 So, today we are going to start with the inequality properties of definite integrals. Inequality properties of definite integrals. Properties of definite integrals, okay. So these properties are the most favorite ones when it comes to comparative exams because many times this requires a bit of reverse engineering thinking backwards. Okay. And that's what makes this topic slightly challenging as compared to the normal properties. So let us get started with the very first property. So let's say there is a function f of x. Okay. Which is basically greater than a function g of x. And lesser than equal to a function, let's say h of x. Okay. For all x belonging to an interval a comma b and of course these all functions are integrable as well. And all these functions f of x g of x h of x are integrable are integrable in the interval a to b. Okay. So if, if three functions f g and h satisfy this condition. Okay. Then please note down, even their definite integrals between a to b will also satisfy the same condition. That means integral of g of x from a to b will be lesser than equal to integral of f of x from a to b will be lesser than equal to integral of h of x from a to b. Okay. Now what is the reason for this property? The reason for this property is very simple to understand from the geometrical interpretation of definite integrals. No idea. Jay will happen during January or after the boards. We are also eagerly waiting for it. Okay. But I have if you ask my speculation, when will the J main first phase happen. I personally feel and many of my colleagues who are in teaching field, they believe that it cannot happen earlier than February end. Okay. That is what is my gut feeling. I may be wrong. People may be wrong. It all depends upon whether national testing agency surprises us. Okay. Yes, the proof is very simple. If you see, I'm just taking a case where all these functions are positive, but they need not be all positive. So if your f of x function is like this. g of x function is maybe like something like this. Okay. And let's say h of x function is something like this. Okay. So as you can see, as you can see in this interval, in this interval from a to b, let me just make the a to b interval. So this is your a and this is your b. This is your b. So in your a to b interval, your f of x is between your f of x is between this is your f of x. This is your g of x. And let me change the color of the pen to the same color as the graph has been drawn. This is your h of x. Okay. So even this area, let's talk about the area or the algebraic area between the function f of x and the x axis. This will be more than this wide area. Let me shade it in vertical stripes. And this will be lesser than this blue area, which I'm shading with horizontal stripes. Okay. So it's very obvious that the algebraic area. So let's say I call this area to be a to I call this area to be a three and I call this area to be a one. Okay. Algebraic area. So even will be between a two to a three. Okay. That's the main reason for this particular property. Now, many times you will get a question where you will be just given a integral. Of course, with the limits and the question setter will ask you that this integral lies between which values. So it may be possible that this property is helping you to solve that particular scenario. I think today morning, I did ask me a question of the same type isn't it. All this falls under this type of question and that questions cannot be solved without looking at the options. Okay. So whenever you're asking such a question that this integral lies between which values. Okay, you should always show the option clearly to the person whom you're asking the question because it depends upon the option also because it requires reverse engineering. But, you know, you will always be able to figure out between which two functions are given function lies so that you can link their integrals with each other. It requires a bit of reverse engineering it requires a bit of backward tracking. Okay. Let's take a question on the same. Let's take a very simple question on the same. I'll take a couple of questions for you. Let's say I want you to prove. So I'm giving it as a proof that question so that you can always, you know, figure out what is the required function so prove that the integral of integral of one by under root of one minus X to the part to win, where the quantity greater than equal to one is always between half to pi by six is always between half to pi by six and the limits of integration is zero to half. Okay. Now you can solve this question only when this information is provided to you directly or indirectly via options. So looking at the options only you can take a call. Right. Now think, think what you can do in order to prove this. It's not a rocket science. Just have a good look at what you have been provided. Pi by six, half. Okay. So the shittish has done a small breakthrough. Yes, you are on the right track just justified. We don't have to solve the integral guy three don't get this question wrong. They're not asking you to solve this integral they're asking you to, you know, bound the answer of this integral. Right. Think about that. Yes, very good. So most of you have done a good job on this. See, let us look at the scenario. If your n is greater than equal to one. Okay. And your x is supposed to be a fraction. Okay, x is somewhere between zero to half. Okay, you can keep open integral zero to half. Okay. Do you all agree that if x is a fraction, and you raise it to a power two, and you raise it to the power of two and N being more than one. Let's say greater than equal to one. Which one will be more or which one will be less out of the two. Which inequality sign I should put x square will be more or actually part two and will be more x square will be more right so can I say this will be greater than okay. Now what I'm going to do I'm going to multiply with a minus sign on both the sides of the inequality. And we all know that when you multiply with a minus sign with on both the sides of the inequality, the inequality switches. Correct. Yes or no. Hey, if x is a fraction, the higher the power, the lesser will be the value, you know, half to the power three is more or half to the part two is more. Half to the part two is more. So two to the power n will be greater than equal to two right so half to the power to x to the part two will be lesser than x to the part two right. Okay. If I add a one that should not affect the inequality sign. Correct. If I take the under root that should not affect the inequality sign. And can I also say that this quantity will be lesser than equal to one also, because you're subtracting some positive quantity from one. Okay, and of course taking the under root. So that will decrease the value and it one will be more than this. Correct. So everybody agrees with this particular scenario. Okay. Now if you reciprocate everything again the inequality will switch. Okay. Now from where did I get this idea because five by six comes when you are trying to put your upper and the lower limits. In this isn't it that only gives you five by six and sign inverse x comes from the integration of one by under root one one by under root one minus x square. So I did a backtracking I did a reverse engineering to figure out this function from this data that was provided to me. So this comes from the integration of one from zero to half. Okay. So all these things are like acting like your hints in the question it cannot be blindly solved. Remember, options have to be looked at. Then only you can solve this question. Okay. Now, once you have figured out that there is a function which is lying between two functions, let me call them as gfx and hfx. And their integral will also satisfy the same property. That means integral of the function from a to b will also be more than integral of gfx from a to b and the center will be lesser than integral of hfx from a to b. Okay. In short, the given integral which I'm calling it as I will be will be greater than integral of one from zero to half. Will be lesser than equal to integral of one by under root one minus x square from zero to half. So this gives us a half and this has already told you this is sine inverse x zero to half that gives you pi by six and there you go. Is it fine. So you need to act as per the options provided to you. And by the way, I would request people who are really serious about cracking J main J advance or any comparative exam, please ring up one of the guys. Let's say if you are from NPS Indra Nagar, ring up Pruchar Parik. We will be more than happy to share his number with you. Of course, after his permission only, because he would be getting calls from so many places. And NPS Rajaji Nagar, please speak to trippan. Okay. NPS HSR people, I think you would be knowing Richard Singh. Okay, Richard Singh, speak to him, and ask him, please get the information about the final months of your preparation because these people have put in their heart and soul. They were 1617 hours of study they were doing in the last few months as what I knew from them. Okay, speak to them, they will tell you how to know balance out things. If you hear from me, you'll think that is repeating the same thing in one day. Yes, in one day 1617 hours of study they were putting right connection ditch. Okay, anyways, so let's take few more questions based on this property. Let's take a lot of your time. Let's take this question. Let I1 be an integral, which is sine x by x. Okay, this integral is from pi by six to pi by three. Okay, I to be an integral, which is sine of sine x by sine x. This is also from pi by six to pi by three. And I3 is another integral, which is sine of tan x by tan x. Okay, this is also from pi by six to pi by three. Okay. Now, which of the following options are correct. Pick the right options. Okay. Choose the correct option. Option number eight. I1 is more than it is more than I3 option number be I3 is more than I2 is more than I1 option number see. I2 is more than I1 and I1 is more than I3. Okay. An option number D I one is more than I three is more than I two. If you want, I can put the polling option on for you all. Think carefully and then solve. So here you need not evaluate these integrals individually. Maybe you will not be able to do that also. Correct. So you need to just compare these integrals depending upon the property that you have just now seen. Oh, sorry. Yeah. Sorry. I'm so sorry. Uh, it's five by six to five by three on. Yeah. Sorry. Slip off. So I've got a response. Very good. Good. Good. Good. Nine of you have responded. Almost two and a half minutes. Yes. Uh, can we stop this poll in the next 30 seconds? 16 of you, I could see you have participated in the poll. Five. Four. Three. Two. One. So I could see. Good. Good. I could see a 50 more than 50% participation. However, I was expecting more than 80. Most of you have gone with option number C, but even these closer, some of you have said D is the right option. Some most of you have said C is the right option. Okay, let's discuss it out. See, first of all, if you see the nature of the integrants involved are of the nature sign something by something, isn't it? Sign X by X sign, sign X by sign X or sign of tan X by tan X. Okay. So first you need to ask yourself what kind of a function is sign X by X. Now, if you remember when we did our limits topic, we had seen that the graph of sign X by X is a ripple like this. And of course, there is a whole here at zero. Okay. This place is pi because, okay. So between pi by six to pi by three, the function is a decreasing function. So this function is a decreasing function in the interval pi by six to pi by three. Correct. Yes. Decreasing function means more the input lesser is the value. Isn't it? Yes, I know. So can I say in place of X if I put a sign X. So which will be, which will have more value sign X by X will have more value or sign of sign X by sign X will have more value. Now you will say, sir, sign X has a lesser value as compared to X in that interval pi by six to pi by three. Isn't it? Isn't it? So pi by six to pi by three X, X graph is above sign X graph. You can check it out. Yes or no. So which of the two has more input? This one has more input. So this value will be lesser. Isn't it? Right? So that means this guy will be more agreed everybody because because sign X has a lesser value as compared to X. In the interval pi by six to pi by three. Isn't it? So more the input lesser will be the value of the function because it's a decreasing function. Isn't it? Decreasing function has a tendency that if you put more inputs to it, it will fall down. Isn't it? Means larger input lesser output. Similarly, can I say that? Can I say that sign of tan X by tan X will be lesser than sign of X by X. Why? Because tan X has more value as compared to X in the interval pi by six to pi by three. Correct? So tan X graph, no supersedes or basically it's above the graph of Y equal to X. Correct? So in light of this, can I make an inequality over here stating that sign X by X will be greater than equal to sign tan X by tan X will be greater than sign sign X by sign X. So is everybody fine with this inequality between the integrants involved? Okay. So using the property, we can say that so will be the integral also. So if you integrate it from pi by six to pi by three. Okay. This integral will be more than integral of sign of tan X by tan X. And in turn, this will be lesser than integral of sign of sign X by sign X. Okay. In short, what I want to say is I one is in middle. I two is the greatest. So this is your I two. This is your I one and this is your I three. So wherever the order says I two more than I one more than I three, I two more than I one more than I three. Yes. So this inequality will be taken into consideration. Is this fine? However, all these inequalities are actually pure inequalities. Other than you take a limiting case. But limiting case happens at zero, which is not in the interval. So you may do away with the impure inequality and state of pure inequality as well. That's also fine. Is it okay? Any questions, any concerns with respect to the solution of this problem? Okay. So we'll take one more last question on this property and we'll move on to another one. There are four integrals, which I'm going to give you I one, I two, I three, I four. Okay. Please note down these four integrals. The first integral is integral of two to the power x square from zero to one. Second integral is two to the power x cube from zero to one. Third integral is integral of two to the power x square from one to two. And fourth integral is integral of two to the power x cube from one to two. Okay. You need to state. True or false for the following statement, which I'm going to write. Okay. Statement A says I one is more than I two. Okay. Statement B says I two is more than I one. Statement C says I three is more than I four. And statement D says I three is less than I four. In fact, if you answer one of them, the other one will automatically get addressed. So let us start with a put your response like this a true or false, whatever you feel like. Okay. And try to give all your answers in one go. Put a, b, c, d and next to it, you put your true, false, whatever you feel it's going to be working out there. Oh yeah, you can also put, let's say t, f, f, t and something like this, whatever. I mean, the first one is automatically for a second one for B, third one for C, like that also. Very simple. I will not be giving you more than one minute for this. This is a touch and go question. Very good. Kinshukh, Archit, Kiran, very good. Vabha, Manikar, very good. Shruti, was that your answer or you still have to give your answer? Okay. Kiran, common sense. These two cannot be at the same parity. Both cannot be true or both cannot be false simultaneously. Correct? Think, okay, Aditi, okay, Kiran. Yes, last 30 seconds. I want everybody to participate. Come on. Should I name people? Gaurav, Gaurav, your participation has gone down. Arjun, Imanchu, Prateek, Shambo. Yes, come on. Yes. Okay. Should we discuss it now? Very good Prateek. Yeah. Thank you. Thank you so much for responding. See, it's very obvious between I1 and I2. See, remember, X is between 0 to 1. Right? So X square will have more value than XQ. Correct? So 2 to the power X square will have more value than 2 to the power XQ. Correct? So obviously I2 will have a lesser value than I1. So I1 will be more than I2. So this option is a true option. Okay. So by default, this will become false. Right? But what is the case between I3 and I4? When your X is between 1 to 2, please note that this time X square will be lesser than XQ. That means 2 to the power X square will be lesser than 2 to the power XQ. Okay. In that case, I4 will be more than I2, sorry, I3. So this will be true and this will, let me write it in white. This will be true and this will automatically become false. So the answer here is TFFT. Is it fine? Any questions? Any concerns that you have? Please let me know. Okay. Now the previous property, I'm going to slightly extend it and make the next property for you. So I'm going to now change the slide. If you have any question with respect to solution of this problem, do let me know. Okay. So property number two, property number two. Instead of sandwiching a function between two given functions, okay, I'm basically trying to scale up the property and saying that if there is a function, okay, f of X, which has a global minima equal to small m in the interval a to b and has a global maxima equal to capital M in the interval a to b. Okay. In short, what I'm trying to say here is that your function f of X is between small m and capital M. Okay. So you need not always figure out the two functions in between which your function f of X lies. Okay. So there may be a case, there may be a type of a problem which can be solved even by this property as well. So if you realize that a function has a global minima of small m in the interval a to b and a global maxima of capital M in the interval a to b. So treat as if your small m is like your g of X of the previous property and your capital M is like your h of X of the previous property. So just like the previous property, you can call it as an extension of the previous property also the integral of the function will basically lie between these values. Okay. Now these values have come from where very simple. This value is nothing, but it is the integral of m from a to b. That's it. M being a constant. So basically it becomes m times b minus a and this has come from integral of capital M from a to b. So again, capital M is a constant. So it becomes capital M times a minus b. Exactly. Exactly. Yes. Okay. So I think today there was a question from Aditya, I was asking integral of 1 by ln X, right? Aditya, I'm not very sure about the options because you had not given me the options completely. Okay. Okay. So now here if you see 1 by ln X function, okay, this function is a decreasing function in 2 to 3 interval. If you differentiate it, what do you get? If you differentiate, you get 1 by ln X the whole square into 1 by X. So in the interval 2 to 3 in the interval 2 to 3, 1 by ln X function is decreasing because if you see the derivative is going to be negative, okay? This is a negative quantity. Correct. That means the function will have a global minima at 2, sorry, at 3 and global maxima at 2. That means 1 by ln X function is between 1 by ln, ln 3 and 1 by ln 2. Correct. Or is it the other way around? It's a decreasing function. So its highest value will be at 2 and least value will be at 3. So I can say that integral of 1 by ln X dx will lie somewhere between 1 by ln 3, 3 minus 2, which is actually a 1 and 1 by ln 2, 3 minus 2, which is actually a 1 again. So this answer will lie between 1 by ln 3 and 1 by ln 2. I hope for most of you are already aware of these values. If not, you should always keep in mind, who will tell me ln 3 value? ln 3. No. You need to remember some value. ln 2, ln 3, ln 5, etc. ln 3 value. What power will you raise on e to get a 3? It has to be more than one year. Yes, it is somewhere close to 1.1. Okay. ln 2 value is 0.69 that you are already aware of. Isn't it? Okay. So from there you can make an estimation depending upon the question. Okay. So I hope this is helpful. Aditya to solve your question. So please proceed from here on. Let's take more questions on this. Let me take a question based on the same. See, all depends on the option. All depends on the option. You have to look at the option and then take a question. Okay. So let's take a question. The question is prove that integral of 5 minus x upon 9 minus x square will always lie between 1 and 1.2 prove that integral of 5 minus x upon 9 minus x square integral limits are from 0 to 2. This value will always lie between 1 and 1.2. A bit of global maxima minima or absolute extrema concept has to be brought in over here, which I'm sure all of you are familiar with done. Very good. Anybody else other than Siddish Pranav also done very good. Guys, let me also tell you one very important thing here. Many of the schools have not even started integration because of the term one exam. And I have a feeling that it is going to be tested for your KVP why so you have an upper hand with respect to your competitors for KVP why because you are very much trained in your integration skills. Okay. So it's very important that you capitalize on it and make an impact in the on the exam. Okay. So please make use of this advantage because many places the school has not even started with integration because they're very much doing your semester one topics. Okay. So let's discuss this. So let me call this function as f of x as of now. So let me find out first of all if there is any local extrema for this. Let's put the derivative as 0. Okay. So let's differentiate it. So 9 minus x square times minus 1 minus 5 minus x times minus 2x upon 9 minus x square the whole span. Okay. Put it to 0. So this gives you I'm just working out on the numerator part because that will be 0. So this will be negative x square. You will get positive 10x. You will get minus 9. This should be 0. Okay. This is as good as saying x square minus 10x plus 9 equal to 0. This is factorizable as this. Okay. So they are two critical points one and nine nine is of no use to us because our limit of integration is zero to two. Okay. So please do not waste your time going beyond the limits of integration. Okay. Now when I'm evaluating the global maxima or minima, what do I do? I check the value of the function at zero. I check the value of the function at any local maxima and of course I check the value of the function at two as well. Okay. At zero. This gives me five by nine undoubtedly at one. It will give me half because it will become four by eight to it will become three by five. Okay. Out of the three out of the three, the least is half. So this is your smaller. Okay. And out of these two, which is more, of course, which is more five by nine is more or three by five is more three by five is more, right? So this is your M. Okay. So in the light of this property, I can say the given integral will be lying between M times B minus a and capital M times B minus a. That means that I will lie between half into two and three by five into two. Which is actually one to six by five six by five is one point two. Okay. So keep this also in your mind in case you are met with a question, which requires you to use this. Okay. You don't always have to figure out a function between which or you don't have to always figure out two functions between which this function is lying. Okay. Maybe your global maxima minima also will do that trick for you. Next property number three. We are still in the inequality properties. Don't worry. So property number three is a very obvious property. It says that if you integrate a function between A to B and take its absolute value. Okay. This value will always be lesser than the integral of the absolute of the function from A to B. Okay. Now this is very obvious. People will say, sir, it can be proved. Let's say there's a function, which is, you know, taking turns like this. Okay. So this is your f of x function and you are integrating it from, let me just keep it away from the origin else become a very special case. So let's say we are integrating it from A to B. Okay. So as you can see, there is a positive part of this particular integral and there's a negative part. Okay. So overall, this answer would be lesser than the area that you would get had you taken a mod of the complete function before integrating. So something like this. So had you integrated it between A to B, the mod of that given function, this is your mod of f of x. All these areas would have been positive. Okay. So this area would be more. Can somebody tell me when will the equality hold true? When does the equality hold true? Anybody? Okay. When the function is completely positive or completely negative. Okay. In the interval A to B. So if the function is completely positive, that means above the x-axis, the mod of that area or mod of that algebraic area would be equal to the area between the mod of the function. Okay. So when it is completely positive or completely negative, that means it is wholly on one side of the x-axis. Okay. Is it fine? Okay. I mean, there's nothing important related to this property. Let's not bother about not very important one, but just keep keep a note of this property as well in case you require it. Next property, property number four. Let me take it to the next slide. If you want, I can give you a small question based on this. Integral of sine x by one plus x to the power eight from 10 to 19. This integral will always be lesser than 10 to the power of minus seven. No, nothing like that. Should there's nothing like that? Could go through origin as well. Special case when, special case, basically I asked the question, when will this equality exist? This is what I was asking, right? Oh, I said that if it is from A to B, let me not make A as a origin, A as a zero point. That's what I said. I mean, as per the diagram, I didn't want to make a very special diagram. That's what I said. Should we discuss it out? We have a lot of things to cover today. So can I say this integral from 10 to 19 sine x by one plus x to the power eight DX? This will be lesser than, okay. Integral of mod of this function itself from 10 to 90. Okay. As per the property that I just now discussed, okay, in light of this property. Now, let's talk about this guy's mod sine x by one plus x to the power eight. First, I would like to start with x to the power eight. Can I say x to the power eight? And I'm looking at the interval 10 to 19. Okay. Can I say this fellow will always be more than 10 to the power eight because the least value x can take is 10. Okay. So between 10 to 19, if I pick up any x, its value will be greater than equal to 10 to the power eight. Everybody agrees with this. In other words, can I also say this? Everybody's happy with this inequality, which I have written of this set of inequalities, which I have written. Now, let's reciprocate it. If you reciprocate it, the inequality is going to switch. Okay. So it's going to become this. Yes or no. Right. So basically what I want to say here is that sign x by one plus x to the power eight that will definitely be lesser than one by 10 to the power eight. So now just compare these two fellows. Okay. And I'm multiplying sign x over here. That means I'm making it even more lesser. So multiplication of sign x to one by one plus x to the power eight will make it even more lesser. So it will be a pure inequality over here. In other words, can I say this will also be lesser than this? So in short, you have got a function, which is basically lesser than this guy. So can I say this integral will be lesser than integral of one by 10 to the power eight from 10 to 19, right? Which happens to be 19 minus 10 into one by 10 to the power eight, right? And that comes out to be more or less nine into 10 to the power minus eight. And this fellow has to be lesser than 10 into 10 to the power minus eight, which means it is lesser than 10 to the power seven, because this is actually 10 to the power minus seven. So that given integral will be lesser than 10 to the power minus seven. Hence. So there are series of inequality, which I used over here. First I started with my property, which I discussed. Okay. And then I use my first property where I said that if one function is lesser than the other. Okay. And both of course are, you know, are integrated between the same limits, then even their limits, even their integrals will follow the same property. And then I further make it made it as 10 to the power this. Okay. Is it fine? Any questions, any concerns with respect to the solution? Can I move on to the next one? Please have a look at it. Please, you know, make yourself convinced. Then only allow me to move further. All right. So no objection. Anybody has very good. I think property number four under inequality. If f square X and G square X are integrable functions in the interval a comma B. Okay. Then the product of f of X into G of X integral modulus from A to B. This will always be, sorry, this will always be lesser than equal to integral of f square from A to B times integral of G square from A to B. All under root. This property is useful whenever your function involves some kind of square rooting or some kind of an irrational function, which can go away by squaring the function. And there is an inequality being asked to you in the problem. Okay. Not a very useful property. Many questions are not many questions have been asked on this property, but it's good to know it because in case you require it, you can use it. So we'll prove this property first. The proof is slightly weird. What I'll do is first of all, I'll take a function like this. Okay. Where lambda is some real number. Okay. So can I say since it is a perfect square, it will always be positive. Okay. So we already know that such kind of an expression will always be greater than equal to zero because you're squaring it's a perfect square. Okay. Now, if this function is completely above the x-axis, so can I say, can I say the area of this function between a to b or integral of this function between a to b? That will also be positive. Okay. Let's expand it. So if you expand it, you'll end up getting something like this a to b lambda square g square x dx minus two lambda a to b f of x into g of x dx and plus a to b f square x dx. So this will be greater than equal to zero. Okay. Now let's do one thing. Let's pull out this lambda square out because lambda is some real number. Now, if you see this entire left side expression, it is actually a quadratic in lambda and a quadratic in lambda is always greater than equal to zero. So when can a quadratic? So please recall your basic concept here of a quadratic inequality and is positive is only positive because is this is going to be a positive quantity because you know, it's it's you're integrating a positive function between a to b. Okay. So if a is positive, can I say this inequality will hold true only when the discriminant here is less than zero. In fact, since it is greater than equal to it will be less than equal to also because it may touch the x axis also. Correct. So in other words, b square minus four ac. Now please note who is acting like your small a who is acting like in fact, I should use my capital alphabet here because small a and small b are already in use in the question. So I'll make it as capital a x square b x plus c. This is capital a is greater than zero right is already true. So your discriminant should be less than zero. So what is the discourse being the role of capital a over here? The capital role is is being played by this term integral from a to b g square x dx. Capital b role is being played by minus two a to b f of x into g of x dx and your capital C role is being played by capital C role is being played by a to b. Let me write it in white only a to b f square x dx. Okay. So b square minus four ac that is your discriminant should be less than equal to zero. So b square is four a to b f of x into g of x dx the whole square. Okay. This should be less than equal to four ac. This is your c and this is your a drop the factor of four from both the sides. So now please note if you're taking an square root on both the sides, this will become more. So this will be less than. Equal to under root of this. Is this fine? Any questions? Any concerns? Any questions? Any concerns with respect to the proof one small question we'll take on this property before we move on to the next concept. So one last question I would like to take up on the inequality properties. We have already spent one hour on this property. So question is prove that prove that integral from zero to one of this integrant under root of one plus x. Okay. This value will always be less than equal to 15 by eight. This value will always be less than equal to 15 by it. Okay. Now as you can see when I'm giving you these questions and obvious question will arise in your mind that how would I know which of the property is going to help me? Is it going to be your limiting of the function between two functions? Is it going to be your maxima minima property? Right? Is it going to be the, you know, the mod property? Is it going to be the property which I just now discussed? So you have to do some trial and error. It is not like, you know, straightforward mechanism that okay, this is the way I have to solve it. So that is all about you figuring it out through reverse engineering. What will work? What will not work? And of course your options will be there to guide you. Oh, sorry, this will be under root of, yes, yes, why not? All chapters, numerical type questions will be there. Why your, definitely integral cannot give an integer answer. Hint is treat your f of x as under root one plus x. Keep your g of x as under root one plus x cube and use the previous property. Just write it done if you're done. So just recall the property mod of integral of the product. So when you integrate the product of f and g from a to b, take the modulus. This quantity will always be less than equal to. Okay. Product of the square of these function integral from a to b individually. So as per our given choice. Our integral of under root one plus x times one plus x cube. Okay. This will lie. So between zero to one square of under root one plus x is one plus x. So please integrate it separately. You all know how to do it. Square of under root of one plus x cube. Is going to be this integrated separately. I hope you all are comfortable. Sorry. I hope you are comfortable integrating these. So this is going to be, I'm just going to do it separately just to not make it very cluttered over there. So it's x plus x square by two from zero to one. That'll give you three by two. Similarly, one plus x cube DX. X plus x to the power four by four from zero to one will give you a five by four. Okay. So this will be less than three by two into five by four under root. This will be less than equal to under root of 15 by hence. Now we are not going to move towards a slightly challenging part of this topic, which is primarily for students. Who are for J main and J advanced exam. Okay. So can we move on to the next concept for this? Any questions that you have here? So now we are going to move towards two famous improper integrals called the gamma function and the beta function. Gamma function and the beta function. Okay. These functions are basically gamma function and the beta function. This functions are basically related to an integral. Okay. And. The integral is actually called an improper integral. What is improper integral? Many people ask me what is the meaning of an improper integral? The improper integral is where your limits are not proper. For example, something like integrating from zero to infinity. Okay. So these are. Or minus infinity to infinity. So if you realize that any one of your limits. Okay. They are not a definite number. They are basically an infinity minus infinity kind of a, you know, limit. Then these integrals are called improper integrals. So they are, they are two important types of. Or two important cases of improper integrals. Okay. You will understand when I, once I define these functions, once I define this function. So. Let us start with gamma function first. Okay. Why I'm discussing this concept is because many a times. You can use these functions to solve certain complicated and time consuming problems. Which otherwise will become very difficult and tedious for you to solve. However, this is a subject matter of a particular type of function. Okay. So let's start with gamma function first. Okay. Why I'm discussing this concept is because many a times. This is a subject matter of your undergraduate syllabus. It is not going to be taken in your 11th and 12th curriculum. It is basically taken in your first or second year, second semester of your engineering, you know, course studies. Okay. Yes. So, yeah, I'm just introducing this concept because in case you think your gamma function beta function concepts, is going to help you solve those questions and I'm going to derive a very interesting result also from the same, which is called the Wallace formula. And in the, in the previous years, people have found these functions to be useful in solving many problems. However, many of the things I will not be proving for you because it is not required. Or it requires some higher level calculus concepts, which is not being, taught to you, which has not been introduced to you in the past. Okay. So at some places I will say, okay, take down this result. It will be useful without actually proving it. Right. So let's get started with gamma function. First of all, I'll come to beta function little later on gamma function. So gamma function is basically a integral. Which looks like this. Zero to infinity. E to the power minus X. X to the power n minus one DX. Okay. As you can see, there is a parameter N. That's why it is basically a function of N. And this N has to be greater than zero. Okay. So this entire integral, when you evaluate it, it's going to give you an answer in terms of, you know, So this entire integral, when you evaluate it, it's going to give you an answer in terms of N. Isn't it? So that becomes that integral itself becomes a function of N. And N should be a positive quantity over here. Okay. It should be greater than zero. Okay. Then this result is basically called as gamma N. By the way, we don't use this notation F. What do we use? Instead of this, we use a symbol, which is opposite of factorial notation. I hope many of you have seen this notation for factorial. Don't get confused. This is factorial N. Okay. This is gamma N. This is to be read as gamma N. Okay. So don't, don't get confused between these two notations. However, this notation is a very old notation of factorial. If you would pick up a very old book of algebra, where they have discussed factorial, you would see that they've used this notation for factorial L. This is not an L. You can say this is a bracket. You can call it as a half of your, the ceiling functions that you have. Okay. Anyways. So don't make a mistake about this notation. Now this result, this result, this expression in fact is asked many at times to evaluate. Okay. So if you evaluate it, but if you know the result for it, it will make your life very simple. Right. For example, if I asked you this question, find, find zero to infinity e to the power x, sorry, e to the power minus x, x to the power five. Okay. How will you find this out? How will you find this particular thing out? You will use product rule. Correct. So you can use your di method there. Right. Treating your x to the power five as your, you and treating your x e to the power minus x as your B. Correct. So let us try to solve this question by using our integration by parts and see how lengthy was it when I'm solving it by integration by parts. Okay. So first of all, I'll just write my DNA. I will be using my di method. I hope everybody is aware of di method. Right. Especially in the case where one of the integral vanishes. So start differentiating it. Start integrating it. Correct. Attach positive negative signs and your integral without the limits. I'm not putting the limits as of now. Your integral definite, indefinite integral would be plus x to the power five minuses, which is minus e to the power minus x x to the power five. So basically I'm multiplying this multiplying this multiplying this multiplying this multiplying this. Okay. So this will give me minus five x to the power four e to the power minus x. This will give me minus 20 x to the power three e to the power minus x. Then I'll have minus 60 x square e to the power minus x. Then I'll have 120 x, sorry, minus 120 x e to the power minus x. And finally I'll have minus 120 e to the power minus x. Okay. Now in each of them you have to find the limit of integration from infinity or zero to infinity. Now for infinity you have to take a limiting case. Okay. So let's say this term I want to find out what is the value of this when x tends to infinity. Okay. So I'll have to take a limiting case. Can anybody tell me what will be this answer? What will be the answer of limit of x to the power five by e to the power x extending to infinity? Come on fast. It should be, you know, one second answer. Anybody. Right. Absolutely. Okay. See exponential function is infinitely more than a polynomial function when x is very large. Remember always remember polynomial divided by exponential and both of them have been, you know, let's say both of them are tending to infinity. Exponential function will be much bigger. Both are infinities. I don't doubt that. But exponential functions are much more than the polynomial functions. So overall this is and will be a zero. Right. So when you put the upper limit everywhere you'll get a zero. Right. Isn't it? This will also give you zero. This will also give you zero. This will also give you zero. This will also give you zero. Okay. And when you put a zero you'll realize that all of them will give you zero other than the last term which is going to give you minus 120. So the answer is actually 120 for this. Now let me tell you, we spend around three to four minutes doing this problem. But how about if I told you the answer to this, the answer to this will actually be five factorial. Right. What did make your life really easy? Now you must be wondering, sir, five factorial. How, how did you get that? Is it because it came out to be 120? You said five factor. No. Okay. So there is a set of results which I'm going to give you. And if you use them, you're going to save a lot of your time while solving these kinds of questions and questions related to them as well. Okay. So let us try to discuss this. I basically wanted you to show a demonstration that if you are going to solve this question like a normal aspirant who has not been introduced to gamma function, you are going to take a lot of time to solve such kind of questions, thereby eating up your precious time of your exam. Okay. All right. So let's discuss it out. Let's discuss it out. So how do I get this five factorial result is what I'm going to discuss with you all. So I'm first going to, I'm first going to use a reduction formula over here. I'm going to prove something very interesting. I'm going to prove a reduction formula applicable to applicable to gamma function. Okay. And what is this formula? This formula is gamma n plus one is n gamma n. Okay. Gamma n plus one is n gamma n. Okay. Let us prove this first. Proof is not very difficult. What is gamma n plus one? As per the definition of the gamma function, gamma n plus one will be zero to infinity e to the power minus x, x to the power n plus one minus one, correct? Which in short is zero to infinity e to the power minus x, x to the power n dx. Okay. Now what I will do, I will use my integration by parts on this. Okay. In fact, I should ask you to prove it because you have already gone through the reduction formula or the recursive formula. So if I use integration by parts on this, this is going to give me x to the power n e to the power minus x integral will be minus e to the power minus x. Okay. Limits of integration is zero to infinity. And then zero to infinity derivative of x to the power n will be n x to the power n minus one integral of e to the power minus x will be to the power minus x. Of course, with a, sorry, there was a minus sign over here. Okay. Of course, with a minus sign over here. So minus minus will become a plus. By the way, this is something which I already told you will be a zero. I hope everybody is convinced with that. Just now I discussed no polynomial divided by exponential upper limit will give you a zero if you put infinity and lower limit is anyways a zero because x zero to the power any number greater than zero will give you a zero. So now this term here is plus n times zero to infinity e to the power minus x x to the power n minus one. Isn't this expression gamma n. Isn't this expression gamma n. So can I not say that gamma n plus one is n gamma n. Yes or no. Yes or no. Okay. Now, if you look at the previous question that I asked you, I asked you to evaluate this integral zero to infinity e to the power minus x into x to the power five. Remember. Correct. Can I say this was actually gamma five. Sorry, gamma six. Is it because please treat this as n minus one. Okay. So I asked you for gamma six. Isn't it. So can I use this recursive relation and say this is five gamma five. Can I use the same relation and say gamma five is for gamma four. Can I not say it is three gamma three. So I'm just using the same relation so on on this I'm applying the recursive relation that I just out discussed. Okay. And then on gamma four I'm applying same property same recursive relation or same reduction formula. So if I continue doing it I get five into four three then this will be two gamma two and this will become one gamma one. Now let's discuss what is gamma one gamma one is something which we can evaluate easily. Gamma one will be integral of zero to infinity e to the power minus x x to the power one minus one which is as good as integrating e to the power minus x. Okay. What is integration of e to the power minus x infinity will give you a zero and putting a zero will give you a one. So this answer will be a one in short it becomes five into four into three into two into one into one which is actually 120 and that's nothing but five factorial as well. Okay. So this is the result which I got in the previous case but of course through a longer way by using our integration by parts. So it's not that I did not use integration by part here but I found out a recursive relation which will or which helped me to solve this question. So please note this down. This is very very important and just add to it that if n happens to be a natural number. Okay. Or in fact you can say a whole number in this case. No, let it be a natural number because then this will become a yeah. So if n happens to be a natural number then gamma n plus one actually boils down to n factorial. Okay. So this result is also very very important. Is it fine? Any questions? Any concerns? You can say so Shateesh. Related to factorial that's why it is having the same symbol. So Shateesh has a very careful observation and he said is it that the reason for this symbol to happen is because it is related to factorial. You can say so. Okay. Let's take a question on this. Let's take a question on whatever we have done so far and you will see how gamma function could be very helpful in solving those questions. No, it will not work if n is not great. This recursive relation will work if n is even not a natural number, but it has to be greater than zero. Why it will be greater than zero? Because if you see Aditya while proving this, I wrote this guy to be zero. This is only possible when n is greater than zero. Else it will not be possible. If n is a number which is negative, it will become an undefined expression even for zero. Okay. But yes, this recursive relation is to even if n is not a natural number. If n becomes an actual number, this gives you a special result that it becomes n factorial. Okay. I'll give you more information later on. After this question, I'll give you some more information about gamma function. Okay. The question is evaluate the value of the integral ln 4 to the power 5 times zero to infinity x to the power 4 by 4 to the power x dx. Evaluate ln 4 to the power 5 0 to infinity x to the power 4 by 4 to the power x. Let's see who gives me the result. See, you can always do it by integration by parts. You can treat x to the power 4 as your d. You can treat 4 to the power minus x as your v. Okay. And keep applying di and do it and then evaluate your upper and the lower limits by using your limits concept. But I would like you to use your result of gamma function to solve this question. Gamma function just saves your time, saves your effort from going into a prolonged version of di. Very good. Janadhan. Who is Janadhan? Hariharan only. No. Is it Hariharan? I'm wondering who is Janadhan. I know issues. That also looks cool, Janadhan. Okay. So I think we are already familiar with this definition. Okay. Now see, let us look at this integral. Very good. I think most of you have got it. Correct. Correct. Correct. Answer is 24. Correct. So in this integral x to the power 4, 4 to the power minus x. Now all of you first recall some basic log properties. Okay. So please recall. 4 to the power x could be written as e to the power ln of 4 to the power x. Which is nothing but e to the power x ln 4. So can I write instead of x 4 to the power x e to the power x ln 4 whole to the power of minus one. Yes or no. So this is e to the power minus x x to the power 4. Now remember e to the power minus ln 4 will also come into picture. Okay. Oh, sorry. I should have written that as well. e to the power minus x ln 4 x to the power 4 x. Okay. Now in order to make it as a gamma function, I have a small substitution that I would like to do here. I would like to call x ln 4 as a T. Okay. So dx will be 1 by ln 4 dt. x will be t by ln 4 also. So now e to the power minus x ln 4 is e to the power minus t x to the power 4 is t to the power 4 by ln 4 to the power of 4 because x was raised to the power of 4. Correct. And dt dx will become dt by ln 4. So or more the less I mean the power is going to become one more over here and it's going to become a 5. Anybody has doubt in this transition and limit of integration is not going to change because as x is 0 t will also be 0 and x is very, very large t is also going to be very, very large. Okay. Oh, no wonder they had given ln 4 to the power 5 outside. Oh, now I understood. So now the actual integral was ln 4 to the power 5 0 to infinity e to the power of sorry x to the power of 4 by 4 to the power x. Now become correct me if I'm wrong into 1 by ln 4 to the power 5 0 to infinity e to the power minus t to the power 4. Okay. This will get cancelled. Now this is clearly gamma 5. Okay. Don't worry about the variable name. It doesn't matter whether you call it as t or x. It's going to be gamma 5. No wonder I say if n is a national number, gamma n plus 1 is n factorial. So here I can write it as gamma 4 plus 1 and gamma 4 plus 1 will be 4 factorial answer is 24. Is this like any questions, any questions, any concerns now related to gamma function. I have some more information that I want you to take down. Okay. Then we will not go into details of important points about gamma function. We already seen that gamma n plus 1 follows a recursive relation. Okay. And if n happens to be a natural number, gamma n plus 1 will lead to n factorial. Okay. So there is a special value that I would like to discuss with you. Let me call this as property 1 itself. Okay. Gamma half. Gamma half is a value which is going to be very useful, at least in many of the questions that we are going to see related to Wallace formula. I'll talk about Wallace formula later on in this discussion, but gamma half is given a special value. In fact, this value, I will derive it also. Gamma half is under root of pi. That is to say 0 to infinity e to the power minus x, x to the power minus half dx. This value is given to be a root of pi. Now we'll prove it. We'll prove it in some time. Okay. Just make a note of this first. Now this proof is slightly overboard. It may go above your head also, but don't worry. I will try my best to make things clear to you. So in order to prove this, I will use first of all a small substitution. Let's call x as y squared. Okay. Let's let's substitute x as y squared. Okay. So when you substitute x as y squared, of course your dx becomes 2 y dy. Okay. Let's see what happens to the entire integral. So this integral becomes that is gamma half becomes 0 to infinity. Please note when x is 0, y is also 0. X is infinity. Y is also infinity. e to the power minus y squared into this term is y to the power of minus one and there's already a 2 y dy. Okay. So this y and y to the power minus one will get cancelled off. Okay. So this is going to give you, this is going to give you twice of 0 to infinity e to the power minus y squared dy. Correct. Okay. Now can I say there's nothing in the name. I can write this back to two times 0 to infinity e to the power minus x squared dy. Sorry x squared dx. Both will be gamma half only. Okay. Let's multiply these two. Let me call this as one. Let me call this as two. Let's multiply one and two. So when you multiply one and two, you end up getting gamma half gamma half square as four times 0 to infinity e to the power minus x squared dx into 0 to infinity e to the power minus y squared dy. Okay. Now when you're integrating, I know such case, you learn it little later on in your undergraduate that you can write it as a double integral like this. Okay. So you don't worry about the proof. It is not going to be asked anywhere. It is just for your extra information, maybe in your first year of undergraduate, you'll study this and then you'll realize that, you know, we were told all these things in our class 12 itself. Okay. Now here is a concept of converting a differential area from Cartesian coordinates to polar coordinates. So please note this down. dx dy in terms of polar coordinates is given as r dr d theta. Now many people ask me, sir, how do you get this result? So there are two ways to get this result. One is by the you please, you know, see, let's say this is your small patch of area on your Cartesian coordinate system. Let's say this is your dx. This is your dy. Okay. Let's try to relate it to a polar coordinates. Okay. Let me make it slightly tilted. Okay. Let's, let's keep it like that. Okay. So let's keep it, you know, approximately, let's, let's call this length as, let's call this length. I'll make it slightly tilted so that it fits in the diagram. Okay. Just slightly tilted. Okay. So let's, let's call this as r. This length as dr. Okay. And let's say I call this angle to be a d theta angle. Let's say this angle was theta. Okay. So this dx dy, you know, length can be approximated as the product of dr that is this length. Okay. Into this length. Now this length is approximately r d theta. Okay. So that is how this solution actually comes. Little later on, you will learn that there is a concept of Jacobian that will be taught to you in your undergrad. From there, you can get these kinds of relations pretty easily. So this is a relation which, which relates differential area in the Cartesian to the differential area in the polar. Okay. So here what I'm going to do, I'm going to put my X as r cos theta and Y as r sine theta. That means in short, I'm going to write X square plus Y square as an R square. Okay. So this entire integral, let me write it down here. So this entire integral, which is gamma half the square, which is this, this entire integral becomes zero to, now I will not put the limit of integration as of now. I just first convert everything in terms of r and theta. Okay. So what I did, I replaced my X square plus Y square as an R square. Correct. And I replaced my X, DX, DY as RDR d theta as I already discussed with you. Okay. Now don't worry too much about getting into the details of this because anyways, this will be a subject matter taught to you in your engineering course of mathematics. Okay. Now what happens to the integration? Now see you are integrating things from zero to infinity, both in X and Y. Okay. So can I say R will go from zero to infinity and since Y is going from zero to infinity, theta will go from zero to pi by two. So remember R will go from zero to infinity and theta will go from zero to pi by two. That's how you will cover everything from zero to infinity in X and zero to infinity in Y. So basically you are covering everything in this zone. Correct. So your R can go anywhere from zero to infinity and theta value can go anywhere from zero to pi by two. So this limit of integration, you can change it from zero to pi by two and zero to infinity. Zero to pi by two is meant for your angle which is just d theta and zero to infinity is meant for this guy. Okay. So this is going to be gamma half the square. Okay. We are almost nearing the end of our discussion. So this is plain and simple. This is going to be four into theta from zero to pi by two will be pi by two only. This integral you can perform separately. Let me perform that integral separately. Zero to infinity e to the power minus r square r dr. You can take r square as a T. So two r dr will be dt. So this is going to become zero to infinity e to the power minus T and this is going to be dt by two. Correct. So this is half e to the power minus e to the power minus T zero to infinity. Infinity will give you a zero. Zero will give you a minus one. So this will become a half. Okay. So now the four and the half will get cancelled off. So what did we get here? We got here. Gamma half the whole square is pi. So gamma half is going to be a root pi. Hence proved. Okay. Again, I am not too worried about you understanding the proof. I'm worried about you keeping this result in your mind. This result itself is important. Yeah. This is going to help you in solving many questions which will be coming for your J exam. Okay. So keep that in mind. That's it. Nothing. Don't worry about getting into your conversion from differential area in Cartesian to differential area in this thing. All these double integration all that is going to be taken care later on. Is it fine? Why did we switch it? Because here we were at a dead end. We could not integrate over here. So in order to facilitate further integration from where we had to switch from Cartesian to Polar. How did we apply the limits? Zero to pi by two you are asking and zero to infinity. So there are two variables involved. One is R, which is going from zero to infinity because you have to cover everything in the first quadrant, isn't it? So in order to access all the points in the first quadrant, your R has to go from zero to infinity. And your angle, the argument of the, you know, the any point has got R and theta, right? So that theta value will sweep all values from zero to pi by two. Understood. How did I write zero to pi by two and zero to infinity? Okay. Now again, don't worry too much about it. We have very less time and we have more things to cover. So just write this result and let's move on. Trust me, these things will not be tested in J. Because J people know that you, I mean, you are not trained. You will not be trained till this extent that you know all these, you know, double integrals and all. Which last step? This step, blue one. E to the power minus T. Anyways, this is gamma one only, you know, zero to infinity, e to the power minus x, x to the power one minus one. This is gamma one only, right? Gamma one will be zero factorial. Zero factorial will be one. So it is one into half. I say we supposed to, I mean, so many ways to think of it. Okay. All right. So please note this down. Now we are going to the next important property. Again, I will not be proving them because the next property is, I should call it three. Next property is no down. Gamma n, gamma one minus n is pi sine and pi. Pies by sine and pi. Or pi and whatever you want to call it. Okay. Where n has to be here in this case, n has to be between zero to one, right? Gamma for negative values is not defined. Even gamma zero is infinity. So please also know that gamma zero is infinity. Okay. It's not defined. So we'll be restricting our n here from zero to one. This is a very interesting, you know, property of gamma function. And this is given a special name because Euler basically proved it. Okay. And this is called the Euler's reflection formula. I will not be proving it. Because it requires your understanding of a little bit of, you know, Maclaurin series and Vistras inequality from where Basel's formula was proved. But I don't want to get into all those things. Just keep this in your mind. Gamma n, gamma one minus n is pi by sine pi n. n has to be between zero to one. Okay. Just note it down. Last year I tried, I proved it, but you know, it was it went above the head of many of the students. So it is not important. Let's not waste time doing this. So what is not important? We'll not invest our time. In fact, I will take a small question and we'll try to do it a little later on. Not as of now. Okay. Let's do it a little later on once we do beta function. Yeah. Pi seems to be so irrational, but it is so well organized also at some times. Absolutely. Alpha no. The next property is something which, you know, is very important for all of us. Okay. So in fact, I will talk about it. I will talk about it. Through beta function. So we'll talk about beta functions first and then I'll come to this. Okay. So all of you have noted this down. So I'll be starting with now beta function concept. Beta function is basically a function which is bivariate, which means it has got two inputs, unlike gamma function, which only has n as its parameter. Beta function is a, you can say by parametric function where there are two parameters m and n. Okay. M and n both are supposed to be positive. Okay. So beta function is defined in several ways. I will give you one of the ways it is defined as integral from zero to one x to the power m minus one one minus x to the power n minus one dx. Okay. Now many people say, sir, you said it is supposed to be an improper integral, but your limits are zero to one. How it is an improper integral. Don't worry. I'll give you more versions of beta functions. In fact, as a question to prove. Okay. There you will realize that beta function is actually an improper integral. Of course, right now in the present definition, it looks like to be having limits, which are very well defined zero and one, but it can also written as an improper integral. So we'll discuss it out through question. So no doubt m and n here are positive values. Beta m comma n is defined as integral zero to one x to the power m minus one into one minus x to the power n minus one. Okay. Now question for all of you. Prove that prove that number one beta function can also be written like zero to one x to the power n minus one one minus x to the power m minus one. In short, in short, your m and n positions are interchangeable. That means it doesn't matter whether you write m comma n or n comma n. The result is not going to change, but this is something which you have to prove right now. Okay. Second thing you have to prove that beta m comma n could also be written as an improper integral. Okay. And of course, from the previous property that you would have already proven, you could also write it like this. As you can see, this structure is now become an improper one. That is why we categorize beta function as an improper integral. So these two questions themselves are basically conversion of one integral to another. So that can, this can come as a question to you in your exam. So examiner knows that, okay, the student might not be aware of a beta function, but can he do this conversion because this requires his basic knowledge of definite integrals, or basic properties of definite integral. So this potentially can be a question to you where the question setter will say, okay, this is a integral. He will not say beta anon. Don't worry. They are not going to use technical terms, which you might not have heard of. You'll say, okay, this is an integral. The first integral is which of the following and he will give you an options in terms of these integrals themselves. So this becomes a conversion of one integral to another integral type of question. So let's prove it. The first one is very easy to prove. I am sure you would have already proved it. How do you prove the first one? A on King's property. Yes. Done first one. So you get a comprehension based question also, right? Yes. Yes. Why not? But of course comprehension, even if they ask it, they will not go to the concept of testing your idea of double integral, whether you can actually do double integral or not. Double integral, triple integral related concepts will not be tested. But yes, they may give you a concept where they put it as a comprehension, a new concept as a comprehension. So first one in the interest of time, sorry, I'll be intervening over here and doing it quickly. So our initial definition of beta function was 0 to 1 x to the power m minus 1, 1 minus x to the power n minus 1 dx. So now use KP, use your King's property. So from your King's property, you can say it is 0 to 1, 1 minus x. Remember King's property is 0 to a f of x dx, can be written as 0 to a f of a minus x dx. So this will become 1 minus 1 minus x to the power n minus 1. So on slightly simplifying it, you'll end up getting 0 to 1. Now this is just an x. And this is actually by the definition, it is beta n comma n. So please note that in beta function, it doesn't matter if you interchange m and n positions. It's not going to affect your result. Now the second one, the B part is important because that can be asked as a question. In fact, I will do it in the next slide, but first of all you do it and let me know by a done that you're done with it. Okay, should we take it up? Okay, so let's take the B property on the next screen. So we need to prove that beta m comma n, this could be either written as 0 to infinity, x to the power m minus 1, 1 plus x to the power m plus n. Or it could be written like, any one of the two you prove it's more or less the same because m and n positions can be interchanged. Okay, so let's do a small substitution in this. So we need to prove this. So in this field do a small substitution. So all of you please pay attention. Here I'm going to put my x as 1 by 1 plus y. Now why did I choose the substitution? So I choose the substitution for two reasons. You will see here that when x becomes 0, okay, what happens to y? When x becomes 0, y will become infinity. Correct, y will go to infinity. Okay. And when x becomes 1, what will happen to y? So put your x as 1 and see what will happen. So 1 plus y becomes 1, so y becomes 0. So the limits of integration are at least taken care of. And not only that, you'd also realize that 1 plus x kind of an expression which we were desiring to get, that also is addressed. Let us see how. So when I put these substitutions, your lower limit will be infinity, upper limit will be 0. And this will be 1 by 1 plus y to the power of m minus 1. 1 minus 1 by 1 plus y to the power n minus 1. dx will become minus 1 by 1 plus y, the whole square dy. Can I say this will become your beta m comma and correct me if I'm wrong. Any questions, any concerns to highlight? Now, first of all, I would address this negative sign over here. I don't want to see any negativity. So I'll interchange my upper and lower limit. So here if you see this part, this is actually y by 1 plus y. So you end up getting y to the power n minus 1. And down here, you will get 1 plus y to the power m minus 1 from the first guy. From the second guy, you'll get 1 minus y to the power n minus 1. And from the last guy, you'll get 1 plus y to the whole square. In short, it becomes 0 to infinity, y to the power n minus 1 by 1 plus y to the power m plus n. I hope that's very obvious because m minus 1 plus n minus 1 plus 2 is m plus n. Okay. Now there's nothing in the name. Nam me kya rakha hai? Sab kuch kaam me rakha hai? Isn't it? Everything is in your skills, not in the name, isn't it? So now we know that. We also know that there is nothing in the changing of the name of m and n. So basically you can also write it as 0 to infinity x to the power m minus 1 by 1 plus x to the power m plus n. Okay. Hence proved. So there are four notations for beta functions. So what is it? The first notation was 0 to 1 x to the power m minus 1, 1 minus x to the power n minus 1. Then in that you change your m and n position. That is second notation. Third notation is 0 to infinity x to the power n minus 1 by 1 plus x to the power m plus n. There also you change m and n position. So fourth notation is obtained from this altogether for notation for beta function. Two of them are in improper version. Two of them are in proper version. Now, the most important connection, the most important property is the connection between beta function and gamma function. So many people say, sir, beta function, gamma function, are they related? Yes, they are related. They're very much related. And that relation itself is a very important relation for us. But I will not prove that relation because it requires, again, double integration and all, which is waste of time as of now for you. Just know the relation because that will be useful. So note down this. Any question, any concerns? Okay. So one of the most important is the property is the relation between between beta function and gamma function beta function. Beta function can actually be written in terms of gamma function. How this is the relation beta m comma n is note down. Gamma m gamma n by gamma m plus m. And here also you can see that symmetricity between m and n is maintained. So even if you switch your m and n position, this value is not going to change. Okay. Proof not required for this. Proof basically goes by double integration. If you want, I can send you the proof on the group. Okay. So if you want, I can send you some proofs of this on the group. Okay. So please note this down. Another thing that I forgot to tell you, in fact, there's another notation of beta function. Okay. And that is not a difficult notation because anybody can figure it out by the use of technometric substitution. So we already learned that this was our expression for beta function beta m comma n. Here put x as sine square theta. Okay. So DX will become two sine theta cos theta d theta. So make the substitution over here. So your beta m comma n will become zero two. Now when X is one, this is going to be pi by two. Okay. This is going to become sine to the power two m minus two theta. This is going to become one minus sine square to the power n minus one, which is cos square to the power n minus one. Okay. And DX will become two sine theta cos theta d theta. Right. So here is another expression for beta function beta function can be written as zero to pi by two twice off sine to the power two n minus one theta cos two n minus one theta. And remember m and n positions can be stopped. That's a property which we all know. So this is another fifth, fifth representation for beta function. Okay. Now we will play with this, you know, expression little bit. First note this down. So please note this down. Now what I'm going to do is I'm going to, I'm going to write this beta function as gamma function, gamma m, gamma n by gamma m plus n. Okay. This is twice off zero to pi by two sine two to the power m minus one theta cos two to the power n minus one theta. And I'm going to do a small change here. I don't like to see the shape of this power. So what I'm going to do, I'm going to call two m minus one as m dash. Okay. I don't like the way it's written as ugly power. I'm just writing it as a simpler power. Okay. Let's say I call it as m dash and I call that as n dash. Okay. So when I make these substitutions over here, it'll become twice zero to pi by two sine m dash theta cos n dash theta d theta. This is going to become now, please note down that your m is going to be m dash plus one by two and your n is going to be n dash plus one by two. So this will become gamma m dash plus one by two. Gamma n dash plus one by two. All divided by gamma m dash and dash plus two by two. In fact, I will do one more step over here. I will send this to down over here. Okay. Right. So now I got a useful result. Okay, so this is a useful result. This says, if you want to integrate sine to the power m theta. There's nothing in the name. So I'm putting back things in terms of m and m. You can write it as gamma m plus one by two, gamma n plus one by two, by two gamma m plus n plus two by two. Okay. So please note this is a down whether you know the derivation or not. That is immaterial. You should know this result down. And if m and n happens to be whole numbers, then, then this formula is called the Wallace formula Wallace. Okay. So we'll see some application of this formula in one of our questions. Now this is one way to solve the question. I'm trying it that if you don't know this approach, you will not be able to solve the question. No, don't wait me wrong. I never mean to say that if you don't know this, you will not be able to solve this. But Wallace formula is just a mechanism by which we make our life simple while solving questions related to this or like this. Okay. So let's take a few questions. Let's take a few questions. Do you want me to summarize everything? I can do that. That one page is good enough for you to keep in mind. So let me summarize everything. People who want to note it down, note it down. Then in the next page, I will summarize everything. Don't worry about it. Okay. So let's summarize. Summary of beta gamma, beta and gamma function, summary of gamma and beta functions. So whatever I'm going to write in the summary, please remember only the definition of your life will be easy. Okay. First of all, the definition of gamma function itself. Gamma n is defined as zero to infinity e to the power minus x x to the power n minus one DX. Second thing, gamma function satisfies a recursive relation. Gamma n plus one can be written as n gamma n. And if n happens to be natural number. Gamma n plus one becomes n factorial. Okay. Some special things value should remember gamma zero is infinity. Okay. I already know gamma one is going to be one. Gamma two is also going to be one because of this property, which I just now wrote. Okay. Gamma three is going to be two factorial. Gamma four is going to be three factorial and so on. You already know this. And a special value gamma half under root pi. Okay. And Oilers reflection formula is something that you should note down. Oiler reflection formula says gamma n, gamma one minus n, where n is somewhere between zero to one is given as pi by a sign and pi. Okay. This is your Oilers reflection formula. Okay. Related to beta function. Beta function. Has the following definitions zero to one X to the power m minus one one minus X to the power n minus one remember m and n they are our positive quantities. It could also be written as X to the power n minus one one minus X to the power m minus one DX. That is to say beta m, gamma n and beta n, gamma m are same. So the positions of m and n can be interchanged apart from it. I gave you some improper integral notation for beta function, which is this. And since m beta m common is same as beta n comma m. This is also true. And of course towards the end I gave you a trigonometric relation, which is zero to pi by two sign to the part to m minus one theta. Okay. Five relation for beta function. And I also gave you the connection between beta and gamma beta m comma n could be written as gamma m, gamma n by gamma m plus n. And in light of these two properties, we ended up getting the famous Wallace formula that says integral zero to pi by two Why am I writing it? Higher level. Let's write it zero to pi by two sign to the power mx cost to the power nx can be written as gamma m plus one by two gamma n plus one by two by twice soft. Just add these two. It'll become m plus n plus two by two. Okay. So if m and n happens to be natural numbers. Okay. So basically, in fact, whole numbers, not, not necessarily natural numbers, whole numbers. It becomes a case of Wallace formula. Okay. Summary of whatever we have learned. Main thing is questions. Main thing is questions. Okay. So first I will take some examples of Wallace formula. Oh, sorry, if you have not noted it down. Sorry, I just changed the slide without asking you. Let's take a few questions and then we'll go for a break. Okay. After the break, we'll do a definite integral as limit of a sum and limit of a sum as definite integral. Okay. That is going to take not more than an hour or so. Can I go to the next slide? So let's take this question. Question is integrate zero to one x to the past six under root of one minus x squared. Integrate zero to one x to the past six under root one minus x squared. Anybody? So this is what I say knowing the formula is not important. Knowing the application of those formulas is important. See, how will you, you know, approach this problem normally you will say, okay, let me substitute excess science data. Right. This is the normal substitution that everybody will think of. Isn't it. So this entire problem will become zero to Pi by two. Sine to the power of six theta. This will become cost theta and sine theta derivative. That is also cost theta d theta. So if I'm not mistaken, it will become cost square theta d theta. Now, if you recall this structure is very much same as what we had given for our beta function. So this can be written as, please note this down. Oh, sorry. Okay. This can be written as gamma m plus one by two. Gamma n plus one by two by two gamma m plus n plus two by two. Correct. So in light of this, this will be gamma seven by two. Correct me if I'm wrong. Gamma three by two and two gamma five. Isn't it. Now, gamma five, most of you will say, sir, I know gamma five. Gamma five is four factor 24. But what about gamma seven by two? You never told me gamma seven by two. Okay, yeah, I don't have to tell you everything. See, use your recursive relation. Gamma seven by two can be written as five by two gamma five by two. Correct. Remember gamma n plus one can be written as n gamma n. And this gamma five by two can be written as three by two gamma three by two. And this gamma three by two can be written as half gamma half. And what is gamma half? Gamma half is root pi. So why do we need to know the value of gamma seven by two separately? Isn't it. Similarly, gamma three by two, as I already told you is half gamma half. So it'll become half under root pi. So overall, let us simplify the result. So I think you'll have 15, not 15. I think this will go off by a factor of eight at least. So five pi by one, two, three, four, five, six, seven, eight. What is two to the power eight? Two to the power eight is 256. This is your answer. Now, just imagine what would have been the, you know, your state. Had you not known this formula and you would have tried to solve this integral by your, let's say your complex number, substitution or, you know, your trigonometric identities. Will it be as simple and as straightforward as this? No, you don't have taken you at least full five minutes to solve the question. Is it fine? Any questions? Any concerns? Okay. Now, one straightforward question I'll give you. Those who want to copy this can copy this. Those who want to attempt the next question, please do so. Finds sine to the power five theta into cost to the power of let's say four theta. Now, this can easily be done by substitution. I'm not denying it. Okay. But let's try to use our Wallace formula to solve it because here our powers are all natural numbers or whole numbers to us to say. Okay. Substitution will work. I'm not denying it. You can substitute cost theta as a T and solve the question. No problem in that. Okay. But let's use our Wallace formula also since we have learned it new. So what will be the expression? Tell me. Okay. This is gamma five plus one by two gamma four plus one by two by two times gamma. Add this added to so 11 by two. Okay. So let's see how fast or how slow is it? This is going to be gamma three gamma three is two. This is gamma five by two gamma five by two is three by two into half root by. And this is gamma 11 by two gamma 11 by two will be nine by two seven by two five by two three by two into half gamma half, which is half root by. So out of this, you can just cancel off to this whole thing to and this whole thing. The answer will be plain and simple eight upon how much is this 63 into five 63 into five is 195. Is it fine? Any questions? Any concerns? Oh, sorry. 63 into five is 350. Good enough. Understood the use of Wallace formula, which comes from your beta gamma function relation. Okay. One last question I'll give you and then we'll just go for a break. One small question I would like you to solve. Can I go to the next slide? Okay. Solve this question. Integrate zero to infinity. Dx by. One plus x to the power four. Nobody. Okay. Let's let's put x to the power four as a Y. Okay. So four x cube DX. Okay. Let's do a simpler activity here. Let's write X as Y to the power one by four. Okay. So DX will become one by four. Y to the power minus three by four. Let's make this substitution. Please note the limit of integration is not going to change because of this substitution of X in terms of Y. So DX will become one fourth. Y to the power minus three by four. And this term will become one plus Y. Okay. Now try to recall something very familiar to us. Beta m comma n can be written as zero to infinity. X to the power of m minus one. Okay. One plus X to the power m plus n. If you compare this, your m will become one by four. Okay. And n will become three by four. Isn't it? If you compare this expression to this expression, your m value is one by four and your n value is going to be three by four because m plus n is going to be actually a one. Right. In short, this expression that you're evaluating is one fourth of beta one fourth comma three fourth. Okay. Let us try to let's write it down in terms of gamma function. So it's gamma one fourth. Gamma three fourth gamma m plus n m plus n is actually one. So that leaves you with one fourth gamma one fourth gamma three fourth. But you'll say, sir, I don't know gamma one fourth and gamma three fourth. You only told me about natural numbers and you only told me about, you know, fractions, which are like multiples of half. Okay. Now here, nothing to worry. If you look at it very carefully, it is like gamma n, gamma one minus n. Correct. And is a number here, which is between zero to one. That's one by four actually. Which formula will help you here? Which formula will help you here now? Write down in the chat box. Oilers reflection formula. Okay. I'll give you pi by sine pi n. So what's the answer? Pi by four. What is sine pi by four sine pi by four is one by root two. So the answer to this question is pi by two root two. So I wanted to take this question just to show you how oilers reflection formula can also be required to solve these kind of problems. Okay. Anyways, we'll take a break as of now. It's already going to be 1130. Right now it's 1123 a.m. Let's meet exactly at 1135 a.m. Okay. We'll have to complete two more interesting concepts related to definite integral, definite integral as limit of a sum and limit of a sum as definite integral. Okay. Enjoy your break. See you in 15 minutes time. So in this part, we are going to discuss how to write definite integral as limit of a sum. Okay. So we are going to look into two aspects. Definite integrals as limit of a sum. Don't worry. This is the basic ab initio method of finding a definite integral just like for derivatives, derivatives. There used to be first principles in the same way for definite integrals. If you want to evaluate it, you can evaluate it by using the concept of limits. Okay. So how do you express definite integral as a limit of a sum and vice versa? Okay. So we are going to, you know, study that to, you can say directional conversion. One is definite integrals as a limit of a sum and limit of a sum as a definite integral. Okay. Now this concept is basically related to your geometrical interpretation of a definite integral. So when you are integrating a function, let's say f of x from a to b. Okay. What are we finding? We are trying to find out the algebraic area between the function and the x-axis, isn't it? So first of all, I'm going to, you know, break this entire function into small rectangular strips. Okay. I'll not be drawing many. I'll just do a dot, dot, dot. Okay. So I'm just making this entire area into thin rectangular strips, each of thickness h. Okay. So this thickness is h and there are n such strips. So this is your first strip. So this is a plus h. Okay. So I have broken down into n strips. Okay. Please note that this is going to be a plus n minus one h. Okay. That means a plus n h is actually equal to b. Right. So what I've done, I have broken down this entire area into n strips of width h. Okay. Right. Now we all know that if n becomes infinitely large. Okay. If n becomes infinitely large, that means your h is almost tending to zero. Then this integral or the sum of the area will basically give me the integral of the function from a to b. Right. Now here a point to be noted is n is tending to infinity and h is tending to zero, but n h is still a finite quantity, which is b minus a, which comes from this. Okay. So please understand zero into infinity doesn't mean zero. Right. It could give you a finite quantity and that quantity in this case is b minus a. Right. So what are we doing here? We are basically trying to evaluate an integral by adding the area of all these small, small, small, small strips. Let me start with, let me start with this strip first. What is the area of this strip? You will say sir, it will be h into f of a, isn't it? h being the width. So this is your h and this height is going to be f of a. Okay. Similarly, what will be the area of this strip? You'll say h into f of a plus h. Similarly, what will be the area of this strip? So you'll say h into f of a plus two h. Okay. And this goes on and on till I reach the last trip over here. Okay. And the area of this trip would be h into f of a plus n minus one h. Right. If you take a limiting case of h tending to zero or n tending to infinity, then this particular limit should give you your desired integral. Right. So this formula, please note it down. So this is a formula by which we can find out integral of a function from a to b by converting it as a limit question. Okay. Now don't worry. I will show you some examples by which you can evaluate it. However, I have been told that in school, they have not taught this part and it is more important from school point of view. So I will not be spending too much time. Yes. It's the basic. The basic. So I am going to put it as an ab initio method or the first principle method of evaluating a integral. So in the school, they have not taught this right. Have they? No. So I will not spend too much time because of course in J you will not be asked to convert it to a limit of a sum and evaluated, but this is for your overall knowledge. My purpose is to give you a complete knowledge, whether it is asked or not. That's a different thing. prerequisites which you already know, but I just like to recall it for you prerequisite is you would require to know your following summations and limits. I think you are already aware of it from your class 11 days summation of all natural numbers from one to N is this summation of squares of natural numbers from one to N is n n plus one to n plus one by six summation of two the whole square apart from it you should also be remembering your sign series I don't know how many of you still remember it sine alpha sine alpha plus beta sine alpha plus two beta sine alpha plus n minus one beta okay so this is it is given by who remembers this is it sine n beta by two by sine beta by two into sine of alpha plus n minus one beta by two similar result for cost as well so I'll write it down here. This is what we used to call it as cost series sine series cost series. This is given by sine n beta by two I'll write it slightly small because I'm just at the corner of the page. Okay, so the same formula is just that now you have a cost over here so this formula and this formula the way to remember it is this part will be same for both of them. Okay, and this part will also be same for both of them is just that instead of a sign over here a cost will come and of course apart from it some limits that you have already learned you know sine x by x extending to zero is a one. Or you can say ln of sorry e to the power x minus one by x is, you know, one, etc. Okay, so limits with limits and your basic special series and your trigonometric series, these are the some prerequisites you require. Okay, we'll take one or two questions not more than that because the other part is more important converting limit of a sum to a definite integral that is more important. So anything that you would like to note down from here prerequisites you already know maybe you can note this down properly so that there's no mistake in copying this. Don't worry n is tending to infinity, h is tending to zero but your n h will be equal to normally n and h looks the same. Yeah, and it will be equal to b minus a. Okay, so all the three will be useful to solve a question. Why there is an edge at the end. Which end you're talking about this edge. Okay, let's take a question very simple question. Let's say I ask you, what is the integration of e to the power x from one to three. Now I know you know the answer. Okay, but I have to solve this as a limit of a sum. So how do I solve it. So what do I'm going to do. First of all understand what is your f of x. What's your a, what's your b. Okay, what's your n h, n h is b minus a which is two. Okay, now start using this formula limit. h tending to zero or n tending to infinity or n h equal to b minus a. You can write all the things below this limit if you want. h into f of a f of a plus h till f of a plus n minus one h. Okay. f of a f of a will be e to the power one f of a plus h is e to the power one plus h. And then one plus two h and so on till one plus n minus one h limit h tending to zero and tending to infinity. So these all are there at your disposal. I mean, depending upon the situation, you know, we are going to use it. Now mostly I've seen we require this guy more and this game. Okay, this is not that needed. Okay. So if you see it's actually a geometric progression with first term as e common ratio as e to the power h, right? So total number of terms is n, but n h tending to b minus a b minus a here is a two. Okay. Now all of you please pay attention. If you see this term it is going to be e to the power n h minus one by e to the power h minus one. Okay. Now wherever an n h is formed, you can easily replace that with a two. So as you can see here, there is a n h getting formed. So you can replace that with a two. Okay. So it becomes h e q to minus e by e to the power h minus one limit h tending to zero. Now there's no n required because n is already gone. Okay. So only h tending to zero is helpful. Now what is this limit? e to the power h minus one by h, h tending to zero or reciprocal of that. That is anyway is going to be a one. So this entire answer becomes e q minus e, which you already know, which you already know. And this is how you basically evaluate your definite integrals by using first principles. Ideally, I should have taught this in the beginning of the chapter, but I thought once I'm doing the limit of a sum as definite integral, I will introduce this concept for you. Is it fine? Is the process clear to everybody? Okay. One more question I will ask you to do it and then we'll move on to the more important version, which is converting limit as a definite integral. That is more important for us. Yes, and his term will always appear somewhere wherever it appears, you have to immediately plug it up with B minus it. Let's take another one. Integrate cos x from a to b by first principles or by ab initio method. Last problem of this type just to give you an idea so that you are aware of the process how to write definite integral as a limit of a sum. Please note that complicated problems cannot be solved by this method. This is just as a beginner, you should know that this is the way to solve it by converting it as a limit of a sum. Okay. So here n h is B minus a for sure. Of course, h is sending to zero and is sending to infinity. So as a limit of a sum, it is h cos a cos a plus h cos a plus 2h. Now, in order to take care of this part, we have already discussed our cosine series. So I can write it as sine. Now remember alpha is a and our beta is h. So sine n beta by two is sine n h by two. Okay. Into cos alpha plus n minus one beta by two. Okay. Remember, h is tending to limit h is tending to zero. Okay. So n is tending to infinity and n h is tending to B minus a. Now, wherever there is a B mine n h coming up, please write it as a B minus a. Now I can see at these spots, I can see B minus a. So this is B minus a by two. Okay. Single edge. Let it be as if as of now here itself. Okay. We'll take care of it a little later on. This is n h minus h by two. H is anyway is a zero and this is B minus a. So overall, overall, this will become, let me write it in a slightly more refined version. This is going to be a plus B minus a by two, which is B plus a by two. Okay. Now, why did I choose to write it like this? Because remember my h is still there in the expression. So sine h by two divided by h by two or a C profile of that is going to be one. So that leaves you with the answer of two sine B minus a by two into cos B plus a by two. Okay. Now, this is nothing but whose formula is this? What is to cause a plus B? What is to cause B plus a by two into sine B minus a by two? What is this? Whose transformation formula is this? Sine B minus. Okay. So just to give you a demonstration of how to write definite integral as a limit of a sum. Now moving on to the last part, limit of a sum as a definite integral. Now, there are certain questions of limits. If you would recall, I would have told you this in class 11th as well. There are certain question of limits, especially of the nature tending to infinity into zero or tending to zero into infinity types. Okay. So these indeterminate forms, they might not get solved by using our, you can say, summation formula or many times they are not even solved easily by your sandwich theorem also. So for those type of questions, there is an approach by which, if there is a possibility, we can convert the limit of a sum as a definite integral. Okay. So what we learned was that in the previous discussion that if you have a integral like this to evaluate, you can write it as h summation f of a plus n minus one h. Okay. And your term, or you can write it like this, r minus one h. Okay, r is going from one to n. And the limit of h is zero and limit of n is infinity. So this is what we had learned in our previous discussion, isn't it? Now I'm going to use a opposite discussion for this and we'll try to convert this as a integral. Now in order to convert this to this, I will tell you an algorithm. Okay, let's discuss an algorithm to do it because by the use of that algorithm, you are going to solve this question or you're going to convert this limit as a definite integral very, very easily. By the way, some of you would already be knowing this, this expression is called the Riemann sum. Okay. So this expression is called the Riemann sum. Riemann sum. Okay, this expression. Okay. So how to write this Riemann sum type of expression as a definite integral. So what is the algorithm involved? So let me explain this through a simple example. I'll pull out a question and I will explain you how it works. Maybe I'll take up a problem from my side because I'm not connected a problem of this nature. See, let us say I want to evaluate this integral. So this limit, let's say limit and tending to infinity. Very simple expression I will take up. Let's take one by this goes all the way till now here what is the approach. Let us try to understand the approach. Step number one, write down the rth term of this particular series that you're trying to sum up. So it's very simple. You have already done your series sequence and progressions chapter. So all of you are well versed in writing down the rth term here, which in this case is going to be r by n squared plus r squared. Okay. So the second step is try to write tr as one by n times a function of r by m. That is, in this example, can I do this, can I write it like one by n. Now this will become our in our and I will do a small activity over it just give me a minute. So when you do that, it gives you one by n, r by n, one plus r by n, the whole square. So this is what I'm trying to call as your function of r by m. Right. Now why are we doing it? Let's try to understand it. See, we have already seen that n is a very large number. One by n is a quantity which will be very small. So one man starts playing the role of your DX. Okay. And second thing that you see here is that your, your X or your R by N, your R by N is going to play the role of your X. So why do we do this is because your R is changing, isn't it? Just like a variable changes, your R keeps on changing. When your R changes from R by N to R plus one by N, this difference. Let me write it in the other way round. This difference should actually become that DX term. This difference should actually become your DX term. That means one by N is your DX term. That is what we have. Okay. So as a result, this entire series where you're trying to sum up this from R equal to one till N and the limit of N is tending to infinity. So we convert this entire expression as now see it one man will become DX. This term will become X by one plus X square. And your summation process will take the form of an integration. Okay. And what about limit of integration limit of integration is very simple since your X is R by N. When R is one X will approximately be zero because one by a very large quantity will be zero. And when your R becomes N, then your X will become N by N, which is a one. So your limit of integration will become zero to one. So as a result, this entire limit question got converted to what it got converted to a integration question. Okay. And this is very easy to integrate. So you can multiply and divide with a half. Take one plus X square as a T. So it'll become ln mod one plus X square zero to one. So when you put a one, you get a link to when you put a zero, you get a one, which is zero. So this becomes your answer is the process clear to you, especially this step where you are writing down our term as one by N times a function of our man. This is where many people get stuck and do mistake. It should be a function of R by N. That means you should be able to write it in such a way that when you substitute R by N as an X, there is no stray N term. There's no stray R term. Stray means something which has not been attended to. Okay. So it should come out perfectly as a function of R by clear is approached clear to you. Then substitute one by N as an X. Sorry, one by N as a DX. Okay. So these two are the important substitution. Okay. So substitute one by N as your DX substitute R by N as your X. Depending upon from where to where you are doing the summation, your limits of integration will accordingly change. Evaluate that definite integral. That would be the answer of that limit. See, from where to where you're integrating, where is your R going from one to N correct. And X is what R by N and being a very large number. So when R is one, that is your first value of R that would be your initial lower limit of your X that is one by N and N being a very large number it can be taken as a zero. And your highest value of R that you're going to is your N. So N by N that is your X is N by N that will become a one. So limit of integration will become zero to one. Now let's say your integration was from one to three N. I mean R was going to from one to three N. Then your upper limit will become three lower limit will remain a zero. Got it. Okay. Now many people can get confused if the limit is going up if the R values going from one to N minus one. If it is going to one to N minus one lower limit will be zero upper limit will still be one because N minus one by N, N is a very large number limits is a one only. Right. So please do not get confused even if the summation is happening only till N minus one. Okay. Let's take a few questions to get hands on on this type of problem. Yes, even if you go till N minus one, it'll give you the same answer. Okay. Let's take a problem. The entire thing is multiplied to N. Yes. And for God's sake, and to start writing these answers as zero. Many people will say, sir, if I evaluate each of the limits separately, I'll get a zero, zero, zero, zero zero zero zero, but remember, my dear, these are all tending to zero terms and they are infinitely many in number. So the answer will not be zero, it will be an indeterminate form. of limit is not going to work on this. So each quantity, each expression, even if you multiply with an n, that will all be 0, tending to 0, tending to 0 quantity. But they are infinitely many in number. So the result here would be, the expression here would be an indeterminate form. It is not going to be 0. Please follow the algorithm which I discussed with you just now. Done? Very good, Siddish. That's correct. Correct, Aditya. Okay. See here, very simple process. Write down your RIT term first. RIT term, if I am not mistaken, please don't make a mistake writing down the RIT term. If you make a mistake in writing down the RIT term, everything will go for a toss. This is the RIT term. Now remember, the last term is 6n square, 1 by 6n square. That will come when your R becomes n. So here, you are going till n. That's how it becomes n plus n to n. And n plus 2n is 3n. 2n plus 3n becomes 6n square. So this tells you that you are summing from R equal to 1 to R equal to n. Fine. So this is step number one. Step number two, as I told you, pull out a 1 by n common because I need that 1 by n as my dx. Okay. So this will become n square. So this will become something like this. Now, this expression I have here, I have to write it as a function of R by n. So divide by n square, first of all, it will become 1 plus R by n, 1 plus 2R by n. Now, what did I tell you? Write your 1 by n as a dx. Write your R by n as an x. So it is as good as you are trying to solve this question. Summation process will become an integration process. And when R is 1, x is 0. And when R is n, x is 1. So it is as good as integrating this. Now, 1 by 1 plus x into 1 plus 2x. I can solve this by partial fraction here. So as a partial fraction, can I write it like this? Correct me if I am wrong. Okay. So integration of 1 by 1 plus x, 1 plus 2x dx is as good as integrating this 0 to 1, 0 to 1. This will be ln 1 plus 2x. Please note that 2 and 2 will get cancelled. This will become ln 3 by 2. 0 will give me ln 1, which is 0 itself. So the answer is ln 3 by 2. Is this fine? So this limit problem basically gives you a result, which is in terms of ln. It comes out to be natural log. Clear, idea clear. Don't worry. We'll take at least two more problems before we close this chapter. By the way, in your normal classes that we have on Thursday for R&R and Friday for the non-R&R people, I'll be starting with differential equations. Next question. Evaluate whole to the power of 1 by n and this whole thing divided by n. n plus 1, n plus 2, n plus 3, all the way till 2n. The product is raised to the power of 1 by n and whole is divided by n. n is tending to infinity. See, right now it is not limit of a sum. Okay. So there's a limit of a product. Right? So you have to take log somewhere to convert product to a sum. So you can start, you can take this as a hint and start with the problem. So first step as I already told you, so even before that, you can do a one small change in the way the question has been written. Include this n within this nth power or nth root. So you can write it like this. You can introduce it as n to the power n. Okay. And take this 1 by n power outside the whole thing. Okay. And one more thing we can do here. We can impart these nn's to each of these factors. So something like this. Now let's call this as l and take the limit, sorry, take the log to the base here on both the sides. Remember, log can enter the limit. Okay. So that will make it log of n plus 1 by n plus log of n plus 2 by n plus. Please note that all of them are to the base of E. This is n plus n. Now think as if this is the limit of a sum or this is a series to infinite terms that we need to sum up. So write down the rth term of this that is going to be ln of n plus r by n. In fact, one very small error that we had done is basically this will become 1 by n over here. Yeah. Sorry. If you take a log, that thing is going to come down first, isn't it? Now here already things have been written in such a way that not only 1 by n has got segregated, but it has also converted the remaining part as a function of r by n. Okay. So this is going to behave as your dx and this part is going to be ln 1 plus x. Okay. Sorry for writing dx before, but this is the rough cut that we have learned. So you are integrating it. Now remember, you are going from r equal to 1 till r equal to n. So r equal to 1 will give you lower limit as 0 and r equal to n will give you upper limit as 1. So whatever answer comes out, that is the answer of limit log to the base of I think that's right. Okay. That should be the answer. So here we can use our integration by parts. So let's take this as our first function, take this as our second function. So it's ln 1 plus x into x minus 1 by 1 plus x into x dx. This is your ln of n. This I can write it as x plus 1 minus 1. So it's 1 by 1. Now note that on putting a 1, you will end up getting 1 ln 2, which is ln 2, and putting a 0 will give everything 0. And here I will end up getting x minus ln 1 plus x 0 to 1. So when you put a 1, you get 1 minus ln 2. And when you put a 0, I think everything will become a 0. So that's nothing but 2 ln 2 minus 1. In short, I can write it as ln 4 minus ln e, which is ln of 4 by e. So note that this result is ln of the limit. ln of the limit is ln of 4 by e. So l becomes 4 by e. This is your answer. So I think only Siddhas got this result. Is it clear? So this is one of the very good problems that you may get in your competitive exams. So not only you are converting it as a limit of a sum by taking a log, you are also integrating it by parts. So many concepts are getting tested. So please give importance to this multi-conceptual problems. Does it find any questions, any concerns? Please highlight. So with this, we are going to take a last problem for the entire chapter. And next class, as I already told you, I'm going to start with ordinary differential equations, ODE. So let's take this question. Evaluate limit n tending to infinity, summation of r square from r equal to 1 till n into summation of r cube. Again, r going to 1 till n by summation of r to the power 6 r going from 1 to n. Very good, Vavav. Anybody else? I mean, they have written it in a very straightforward manner. So I don't think so there is a problem in addressing it. Okay, let's discuss this out. So my purpose would be to generate a 1 by n and functions of r by n. So can I write the first summation as r by n the whole square? And of course, I have to have an n square to compensate for it. Second summation, multiply with n cube r by n the whole cube r going from 1 to n. And down in the denominator also r by n to the power 6 into n to the power 6 r going from 1 to n. Okay, now this expression will so beautifully take care of each other. See here, what I'm going to do is I'm going to divide with n to the power 7. Okay, n to the power 7 I'll split this up as n cube here and n to the power 4 over n. So this automatically becomes 1 by n summation of r by n the whole square from r equal to 1 till n. And of course, your limit on n is infinity that I forgot writing here limit on n is infinity. Similarly here, I'll end up getting 1 by n summation r by n whole cube r going from 1 to n. And down in the denominator I have 1 by n summation r by n to the power 6 r going from 1 to n. Now time has come that we write these as a definite integral. So this entire thing becomes integration of x square from 0 to 1. Correct me if I'm wrong. So 1 by n behaves as your dx, r by n whole square is x square. r is going from 1 to n means x is going from 0 to 1. Okay, similarly, the other one is 0 to 1 x cube dx and denominator is 0 to 1 x to the power 6 dx. So this integration is x cube from 0 to 1, which is 1 by 3. This is x to the power 4 by 4, which is 1 by 4. This is x to the power 7 by 7, which is 1 by 7. Answer is 7 upon 12. Okay. Any questions, any concerns with respect to this? So next class, as I told you, I'll be starting with differential equations. And it would be my effort to complete this before your KVPY session. Exactly, Aditya. Had we known the summation of 6th power of natural numbers, you could have used that also to solve it. But unfortunately, we have not derived till that extent. That is why this method was helpful. Okay. So it's not that if you know your definite integrals, you can or you may not be able, I mean, it is because you could not use your summation process. No. Definite integral can also help you where your summation process will also work or might also work. But its domain is much wider. It can be used to solve even when your summation is not working out, when your summation formulas are not there. Okay. Thank you class. Bye bye. Take care. All the best for any exams that you have. And I think tomorrow your schools are also reopening. So please complete your pending work. Thank you. Bye bye. Evening you have a chemistry class. Okay. I think message would have already gone. Let me just check.