 This conference will now be recorded. All right, so I can see Hamsini there. Okay. Oh, okay, fine. Yeah, I think so. She also joined. All right. So it's exactly 3pm by my watch. So let us begin the session. So I'll be continuing with the further properties and today is going to be the most important of all properties. I'll be starting with the Levini's rule. Okay. So today I'll be starting with the Levini's rule. As you all know, Levini's was a co-inventor of calculus along with Newton and he gave a very important rule of finding the derivative of a function which has been integrated. Okay. So that's why we call this rule as derivative under the integral sign. Derivative under the integral sign. You can see the screen and I'm all available to all of you. In case of any query, any concern, please do type it on the chat box. You can also speak. I have not muted anyone. Okay. All of you are free to speak today and ask questions. Great. Yeah. So what is this rule? What is this property? Let's say you have a function. Okay, which you are integrating with respect to T. But the limits of these functions are functions of X. So let's say you are integrating from Phi of X to Psi of X. Okay. Where Phi X and Psi X are some functions of X. Okay. And you want to differentiate this result with respect to X. Of course, so that if you integrate this and put your limits which are already a function of X, your answer would also be in terms of X. Right. Correct. And if you want to differentiate that result, this rule actually helps us to find that very efficiently. If you don't know this rule, what would you otherwise do? You would actually integrate this. Correct. Put the upper limit, put the lower limit, subtract. Right. So in the primitive of the function, you will put the upper limit, you'll put the lower limit, you'll subtract and whatever you get, you are going to differentiate that. But this is time consuming because many a times you'll get such functions which are difficult to integrate and many a times you don't need those results also. So why would you go through a longer process to achieve it when there is a shorter one available? And that's what this rule says. Okay. So let me do says that the result is going to be F of Psi X into derivative of Psi X minus F of Psi X into derivative of Psi X. Okay. Hello, Boris. Good afternoon. And wish you all a happy Ganesh Chaturthi today. For Pooja and all is done. Okay. Yeah. Thank you. Yeah. So this is what this rule says and you can see in this rule that there is nowhere I have integrated the function. With F itself. Okay. So what I did let me explain what I've written over here. I first put the upper limit in place of T. Right. So Psi X this is pronounced as Psi. This is Phi Psi and Phi. Okay. So this upper limit was put in place of T in the function and it was multiplied to the derivative of the upper limit. Okay. With respect to X of course minus then the lower limit was put in place of T and multiplied with the derivative of the lower limit. Right. Okay. Now first of all from where does this formula actually come? Right. What is the proof for this formula? The proof for this formula is very simple and straightforward. It actually comes from your understanding of fundamental theorem of calculus and the use of chain rule on that. Okay. So let's say there was no derivative. Let's say I had asked you to evaluate integral from Phi X of Psi X of F of TDT. Correct. Now assuming that your capital F let your capital F of X be the primitive be the primitive of small f of X. What is the meaning of primitive? It means it means if you differentiate your capital F of X you're going to get your small f of X. Okay. So capital F of X is called the derivative capital F of X is called the primitive. Okay. So this result would have been capital F of Psi X minus capital F of Phi X. Right. No doubt about that. Hi Ritu. Good afternoon. Happy to meet you. Okay. Now we need to differentiate this result. So let's say we are finding the derivative of this result. Okay. So whatever is the answer of this integration, I'm going to differentiate that with respect to X. Okay. Remember your answer is a function of X. Okay. So you are differentiating with respect to X. So it just means you're trying to differentiate capital F of Psi X minus capital F of Phi X. Yes or no. Now if you differentiate such a function, we need to use our chain rule for that. So we write it as F dash Psi X into Psi dash X minus F dash Phi X into Phi dash X. So far so good. Okay. No problem in understanding this. So this is through fundamental theorem of calculus. We all know this is FTC fundamental theorem of calculus. Okay. So what I did, I wrote the fundamental theorem of calculus. I differentiated both sides with respect to X. And on the right hand side, I applied chain rule. So this result comes from the application of chain rule which we all know very well. Right. Now see what I'll do. We already know that capital F is the primitive of small f of X. Correct. So this formula is known to us. Okay. So whatever we put as X here, the same will be reflected here also. Right. So as you can see here, you're finding F capital F dash Phi X. So if you place your X with Psi X, wouldn't this become F of Psi X? Correct. So what I've done is this X which I'm showing with the arrow sign, see the motion of my pen, I've replaced it with Psi X. So my result will also be small f of Psi X. Correct. And this I'll copy as such Psi X. Again, this will become F of Phi X. And this will become Phi dash X. Is that fine? Okay. And this is what your Leibniz rule actually says. This is what we had actually written over here. Okay. As you can see the result which we have written over here and hence proved. Okay. So the process was very clear to you. So at any stage of this evaluation, you never needed capital F of X. That means you never needed the primitive of small f of X. So never needed to apply integration at all. Right. So this becomes so simple to evaluate that you don't have to take a longer route. Okay. Now I'll give you an example to understand this. Let us have an example. So far. Okay. I cannot see this way. Okay. Anyways, fine. Any questions so far guys? No. Okay. Let's move on to the example on this. I'll give you a simple example to understand this. Let us say your function is Y which is integral of 1 by log T dT. And you are integrating from X square to X cube. You are integrating from X square to X cube. Okay. X is a positive quantity here. And the question is find dy by dx. Find dy by dx. The question is find dy by dx in this case. Okay. Would you like to try? Everyone would you like to give it a try? Just try it out. I'll rewrite the formula for you over here if you have not copied it. It says that the derivative of the integral from Phi X to Psi X Phi X and Psi X being some functions of X of a function f of T dT is given as f of Psi X into derivative of Psi X minus f of Psi X into derivative of Psi X. Okay. So please do give it a try. And once done, just type in done on your chat box so that I can discuss it. It should just take you one minute, not more than that. I think lesser than a minute. So use Leveney's rule. Done. Okay. So what I'm going to do is as I already written in the formula, first I'm going to put upper limit in place of T. Okay. So when you put upper limit, so it will become something like this. Divide by dx is equal to one by log upper limit. When you put it will become log x cube into derivative of x cube derivative of x cube is 3x square minus now put the lower limit in place of T, which is log x square into derivative of x square, which is going to be 2x. Okay. Now you can simplify it the way you want to. So 3x square and this becomes 3 log x. We all know the log properties. This becomes minus 2x by 2 log x. We can cancel off 3, 3 factor and here and I can take a 1 by log x common and of course we can take an x common on top. So it becomes x x minus 1 by log x. Okay. And the reason why they have given x should be positive is because they want log to be a defined function. Okay. So this becomes your answer. Remember one by log x is a non integrable function. That means the integration of 1 by log x cannot be expressed. Okay. So there are some lists of functions which you cannot integrate like sine x by x x by x e to the power x square e to the power minus x square x square by 1 plus x to the power 5 1 by log x. So this belongs to that group where you cannot find the integration of this. So 90% of the people would who do not know these formulas will be like breaking their head how to integrate this and then I can do the process of differentiation. Okay. So no need to go through the process of integration at all if you are following the Leibniz rule. Okay. Hope it makes sense everyone. Any question please do ask because this rule is very, very important. Some of the most difficult questions have been framed on this and students are sometimes a rule is that Leibniz rule would be required for this. Okay. Now I'm going to give you a general idea or the general overview of the Leibniz rule. Okay. Generalization of the Leibniz rule. So if you have a function which is a multivariate function that means it has x also in it and it has t also in it. Right. What is a typical example of a multivariate function like this? Something like see x squared t plus t by x to the power 4. Okay. So this is a function which is both in terms of x and t. So this is a multivariate function. Okay. In the prior example which I cited it was a single variable function. Right. But this time my function let's say it has more than one variable in case let's say we can say it has two variables x and t. Okay. Now if I want to if I want to integrate this function if I want to integrate this function. Okay. With respect to t from Phi x to Psi x mind you you are integrating it with respect to t not x. So x would be treated like a constant. Are you getting my point? So integration is happening with respect to which variable integration is happening with respect to t not x. So while the process of integration is happening x would be like a constant x would behave like a constant because it is being integrated with respect to t. So t is a variable here. Okay. But remember your upper and lower limit are functions of x again Phi x and Psi x. Right. That means after integrating wherever you see the t you have to put these functions of x in place of that. Okay. So overall this entire thing overall this integral would be a function of x. Let's say that function of x I am differentiating with respect to x. Okay. So how would I do this? So basically Leibniz rule in such a situation gives you a slightly more complicated result it says it will be integral from Phi x to Psi x of partial derivative of this function fx, t with respect to x whole integral with respect to t plus f of Psi x f of f of x, Psi x into Psi dash x minus f of x, x, Phi x into Phi dash x. Okay. And this result I had shared with you in the morning today. Okay. Did you get an opportunity to see that? How many of you got the opportunity to see that derivation? Probably you would have not understood anything from that derivation. But now once you know what is the Leibniz rule, please go and visit that derivation once again. So in the interest of time it is pretty long. I think you would have seen the derivation. It has run into two pages. Right. Correct. So in light of that, in light of that time saving effort, we will not discuss the derivation once again because that is not required first of all. And secondly, it will take a lot of time. So better to see some problems based on this. But I will appreciate some problems based on this application. Okay. Now everybody knows the partial derivative with respect to x partial derivative. When you're doing you have to treat only x as a constant. Sorry, x as a variable. Everything will be like a constant. So when you do partial derivative with respect to x only x will be treated as a variable. Everything will be like a constant. Okay. And actually the first rule that we did was also the same thing. But do you recall that in my first rule there was no x. There was there was no x. So when you applied Leibniz rule on this. Okay. When you applied Leibniz rule on this the first term here actually when you do the partial derivative of a function which is in terms of t only. This actually became a zero term because there is no x over here. So there's no point in writing this term in that rule. So this guy actually vanished. Okay. And the rest of the terms. Yes, they didn't make appearance. So it was f of psi x into psi dash x minus f of phi dash into phi dash x. So the first was actually the same rule. And by this rule is the generalized form of the Leibniz rule. Okay. Is that fine any questions regarding this form? So please note down this formula. Okay. There's an additional term over here. Remember you have to partially differentiate with respect to x and whatever remains you have to integrate with respect to t from phi x to psi x. And these two results are more or less very similar to what we did in the first form of Leibniz rule. Any questions you can unmute your mic and speak. No problem. Okay. I'm sitting here. Ritu. Paras. Ardhara. Rashmi. Hello. Rashmi. Good afternoon. Happy to you. So no question. Yes. Good. So let's let's solve problems on them. Let's at least for one one-half hours will take a lot of problems on Leibniz rule. So let me begin with this question. If if integral of t square ft. Okay. From sine x to one is one minus sine x. Okay. Where your x belongs to the interval zero to pi by two. Where your x belongs to the interval. Where your x belongs to the interval zero to pi by two. Okay. The question is find f of one by root three fine f of one by root three. Remember my functions have my function has not been given to me directly. They have given to me that there is a integral of the function from sine x to one. Okay. And that to the function has been multiplied to t square and this is one minus sine x. Just try out. Once you're done peace type in your answer on the chat box. I'll give you three minutes to try this out. Hint is differentiate both sides with respect to x apply Leibniz rule. That's my hint. Yes. Anyone else can you all hear me? Yes. Yeah. Yeah. Okay. Let's solve this problem. See just have to apply Leibniz one sides. So let's differentiate both sides with respect to x sine x one. We have t square f of t dt and this also if you differentiate with respect to x now. So when you when you apply Leibniz rule remember you have to first put the upper limit in place of t and differentiate the upper limit also right now since this upper limit is a constant you know that ultimately when you differentiate that constant it is going to make all the terms 0 right. So don't even waste time doing that just write a 0 minus put the lower limit in place of a T. So it becomes sine square x into f of sine x. Okay. Into derivative of sine x which we all know is causing right on the right hand side we get a 0 minus cos x. Is that fine? Okay. Simple enough. Now since your x lies between 0 to pi by 2 you know cos x is not going to become 0 right. So you can cancel off cos x factor in fact minus cos x factor from everywhere. So it leaves you just with sine square x into f of sine x equal to 1 right. That means f of sine x becomes 1 by sine square x. Correct. Guys let me tell you here whatever you have put over here the same thing square is being reciprocated as the output of that function right. So for example if I put this as y right you will read this as 1 by y square. That means your function is behaving like this. So whatever you are inputting into the function it is just reciprocating the square of that input. So if I put 1 by root 3 into my input what will it do to it? 1 by 1 by root 3 square that's nothing but 3. Correct. So this is the answer over. Are you getting the point? So there is no need to actually know the function to you know solve this. You can easily apply Lemney's rule and it makes your effort so so less in this. Is this clear? Please type CLR on your chat box if it is clear. Oni Hamsini and Adara, Simukta, Rashmi, Shea. Okay great. Okay well I am sure you would like to solve more problems on this. So let's move on. So next question let there be a function defined from real numbers to real numbers. Okay and this is a differentiable function. Okay the function is derivable. Okay. This function satisfies the following characteristics. F2 is 6. F dash 2 is 1 by 48. Okay. Then question is evaluate. Evaluate. Limit X tending to 2 integral of 40 cube by X minus 2 dt from 6 to f of X. So what do you have to evaluate? You have to evaluate the limit of this particular function which itself is an integral of 40 cube by X minus 2 respect to dt and limits of integration are from 6 to f of X. Okay please give it a try and type in your response once you're done. Once done please feel free to type in 4.5. Okay. Shea also says she got the same. Well good. Anybody else who wants to respond? Rashmi, Paras. Okay let's evaluate this. So basically see here you are integrating with respect to t right. So X is just like a constant. Okay. So this entire expression you could actually write it like this. Limit extending to 2. Okay. Integration of 40 cube dt from 6 to f of X. And you can treat as if you're dividing the whole thing by X minus 2. Correct? Okay because X minus 2 has no role to play because you are integrating with respect to t. Correct? Now if you see this carefully it is actually a 0 by 0 form of indeterminacy. Why 0 by 0 form? Denominator is definitely 0 because the moment you put X as 2, 2 minus 2 becomes 0. Whereas numerator also becomes 0. How do I know that? Simple. If you put 2 in place of X it becomes integral from 6 to f of 2. And f of 2 is already known to me as 6. Right? So it is as good as integrating from 6 to 6. Correct? So anything integrated from the same point to same point will give you a 0 result. Right? That's why I'm so sure that it's a 0 by 0 form. So if it is a 0 by 0 form we can actually apply the Lopital's rule. Yes or no? Okay? So if you apply Lopital's rule, just differentiate the numerator with respect to X. Correct? So you are differentiating the numerator with respect to X. So for that you need Leveny's rule. So basically you are doing this. Let me write it in a plain and simple words. D by dx of 6 to f of X, 4t cubed dt divided by 1 minus 0. Okay? So applying Leveny's rule here it becomes 4 f of X whole cube into derivative of f of X. And we don't have to put anything like 6 because ultimately derivative of 6 is going to be 0. Okay? So you can avoid writing anything else because it will be ultimately 4 into 6 cube into 0. Okay? So this entire thing becomes a 0. So whenever there's a constant you don't have to write that term because you are going to multiply with the derivative of a constant at the end which is going to make everything 0. Right? So ultimately this is your answer because denominator you have only one, right? Okay? Now put the value of X as 2 because ultimately after applying Leveny's we have to put the value ultimately after applying the Lopital's rule we have to put the value of X as a. So it becomes 4 f of 2 whole cube into f dash 2. Okay? Now f of 2 is 6. So that's going to be 6 cube. And f dash 2 is going to be 1 by 18 if I'm not wrong. Let me just drag up and check. Oh yeah 1 by 48 I'm sorry 1 by 48. Okay? So this goes by a factor of 12. In fact this is 36 into 6 by 12. So this goes and ultimately your answer is 18 not 4.5. Okay? All you girls who got the same answer you were all wrong and you realize where you went wrong. I'm sure it must be some silly mistake here and there but all of you doing the same mistake is quite surprising. What happened Shreya? So 18 is the answer. My smiley is for all of you. Okay? Let's have another question. Integral of cos of t square. If integral of cos of t square with respect to t is same as integral of sin t by t from 0 to x square. Then prove that then prove that dy by dx is twice of sin x square by x cos y square. No why should it be f of x minus 2? Why are you changing your x to f of x? Let x be x. Yeah exactly. You forgot to differentiate the denominator. That's correct Shreya. Never mind Anthony. Let's try out this problem. This pretty straightforward problem. Again I'm giving you a hint. Differentiate both sides with respect to x. Okay? Just type done once you're done. Done. Great. Let's discuss it. So what I'm doing is I'm differentiating both sides with respect to x. So this side will become cos of y square into derivative of y square with respect to x which we know is dy by dx. And as the lower limit of integration is a constant which is 0 in this case. You don't have to bother writing the second term also right. Please don't waste time writing the second term because you know you're having a derivative of a constant at the end. Well this will be sin of x square by x square into derivative of x square which is 2x. And again I will not bother to write the next term. Right? Well this is the end of the game. So xx gets cancelled. Get this cos y square in the denominator. So this becomes 2 sin x square by cos of. In fact there's 1x left over here x cos of y square. Done. And it's proved. Okay. Simple. Okay. Let's take another one. Guys I want to do a lot of questions on this because you know this concept is not there in your CVSC. And secondly Je likes this concept a lot. Okay. And this is the most challenging of all the properties. I would not say difficult but yes most challenging of all the properties. Okay next question is if x is equal to integral of. Okay. If x is equal to integral of 1 by under root 1 plus 90 square from 0 to y. And d2y by dx square is Ay. D2y by dx square is Ay. Question is find the value of A. Find the value of A. A is a constant. Okay. So here A is a constant. A belongs to a constant. Try this out. Type in your response in the chat box once you're done. Okay shares is 9. I would not say right or wrong as of now. Let's see what others have to say. 9. Absolutely. Okay. Correct. 9 is the right answer. Okay. Okay. So again we have to apply Leibniz rule here. So let's differentiate with respect to y this time. Okay. So let's find dx by dy. So dx by dy is going to be 1 by under root of 1 plus 9 y square. Okay. Okay. Into derivative of y with respect to y which is 1. And I would not even bother to write the next term because that contains the derivative of a 0. So which is going to be 0. So this implies dy by dx is nothing but under root of 1 plus 9 y square. Right. Now d2y by dx square means what? D2y by dx square means you have to differentiate dy by dx again with respect to x. Okay. So we can all do that because it's a simple cakewalk for everyone here. So that's going to be 1 by 2 under root of 1 plus 9 y square into derivative of 9 y square with respect to x which is 18 y dy by dx. Right. So simplifying this further it becomes 9 y into dy by dx. dy by dx is your 1 plus 9 y square. As you can see here dy by dx is already under root 1 plus 9 y square. So this and this gets cancelled giving you the answer as 9 y. And if you compare this with a y your a is nothing but 9. Absolutely correct guys. Okay. Hope you are finding this concept more familiar and now I'm going to take up a problem which is slightly more challenging. Let's say I have a function defined from the interval 0 to infinity to 0 to infinity. So this is the domain and co-domain of the function. Okay. And this is a differentiable function. Okay. So this is a derivable function. Okay. And this function satisfies a functional equation which goes like this. Okay. It says x times integral 0 to x 1 minus t f of t dt is equal to 0 to x f of t dt. Okay. For all x belonging to the interval 0 to infinity. Okay. It's also given that f of 1 is 1. It's also given that f of 1 is 1. Okay. Question is find the function. Find the function. What is the function you have to state? Hope the question is clear. So we are dealing with a function which is a derivable function and the function is defined from 0 to infinity to 0 to infinity. And it satisfies a relation x into integral of 1 minus t f t dt from 0 to x is equal to integral of t f t from 0 to x. And we know f of 1 is 1. We have to find what is my function in this case? What is the function in this case? I know you will not be able to type the function. So you can just unmute yourself and speak out the function to me. If you feel you have no idea how to proceed with this, you can also type that so that I know that nobody's everybody has given up and have to start solving this. But if you're trying well and good, keep trying. All right. So I think enough time has been given. Let's try to look into the solution for this. See the first thing is I will differentiate both sides with respect to x. Okay. This is one of the functions. That's the only result that is left to differentiate both sides with respect to x. Now, please mark here that there are two functions multiplied over here. Okay. So x into this. So I'll be applying product tool while I'm using the derivative on the left-hand side. So first, let me differentiate x. So that's one and this other things I'll copy as such. So 0 to x 1 minus t f of t dt. Okay. Next is I'll keep this x as such and then I'll differentiate this by use of the Leibniz rule. So that's going to be 1 minus x f of x into derivative of x. That's going to be 1. I don't even bother writing the next term because that's going to be 0 anyhow. Right hand side will be x into f of x. Okay. So far so good. Nobody has any problem in this. Let's try to bring this term to the right-hand side and let's try to group up f of x together. So 0 to x 1 minus t f of t dt is going to be, let's take x f of x common. So that's going to be 1 minus 1 minus x. That's going to be x square f of x. x square f of x. Okay. So finally you see, let me change the ink. Finally you see you are left with this expression 1 minus t f of t dt. This is equal to x square f of x. Okay. Now again apply Leibniz rule. Let's apply Leibniz rule again. Okay. So it will become 1 minus x f of x. Here I will write x square f dash x plus 2x f of x plus 2x f of x. Correct. So let's bring again terms having f of x together. So 1 minus 3x f of x is equal to x square f dash x. Okay. Now let's do one very interesting thing over here. Let us bring f of x and f dash x to one side, put a dx and integrate both sides. Okay. This is a critical step over here because most of the people would not be anticipating the step to come up. So now when you do that, the left hand side is just integration of 1 by x square minus 3 by x. And the right hand side, you know that if you take f of x as t, I'm sorry, f of x as t, f dash x dx is going to become a dt. That means it is just like integrating dt by t, isn't it? So finally the result will be minus 1 by x minus 3 ln x. And this is going to be ln mod t. Mod t is nothing but your function itself. So I'm putting the function back. And of course, don't forget the constant of integration. Guys, just a word of advice from my side to all of you. Whenever you are performing an indefinite integral, never ever forget the constant of integration. Okay. You may feel that it's a unnecessary thing, but remember, you will not be able to get a correct answer if you are not using that. Good. Now, I don't need a mod here because if you see the function definition is from 0 to infinity to 0 to infinity, that means the function is already attaining positive values. So it is only defined for positive values. So mod is redundant. So can I remove this mod from here? Let's drop this mod. I don't need this mod. Okay. Because f of x is already positive, right? Because f of x is already positive. Okay. Now, how do I get this C? To get the C, you can see that we have been provided this condition. f of 1 is 1. Okay. This is going to help us to get the value of C. So what I'm going to do is I'm going to put x as 1 everywhere and my f of x is going to be 1. Okay. So this gives you the value of C as mind you, ln 1, ln 1 becomes 0. So C is actually negative 1. As you can see, there is a value of C that we are getting. So had you ignored it, you would have got a wrong result. Right? So ultimately, your function is... Let me rewrite this function here. So ultimately, your function is minus 1 by x minus 3 ln x is equal to ln of f of x, ln of f of x minus 1. That means ln of f of x is nothing but 1 minus x minus 3 ln x. Let me bring the ln's together. So ln of f of x plus 3 ln x is equal to 1 minus 1 by x. And we all know that this is going to be f of x into x cube. Correct? Isn't it? Because 3 ln x is nothing but ln of x cube. And ln a plus ln b is ln a b. So I've skipped a step in between but hope you are able to understand this. If not, please let me know. So I can say f of x into x cube is take antilog. So it will become e to the power 1 minus... Oh, why did I write an x cube here? I'm sorry, that's an x. Yeah. So it's going to be e to the power 1 minus 1 by x. So ultimately, your f of x is just divided by x cube. So 1 by x cube e to the power 1 minus 1 by x. Okay? So this becomes your function. Is that fine? Are you able to connect now? I would say this problem to be a J advance problem. It is not a J main problem. It is higher than the J main level. Okay? So please type CLR on your screen and this is clear. Awesome. Great. Great. Okay. Oh, Aditya is also here. Now I'll take an example where you have the generalization of the Leveni's rule being used. So if y is equal to integral of f of t sin kx minus kt. Now see here, in this case, they have given an x also in the function. So it's a multivariate function. Okay? Then prove that, then prove that d2y by dx square plus k square y is equal to k times f of x. In this, you have to use the generalized form of the Leveni's rule. Once you're done, please type down on your screen on your chat box. Great. Share is done. Let's wait for one more person to finish. Then we can discuss this. Okay, let's discuss this. So first, we'll apply this formula of Leveni's rule. I'm writing it down for the benefit of all of you. The formula is, remember here you are integrating with respect to t while you're differentiating with respect to x and this function contains both the variables. Okay? So this is going to be integral from phi x to psi x of the partial derivative with respect to x of the function f of x comma t. Okay? Plus f of x comma psi x minus, sorry, psi dash x minus f of x comma phi x into phi dash x into phi dash x. So this is the most generalized form of the Leveni's rule. The proof of this was already sent in the morning to you. So please have a look at it after today's class. So now, this is my multivariate function because it has t also in it and it has x also in it. Right? So first is we have to do a partial derivative of this function with respect to x. Remember when you're doing partial derivative with respect to x, t must be treated as a constant. Okay? So f of t will also be a constant. Correct? And the derivative of a sine is cos kx minus kt into the derivative of kx minus kt, which is k. Okay? Let's see what are the other terms plus, now you put your x in place of t and differentiate with respect to, differentiate x, which is going to be zero anyways. Correct? And then you have to put a zero and differentiate with zero that is also going to be zero anyway. So if you are going to write both the terms, both these two terms will actually be zero each. Is that fine? Why the first term is zero? Because this function will become a zero because when you put t as x, kx minus kx will be zero and sine zero we all know is zero. Why does this term become a zero? Because you are ultimately having a constant over here whose derivative is anyway zero. Right? Is this fine? No problem in finding dy by dx, so this is your dy by dx so far. Let us apply Leibniz once more. So let us differentiate both sides with respect to x once more. So again the same thing. This is my new function. So phi x to psi x partial derivative of this will be k f of t. This will become minus sine kx minus kt into a k. So k square will come over here. Right? Plus put the upper limit. In fact my upper limit here is x. So let me just write x and zero x and zero. So upper limit is x that is going to make this function k f of x k kx minus kx into one. And the other term I don't even have to write because that's going to be zero anyhow. Okay? So this is going to give me d2y by dx square minus k square zero to x f of t sine kx minus kt. This will become k f of x. Remember cos of zero is going to be one anyhow. This term if you recognize, this term is actually your y itself. So this is nothing but minus k square y plus k f of x. Bringing it to the other side we end up getting d2y by dx square plus k square y is equal to k f of x. Okay? Hence proved. Is that fine? Any questions? So Levne's was applied twice and probably this is the first example where I gave you an instance where you had a multivariate function appearing in the question. Okay? Now in all these questions we pretty much have an idea that okay Levne's rule is going to be used. But let us take a question like this which will probably surprise you. Evaluate integral from zero to one x to the power b minus one by ln x. Solve this question. It's a pretty simple question. So basically b is a quantity here which is greater than equal to zero. So I have given integration for integration of x to the power b minus one by ln x from zero to one. Any idea? Any idea how to do it? Please type no idea if you don't have any idea how to proceed. So that I know how many of you are just sitting idle. Great. Everybody's fine. Sure. I'll give you two minutes. Try it out. Let's see whether you're able to make some progress. b. No, b is not the right answer. Okay. So let me just solve this first. Guys, by looking at this problem, you would have realized by this time that your answer would be in terms of b only. Right? Correct. So let this be your function of b. Okay. You can write ib or let me make it as a function only so that you can relate to it. So the answer would be a function of b only. Right? Everybody agrees to this. Everybody agrees to the fact that when you integrate this, of course you don't know the value of b yet. You can only substitute for x. So everything will be in terms of b. So this answer would be a function of b. Correct? So now let us differentiate both sides with respect to b. Okay. Now it is pretty surprising why because b here, as you assume to be a constant, it is actually a parameter because as your b changes, you will have your answer according to the b value there. Isn't it? So it surprises a lot of students when they listen to this expression that, oh, you are differentiating with respect to a constant. No, it is not actually a constant. It's a parameter for a different, different b for a different, different question be will be different. Okay. Now, if you differentiate this with respect to b, remember x would be playing the role of your T now. Are you getting it? So this has to be treated as a function of b and x. Correct? So you have been given something like this. You have been given integral of b f of x, which are integrating with respect to x, but differentiating with respect to b. Are you getting this point? So there is a role change now. You're not differentiating with respect to x. You're differentiating with respect to b. But you are integrating with respect to x. So what role T played earlier? Now x is playing that role. And what role x played earlier b is playing that role? Are you getting this point? This is very critical. Understood. Please type clear if it is clear to you what I'm trying to say. Okay. So now when I apply my Leibniz rule, I have to do partial derivative with respect to b of this expression. Okay. And other terms will not appear because both the numerator and denominator are like constants. So these two terms would be zero minus zero. So there's no point writing them out. Okay. Now, what is the derivative of this term with respect to b? Remember, x is a constant. So when x is a constant, it is just like 1 by ln x is just a constant for you. And derivative of this term is like constant to a power of a variable. Constant to the power of the variable will be nothing but this, right? It's just like saying what is the derivative of a to the power x? a to the power x derivative is a to the power x ln a. So here your x is your constant, b is your variable. So it is constant to the power variable ln constant. Yes, I know. Now cancel this out. It just boils down into integration of x to the power b, which is nothing but x to the power b plus 1 by b plus 1 according to your power rule of integration. Put your limits. It becomes 1 by b plus 1. Correct. Now the problem is not over yet. You have found out f dash b, my dear friends, getting my point. We need to know f of b. So for f of b, you would say, hey, it is simple. I can just integrate this with respect to b. Okay, which is nothing but ln of b plus 1. Okay, now I'm writing b plus 1 because I know b is positive. Okay, I don't need to put a mod around it. But remember to put a constant of integration. Many of you would forget this constant of integration, which is going to cost you dearly. Okay, so don't forget that constant of integration. Fine. Now, how do I find that constant of integration? That's the next question. Now, if you can see the question very clearly here, you realize that f of 0 will be 0. Why? I'm making a claim here that f of 0 would be 0. Why? It's obvious because it will be now integration of 0 to 1, x to the power 0 minus 1 by ln x. And x to the power 0 is 1 itself, so it's going to be 1 minus 1 on the numerator. Okay, so this is going to help me find the value of c. Now, let's use the fact that f 0 is 0. It implies f 0 is equal to ln 0 plus 1 plus c. This is 0. ln 1 is also 0, so c is also 0. So here, by mistake also, if you would have forgotten c, you were safe actually. But don't expect that to happen every time. So, here we are. This is my answer. This question, let me tell you, surprises a lot of students because this is the first time they realize the importance of and the strength of Leibniz' rule. Any question with respect to this? Please type no q if there's no question. At least everybody is attentive listening to this. Great. Let's talk about the next question. I know you would be more inquisitive to solve questions of the similar type. So let's move on and do one more. In fact, two more we'll do today. So 0 to pi by 2 and the value of this integral. I'll give you the options. Options are option A, which is pi times under root of ln x minus 1. Option B, pi times under root of 1 plus x minus 2, pi times, sorry, under root of pi times under root of x, under root of 1 plus x, I'm sorry, minus 1 and option D is no tau. Remember, you're integrating with respect to theta. So your answer would be in terms of x. Just type the option number, which you feel is correct. Okay, let's discuss this. So we all know that this is going to be a function of x, because x is just like a constant and you are integrating with respect to theta. So ultimately, you'll substitute the upper and the lower limit on theta. Now, let us differentiate both sides with respect to x. So according to the Leibniz rule, this will be partial derivative with respect to x of log 1 plus x sin square theta by sin square theta d theta. And we don't have to put the terms after this because you know your upper and the lower limit, both are constants. So the derivative of constants are going to disappear. So partial derivative of this means you are treating theta to be constant and only x is treated as a variable. So this will be 1 by 1 plus x sin square theta into derivative of this is going to be 0 plus 1 sin square theta. So sin square theta, sin square theta will be gone. So 0 to pi by 2 d theta. So ultimately, it boils down to integrating 0 to pi by 2. 1 by 1 plus x sin square theta with respect to theta. Now, how do I integrate this? There are a lot of ways. I can divide the numerator and denominator by cos square theta. So 0 to pi by 2. So let's divide the numerator with cos square. So it will give me sin square theta d theta. This will again give you a sin square theta and this will give you x tan square theta. So this gives you integration from 0 to pi by 2 sin square theta d theta. And this is going to be 1 plus x plus 1 tan square theta. Let's take tan theta is t now. So sin square theta d theta will be equal to dt. So this will become integral from 0 to infinity dt by 1 plus x plus 1 t square. Guys, remember here, x will behave just like a constant because integration is happening with respect to t. So pull out our x plus 1 out. So it becomes 0 to infinity dt by 1 by x plus 1 plus t square. You can write this as under root square. Okay, so we all know it's a square plus t square form, which is a standard integral. So that will give you 1 by a. 1 by a means under root of x plus 1 tan inverse x by x. x by will be this. Okay, upper and the lower limits. Let's put it. So this and this will get cancelled. So it will be 1 by root x plus 1. When you put an infinity, it becomes a pi by 2. When you put a 0 in place of t, it becomes a 0. So ultimately, you get this expression. But remember, this is f dash x only. I have to find f of x. This is my f dash x. So f dash x is 5 by 2 under root x plus 1. So to find f of x, it's pretty easy. We have to integrate with respect to x, right? So what will this become? This will become pi by 2 integration of x plus 1 to the power of negative half, which is nothing but pi by 2 x plus 1 to the power half. Okay, divided by half. So into 2. Plus a C. Don't forget the C. I'm telling you this C is very, very important and many people will forget it. So this will become pi. You can write this as root of 1 plus x if you want. Plus a C. Now, how do I get C? How do I get C? Do you recall that f of 0 in this case, if you put your theta as... If you put your x as 0, okay? Your function will become integration of 0 to pi by 2 log 1 plus 0 because your x will be 0 here, okay? Divided by sine square theta, which ultimately will boil down to 0 itself because log of 1 is 0. So what do we know from here that f of 0 is 0? This will help me to get the value of C here. So let's go down here. So this is your f of x. So let's put x as 0. So this will be 0 itself. This will be pi plus C. So C becomes minus of pi. C becomes minus of pi. So put it back. Put this back over here. So ultimately, let me write it down. Ultimately, your answer is f of x is pi under root of 1 plus x minus pi. Take pi common. So it's under root 1 plus x minus 1. Let's see which option matches with this. Pi under root 1 plus x minus 1. Oh, option A is the right option. Option A matches with it. Is that fine, guys? So I'll just repeat the roadmap. The roadmap is... I know at the end of the day, this integral would be a function of x, okay? Then I differentiated both sides with respect to x by using a Leibniz rule. Once I differentiate it once, I got an integral which is right here in front of you. This is easy to evaluate. This was easy to evaluate. Okay? Once you evaluate it, I'm sure you know your indefinite concepts very well by this time. So once you evaluate it, you know the value of f dash x, okay? Which is right over here. So this is the stage we reached at. Then we integrated this once more to get f of x, because ultimately that's what we are seeking for. And to get the value of constant of integration at this step, I used the fact that f of 0 is 0, and I got that constant of integration and thereby completing my f of x, which is the required integral. Is that fine? Does it make sense? Any doubt? If it is clear, please type CLR on your screen. Next question is prove that. Any idea how to do this? Okay, I'll just give you a hint. I think we had... In fact, let me ask this as a question to you. How would you integrate this? Forget about this square. Without this square, how would you integrate this? So you would say pretty simple. How would you divide by cos square? Just like the way we did the previous question. So if you divide by cos square, it becomes secant square x dx by a square tan square x plus b square. Okay? Then of course you would take your tan x to be t. So secant square x dx will be your dt. This would be from 0 to infinity, dt by a square t square plus b square, which is nothing but... You can treat this as 0 to infinity, b square plus at the whole square. Okay? So something like the form of the special integral, a square plus x square. We know this result is 1 by 8 tan inverse x by... So this will be similarly 1 by b tan inverse 80 by b divided by... So this will become ab. Okay? Now put the limits of integration 0 to pi by 2. So the moment you put a pi by 2, I'm sorry, 0 to infinity, not pi by 2. Pi by 2 is already gone. We have to put 0 to infinity. So when you put infinity, you'll get pi by 2. When you put a 0, you'll get a 0. So the result is pi by 2 ab. Now you must be wondering why am I doing this? Right? What it has to do with this question. Okay? My question was this. Why did I sit and evaluate this? Okay? Now there's something very interesting which I'll show you right now. So I can actually get this result by differentiation. Right? Differentiation. So I'm going to show you that next. Now that I know that integration 0 to pi by 2 dx by a square sine square x plus b square cos square x is pi by 2 ab. Okay? Let us differentiate both sides with respect to a. Okay? Let's differentiate both sides with respect to a. Okay? Of course you'll be using Leibniz rule here. I'm treating my bx everything to be like another variable and a is my, you know, variable of integration. Sorry, variable of differentiation. So I'm differentiating with respect to a here. Please make a note. Right? So as per Leibniz rule, this will be 0 to pi by 2. Partial derivative of this with respect to a. What will be the partial derivative of this with respect to a? So let's say 1 by, please note sine x, b, cos x, etc. All b will be like a constant. So minus this square into derivative of this term, which will be 2a sine square x. Okay? Are you getting this point? Okay? Other terms will not appear. And I have to differentiate this side also with respect to a. So that will be minus pi by 2a square b. Is that clear? Is that clear? Okay. So I'll drop the minus 2a terms from both the sides. So this I can write it as 0 to pi by 2 sine square x dx by a square sine square x b square cos square x square and drop a minus 2a term. So that will become pi by 4a cube b. Pi by 4a cube b. Yes, sir. Oh, somebody spoke. Okay. Okay? In a similar way, if I differentiate both sides with respect to b. Okay? Let's go back to the original expression 0 to pi by 2 dx by a square sine square x plus b square cos square x and let us differentiate this both sides with respect to b. So in that case, what will I get? I will get minus 1 by a square sine square x plus b square cos square x whole square into derivative of this will be 2b cos square x and this side I'll get minus 2 by a b square. Is that fine? Now drop the factor of minus 2b both sides. So it'll become 0 to pi by 2 or why I'm writing cos b. I'm sorry. Yeah, cos x. Okay? So it'll become cos square x dx by a square sine square x plus b square cos square x whole square and on the right hand side I'll have 4 by pi a b cube. Right? Now let us add 1 and 2. Let's call this result as 1 and 2 and let's add them. Let's add 1 and 2. Any question? Okay, let's add 1 and 2. So when you add 1 and 2, it becomes 0 to pi by 2 numerate since denominator is same. You can just simply add up numerator on the left hand side. So denominator for both the terms is this. So you'll have sine square plus cos square and on this side, 4 a cube b and pi by 4 a b cube. Right? So let us take the LCM of pi a cube b cube. So you'll get pi b square here and pi a square over here which is nothing but pi a square plus b square by 4 a cube b cube. This is my answer. Is this what we wanted to show? Yes, absolutely. Is this what we wanted to show? Is that fine? Again, see the beauty of using derivatives with an integral sign. And how efficiently you can solve this problem without having to go into the nitty gritties of complicated substitutions or by using a kind of product tool etc. Is that fine? Okay. Yes, sir. So now we'll move on to the inequality properties. So now we are jumping on to some other properties, the inequality properties. Okay. So let's take up the inequality properties of definite integral. First property, which is a very obvious and trivial one. Okay. So let's say we have two functions, f, g and h. Okay. And these functions are defined and continuous in the interval a to b. Okay. So they are defined and continuous in the interval a to b. And these three functions are related by this relation. That is g of x is lesser than f of x and f of x is lesser than equal to h of x. Okay. Now the similar property will also hold on their integrals as well. That means if you integrate g of x from a to b, this will be lesser than integral of f of x from a to b. And this will be lesser than equal to integral of h of x from a to b. Are you getting this point? Now the proof of this is pretty simple. I mean, you can just look at the geometrical proof for this rather than anything else. So you know your g of x, h of x is on the top. Let's say this is your h of x, then is your f of x and then let's say there is your g of x. Okay. f of x and this is your g of x. Okay. And you're trying to integrate from x equal to a till x equal to b. Okay. As you can clearly see that the area within, the area below h of x will be higher than the area between f of x and that will in turn be higher than the area between g of x. Correct? So you can say, let me name this figure as a b c d e f. So the area of, the area of, let me name this triangle, let me name this curvilinear structure as a c d b. That would be higher. In fact, let me start from the low order a f b would be lesser than equal to area of a b e b and this in turn will be lesser than the area of a c d b. Okay. And they represent nothing but the respective integrals. So this is integral of f of x, sorry, g of x. This is the integral of g of x and this is the integral of f of x and this is the integral of h of x. Is that fine? Well, what kind of problems can I get on this now? Sometimes the problems on this becomes very dicey. I mean, you have to apply your trial and error method. Okay. And it becomes irritating at times. Okay. For example, let's take this question. Prove that, sorry, prove that integral of lies between 0 and 1 by 8. Okay. So many times we need to, you know, react to this as per our, you know, requirement of the options also. Here your x is lying between 0 and 1. That's it. I don't need to give that because the limit of the constant of integration, sorry, the upper and the lower limits of integration already say so. So this information may not be given to you because we all know we are integrating from 0 to 1. So your x has to be between 0 to 1. Now, let me solve this question because you'll get an idea. Of course, when you are between 0 to 1, all these terms are positive, right? Correct. So this function itself is a positive function. I can say this will be greater than 0. Correct. Now, when x is between 0 to 1, okay, you would all appreciate that x to the power 7 will also be between 0 to 1. Okay. Sorry, why I'm reading x to the power 7, x to the power 8 because we have x to the power 8 in the denominator. Correct. So 1 plus x to the power 8 will lie between 1 to 2. Correct. So root of 1 plus x to the power 8 will be greater than 1 and less than root 2. Correct. Okay. So can I say this quantity, if you reciprocate it, can I say x to the power 7 by under root of 1 plus x to the power 8 since this quantity is greater than 1, this entire quantity will be less than x to the power 7. Yes or no? Because if you're dividing any quantity by a number which is greater than 1, the resultant value would be lesser than the numerator quantity. So this function is bounded between this. Basically what I did, I did a kind of reverse engineering. When I saw 0, 0 came into my mind and when I saw 1 by 8, basically what came in my mind was integration of x to the power 7 from 0 to 1. This will also give you 1 by 8. Yes or no? And hence the result that if you integrate both sides from 0 to 1, okay. So your desired expression will now appear over here. So this term would lie between 0 to 1 by 8 and hence proved. So here because the question was given to you to prove it, you could think of it. But many a times you have to look at the options to see what they're actually asking you. Are you getting my point? So it requires a bit of practice. Let's take one more. Let's take this question. Prove that integration from 0 to half, this lies between half to pi by 6 for n greater than equal to 1. Think about it. Just try to figure out how would you get that half and how would you get that pi by 6 and reason it out. If done, please type done. Let's see here. You're dealing from 0 to half means you're dealing with fractions, correct? So any fraction which is between 0 and half, if you square it, right? It will be a value which is lesser than X. So this guy will be lesser than X, right? Yes or no? And if you raise to a higher power cell which is greater than 1, can I say this will be lesser than this, isn't it? So ultimately what I'm trying to say is if you raise a fraction to a power greater than 1, the resulting quantity will still become lesser and lesser, right? If you put a negative sign in front of it, the inequality will switch, right? Put a 1 that will not affect the inequality. Put a under root that will not affect the inequality because both are positive quantities, okay? Can I say this quantity will be less than this quantity and this in turn will be less than 1 for sure? Yes or no? Of course, when it becomes 0, it will become equal so we can put equality also here. Now, let us reciprocate this. If you reciprocate this, it will become Ulta. So this guy will come over here, one will come over here. I'm sorry, the inequality sign, I have also changed the inequality sign, that will remain the same. This will be greater than, this will be greater. Okay, yes or no? Now, treat this as if this is your f of X, this is your, you know, whatever is your g of X, this is your h of X, okay? So even the integration will also follow the same inequality sign, that means integration of this will always be greater than equal to integration of this. Will always be greater than integration of 1, okay? From 0 to half, 0 to half, 0 to half, right? This is very simple to integrate. This is sign inverse X, 0 to half. This is something which I don't need to bother because this is asked in the question, the range of this interval. And this is nothing but integration of X from 0 to half. So that gives you pi by six, greater than integration of this from 0 to half. And I think this is the same thing that has been asked in the question. And I've just written the inequality sign slightly opposite. So ultimately, the result is the same. We can just flip the order in which you have written. So half is less than equal to integration from 0 to half, dx by under root 1 minus X to the power 2n less than equal to pi by six. Is that your race? How it works? Any questions so far? No doubt, sir. No doubt, okay, great. Okay, then let's have this question. We have an integral I1, which is integration of sine X by X from pi by six to pi by three. And this is another integration from pi by six to pi by three, which is sine of sine X by sine X. And we have another integration I3, which is sine of tan X by tan X, sine of tan X by tan X. Then which of the following options is correct? Then select the correct option. Option A. I1 will be greater than I2 will be greater than I3. Option B. I2 will be greater than I1, and this will be greater than I3. Option C. I3 will be greater than I2 will be greater than I1. Option D. I3 will be greater than I1 will be greater than I2. So Shreya says option A, okay. Is it option B? Aditya says B. Okay. What about others? Okay, let's discuss this guys. See, if you see this, the structure actually looks like sine t by t kind of a structure, right? Where first your t is x, here it is x sine x and here it is tan x, correct? Okay. Now, this function, I'm sure most of you would have seen this function, you know, when you were doing limits chapter. It's like a wave, okay, kind of a thing. Okay. So in the interval 0 to pi, this function is a decreasing function. I'll better show you on GeoGebra also, how it works. Hope you can see my screen here. So y is equal to sine x by x. Okay, as you can see here on your screen, it's a kind of a ripple. Okay. And let's have a x equal to pi also. So from 0 to pi, you can see this function is decreasing. Okay, this function is falling down as you can see it. So it is showing a decreasing characteristic. Yes or no? Do you agree with me? It's showing a decreasing characteristic. If it is showing a decreasing characteristic, then can I say that if you put a higher value of the function, if you put the higher value of x, sorry, t, the function is going to be lesser. Okay, because when a function is a decreasing function, for a higher value, the function is going to fall down more. So can I say this function, if I put input of x, sine x, x and tan x, we all know that they are related to each other by this. Sine x is always lesser than x and x is always lesser than tan x. So when you're putting this in the function in case of t, that is sine of sine x by sine x, sine of x by x and sine of tan x by x, can I say, oh my bad, I should not put this. Now you tell me how the inequality will be placed over here. For the most heavier one, the value will be the least, correct? And the most lesser one, the value would be the greatest. So this is how this function would be related to each other. And so would be their integrals, correct? So if I integrate this from pi by 6 to pi by 3, this also from pi by 6 to pi by 3, this also from pi by 6 to pi by 3, okay? This is how this integrals would be related, by the way, this is I2, this is I1 and this is I3. So I2 would be greater than I1, I1 would be greater than I3. So option number B, Aditya was absolutely correct. Is that clear? Please type CLR if it is clear. Yes, sir. Anybody else? Clear? Awesome, great. Moving on to the next property in inequality, property number 2. If there is a function f of x, okay, defined in the interval a to b and this function f of x has a global maxima of let's say capital M and a global minima of a small m in this interval a to b, then the property says integral of f of x actually will lie between mb-a to capital Mb-a. Can you prove this? Done? Very simple. See just now we learned that when f of x lies between g of x and h of x, so will there integrals also, correct? So will there integrals also. Now, apply the same concept over here. Since you know that small m is your global minima, your function will always be greater than your global minima and always be less than your global maxima, correct? Here actually your g and h roles are being played by small m and small m capital M respectively, right? So if you apply the property, same property here, I can say integral of f of x from a to b will be also be applied to this and these are constants, right? So you just have mx and you put the limits of integration from a to b. So this gives you answer as mb-a is less than equal to integral of f of x from a to b, plain and simple. Graphically also let us try to understand this. See, let us say your function has something like this characteristic, okay? So let's say here, okay? So this is a starting point a and this is your ending point b, okay? So this is your global maxima of this function and this is the global minima of this function, okay? So what does this situation say? The situation is saying that if you make a rectangle like this, this rectangle will always have a higher area than the area within the yellow graph and if you make a rectangle like this, let me change my pen color to blue. If you make a rectangle like this, this area will always be blue area and let me do it as white area. And let me make a yellow area also. So what it's trying to say is that the white area will be greater than the yellow area and the yellow area will be greater than the blue area. This is what this inequality tries to suggest, okay? Let us take questions on this property. What type of questions are asked on this? I'll just take one question. Prove that this integral 5 minus x by 9 minus x square dx from 0 to 2 lies between 1 and 6 by 5. Once done, please type done on your chat box. Yes sir, done. Done. How about others? Ardira, Rashmi, Samitha, Hamsini, okay? All you need to do is figure out the global maxima and global minima of this function in the interval 0 to 2. So if you see this function, let's try to figure out where does the global maxima and the minima of this function lie. For that, we need to put the derivative of the function as 0 and see what roots come out from this. So if you differentiate this, you will be getting 9 minus x square into negative 1 minus 5 minus x into negative 2x by 9 minus x square whole square equal to 0. Okay, let's simplify this. So we'll get x square minus 9 plus you will have plus 10x, okay? And you'll have minus 2x square. So equal to 0. So this gives you minus x square plus 10x minus 9 equal to 0. I'm sure this is factorizable, right? 9x minus 1 and x minus 9. So 9 is beyond our, this thing, interval 0 to 2. So I'll be only worried about 1, okay? So 1 is the critical point, 1 is a stationary point. So what I'll do, we all know how to find out global maxima and global minima, right? What we do is we individually put the values like 0, 1 and 2, because this is my interval, right? My interval is from 0 to 2, okay? And see manually which of them gives you the maximum value and which of them gives me the minimum value, okay? So 0 gives me 5 by 9, undoubtedly. 1 gives you 4 by 8, which is half, correct? And 2 will give you 3 by 5. Yes or no? So which is the maximum, which is the maximum of them? I think this is the max and this is the min. So this function lies between half and 3 by 5, okay? So I'm not writing the function here. So what I can say is that the integral of this function should lie between half 2 minus 0 and 3 by 5, 2 minus 0, which is nothing but it should lie between 1 and 6 by 5. So whenever any quality relations are being asked, please be aware of this property as well. It is not always necessary that you have to find two bounding functions like we did in the previous examples, okay? Sometimes they can ask you a pretty simple straightforward one where you have to just figure out the global maxima and the global minima of that function in that given interval of integration, okay? And then apply this property. So it becomes icy because people ask me, sir, how would I know? Because you know, whether we have to find two bounding functions or whether we have to do it global maxima and global minima property, see the idea is only to try an error. There is like no straightforward. That's why these type of questions become most unpredictable ones, okay? Try this out. Prove that. By the way, let me tell you under root of 3 plus x cube cannot be integrated. I mean, it cannot be expressed as a integrable function, okay? You cannot integrate under root 1 plus x cube type of function. This cannot be integrated. Cannot be integrated means you cannot express it. You cannot express the answer of that. It's inexpressible integral. Guys, many of your parents have complained that you are not serious during an online session. Guys, this is unacceptable at least at this stage of your career when there's just three to four months left to write the most important exam of your life. If you're not serious, you know, at this stage, nobody can make you serious at this time, okay? Whether it's an online offline, it's a matter of getting into a good college with a good branch that is going to decide your future prospects in your life. So nobody should tell you to be serious at this time. You should be self driven, right? It's very saddening and disheartening to listen from the parent that don't keep online sessions. My child is not serious in that at least for a 12th grader, this is not acceptable. I can understand an 11th grader who takes a little bit more time to become serious once he starts realizing that things are going beyond his head now. But right now, just three, four months are left. And moreover, the problem solving sessions are only taken online because we don't have to write the question. We can just project the question and discuss about it. Okay? So going forward, I should not hear these complaints that you are not serious in an online session because you are, if you are not serious, you know, you are being dishonest with yourself. You are harming yourself, nobody else. Right? You are here to learn. And if you are a person who is keen to learn, you should be very, very attentive and serious during any type of session, whether it's online offline. Offline, you guys are definitely, I've seen you guys, you're very disciplined. You take notes very seriously. Take the class very seriously. But online also, you should show the same behavior. You're not studying for, you know, your parents, you're not studying for me. You're studying for yourself. Yeah, any idea how to do this? I can just, just go from that global, maximum, global, minima perspective. See, if you see this term, f of x is equal to under root of 3 plus x cube. Okay? If you differentiate this, you get 1 by 2 under root 3 plus x cube into 3x square. Okay? Now, this will always be positive for sure. That means this function, if your f dash x is positive, what does it imply? What does it comment about the nature of the function? That means f of x is always an increasing function. If it is an increasing function, its minimum value will be obtained at the initial point, which is 1. So minimum value will be g1, sorry, f1. Right? f1 will be what? Under root of 3 plus 1 cube, which is nothing but 2. What is gmax? gmax is going to be the last point, f3, which is nothing but 3 plus 3 cube. And that's how that root 30 makes its appearance. Correct? So since this function lies between, since this function lies between 2 and root 30. So we can say by the property, which we just now discussed, that the integral of this function from 1 to 3 will lie between 2 into 3 minus 1 and root 30 into 3 minus 1. That's nothing but 4, 4 to 2 root 30. 4 to 2 root 30. Getting the point? Yeah. Awesome. Great. So we'll now make a move to the next inequality properties. This property says, I think this is the third one in the line. This property says integral of a function from a to b, if you take the mod of it, this value will always be lesser than the value that you get when you integrate the mod of the given function. Okay? Geometrically speaking, it is very obvious. Right? Let's say I have a function, which is like this. Let's say like this. Okay? And you're integrating it from a to b. Let's say this is your b point. Okay? Now this area, I'm just hypothetically assuming that this area is 5 and this area is 3. Correct? So if you do integration from a to b f of x dx, what answer do you expect? You expect to get 5 minus 3. That is 2 as your answer. Correct? Okay? And if you mod the function, the very same function, if you mod it, you know its graph is going to be like this. Correct? So you are integrating from a to b. So this will be 5 and this will be 3. So if you integrate the mod of this function, you are always going to get a positive answer that is 5 plus 3, which is 8. So this term will always be less than equal to this answer. By the way, may I know when will the equality hold true? When will the equality hold true? When the function is... When doesn't go to the negative part? That's fine. Aditya, it can be completely positive or it can be completely negative also. That is also fine because you're ultimately modding your answer. Right? So equality will hold true when the function is completely below the x axis or above the x axis, not partially above and partially below. A mathematical way to prove it is very simple. We all know that mod of anything will actually lie between negative of that quantity and that quantity itself. Yes or no? At the same time, I can also say this function will lie between the negative of its mod and this. Okay? Any function will always be greater than negative of its mod and the mod of itself. Now this actually brings us to the property number one that we discussed. So this is sandwiched between or this is bounded between these two functions. Right? So integration of this from a to b will also be bounded between integration of mod of f of x and this function. Yes or no? So it is the same way as if you are trying to say that mod of this quantity. See, it's like saying if x lies between minus a to a, what does it mean? Mod x is less than a, right? Treat the same situation over here also. Treat this as your minus a, treat this as your x, treat this as your a. Getting my point. So in your mind, let me write capital A, capital X and capital A over here so that you don't get confused. So treat this as your minus a. Treat this as your a. Treat this as your x. So it is saying something like this. Capital X is lying between minus a to a. So this is as good as saying mod x is less than a. So mod of this quantity is your this expression. This will be less than a is your this expression and hence the property. Okay. And hence the property. Fine. What kind of questions can you get on this? Let's try to look into a question. Let's take this question. The minimum odd value of a, the minimum odd value of a a greater than one for which for which integral of sin x by one plus x to the power a dx from 10 to 19 is less than one by nine is equal to options are one, three, five, nine, minimum odd value. Any idea anyone? Please type done once you're done. Okay. Let's discuss this. See, sin x we know is less than x. Correct. Sin x is less than x. Okay. Sin x is also less than one. Correct. So let's take that as okay. This is also less than one. Okay. Of course, we're talking about the interval 10 to 19 sin x is always less than one. Okay. So I can say this term will always be less than this term. Correct. Yes or no? And one by one plus x to the power a as you can see your x ranges from 10 to 19. Right. So if x lies between 10 to 19, let me use square brackets and you're talking about a being an odd value odd natural number. Can I say this will be always be less than one plus 10 to the power a because the smaller this is the bigger is the value of the quantity because your denominator will be as less as possible. Right. And this will always be greater than one by one plus 19 to the power a are getting my point. So now see the chain of events. This is less than this and this is less than this. So can I say sin x by one plus x to the power a will be less than one by one plus 10 to the power a. Right. And so will be the integration. So if I integrate sin x by one plus x to the power a from 10 to 19, this should be less than integration of one by one plus x to the 10 to the power a which is actually a constant altogether. This entire thing is a constant. There is no x over it. Right. So if you evaluate it, if you evaluate it, it becomes is don't forget to write DX in your this thing. Answers script. Some teachers are very strict. They'll deduct marks straight away C and DX people forget to write. Okay. So this will be nothing but it's a constant times x. Correct. So it'll be this so far. So so good. Correct. So this can be written as 19 minus 10 by one plus 10 to the power a 10 to the power a. Hope this looks like any. Correct. Now, can I say if this integral that we seek for should be less than one by nine. Here you're saying this answer should be less than one by nine. Can I say this this term should also be less than one by nine. That means you're trying to say that nine by one plus 10 to the power a should be less than one by nine. That means one plus 10 to the power a should be greater than 81. That means 10 to the power a should be greater than 80. So which is the minimum a for which this will be greater than 80. So the minimum a cannot be one. It cannot be it can be three. So a has to be in fact to only it'll become but they're asking the odd number. So a equal to three is your answer. So option number B is correct. So see the types of questions which they have started framing off late in J. It is no longer a straightforward application of indefinite or the properties which you're learning right now. They want you to think out of the box. They want you to apply those concepts in a slightly different way. Let us move on to the last of these inequality properties. This property says that if f of x square and g of x square are integrable functions in the interval a to b then mod of the product of f of x into g of x integral will be less than equal to under root of a to b a to b g square dx. So what is trying to say is that mod of integral of the product here you are multiplying f of x into g of x and then integrating from a to b the mod of this answer will always be less than under root of the product of these two separate integrals that is integral of f square and g square both from a to b under root. So first I will prove this property. Sorry guys. I'm not going to give you any break today. We just have 15 minutes of class left and I need to be a little bit faster also with respect to completing this chapter because some few more things are left off which I plan to take next class. No problem. No problem. Good. So let us try to discuss this proof. Let's say I have capital f of x as let's take this function as f of x minus some lambda g of x where lambda is some real quantity square. Let's say lambda is a real quantity. So I just assume a function like this. So let f of x be a function which is a perfect square and it is a perfect square of what it is a perfect square of f of x minus lambda g of x where lambda is some real number. Okay. Now we know that any square of a number will always any square of a function will always be greater than equal to zero right and if a function is always positive so will be its integral. So I can say the square of this dx from a to b that will also be positive. Right. Let's open the brackets. So let's square it up and open the brackets will become f square x okay plus lambda square g square x minus two lambda f of x g of x. Okay. This would be greater than equal to zero right now let us open this integral into different integrals. So first I will take this lambda square term so lambda square since it is a constant will come out a to b g square x dx okay then minus two lambda a to b f of x g x dx then finally you have a to b f square x dx this should be greater than equal to zero okay. Now ultimately these are constants you can see that these terms will be constants only right because after evaluating the integral they would just come like a constant to you am I right. So what I'm going to do is I'm going to treat this as capital A capital B and capital C correct. So ultimately you see something like this a lambda square minus two b lambda plus c greater than equal to zero okay no doubt so far now this is like a quadratic in lambda this is like a quadratic in lambda and if a quadratic if a quadratic is always positive what does it mean what does it comment about the nature of the roots first of all a is positive here because your a is up a is integration of a positive function correct. So if this is always positive means the graph doesn't have a real root it doesn't touch the x-axis it doesn't cut the x-axis so can I say discriminant here must be less than zero right in fact less than equal to also I can consider because here I'm saying it is greater than equal to zero see why I'm saying that is because if it is always greater than equal to zero it can be hanging like this or it could be touching like this but it will never cut and go down that is the meaning of saying that your value of this is always greater than equal to zero am I making sense guys please respond yes no yes sir are you there as a representative of your yes okay so discriminant less than s means what b square in fact discriminant here will be b square means four b square minus four ac should be less than zero that means you are trying to say b square would be less than ac correct yes or no if I take the under root on both the sites can I say mod b will be less than mod ac you don't have to take a mod here because under root is always positive so mod b is under root of ac correct what is mod b then just try to recall from here b is this so mod b is like saying mod from a to b f of x gx is less than ac under root ac under root ac is it is hidden actually a is this guy c is this guy so it would be from a to b f square into a to b ac I'm writing so this is your term and hence the property is that fine again proof is not required but you should know how quadratic equation is actually helping you to come up with this property let's take a question on this prove that integration from zero to one under root of one plus x one plus x cube dx will always be less than under root of I'll just say less than equal to under root of 15 by 8 I've done once you're done this should be a straightforward because you've already seen the property it's just the application of the property nothing else and again write down the property for you mod of f of x into g of x is always less than under root of a to b f square x dx into a to b g square x dx okay so here you can treat your under root of one plus x as your f of x okay and under root of one plus x cube as your g of x okay please note you don't have to take a mod because it's going to be a positive answer because this is already a positive quantity and you're integrating from a lower value to higher value so that will be a positive answer so this mod becomes redundant so 0 to 1 you can write this if you want like this okay because both are positive quantities individually as well this will also be this will be less than 0 to 1 square of this square of this will be 1 plus x into square of this quantity which is 1 plus x cube so this will give you integration under root of integration of this is x plus x square by 2 okay from 0 to 1 into this is x plus x to the power 4 by 4 0 to 1 so 1 putting 1 will give you 1 plus half that is 3 by 2 and putting a 0 is 0 so I don't care about that value and this will give you 5 by 4 minus 0 okay so when you multiply ultimately it becomes 15 by 8 so this result can never be exceeding 15 by 8 hence proved does it make sense clear please type clr if it is clear so guys before we end today's session I have a question for all of you which anyways I will take more examples in the next class and these type of questions have been you know asked in j off late where they want you to convert one integral in terms of another okay they want to see how good you are in your substitution okay let's take a question let's take a question on this let me ask you though it is an objective question I am making it as a prove that question because I don't want to write too much right now so prove that e to the power x t e to the power minus t square dt is e to the power x square by 4 0 to x e to the power minus t square by 4 dt so they are not asking you to evaluate this integral they are just asking you to convert one integral to another integral so what they will do is they'll frame a question like this this integral is equal to which of the following integrals we don't have to find them separately you don't have to find you know the integral mentioned in the question and the integrals mentioned in the answer and see which results are same by some subtle substitution you need to convert one to the other how will you do this okay let me do this see let me start with the left hand side okay 0 to x e to the power x t into e to the power this so basically I can do this here okay I can combine the ease together now note that where I have to go I have to prove the right hand side so what I'll do is I'll take an e to the power x square by 4 forcefully out okay but for this I have to compensate with minus x square by 4 inside like this getting the point which is nothing but e to the power x square by 4 okay let me write it as 0 to x here minus sign common so it will be like x square by 4 minus x t plus t square should not write dt on top it should be dt so far so good I'm just trying to inch towards the desired expression now this is nothing but 0 to x e to the power minus x by 2 minus t the whole square as you can see this is a perfect square this is a perfect square this is x by 2 minus t the whole square okay now we are very much close but we just have to show that this this will boil down to this term e to the power x t minus t square how will I show that I'm sorry this will boil down to this term e to the power minus t square by 4 how will I show that simple let's do one substitution let's put x by 2 now see this has to be basically treated as t by 2 right so you can actually put a variable y by 2 in base of that okay so x by 2 minus t I'm putting as y by 2 such that I generate a y square by 4 term as given in my right hand side okay now minus dt is same as dy by 2 so this integral will convert to e to the power minus y square by 4 dt will be minus dy by 2 minus dy by 2 note I have put a minus dy outside what about the limits of integration when t is 0 y is going to be x correct and when t is x y is going to be what is y going to be minus x okay see here when I put t as 0 y is going to be x and when I put t as x y is going to be minus x so this happens to the limits of the integration okay now since this is an even function remember even function property and first of all let me swap the position of the upper and the lower limit I don't want a negative sign in my expression so minus x to x e to the power minus y square by 4 dy by 2 now since e function this is an even function try to recall this property minus a to a f of x dx if this function happens to be an even function it becomes 2 times 0 to a f of x dx I'm going to apply the same over so 2 times e to the power x square by 4 0 to x e to the power minus y square by 4 dy by 2 to get cancelled now there is nothing in the name that means I can put back my y as t nothing in the names I can rewrite this as 0 x e to the power minus t square by 4 dt also right there is nothing in the name okay but please don't write such terms in the exam this is just I you know just to explain your writing systems okay which is nothing but your RHS which is nothing but your RHS this is what I wanted to sorry this is what we wanted to prove here okay so I started with LHS and I ended up in RHS next class when we meet the agenda would be following we'll do more questions on integration by integral substitutions or converting one integral to another integral by typical substitutions number two I'm going to use the limit of a sum definite integral as limit of sum definite integral as limit of a sum I'm going to do vice versa also limit of a sum as definite integral this is called the reman sums reman sum okay and finally I'm going to talk about improper integrals improper integrals with special emphasis on the Wallace formula this will be the agenda for the next class I'm sure I'll be able to do this and start the new chapter with you that you wanted to start with vectors next class the exams are exams will be on so it will be again an online session okay all right guys thank you so much over and out from my side bye bye if you have any questions thank you sir thank you so much