 Okay, so let's start with the problem. I think last class we ended up at even odd property if I remember it correctly, right? So I'll resume today's session with a question on that, okay? And then we'll take up more properties. So initial warm-up session, let's start with some question, question number one. Evaluate integral of f of x to the power n plus x to the power minus n ln x whole divided by 1 plus x square dx, okay? So the question is evaluate the integral of f of xn plus x minus n times ln x whole divided by 1 plus x square, okay? Time starts now, two minutes to solve this question. Let's see if everybody has joined in. Good evening, sir. Hey, good evening, Ashutosh. So you started just now? Just now we started. This is the first problem that I've taken up. Where is Bharat? Hi, Udushi, good afternoon. Afternoon, sir. Yeah, we just started. This is the very first problem that I've given them. We have to evaluate the integral 0 to infinity f of x to the power n plus x to the power minus n ln x by 1 plus x square. Sharavanand. One more minute. Hi, Omkar. Good evening, sir. Anyone? Okay, let's take this up. Let us substitute ln of xs t, okay? So x will become e to the power t, correct? Now, let's first decide on the limits of integration. When x is 0, t has to be minus infinity, correct? And when x tends to infinity, t also has to tend to infinity, okay? So this is your lower limit. This is your upper limit, okay? So dx would be e to the power t dt. Let us substitute in this. So it will be minus infinity to infinity f of x to the power n means e to the power nt plus e to the power minus nt, okay? ln x is t and dx is e to the power t dt divided by 1 plus e to the power 2t. Got the point? Okay? Now, minus infinity to infinity f of e to the power nt plus e to the power minus nt. This you can write it as t dt by e to the power minus t plus e to the power t, correct? If you see carefully, here you have three functions. This function, now you comment whether this function is even or odd. Even, right? It's an even function, right? Because if you change your t with minus t, these two terms just switch their positions. That means overall this function doesn't change. So this is an even function. Similarly, what is this function which I am circling right now? Even again. Even if it's odd, so zero. Yeah, so this guy is odd, right? And you're integrating it from minus a to a. It doesn't matter if it is infinity or minus infinity, okay? Let me write it odd, okay? So everything becomes zero, okay? Simple problem. It just scared you because I involved all those functions and all just to scare you up. Let's take another one. Any question with respect to this? Lot of question mark. Who's lot of question mark Bharat? Next question I will just take in pair because the approach to solve this is both the same. Show that, show that number one, the integral of sine x from zero to infinity is actually one. And simultaneously we'll also try to prove that integral of cos x from zero to infinity is zero. Can you give it a try? It may not have anything to do with the even odd property, okay? So don't try to think in line of that. Just think it in a neutral sense. So don't you just integrate it normally and then put the limits in? Yeah, but what kind of infinity? Yeah, what would you put for sine of infinity or cos of infinity? Yeah, cos infinity, I can't find one. My bad, my bad. See, it has to be somewhere between zero, one, something like that, right? Sir, can we consider sine of one by x integrating that from zero to infinity? Sine of one by x. Yeah, so sine x dx integral from zero to infinity. So we can just say that replacing x, we could just say that sine of one by, it's equal to sine of one by x, oh wait, never mind, sorry. Any idea, anyone? Sir, can you convert it to a tan x form? Okay, try it out, Siju. Sir, shouldn't the first integral be equal to zero and the second be equal to one? Okay. Can you send me the working for it? Just looking at the graph. How can you come in looking at the graph? The areas above the x-axis and below the x-axis will cancel out. And looking at sine which is an odd function, I think it will be zero. Sir, I have a question. Sir, I got it. Can you do it from like 0 to pi then pi to 2 pi like that till infinity? Sashijan was right. But that's fine. But how will you end up concluding it? Sir, tan x is equal to t, x is tan inverse t. You put in the limits, take t from, go to that. Try it out, no issues. Yes, sir. And then you take the limits from zero to infinity, you will get tan inverse of infinity as pi by 2. You get sine of pi by 2 plus, oh god. Sir, can you do it like what I said from 0 to pi then pi to 2 pi and pi to 3 pi like that? Yeah, that is fine. Any ways you can figure out from the graph. But what will happen when you go to infinity? Sir, wait. So just give us a bit more time. Fine. We need to get it in the form of some tan inverse. Just try it out, Bharat. Let's see if you get any success with that. So did I miss anything else? He secured clear on the second problem today. First we did. Sir, only these two problems? No, no, no, no. There was a problem before this also. I'll show you the board. Sir, yes, sir. I'll just quickly take a screenshot. Yeah, this was a problem. Okay, yes, sir. I'll solve it later. Yeah, can I help you out now? Wait one second. I think I got it. I think I got it. One second. Sure. Sir, if I want to send you the solution, how do I do it? You just WhatsApp to me. Okay. Sir, if you prove the first one, then the second one is obviously true, right? I mean from graph. Okay. So because everything is getting phase shifted by pi by two. And then we know that the area from zero to pi by two is one. So basically an area of one goes away. I mean before zero. The area of one just goes away before zero. So if you approve the first one, then the second one is. Okay. Okay. I'll send something. Fine. Just a minute. Just last a minute. Sir, we have to prove it without graph. Sir, can we take it as a series and then like everything cancels out in the end? If you take or break it up into smaller parts. Yes. I'll just send you a picture one second. Very good. So would it be valid if we took X as tan inverse of tan X substituted tan inverse of tan X as you and then integrated it between zero to pi by two. Okay. Can we just use graph and like prove it because it's pretty obvious. Yeah. You have any way to solve it? Yes. We have to use the expansion. So from first. So okay. Fine. So for the first one, integral of zero to infinity sign X DX. So now I'll take the integral integral of zero to infinity minus sign minus sign X DX. Now we know that minus sign is just sign which is phase shifted by pi. So therefore, it swipes an area of two because from zero to pi the area is two. So we know that two is the sum of, you know, sorry, two is equal to sign X integral of zero to infinity sign X minus integral of zero to infinity minus sign X and then when you add them up, you'll directly get integral zero to infinity sign X equal to one. And then after that, when you phase shifted by pi by two, now we know that the area covered when you phase shifted by pi by two is one. So you have to just subtract one for cost. You'll get zero. Great. I mean, I liked all of your ideas. So what I'll do is I'll solve it in a slightly rigorous way. Okay. Most of you have gone through graph and a bit of substitution, which is very good. Okay. I'm still getting answers. Yes. Very good. The same word with us. She has also done very good. Fine. So, yes. So let us try to solve this simultaneously. So what I'll do is first let me call one integral as e to the power minus sx sign X. Okay. Okay. Just just follow what I'm doing. Okay. I'll connect it to these two integrals in a while. And let me call another integral J as e to the power minus sx cos X. So what is S? S is something which will be tending to zero because as it tends to zero, it will just leave you sign X. Okay. Why I'm introducing this e to the power minus sx because this will make this problem integrable actually. Okay. So wait, you're taking the limit of the integral. Yeah. I'm not just, I'm not putting any limit of integration. I'm just saying that let us have a function like this. Of course this answer will be in terms of s. Okay. In terms of s and X both. Okay. And then I'll do the limits on X. So it'll leave me something in terms of s. So when I put, let's say zero to infinity, this will end up giving me a function of s. Right. Okay. There if I take s tending to zero, then this will end up giving me integration of e to the, e to the power minus e to the power zero actually into sign X dx. Correct. Giving me your integral, which is required. Is that fine? Yeah. Yeah. Yeah. We just saw your solution. That was good. That was absolutely correct. Okay. So this is my intention. So what I'm going to do is I'm first going to evaluate I by the use of integration by parts. Okay. So let us apply integration by parts on I. Okay. So when we do that, we end up getting I is equal to take this as your second function. We take this as your first function. Okay. You. So are you going to take other way around also doesn't matter because both are of the same difficulty level in integration. So this is going to be e to the power minus sx integration of sign X is minus of costs. Okay. Minus integral of derivative of this, which is going to be s e to the power minus sx into integral of sign X, which is minus costs. Okay. So we realize here that we end up getting something like s and this term is actually a j. By the way, there's a minus once more because minus comes from both sign integration and this that doesn't matter in the end. Wait, wait, wait. Let's do it properly. Yeah. Yeah. So I is in terms of j. Similarly, if you apply integration by parts on j. Okay. What does it give you j is equal to e to the power minus sx into sign X. Okay. Minus derivative of this, which is minus s e to the power minus sx into sign X, which is nothing but j is equal to e to the power minus sx sign X plus s I now from these two solutions from these two equations, we will try to figure out what is my I. Okay. So let us try to solve it by using our elimination method. That means I will put my j here. So it will become minus e to the power minus sx cos x minus s e to the power minus sx sign X plus s I. Okay. So this is going to give me one plus s square I minus e to the power minus sx cos x minus s e to the power minus sx sign X. Okay. So ultimately my answer for I will become negative e to the power minus sx cos x plus s sign X by one plus s square. This result could have been directly been also written because we have already done in our indefinite integral e to the power AX sign BX form right. Yeah. The Euler's thing that imaginary thing would be. Yeah. Yeah. The same approach I'm taking here is just that now my role of I is being played by S. Now, next is your J term. By the way, if you want to evaluate I under the limit zero to infinity. Okay. What will your answer be? I can see that when I put infinity because of this term everything will become a zero. Correct. Yes or no? Yes sir. So everything will become a zero. So minus zero then minus minus means plus and when I put a zero I'll get a one. One plus zero by one plus s square. Correct. So if I'm not wrong my I becomes one plus s square. Correct. And remember I was actually let me just put the limits of integration also here. Okay. I've already put here. Let me put here also. Okay. Now my required integral is nothing but the same I but I just have to take the limit on ascending to zero. That means I have to take limit ascending to zero on one by one plus s square that is going to give me one. So that is how we get the answer is one. Many of you have sent solutions which are, you know, very correct. You have found a recurring terms and you have renamed it as I again. Okay. Correct approach. Now, cause x integration from zero to infinity. So is there a fault with taking x as tan inverse of tan x and then compressing the limits like that? That approach I have to just inquire because when I'm dealing with inverse trick functions are these functions change their definition. So we have to be very careful. So I'll just look into that solution as well. Just give me some time. Now regarding j. Okay. If you use the formula j will directly come out to be e to the power minus sx sin x minus s cos x by one plus s square. Okay. And our required integral that is zero to infinity cos x is nothing but limiting case on j when s is sending to zero. Okay. Oh, first of all, I didn't put the limits of integration on this again zero to infinity infinity will give you everything zero. Okay. Minus putting zero will give you zero minus s by one plus s square. That is nothing but s by one plus s square. And here we have to take the limiting case on that and that becomes actually zero. I guess. Yes. Okay. A lot of different approaches are there. So I thought I would introduce you to this method as well so that it's a bit of extra learning also for you. Okay. So but here you guessed the function, right? Huh? Yeah, I took a different function, but I made that functions came ultimately by choosing certain a certain value of s. So I changed my parameter to make my function resemble my original question. Okay. Okay. Great. Next question. Yeah. This question goes like this. If you have a function f of x which satisfies this equation. Okay. So this function itself is made up of x plus integral of x y square plus x square y f y d y. Remember you are integrating here with respect to y. Okay. So the answer for this part will still be a function of x. Okay. Question is find the function itself or find the expression for the function in J. In J they would say find a particular value. They'll put some particular value and they'll ask you the value. Try this out. Three minutes. Sir, while, oh wait, we can't different. We can't take x as a constant and do it. While you're integrating it is a constant. No worries. But when you're differentiating it, you have to be careful. If you want to differentiate throughout, which actually is not required because I've not talked about Leibniz rule yet. Sir, over here do we treat, we can't treat x as a constant, right? You have to treat it as a variable itself. See, when you are integrating here, you are integrating with respect to y. Here x is a constant then. Okay. This y is not the y that you are thinking. This y is not f of x. There's no, this is not a relation here. This y x are just inputs. Getting my point. So don't unnecessarily link y and x unless until it is stated in the question. Treat x and y as independent inputs to the function. Sir, is it x plus 4 by 5? Sorry, your voice was breaking. Come again. Sir, is it x plus 4 by 5? No, Sukheer. The answer is much more scary. Okay. The answer has got 119 terms and all. Sir, what? 119 terms. 119 as a term in it. Not 119 terms. Oh, yes. Okay. Any idea, anyone? So one second. Can we divide and multiply by 2? It is x squared y. Gaurav, any clue? Sir, multiply and divide by 2. So you get 2xy squared plus 2yx squared, which is the derivative of x squared y squared with respect to... Yeah. I got it. Vidush, any idea? Sir, what is dy by dix of 2 completely unrelated quantities? I didn't get your complete question, Bharat. Dx of 2 completely unrelated quantities. Zero. No, unrelated. Zero, yeah. See, I think you're just complicating it. If you see this function carefully, okay? If you just break this up like this, fydy and 0 to 1 x squared y fydy. Okay? Since your integration is done with respect to y, if you take this x out, okay? It's something like this. Is that fine? Now, ultimately, this is going to be some constant, isn't it? This is going to be some constant because you are integrating with respect to y and putting the limits of integration, which are constants. So ultimately, this is a constant. Ultimately, this is also a constant. Correct? Now, this actually gives you an idea about the structure of the function. The function is actually something like x 1 plus 0 to 1 y square fy and x square into 0 to 1 y fydy. Sir, by mistake, I took the original function as x plus c, I forgot the x square. Okay. So this gives you an idea that this function is of this nature. Correct? And then you substitute in y. Yeah, I'm calling this as a, I'm calling this as b. Okay. So your function is of this nature. It's a, it's a polynomial with some coefficient with x and some coefficient with x square. Fine. Now how do I get these coefficients? That's very simple. If you see a itself is 1 plus integral 0 to 1 y square fy. Fy means you just have to replace your x with y in this particular thing. Isn't it? So it is nothing but ay plus by square. Correct? So let's evaluate this. So it becomes 1 plus. Ay cube integration will be ay to the power 4 by 4. So it is a by 4 b to the b by 5. Is that correct? So a is equal to 1 plus a by 4 plus b by 5. So we'll get two equations and we can solve for a and b. Absolutely Bharat. Absolutely. So in other words, basically let's multiply with 20 throughout. So you end up getting 20 a is equal to 20 plus 5 a plus 4 b. Okay. Which is nothing but 15 a minus 4 b is equal to 20. Equation number one. Okay. In a, in the same way we can say b is nothing but integral of 0 to by y fy fy is ay plus by square. Okay. And this is going to give me a by 3 if I'm not wrong. And b by 4. Right? In other words, we'll get 12 b is equal to 4 a plus 3 b. Okay. That is 9 b is equal to 4. I'm writing small. Yeah. 9 b is equal to 4 a. That's your second equation. Okay. Can we just solve for a and b from here? We can do some substitution if you want to. So let's, let's find a first. So 15 a minus 4 b you can write it as 4 a by 9 is equal to 20. So multiply throughout with 9. This will give you 135 minus 16 a is equal to 180. Oh, no wonder we are getting 119 over here. 100 n is by 190. Okay. And b will be similarly b will be 4 a by 9. 25 by 119. 80 by 119. Sorry. Sorry. So your function will be now a x. A x is 1180 by 119 x plus b x square. That is 80 by 119 x. This is your answer. So is there a way to solve linear equation and three variables quickly? Linear variables. Linear equation and three variables. Three equations. Because in physics it may come a lot. Kramer's rule. So no, like, which is faster because certain physics some current values come as one by seven and things like that. If you just send me the question early, I'll be able to help you in a better way. Okay. Final C. Okay. Do you want another problem of the same thing? Or is it fine? So I think it's fine. Fine. Everyone can give you if you want. No, no. I mean, are you feeling confident about the approach or do you want me to give you another problem so that you are confident? That's fine. We'll find the approach. I see that definitely as a constant that was our mistake. Nice. Okay. Um, next question is, I'm sorry, I'm yet to start with the new property. Just wanted to give you this question. If you have a integral UN, which is zero to pi. One minus cost of NX by one minus cost X DX. Okay. And is a positive integer. Uh, it can be zero also. So N is, let's say a whole number. Okay. Then show that I'm giving you as a show that question, but it can be as a option based question also show that UN plus two plus UN is twice of this. In other words, UN plus two. Alternately, this have this is in arithmetic progression. Okay. And hence evaluate. Hence you are hence means using this is it. Evaluate integral of sine square and theta by sine square theta from zero to pi by two with respect to theta. Okay. I'll give you three four minutes to solve this. Time starts now. At least first part is a few seconds almost done. Almost one second one second. Okay. Oh, I think I got the first part. Got the first part. Can you send me the solution? Yes. That's correct. Anybody else other than Ashutosh for the first part. Does it, does it devolve into showing that everything equals zero and so you will get. It's just trigonometry. Yeah. It's just pure trigonometry. Nothing else. Only trigonometry and then just showing that I think equals zero. So it's a nappy. I think I got something that shows that it's equal to zero, but I'm not very convinced with the proof that I got. So. Okay. We'll do an activity here. We will split this term like this. UN plus two minus. UN plus one. Okay. Minus UN plus one minus UN. Okay. In fact, that would be easy to find because they are of the same recursive nature. Correct. If I'm somehow able to show that this is zero, that means I'll be able to prove that this is true. Correct. Okay. So let us focus on evaluating. UN plus two minus. UN plus one. Okay. All you need to do is limit of integration is not a function of n. So it will remain zero to pie. Okay. So this will become one minus cos n plus two x minus one minus cos n plus one x whole divided by one minus cos x dx. One one term will get cancelled off. So this will become zero to pie. This will give you cos n plus one x minus cos n plus two x. Okay. So this will become cos a minus cos b formula. We can use to sign a plus b by two. That will be I think n plus three by two. Okay. Into sign b minus a by two, which is going to be x by one minus cos x also you can write it as two sign square x by two. Okay. So two two gone and one of the sign x by two is also cancelled off. So this is the expression that we have got for UN plus two minus UN plus one. That is sign n plus three by two x by sign x by two. Can I say by the same logic, UN plus one minus UN could be written as zero to pie sign. You just replace your N plus one with an N. So this will become N plus half. Let us subtract these two results. Let us subtract these two results. So UN plus two minus UN plus one minus UN plus one minus UN is nothing but zero to pie sign n plus three by two x minus sign n plus half x dx by sign of x. So we can just say n plus three by two x is just m times x. Yeah, you can use your transformation formula here. There is two cos a plus b by two. Sign of x. Huh. In the denominator. What what what happened in the denominator. Sign x by two. Sorry. Yeah. So this will become a sign x by two divided by a sign x by two. Is that fine. Which gives me a very simple integral to evaluate that is zero to pie cos of I'm sorry. Two cos n plus one x. Which is nothing but two times putting five will also give you zero putting zero also will give you zero everything is a zero over here. And hence the road map which I had in my mind that is this road map is successfully proven here. That means UN plus two plus UN is equal to two UN plus one. Okay. Now how do I use this result. To evaluate this integral zero to pie by two sine square and theta by sine square theta. It's the same thing as the given integral. Absolutely. If you see. If you approve that it is the AP. Can I say you zero zero. Hello. Do you want to say anything. Hello. Okay. You one will be what you one will be zero to pie one minus or six by Cossack's which is actually pie. It means it tells you that you two will be to pie that until UN will be in pie. Because an AP. So if it is an AP if you see this expression this is nothing but zero to pie by two. Okay. Put two here. This is one minus Coss two and theta by one minus Coss two theta. Okay. Just put two theta as X. Okay. Just put two theta as X. So it will be DX by two. Okay. So this entire integral will become zero to five one minus Coss and X by one minus Coss X DX by two. Right. This is nothing but UN. This is nothing but UN. So your answer is half of UN. That is nothing but half of N pi. So your answer is half of N pi. So this integral is. Is that clear? Yeah. Yes. So we prepared for these kind of questions. Okay. So not normal questions will not be asked. In the sense that you don't have just apply your property and solve it. Can we apply King's property here? I was getting like the I was getting like an even odd sort of thing with the King's property once I shifted origin to pi by two. Okay. That's only for the for that when you put you one that time you can put King's property saying that it's symmetric about pi by two. Not necessarily. If you just say that one final just work on that and get a little bit. Let's have a concrete thing to discuss about. Okay. Next question. Evaluate limit and tending to infinity summation from K equal to zero to N and CK by N to the power K K plus three. Guys, let me tell you one thing. Integration and summation positions can be interchanged. Okay. Because both are summations at the end of the day. So but integrations are continuous summation, right? Yeah. You have to apply summation only because it's blocks like we can't take continuously. Right. That depends upon the question at hand also if your quantity is changing by very minute minute amount. For example, in this case, even though there's a summation and it's ending to infinity earlier, changing it very, very continuously. That means it's not a discrete change. You may use, you may interchanged. That's what we use actually in the part which we call as limit of a sum as definite integrals, which probably I'll be starting today or maybe next class. We haven't done limit as a sum of different integrals here. We have a lot of things to cover. Just one second. Sorry. Anybody saying anything? So we have to find the integral of that. You have to evaluate this. No, there's no individual person in the question, but I'm just saying in general. Okay. Any idea? Anyone? They're almost done. So what is the expansion of e power of kx? Is it k by zero factorial plus k times x by one factorial? e to the power k. When k is also there. Okay, we just replace x with xk. Just replace this with kx. Yes, sir. No, Shavan. Answer is not that. So one minute. Yeah. Let me handle this. This is almost done. Okay, fine, fine. So I'll stop here. So is it e minus one? Answer is e minus two. Wait, what? Two. Yeah. Check if you have missed out anything. Okay. So see here, you can write this as nck by n to the power k. Okay. Okay. Now this sum I have just segregated it out because. Yes, sir. Found a mistake. Got a mistake? Yes, sir. My C value. See, I told you C is not a thing that you can ignore. It is very important. No, sir. I didn't ignore it. I didn't ignore it and I still got it wrongs. Okay. So can I say this term. Is actually integral of x to the power. K plus two. From zero to one. Introducing this term. Oh, I didn't understand that. If you integrate it by power rule, what will you get x to the power k plus three by k plus three. And when you put one zero to one, yeah. So what I do now, instead of this one by k plus three term, now I will introduce integral zero to one x to the power k plus two dx. Okay. So far so good. No problem. Okay. Now what I'm going to do is I'm going to take this limit of integration completely out. So and introduce these terms inside x to the power k. I'll write it as x square into x to the power k by n to the power k. So this term, I'll take it together. Okay. Thing into dx. Okay. Now, since the summation is applied on n, I can even pull out a x square to this extent. Okay. Limit and tending to infinity and summation k from zero to n and ck. This you can write it as x by n to the power k. Does this remind you of a binomial, a general term of a binomial expression. Isn't this summation, a summation of expansion of this term? So basically this term is nothing but it's this binomial expansion. Okay. Limit and tending to infinity. Okay. And what is this? This is a limit that you have already done one to the power infinity form. Correct. Can I say this will be nothing but e to the power x. So ultimately, what are you doing? You are evaluating integral of x square e to the power x. Let's use the I method. The voice is breaking up. No one can hear what he's saying. Probably he's using his phone or something or iPad. See, this is okay. Sure. I'll explain the previous step. See, this is nothing but what you're writing. You're writing nc0 x to the power n to the power 0. Then you're writing nc1 x to the power n to the power 1. Okay. Let's say you go till ncn x by n to the power n. Basically, what you're writing, you're writing a binomial expansion like this. Isn't it? Yes or no? So this summation thing, this entire thing is actually 1 plus x by n to the power n. Okay. And when your n is sending to infinity, it's like 1 to the power infinity form. Okay. 1 to the power infinity form. How do you find out the answer? This is just take it as e to the power, take this as fn by gn n tending to infinity. Correct? So e to the power x by n. Okay. g is 1 by n. So divided by 1 by n will make n by n on top so it will become e to the power x. So that's how we ended up getting this e to the power x term. Got it? Okay. Fine. So let us apply di rule over here. So plus, minus, plus, minus. So your answer will be x square e to the power x minus 2x e to the power x plus 2e to the power x. Okay. So it becomes e minus 2e plus 2e. Correct? Minus. When you put a zero, you just get a 2. Okay. So your answer is e minus 2. Make sense everyone? So we can also expand nck and then solve. Yeah. That method also I understood what you were trying to do. Sir, but how do we exactly think that writing 1 by k plus 3 as integral 0 to 1 x bar k plus 2? Actually, I got the hint is from this term. So I got the most tangential hint. I wanted to create a binomial term because of nck. And I wanted to create a term which also has the same power as this. Sir, or if we, if at all we like need the, okay, if we can, if at all we expand nck and do it, like, and we just write down the first few terms. We can, we can see the pattern. You'll get 1 by 0 factorial times 3 plus 1 by 1 factorial times 4 and then you can. Something like the Poisson distribution. Something like the Poisson distribution, right? Proof. Somewhat, yeah. Okay. Is this, okay. I was just expanding this and I evaluated that integral as a limit of something. So I'm going to start with the new property today. I don't remember the number. So I'll just write property. This property goes like this. If you integrate a function, okay, continuous function in the interval a to b, then this could be written as b minus a 0 to 1 f of b minus x a plus a b x. Okay. Can somebody give me the proof for this geometrically? I'm sure you can prove this by non geometrical method, but I would appreciate if somebody can tell me what exactly is happening here. As you can see, your limits have shun from a to b to 0 to 1. That means instead of an unknown values of a and b, it is always a fixed value 0 to 1. But of course, there has been other changes brought about in the question in the function. Can somebody explain to you what is happening here? So you have a function from a to b. So basically we are shifting the function. So we're flipping and scaling it, right? Yes, absolutely. So let's say you have a function like this. And shrinking the graph by a factor of b minus a. It may be shrinking, it may be expanding. Yeah, so then basically you will have f of b minus a times x plus a. Correct. So, Sukirt has correctly summarized it. So let's say we are trying to find out integral of f of x from a to b. That means we're looking for this area. Let me call this area as a. Now what is happening here in this function? If you see here, this function is obtained by two step procedure from y equal to f of x. Let us go to y equal to f of x plus a. That means I am shifting the graph a units to the left. Correct. If I shift the graph a units to the left, this is what will happen. This will become 0 and this point will become a b minus a point. Correct. So same function. I'm shifting a units to the left. Now what I'm doing is I'm replacing my x with, I'm replacing my x with b minus a x. That means I am fiddling with the periodicity of the graph. That means I'm fiddling with the, I'm contracting or expanding the graph along the x axis. So as to say, okay, that depends upon what is your b minus a actually. If your b minus a is greater than one, you are contracting the graph. But if your b minus a is less than one, you are expanding the graph. Okay. Doesn't matter. Let's say I'm contracting the graph. So what I'm doing is I'm contracting the graph by a factor of b minus a. Okay. Now contracting about what? Contracting about y axis. If yes, how do I know that? Sir, it's contracting about the x axis. No, no, contracting, but which part will not move at all? So basically the width is changing. Will it be like this or will it be like this? How do I know that? Okay. Now for that very simple, when you put zero x equal to zero, this also gives you f of a and this also gives you f of a. Correct. That means this part doesn't move anywhere. So this part contracts by a factor of b minus a and it comes to let's say this position like this, which is actually one. Correct. Now this problem is saying that let's say this area is a2. This particular property is saying that a1 is actually b minus a times a2, which is actually correct because if you contract it by a factor of b minus a, then this area will be b minus a times lesser than the original area. So in order to make them same, you have to compensate by multiplying with b minus a. Okay. Which is actually correct. Are you getting this point? So this property is basically helping you to convert two let's say integrals which are at different limits of integration as one limit of integration that is zero to one. So that you can add or subtract or play with these two functions. Getting my point. I call this property as expansion contraction property. So I've given some name to each of the property probably you'll not find the same name in your textbooks. So I've called this expansion contraction property. What kind of problems can we get on this? Let's have a look at a problem. Not a very widely used property. But again, a question can come on this. So let's take one simple question on this. Evaluate. Let me just bring the property in front of your eyes. Evaluate. Yeah. Evaluate. Yeah. Integral from minus four to minus five e to the power x plus five the whole square plus three times integral of e to the power nine. Times x minus two by three whole square from one by three to two by three. If you have copied this down, I can shift the screen up. But if you want to refer to this, I have to keep the screen at this position. So would it be zero by any chance? Wait. Anybody else? So zero. There is definitely zero. Definitely zero. Okay. The answer. Yes, zero. Santosh is not correct. Zero is correct. I'm saying to Santosh. Santosh has privately mentioned. Yeah. Answer is zero guys. So basically since there are two separate functions and they cannot be evaluated separately, we cannot integrate e to the power x square. Okay. This is a non-expressible integral. Okay. So what I'll do here is I'll try to bring the limits of these two integrations, both as zero to one by the use of the expansion contraction property. So B minus a here would be minus one. Okay. Replace your x with B minus ax, which is minus ax plus a, which is actually minus four. So this is what you'll end up. Okay. This becomes three times B minus a will be two third minus one third. Yeah. That's correct. Santosh is zero. So this will become e to the power nine one third x, which is x by three plus a is one third minus two by three, the whole square. Okay. So remember in the function, you're replacing x with B minus ax plus a. This is what we are doing in all the functions. Okay. So this gives me minus zero to one e to the power. You can call it as one minus x or x minus one. The whole square doesn't matter. Our three into one by three will be again a one zero to one. This again will be e to the power nine x by three minus one by three, the whole square. That's nothing but minus zero to one e to the power x minus one whole squared ex plus zero to one e to the power nine can get cancelled with the three square coming from the denominators, which actually ends up canceling both of them, giving you the answer as zero. Okay. Simple one. Next property. If f of t is a is an odd function. If f of t is an odd function. Okay. Then our integral function like this a to x f of t dt. Okay. Where a is any real number. This will always be a even function. And the inverse is to write if it's an if it's an even function, then that integral will be an odd function. Definitely. Can we prove this? Can we prove that? Phi of x will be an even function. If it is expressed as integral of an odd function from a to x. We are in the interest of time since we have less time left and not for the class, but for completion of the syllabus. So let me do it. So if you want to prove that this function is even function, it is necessary to show that. So we have to prove that Phi of minus x should be Phi of x. Isn't it? Okay. Start from the left hand side. Just substitute t as minus t and then just watch it go through and it will become just Phi of x again. That's what I'm going to do, but I'm going to do it in a very methodical way. So we can also say that when you differentiate even you get odd and when you differentiate odd you get even. So you're just doing the same thing, but then the lower limit a will not really matter. Yes. Agreed. The A has a role to play. You'll come to know in the next part of the property. So yeah, what I'm going to do is first I started putting, first I started my left hand side by putting x as minus x. Then what did I do? I've split this limit of integration as a to minus a and then from minus a to minus x. What will happen to this term? Zero. Because it's an odd function. Here if you put t as let's say minus y. So dt is minus dy. So limit of integration will be a to x f of minus y minus dy. Since the function f t is an odd function, since this is an odd function f of minus y will be minus f of y giving you the expression as this. And there's nothing in the name. So you can call it back as f of t dt giving you the five function back. Okay. Since the function doesn't change, despite changing x with minus x, it's an even function. Okay. Next part of the property is do you want to know down the proof? Anyone? Second. Anyway, I'll share the notes with you. Yes, sir. Okay. So just cut it up into two pieces. You just cut that into two pieces and then just evaluated each. Yes, yes. I can, let me say a part of the property. Let me call, let me talk about now the B part of the property. This probably says that if f of t is an even function. Okay. Then pi x function, which is integral of f of t from a to x. This will be odd only if integral from zero to a f of t dt is zero. Should we prove it? Please note only when this integral is zero or you can say this will be alternately, you can say this will be odd function always. Zero to x f of t dt. This will be odd function always. So either this or this condition must be satisfied. Of course, you can say if you're a zero, it'll automatically become integral from zero to zero. Okay. So this is extended version or you can say this is the most, this is the more generalized version and this is the most specific version. So the difference in the previous and this properties, there a could be any real number for it to be even. But here it has to be zero for it to be odd or integral of the function from zero to a must be zero. Then only you can call this as odd. Are you getting the difference between those two properties? Can you prove this quickly? Let's write out in the interest of time. Let me do it. So phi of x, I replaced my x with a minus x. So it becomes a to minus x f of t dt. The same thing that we did in the previous one, I'm just going to change a break the limit of integration from a to minus a and then from minus a to minus x. Okay. So this gives me two times zero to a f of t dt. Okay. And here I put my t as minus y. That means dt is minus dy. So it becomes a to x f of minus y minus dy. So minus dy. Since this function is even function, it doesn't matter whether you have y or minus y inside it. There's nothing in the name of the variable also. So let me drop y from everywhere and let me put an x. Sorry, let me put a t. So as you can see, phi of minus x becomes two times zero to a f of t dt. And this is nothing but minus of phi x. Now, if you want your function to be an odd function, this term should not have been there in the first place. Correct. This should have been zero. So this implies if your phi of x has to, sorry, phi of minus x has to be minus of phi x. Then it implies zero to a f of t should have been equal to zero. That's what this property says. If you want this function to be odd, then integral of f of t from zero to a should have been zero. Or in other words, just write it from zero to x f of t dt will be on our odd function. So either follow this or you follow this approach. Is the idea clear? Let's take a small question on this, very small question. If I have a function zero to x ln one minus t by one plus t dt. So question is f of x is option number a odd even even. Neither neither data and stuff be okay. Simple. This is odd. Yeah. This is odd. So integral of this has to be done. Okay. Next question we'll take on this another small one. Sir, pizza problem, sir, please. Definitely I'll give today pizza problem, but not now. Sir, you don't even probability you'll get pizza problem. I'll give ya. Don't worry. Okay. It is known that it is known that f of y is an odd function. It is known that f of y is an odd function in the interval minus t by two to t by two. Okay. And this function has a period of t f of y is periodic with t. Okay. Find the period of a to y f of t dt. Sir, will it be t by two? Is it t only? Huh? Is it t only? Is it t by two? Period is correct. It's t only. Sir, can you assume f of y is cos y? Sin y. Cos is not an odd function. Sin y, sin y. Can you assume f of y is sin y and then do it? Okay. You can do it normally. Yeah, just t. Damn it. Why? Yeah. Let me give this as a proof that question. Prove that the period is t. Okay. Now you not try to prove it. At least prove that itself that the period is t. Prove that this is this has a period of t. Remember this is an odd function only in this interval. It is a property, right? B minus a times f of... Sir, we can just say that the first two statements imply that it's odd function throughout, right? Mm-hmm. That's what I did. Anybody who wants to take an idea? I don't think it's like a... I don't think it'll be proper sin x though. If it's an odd function, the minus c divided by two, then it'll be just like a cascaded sign. There'll be a discontinuity every time. Yes, but it's still an odd function throughout. That's what I'm saying. We'd have to break it up in terms of the period of f of t. So if we can break it up... Oh, wait. Okay. So if we can break it up... Minus c divided by two to t divided by two, the difference is t. And then it's saying it's periodic with t. So everything's going to repeat. So two things are quite obvious here that f of minus y is minus f of y in this interval. Minus t divided by two to t divided by two. And this function is periodic with t. Correct. So these two informations we have at our hand. Only these two are given. Okay. Let me try this out. Let's call this function as g y. Okay. So if this function is periodic with t, I should be able to show that this is g y itself. Okay. I have to show this. Okay. So let me start with LHS. So g y plus t is nothing but a to y plus t f of t dt. Okay. So now I'm going to break this limit of integration as a to y and y to y plus t. Now a to y, I will not disturb it because if others, other terms cancel, my job is done. Right. Okay. So now what I'm going to do is since t by two has been referred to what I'll do, I'll first take it from y to t by two and t by two to y plus t. Let's say in this integral only, I will not disturb the other two. In the last integral only, what I'll do is I'll make a substitution. I'll put another variable. I'll put this t as let's say x plus capital D. Okay. So dt is dx. Nothing is going to change. So this will become f of x dx. Limit of integration will become from minus t by two to what will happen here? Why? Right? Yes or no? Let me copy the others as such. So this I'm copying here. This I'm copying here. By the way, there's nothing in the name. So you can start calling this back as t again. So isn't this like saying, isn't these two integral like saying you are integrating from minus t by two to t by two? Okay. And since this is an odd function, integration from minus t by two to t by two will vanish off. So this will entire thing will vanish off to zero, leaving you only with this expression, which is actually your g of y. So yes, I started with my left hand side and reached my RHS and hence shown. Is that clear? Now, since I'm talking about these periodic function, the next property is basically the definite integral properties for periodic functions. So let us take that. By the way, any question here? Anywhere please? If you didn't know the period of steel, it wasn't the show that question. See, if it wasn't the show that question, what I would have done from the options I would have taken the minimum of the period and try to prove it. For example, if it is 2t, 3t, etc., I would take t and start. Okay. It has to be a bit of trial and error because finding the periodicity is not a very simple task. Understand, if this was not a show that question, probably it would have taken a little bit of trial and error as well. But it wasn't a show that question in the beginning. I mean, you made it a show that question. Yeah, I made it a show that question. It was an option-based question actually. Sir, but you can directly do it also. I mean, we can see that the difference between t minus t by 2 and t by 2 is t. And it's an odd function. So it's not going to replicate itself. So the period cannot be less than t. So therefore, and you're integrating the odd function, it became even. So it will be t. Period cannot be less than t. Yes, sir, because it's an odd function. I mean, either period is not defined or... Right. This duration is t. So it can't be less than t. Okay. Yes, sir. How do you know the possibility of being greater than t? No, not great. No, sir. But then now it's an odd function. And when you're integrating it, you're going to get an even function which will have period of t. It should have the same period, basically. Because it's a rate of change of slope, all that. Even the slopes will also follow the same period. So it will not be greater than t. So the problem is it should be either t or less than t, but it's not less than t because it's an odd function from minus t by 2 to t by 2. Yeah, that's what. So we'll start the trial at t first, right? Yes, sir. So it should be t. Like it can't be anything else. I did not understand why it cannot be greater than t. So because now when you're integrating it, you're integrating the slopes, basically. And the slopes also, the rate at which the slope is changing all of that is following the derivatives, integrals, everything for the period t. That seems to be intuitive to me. Yeah. That is intuitive. So we can't base our judgment on that intuition. So next property is the periodic properties. I call it as periodic properties, but it's like definite integral properties for periodic functions. Okay. So there are few of them that we need to take into account. If you integrate any function from 0 to nt, okay, where n is some integer of a periodic function f of x. So f of x is basically a periodic function with a period of t. Okay. Everybody knows the meaning of period of a function, correct? This is the definition of the period of a function. So period of a function is always a positive real number such that the function repeats itself for all values of x in the domain of the function for every change of t in the input value. So this is going to be integral of f of x from 0 to nt is going to be n times 0 to t f of x dx. Okay. Needless to give you the proof, you can make out how. Okay. Because the same area will be repeated n number of times. If you integrate from a to a plus nt, can I say it is as good as the previous answer only? Yes, sir. Yes, sir. If you integrate from a to b plus nt, can I say it will be a to b f of x dx? Plus n times, sir. n times 0 to t f of x dx, correct? If you integrate from nt to mt. Okay, by the way here a can be any real number. Here a and b can be any real number, but n has to be integer. Here m and n both are integers. Then it will just be m minus n. Minus n integral 0 to t. And finally the father of all the properties a plus nt to b plus mt. And this is the father because all the other properties can be evaluated from here by choosing some special values of a, b, m and n. So this will be a to b. Sir, this is the most generalized form. This is the most generalized form, yes. m minus n 0 to t f of x dx. Okay, so I'll request you to remember this. That will help you to solve all the questions of this nature. Is that fine? Can we take questions on this? Sir, one second. Shall we? Everybody's done copying it? Yes sir. First question is evaluate minus pi by 4 to n pi minus pi by 4 mod of sin x plus cos x. Easy question to begin with. Sir, is it 2 root 2n? 2 root 2n. That's correct. Yes sir, 2 root 2n. Yes sir, 2 root 2n, got it. Others? Niranjan, Ashutosh, Omkar, Shawan. One second sir. Vidushi? One second. So one second. So this resembles that a to a plus n t form. Correct? This is your a. By the way, this function is nothing but root 2 sin x plus pi by 4 mod. Okay? This is periodic with a period of pi. So you can write this as 0 to n pi root 2 sin x plus pi by 4 whole mod. Okay? So this function graph as you can all see will be starting from minus pi by 4 like this. Okay? So this is minus pi by 4. This is 0. This is 3 pi by 4. And this part is pi. Okay? So I can say this function will behave as from 0 to 3, 3 pi by 4, it will behave as root 2 sin x plus pi by 4. But from 3 pi by 4 to pi, it will behave as negative root 2 sin x plus pi by 4. Okay? I'm sure you can evaluate this very easily. Put the limits of integration that I'm not going to do. The answer is 2 root 2n. So make sure there's a mod function and you have to change your definition. So from 0 to 3 pi by 4, the definition is positive. From 3 pi by 4 to pi, the definition is negative. Okay? Let g of x be a continuous and differentiable function such that integral of integral. So basically they are integral within an integral is there. Integralception. You can write it like this. This is 0. And g of x is 0 when x lies between 0 to 2, where this is the greatest integer function. So these brackets are not ordinary brackets. They are greatest integer functions. Then this function g of x equal to 0 when x lies between 0 to 2 has exactly one real root, at least one real root, no real root, none of these. Sir, is the answer okay? Second? Graphically interpreted, it's very easy to solve. Is it a? A. Sir, don't tell if it's right or wrong. Okay, I don't know. Can you message me privately? Bharat? Sir, I just did it intuitively, sir. Okay. Sir, can you give us two minutes, sir? Yeah, yeah, sure. Third, a bit inside will be a constant. And so we're just saying that integral of that is... Then it's b, isn't it? You got my response? Yeah, we'll explain that, we'll explain that. It's actually a kind of intuitive only, but of course you can justify it with a lot of reason. Santosh, where are you? Oh my God, okay, I will just message you. Santosh is here. Santosh, you are correct. Niranjan, you are correct. Srijan, you are correct. Okay, answer is b. See, it's very obvious. Yes, sir. This function 2x square minus 3, okay? If you talk about your x belonging to root 2, 2, and a root 5 by 2, okay? 2x square means 2 into 2 minus 3, that's giving you 1. A root 5 will give you 2 into 5 by 2 minus 3, that's giving you 2, okay? That means it is somewhere between 1 and 2, right? So gif will be 1, okay? That means if you integrate the gif of this function from root 2 to root of 5 by 2 dx, it will always give you a positive answer, right? Okay. Now, you can treat this as a theorem also that if you have any function, let's say I give you a function f of x whose integral from a to b gives you 0. It means that this function will at least cut the x-axis once between a and b. That means f of x equal to 0 will have at least one root belonging to a to b. Why? Because if there is some positive part, there has to be some negative part also, correct? So let's say a to b is like this and the net area is 0. That means these two areas are compensating for each other. That means there has to be a point between a to b where the function has a root for this equation, right? Yes or no? Now, if you treat this entire thing to be your f of x, okay? That means there is a positive function, okay? Let me call this value as k value, k is positive. So there is k g of x integrating from 0 to 2, okay? Now, since k cannot never be 0, so you can, it's never negative, so you can just take it outside, okay? This guy, if it is 0, it implies that g of x equal to 0 should have at least one root between 0 and 2. Option number b is the right option, simple as that. So even if the first integral would give a negative value, it wouldn't matter, okay? Yes, because this is a fixed value that you get from it, absolutely. What if it is a multi-correct question? So answer all the correct answers, options. It will still be b only. Yeah, it can't be a because we don't really know. It can't be c because we got at least one root, it can't be d because we're getting b, so it has to be a pb. Yeah, the value of x satisfying this equation. Please note all the square brackets are gif functions and all the curly brackets are fractional part. Sir, is this single correct or multi-correct? Yes, sir. What is it? It's single correct. Damn it, it just became... Cams, cams, cams. Scam mode on. Sir, I hope my scamming work will just... Is it c? I will answer privately. Okay. Sir, is my answer correct? One second. Yes, your answer is correct. Scamming success. Do this in a rigorous way, man. No suitcase, no home car, that's rather strong. Yeah, yeah, we'll discuss that, don't worry. Santosh, you are correct. Okay, let's discuss. Yeah, let's discuss this. Okay, first let us use the properties of gif and all. 0 to 2, you can write this as this plus 14 into 2, which is 28. Okay, since 14 is an integer, you can always bring it out of the gif. Okay, okay. Now here, there's something very important that we need to understand. This is actually an integer at the end of the day, right? Yes, sir. Yes, sir. You know that this guy is not going to change its value beyond 1. And gif also won't change its value beyond 1. That means this guy will never change its value, so it's like a constant function for you right now. So you're integrating a constant function from 0 to a value which is less than 1. Are you getting my point? What I'm trying to say is that this is not going to change its value from 0 to a fraction part of x. Let's say x is 3.7. Correct. So if I integrate from 0 to fractional part of x, which is 0.7. Okay, this guy which is going to be 3 plus 14 always from 0 to 0.7. Getting my point? So this is as good as saying 14 plus gif of x times fractional part. Okay. So what about the right-hand side? Can I say right-hand side can be written as 0 to 28 plus 28 to 28 plus 2 fractional part of gif of x? Yes, sir. Okay, now we all know that this is periodic with 1. So x by 2 will be periodic with? Yes, sir. Can I say this is 14 times 0 to 2? Okay. And this is again 0 to 2 is the period. So can I say gif of x times 0 to 2? So this will be equal to, by the way, this is ordinary bracket. I'm so sorry, I used a curly bracket. In a land where you are having a special meaning for those brackets, we have to be very carefully choosing the brackets. This is how we should be writing it. Correct. Can you further simplify these values? 0 to 2, this can be 1. So this into 1. Do I need to explain how it is 1? No, sir. Okay. Similarly, this will be this. My daughter has woken up. Now there are two possibilities. Either 14 plus this is 0. That means this is minus 14. Or this is 1. Right? But this cannot be 1 for sure. So only this is the possibility. So if gif of a function is minus 14, that means x should lie between minus 13 to minus 14, excluding this, including this. Is there, is this in the option, by the way? By the way, this is not the right way to write it. 14 minus 14 is less. Is this there in your options? Yeah, option A. Is that clear? Sir, but over here scamming it is much, much easier, right? Yeah, I know that. But learning stops if you scam. Yeah. Sir, should we always rely on the rigorous proof or what? See, I always told you while you are doing assignments, while you are doing practice, rigorously you solve it. Make your life as difficult as possible. But when you are doing it in the exam, go to the other mode, make your life as simple as possible. Sir, but then it is very hard to switch between those two modes. No, it is easy. You have to be just honest with yourself, okay? Next question. Evaluate gif of sine x plus cos x from 0 to 2n pi. Sir, do the brackets mean gif or? Gif, yes. Sir, won't it just be 0? How will it be 0? You are not right. You will only get two attempts to answer. Wait, what? Oh my God! Sir, we have to find the periodicity of this. No, no, you have to evaluate this. Evaluate this. That's not right. Sir, one second sir. Sketch it, sketch it. You will get all your answers. Sir, is it root 2 minus 3? Sorry, root 2 minus 3 times n. No. Message me privately. Sir, could we have two more minutes? Sir, silly mistake one minute. Sir, I am getting an answer independent of n. No, the answer is very much dependent on n. It will be dependent on n. No, sir, one minute. Sir, one minute sir, just one minute. No, that is also wrong. No, Gaurav, wrong. Ashutosh is correct. Sir, is it minus of root 2 plus 1 times pi by 2? Wrong. Why are you getting such complicated answers? Okay, fine. Is it just pi? Wait, is it minus n pi? Gaurav Sushant? Yes, sir. Sir, is minus n pi correct or wrong? I don't know. Sir, could we have two more minutes? Sir, another silly mistake. Damn it. Sir, I didn't get any response. It will be minus. Sigh to Sushant. You ping me, I'll reply you also. No, Sharvanan. Okay guys, you have been taking a lot of trial and errors on this. See, sine x plus cos x. We all know it's nothing but root 2 sine x plus pi by 4. Correct. Once you draw the graph for it, everything is clear. Okay, let me do it on Desmos. Or what do you call it? GeoGibra. Things will be faster. Where is my GeoGibra? Yes. So, y is equal to floor of sine x plus cos x. Okay. This is my units in terms of pi, x axis, label in terms of... Okay, now see it. It's very obvious that this function is periodic with 2 pi. Correct. This function is periodic with 2 pi. So, first step that you could write here was, this integral will be nothing but 10 times 0 to 2 pi gif of sine x plus cos x. Okay. So, if you just figure out what is happening between 0 to 2 pi and whatever answer I get from there, you just multiply it with n. My job is done. Right. So, let's try to figure out from this graph. I'll zoom it so that you can all see. Yeah. So, my play area is only from 0 to pi. Guys, I'm not able to read your response right now. So, please hold on your response. So, I'm integrating it from this point to this point. Okay. By the way, I'm not taking the effort of telling you how to draw a GeoGibra function. You already know that GeoGibra function is drawn by first plotting the function. Make parallel lines at unit distance apart starting from y equal to 0 up or down both sides. Making that part of the graph which is sandwiched between two lines to fall on the line below to the floor line. Okay. Now here, if you see this part of the graph is a ladder. By the way, this position, what will be this position? Okay. This position is nothing but a pi by 2 position. Okay. So, from 0 to pi by 2, it is following this particular step size. That step size is of thickness 1. So, this area is pi by 2. Okay. From 0 to 3 pi by 4, this position is 3 pi by 4. Your area is 0 because the graph has fallen on the x-axis. Correct. From 3 pi by 4 to pi, this area. Okay. That's nothing but minus pi by 4. You'll all agree with me. Correct. Again, this part to this part. So, pi 2. 3 pi by 2. Right. This area is going to be negative 3 pi. Now, this part. This part is 7 pi by 4. So, the negative. So, only just be minus pi that area. Sorry. That region which you told. 3 pi by 2 minus pi is just pi by 2 into minus 2 which is minus pi. Yeah, minus pi. Which part? The third block. Even the fourth part is just minus pi by 4. Yes, sir. Now, if you sum up this area, what will happen? Minus pi by 4. This will cancel out with these two gentlemen. Okay. Minus pi only. So, your answer for this part. Sorry. Your answer for this part is going to be n into minus pi, which is nothing but minus n. Done. But since I'm doing it with a graph, I never redefined the function. But please redefine the function. That means this function. That is gif of sine x plus cos x. This will behave as 1 when your x is from x is from 0 to pi by 2. It will behave as 0 when x is from pi by 2 to 3 pi by 4. You can include 3 pi by 4 also. Okay. It will behave as minus 1 from 3 pi by 4 to just from that graph you can actually write down. Okay. Then you can say it's minus 2 from pi to 3 pi by 2. From 3 pi by 2 to 7 pi by 4 minus 1. It will behave as minus 1. Then 7 pi by 4 to 2 pi. And from 7 pi by 4 to 2 pi it will behave as 0. And you can accordingly choose your limits of integration as per the different definitions it takes. Is that clear? Simple. Next we are going to talk about the Leibniz rule. There are other properties also, which I will take after Leibniz rule. So Leibniz rule is finding the derivative of an integral. Or you can say derivative under integral sign. This rule has two things under it. First I will write the simpler part of this rule. This rule says if you have a function f of t which you are integrating with respect to t from say phi of x to psi of x phi and psi are some functions of x and you are differentiating the result. Then this result is f of phi x phi dash x minus f of psi x psi dash x. So this rule makes your life pretty easy. You don't have to find the integral and differentiate it. You can directly deal with the integrand. So there is no where I am evaluating the integral of small f of t. Sir, then should it be integral under derivative? You said derivative under integral. Isn't it integral under derivative? Derivative under integral sign that's fine. Now, how do you prove this particular thing? Very simple. Let's say f of t is the primitive of this particular function small f of t. Primitive means when you integrate f of t you get capital f of t. Anti derivative primitive whatever you want to call it. Okay. So we all know by f t c that if you integrate any function from phi of x to psi of x you put the upper and lower limit in the primitive and subtract the result. This is what we have learned from our f t c fundamental theorem of calculus. So if you differentiate both sides with respect to x means you are finding the derivative of the right hand side. Now if you differentiate this side you will be using your chain rule for it. And if you differentiate this guy you will be again using chain rule for it. Okay. And from this very definition that capital f of t is the primitive of small f of t. Can I write this term here as f of psi x. This I am copying as such psi dash x. Similarly this term could be written as small f of phi x into phi dash x. That is how your rule comes out. This formula. Now there is an extension of this rule which is a slightly complicated version and the proof for this I have already sent to you on the whatsapp group. You can just search for the Leibniz rule pdf. It will be there if you have not downloaded it. Please ask me again and send the proof. That rule is rule number 2 if you have a bivariate function f of x, t that means that function has both x and t variables. Where you are integrating with respect to t let's say from phi x to psi x and differentiating with respect to x. Then how does this rule change? It is different from our previous rule. In our previous rule we only had a single variable function f of t where we are integrating with respect to t and differentiating with respect to x. Here this function contains both the things. t with respect to what you are integrating and x with respect to what you are differentiating. Getting my point. So the rule changes slightly. The rule says you have to integrate from phi x to psi x the partial derivative of this function with respect to x that means with respect to whatever variable you are differentiating you have to do the partial derivative and of course you have to integrate this result again with respect to t minus the same thing continues what we had learned in the previous thing in place of t you write the upper limit that is psi x into derivative of the upper limit it's not minus it's plus here minus f of x is kind of like chain rule this is actually a combination of mean value theorem and change of limits formula okay let me pull out the proof for this sorry I didn't mean value theorem coming here yeah I'll just send you the proof for this once again on the group all of you have your mobile phones in front of you yes sir yeah no sir no sir okay let me do one thing let me send it to you can send it on the group I'll email it to your engine I've just posted it on the JEE advanced group the engine on send it to you thank you do you want me to open that up explain you or can you read it everybody everybody just take one minute to look at that proof Niranjan you don't have a phone with you right no sir send it to me on email okay so basically it's a use of first principles it's a use of change of the limits of integration then it's the use of flag ranges mean value theorem all these things are used to prove it so record this session I'll just open the proof in front of you just give me a second I'll just load whatsapp here can you see it now can you all see it now yes sir so see first of all I call this integral as IEX comma T okay this integral I name it as IEX comma T just follow my cursor over here then the derivative of this by first principles since you're finding the derivative with respect to X you only need to change X by delta X rest of the variables will remain same getting it all of you can read it yes sir okay now what I did was IEX plus delta X I just put the integral back so this was your integral this was your integral so X plus delta X means you're changing your limits like this so this integral over here which I'm showing with my cursor is your this integral in yellow and this integral over here in the first principle formula IEX comma T is this integral correct now I changed the limit of integration so from phi plus delta phi then I went to phi then I went from phi to psi then from psi to psi plus delta psi okay and finally this integral I copied over here in white you can see in white I have copied the the second integral over here okay then what I did I clubbed the position I clubbed these two integrals second integral and the last integral okay and first and third I clubbed it together okay by the way for the first one I switched the position of the upper and the lower limit okay since these two limits of integration are same I clubbed it as a single function here okay and these two integrals have started calling it as I1 and I2 wait can you repeat what you just said see what I did was you can see in my cursor right yeah the club the first and the second and the fourth second and the fourth I clubbed it first and the third I did it like this okay now these two limits of integration are same correct so I clubbed the first one and here this integral I called it as I1 and this integral I called it as I2 see I have written let's say I1 and I2 okay now then I use the mean value theorem recall mean value theorem says that this everybody knows the mean value theorem okay now I applied it to this particular function so this function I1 which is psi to psi plus delta psi okay I wrote it as this by FTC this is your FTC nothing else FTC okay but from this formula can I say if you treat your capital F psi plus delta psi comma t as your f of b and this as your f of a can I say it is f dash according to this rule itself the Langrange's mean value theorem itself f dash ct difference in the argument difference of b minus a that is psi plus delta psi minus psi correct so from FTC I took it to mean value theorem here okay now capital F is the primitive of small f so the derivative of capital F will be small f c comma t and this difference is delta psi okay now my logic here is that here c has to be psi because you are choosing a value between psi plus delta psi and psi itself which has to be psi itself because you can't choose any value because this interval itself is very very small so I did one thing I replaced my c again with psi ultimately what happened is this entire thing became f of psi t delta psi okay similarly the same result is applied to here it becomes f of phi t delta phi okay then I use this result back into my given expression so I replaced it back over here as you can see okay now by first principle we had a delta x down over here so I introduce it to both the terms okay and this term is nothing but the partial derivative of f with respect to x which is right over here and this term is nothing but f psi t and this is d psi or you can say psi dash okay this is phi dash and that's how the formula actually came about so normally we don't tell this formula in the class and all because many people don't understand the concept of the first principle for partial derivatives okay this proof is already sent on the group you can have a very deep look at it and try to understand all the parts of it and if at all you are feeling difficult in understanding it do let me know is that fine? yes sir okay so this is the rule which I have given to you now what type of questions can you get on this let's start with this sir no break right? what is the meaning of break? nice we break our butts and be like Gaurav sorry Gaurav Gaurav is yawning I did comments no I was not yawning okay let me just pull out a question immediately 10 this is my first question to all of you 0 to pi 1 plus a cos x by cos x dx here a is the value whose mod is less than 1 okay please evaluate this integral in terms of a of course a is the parameter here so your answer will be in terms of a any idea? I am sure many of you one minute how is it that a less than 1 can let us substitute another sin of cos what do you mean so one minute sir your angel is furious he didn't get a break today what is coming for me for English tomorrow? we have English tomorrow I want to revise portions I want to just listen what do we have for English tomorrow? how many chapters? 20 thank you sir you will leave your friend at 5 o'clock yeah bro at 5 he said 445 not even 5 he said come there same difference he started 5 almost a so how do you take it 1 by 1 plus this is a way sir the skin's property useful here that's what use anything you want property of the run tree welcome to state in simple words just because you don't understand then doesn't mean then always English exam try to understand sir this 2 minutes no sir you understand 2 minutes anyway may I do it? sir this is a one minute okay one minute sir sir is it 0? no 2 minutes of a not 0 it's 0 only when a is 0 by the way yeah it will be 0 and a let me solve this problem I will give you a chance to solve the other one so do we have to differentiate and integrate in some way yes Ashutosh we will discuss it don't worry okay so by this time you would have figured out that this is going to be some function of a correct let's differentiate with respect to a on both the sides that means d by d a of this expression now don't get surprised because even though a is a constant I am treating it as a function of a so you can always differentiate with respect to a now what role actually x played for you in the theory part a is playing for you x is playing t is role and x is playing t is role okay so according to Leibniz second rule that we discuss it will be 0 to pi partial derivative of this with respect to a so do by do and that will be 1 by 1 plus a cos x cos x by cos x this integral with respect to dx other terms won't even appear why other terms won't even appear because of when you differentiate pi and 0 you get 0 right because since both of these are constant constants we are not getting 0 minus 0 I did all this then I ended up with a simple integral and you are stuck now this is nothing but cos x is like a constant so 1 by cos x derivative of this is going to be 1 by 1 plus a cos x now remember you are differentiating with respect to a so cos x will be like a constant so cos x will come like this so this is gone this is gone this is what Ashutosh was also stuck at and he is asking how to integrate this right Ashutosh no sir that's me Sukiv sir can't you just take cos x by 2 tan squared x by 2 by tan squared half angles of tan I was stuck here when you were also stuck here Ashutosh also got stuck here so you can break this up as a 1 minus tan squared x by 2 that means you are not practicing your indefinite integrals yeah sir I sadly forgot this part take tan x by 2 st that means secant squared x by 2 into half dx is dt by the way this term is nothing but 1 by 2 1 plus t what am I writing 1 plus t squared by 2 dx is equal to dt so dx is equal to 2 dt by 1 plus t squared so changing this limit first of all when x is 0 t will also be 0 and when x is pi t will go to infinity okay and dx will become 2 dt by 1 plus t squared and you will have 1 plus a 1 minus t squared by 1 plus t squared multiply this word with 1 plus t squared so it will become 2 times dt by 1 plus t squared plus a 1 minus t squared okay this is t okay so let me first collect all the t terms together 1 by a and this will become t squared 1 minus a correct yeah cool so now now you take a 1 1 plus 1 minus a you can take out common 1 minus a you take out common it will become t squared plus 1 plus a by 1 minus a you can do a under root and square thingy right so your answer will be 2 by 1 minus a into 1 by a 1 by a will be this into tan inverse of t by a t by a would be this plus a c this will be f dash correct why I am putting a c limit of integration is already known 0 to infinity so when you put an infinity you get a first of all this is nothing but 2 by under root of 1 minus a square undoubtedly putting infinity will give you a pi by 2 putting 0 will give you a 0 okay so your answer will be pi by under root of 1 minus a square but this is your f dash a I have to find f of a right let me write it over here yeah and now we have to integrate with respect to a so f of a would be integral of this with respect to a that's nothing but pi sin inverse a right but put a constant of integration now how would I find this constant of integration we can use the fact that f 0 is 0 right why because if you integrate 0 to pi log of 1 by cross x it's going to be 0 itself so f 0 is 0 means f 0 is pi into 0 plus c so c is also 0 that means your answer is pi sin inverse getting my point so this cannot be done without the use of Leibniz rule is there any other way to do it just like brute forcing it I don't think so you will be able to do it easily right I am not denying it also right now because I have not tried it but there are a lot of integrals for example the next one also it doesn't give you any clue that you can't solve it without Leibniz rule you have to just you come to know from practice that we can't solve such questions without the use of Leibniz rule integration from 0 to pi can we solve everything using Leibniz rule like most of it see the indication is a presence of a parameter over there for example in this question also you see that I will introduce a parameter over here c sin theta where c is some parameter so evaluate this when such kind of thing occurs normally it gives you an indication that Leibniz rule may be applicable over here because normally we cannot solve such kind of integration with our brute force or our properties you may try it out actually do this question for c theta times tan inverse c sin theta with respect to theta from 0 to pi by 2 so tan inverse c sin theta right tan inverse c sin theta yes tan inverse of the whole thing c sin theta okay I have a response which is struggling with indefinite integrations pi by 2 lmc that's not correct Gaurav that's not correct I am at the end of the last integration okay take your time anybody else apart from Gaurav who wants to share the response Vidushi sir Santosh doing sir is it correct it's probably wrong that's not correct Gaurav that's not correct is my answer correct or is it completely wrong did you send your answer to me yes sir over what sir sir I didn't get it at all Sukirt sir I am still doing the indefinite integrations I have to practice that sir is my answer correct did you send it to me when I send it to the zoom just send it privately to sir over whatsapp it's a long expression oh you have send it on the whatsapp sorry sorry sorry I didn't look at it no that's not correct Bharath so is this a problem how far off am I see Bharath how can sin theta appear in your answer you are integrating with respect to theta and you have your limits for theta right so how can your answer be in terms of sin theta oh my sir I sent you something else we are on the chat or on the group or chat chat sir no no that is also not correct sir I am still stuck in the integration of indefinite do you realize it's the function of C yes let's differentiate with respect to C with respect to C on both the sides for that you have to do a partial derivative with respect to C first of cos theta tan inverse C sin theta ok this whole integration with respect to theta ok now cos theta is like a constant so it will remain as such this is going to be 1 by 1 plus C square sin square theta into derivative of C sin theta C is the variable sin theta is a constant so sin theta will come like this so sin theta cosic theta will take care of each other sir and I am stuck here that's all come on nobody can be stuck over here it's division by cos square both numerator and denominator oh that oh my god so it just becomes secant square theta d theta by secant square is 1 plus tan square plus C square tan square correct so 0 to pi by 2 secant square theta d theta this is going to be 1 plus C square plus 1 tan square theta correct this is your f dash c right everyone are you there with me so take tan theta as t so secant square theta d theta is your dt so f dash c is equal to 0 to infinity dt by 1 plus C square 1 t square okay I am sure you should be able to handle this so 1 by C square plus 1 if you take out common you will have 0 to infinity dt by t square plus 1 by okay so 1 by a will give you 1 by and a root of C square plus 1 square into tan inverse t when you put this thing you will get a pi by 2 this is your f dash c oh god of whatever I will try to do now to complete this process I just have to integrate with respect to C so pi by 2 dc by and a root of C square plus 1 that is going to be pi by 2 ln C plus and a root of C square plus 1 plus a constant of integration k okay now how do I find k from the simple fact that f0 will be 0 f0 will be 0 so f0 will be 0 means 0 is equal to pi by 2 ln 0 plus 1 which is 1 plus k that means k is also 0 that means your answer is pi by 2 ln mod C plus and a root C square plus 1 oh my this is I hard to kill is that fine Sukith Sreejan everybody yes sir yes sir yeah thanks sir okay one last problem pizza sir probability pizza probability pizza you want you don't realize it sir is never going to give the pizza it's fine we want the problems rather than pizza why not yeah I am going to definitely give you pizza someday I am definitely going to treat you with pizza if we get it right huh if we get the answer correct then yeah pi equals 0.001 in that case anyways also I am going to give don't worry that's if we get at and that's worse than getting a pizza problem that has low probability yeah probably someday you will get a pizza where is my collection of yes this is my question sir you are giving of the same type of a random question what what what are you giving a random question from the chapter or based on this topic I am giving on based on this concept only okay fine why what happened no no I just asked he is asking if you are giving a pizza problem from this concept or from the other concepts not this concept only okay question is let let f be a real valued function defined on the interval minus 1 to 1 okay I don't know to where it will go it will go to its let's say range whatever we just know that it is defined from minus 1 to 1 it's a real valued function such that this function satisfies this equation f e to the power minus x f of x is 2 plus integral 0 to x under root of t to the power 4 plus 1 dt for all x belonging to the domain of the function okay then then find f inverse dash at 2 find f inverse dash at 2 question is clear to everyone in this case the variable within definite integral matters yeah of course yes it matters you are integrating with respect to t, shawan means t to the power 4 plus 1 under root will be maturing is that your answer Santosh you wrote as x equal to something no Varad that's that's wrong that's wrong how far off am I I can't comment on being far or near it's a different answer altogether it's a fraction by the way the answer is a fraction oh nice okay but whatever you have written Santosh you are on the right track I understood your emotions our physics teacher used to keep saying that of now kusamsho anyone who is willing to take a chance it's a pretty messed up fraction no ya kusamsho it's a very resolved question it's not like too much of trick in that I'm going to take a very wild guess is it minus 1 by 2 that's wrong so 1 by e square that's wrong okay in the interest of time you just have 1 minute left and you have exams tomorrow also so shall we do it first of all okay see what is the how do you get f dash if you all know that for the purpose of convenience let me call f dash as g okay we know that g f of x is x correct okay so if you differentiate both sides this is what we get that means g dash f of x is equal to 1 by f dash x everybody knows this because we have already done this in the derivative chapter okay now this expression and if f of sir that is what I was doing now if you put this as 2 for what value of f of x f of x becomes 2 you can clearly take a guess from here that f of x becomes 2 when x is 0 correct yes sir that's exactly what I did I understood that you were trying to do that only so basically when you're doing this you're trying to evaluate 1 by f dash 2 f dash 0 do you agree with me everybody okay so my role is to find f dash 0 for that I'll use lemne's rule here simple so when you use lemne's rule over here first of all left hand side derivative you just need product rule correct so when you apply product rule over here it just becomes minus e to the power minus x f of x plus e to the power minus x f dash x right hand side will be 0 lemne's rule on this will be simply putting x in place into 1 rest of them would be 0 you don't have to write it down 1 by 3 sorry 1 by 3 1 by 3 is the right answer I added instead of subtracting I subtracted instead of adding too many mistakes you're making I'm sorry f of 2 I'm very scared that you're going to make these kind of mistakes okay so f dash 0 is 1 plus f of 0 what is f of 0 my dear students f of 0 is 2 so 1 plus 2 is 3 so your answer is 1 by f dash 0 answer is 1 by 3 done getting the point Sukirt yes sir and don't miss the animation of this yeah work on your accuracy Bharat I'm seeing almost 50% question you get wrong because of your accuracy fine guys so let's find up the session right here and all the best for tomorrow's English exam please start the syllabus now thank you sir thank you have a good night okay bye bye bye bye