 So we have now reached the last phase of our definite integrals chapter where we are going to evaluate definite integrals as as limit of Limit of a sum. Okay, so now we are going to evaluate definite integral definite integrals as Limit of a sum. Okay, we also call it Evaluating definite integrals by first principles Why it is so thick? by by first principles Okay, and People who would have done a piece they would know that this is actually called the Riemann sum Okay, so those who have written a piece Param is there. No, Param is not yet joined in anybody who has written a piece AP calculus whether AB or BC, this is what they call as the Riemann sum Okay, so this is an art of evaluating a definite integral of a function from A to B By breaking it into Small sub regions Okay, so let's say you are you are trying to evaluate the area under this function This function is defined from A to B. So let's say for the purpose of discussion I assume a continuous function and We are trying to find out the area from A to B Okay So this region from A to B is further divided into sub regions Okay, now there are several ways to divide the sum region. You may divide it in terms of rectangles Okay Now the rectangle also can be made in two ways. You can have a rectangle whose edge On the left side touches the triangle Sorry touches the curve for example something like this Okay, okay, so let's say you are making rectangles like this. Who's you can see left hand The left hand of the rectangle is always touching the curve You could also make a rectangle in such a way that the right hand touches the curve Okay, so these are called the left hand Riemann sum right hand Riemann sum But you don't have to know these names because unfortunately or fortunately CVSC or JEE is not so much interested in what is left Riemann sum and right Riemann sum In you can say in international books, you will also find reference to trapezoid rules to find it So instead of making rectangles, they will make trapeziums. That would be a more approximate way to get the answer Okay But anyways when we are taking the limit of this as these size of these rectangles become very small All of them will give you the same result Okay, so these rules make more sense Or the difference between taking a left hand Riemann sum right hand Riemann sum a trapezoid rule is more evident when you have Taken a very thick size or you can say you're trying to approximate the integral by making few rectangles Okay, but right now for our Evaluation since we are aware of the concept of limits will be using it in addition to the Riemann sum To find out this integral. So what do we do as I have already discussed we start breaking We start breaking this entire region into sub regions This sub region will have a thickness of H Okay, remember this H is a Quantity which will keep very very small. It is attending to zero quantity. Okay, if you keep it bigger your area would be Further away from the actual value of the integrals So if you want your answer to be closer and closer to the actual integral your H That is the thickness of these rectangles should be tending to zero and how many sub regions we make out of it We make n rectangles over here and This n would be a very very large quantity. You can say tending to infinity Okay, so this is your first rectangle. This is your second rectangle and so on till your last one, which is your nth rectangle Okay So in order to evaluate this integral, we will literally take the area of each of these sub regions and add them So what do you think is the area of the first sub region? Now remember this height is f a Okay, and thickness is H. So the first sub region area is H into f a so this is the area of the first sub region Okay, what are the area of the second sub region you will say similarly it will be H Now this point this point that you see here that will be a plus H Okay, this point here That will be a plus 2h if I continue the same trend can I say B will actually be B we actually be a plus NH. Now, this is actually a very important Finding for us. We'll use it Little later on. Yeah, let us continue. What is the area of the second sub region? It will be H into f of a plus H Okay, third will be H into f of a plus 2h If I go to the last sub region that is your nth region your area would become Your area will become this Okay Are you finding any problem disturbances in the net connection? I mean, is my voice audible to all of you? So it's breaking. It's breaking not Okay, so this will go to the nth sub region now In order to find the area, we will use the fact that We have to take the limiting case of this as H tends to zero and n tends to infinity But normally what do we do is we drop this condition and tending to infinity But instead use the fact that NH Now come back to here NH from here will be what NH from here will be B minus a So we use the fact that NH will be equal to B minus a Now this is very surprising because n is tending to infinity and H is tending to zero And we are saying we're trying to say that a quantity tending to zero into infinity is a finite value That is quite possible because zero into infinity or tending to zero into infinity is actually considered to be an indeterminate quantity and An indeterminate quantity can take a finite value Okay, there's no issue with that Okay, so please make a note of this So basically you're taking a limiting case You are taking a limiting case of the left-hand remansem to get to this answer now Many people ask me sir. Can I take a plus H a plus 2h and go to a plus NH? Yes, no issues with that both of them will give you the same result both of them will give you the same result Don't worry about that Okay, so I'll just summarize it limit H tending to zero and H is equal to B minus a H times F a F a plus H F a plus 2h All the way till you reach F a plus and Minus one is that fine So before we start solving questions, there are certain things that we need to remember from our You know previous chapters So as you can see two things are involved here. One is the concept of limit Other is the concept of series So you should be good with your limits basics of limits and it should be good with your Series that is the special series sign series cost series even arithmetic geometric progression all those things Okay, so I'll quickly write down those formulas for you so that when we are trying to solve the problem You don't feel the need to go back to those formulae. So some important formulae for this concept the first formulae is your summation of Natural numbers from one to n. This is something which you already know You have been using it extensively Summation of square of natural numbers from one to n Okay, summation of cube of natural numbers from one to n Okay Next is your GP GP also many people they keep forgetting the formula, okay sign series Sign series that is sign alpha Sign alpha plus beta Sign alpha plus 2 beta. I'm sure You know this but somehow since you're not using it very frequently you may have forgotten it So I'll just quickly write down the formula. This formula is sign and beta by 2 by sign beta by 2 Into sign of arithmetic mean of the first angle and the last angle so alpha and alpha plus n minus 1 beta by 2 That'll actually give you alpha plus n minus 1 beta by 2. Okay, this is your sign series Then is the cosine series a very similar looking series is cosine series You don't have to remember too much if you already know your sign series well. Oh Sorry, and again once again, I want to remind you if you have not taken your Monthly test yesterday, please do so today Yeah, in cosine series only change happens is the sign here becomes a cosine So it is cosine n minus 1 beta by 2 2 is only below n minus 1 beta. Okay, it is not below alpha. Is that fine? Let us try to apply these formulae. In fact, we'll also revisit our limits also I'm sure you all know it. So one of the most frequently used ones is this guy H tending to 0 e to the power h minus 1 by h that is equal to 1 and of course, you know your Trignometric standard trigonometric limits Okay, now many a times we may have to use Different concepts to solve the limits. It may not be very straightforward. We'll take few questions just to understand how it works Let us begin with the first question So I can use Oh, yeah, sure. Have you started differential equations in school? So today after this session will take up ODE So can you move a little to the right right? Yeah, sure Thank you. Okay No one wants me to go to the previous slide Yeah, this is the most important formula. Please make a note of this so we are using the limiting case of left remand some To find out the integral of a function from a to b These concepts would be used in both directions Not only we need will it'll help us to evaluate an integral as limit of a sum But also limit of a sum as an integral. That is something that Jay asked very commonly This is something which is more of a subject matter of school exams and I'm not sure whether CBC has dropped it as CBC drop this concept. Yes. Yeah. Oh my goodness Okay, anyways, so we'll not spend too much time on this because This is just for your knowledge that we're doing it. So let us say we want to evaluate We want to evaluate Evaluate integral of integral of a function from one to two x squared ex as limit of a sum as Limit of a sum Since this is the first question, I'll help you out with this. So we'll be a little bit faster also So here our f of x is x squared. Of course a is one b is two and NH here will be b minus a which you all know is going to be two minus one, which is one Okay, so see this is how we deal with this. So we take this as a Limiting case of H tending to zero NH equal to one. Okay now H F a F a means one square Isn't it? If you see the formula formula had a F a Then it has F a plus H F a plus H means one plus H the whole square Then one plus two H the whole square and so on till one plus n minus one H the whole square Okay, so limit H tending to zero now this term will be Opening up as one one plus two H plus H square Then you'll have one plus four H plus four H square and So on till it goes to One plus twice of n minus one H plus n minus one square at square Okay, if you see you have actually written You have actually written one n number of times So write an N for that then you have Then you have two H. Let's say I take two H common from all these terms So if you take two H common, you'll have one two all the way till n minus one correct me if I'm wrong Okay, and we'll take H square common from all these terms which I'm showing with a white tick So you will have an X square and you'll have one square two square till n minus one square. Okay Now Summation from one to n is known to us So we can also find out summation from one to n minus one. So it will just be n minus one Into n by two and there's a two H over here. So into two H. Similarly, this will be n minus one n Two n minus one by six And there's an H square over here. So we'll write it's good. Okay Now something very interesting is going to happen over here. All of you please Pay attention This H. I'm going to introduce inside This H. I'm going to introduce inside Okay, let us simplify Whatever extent we can In this expression, you would realize that There will be as many H's as there are factors of N Interestingly, yes. So as you can see, NH is getting formed over here Okay, here there are two factors of N. So there's two H. There's H square over here. So what I'll do is I'll split this H square By giving one H to this and one H to this Something like this Okay, and it'll always happen you will have as many H's as there will be factors in N So you have to split that H equally among those factors Why are we doing that so that we can make use of this NH equal to one Okay, here also you'll see One two three factors of N are there and there's H cube. So split the H equally among these Okay Now wherever you have a NH Put a one there and where you have a single H put a zero over there So this will become one. This will become one minus zero into one This will become one minus zero one two into one minus zero by six Okay, so if you evaluate this you get one plus one plus one by three which is equal to seven by so we all know integral of x square from One to two is going to be Seven by three. So this is how you got the answer but by using your first principles Now many books will teach this or many Lectures will talk about this in the beginning of the course, but normally I talk about this in the end of the course because I want to talk more about limit of a sum as a definite integral not the other way around So these concepts are primarily important for your basic understanding I think even in school they have not entertained this so we'll take just few more questions not more Any questions here is the approach clear how to work around with this? Yes Sir, would there be some case where the sum is easier than the integral? I'm sorry? Sir, are there any cases where the limiting sum is easier than the definite integral? No, no Definite integral is If you are using your methods of definite integral that will always be convenient than doing it by first principles. It's like How easy is it to find the derivative of a function by first principles? And how easy is it to find the derivative of function by methods of differentiation? So first principle is always tedious Okay. Yes. Yeah So let's take this one evaluate integral from a to b e to the power x So I'll take one of exponential one of a Tecnometric and I'll take one of irrational function Done very good. Let's discuss it so Limit h tending to zero You'll keep n h as b minus a And it will be h f of a f of a will be e to the power a When is the nps in the other having a semester exams? The same So we'll reduce the duration of the class during those periods We'll not have four hours. Well, you can have two and a half hours or something Will that be fine or is it are you happy with four hours? What is fine four hours four hours Rahul, what about you? I mean it's not bad, but yeah, I mean Probably an hour less Okay, we'll we'll keep it an hour less three hours. Okay Okay, so I'm sorry. Yeah, so basically you can see here that you're dealing with a geometric progression who's um First term is e to the power a and common ratio is e to the power h So it'll be one minus e to the power n h By the way, you can perceive it as if I'm doing this Okay by One minus e to the power h Okay Now all of you, please pay attention to the fact that when h tends to zero. This is your standard Exponential limit. Okay, and this will become a one So we'll have e to the power one into e to the power a One minus e to the power n h e to the power n h is actually e to the power b minus a So when you open the brackets, you get e to the power a Minus e to the power b as your answer Correct There will be a minus one here. So there'll be this yes There's a minus one because it is one minus e to the power h So it'll be e to the power b minus e to the power a so this is what will be your answer. Okay Try this out integration of Evaluate integration of sine x from A to B As a limit of a sum So today we are going to start with differential equations and Differential equations will take three classes Then one class for area under curves And one class for two classes for probability And around three classes for vectors so around Two months of classes are left So we left a few chapters and leavens also right like triangles and a bit of That we will revise when we are doing the crash course Okay, sir triangle is the property of triangle is the only thing right There's also something in coordinate geometry, right? No coordinate geometry. I think we did a lot of things pair of straight lines. We did right Yes, sir. We did all that right. Okay. Yeah okay, so Again here your function is f of x is your sine x Okay, n h is b minus a So your answer would be limit h tending to zero n h is b minus a sine a sine a plus h all the way till sine a plus n minus one h Okay So this will give you Now here your alpha is a and beta is h So this will give you sine n h by two by sine h by two Into sine of arithmetic mean of the first angle and the last angle which is a plus n minus one h by two Okay now There's an h over here, which I have forgotten sorry about that Now if you just provide this with a two and this with a two This would become a one because it's the standard trigonometric limit This would become a one Okay, rest of the places you will conveniently find an n-h combination So your answer will be two sine n h by Two which is b minus a by two. This will be sine A plus this is n h minus h n h minus h means b minus a minus zero by two Okay So this is a two sine b minus a by two into sine a plus b by two Okay, this is clearly the formula that we know for cos What cos? a minus cos b Okay, which we very well know that this will give you cos a minus cos b because The integral integral of sine x is minus cos So from a to b it will give you minus cos b minus of minus cos minus of minus cos a So that is cos a minus cos b. Is that fine? Now these are easier ones, but there are few where you will get stuck For example, let's take this question evaluate integral of One by root x from a to b By limit of a sub by using Limit of a sub How do you do this? Okay, so you would realize that when you're writing it I'll also help you so that we don't waste too much time doing this So when you write down the expression you have h times let me write Limit h tending to zero and n h is equal to b minus a h times one by root a One by root a plus h One by root a plus two h and so on till you reach one by A root a plus n minus one h. So how do you sum this guy? Okay So we can use the binomial approximation Binomial approximation, okay So in order to sum this I will use Sandwich theorem Okay, so I'll try to sandwich this function between two such functions whose sum can be easily be found out Okay, so what I'm going to do is all of you please follow here Can I say can I say the expression of Of this kind is of the nature one by root r. Okay, and r is going from a to a plus n minus one h Okay, so let me do one thing Can I say this expression This expression will always be greater than Always be greater than two by Under root of r plus h plus root r Correct Yes or no And can I say this will always be less than two by Under root of r minus h plus root r Can I say so? Can I say so Do all of you agree with this or not? Now see you can treat it as if you have divided by two throughout. Okay, let me do this. Let me divide by two throughout Now does it make sense? Because the denominator here will be more than slightly more than this So this expression will be lesser than this expression Similarly here denominator is slightly lesser than Two root r. So this term will be more in value than this term Okay Now why have I chosen these two terms is because I'll show you something very interesting If you rationalize this term both on the left and the right you can say This term What happened to my r? Yeah, this term Will be under root of r plus h minus root r By h And this is under root of r minus h minus root r by By or you can just flip the positions here This will become a minus h. Am I right? Yes, minus h. We'll do one thing. We'll just flip the positions So we'll do a root r minus A root r minus h by h Okay Now let us start putting the values of r as a So let's put r as a a plus h etc. So when you put a you get a plus h minus a by h This will be one by two root a This will be root a minus root a minus h by h When you put a plus h You will get under root of a plus two h minus under root of a plus h by h Two times a plus h a Plus h minus root a by h and so on so dot dot dot. I'll just go to the last term The last term was a plus n minus one h Okay, so what will happen to this term? This will become under root of a plus n h Minus under root of a plus n minus one h by h This will be dot dot dot One by two under root of a plus n minus one h And this will be under root of a plus n minus one h minus under root of a plus n minus two h by h, okay Just add them all Just add them all Okay, now here you realize that when you're adding a lot of cancellations will happen And you will be ultimately left with this guy and root a over here So this will give you under root of a plus n h Minus under root of a whole by h And this is one half of the required sum whatever we needed Okay Similarly here, what will happen all these terms will start getting cancelled and you'll left with this guy and this guy Correct. So that will leave you with under root of a plus n minus one h Minus under root a minus h by h Correct me if I'm wrong Okay Now let us take the limit of The end points or the sandwiching functions. Let's see where do they tend to so I'll just repeat this Whole thing again in the next page Let me just take a copy Let me take a copy of this right now here your s s will be twice of under root of a plus n h minus root a Okay twice of Twice of under root of a plus n minus one h Minus under root of a minus one h, okay now multiply h Multiply h and take the limit of it as h tends to zero And n h is b minus a so limit H tends to zero n h is b minus a Limit h tends to zero n h is b minus a okay Now you will see that The moment you put n h as b minus a this will become under root of a plus b minus a minus root a Which is actually indicating two under root b minus root under root a okay Now this is something which we don't know so let me you know copy it as is it and this will also become two times Now this will be under root of a plus n h minus h n h is again b minus a and h is again a zero Okay, and this is again h is zero. So ultimately it will also give you under root of two times under root of b minus under root of a What does it mean that if the limit of these two terms Tend to the same value by sandwich theorem. I can say that By sandwich theorem, I can say that The limit of the middle function, which is the sandwiched function That will also be That will also be Two under root b minus under root Okay, so this becomes your answer and this is what we were looking for right? This is what we were looking for Okay So many a times you need need to apply sandwich theorem also to get your job done Okay, anyways, we'll not waste too much time. We'll directly jump to the the part which is more important from j point of view Can I go to the next slide or you want to copy something here Mehul, should I go to the next slide? Yes, sir Okay Now limit of a sum as definite integer limit of a sum as Definite integer in this concept. We are mostly going to talk about Some unfinished agenda of limits Now, let me tell you with this concept your limit concept is completed We we had not completed the concept of limit if you remember we started it in the month of march But even that period we could not complete it So with this session your concept of limit will be completed where we'll be mostly focusing on Tending to zero into infinity kind of Forms Okay, how do I deal with this? When it comes to you know finding this kind of a limit by use of definite integral By use of definite integral. So just now we learned that we could write limit of a function from a to b as Sorry integral of a function from a to b as limit h tending to zero summation of f of a plus r minus 1 h r going from Zero to sorry r going from 1 to n. Isn't it? This is what we had learned a little while ago. Yes, I know Now the same thing I will use in a reverse direction To convert a limit problem into a definite integral We will convert it to a definite integral. So I'll tell you the working rule for it So let's say You have been given a kind of a limit. I'll take a very simple example Let's say we have this question evaluate Try to understand through this example. How do we solve such kind of questions? Let's say we have evaluate limit limit end tending to infinity end tending to infinity A question like this Now, how do I evaluate this limit? How do I evaluate this limit now all of you please pay attention? I'm giving you the working rule over here Okay, I'm giving you the working rule So what do we do first? We write down the r-th term of this particular series. So what do you think is the r-th term of this series? r-th term of this series will be 1 by under root of 4 n square minus r square Correct Okay, so first term is find the r-th term of the series Okay second step is from the r-th term From the r-th term Take out 1 by n As a factor. So if I take 1 by n as a factor from it, you obviously will know that you have to take n out from the denominator So when you take n out from the denominator It will leave you with This expression Yes or no Okay, okay Is this step clear first r-th term you have written from the r-th term you took 1 by n factor out So 1 by n factor has come out And because 1 by n factor has come out from this you'll be left with this Okay now Wherever you have an r by n term Write an x for it Wherever you have 1 by n term Write a dx for it Okay, so what has happened? Your r-th term has actually got converted to I like it as a fourth term It has got converted to 1 by under root 4 minus x square And this one man will become a dx Okay Now summation means integration And you're summing from r equal to 1 Till let's say r equal to n That means if you put r as 1 over here x will be a 0 And if you put n over here x will be n by n which is 1 In other words This integral can this limit can be evaluated By evaluating this integral Are you getting my point? And I'm sure you can evaluate this it is 1 by under root a square minus x square form That is sin inverse x by Okay, that is something that you know you can always do on your own So is the process clear my my purpose is to explain the process to you Result you'll definitely be able to find out Okay Are you getting my point? So the agenda is here in short that you have to make your r-th term as 1 by n into some function of r by n This is what you need to do Okay, and in this you have to replace your 1 by n with a dx And r by n with a x right And one more thing here Your limit of integration will be decided by from where to where you are summing up So let's say if you're summing up from r equal to 1 to let's say 2n Then your limit of integration will become 0 to 2 Are you getting my point here? Okay, so if you're if you're let's say summing up From r equal to 1 to let's say, you know, um Pn Okay, then your limit of integration will become 0 to p of f of x dx Are you getting my point here? We'll take more questions somewhere the entire process will be much more clear to you Okay, let's take few questions to understand this Can I go to the next page? note that Some of these integrals which are of the type tending to zero into infinity cannot be evaluated by your You know the initial Processes that you were aware of Okay, you have to use definite integrals for it. Oh, sorry. This is picked up from your Arihant book Amitagrawal Arihant book Let us try to target each of these questions so that we are Convenient with this process. Let's say you want to do this now You can evaluate this by your summation process also no doubt Okay, but for a change, let us do it by definite integrals concept, right? So it says that I can take n square, you know lcm and I can evaluate 1 plus 2 plus 3 till n minus 1 And then I'll use my summation and you get my answer. Yes, of course you can do it by that way Okay, but let us for a change do it in a different way So for the first question, what is your tr? Let's go step by step. What's your tr? If you look at it, what's your tr? R by n square, okay And thankfully this is your step number one. Thankfully if you take your 1 by n Separately you'll get an r by n Okay Third step, what did I say r by n is x 1 by n is dx Okay, so this is your tr And if you're summing it up from r equal to 1 see r is 1 over here And you're summing it to n minus 1 So if you're summing it from r equal to 1 till n your limit will become see when you're putting When you're putting r by n as x You only tell me that when you put r as 1 what happens to x value Won't it become 0 because n is a very large number? Okay, so lower limit will be 0 And when you put r as n minus 1 you will say the limiting case of this is 1 actually so your upper limit will become a 1 Correct, so evaluating this is as good as getting solving this integral Which is going to become x square by 2 from 0 to 1 which is half Okay, so the answer to the first question will be half And you'll check that even if you use your series Your sum of natural numbers you'll get the same answer No difference in the answer Is it clear, but when it comes to this type of question Right, you would realize the need for a definite integral because your series and all will take a backseat So I would like you to attempt the second one quickly Please give me the answer for the second one Put it on the chat box Very good Mahit And for god's sake don't write these answers as 0 many people say sir each of them each of them are tending to 0 right So 0 plus 0 plus 0 plus 0 only no it is actually 0 written infinitely many times In fact these quantities are very small quantities, but written infinitely many times And hence they are indeterminate. They are not 0 Had there been a finite number of terms, let's say you're summing up till any end tending to 1 lakh Then I would write 0 But it has not been summed to finite terms it is summed to infinite terms Very good. I'm I'm getting answers from everybody So your nth term would be sorry rth term would be 1 plus n 1 by n plus r, right Next step, what did I tell you take a 1 by n out if you take a 1 by n out it becomes 1 plus r by n Now this is what I am calling as a function of r by n This is actually should be written as a function of r by n because we are going to substitute r by n as an x okay, so Your this term will be 1 by 1 plus x dx. Okay now When you are summing things means you are integrating now from where to where you are summing you are summing up from 1 to n check here this term is actually n plus n so your r is going all the way till n isn't it So if your r is going all the way till n you're summing from 1 to n That means if your x is r by n when you put the lower limit you'll get a 0 when you put the upper limit You'll get a 1 so it is integration of this from 0 to 1 so it is ln mod 1 plus x 0 to 1 when you put a 1 you get ln 2 when you put a 0 you get ln 1 which is anyways ln 2 only Which you're saying how ln 3 and all yeah Okay Try the third one If you want I can copy the question once again in the next page so that you have space to work on in fact I will have space to work on Not you Where is where is definitely Very good Chaitanya he are Richard Pollack Good good now all of you are giving me Correct answers very good So in the third expression your series goes like n by n square plus r square As I already told you you have to pull out a 1 by n from the given expression because that will play the role of a dx So that leaves you with n square by n square plus r square But we need to convert it as a function of r by n So these are the some things that we need to uh look out for So it'll be r by n the whole square So now your Areate term will be 1 plus 1 by x square dx Okay summation means integration And you're summing from r equal to 1 till n if i'm not wrong. Yeah Summing to my equal to 1 so this is 0 to 1 So this is tan inverse of x formula from 0 to 1 which is pi by 4 minus 0 which is pi by Absolutely absolutely So now let us work on the fourth one Very good. Hiya Chetanya Mahit Very good So this is also very easy. Uh your rth term here would be r to the power p by n to the power p plus 1 Okay We are supposed to take a 1 by an out if I do that I'll get r to the power r by n to the power of p So your rth term is like x to the power p dx Okay Summing it up from r equal to 1 till n is like integrating it from 0 to 1 So that is x to the power p plus 1 by p plus 1 0 to 1 your answer will be 1 by p plus 1 Excellent excellent So we'll now take up some complicated questions on this Can I move on to the next page? Let's take this one Integrate Oh, sorry not integrate evaluate limit n tending to infinity n times 1 by n plus 1 n plus 2 1 by n plus 2 n plus 4 1 by n plus 3 n plus 6 Da da da da to 1 by 6 n square 1 by 6 n square Thank you kritika for sharing the schedule with me So when is your Gannett exam Gannett is on tuesday Tuesday the 13th So they have completed the syllabus. It means no what is left. I can't see any chapter left. What are they doing now? Right now the 3d Watch our three oh vectors only is there 3d is left. Yeah, they'll just Brush through like anything Okay, so I've got responses from four of you. Let me see Oh, no response actually. Yeah, yeah, ruchir carry on carry on. You're on the right track down Very good, ruchir Very good, padit ruchir sing. Yeah Okay, guys, uh enough time given. Let's Write down the arid term arid term if you see very closely. It is actually 1 plus n by n plus r n plus 2 r Okay, this actually indicates 1 by n plus n n plus 2 n That's why it becomes 6 n square. Okay, so they they might not they might not write it in a very Obvious way so that you know, you have some problem in getting the series now Take a 1 by n out the customary move that we do always Okay divided by n square both in the numerator and denominator Giving one end to each of the factors will be r by n And 2 r by n. Okay Now the moment we have got this the process becomes Very predictable from here. That means your given limit is going to be 1 by 1 plus x 1 plus 2x dx Okay summation of this summation of this Is integration and you're integrating from r equal to 1 to n. So this is 0 to 1 Okay, now, this is something which you can easily evaluate by use of partial fractions. So I'll not Waste time in that I'll just directly write down the result. So when you break it up as partial fractions This is what you will see so minus ln mod 1 plus x and ln mod 1 plus 2x remember 2 and 2 will get cancelled off. So you don't have to write it So this whole thing limit is from 0 to 1. So it's minus ln 2 Plus ln 3 Another would be 0. So that is ln 3 by 2 good enough Any question any concern here? He island ln uh ln 3 minus ln 2 you'll get. Yeah Okay, next question This question you may have seen several times in past limit n tending to infinity n factorial by n to the power n Hold to the power of one man. I'm sure many of you would have seen this in your limits chapter also Okay, now the time has come that we solve this question You have to take a log here. Please remember without log It will not come in the form that you would like it to be in So don't shy away from taking log in a limit Correct, which is park Yes, yes, uh, which is saying that's correct. You're on the right track Then you have to take n t log of that So let's solve this in the interest of time. First of all n factorial Uh, no kirtan Okay, and this n to the power n also you can distribute it like this correct, which is Okay, so let's say I call this limit as l Then take log to the base e on both the sides when you do that When you do that you'll get one by n log of all these quantities You can write it like this It's very obvious now that your rth term here in this particular series will become one by n log r by n So it is very much in the form that you would like it to be in so it is log of x in fact ln of x dx Okay, summation means integral And you're summing it from r equal to one till n that means your integral is from zero to one Okay, but remember when you evaluate it you get ln l You get ln l Okay, what is log x integration? x ln x minus x x log x minus x x ln x minus x absolutely Zero to one now when you put a one Yeah, when you put a one Log will become a zero and when you put a zero x will become a zero so in either case there will be a zero over here Okay, so when you put a one there's a zero minus one and when you put a zero any way to be zero minus zero So the answer is minus one So your ln l ln l is equal to minus one so l is e to the power or you can say one by e This is your okay very famous question keeps coming in Many limit exercises See it's subjective to you. You can try that out also evaluate Limit n tending to infinity one by n square secant square one by n square two by n square Secant square four by n square till one by n Secant square one Oh, once again, Advik wants me to go to the previous slide Just a second Advik Yeah, do you want me to drag it somewhere to the top? Okay Okay, thank you. Okay, Richard R.P. Richard Parik Advik asked check you're working Okay Two different types of answers I'm getting Okay, so let's look into this So tr here is if I'm not mistaken tr is one is r by r by n square secant square r square by n square correct If you pull out a one by n factor it becomes r by n So tr is x secant square x square Okay integration summation means integration and you are integrating it from Zero to one because you're summing it from one to n Now first of all here First of all here, I will take my x square to be t So x dx will be half of dt So it's because basically this is zero to one again Secant square t dt So that is nothing but tan of t from zero to one Now tan one doesn't mean pi by four some of you have Misinterpreted it. Okay. So it's tan one minus tan zero which is actually half tan one Okay, it's not pi by four. It is one tan one radian tan one radian is tan one only you can't write one for this I know I know it happens it happens Okay, try this one out. Can I go to the next page? Yes, sure correct Richard Parik Is that difficult? I think only one one of you have got the answer so far Okay, let's discuss this So what I'm going to do here is I'm going to do some minor manipulations Okay, this guy in the denominator I let me let me write down the denominator first This guy in the denominator needs a one by n right, so I'll provide it with one by n Okay Okay, I'll also provide it with r by n So if you see closely if you see closely I have provided it with one by n square Okay So I'll do the same thing over here. I'll provide this with one by n Okay, and I'll provide this also with one by n Now the problem is I don't have a function of r by n over here. What to do? Okay, so for that there is a simple trick we can do we can do r by root n over here and root n by r over here So all the n's are getting compensated now and not only that you also get a function of r by n everywhere. Okay, so The algebra of limit allows me to evaluate these limits separately Okay, so I can evaluate this limit separately root of r by n This limit separately on a root of n by r Okay, the denominator which you think that's correct. Your answer is correct Now directly, let me write down the function without much issues. So this is 0 to 1 root x Into 0 to 1 1 by root x here we have 0 to 1 x Is that fine? So this guy will give you x to the power 3 by 2 divided by so it'll be 2 by 3 Okay, this is x to the power minus half plus 1 which is 2 root x Okay, and this term will give you Again 1 by 2 So this will give you 8 by 3 as the answer So only two of you could get it Both the ruchits Let's take Let me put the poll on Okay, only one person has answered so far Last 30 seconds Okay, five four three two One, please vote please vote Okay, just 12 of you have voted so far Mostly you say option a Option a See Again, you take a log of this limit l. Let's say this is l Correct So when you do that you end up getting You end up getting limit Entending to infinity 1 by n Okay, and log of these quantities. So you'll have 1 by n a log of sine Pi by 2n plus log of sine 2 pi by 2n And so on And this will go all the way till all the way till log sine n minus 1 pi by Pi by 2n. So yeah, so basically it is going till This right Okay, please make a note that there was an n minus 1 pi by n over it Correct. So if you write this nth term over here Your nth term would be 1 by n log log r pi by 2n, sorry sine r pi by 2n my bad sine r pi by 2n Okay, and when you're summing this up Means you're integrating this now summation is happening from r equal to 1 Till 2n minus 1 am I right? If r is becoming 2n minus 1 Correct. So your limit of integration will become from 0 to 0 to 2. So what I'll do first I'll Write this as dx and I will write this as x 0 to 2 Got the one. It's a y is it till 2 into n minus 1 So it is going till n minus 1 pi by n, you know, so if I put a 2 here, yes Oh, yeah, I didn't Yeah, there that's where your limit of integration will become wrong Okay, that's a very important part now We can take here pi x by 2 as a t. Let's say Pi x by 2 is a t. Let's say So dx is 2 by pi dt Right, so it'll become log of sine t 2 by pi dt Remember on the left side you have ln of l Now what about the limit of integration when x is 0 t is also 0 when x is 2 t is pi Okay, here we can use half the limit property. So it'll become 4 by pi 0 to pi by 2 ln of sine t ln of sine t from 0 to pi by 2 I told you to remember this result. It'll really save a lot of time for you What was the result? The result was pi by 2 ln half Isn't it? So ln of l is going to be this so pi pi gone So ln l becomes 2 ln half 2 ln half is ln 1 fourth So direct comparison will tell you l is 1 fourth Option c is correct. Janta was wrong Okay, so please note remembering these results really saves a lot of time for you Else you'll have to sit and do all those Halfing the limit and you know using king's property and all this stuff Is that fine any questions any concerns? Now remember summation and integration They are interchangeable Sometimes question on this concept also comes in J exams Okay, please understand summation and integration are both the same things So their positions are sometimes interchangeable Let me take a question on the same Uh, if you permit can I go to the next slide? Yes, sir Thank you Find or evaluate Limit n tending to infinity summation From k equal to 0 till n n c k By n to the power k k plus 3 How will you do this question any idea? Okay Let us discuss this Okay here Let's discuss this So this term that you have you can write it as limit and tending to infinity 1 by k plus 3 Times n c k Into 1 by n to the power k Now what I'm going to do is this guy over here. I'm going to replace it with integration of x to the power k plus 2 From 0 to 1 is that fine? No problem so far Just have a good look at it and tell me if you have any problem in writing this step So now what I'll do is Oh, where was the summation? Yeah summation symbol I forgot Yeah So now what I'm going to do is I'm going to write this like this Integration 0 to 1 Okay summation will come inside dx Okay, this concept works because summation And integration both basically both positions can be interchanged Okay, now if you look at this it basically says x square you can take outside because x square is not a part of the summation so x to the power k By n to the power k n c k Now I'm sure this must ring a bell in your mind. This is nothing but summation n c k x by n to the power k Recall your binomial days Recall your binomial days What is this basically? This is a binomial term correct So it will be x square 1 plus x by n to the power of n Okay, now limit also can be introduced inside here Limit and tending to infinity Limit and tending to infinity Now this term If you evaluate the limit you all know that it is going to give you e to the power x. It's a standard You know 1 to the power infinity form that you have also seen in the past. So this guy is e to the power x Okay, 1 to the power infinity form So ultimately your problem boils down to evaluating integral of x square e to the power x from 0 to 1 okay Now I can use my di method for here because I have to do integration by parts more than once. So I prefer using di method So 2x e to the power x 2 e to the power x 0 e to the power x Plus minus plus you don't have to put a sign here. So just multiply these Tic-tac-toe method. I hope you all remember tic-tac-toe method So your answer is going to be x square e to the power x Minus 2x e to the power x plus 2 e to the power x from 0 to 1 When you put a 1 you get e minus 2e plus 2e And when you put a 0 you end up getting 0 0 here and you get a 2 here. Have I missed out anything? Do let me know Correct anything I missed out I hope not So this gives you answer as E minus 2 that is your answer for this question. Is that fine? So go to the left of it? Yeah, yeah sure So wherever summation is there you can put the limit next to it which I had forgotten Limit and tending to infinity k equal to 1 till n See I'll explain once again So What I did was I introduce a integration over here, which is this term Okay Then what I did was 0 to 1 dx integration I brought it outside and the summation process I introduce inside Okay like this Once I did that remember x to the power 2 can be pulled out From the summation process because summation is applied to k So only k related terms I have kept inside Okay, but how can you just switch the integration and the both are both are summation process And both are integration Both are summation process so both are both mathematically mean the same thing Okay, so I've forgotten limit and tending to infinity and I hope you can figure it out Now once you get this term This term is a familiar term to us It comes from 1 plus x by n to the power n expansion binomial expansion, isn't it k should be starting from 0 in this case if I'm not yeah, k should be starting from 0 in this case Yeah, zero zero so I was writing one by mistake So this is a binomial expansion for this So I replaced it back over here But at the same time limit of this as n tends to infinity will become e to the power x From your standard 1 to the power infinity form So ultimately the whole question boils down to integrating x square e to the power x dx 0 to 1 A similar question came in nicer exam nicer. I said, you know for Getting into pure sciences. This question was there actually clear, which you're saying Yes Okay So with this we close this topic