, remember one more important thing, never forget the constant of integration, never forget the constant of integration. It's going to change the game for you, okay? I've seen these common mistakes made by students. First of all, they don't write DX while integrating. Secondly, after integrating, they give no respect, no value to constant of integration, which is going to be very, very costly or dear to you when you're doing the differential equations chapter, okay? So don't forget the C. By the way, some of you would be surprised why did I write ln B plus one and not ln mod B plus one because we already know that B is positive. So B plus one is also a positive. So mod becomes redundant there, okay? So far, so good. Now what is C here? To get this C, all of you please pay attention to this fact. Do you realize that? Come to this expression. Do you realize that when your B is zero, your F of zero will be also zero? Why? Because then you'll be integrating from zero to one, X to the power zero minus one by ln X. Correct? X to the power zero is already one. So one minus one will go zero and everything will become a zero. So this becomes our boundary condition to find the value of the constant of integration over here. So let's put that. So let's put B as zero and F zero will also be zero. So this means zero will be equal to ln zero plus one plus C, okay? Which is nothing but zero is equal to zero plus C. So luckily here, C was also zero, okay? That means your answer to this question is nothing but ln of B plus one. So this becomes your answer, okay? So certainly this is a new type of problem that you have seen. Here you actually differentiate it to integrate something, isn't it? So direct integration was not possible. So you applied your Leibniz rule first. You converted it to such an integral which was easy to evaluate and then you anti-differentiated it and got your answer, right? Isn't it a new type of problem which you have seen? And I'm sure you would like to try one more, okay? So let's see another problem. By the way, Ramya, are you able to hear me now? Oh, how would she hear this because Shri Ramya, are you able to hear me? Can you hear my voice? Yes, she can, okay, awesome. So here comes the next problem for you guys. Let's see who's able to crack this. Evaluate, evaluate. Integral zero to pi by two, log to the base E. Remember if I don't mention the base, take that to be log one plus x sine square theta, whole divided by sine square theta, d theta, okay? Here x is a positive quantity. Here x is a positive quantity. For a change, I'll give you an option in this problem. So it's pi under root of one plus x minus one. Option B is pi under root of one plus x minus two. A root pi under root of one plus x minus one. And option D is no tau, no tau. Okay, two minutes to try this. Nikhil, actually I gave the option so that you know you are going on a right track. Oh, sure, I'll go to the previous page. Sorry, I did not read your chat. Okay, all right, just give me a second. Ramya wants to see the previous board. Sorry, Shiram wants to see the previous board. Shiram, just let me know once you're done. Just let me know when both of you are done. I'll just take a picture of it if you want. If you want to look at it for a longer time, just take a picture. Okay, now look at this question. I'm sure nobody could have even imagined that this would be solved by the use of Leibniz rule. I'm sure 90% of the people who get this question for the first time would spend at least five minutes of their time figuring it out which property is going to fit in over here. And after all the trial and error and disappointment, then they would, okay, let's apply Leibniz rule. But here, since you know that we are talking about this topic, you can start with that. Okay, Nikhil says C. I'm not saying right or wrong to it. Let's wait for at least three people to answer. Guys, don't get confused. It's a multivariate function. One is theta, other is X. Remember, you are integrating with respect to theta. Okay, so few people are going for option A. Okay, fine, then let's try it out. Okay, so let us first differentiate. Now, you know that after integrating with respect to theta, you will end up getting something in terms of X, correct? So let me call this as a function of X, okay? So let's differentiate both sides with respect to X, okay? So if you're differentiating both sides with respect to X, according to the Leibniz rule, you will first do the partial derivative of this with respect to X, okay? So you are partially differentiating this with respect to X. And whatever you end up getting, you have to integrate that with respect to theta, okay? Remember, I'm not writing the other terms because they will anyways be zero because your upper and the lower limits of this integration are constants, right? So F dash X would be equal to, okay? So what is the derivative of this with respect to X? Just once again, I think Shugosh is having a problem. People who are watching via phone, how do you make sure that you can hear? What is the audio settings that you do for that? Is there anybody who is watching through phone? Yeah, so if you do a partial derivative, this will become one by one plus X sine square theta, okay? Correct? And you have to differ, and one by sine square theta is anyways a constant, correct? And you have to differentiate one plus X sine square theta with respect to X. So that will give you again a sine square theta. Is that fine? That'll again give you a sine square theta. So this sine square and this sine square will get canceled off. And basically you're left with integral of one by one plus X sine square theta with respect to theta, right? So F dash X, you need to integrate zero to pi by two, one by one plus X sine square theta with respect to theta. So how do I integrate this? Very simple, I would first divide the numerator and denominator by cos square theta. So basically I'm applying my knowledge of indefinite integrals for that. So let us divide my numerator and denominator by cos square theta. So that's going to become zero to pi by two secant square theta by secant square theta plus X tan square theta. Okay? Let's write everything in the denominator in terms of tan square. So it will become one plus X plus one tan square theta. Let's take tan theta as some other variable U. So secant square theta d theta is going to be du, correct? So F dash X is going to be integral du by one plus X plus one U square. Limits of integration will become zero to infinity. Okay? Now, basically it is a standard integral which you have already seen before, du not dx. One plus root X plus one U the whole square. So treat it as if you are dealing with one by A square plus X square integral, which is nothing but one by A tan inverse of X by A divided by the coefficient of U over here. Remember reverse chain rule, correct? Plus a C, but C you don't need to write because you have to put the upper and the lower limits of integration, which is zero to infinity. Okay? Everyone is clear so far? Is everyone clear so far? Okay, great. So when you put the upper limit, F dash X becomes one by under root of one plus X tan inverse infinity is going to be pi by two. Okay? And tan inverse zero is going to be zero. So that becomes one by two under root of one plus X. But this is just half the problem done because we have to find F of X and we have just now found F dash X. So what we have to do next is we need to integrate both sides with respect to X to find the value of F of X. And everybody knows how to integrate this. This is going to be pi by two into two under root of one plus X, correct? And a constant of integration, guys. This is very important. Never miss out on the constant of integration because not every time the value of the constant of integration is zero. You were lucky last time that it was, but may not be this time. So this is going to be pi under root one plus X plus C. Now how do I get C value? Now let us apply a bit of reasoning over here. Do you all agree with me when I say that F zero will be zero? Now I'll say again. Yes, again. Why? Because when you put X is zero over here, it becomes log of one plus zero by sine square theta. And log one will be zero anyhow, correct? So by using this, I can find out the constant of integration over here. So F zero is equal to zero implies zero is equal to pi under root one plus zero plus C. That means C is equal to negative pi, okay? Which means your function is going to become pi under root one plus X and this C is minus pi over here. So just plug in there, take pi common and here comes your answer. Let's see which of the options match the answer that we have got. Oh yes, it is option A. I think few of you replied with option A. If I remember Preeti and Laxia, where the one who gave A has the answer. Awesome, very good. Did you guys guess it or was it actually solved rigorously? Okay, anyways. Let's talk about one more question and then we'll move on to the next property which is the property based on inequalities. Guests, obviously. Okay, so here is a question for all of you. It is known that, let's say it is given that integral of one upon A square sine square X plus B square cos square X with respect to X, okay? Is pi by two AB? By the way, this can come as a question in itself in the school exams. I hope you know how to solve this. You have to divide the numerator and denominator by cos square, take tan XST and then it gets converted to a standard integral which you know how to evaluate. Okay, now this result here, by the way, is given to us. So we don't have to find it. Now using this proves that, using this proves that integral of dx by A square sine square X, B square cos square X whole square is pi A square plus B square by four A cube B cube. Please note, there is a square over here. It is different from this expression. They're not the same. There was no square in the given expression but there is a presence of a square in the denominator. Here, okay, anybody who's trying? Anybody who wants to buy a minute or so? Okay, fine. I'll give you a minute just to figure out what's going on. Fine, fine, fine, take your time, guys. No, no, it's A square plus B square. It's A square plus B square by four A cube B cube. Any success, guys? Those who are trying? Okay, so let's talk about it. So let's say I call this as a function of A and B, okay? Because your answer is in terms of A and B, right? Now, if you see this expression, this integrand, this is called the integrand. Whatever you are integrating is called the integrand. If you see this integrand, it has got A in it, X in it and B in it. So it has got three variables. I would not call them all variables. I'll call X as the variable but A and B are parameters. But still, they can be used, they can be, the function can be differentiated with respect to those parameters, okay? So basically what we are going to do here is I'm going to apply Leibniz rule by differentiating both sides. Now you will ask with respect to what, okay? So first of all, I will differentiate both sides with respect to A, okay? So let's find the derivative of this with respect to A, fine? So derivative of this with respect to A, if you do, you have to apply the Leibniz rule. Now according to the Leibniz rule, it will be partial derivative of this with respect to A and integral of that result with respect to X. So what are the partial derivative of this with respect to A? So partial derivative of this with respect to A would be, you have to apply chain rule first, into derivative of this part with respect to A, which is going to be two A sine square X. Remember B and X are treated as constants. B and X are treated as constants. Why? Because you're doing a partial derivative with respect to A, is that fine? Right? The integral of this result you have to do with respect to X. Now on the left-hand side, when you differentiate pi by two AB with respect to A, you will end up getting minus pi by two A square B, correct? On the right-hand side, you get zero to pi by two minus two A sine square X by A square sine square X plus B square cos square X the whole square, okay? Anybody having any problem in understanding this step? I'm sure you would have understood this step properly. So what I did was, I just did a derivative with respect to A on both the sides. On the left-hand side, this result is clear. And on the right-hand side, I use my Leibniz rule formula to get the derivative, okay? Now one more thing that we can do over here is we can drop the factor of minus two A from the right-hand side to the left-hand side. So it'll become pi by four A cube B is equal to integral from zero to pi by two sine square X by A square sine square X plus B square cos square X whole square, okay? Now I'm going to do exactly the same thing by differentiating this time with respect to B. So similarly, I'm going to do this derivative of the same thing with respect to B, okay? So pi by two AB is equal to zero to pi by two DX by A sine square X plus B A square sine square X plus B square cos square X. I'm going to do the derivative with respect to B on both the sides, okay? On the left-hand side, I will be getting minus pi by two AB square while on the right-hand side, remember by use of Leibniz rule, first I have to do zero to pi by two partial derivative with respect to B of this expression, okay? And whatever is the result, I have to integrate. And other terms would be zero minus zero. So I'm not even bothering to write that. So partial derivative of this with respect to B would be minus A square sine square X plus B square cos square X whole square into the derivative of A square sine square with respect to B would be zero and the derivative of this would be two B cos square X. Remember you are differentiating with respect to B. So cos square X will be treated like a constant. Integration of this from zero to pi by two DX, okay? So drop the factor of minus two B from both the sides. So you'll get pi by four AB cube will be zero to pi by two DX, I'm sorry, cos square DX by A square sine square X, B square cos square X, the whole square, okay? Let me call this as one. Let me call this as two. Now I would like all of you to spend a minute or so figuring out what has been done in case you feel that you want me to explain some place once again, you can unmute yourself and speak. Nobody has any doubt, everybody is fine with it. Just type fine on your chat box so that I know you are fine with it and we can proceed further. Otherwise I don't know how to zoom out in this, okay? Everyone is fine. Now the last and the crucial step, so I'm going to add one and two. Let's add one and two. So when you add the left hand sides, you get zero to pi by two, sorry, right hand sides, zero to pi by two, sine square X, remember the denominators are same in both the integrals, integrants. So you can add the numerator because they have a common LCM like this, correct? And on the left hand side, the terms were pi by four A cube B and pi by four AB cube, correct? So please note here, this term would now be simplified as integral zero to pi by two one by or DX by A sine square X plus B square cos, sorry, A square sine square X plus B square cos square X whole square, where on the right hand side, I can take a pi by four AB as common. So you end up getting one by A square plus one by B square. For the simplifying, it becomes A square plus B square by A square B square, which is nothing but pi A square plus B square by four A cube B cube, right? This is what we wanted to show over here, correct? So hence the result over here, as you can see, this is the result that we wanted to show, pi A square B square by four A cube B cube. So this problem, I found it a very special one because it gave you a very wide perspective about the use of Leveny's rule. How you can actually differentiate with respect to, you know, A with respect to B, which actually are parameters in this question. And one would not have even fathomed that such an approach can be used to solve a problem. Oh, no, no, no, it's not a mains level. Slightly higher than mains level. Okay? So today we are going to begin with another concept which I call as the inequality properties. The inequality properties of definite integral. Inequality properties of definite integral. So let me start with the very first property and under this inequality property. So this says that if in an interval, A comma B, if there are three functions, F, G and H, this satisfies this inequality. Okay? So if there are three functions, F, G and H, which satisfies this inequality, okay? Then even their integrals from A to B will satisfy the same inequality. That means integral of G of X from A to B will be lesser than equal to integral of F of X from A to B will be lesser than equal to integral of H of X from A to B. Okay? Isn't this property very obvious when you try to prove this geometrically? So if you have three functions, let's say like this, okay? So I have my interval A to B over here. So let's say G of X is like this, okay? F of X is like this and H of X is like this, okay? So you can see your G of X is the bottom most function. F of X is here and H of X is here. So F of X is greater than G of X and H of X is greater than F of X. So this property says that even the area, even the area under these functions will follow the same nature. So this area, okay? Let me call it as A1. This area, let me call it as A2. And let me shade a blue area, okay? A3. So it is very obvious that A1 will be the least of all of them. A1 will be lesser than equal to A2, A2 will be lesser than equal to A3 and hence this property. Now off late, J main and J advance, both these exams have started asking questions based on inequality properties, okay? And normally this type of questions are seen to be very difficult by the students because you need to guess two functions which bound this given function F of X. And not only you need to guess them, you need to also see that whatever you have guessed, they are integrable, okay? Let me just give you an example on this. Let's have an example on this. Let us say I ask you to prove that I know I'm giving you a prove that question so that your life is easy, but normally if it is not a prove that question, you need to check from the options, okay? Prove that integral of X to the power seven by cube root of one plus X to the power eight lies between zero and one by eight. I'm sure you would be facing a problem over here because you don't know how to proceed. Oh, done, very good, Nikhil, awesome. Okay, so see here, first of all, since you're integrating between zero to one, it is very obvious that your X lies between zero to one. Great, Shahra, great, awesome. So can I reason out one thing? One plus X to the power eight would be definitely greater than one, correct? Yes or no? Yeah? So can I say cube root also will be greater than one? Correct? Okay. So one by one plus X to the power eight to the power one third will be less than one, right? Correct? And if you multiply with X to the power seven, both the sides, you know, you are between zero to one. So X to the power seven is going to be a positive quantity. So multiplication can take place on both the sides without affecting the inequality. So even this relation will hold true, right? Now, all of this is basically reverse engineered. I actually, you know, work this problem backwards by looking at the options because this I actually had a feeling was obtained by integrating this term, correct? Because when you do that, you get X to the power eight by eight, and when you put the limits of integration, it results into a one by eight. So I'm just doing backtracking. I'm just going backwards. And obviously this is positive, so this is greater than zero. So what you did, you were able to sandwich this function or you're able to bound this function between zero and X to the power seven, correct? So something like this. So you basically achieved this step. So this property says that the same relation would also hold on its integrals. So integral of zero from zero to one DX would be less than equal, will be less than equal to or less than zero to one integral of this term and will be less than zero to one integral of this term, correct? So this is obviously zero and this is your desired integral and this will be one by eight. Just now we have seen this. So this is the required proof for this. Is that fine? Let's take another one. So hope you can see this question. Prove that integral from zero to half DX by one by X to the power two N will lie between half and pi by six for any greater than equal to one. Guys, how have your practicals gone? I'm sure all of you would have done well in that. Good to know that, Nikhil. Good, good, good. Good, Kathy, good. Except comp, nevermind, better like next time. So thankfully we have been given some hint over here because the question says prove that. Oh, Shiram is already done. Good. Let's give others a bit of time, Shiram. Let's wait for another minute. Nikhil also done. So one thing for sure is if you have an upper limit of half and you have a pi by six, that means somewhere sine inverse must be appearing, right? And sine inverse appears when you have integral of one by under root one minus X squared. So you have to reason it out like that. Good, good, so let's talk about it. So remember you are dealing with X which lies in the interval zero to half. So you're dealing with fractions, remember that. You're dealing with fractions, okay? So if you're dealing with fractions, can I say X will always, or X to the power two n where n is greater than equal to one will always be less than, let's say X squared, less than equal to I can say, right? So if you take higher powers on a fraction, the value is going to just fall down further, right? If I put a negative sign, can I say negative X to the power two n will be greater than equal to negative X squared, correct? If I'm going to add a one, the inequality is not going to change, but can I say definitely that this is going to be less than one, correct? If I take the under root on all the sides, nothing is going to change because all our positive quantities, correct? If I'm going to reciprocate them, inequality is just going to flip. The inequalities are just going to flip, correct? So here I have done one thing. I have bounded my desired, I have bounded my given integrand between two known functions, one and under root of one by, one by under root of one minus X squared. And both these integrals are well known to me. Both these integrals are well known to me. So I can say integral of one from zero to half will also be less than integral of this given integrand will also be less than integral of this, okay? And these results are super simple to evaluate. This is going to be half. And let's say I call this as I. This is going to be sine inverse X from zero to half, which is going to be pi by six. So I is going to be less than pi by six, but greater than half, hence true, okay? Next question. Next question, I think you can read it on your screen. It says that there are three integrals, I one, I two, I three. As you can see, they are given as on your screen. This is divided by, okay? Now what you have to do in this integral is you have to arrange them in the decreasing order of values. Well, let me give you the option so that you can type the response very easily. Else you'll have to type a lot of things. So first is I one is greater than I two is greater than I three. Option A, option B, I two is greater than I one is greater than I three. Option C, I three is greater than I one is greater than I two. And option D, I three is greater than I two is greater than I one. Look at the pattern of the integrand, sine X by X, sine of sine X by sine X, sine of tan X by sine X. You'll come to know the nature. See, if you're talking about pi by six to pi by three, yes pretty, it's an increasing function, okay? And she says B, Shiram also says B. Let's wait for a few more people to respond. Just two more responses I need before I start solving this. Sine X by X, I'm sure most of you remember the graph of it. So let me just take this graph up in front of you on GeoGebra. Y is equal to sine X by X, okay? Okay, let's have X is equal to pi by six and X is equal to pi by three. So just figure out this part of the graph. Let me just zoom in a bit, zoom in much. Okay, so here you see this is the part of the graph which we are talking about and it's clearly decreasing in this part, okay? Many of us, we are already aware of sine X by X graph, right? The whole limit question on tignometry was basically based out of this sine X by X graph, okay? Now how do I make use of this while solving this question? So sine X by X, no doubt that this is a decreasing function for all X belonging to pi by six to pi by three. There's no doubt whatsoever regarding this, correct? Now, if this function is fed something which is higher in value, its value is going to be lesser as compared to the value that it will give for a lesser input, correct? So what I'm trying to say here is that if you feed a higher input, the value is going to drop, isn't it? Now, out of this X, out of this X sine X and tan X, right? Can I say sine X is the least and then we have X and then we have tan X? Yes or no? That means if I feed sine X in place of X, both the places, can I say its value will then be the greatest of all? Then what I get by feeding X and what I get by feeding tan X? So think as if sine X, X and tan X, these are inputs. These are inputs to this function, correct? So higher the input, lesser will be the value because the function is a decreasing function, correct? No doubt about that. So what I figured out is the relationship between these integrants. As I can see, this integrant is the max of all of them. So integral of this from pi by six to pi by three, sorry, I wrote pi by six two times over here. Let me write it pi by three. We'll also follow the same inequality, correct? That means I three, I two would be greater than I one would be greater than I three, correct? Because this one I three is the least. This is your I one and this is your I two. So I two would be the greatest followed by I one and then finally I three. So yes, option number B is correct. So advaith, preety, shiram, all of you were correct. Congratulations, good. So seeing as to how pi by three and pi by six are included in the interval, you could just plug in one of those values and get the value returned by the functions. That also could be done, but again, just wanted to solve this question in a rigorous way. Of course, you can use that to save your time. Moving on to the property number two under this inequality. So we are still working on the inequality properties. So we are talking about property number two here. Okay, fine, fine. So let's say we have a function. We have a function, f of x, which has the greatest value or we can say maximum value to be more precise or it has a global maxima, so as to say. It has a global maxima of capital M in the interval a to b, correct? And has a global minima of small m in the same interval a to b, okay? So this property says if f of x has a global maxima of capital M in the interval a to b and a global minima of small m in the interval a to b, then the property says integral of f of x from a to b would lie between small m times b minus a and capital M times b minus a. Yes, yes, yes, we'll have revision class of conics also, but how come that struck you right now in the middle of a definitely integral class? We'll revise all the important topics of algebra, trigonometry, calculus, coordinate geometry, everything would be included, no problem. Guys, prove for this, prove for this. I'm sure you'll be able to prove this within two seconds. This property has to do something with the property number one of inequalities, which we just now did. Just type done if you're able to prove this on your own. Done, done, very good. See guys, it's very obvious. If small m is the global minima and capital M is the global maxima, that means you know your function is sandwiched between these two values, isn't it? So instead of having a function, now these are constants, but nevertheless the properties still hold true, right? So even if you integrate both sides, it's from a to b, a to b, a to b, the inequality still holds true even in this case. So this is nothing but m times b minus a. Of course, this is the required integral and this is going to be m capital M times b minus a, right? Hence proved, very simple. Let's take a question on this. Let's take a question on this. So this is also one of the possibilities that you need to keep in mind by solving questions, that it may be a question which is based on the maxima minima values, okay? So prove that this value lies between 1 and 655. See, now that you know that it's based on global maxima and global minima, right? Had this question been projected to you without discussing that theory prior to it, you would have actually tried to figure out a function which is enclosing this or bounding this given function, right? So guys, that's what makes this topic a little bit dicey and unpredictable. You may waste a lot of time, you may not know that, oh, Leibnizu is going to be applied, or you may not know that by solving these inequality type of questions, you have to figure out two functions or you have to figure out the maximum minimum values. So yes, a bit of trial and error goes in and that's what makes this topic difficult in J exams. Done, oh, that was fast. It doesn't matter, even if you take non strict ones, let's say I include, it should not matter. Yes, of course, now that you know that, we have to deal with very, very, very, very, you have to deal with global maxima and global minima. You need to find small m and capital M in this case. That's all what is required. So, who else other than Adwet? Done, done, done, okay, great, awesome. Shall we discuss this? So now we have a function, the function is five minus x by nine minus x square. So to find the global maxima in the interval zero to two, I have to first find out the local maxima for that you have to find the derivative of the function and that's going to be nine minus x square times negative one minus five minus x times negative two x by nine minus x square whole square. So that's going to give you x square minus nine plus two x five minus x by nine minus x square the whole square. So that gives you three x square, sorry, minus x square plus 10x minus nine by nine minus x square whole square equal to zero. And I'm sure this is factorizable. You can just send the negative sign on the other side, make this as positive over here. So this is nothing but x minus one into x minus nine by nine minus x square whole square equal to zero. So only possibility is one and nine, but nine is rejected because this doesn't belong to the interval zero to two. We are only talking about zero to two interval. Guys, never inquire beyond the interval of integration. It's just a waste of time. So don't go beyond the interval of integration. Great short answer. So now one, one is the only point that we have got. So let us analyze F zero, F one and F two. Remember for finding the global maxima and minima in the close intervals, I discussed with you in the application of derivatives chapter that we see the value of the function not only at the critical points but also at the end points of the interval which in this case is zero and two, correct? So when you put a zero, what do you get? So in this function, if you put a zero, what do you get? Five by nine, okay? When you put a one, what do you get? Half, four by eight, which is half, okay? And when you put a two, you get a three upon five, correct? Yeah, so out of this, I can see clearly that the maximum value is three upon five. That's point six, okay? So this is your capital M, okay? And the minimum value here is half. So this is your small M, correct? So according to the property, the integral which is zero to two, five minus X by nine minus X square DX should lie between half B minus A and three by five B minus A. So that's nothing but one, copy this, six by five. So you can also write non strict inequality here. So let's move on to the next property which is your third property in the same series, okay? This property says, if you take the mod of a function which was integrated from A to B, this will be mod of, this will be less than equal to mod of the function from A to B, okay? So mod of the integral will always be less than integral of the mod of the function. That's what this property says. Now intuitively speaking, this property is very obvious. So let's say I have a function like this. So I have a function like this, okay? Let's say it goes from here to here, okay? So you are integrating it from zero to let's say five and this value is let's say two, this value is five, okay? This area is let's say three and this area is let's say one, correct? So what are the integral of this function from zero to five? Okay, you will say three minus one which is two. So even if you take a mod, it remains a two. But once you mod this function up, so this is the graph of f of x. Once you mod this function up, the part of the graph which was below the x axis will now come up like this, okay? Same, this is two, this is five and this area magnitude wise remains same and this also remains same. So this time the integral of mod of this given function from zero to five is going to be three plus one which is four and obviously four is higher than two. So this is a very intuitive kind of a proof. How do you prove this rigorously? So rigorously proof is easy. We know that any function f of x will always be higher than negative of its mod but lesser than mod of f of x, correct? Yes or no? For all x belonging to A to B, for all x belonging to A to B, is this property very clear? That a function will always be higher than negative of its mod, correct? And always be lesser than the mod of that function, correct? So this is basically your property number one. So your integral on this from A to B will also satisfy the same property, correct? My property number one in this case, let me make in some space by raising this extra stuff over here. Now imagine as if this is x and this is negative A and this is positive A. So you have something like this, correct? Imagine that this middle thing is capital X and these two quantities are negative A and plus A. Of course they are negative and positives of each other. So any quantity which is between negative and positive A, can I say mod of that quantity will be less than A, correct? This is a modulus inequality, right? So the same way I can say mod of this quantity that is mod of A to B f of x would be less than equal to A and this is your A. So A is zero A to B mod of f of x. Hence proved, okay? Just remember this property, not very well known but some questions may come on this property. Moving on to the fourth property here. This property says if f square x, f square x means square of f of x. Let me write it in a language that we all can understand. f of x square and g of x square. Normally we refer this term as f square also, f square x and this has g square x. So let's say these two are integrable functions. So these are integrable functions, okay? In the interval A to B. Then the property says integral of A to B f of x into g of x. Mod will always be less than equal to under root of f square x dx. Let me put this into brackets. Into g square x dx. So basically what you're saying, if you multiply these two functions, take its integral from A to B and mod it, its value will always be lesser than under rooting the integral product of the square of those functions integrated. This is product over here. This is an important property. Okay, so first we'll prove this. So first we'll prove this property and then we'll take up a question on this. So let's prove this property. So first of all, I'll create a function capital f of x which is made up of the square of these two functions linearly connected to each other. So let's assume that there is a function capital f of x which is f of x minus lambda g of x whole square. Here lambda is some real number, okay? Here lambda is some real number. Now since this is always a square, can I say this term will always be greater than equal to zero? Right? Can I use my property number one? Since this function is always greater than zero, integral of this function from a to b will also be greater than equal to zero, right? From property number one of inequalities, not the original property number one, property number one of inequalities, okay? This will always be greater than equal to zero. So let us write this term over here. Instead of capital f of x, let me replace it with this term. So f of x minus lambda g of x the whole square, okay? So if you expand it, it becomes f square x plus lambda square g square x minus two lambda f of x into g of x dx. So what I'm trying to claim is that this term will always be greater than equal to zero. Okay? Now treat this like a quadratic in lambda. So this is like a quadratic in lambda, right? And any quadratic equation we have all seen in our last year curriculum of theory of equation. If any quadratic equation like this is always positive, it can only happen when your a is positive and the discriminant is less than equal to zero, right? That means this graph is always hanging in the air or just touching the x-axis. That's what is the meaning of greater than equal to zero. Correct? So either it will have equal roots, okay? Or it will have non-real roots. That means your discriminant will either be greater less than or equal to zero, correct? So in the same way, if you treat this as lambda square, lambda square is like your x. So this is your a. You can treat this as your a term, okay? You can treat this as your minus two b lambda. You can treat this as your b term and c term is your this term, okay? And this is greater than equal to zero. So my a is nothing but I'm writing it down for you here once again. This is my c term. So my a term here is integral of g square. b term is a to b f of x g x and c term is a to b f square x dx, okay? Now a should be greater than zero. That is always there. Okay, sure. I'll just scroll up. Rithi, is it clear? b square minus four ac, b square minus four ac if you do, this is playing the role of b over n. So four b square minus four ac should be less than equal to zero. That means b square should be less than ac. That means take the root on both the sides, okay? So mod b should be less than under root ac. Now what is mod b? What is mod b here? Mod b is the mod of this expression. So mod of integral of f of x into g of x from a to b will always be less than equal to under root of ac. Ac will be integral of a to b g square into c is integral of a to b f square and hence this property, hence proved. Is that clear? Just type clr on your screen. If it is clear, great. Let's take a question on this. Rithwi, you need to mute yourself. So let's take this question. Prove that this integral cannot exceed under root 15 by eight. No problem, Advait. That's a correct statement. God knows. There's nothing wrong about that. Anyone who can find the answer for this, please type done. Thwi is absent. Andrew is absent. Done. Okay. It's very simple, guys. Very, very simple. Treat your f of x as under root of one plus x. Treat your g of x as under root of one plus x cube. Correct? And apply the property that we just now learned. That mod of a to b f of x g of x dx will always be less than equal to under root of a to b f square a to b g square. Okay. So now fit in these expressions. So mod of zero to one, f of x is under root one plus x, g of x is under root one plus x cube. This integral will always be less than equal to square of this from zero to one, which is nothing but integrating one plus x into square of this is nothing but integral of one plus x cube from zero to one. Now this term is already positive. So this mod is redundant over here because you are integrating two positive functions from zero to one. Okay. So the result will definitely be a positive quantity. Okay. So this will be a desired integral. Let's evaluate this. This is nothing but x plus x square by two from zero to one into this is nothing but x plus x four by four from zero to one, which is one plus half, which is three by two. And this is five by four under root, which is going to be 15 by under 15 by eight under root. So this value is the max value that this integral can actually take up. Correct. So the maximum value that this function can take is 15 by eight. So it cannot exceed 15 by eight. Is that fine? Great. So with this, we come to end, come to the end of the inequality properties. Now I'm going to talk about, lately we have seen that in J. They have started asking questions where they want you to convert one integral to another. Okay. So you would not see a question where they are seeking for an answer to that integral, but they would be seeking for an answer of an integral in terms of another integral. So those type of questions we are going to take up now, which I call as integration by typical substitution. So these are definite integral integrals. Which can be evaluated by typical substitutions and these substitutions are very much dependent upon what you want to achieve. So there's no theory involved over here. It is just looking at what you want to convert it to. We actually make those substitutions. Okay. So I'm going to jump to a problem directly with respect to it. Let's say we have this question. Let A be the integral of log x by one plus x cube from zero to infinity. Okay. Find the value of x log x by one plus x cube from zero to infinity. Find the value of this expression. As you can see, I'm rolling my pen on it in terms of A. So you don't have to find the value of the integral. This you just have to express it in terms of this integral. Can you try it out? Let's see how good you are in substitution. You just have to tell me how is it related to A? Is it A itself or it is 2A or it is, you know, 3A or A by 2, whatever. In terms of A, only you have to express it. Any idea how to do this? How many of you are trying? Ideally, you should all be. Sure, sure. Take your time. Okay. So here the idea is, okay. Yeah. Okay. Fine. Fine. Minus A. Wow. Sure. Well done. That's correct answer. Okay, guys. Let's check. So let's say the given integral here is, uh, let me call it as I. Okay. So I is zero to infinity x log x by one plus x cube. Fine. Now I have to somehow write this in terms of A. Correct. So what I'm going to do is I'm going to write it as one x plus one log x minus log x by one plus x whole cube. So overall the expression still remains x log x by one plus x cube. But the benefit of doing this is I can segregate this into two integrals. One is by the way, x plus one divided by x cube plus one. You all know that it is factorizable. Okay. We will take care of that little later on. Whereas this integral is what we have been provided with. Correct. So I can directly write this as your A. So it's this integral minus A. So I is this integral. By the way, I'll just simplify this also. This will become x square plus x plus one. I'm sorry, x square minus x plus one. Now this is also something which I don't know how to evaluate. So let me take it separately. Let me call this integral as I one. So let's try to evaluate. What does I one give you integral of log x by. x square minus x plus one. Okay. Here what I'll do I'll make a substitution x is equal to one by t. That means dx is minus one by t square dt. Okay. Now I'm done the substitution is because I don't want to disturb my limit of integration. Neither do I want to disturb the presence of log t or log x over here. Remember such a substitution will just give you minus log t on the numerator and the denominator limits will just be swapped. So this will become infinity. This will become a zero. This will become negative log t. Whereas in the denominator, you have one by t square minus one by t plus one and dx will become minus one by t square dt. Right. Correct. So now if you simplify this further negative negative will take care of each other. So we have log t dt t square. If you multiply in the denominator, it's going to give you one minus t plus t squared. Is that fine? Correct. Now, if you switch the upper and the lower limit position, it becomes negative of this term. Correct. Isn't this like negative i1 itself? See you started with i1 and this is negative i1. Remember there is nothing in the name. There is nothing in the name. So you can still call it as zero to infinity log x by one minus x plus x square. So it is your minus i1. So i1 is equal to minus i1 means 2 i1 is equal to zero. That means i1 is zero. Yes or no? Okay. If i1 is zero, this term is zero, which means your i will be equal to negative a. That's it over. This is the answer. Is it fine? They just so this is something which you need to practice a bit in order to be more confident about. It doesn't come very, you know, obviously and very conventionally to us. Let's take few more questions. Let me give you another one. Yeah, let's take this question. Show that integral from zero to one. This is zero. Actually may be slightly hidden zero to one log x by one plus x dx is integral of negative zero to one log one plus x by x dx just type done once you're done with it. Any idea? So zero to one log x by one plus x. So limit of integration as you can see has not changed. Okay, and you have a term log one plus x and an x coming over here. Anybody can guess that this one by x. See this is this is basically made up of one by x into log one plus x correct log one plus x comes when you are integrating one by one plus x and this comes when you're differentiating log of x and both of these terms are present in your question. Both of these terms are present in your question over here. Correct. So it's obvious that you have to apply integration by parts and for that you have to choose your log x as your first function and one by x as one by one plus x as your second function, right? Is that clear? So treat this as your you treat this as your P. So this becomes you times integration of one by one plus x is log one plus x. Okay, minus integral of derivative of this is one by x integral of this is log one plus x. Please note here you have to put the limits of integration but here you have to substitute the limits of integration in the entire result. Correct. So this one if you put this will become zero zero if you put this will become zero. So it is zero only and this is your desired integral that you wanted to convert into and therefore the results so integral of zero to one log x by one plus x dx can also be written as this. So okay, hence, please type CLR if it is clear. Oh, Andrew joined us now. Welcome Andrew. Okay. So let's take another one integral of e to the power t by one plus t dt from zero to one is a then find integral of integral of zero to one e to the power t by one plus t whole square in terms of a minus a. I'm sure you are influenced by the last two results but this time they kill. No, it is not minus a is very much similar to the previous problem very much limits of integration have not changed. One function has been differentiated. Other function has been integrated. So it tells volume about yes. Yes, pretty by parts. Absolutely. Yeah, please do that and let me know the answer. I'll give you a minute or so. Please feel free to type in your response in the chat box or speak it out on your phone a log t. How does t appear in your answer? Aren't you substituting the limits on t? So your answer will be completely in terms of a and constants. E by two minus one minus a. You're very close. I think there's a mistake in the sign. The entire answer is negative of that. Pranav who is signing by the name of Pranav? Oh, you are new to the session. Okay. Oh, no, two a by minus two is also not correct. All right. I think just a silly mistake has happened from all of you. Else's problem was very easy. So basically you are trying to integrate this term by integration by parts. Okay. So you're trying to integrate this by integration by parts. Right. So let us take one by one plus T as my first function and e to the part T as my second function. Okay. So let's call this as my first function. You this as my second function V. Okay. So by integration by parts, you first have to keep this as such integration of e to the power t is t limits of integration. You can put right over here minus integral from zero to one. The derivative of this will be minus one by one plus t square into e to the power t dt. Right. So minus minus you can take care of it will become a plus. So integral of e to the power t by one plus t dt is this remember this is your a many put a one over here you get e by two and many put a zero over here you get minus one and this is your plus zero to one e to the power t by one plus t square integral. Is that fine? So this is my subject of the formula. So make this the subject of the formula. So make this the subject of the formula. Okay. So take this other terms on the other side. So it'll become a plus one minus e by two is the desired integral. So Ramya was very close. I think she just made a sign error Ramya. Just check out for your science. I'm sure you would have, you know, not consider the negative sign from there. Yes, it was a sign error. Okay. Let's take another one. So the last part of the question is not that clear. So I'll write it down again. So you have if f of x is given to you as x plus sign x, you have to find the value of integral from pi to two pi inverse of the function. You have to find out the integral for inverse of this function. I'm sure all of you have done inverse in your functions. So your function is x plus sign x. You have to integrate the inverse of this function from pi to two pi. So I'm assuming you guys have holidays for the next four days, Sunday, Monday, Tuesday, Wednesday. Make use of this time for your semester exams. And don't completely stop the JEE preparation during the semester exams. Somebody asked me, sir, how should we distribute the time when we are having semester exams? Give 30% of your time to JEE and 70% to your school right now when your exams are around the corners. And then flip the issue. Yeah, yeah, yeah. Sure. Guys, let me tell you one thing. Here you don't need to find the inverse of a function to do this. If anyone of you is trying to solve this question by finding inverse of F stop right there because that that is not the right way to solve it. Okay. In fact, finding the inverse itself may be very challenging. The answer to the previous question PT was a plus one minus e by two a plus one minus e by two. Okay. Anyone any progress in this? Okay. See guys, let me solve this question. Let's do one thing. Let's substitute X as the function itself. Okay. So what I'm doing, I'm substituting the variable in this integrant as the function itself. That means DX is F dash T DT. Okay. So when you make that substitution. When you make that substitution. Yeah. What happens to the given integral F inverse of F of T will be T. Okay. Because you all know F inverse F of T is going to give you T. Remember, it's an identity function. So it'll give you a T and DX is going to become F dash T DT. Okay. Now, what about the limits of integration? Let's take the limits of integration. When you put X as when you put X as pie. Okay. What will your TV? What will your TV? Can I say your limits of integration will be F pie to F to pie? Hello. I'm audible to everyone. Can you all hear me properly? Yeah. Oh, sorry. Probably because of the internet issues. My voice would have been cut. Okay. Is this clear? The limits of integration. Are they clear to you? Sorry. It would be F inverse. Is that fine? Now, what is the F inverse of pie? F inverse of pie means which value will give you a pie. So if you relate this to pie, can I say this will only happen when X is equal to pie? Yes or no. So F inverse pie will actually be pie itself. Right. And when does when do you get to pie as your answer? That means if you equate this to to pie, what should be the possible value of X? Can I say X will be to pie in this case also? That means this upper term would also be to pie. So F inverse to pie would also be to pie. Right. So this integral over here. Can I write it for as integral from pie to to pie t f dash t? Anybody having any problem so far? Please stop me right here and get your doubts or questions or concerns addressed. Anybody having any problem? If you have a problem in the limits, let me explain once again. So if you put your excess pie over here, T will become F inverse pie, isn't it? T is nothing but your F inverse X. Correct. So when you put excess pie, T will become F inverse pie. That's why I wrote this lower limit as F inverse pie. When you put excess to pie, T will become F inverse to pie. So this is your this thing. Okay. And how do you find F inverse to pie F inverse to pie means what should be the value of X such that the answer to this function is 2 pie. So the value of X should be 2 pie in this case lower limit. Also, you can find it as pie itself. No doubt now. No questions. Now from here on, I'm going to apply again integration by parts. So let's take this as our you and take this as our week. So integration of this will be you times integration of F of T DT will be F of T lower limit is pie upper limit is 2 pie minus integration from pie to 2 pie derivative of T is going to be one integral of F dash X is going to be F of T DT getting this point now the F of T F of T here I can put it as T plus sin T. Because F of X is already given to me. F of X is already given to me in this question. Look here. F of X is X plus sign X. So F of T will be T plus 90 getting the point now. Let us integrate this. So this will be 2 pie F of 2 pie minus pie F of pie and the integration here will be T square by 2 minus of cost T from pie to 2 pie. Okay, so we just now saw that F of 2 pie is also 2 pie and F of pie is also pie. So it becomes this minus here will become 4 pie square by 2 here we get 4 pie square by 2 I can take a half common. So it is 4 pie square minus pie square and this is minus cost 2 pie minus cost pie. Let's simplify this further. So 4 pie square minus pie square will be 3 pie square. Okay, and here you will get half of 3 pie square. Okay, this is going to be 1 plus 1 which is going to be 2 so the entire answer will be 2 minus 3 pie square by 2. This is going to be your answer. I'm sorry, 2 plus 3 pie square by 2. How many of you got close to this answer? Okay, Nikhil says he got very close to this answer, but he's squared or you found inverse and did that. Okay. Can you share your working with me privately? So great guys. Now we are going to talk about a different concept which is expressing definite integral as limit of a sum. In fact, most of the books will do this part of the chapter in the beginning itself. Definite integral as limit of a sum. See guys, this particular concept comes directly from the very basic definition of an integral. So when you say you are integrating any function, okay, from A to B, from A to B. Okay. What are you doing here? You're trying to make at a distance of X. Let's say, let me make take a different color at a distance of X. Let's say you take a thin rectangular strip of width. Let's say DX. So this with his DX. Okay. Of course, since you're taking it at a distance of X, the height of this trip will be f of X. So this becomes the area of this trip. Correct. And you are trying to sum up this area from A to B. That is what we call as the integral. So instead of summing it by using integration formula, we can actually literally sum it up. Okay. So we can start making strip right from here, right from the first step over here. Okay. So let us make a very thin step over here. Okay. Please note that you are dividing this entire A to B interval into thin strips. Okay. And those strips are off with H. Okay. So let's let's not do DX. Let's take that strip of length of thickness H. So this thickness is now off thickness H and this H is almost tending to zero. Okay. And you are dividing it into infinitely many strips. So let's say you are dividing it into n strips and n is tending to infinity. Okay. So as to say that if this point is a this point is a plus H and you're going all the way till this point. Okay. Let me call this point as a plus n minus one H. Okay. That means B is your a plus an H. B is your actually a plus an H. So what you're doing the first trip is right over here. In fact, you can make a strip with the lower edge over here rather than upper edge. So this is your first trip. Those who had studied AP calculus with me would know that it is called the left sum also. Okay. Normally this is studied under the remand some concepts in advanced placement calculus. So there are two ways you can make this rectangle one by taking the left edge means I'm going to say left means your left your left hand is touching the curve. Okay. So next curve will be like this. Okay. Next curve will be like this and so on people who did advanced placements calculus. I think few of you did. Okay. So in this case, the first rectangle would have the area H into f of H. So the one which I'm showing in the blue sorry in the yellow correct height is f of H. I'm sorry height is f of a and thickness is H. So the first yellow strip that I'm showing over here on your screen will have an area H f of a the second step over here. Let me change the color. The second step over here, which I'm showing with a white shading. The area of that will be H into f of a plus H. Okay. The third strip which I'm showing with orange color. The area of that will be H into f of a plus 2 H. Remember this is your a plus 2 H. If you continue doing it, if you continue doing this till your last strip over here, which is I'm shading in blue, the area of that strip will be dot dot dot H into f of a plus n minus 1 H. Yes or no. If you sum this up, if you sum this up. Okay. It's already summed up. So we don't have to write the summation symbol. So if you sum this up and take the limit of this as H tends to zero and n tends to infinity, this entire result will boil down to give you the integral from a to B of f of x dx. So basically what I've done, I have converted a definite integral as limit, see limit of a sum, right? Here your H is tending to zero and n is tending to infinity. But remember a very interesting thing is happening here. Sorry. Even though n is tending to infinity and H is tending to zero, your n H is tending to n H is actually a fixed value B minus as you can see from here. If you make n H the subject of the formula and it will become B minus. So normally we take this and this into our account. This we normally ignore. So it is sufficient to write this entire formula as limit H tending to zero and n H value is equal to B minus a of H times f of a f of a plus H f of a plus to H all the way till you reach f of a plus n minus one inch. Is that clear guys? Any question with respect to the derivation of this? Before solving the problem, there are some prerequisites. There are some important expansions, some important formula. These formulas will help you in solving these type of questions. The first formula, I would I'm sure all of you know summation of r from r equal to one, r equal to one till n that is your nothing but your sum of all natural numbers from one to n. What is that n n plus one by two summation of r square from r equal to one to n. What is that n n plus one to n plus one by six summation of r cube from r equal to one till n. What is that n n plus one by two whole square? Okay, apart from this, you need to know your sign series and cosine series. Okay, what is sign series? Sign series is sine alpha, sine alpha plus beta, sine alpha plus two beta and so on till sine alpha plus n minus one beta. What is this? This is nothing but sine n beta by two divided by sine beta by two into sine of alpha plus n minus one beta by two. If you have forgotten these formulas, please make a note of it because these are going to be used in the problem solving of converting definite integral as limit of a sum. Okay. Next is the cosine series. So first note this down. Then I will write down the cosine series. Cosine series is very much similar. You just have to replace this sign with a cos. Okay, so never mind. I'll write it down once again for you all. Cos alpha, cos alpha plus beta and so on till cos alpha plus n minus one beta is nothing but sine n beta by two divided by sine beta by two into cos of alpha plus n minus one beta by two. Is that fine? So these formulae should be well known to you. The one which I'm showing with the smileys. Okay, so let me scroll down. I think most of you would be still in the process of copying it. Okay, apart from this, yes, we need to know our basics of limits like limit of e to the power, you know, x minus one by x extending to zero is one. Okay, et cetera. So these are required sine h by sine x by x extending to zero is going to be one. So all these things should be known to you. Yes, yes, yes, yes, yes. I'll be sending these notes on the group as well. I'll be sending you the link to the video also recorded. So right now the videos are getting recorded. Okay, let's let's directly jump to problem solving. I'm sure you would have done a lot of problems based on this in your school. So I'll start with a easier one. This one evaluate e to the power x integral from a to be using limit of a sum and CRT problem. Just let me know. Once you're done, just type done if you're done. So it's very simple. First of all, you're using the formula limit h extending to zero and an h equal to b minus a in this. Correct. H times f a f a is e to the power a f of a plus h is e to the power a plus h then e to the power a plus two h all the way till e to the power a plus n minus one inch. Okay. All of you would appreciate over here that it's actually a GP. Okay. And the GP's first term is e to the power a and common ratio is e to the power h. So according to the formula that we have learned the sum of n terms of a GP is a r to the power n minus one by r minus one. Right. So here what will happen e to the power a is your first term common ratio is e to the power h. So e to the power they're all together n terms. So e to the power n h minus one by e to the power h minus one. Okay. Remember your h is sending to zero and your n h is b minus a. So wherever you see a b n h for example, here you see an n h you write it as b minus a. Okay. So it becomes h e to the power a e to the power b minus a minus one by e to the power h minus one. Now your limit is tending to zero. Now what happens to this term as we already discussed this term will become a one. So we'll have e to the power a into e to the power b minus a minus one which is nothing but e to the power b minus e to the power a. Okay. And we all know from our normal formula also that this is going to be e to the power x from b to a which is e to the power b minus e to the power a. Okay. I'm sure you'll find this easy. This is the most scoring part in your school exam as well. Let's take another one. Let's integrate sine x from e to b. So here you would require the sine series. Okay. So limit h tending to zero and n h is again b minus a. So the terms would be h times sine a sine a plus h sine a plus two h and this will go all the way till sine a plus and minus one h. Okay. So according to the science series, the answer for this will be sine n beta by two by sine beta by two into sine alpha plus n minus one beta by two. Right. Clear. Now wherever there's an n h, please write a b minus a first of all there. So I can see n h by two over here which is nothing but b minus a by two. Okay. Let's sign h by two remain an h by two. This will become sine a plus n h minus h. N h minus h means this term is n h minus h by two. Now this h is zero and this n h is b minus a. So it's b minus a by two. Is that fine? Is that fine? Now here also I can do one small activity. I can make this as one because your limit on h is tending to zero. Okay. So it becomes two sine b minus a by two into sine. If you simplify this by taking the LCM of two, it will become b plus a by two. Okay. We all know the formula for two sine a sine b. What is that? Cos of a minus b minus cos a plus b. So if you use that will become cos a minus b. If you do a minus b, it will become a correct. Minus cos a plus b will become a b and that's your answer because we all know integration of sine x from a to b would be nothing but negative cos x from a to b. So when you put the upper limit, you get a minus cos b lower limit. You get minus of minus cos a which gives you cos a minus cos b. Is that fine? We'll quickly take another one in this. Let's see how many of you get this. Hmm. Any idea? Are you facing some trouble while doing it? So most of you would have started very well on this. So n h is equal to b minus a and the terms that you would have written is h times one by root a one by root a plus h one by root a plus two h and all the way till you go one by root a plus n minus one h. But my predicament here is how do I sum this series? I don't know a formula for summing up such terms. Okay. Now guys, you'll be surprised to know that I'm going to solve this problem by the use of sandwich theorem. Those who were with us last year would recall. What is the sandwich theorem? If you don't, I'll just quickly brush it up. If it is known that a function is sandwiched between let's say g of x and h of x in the neighborhood of x equal to a and you know the limit of g of x as extends to a is l and you also know the limit of h of x as extends to a is l then the limit of this guy f of x also as extends to a will also be l. Since we're evaluating limit over here, I would try to sandwich each of these terms between two such terms which are summable or who summation I can find out. Right? So this is a very good problem actually. So what I'm going to do is I'm going to pick up any general term. In fact, I'm going to pick up a term like yeah. So all of you please pay attention here. Yeah. So what I'm going to do is first of all, I take a term like let's say to root R. Okay. So in fact, I can take a root R also not an issue. Let me take something like this. Okay. One by root R. So I picked up any general term as one by root R. Okay. Now I'm going to do one thing. I'm going to multiply and divide with two. Is that fine? Now can I say this term will always be greater than one by this term? Remember, I have increased my denominator here. So this value is going to fall down. Right? And this term will always be lesser than two by under root of R plus. In fact, you can say R minus H minus root R. Sorry, plus root R. Do you all agree with me on this? So here your numerator is slightly heavy. Here your numerator is slightly lighter than this numerator. So where the numerator is slightly heavier, the net value would be lesser and where the denominator is slightly lighter. The net value will be more. Is that fine? Everybody's clear about this? Fine. If this is clear, you would also be clear of the fact that you can rationalize this. That means you can write it like this. And this is already one by root R. Yeah, yeah, yeah. I'll repeat once more pretty. Don't worry about it. See what I said was forget about this to two to remain same everywhere. Now this to root R is like root R plus root R. Correct. Now one of the root R I have kept it like this other root R. I have made it slightly more by adding an H H is of course tending to zero. Okay. So the entire value will be lesser than this value and on the right hand side, I've done all the opposite of it. I've subtracted H from R so that denominator is lesser than the previous one and hence the net value will be higher than the previous one. Clear pretty. So now let's cross this out. Okay. So can I write this as two under root R plus H minus R less than one by under root R and this can I write it as so there will be H also. So can I write this as root R minus root R minus H by H. Now start summing this up start summing this up. That means start putting the values over here. Let's say I put one by root A. So can I say this will lie between two under root A plus H minus root A by H and two under root A minus under root A minus H by H correct. Then put under root A plus H. So this will become two under root A plus two H minus root A plus H and this will become two under root A plus H minus under root A. Okay. Keep doing this. Keep doing this till you reach one by root A plus and minus one H. So can I say the last term over here that you would get is under root of two under root of A plus NH minus A plus and minus one H whole divided by H and this side you'll get under root of A plus and minus one H minus root A plus and minus two H whole by H. Okay. Let's add from here onwards. Let's add everything. Let's add everything. If you add everything you end up getting this term. Okay. So I'll do it over here. So one by root A one by root A plus H all the way till one by root A plus and minus one H would be between. It will be between now on the right hand side if you see sorry on the left hand side if you see. Who batch will be common and these terms will start cancelling out correct. So ultimately your this term will survive and root A will survive. Correct. So it will become two by H root of A plus NH normally people complain that my H and N look the same. So I'll make special attention to write it NH minus root A. Okay. And on right hand side if you see again cancellation of terms are happening this will get cancelled. This will get cancelled. Ultimately what will be left with is this term and this term. Okay. So let me write it over here. Who by H under root of A plus N minus one H minus a minus H take H on the other side. Take this H and of course the other H also. So this H also take it on this side. Now this is my required limit. So let us evaluate the limit of both the sides first. So let's evaluate the limit of this as NH becomes B minus a. Let's see what happens. So it becomes to under root of a plus B minus a minus root a a gets cancelled. So isn't it to under root of B minus under root of a. Let's evaluate the limit of this side also. So it'll be to under root of a plus NH will be B minus a minus H will be 0 and this will be again a minus 0 right and this will further simplify again to B minus root a that means left hand limit and right hand limit both are not sorry left hand and right hand. I'm saying the function on the left and the function on the right both are giving you the same value that is to under root B minus a so the limit of this also as H tends to 0 and NH become B minus a would also be 2 times root B minus root a. And we all know that if you integrate one by root x from a to B you'll get to under root x and when you put the limits it becomes to under root B minus under root a. So a very simple problem if a formula is used but an equally complicated problem if you don't use if you use your first principles or you can say limit of a sum. Okay. So this actually reminded you of that's it. That's the only thing you have after the man some nothing is left. No only when we only use in these those cases where the limits are not directly evaluatable. Okay. In this case it was not. That's why I use the sandwich theorem to find this limit. Okay. So guys quickly without wasting much time we're going to move on the last part of this chapter that is limit of a sum as definite integral CLR SCR still in the computer mode limit of a sum as definite integrals. Now most of the limit problem which can actually be your solved by converting it to a definite integral mostly of the type 0 into infinity form which are directly not solvable by use of our limit fundamentals. Okay. I'll start with a question. Okay. Probably that will make more sense to you. So let me let me explain with a question. How do you solve such what type of problems can you get and how do you solve it? Let's say this question. We want to evaluate this limit as n tends to infinity one by n a one by n a plus one one by n a plus two and so on till one by n b. Okay. By the way, if you see the pattern, the last term over here is one by n a plus n b minus a actually. Okay. Now, do you have any funda how to how to find this limit? Those of you who think that all of these terms are zero actually because n is tending to infinity and zero added infinitely many number of times will end up giving you zero. Let me tell you you are making a big mistake by doing it. The answer for this is not zero again. I'm telling you infinity zero into infinity doesn't mean exactly zero. It is tending to zero into infinity. So each one of these terms are very very small. They're tending to zero, but they're actually summed up infinitely many number of times. It doesn't mean it'll end up giving you zero. It can give you anything. It can actually be a it is actually an indeterminate form. Right. You can never underestimate the power of, you know, let's say few drops which are present in infinitely many quantity. Okay. Drops present small drops whose volume is almost zero when present in infinite amount of quantity can even lead to an ocean. So please don't be under the impression that in all of them are zero limit in itself. So adding them infinitely many of number of times will give you zero. No, it will not happen. It may not happen. Okay. Remember if you're adding for finite number of terms, let's say if your n was tending to 100 then of course you can write zero. There's no doubt about it. But right now your number of terms themselves are infinite. So how do we evaluate it? We can't even sum it up because there is no direct formula to sum this up also. Can I? There's no direct formula to sum this up. So how do we do this question? Now all of you listen to this. The method to find out such limits can be. We can use definite integrals to actually find out such limits. What is this theory actually all about? We learned a little while ago that when we're integrating f of x dx, we actually use a summation process, right? We summed up f of a plus n h. Okay, from or you can say r h in fact or r minus one h whatever you want to call it from r equal to one to n, correct? And the limit of this n was tending to infinity, correct? This was what we learned as definite integral as limit of a sum. Now I'm going to use the reverse process of this. I'm going to use the reverse process of this. I'm going to now convert this term. Let's say I take this rth term of this particular series. So rth term will be one by a plus n a plus. Let's say r minus one or let's say take r, okay? And you can say r is going from zero to b minus a, correct? So this is your sum that you're trying to find out, correct? r equal to zero to b minus a, correct? So what I do next is I try to compare. I try to compare this with the initial theory that we had discussed this concept where I convert my rth term as one by n times a function of r by n, okay? So when you sum this up, when you sum this up from r equal to one till n, from r equal to one till n and take the limit of this as n tends to infinity, take the limit of this as n tends to infinity. Remember this term here would start behaving as dx, okay? And this term here would start behaving as x. And this term here would start behaving as the integral. Now limit of integration will depend from where to where you are integrating it. So this will take the form of integration of f of x dx from a to b. Now how to find a to b? I'll talk about it through this example. Okay, let us take this example. In this example, if I take one by n common. So from this TR term, let me just erase the other things. This TR, I have to express it as one by n f of r by n. So what I'll do is I'll take one by n term common. So n will come up, okay? Divide by n like this, correct? And just make a role change. This will become x, this will become dx. So this entire expression becomes 1 plus a plus x dx, okay? And you're trying to sum this up. That means you're trying to integrate this, right? From where to where you're trying to integrate this from r equal to 0 to r equal to b minus a, correct? In fact, it is nb minus a. If I'm not wrong. Yeah, it is nb minus a, okay? Now all of you see here, in order to find the limits of integration, what about the limits of integration? Since your x is r by n, put r as 0 first. When you put r as 0, this will become a 0. So lower limit will be 0. Then put r as, put this position as nb minus a. So nb minus a by n, you just have to plug it into this formula. It will become a b minus a. So this will become integral from b minus a. So this entire summation can actually be done by evaluating this definite integral. So this will become ln a plus x from 0 to b minus a. So when you put that value, it will become a plus b minus a minus ln a. That is nothing but ln b minus ln a. That's ln of b by a. I'll give you one more example. Let's try to evaluate this. Limit n tending to infinity, 1 by n square, c square 1 by n square, 2 by n square, c square 4 by n square, and it goes all the way till 1 by n, c square 1. c square 1 is basically n square by n square. And this is actually going to n by n square. How to evaluate this limit? Now just listen to the process. All of you just follow the process. Guys, don't send me private chats because I will not be able to read them right now. If you want to ask any question, you have to send it on the group itself. Okay, all of you please follow these steps. If these steps are clear, you'll be able to solve all these problems. Step number one, write down the rth term of the series. Do you all see that the rth term would be r by n square, c square, r square by n square? As you can see, this is 1 square. Okay, sorry, this is 1. This is 1. But this is 1 square by n square, 2 by n square. This is 2 square by n square and so on. Okay. Step number two, from this rth term, pull out a 1 by n out. So whatever is left, this is left. Correct. Step number three, replace your 1 by n with dx. Replace your r by n with x. So if you do that, this entire thing gets converted to x, c square, x square dx. Now, all you need to do is find your limits of integration. Step number four, finding the limits of integration. That is very simple. Your x is r by n, right? And your r is going from, you can see that your r is starting from 1 and going till n, right? R is going from 1 to n. So r by n will go from 1 by n to n by n. 1 by n is approximately 0 because n is very large. n is very large, n is tending to infinity, correct? And n by n is actually 1. So your x is going from 0 to 1. So you all have to do is your answer would be integral of this from 0 to 1. Understood this. This single problem will be able to explain everything very clearly to you. Now, all of us, we know how to evaluate it. Let's take x square s t. So 2x dx will become a dt. So x dx will become a dt by 2. Okay. So this will get converted to dt by 2 into secant square t integral from again 0 to 1. That's nothing but half tan t from 0 to 1. That's nothing but half tan 1. Is that clear guys? Is that clear? So this is going to be your answer for this question. Just follow these four steps. All you have to do is follow these four steps. Step number one, write down your rth term of that limit. Pull out a 1 by n. Okay. You will be left with a function of r by n. So as to say. Put r by n as x everywhere. Okay. Put 1 by n as your dx. And see from where is your r going to from where to where it is going to, it's going from 1 to n in this case. So r by n will go from 1 by n to 1, which is 0 to 1. Okay. One more problem will take up and then we'll wind up this session. Find the limit. And tending to infinity. 1 by under root 4 n square minus one. Try this out. Done. Okay. Okay. Let's discuss this quickly since we are running out of time. So rth term, step number one, write down the rth term. Step one. Okay. Our term is under root 4 n square minus r squared. No doubt about it. Okay. Step two. From your rth term, take a 1 by n common. So it'll become n by under root 4 n square minus r squared, which actually you can write it as 1 by n under root. Sorry. Under root 1 by 4 minus r square by n square. Okay. Step three. Replace 1 by n with dx and replace r by n with x. So it is 1 by under root 4 minus x square. Okay. Step four. Choose the limits of integration. Now, since you are. Integrate your r is going from 1 to n. Your r by n will go from 1 by n to n by n. That means your x will go from that is to say x will go from 0 to 1. So ultimately you are integrating this from 0 to 1. I'm sure you know how to integrate this. Yes, this is going to be 2 square minus x square 0 to 1, which is nothing but sine inverse x by 2 from 0 to 1. So that's sine inverse half minus sine inverse 0, which is 0. So answer is pi by 6. Pi by 6. Absolutely. Is this clear guys? So well, we have ended up definite integrals quite a big chapter, but very interesting, especially the Leibniz rule would be very interesting to you. And this is the last part, which is converting limit of the sum as a definite integrals. So I'll be sending you assignments. I think I'll only send you on the group or try to attempt all of the all the questions. And all the best for your upcoming semester exams next class also will be online. And that session will also be recorded. Okay, so where is your maths? Which day is your maths? 16th of September. So I think we'll get one more class before that. So I'll do vectors in that class. So I'll do vectors in that Saturday's class. Okay. Thank you guys. Thank you so much for joining in. Bye bye. Have a good day.