 All right, so I'm going to start with definite integrals So last time when we had discussed definite integrals, we mostly talked about Leveny's rule and Inequality properties. Okay. However, that is not the end of the properties. Unfortunately, there are a few more to be done So I'm slightly doubtful whether we'll be able to start a vector today. Okay, so let's see I'll try my best to start So meanwhile, let me start with this property. Okay, I don't remember the property number actually, so Let's look into this property. This property says that if f of t is an odd function if f of t is an odd function, okay, then an integral function like phi x which is defined as Integral from a to x f of t dt Okay, here is some constant Okay, here is some constant a is a constant here then this function is An even function. So this integral function that you see here this integral function. I am just putting it under a box This integral function will be a even function. That is what this property says Okay Can we quickly prove this the proof for this is very easy? Let's try to prove this Now if you want to show that any function is even function What do you actually do? What do you actually do? You actually replace your x with a minus x correct and you try to see whether The function gives itself back or not. That means is We are trying to inquire whether is phi of minus x is equal to phi of x or not because if it is then only we can prove It's an even function Correct. So if you have to change your x with a minus x in this You are replacing the sign of x with a minus x isn't it? Okay now let us use some properties to Figure out whether this is giving me the same phi of x function back or not So what I'm going to do is I'm going to Take this as a to minus a f of t and Minus a to minus x f of t. Okay. So what I did I just broke the limits of integration Which was from a to minus x at minus a so the first integral is a to minus a f of t dt And the second integral is minus a to minus x f of t dt Okay now since Since f of t is an odd function since f of t is an odd function Can I say? A to minus a or minus a to a f of t will ultimately become a zero Remember the even odd property. So please recall the Even odd property which we had done This property said that if you are integrating any odd function from minus a to a or a to minus a doesn't matter The result would be zero Correct. So this term over here goes to zero right So what are we left with ultimately we are left with? minus a to minus x f of t dt Now what I'm going to do here is I'm going to just make a interim change of the variable m So let me put t as minus y Okay, so this implies dt is negative dy Dt is equal to negative dy So if you make this change over here, it becomes f of minus y and dt is negative dy What about the limits of integration? Can I say limit of integration will now become from a to x because the moment you put t as minus a y will be a And the moment you put t as minus x y will be x Now given that it's an odd function Can I say f of minus y is minus f of y? Because f of x is an odd function correct, so can I write this as negative a to x negative f of y dy Negative negative will go off and it becomes a to x f of y dy And we know there's nothing in the name. I can put it back to t. It doesn't matter And you can see that you are back to the same function phi of x Okay, so we started with phi of minus x as you can see here and We ended the proof at phi of x right which clearly implies That phi of x is an even function Please note that here this a could be zero also. It doesn't matter whether we have a as zero or a non-zero quantity This relation will always be true So if you are integrating any odd function from a to x where a can be any quantity zero non-zero doesn't matter It will always result into an even function It will always result into a even function. Let's take a quick question on this. Just one question is sufficient for this concept So if you have a function Which is zero to x log of 1 minus t by 1 plus t dt Okay, then Then f of x is option a and odd function Option be an even function option C Neither even or odd Option D data is efficient. How will you solve this? Good afternoon Likith Paras and my dear friend Aditya is missing. How was the exam Paras? Paras Likith. How was the exam? You can unmute and talk to me. How was the paper? Good, sir Good. How good? Okay At least you just did justice to your preparation, right? Okay, Shreya says even This problem is a very simple problem. It's actually a one-second problem. Okay, so if you see this function, this is your f of t, correct? Right and it's clear that f of t If you replace your t with a minus t it will become log 1 plus t by 1 minus t Which is nothing but negative log 1 minus t by 1 plus t and Isn't this negative of f of t? correct Which means f of t is an odd function Okay, and integral of an odd function from a to x a can be zero non zero doesn't matter This will always result into this answer will always be a sorry this result will always be a even function So f of x I should not call it as f because f was already used. I'm sorry for using the same name Let's use some other name G. Let's say. Yeah So G of f of t in this case Sorry G of t in this case is an odd function So f of x is an even function. So option number B is correct Is that fine easy one? Now you don't have to sit and evaluate this limit. Okay, sorry You don't have to sit and evaluate this integral. Okay, without you all evaluating these integral You can use this property to reach to answer directly You know 90% of the people who don't know this property would sit and evaluate this and then figure out whether the Integral is an even function and odd function right this saves a lot of time if you can use this function. So we'll move on to The twin of this property this says that if if f of t is an even function if f of t is an even function then Then Integral function like this Why I'm calling it as an integral function because this function is dependent on an integral. So you're evaluating an integral Integral from zero to x my dear not a to x this time. This will always be an odd function so the difference in the previous property and this property is that in the previous property you had f of t as an odd function and your Integral function was from a to x f of t and that answer was a Even function here f of t is an even function and Integral of that from zero to x is an odd function. That is the difference. So please appreciate this difference. Okay now here Please note that If you are claiming your phi of x that is a to x f of t is Is an odd function then Please remember your integral from zero to a must be zero This must not exist Because if this exists Then this may not be an odd function See, I'm moving my cursor. This may not be an odd function if this condition is not satisfied Are you getting my point? So, please remember this also So normally if f of t is an even function then zero to x f of t dt will be an odd function But if you are claiming that a to x f of t is an odd function Then remember zero to a f of t dt must give you zero else it will not be an odd function Am I clear? Am I clear on this statement? Please type clr on your chat box if this sounds okay to you Clear sir Clear Oh Aditya is here Hi Aditya how was the paper? It was good sir Good sir, you don't sound like that It was like good enough Okay, I know you all guys are tired after three hours of paper and then again a three hours of class But this is how class 12 goes No sir, I was like taking a power nap so I just woke up Okay Now you are powerful enough Okay, so let me prove this Let me prove this first The proof is very simple So if you want to prove that your phi of x is an odd function Basically we need to prove that phi of minus x is minus of phi x Correct Yes sir Yes sir Okay So phi of minus x would be what? Phi of minus x would be zero to minus x f of t dt Now very simple Let t be minus y Just once again Yeah So if t is minus y dt would be negative dy Correct So f of negative y And dt is negative dy From limit of integration will now be Zero to x Getting the point Now f of minus y is negative f of y because f of x is an odd function I'm sorry By the way it's an even function right so this will Yeah f of x is an even function Correct So I can write this as Negative zero to x f of y dy itself So I'm just replacing f of minus y with f of y Now there's nothing in the name I can just put the y back as t Correct And this is nothing but negative of phi x Isn't it Which clearly implies that phi of negative x Is negative of phi x Which implies phi of x is an odd function Is that clear So guys next we are going to talk about Some improper integrals I'm not going to take any question on this Because the concept per se is clear to you Through the previous example So all of you please be very attentive In the next half an hour of this session Where I'm going to talk about improper integrals Okay This concept is basically Slightly above J main level Okay But normally when questions come on this People are clueless how to go about it So I'm going to give you some tips and tricks How to solve it Many a time these questions are Are asked in J advance exams Okay But if you know these tricks Probably it will make your life better Okay So I'm going to introduce to you Two improper integrals today One is called the gamma function And second is called the beta function They both are related Okay so if you know one properly Other could be expressed in terms of the other Okay Now Let me start with gamma function These are two special improper integrals It's not like these two only make the improper integrals What is improper integral? Improper integral is where One of your limits is either infinity Or minus infinity like that For example if I say something like this Integration from 0 to infinity Okay Of any function Since your upper limit is infinity Right Both the limits can also be infinity Or minus infinity For example something like this Okay These are typical examples of improper integrals Where you don't have a finite value Of either the upper or the lower limit Okay Now these two are of special mention It's not like all improper integrals Come under gamma and beta functions These two functions are some special type of improper integrals And why I'm dealing with this is because There is a special formula which is coming up In this today's class Which is called the Wallace formula Which is based on this So it is important to know a bit of their preview So that we understand that formula very well Now what is a gamma function? So let's say Gamma function Let's say n is a positive integer So n is a positive number Not integer Okay Then an improper integral of this type 0 to infinity e to the power minus x x to the power n minus 1 Okay This is called gamma n The symbol is like this Symbol is just like a square root symbol So gamma n is this integral Okay this integral is given a special name And that name is gamma function Okay So x here are all x should be from positive x should be all positive rational numbers here Okay Now of course we don't have to evaluate this For all values of n So some special values you should know For example you should know what is the value of gamma 1 I would request you all to tell me what is gamma 1 I'm sure you can tell me Gamma 1 means what? You are integrating it from 0 to infinity e to the power minus x x to the power 1 minus 1 This is gamma 1 Which is 0 to infinity e to the power minus x x to the power 0 Which is 0 to infinity e to the power minus x I'm sure you can evaluate this What is it? e to the power minus x minus 1 Right What happens when you put infinity in place of x over here Who will tell me what is e to the power minus x as x tends to infinity It tends to 0 It tends to 0 absolutely Because it is as good as saying 1 by e to the power x Right So any number whose modulus is less than 1 raised to an infinitely big power will end up giving you 0 Okay So that's why this answer is 0 So it will become 0 minus e to the power 0 That's going to be 1 So gamma 1 is 1 No doubt about it Okay, tell me what is gamma 2 Sir, I got it as e to the power minus x into 1 minus x No, no, no, your answer will never be in terms of x Because you are doing a definite integral, right? Your answer will always be a number See, it is 0 to infinity e to the power minus x x to the power 2 minus 1 2 minus 1 is 1 only, right? So ultimately you are integrating this Okay So what will be used to integrate this? Integration by parts You will use IPP here Correct, Aditya? So if you integrate it by parts You can choose this as your first function Choose this as your second function So x into integration of this is minus e to the power minus x Derivative of x is 1 And integration of this is again minus e to the power minus x So which is nothing but minus x e to the power minus x Minus e to the power minus x from 0 to infinity When you put an infinity Okay Now here is a call that we need to take Here is something which is infinity to the power tending to 0 Okay, so let me write it like this Let me write it as minus x e to the power minus x Limit extending to infinity Okay This would anyways become a 0 Okay Let me just make it equal to Okay, minus When you put a 0, this becomes 0 And this becomes minus of 1 Okay So this will become minus x e to the power minus x Limit extending to infinity And this will become plus 1 Okay, let's evaluate this To evaluate this we can use L'Hopital's rule Okay, so use L'Hopital's rule So differentiated it becomes 1 This becomes e to the power x Limit extending to infinity Now as x tends to infinity This will become a super large number So 1 by a super large number will give you everything as 0 over here So answer is again a 1 Is that clear? Yes sir Yeah, Hamsini, is it clear? Great Now, I don't want you to evaluate a little further I just want you to find gamma 3 That's it Then I'll tell you a basic pattern that these gamma functions follow Gamma 3 you realize that your answer will come out to be 2 Let's check Yes sir Okay, by the way, in order to do integration by parts You can use DI method over here Do you remember DI method? Who doesn't remember DI method? Tic-tac-toe method, DI method Do you remember we have done that? I'll do it over here So gamma 3 Which is 0 to infinity e to the power minus x x square So if you want to apply integration by parts over here Take this as u, take this as v And then we follow the DI method In the DI method we write the v function under i u function under d So keep differentiating it till you get a 0 Keep integrating it so minus e to the power minus x Again e to the power minus x Again minus e to the power minus x Put an alternating plus minus plus sign here plus minus plus minus Then start multiplying this crosswise So your answer for this will be minus x square e to the power minus x Okay This will be minus 2x e to the power minus x Again minus 2 e to the power minus x From 0 to infinity Now let me tell you that when you put infinity Everything here will become a 0 You can try out taking the limit also Okay And when you try putting a 0 You will get this 0, this 0 and you get a minus 2 over here So that answer will actually give you a 2 Correct Now time to know some properties which will make your life very easy So I'll write this as properties of gamma function Properties of gamma function The first property that you basically would observe From whatever we have done so far is Gamma n is basically nothing but n minus 1 factorial Okay If n happens to be a natural number Okay This is something which you'll observe through pattern When you do lot of Let's say you found gamma 3, gamma 4, gamma 5 You'll realize that this is the property which it follows Many books will write it as gamma n plus 1 is equal to n factorial Both are the same thing You're just replacing your n with n plus 1 Correct So if somebody says What is the integral of Let me give an example Integral of 0 to infinity e to the power minus x x to the power 5 let's say Right What is the answer for this? The moment you realize Oh this is nothing but gamma 6 Isn't it? Please remember this term is your n minus 1 term So n is your 6 So it's gamma 6 Gamma 6 will be 5 factorial which is 120 Right So this is the property which is very important Next property which is important is Gamma n plus 1 Actually follows a recursive relation Which is n gamma n I think we can easily prove this I think we can easily prove this Anybody who wants to try the proving of this Prove that gamma n plus 1 is n gamma n That means prove that 0 to infinity e to the power minus x x to the power n Is n times 0 to infinity e to the power minus x x to the power n minus 1 So the factorial thing itself The factorial thing actually has come from this property This is the father of this property Okay So we cannot prove the father from the child property Just let me know once you're done with the trying Integration by parts is the key for this Hint Just type done on your chat box once you're done So that I can discuss it Done anyone? Okay gamma n plus 1 is 0 to infinity e to the power minus x x to the power n dx Take this as your first function Take this as your second function Correct So x to the power n Okay Integral of this is e to the power minus x with a minus sign Okay from 0 to infinity Minus Derivative of x to the power n is n x to the power n minus 1 Integral of e to the power minus x is e to the power minus x with a plus sign Okay 0 to infinity Okay Now Minus x to the power n by e to the power x from 0 to infinity This anyways is n gamma n Correct Just pull this n out Okay and the remaining term is gamma n isn't it Now here exponential function as Let's evaluate this as limit n tends to limit x tends to infinity So what is the answer for this Okay minus Anyways it's going to be 0 when you put a 0 Okay What is this term? This term has a polynomial by an exponential Now remember when x is going towards infinity Exponential grows faster than a polynomial function It grows much faster than a polynomial function Okay Therefore this result would have a very heavier base as compared to the numerator So this term will become a zero term Leaving you with n gamma n Okay So therefore we can say gamma n plus 1 is n gamma n Is that fine Next property, property number 3 is Very important expression is this gamma half Gamma half is root pi Gamma half is root pi Now the proof for this I will not share with you Because it involves undergrad concept Which involves the use of Jacobians and all Okay so I will not be talking about the proof for this But knowing this value is helpful Because it can solve lot of problems for us Okay It's a helpful result So please just remember that gamma half is root of pi Okay Anybody who is interested in knowing this proof Okay I will send you the proof separately So this proof involves converting Cartesian coordinates to polar coordinates And then using Jacobians to write a relationship between dx and dy In terms of the polar coordinates differential Okay so that's why I am not talking about it Because there are so many things which you don't know right now Next important thing that we need to know here is How is this helpful in solving many types of problems So we will look at some illustrations on gamma function By the way this can come as a direct question also to you Because there is nothing that except for the last property that we saw There was nothing which was beyond your scope Okay Let's evaluate few things Find first question Find 0 to infinity e to the power minus x x cube I am just giving you 10 seconds for this Before the count of 10 is over your answer should be there Sir 24 No Sorry sorry sorry Sir 6 6 Yes This is n minus 1 aditya Okay n minus 1 is the That means its answer will be gamma 4 Gamma 4 will be 3 factorial Answer is 6 Okay good Next question Find 0 to 1 log of 1 by x To the power of 5 dx Try this out Now you must be wondering how is it related to gamma function Remember the last class The last part of the Class that we had we had discussed about how you can convert one integral to another By use of proper substitution And there are lot of hints in the question to know what substitution will work For example you have to convert this into something of the nature 0 to infinity kind of a thing Correct So this integral has to be converted to such an integral Correct Okay So you know that 0 and 1 by x right Somehow they will be helping you to get infinity Log 1 is also helping you to get a 0 Anybody can suggest me what substitution should I make here Log of that thing as something Right very good So share is suggesting Let's substitute log 1 by x as t Right that means x is e to the power minus t dt See we also generated that exponential term which we desperately need in a gamma function Correct This term is what we require correct So dx is nothing I'm sorry dx is nothing but minus e to the power minus t dt Okay so first of all dx we can replace it with Let me write once again Integral minus e to the power minus t dt And log of 1 by x is t Correct To the power of 5 here in this case What about the limits of integration? Let's check out When x is 0 When x is 0 you realize t has to go to infinity Then only e to the power minus infinity will become a 0 Correct And when x is 1 t has to be 0 Indirectly if you swap the position of the upper and the lower limit And observe this minus sign You end up getting something like this And this term is nothing but what? This term is nothing but what? Gamma 6 Gamma 6 And gamma 6 is what? 5 factorial 5 factorial means 120 Do you see that? See what I'm saying is This is nothing which is beyond you After this substitution you would have any ways done integration by parts here right? And that would have wasted a good amount of time for you Let's say 2 to 3 minutes will go on doing integration by parts Or let's say 2 minutes will go This made your life easy because you already know this type of function Okay So you save a lot of time that's why I told you Knowing this will actually be like a tool for you to save time in your J exam Okay Next we'll move on to the second of the improper integral which we're talking about That is beta function Okay Beta function written as bm,n Okay and it is defined as integral 0 to 1 x to the power m minus 1 1 minus x to the power n minus 1 Here m and n are some positive numbers Please note If you make the substitution, if you make the use of King's property If you use your King's property The same expression You can write it as 1 minus x to the power m minus 1 And 1 minus 1 minus x to the power n minus 1 Okay Which is nothing but 0 to 1 x to the power n minus 1 1 minus x to the power m minus 1 Isn't it? Which means that even if you swap the position of m and n It doesn't make a difference to the result Right That is a very important property that beta m,n is same as beta n,m Now a very interesting result here is that These two are not the only expressions for beta m,n So one expression we saw over here Another expression we saw over here This is the third expression also Okay and that expression is You can write your beta m,n also as 0 to infinity x to the power m minus 1 By 1 plus x to the power m plus n Now let me ask you this as a question Can you show that? Or can you show that? Please prove that 0 to infinity x to the power m minus 1 By 1 plus x to the power m plus n Is same as 0 to infinity x to the power n minus 1 By 1 plus x to the power m plus n And is equal to 0 to 1 x to the power m minus 1 1 minus x to the power n minus 1 Can you show that? While proving this is very easy Because you already know that beta m,n is same as beta n,m So replacing your n with m Doesn't make a difference to the result But what I am interested more is How do you prove this to this? How do you convert an improper To a proper looking function like this? Here your substitution skills is going to be tested How good you are in substituting Any idea? Look at the limits of integration Those are very important identification Or hints that you can actually use For your choice of the substitution Anyone having any idea? If you are trying please type Trying sir Okay I think no movement has been seen so far With respect to this problem So let me use here The substitution x is equal to 1 minus t by t This is in light of a lot of things For example I see 1 minus x also appearing Secondly as you can see that When you put your t as 0 x is infinity and when you put your t as 1 Correct x is 0 Correct So in light of all these things I have chosen this substitution Let's see what happens So first of all what is dx Let's find dx here dx would be nothing but minus 1 by t square dt By the way you can write this as 1 by t minus 1 So if you differentiate it dx is equal to minus 1 by t square dt Correct Now let us substitute this in our given expression So x to the power m minus 1 is like 1 by t by t to the power m minus 1 So m minus 1 t m minus 1 will come here 1 plus x, what is 1 plus x? 1 plus x here is 1 by t Correct So you can write it as 1 by t to the power of m plus n Correct dx is nothing but minus 1 by t square dt So far so good What about limits of integration? When you put x as 0 When you put x as 0, t is 1 When you put x as infinity t has to be 0 Any problem with respect to this? Now If you simplify this First of all remove this minus sign and switch the position of the upper and the lower limit Secondly If this n goes up it will become 1 minus t to the power of minus n minus 1 and t to the power 1 minus m dt by t square Anything which I have missed out please highlight Oh I am so sorry This will go up and become t to the power This is 1 by t by the way This is 1 by t One second So slight mistake here This will go up and become t to the power m plus n 1 minus t to the power m minus 1 by t square Or you can say t square and t to the power m minus 1 will become m plus 1 Now again adjusting the powers here It will become 0 to 1 t to the power n minus 1 1 minus t to the power m minus 1 dt Now there is nothing in the name You can convert it back 0 to 1 x to the power n minus 1 1 minus x to the power m minus 1 And you know your positions of m and n can be swapped So if you swap the position of m and n This becomes 0 to 1 x to the power m minus 1 1 minus x to the power n minus 1 dx Is this what you wanted to prove? If you check above over here This is what we wanted to prove See here, this is it and this is it Watch here, this is it They are same Now this can be a potential question for you to convert one integral into another Now these GE guys know that beta gamma function probably will not be used directly because that is not written in your syllabus But they would frame a question involving these functions because it incorporates nothing but it's the use of proper substitution to convert one integral to another That's why I took up this concept today So that you are prepared even if such special functions are asked to you What is most important now is the fact that there is a relationship between beta function and gamma function And that relation is beta m n is actually gamma m gamma n by gamma m plus n This is the most important property that we will be using this in our Wallace formula So gamma and beta functions whatever we had studied are related by this relation with each other How this is useful? I will take up this after a small problem So let's solve this question first Integrate 0 to 1 x to the power 6 under root of Let's try it out How many of you are trying to solve this by the use of trigonometric substitutions I mean nothing wrong in doing that I just wanted to know how many of you substituted x as sin theta Let me walk the talk How many of you did x equal to sin theta That means dx is equal to cos theta d theta So if you do that it ends up giving you 0 to pi by 2 sin to the power 6 theta This is going to become under root of 1 minus sin square which is cos theta And dx itself has a cos theta with it So cos square theta d theta Okay overall this becomes the expression Okay Now what do you do here So here you have to use Since both of these powers are even if you recall we had done a similar case in indefinite integrals Please turn your notes where I had discussed with you the type sin to the power mx cos to the power nx where m and n both are even integers So in that case the approach was either use your complex number approach or you use your trigonometric identities Now both of these scenarios are going to make this a very long problem Isn't it? Yes or no There should be a sad smiley here but unfortunately because there is no other smiley other than this guy Okay Yes very long problem Okay so we would not even fathom to do it by this approach On the other hand let me show you how my beta and gamma functions are going to make my life super easy while solving such problems Okay So here what I will do All of you please see this carefully I am going to make a substitution Let x square be t That means 2x dx is equal to dt So x dx is going to become dt by 2 So borrow 1x from here So write it like this 0 to 1 x to the power 5 under root 1 minus x square x dx and replace this x dx with dt by 2 So can I say limit of integration will not change This will become t to the power 5 by 2 This will become 1 minus t to the power half and there will be a dt by 2 Now try to compare this Try to compare this with 0 to 1 x to the power m minus 1 1 minus x to the power n minus 1 That is actually the structure of beta m comma n So here m minus 1 is 5 by 2 and n minus 1 is half So m is 7 by 2 and n is 3 by 2 So basically this thing nothing but half beta 7 by 2 3 by 2 And according to this relation which I had discussed with you right now According to this relation that is why I told you this relation is very important You can write this term as half gamma 7 by 2 gamma 3 by 2 by gamma 3 by 2, 7 by 2 will add up to give you a 5 Isn't it? So please remember I have just used this particular formula So if you replace here 7 by 2 and here 3 by 2 it will become gamma 7 by 2, gamma 3 by 2 by gamma 7 by 2 plus 3 by 2 which is gamma 10 by 2 which is gamma 5 Now you must be wondering how do I get these values Fine I got this expression but how would I get their values Now here I would like you to take your attention to this property Try to recall this property which we had done a little while ago gamma n plus 1 is n gamma n This is a very important property the recursive property In the light of that, can I write gamma 7 by 2 as 5 by 2, gamma 5 by 2 5 by 2, gamma itself can be written as 3 by 2, gamma 3 by 2 and gamma 3 by 2 itself could be written as half gamma half and gamma half is root pi we just now discussed gamma half is root pi So this is the value of the gamma 7 by 2 function which is nothing but 15 by 8 root pi Similarly you can evaluate others as well Let me write down the final result half This is nothing but 15 by 8 root pi gamma 3 by 2 is half root pi and 5 factorial is 24 5 factorial is 24 So ultimately you will see your answer is 15 into pi by 4 into 8 into 24 You can cancel out a factor of 3 from here So it's 5 pi by 64 into 4 64 into 4 is 256 This is your answer This is much shorter than Please let me tell you this is much shorter than you applying your trigonometric identities on that Are you getting it? Yes sir Now something which is very associated with this result is Just now we talked about sine to the power 6 cos to the power 2 integration So I will tell you something very interesting Just now we studied that gamma function beta function is nothing but integration from 0 to 1 x to the power m minus 1 1 minus x to the power n minus 1 And we learnt that this is gamma m gamma n by gamma m plus n correct? Just now we learnt this Now I will show you something interesting Let's say I put my x as sine square theta I put my x as sine square theta So can I say dx will be 2 sine theta cos theta d theta Let me make this substitution over here So this will become sine to the power of 2 m minus 2 1 minus sine square is cos square so cos to the power 2 n minus 2 Okay And we have 2 sine theta cos theta What will happen to the limit of integration? When x is 0 theta will also be 0 And when x is 1 theta is going to be pi by 2 correct? Let me further simplify this So this becomes sine to the power 2 times sine to the power 2 m minus 1 theta cos to the power 2 n minus 1 theta So this is given to us as gamma m gamma n by gamma m plus n Bring this 2 down If you bring this 2 down we have a formula over here We got gamma m gamma n by 2 gamma m plus n Okay If you further want to refine it That means put your 2 m minus 1 as m dash And put your 2 n minus 1 as n dash That means your m is m dash plus 1 by 2 And n is n dash plus 1 by 2 and substitute this over here You would realize that you end up getting this There is nothing in the name so I can put back as m So sine m theta cos n theta d theta is 2 times Please replace your m with m dash plus 1 and m dash again with m So this will become like this Why did I put it 2 up? 2 will be down Some of these 2 will be m plus Okay So if m and n are integers over here this formula is called as the Wallace formula This formula is called as the Wallace formula So wherever you see such kind of integral you may directly jump to the use of Wallace formula So please make a note of this All of you please make a note of this Very important formula called as the Wallace formula Note it down Let's have a question Evaluate 0 to pi by 2 sine to the power 4x cos to the power 6x 0 to pi by 2 sine square x cos to the power 3x 0 to pi by 2 sine to the power 5x cos cube x If you are done just type 1 or you can speak it out also First one pretty easy This is your m, this is your n So your answer will be what? Gamma m plus 1 by 2 which is 5 by 2 Gamma n plus 1 by 2 7 by 2 by 2 Just add these 2 Now gamma 5 by 2 Again just write it pretty quickly 3 by 2 into half into root pi Gamma 2 is going to be 5 by 2 3 by 2 half root pi divided by 2 into 5 factorial Pretty simple So I am sure you can cancel a lot of term this 5 factorial you can write it as 120 So on the numerator you have 3 into 3 into 5 pi by denominator you have 2 to the power 5 into 2 into 120 So this will go by a factor of 8 I am sure So your answer is 3 pi by 2 to the power of 9 that is nothing but 3 pi by 512 Done See it becomes so easy No need to use any complex number approach or trigonometric ratios and identities any substitution nothing is required I know it is a very special tool that you have at hand but this is only to enable you to save a time Please don't use this in your school exams Don't start writing gamma function and all in your school papers please Okay next one Fast So 2 means 2 plus 1 by gamma 3 by 2 This means gamma 4 by 2 which is gamma 2 by 2 gamma some of these 2 So gamma root 3 by 2 is going to be half gamma half which is root pi gamma 2 is 1 so leave it This will be 2 into this is going to be 5 by 2 3 by 2 half gamma half so this and this gets cancelled So your answer will be nothing but 2 by 15 Done Last one you will solve it guys I have solved a lot of them I would not move forward till I get answer from at least 3 of you Shreya says 1 by 12 1 by 24 I got 1 by 24 Shreya has also changed it 1 by 24 Hamsini, Samyukta, Ruthu, Ardhara Rashmi Malikhit also 1 by 24 Let's discuss this This is gamma 5 plus 1 by 2 which is 3 3 plus 1 by 2 which is 2 by 2 gamma 5 So answer is going to be 2 into 1 by 2 this is 24 gone 1 by 24 done That doesn't even take me 5 seconds if I know these gamma functions Awesome So now we are going to move on to the last leg of this chapter which is expressing definite integral as limit of a sum This definition actually comes from an understanding of definite integral Many of the books will do this part early in the chapter But I decided to do it in the end because I can club this with limit of a sum as definite integral So we will go in both directions We will do definite integral as limit of a sum and limit of a sum as definite integral So the meaning of integral f of x from a to b is what? It is like finding the area under the function f of x So this is your graph and you are integrating from x equal to a till x equal to b So how do you integrate this geometrically? You choose a strip thin strip you choose a thin strip of with dx at a distance of x Thickness is dx height is f of x So area of this strip or differential area of this infinitesimally thin strip is going to be f of x into dx And if you integrate it from a to b you end up getting your area under this curve Now the same thing I will do manually not by taking a dx but by taking a width of thickness h h tending to 0 So what I will do is I will start with a very thin strip right in the beginning over here This is a rectangular strip whose height is f of a and thickness is h correct? So the area of that small strip let me call this as strip number 1 is going to be what? Strip number 1 area is going to be f a into h or you can say h into f of a If I make a second strip over here remember now the height of this strip will be a plus h and thickness is still h So this will give me the area of this second strip correct? If I keep doing this till I reach my nth strip over here please note that this is my nth strip now what is n here n here is infinitely big quantity that means you are breaking this area into very thin strips but those number of strips are infinitely many it's like the thread in your cloth okay? There are infinitely many threads and each thread is almost of thickness 0 which results into the area of that cloth isn't it? So in the same way these rectangular strips are like those threads in that cloth and there are infinitely many of them and each of these threads is almost tending to 0 thickness so h is tending to 0 correct? So who will tell me what is the area of the nth strip that I'm going to find out it's going to be h into f of a plus n minus 1h yes or no? So if I add all of them can I say I'll end up getting my integral of f of x from a to b correct? and on the right hand side if I take the h term common it will be f of a f of a plus h f of a plus 2h and so on till f of a plus n minus 1h remember here h is tending to 0 n is tending to infinity okay? Now something very interesting guys since this is a okay? this is a plus h this is a plus 2h this is a n minus 1h and b is actually a plus nh so there's a very important finding here b is actually a plus nh which means nh is nothing but b minus a now this expression is quite surprising because n is infinity h is tending to 0 so you're trying to say 0 minus infinity is a fixed number yes of course it can happen because it's an indeterminate form it is an indeterminate form correct? so here one more thing I would like to add instead of writing n tending to infinity I will basically write nh is equal to b minus a because I don't want two limits variable to be involved unnecessarily this expression is called finding the finding definite integral as limit of a sum or also called finding the definite integral through first principles okay now before I even give you the first problem to solve on this there are some important results which we need to remember I'm sure you would all remember that but I just want to reiterate it okay in case you have forgotten it okay first result summation of all natural numbers from 1 to n what is it summation of square of summation of square of all natural numbers from 1 to n this would be used that's why I'm giving you these results so you must be wondering how come I started talking about progressions all of a sudden they are the prerequisites they would be required while solving definite integral as limit of a sum because ultimately you are summing correct? and when you are summing you need summation results apart from this sign series sign series is where you are having sign of angles which are in AP so let's say you are summing this note down this result sign n beta by 2 this 2 is only below n minus 1 n beta okay not next to alpha please note this down next cosine series also only difference from sine series is there is a cost term coming over here so this sign is replaced by cost everything is did to the same don't forget these formulas apart from that you should know your standard limits also right so limit of e to the power x minus 1 by x as x into 0 is 1 limit of sine x by x extending to 0 is 1 let's take questions on this n crd evaluate e to the power x integral from a to b using limit of sum done? once you are done please type done okay anybody else shia says done anybody else okay so it's very simple so integral from a to b e to the power x dx you can write it as limit h extending to 0 and your n h wherever it appears you have to put a b minus a so it goes like this f of a would be e to the power a f of a plus h would be e to the power a plus h f of a plus 2h would be e to the power a plus 2h and it goes all the way e to the power a plus n minus 1h if you see here it's a gp okay it's a gp whose first term is e to the power a and common ratio is e to the power h so we all know the gp formula that's the first term the common ratio to the power of n which is e to the power nh minus 1 by common ratio minus 1 now wherever you see an n h let's quickly place a b minus a there so it becomes something like this now what about this term h by e to the power h minus 1 as h tends to 0 it's just the reciprocal of that standard exponential limits which I gave you a little while ago so e to the power x minus 1 by x as x tends to 0 is 1 so it's h by e to the power h minus 1 h tending to 0 so this is going to be 1 e to the power a e to the power which is nothing but e to the power b minus e to the power a and we all know that this is the result when you integrate e to the power x from a to b right it's nothing but e to the power x put upper limit okay so this is what we got from the limit of a sum expression as well is this clear guys is the approach clear please don't make any mistake in you know writing the formula secondly note that wherever an n h comes you have to put a b minus a there right so there's a small rule of thumb which I'll tell you whenever you see an n h expression okay where a complete pair of n n h is getting formed we hesitate to put a b minus a there but wherever there's an h alone wherever there's an h alone there's no end to pair up with it then you put that h as 0 or evaluate the limit of h tending to 0 as the case may be let's take this question evaluate integral of sin x from a to b using limit of a sum just type done once you're done share is done done okay let's discuss this so again the same integral from a to b sin of x you can write it as limit h tending to 0 and n h equal to b minus a h times f of a f of a would be sin a sin a plus h sin a plus 2 h all the way till sin a plus n minus 1 h now here we can use our sin series so your alpha your alpha is a beta is h so sin n h by 2 by sin h by 2 sin a plus n minus 1 h by 2 okay now as I told you wherever you see an n h put a b minus a immediately so this n h I can put a b minus a by 2 sin h by 2 remain sin h by 2 sin this you can write it as a this term here this term here I'm just writing it down here it's n h minus h by 2 which is nothing but b minus a minus 0 by 2 so this entire term is b minus a any doubt so far all these steps are clear so this becomes this becomes sin of a plus b by 2 okay and what I'm going to do here is in order to neutralize this I'm going to put a 2 and a 2 over here says that sin h by 2 by h by 2 or reciprocal of this becomes 1 so your final answer looks like this sin b minus a by 2 into sin b plus a by 2 okay now let us use our transformation formula we all know that we all know that 2 sin a cos b is nothing but cos a minus b minus cos I'm sorry I'm sorry I'm sorry sin b 2 cos a minus b cos a plus b so it's going to be cos a minus b is going to give you an a and a plus b is going to give you a b so it's just cos a minus cos b that's going to be your answer please note when you have been asked to do it by limit of a sum please directly do not integrate it you will get zero marks in your exam okay because see it becomes a super easy problem if you're using your direct formula but that is not what is the purpose of the problem the purpose of the problem is whether you know how to write it as a limit of a sum so in your exam if it is mentioned write it as a limit of a sum do not directly write the formula and do it you will not get any marks okay don't even expect some grace marks to be given to you let's take this question evaluate dx by under root of x for me to be any idea how to do this this problem is slightly different from what we did in the previous two problems okay and by this time you would have realized that because there is no way you can actually solve this up by using your regular summation series okay so basically when you write this as a limit of a sum what do you end up writing you write something like this limit h tending to zero okay and n h is b minus a and the expression that you write is h f of a f of a is 1 by root a then f of a plus h f of under root a plus to h and so on till correct now how do you sum this up that is the problem how do I sum this up any idea anyone in order to do this in order to do this I would like to recall for you sandwich theorem how many of you remember sandwich theorem that's how you get that sin x by x thing no yes exactly in sandwich theorem if you are able to let's say there are three functions gx fx and hx you know that this function f of x is sandwiched between these two guys in the neighborhood of x equal to a okay if you know this that your f of x is sandwiched between um g of x and h of x as extends to near the neighborhood of x equal to a and you know that the limit of g of x as extends to a is l and limit of h of x as extends to a is also l then then the limit of the sandwich function f of x as extends to a this will also be l without actually evaluating it now you must be wondering how am I planning to use this to solve this limit so all of you please focus on this approach so I'll come back to this page I'll just go to the next page because I need a lot of space to write consider my following argument if you see each of these terms what of the nature 1 by root r correct each of these terms was of the nature 1 by root r right 1 by root a, 1 by root a plus h, 1 by root a plus 2h etc they were of the nature 1 by root r so now what I'm going to do is I'm going to multiply and divide with this numerator and denominator by a 2 so 2 here, 2 here ok can I say this term would always be greater than equal to 2 by under root r plus h plus root r please note h is a very small quantity but nevertheless this denominator is heavier than this denominator so if a denominator is heavy the entire value comes a little down comes a little less in the same way can I say this term would slightly be lesser than 2 by root of r minus h plus root r now here your denominator is slightly lighter than this root this 2 root r see this quantity is lesser then 2 root r and this quantity is greater than 2 root r so when your denominator is heavier the overall expression drops down when your denominator is lighter overall expression slightly becomes more right so far so good any problem in understanding these two clear please write clr if it is clear clear sir clear only 1% has written clear 2% has now now how do I plan to use this let me rationalize it so if I rationalize it do you all recall that it will become root r plus h minus root r by h here also I am dropping the 2 2 factor here also if I rationalize it it becomes by minus h can I just reverse the position of these two can I just swap the position of these two and get this negative sign out of the picture so let me do that is that clear ok now what I am going to do is I am going to bring this h up I am going to multiply throughout with h now start putting the value of r as 1 by a so let's say I put r as a ok so this will become 2 under root a plus h minus root a this will become 2 under root a minus under root a minus h ok then put your value as a plus h so here it will become a plus 2 h minus a plus h here it will become 2 under root a plus h minus root a ok keep on doing this till you reach h by under root a plus n minus 1 h so this expression will be 2 times under root a plus n h minus under root a plus n minus 1 h whereas this side will become under root a plus n minus 1 h minus under root a plus n minus 2 h any doubt so far let us add them on the middle side of this expression you will get h times 1 by root a 1 by root a plus h till 1 by root a plus n minus 1 h this is actually our expression that we required correct I will go back to the previous board and show you this was the expression that we were actually looking for whereas this side you will see cancellation of terms will happen ultimately leaving you with under root of a plus n h minus root a again cancellation of terms will happen leaving you with this term minus this understood so treat as if this was your function and this was sandwiched between h of x and g of x ok you want to find the limit of this guy correct we want to find the limit of this guy so let us see what is the limit of h of x as h tends to 0 so let us find out limit of to under root of a plus by the way n h is b minus a so you just write a b minus a ok this will become a very simple case it will become to root b because a and a will get cancelled minus root a this let us find out the limit of this I will write it in yellow here also same thing plus n h minus h h is 0 so I will put a so again here h n h you can put this n h you can put as b minus a and this h alone is 0 and this is root a so a and a gets cancelled ultimately you get up you get the same expression back ok so it means that both the sandwiching functions are tending to to root b minus a so the sandwiched function will also become to root b minus a so this limit here I am going back to the previous problem this limit here is going to be to under root of b minus root a which is obvious because we know from our formula that 1 by root x dx from a to b gives you to root x from b to a which is nothing but to root b minus to root a that is 2 times root b minus root a is this clear guys any question so this is the use of sandwiched theorem that probably you would have done last year ok so directly use of summation is not helping us so we have to fall back on sandwiched theorem to do this any questions please type no questions if you don't have any questions no questions ok now we will be talking about the reverse process limit of a sum as definite integral so this is a reverse of the process that we just now did so in the previous process what did we do we converted definite integral to a limit of a sum now we are converting limit of a sum as definite integrals so there are a lot of limits question I don't know whether you have come across them or not where the limit is of the form or indeterminacy is of the form 0 into infinity and you are not able to apply any standard summation process neither are you able to apply any standard limits to solve them ok how to deal with those kind of problems that is what we are going to study in this particular part now I would try to explain this to you by taking a simple example ok I will try to take an example to explain this concept to you let me just pull out an example for it suitable example so that this thing is clear in your mind let's say we want to evaluate limit as n tends to infinity 1 by n square 2 by n square 3 by n square all the way let's say 1 by n by the way 1 by n is nothing but n by n square it's a short form of writing it how do you evaluate this now here you would say sir it's very simple I would just now how many of you think that the answer to this would be obtained by using this approach this is 0 because n is very big n is infinity this is 0 this is 0 everything is 0 so the answer will be 0 how many of you think this I don't think that sir of course not sir of course not and you should not also think like that because let me tell you if you are thinking like this then there is a very big conceptual error that you are making here each of these quantities is tending to 0 ok who wrote that no oh Likith is controlling yes the Likith how can Likith is he is an artist he is an artist he is an artist Likith how did you get the control of my desktop so even we can do that sir like all of us can do it but I have to make you the presenter right I didn't do that sir ok never mind see each of these quantities is tending to 0 right but there are infinitely many of them right so you cannot underestimate a small quantity which are present infinitely many number of times it's like a drop of water a small drop of water present infinitely many number of times can give you an ocean also correct so never think that they are equal to 0 they are tending to 0 here the entire idea is this is actually an indeterminate form an indeterminate form can take a lot of values ok but here you would say sir I don't need a definite integer to solve it because even if you if you don't allow me to do that actually should not be used let's say this method is ruled out I can actually sum this up right 1 plus 2 plus 3 all the way till n by n square correct and we can evaluate this limit by using our formula of summation of national numbers from 1 to n ok and what do we do next we divide by n throughout so it becomes 1 plus 1 by n by 2 and as n tends to infinity this term will become 0 so the answer is half actually not 0 fine now there is another approach that we normally take here it works fine but there are so many questions which I am going to give you next where your summation will fail where your summation will fail but since I have taken this as an example to explain how limit of a sum can be converted to a definite integrals ok let me explain you something very interesting so here let me go by algorithm so that you follow that algorithm to solve all the problem the first step of the algorithm is identify the rth term of the series who will tell me what is the rth term quickly tell this is the first term this is the second term this is the third term this is the nth term so what is the rth term r by r square r by n square oh yeah step number 2 from your rth term pull out a 1 by n common always always so write it like this understood so what is the second step from your rth term pull a 1 by n out always in whatever problem that you are going to come across this step is mandatory right step number 3 put your r by n as x or replace your I should not say replace I should say replace your r by n term with an x and 1 by n term with a dx correct so what happens tr becomes what in this case x dx okay now what are you doing you are trying to find out summation of tr from r equal to 1 till n and n is tending to infinity this is what we are trying to do correct now this term would be replaced by this term would be replaced by an integration sign because summation is an integration process or vice versa where limit of integration is decided by from which r to which r you are summing up now see here in this present question all of you would have understood by now that you are summing from r equal to 1 so this summation is from I can write it like this this summation is from r equal to 1 till n correct and you are trying to sum this term okay where n is tending to infinity so now here you have claimed your x to be r by n see here you have claimed your x to be r by n so this will help you to decide the limit of integration how that we will see in step number 4 in step number 4 which is the step of deciding the limits when your r is 1 let me write it like this when your r is 1 x is equal to 1 by n which is nothing but 0 and when your r becomes n x will become n by n which is nothing but 1 so this will become your lower and the upper limit of integration so this is your lower limit of integration and this is your upper limit of integration okay so ultimately what I am asking you to do is I am asking you to convert this entire problem I will go back on talking I am going to ask you to convert this entire problem to integration from 0 to 1 this is your x this is your dx and that is your answer so which we know that it is x square by 2 put your upper and the lower limit it is going to be half the same answer that we got over here are the steps clear to you yes sir I will repeat once again write down the rth term okay pull out a 1 by n from the rth term put r by n as x put 1 by x 1 by n as dx depending from which r to which r you are integrating or what are your limits on r since your r is x by n sorry since your x is r by n you can find your x accordingly anybody has any doubt if you have no doubt I am going to give you the next question in fact I am going to give you at least 5-10 questions on this concept because this is very important find limit n tending to infinity 1 by n plus 1 1 by n plus 2 1 by n plus 3 all the way till 1 by 2n done any idea how to do it how many if you are trying the steps which I told you step number 1 write down the rth term I am sure you would have all written this rth term correctly step number 2 take 1 by n common from the rth term now try to convert this as a function of r by n okay try to write this as a function of r by n that means I should see constants and r by n terms here divide by n throughout so it will become 1 plus r by n so here this is your function of r by n guys this is a very critical step so 1 by n after you have taken out whatever is left write that as a function of r by n because ultimately you have to put r by n as your x isn't it so this becomes 1 by 1 plus x because you have to replace your r by n with an x replace 1 by n with a dx on that replacement you will get 1 by 1 this is 1 by 1 plus this is your x this entire term is your x okay and this term will become your dx now what about the limits of integration if you see you are integrating from r equal to 1 by the way this term is n plus n so you are integrating from r equal to 1 till n okay so when your r is 1 x will be 0 so this will be 0 and when your r is n x will be 1 so this will be 1 so your problem boils down to solving this definite integrate which we all know is ln 1 plus x so with the limits of integration it becomes ln 2 minus ln 1 which is nothing but ln 2 so the summation of this or the value of this limit is nothing but ln 2 guys let me tell you without the use of definite integrals it would be almost impossible to evaluate such kind of limits right and especially the ones of the type 0 into infinity so whenever these kind of limit problems are found out in your questions keep the options alive that you can use your definite integral also to solve it first try your normal approach first try your formula first try your summation if it works then well and good if it doesn't then there is possibility of using definite integrals there and just follow these steps are you finding these steps difficult to follow I don't think so let me ask you another one if you are confident about it we will keep having problems after problems so let's say limit n tending to infinity n by n square plus 1 square n by n square plus 2 square and so on till n by 2 n square this time everybody should solve it I am giving you 3 to 4 minutes to solve it I am sure you can all do it sir I am getting 5 by 4 5 by 4 okay 5 by 4 anybody else 5 by 4 let me tell you is the right answer very good step number 1 nice bro paris step number 1 write down the edit term sorry edit term step number 2 pull out a 1 by n common so it will leave you with n square by n square plus r square now you have to convert the remaining part which is after taking 1 by n whatever is remaining part you have to convert it as a function of r by n only right now it is not in that shape so I have to bring it to that shape by taking n square in the denominator so it becomes 1 by 1 plus r square by n square which is like saying this okay now step number 3 role change wherever there is a 1 by n put a dx wherever there is an rn put by x so it becomes 1 by 1 plus x square correct what about the limits of integration see here r is going from 1 till n if r is going from 1 till n remember x is r by n so when r is 1 x is 0 x is 1 so limit of integration is 0 to 1 so isn't it integration of tan inverse x from 0 to 1 which is pi by 4 absolutely correct awesome see basically this series goes like this next term is 1 by n plus 2 n plus 4 next term is 1 by n plus 3 6 and so on that's why towards the end it becomes n plus n and n plus 2n that's nothing but 2n into 3n which is nothing but 6n square that's how this 6n square over here comes up okay so this is the weight in which it is moving on done what is the answer people have done please type in the answer I forgot to substitute sir I am getting a very weird answer it's like I can't type it can you send me the photograph yeah I will sure 2 pi by 6 root 3 okay yes sir I've sent it the book answer is log of 3 by 2 please note there is a n also waiting outside here this n so if I ask you step number 1 tell me the rth term what will you say n by n plus r n plus 2r correct step number 2 oh let me check oh yeah yeah see I think it could be further simplified had used partial fractions it would have been much better step number 2 take a 1 by n out so it becomes n square by n plus r n plus 2r okay now here an important part is you need to express this as a function of r by n this you need to express as a function of r by n okay so divide by n square but when you divide by n square you give 1n to this so divide this by n square divide this by n and divide this by n so it becomes 1 plus r by n 1 plus 2r by n okay now replacement let's replace 1 by n with dx r by n with x okay so it becomes 1 by 1 plus x 1 plus 2x dx limit of integration will be again 0 to 1 because your r is from 1 to n so ultimately you are evaluating this definite integral so 1 by 1 plus x you can break it up as partial fractions 1 by 1 plus x plus b by 1 plus 2x okay in fact this can be compared to 1 put x value as minus half this will become b by 2 so b is equal to 2 put x value as minus 1 so a value is minus 1 so this could be written as minus of 1 plus x b value is 2 integration of dx by 1 plus x 1 plus 2x is as good as integration of this term okay limit of integration 0 to 1 so minus ln 1 plus x 0 to 1 so this will become minus ln 2 and this will become ln 3 minus ln 1 which is actually 0 so the answer is ln of 3 by 2 okay so once you know your denominator is factorisable the better approach would have been to convert it as a partial fraction problem and then do it is that fine everyone yes sir everybody just type fine if it is fine not clear samyukta fine f9 fine simple one limit n tending to infinity 1 by n a 1 by n a plus 1, 1 by n a plus 2 all the way till 1 by n b samyukta has written fine factorial where is on this sir i did it sir for vectors i will send you some videos please do watch it done what is the answer Shia i got ln of 1 by a plus 1 come again come again ln of 1 by a plus 1 ln of 1 by a plus 1 that is actually 1 minus ln a right yeah there is no b in the answer but there is a b in the question i will try check your working once again oh okay i got my mistake sir don't worry sir i am not getting what the upper limit could be why not see let's go step by step step number 1 r at term would be what see it doesn't matter even if you start with 0 it doesn't matter so let's say n a plus r r starts from 0 all the way till okay now what is the last step when you say n a plus r becomes n b that means r is n b minus a right so your r is going from 0 to n b minus a this is very important meanwhile let me complete step number 2 pull out n so it becomes n by n a plus r which you write it as 1 by n 1 by a plus r by n does everybody would have got it any doubt so far now step number 3 replace 1 by n with dx replace r by n with x okay now limits of integration also i will write it as a and b okay so a is nothing but 0 by n which is 0 okay i should not write a and b here actually let me write upper and lower limit because b is already used and a is already used in the question sorry lower and the upper limit upper limit will be what r by n and you are putting r as n b minus a so n b minus a by n so upper limit will be minus a getting the point Paris it's a record now got it so basically your entire expression gets converted to dx 1 by a plus x integral from 0 to b minus a which is nothing but ln ln of a plus x put the upper limit b minus a and lower limit 0 so upper limit will give you ln b lower limit will give you ln a so it's ln of b by n is that fine so if you have any problem understanding how do we evaluate upper and the lower limit let me tell you once you have put your x as r by n you just have to put r as the initial value of the summation and r as the final value of the summation and then you have to check your upper and lower limit please try this hint is first take log any idea anyone yes sir sorry if I want to include this n within this bracket can I say it will come inside as n to the power n if I want to put this 1 by n as a power on the entire expression let's say I call this limit as n so basically you can give one n to each one of them take log because ultimately we have to convert it as a process of summation isn't it so log of this I can individually write it as log 1 plus 1 by n log 1 plus 2 by n log 1 plus 3 by n and so on till log of 1 plus 1 by n which is unbanned now from here onwards you start applying your normal steps find out the arid term take a 1 by n common I think 1 by n is already taken out common because by our problem itself 1 by n is outside so step 1 write down the arid term arid term is 1 by n log 1 plus r by n now you can do it I think complete it and tell me the answer step 2 replace 1 by n with dx r by n with x limit of integration would be a would be 0 b would be n by n which is 1 so is 0 to 1 log 1 plus x dx now to evaluate this we have to use integration by parts but we already know the result of log x minus 1 no so we already know how to integrate log x log x integration is x log x minus 1 same way this will be 1 plus x log 1 plus x minus 1 put your limits of integration so when you put a 1 you get 2 log 2 minus 1 it becomes 0 sorry it becomes 1 log 1 minus 1 which is nothing but 2 log 2 minus 2 and this will give you a 0 plus 1 so answer is 2 log 2 minus 1 which is log 4 minus 1 which is actually log 4 by e now the problem is this is the answer for log of the limit so log of the limit is log 4 by e so L is 4 by e is that clear next sir is that an error in the question the 3 cube thing yeah this is square that's a typo now the problem is even if you know the result for these two we are not aware of the result for this right now the time to actually find that result out so don't try to attend this question by the use of your special summations that you have learned see basically you are doing summation r square here and this is summation r cube and this is summation r to the power 6 let me do one thing let me multiply this with an n square and provide this with a summation of n square and let's divide throughout with n to the power 6 if you divide by n to the power 6 it will give me 1 by n sorry this term will not be there correct now since these two term are falling short of a 1 by n so I will introduce a 1 by n 1 by n here correct now limit has a property that you can take the limit individually in all the terms so this into this divided by this so this appears to be your dx and this is x square so it is integral of x square dx okay this is nothing but integral of x cube dx this is integral of x to the power 6 dx I am sure limit of integration you would all have figured out it is 0 to 1 okay so this answer is 1 third this answer is 1 fourth this answer is 1 seventh okay so the answer is 7 by 12 is that fine yes sir yes sir now I would like to end this chapter with a small test for you all okay so I will give you a question okay you guys need to solve it and tell me the response before it is 5 o'clock so the question goes like this first online test the question goes like this if you have a function this satisfies this functional equation okay find the function any idea guys sir last 40 seconds oh okay no sir don't know good try others anybody who could find success okay just see here carefully if you just split open this okay the answer is hidden in the question itself if you split this up it becomes since you are integrating with respect to y, x is just like a constant so x 0 to 1 y square f y dy and x square y f y dy so what I have done is this term I have just opened it up clear now anyways this is going to be a constant isn't it let's say I call it as a this is also going to be a constant let me call it as b okay that means what I will do now I will take an x common here so that it makes my life easy because x is here also x is here also so I will take an x common okay now this entire thing is a constant let me call it as a which actually tells you that your function is of this nature a x plus b isn't it a x plus b x square correct getting the point are you convinced till this stage are you happy till this stage when you when you integrate something and you put the limits of integration ultimately that gives you a number right that gives you a constant so what I am saying is that this constant is another constant so overall the structure of your function is just like a constant into x and b into x square okay now how do I get this a and how do I get this b that's a different thing so let's just try to find them out now what is a itself a itself is 1 plus integration 0 to 1 y square f y now f x is this so can I say f y will be a y plus b y square see if this is your f x f y will be what just change the name of the variable from x to y isn't it now this is something which you can easily integrate so 1 plus a y square integration will be a y to the power 4 by 4 b y to the power 5 by 5 okay put 0 to 1 it becomes 1 plus a by 4 plus b by 5 so this is your a so first equation I got yes or no yes now what is b itself b itself is let me just increase yeah b itself is integration of y f y f y is again a y plus b y square okay from 0 to 1 so a y square is a y cube by 3 and a y cube is b y 4 by 4 from 0 to 1 so that means nothing but a by 3 plus b by 4 so second equation that you end up getting is b itself is a by 3 plus b by 4 this will help us to get the value let me call it as 1 and 2 from 1 and 2 we can get the value of a and b and we can substitute it back over here as you can see okay so let's solve the second one first so 3 b by 4 is equal to a by 3 that means b is 4 a by 9 put it in the first one so a you can write this as 3 a by 4 is equal to 1 plus b by 5 b by 5 will be 4 a by 45 correct so take the age together so 3 by 4 minus 4 by 45 is equal to 1 so this is going to be 135 minus 20 by 180 okay so a is equal to 180 by 115 correct sorry sorry 16 16 so 119 okay so what is b itself b will be 4 a by 9 4 a by 9 180 by 119 this is 20 so this is 80 190 so your function is as you can see your function is ax plus bx2 so ax means 180 by 119x plus bx2 is 80 by 119x2 okay so guys this is end of definite integrals okay but subject to more practice from your end with respect to vectors I will send you a video so you watch that video I think vector that you have for your semester exam is pretty easy you will have just dot product and cross product and all which you have already done in physics a little bit of co-linearity and co-planarity concepts that you need to understand for that I will send you a video you can watch it okay so thank you so much over and out from my side bye bye have a good day ah who is that I was I was about to save that okay guys bye bye thank you bye bye