 What if old pictures to someone I can send it right now? Oh, please do so. Okay, so let's start. I think it's already 4 o'clock. Definite integrals. I hope you can all see the screen and I'm audible to all of you. Yes, sir. Okay, so Shant has also joined in. Definite integrals. A very interesting chapter. Almost 70 to 80% of the question which is related to integration asked in cognitive exams are based on definite integrals because of course the options are plain and simple. So as the name itself suggests definite integral, that means the answer is the definite quantity. Okay. Now, first of all, we need to understand the first formula that we are going to talk about is basically the fundamental principle or fundamental theorem of integral calculus. It says that if you have a function whose derivative is f of x. Okay. So as to say that integral of f of x is capital f of x plus c. Okay. Remember, capital F is called the primitive or antiderivative, whereas small x is called the derivative. So the fundamental theorem of integral calculus says this result is going to be f of b minus f of a. And since this is a very different result, it's a numeric value. That's why such kind of integrals are called definite integrals. Okay. Quickly talking about the geometric interpretation when I say I'm integrating f of x from a to b. Okay. And let's say this is my graph of the function. And I'm integrating from a to b. So if this is your function f of x and I'm integrating it from a to b, it actually gives you the algebraic area between the function and the x axis. What do I mean by algebraic area? That means let's say this area which is above the x axis, this is called a positive area. Now, the question will arise in your mind. Why do we call this area as positive area? Very simple. If you look at the expression, you are actually integrating this, correct? So at a distance x, if you choose a very thin rectangular strip of dx. Somebody has pinged me just once again. Oh, okay. No problem. So if you choose at a distance of x, a very thin rectangular strip of the thickness dx, okay? Of course, the height of this will be f of x itself. Then f of x basically represents the height, d of x represents the width, okay? So this gives you the differential area or the infinitismally small area of the strip, okay? When you integrate this, basically you are adding all these small areas. Now remember, since height and width both are positive here. Now height is positive I can see because you are measuring in the positive direction up. How do you know width is positive? How do you know width is positive here? So we are going from left to right. Yes, if you are going from left to right means your x is increasing. If your x is increasing, dx will be positive, right? dx is positive means x is increasing, okay? So in this way, the entire area that comes out in this part, let's say I call this part as c over here. From a to c, the area is a positive area. Whereas when you find this area, okay? Here f of x itself, sorry, this is from c, okay? But if you find the area from c to b, your f of x is a negative quantity because the height here is negative, okay? So this becomes a negative quantity while this is still positive, okay? Thereby giving you a negative answer over here. So this entire thing is positive whereas this entire thing is negative. So this area comes out to be negative. So integral of the function from a to b gives you the algebraic sum of the area between the function and the x axis, okay? Many people say it gives you the area under that curve. That is not completely correct actually, okay? It doesn't give you the area but however it helps you to find the area. In order to find the area, you have to cleverly choose your points. You have to cleverly choose your signs of that particular part of the curve. Okay? Now, when you look at this formula, it actually reminds you of the fact that you first have to find out the primitive through antiderivative or use of indefinite integrals and then put the limits of integration like this, okay? Remember all the methods that you have learned for finding indefinite integrals will work fine over here, correct? Your method of substitution, etc. Everything will work. Integration by parts. That is also going to be working in this case. So what I'm going to do is before I introduce you to further parts of indefinite integrals which is basically a property-driven topic. This is a property-driven topic, okay? I'm going to ask you few questions just on the basis of how much you understand when you say integration of f of x from A to B is f of B minus f of A, okay? So I'll begin with a question. Okay, let me begin with this question itself. Sorry. Okay, let me start with this question. This is a very strange question. It says evaluate this directly as well as by substitution of x is equal to 1 by T and examine as to why the results do not tally. Okay, if you look at this question, would you all like to try it first? Yes, sir. Sir, can we answer? Yes, yes. Sir, because when we substitute x equal to 1 by T, the limits also change from minus 1 by 2 and 1 by 2. When we solve the integral, for T, we get the integral as half tan inverse 2T and for x, we get 2 times tan inverse of x by 2. Since the limits are different for both, we get a different answer. Limit? If it is different, it doesn't mean the answer would be different. When we substitute x as 1 by T, x goes from minus 2 to 2. But when we do that, T will also have to become 0 at one point, which can't happen. Exactly. The answer is there is a point 0 somewhere in between. So when you say T is equal to 1 by x and you can see that your limits of integration go from minus 2 to 2. That means there is a point where x becomes 0. There your T will become undefined. That means the function will suffer discontinuity at a point between minus 2 to 2 if you follow this substitution. In such a case, we cannot use this substitution. Getting the point? A very important learning from here is that whenever we are using a substitution, make sure that nowhere between that interval of x, the newly substituted variable becomes discontinuous. Getting the point? If you put it directly, minus 2 to 2 dx by x square plus 2 square, your answer is going to be half tan inverse x by 2. If you put your limits of integration, this is how we put the limits of integration. When you put a 2, it becomes half tan inverse 1 is pi by 4 minus of minus pi by 4. So I think this gives you pi by 4 itself as the answer. But the moment you put x is equal to 1 by T, dx is equal to minus 1 by T square dt. What will happen? Limit of integration of course will also change minus half to half. This will become minus 1 by T square dt 4 plus 1 by T square. Which is nothing but minus dt by T square plus 4 minus half to half. So this gives you the answer as half tan inverse of T by 2. Yes or no? So it's 4T square plus 1. Oh sorry, 4T square plus 2T. So it's 2 tan inverse 2T. And it's a minus also. What happened? Nothing sir. So when you put a half, it becomes a minus pi by 4. Correct? And again when you put a minus half, it again becomes a minus of minus of minus pi by 4. So it becomes minus pi by 4. That means your answer becomes negative which is impossible. Why? It is impossible because we know that this function is a positive function. 1 by 4 plus x square is always a positive function. That means it's always above the x axis. So how can the area under that become negative? Now this discrepancy has arisen because of a faulty substitution over here. Because my function becomes discontinuous at some point between minus 2 and 2 which is at 0 point. So hence, please be careful when you are using substitutions to solve the questions in definite integrals. By the way, just to clear that myth, if substitution changes, that doesn't mean your answer is going to definitely change. In fact substitution will change, the limits of integration will change every time you substitute x with a new variable. Okay? So in definite integral, we have to change the limits of integration for each new substitution. Okay? Let's take another question. Hope you can read the question properly. Find the value of integral of 0 to e to the power x square plus 2x minus 1 by 2 by x plus 1. Plus integral from 1 to e x ln x e to the power x square minus 2 by 2. Paper was quite easy, I saw this. Just two minutes to answer this. Just type in your response so you can also speak it out. Okay, any idea anyone? One second, one second. Okay, time up. So what I'll do here is, I will not disturb the second integral. Okay, let the second integral be integral from 1 to e x log x e to the power x square minus 2 by 2. Okay? While in the first integral, what I'll do is, I will substitute x plus 1st. Correct. So dx will be equal to dt. Okay? Limit of integration will change from 1 to e. Hope everybody knows how to change the limit of integration. It's very simple. When you put x as 0, t becomes 1. So lower limit will become 1. When you substitute x as e minus 1, t will become e. So upper limit will become e. Okay? So e to the power, this term here is nothing but x plus 1 the whole square. This is x plus 1 the whole square minus 2. Correct? So it's t square minus 2 by t. Okay? dt. Is that fine? Now, so divided by 2. Yeah. Now, if you see the limits of integration for both of them has become the same. Okay? And let me tell you in a definite integral, it doesn't matter the name of the variable. Okay? So you can call again t as x. Now, this is very surprising that you can, I'm again calling t as x because there's actually nothing in the name. Ultimately, you have to integrate that function and put the limits of integration. That's why the name of the variable at any instant can be changed to anything you want. Okay? So t here, I will put everywhere x doesn't matter. So what I'm going to do is I'm going to just change them to x again. So wherever there was a t, I put x. Okay? So, but this shouldn't be fair, right? I mean, because in the beginning of substituting something as x function of x is t. But this is getting out of value. So what is happening is treat this as a new problem itself, don't have any history to it. Okay? Once you have made the substitution, you have changed the limits, treat it as the beginning of the problem. See, if you want, you can keep on keeping different names, but it's just going to make your problem more complicated. I'm just using the same name just in order to bring it back to the same old function. Okay? Ultimately, you are going to put the limits of integration. So the variable name should not matter at all. Getting my point? So when you're talking about conversion of t back to x, think as if there was no previous relationship between t and x. Start with a new problem per se. Okay? So this becomes the same function. Okay? So you can combine these two functions and write it as single function. So e to the power x square minus 2 by 2. 1 by x plus x log x. Okay? Integration from 1 to e. Now, how do you evaluate this? How do you evaluate this? If you look at this very carefully, you would realize that this is the exact differential of log x into e to the power x square minus 2. Okay? If you differentiate this, you would realize you will end up getting the integrand over here. Write out. If you keep this as intact and differentiate log x, you get 1 by x. If you keep log x and this derivative differentiate this, you'll get log x e to the power x square minus 2 by 2 into 2x by 2, which is actually x. Okay? So this is an exact differential over here. So the answer for this will be just log x into e to the power x square minus 2 by 2. You just have to put the limits of integration now. So when you put an e, you get 1 e to the power e square minus 2 by 2. Okay? And when you put a 1, it becomes a 0. This is as we were saying root e to the power of e square minus 2. I think such an option is present in option number D. Is this clear? So one thing that we learned over here, something which was surprising that there is nothing in the name of the variable being used. So if you say integration of f of x from a to b or integration of f of t dt from a to b, integration of f of z dz from a to b, they are all the same. That is nothing but capital F of b minus capital F of a. So in the primitive of the function, you have to substitute upper limit and substitute lower limit and take the difference. Let's take another question. Find the value of integral of big pi x plus r. That is a product of x plus r, r from 1 to n times summation of 1 by x plus k, k from 1 to n. And this product you are integrating from 0 to 1. Two minutes to solve this again. In fact, two minutes is also a big time for this. Should solve it within one minute. I think Omkar needs to mute his mic. Done? Yes, sir. Okay. D. Fine. Okay. So let's discuss this. See here, when you look at this symbol, big pi of x plus r times summation of 1 by x plus k. Okay. If you expand it, it will be nothing but x plus 1, x plus 2, x plus 3, dot, dot, dot, dot, till you reach x plus n. Times 1 by x plus 1, 1 by x plus 2, till you reach 1 by x plus n. Okay. I don't know how many of you are able to recognize that this is actually the derivative of x plus 1, x plus 2, all the way till x plus n. How many of you are able to identify this? Remember, when you're differentiating this, that means you're using product rule, you differentiate one at a time. Correct? So let's say you differentiate the first term and keep others as such. That means you're going to get the same product with the first term. So x plus 1 will automatically get cancelled off and hence x plus 1 will automatically disappear. Right? So when you differentiate it, you just first differentiate the first term, keep others as such. Correct? Okay. Then you keep the first, differentiate the second, which is again 1 and then keep the others as such. Okay. And keep on doing so. And the last term would be x plus 1, x plus 2, all the way till x plus n minus 1. Correct? Now, if you take x plus 1, x plus 2, etcetera, common from all of them, the first term is just going to be 1 by x plus 1. Second term is just going to be 1 by x plus 2, etcetera, till 1 by x plus n. Okay? So this entire expression is just the integral of an exact differential. Correct? Okay. So this dx integration you are doing from 0 to 1. Okay? So basically you are integrating d of something. So answer is that something itself. So this is going to be your answer. And you just have to put the upper and the lower limit in that. So when you put a 1, you get 2 plus 3 all the way till n plus 1, which is n plus 1 factorial. And when you put a 0, you get a n factorial. So this. Okay? If you take n factorial as common, it will be n plus 1 minus 1. That's nothing but n into n factorial. An absolutely correct decision. Answer is d. Clear? Everyone? Next. If there's a function defined from R to R as sine x plus x, find the value of integral of inverse of this function from 0 to pi. Two and a half minutes to do this. Time starts now. I'm sorry. Is that greatest integer? No, no, no. It's an ordinary bracket. Greatest integer will always be listed out here. They would always say where this represents the gif function. If nothing like that is mentioned, you have to treat it as an ordinary bracket. So basically I'm just testing you on your substitution skills right now. I have not even started with the first property. Okay. Shrijan has already replied. I need two more answers please. Okay. Shavan has also said. Given his response. One more person I need. So one second. Okay. Fine. Okay. So most of you have almost replied. Let's solve this. Now many of us would be trying to find the inverse of this function to solve this question, but that is not required. Okay. What you can do is you can substitute f inverse x itself as t. Okay. So f inverse x itself as t. That means you're saying x is f of t. That means you're saying dx is f dash t dt. Okay. Let us put this in our given expression. So t f dash t dt. Okay. Now what about the limits of integration? When x is zero, t will be f inverse zero. Correct. Correct. So now how do I find the value of t? f inverse zero means f inverse zero means what value of x will give you this answer as zero. It's obvious that that answer would actually be zero. So f inverse zero will also be zero. So this will become zero. Next f inverse pi f inverse pi means what value of x will give this answer as pi. It is quite obvious that pi will be that particular value. Okay. So the limit of integration is not going to change anyhow. Fine. Now, in order to find the integral of this, I can use my integration by parts. Correct. By the way, if this is your f of x, f of t would be what? f of t would be sine of t plus t. So f dash t would be what? f dash t would be cos of t plus one. Oh, I got a very simple expression t into cos of t plus one dt again, zero to pi. Okay. So now t into cos of t, I can use by parts. Okay. And other term is going to be just t square by two from zero to pi. Now, for this, I will use my tic-tac-toe method or di method. So t and this is cos t. Differentiate this one, zero, integrate this sine of t. Integrate sine t again minus cos t. Okay. Just put these signs. I hope you still remember these methods. Very useful, saves a lot of time. So this will become t sine t plus cos t from zero to pi. This will just become a pi square by two. When you put a pi, this will give you zero while this will give you a minus one. Okay. And when you put a zero, it'll give you zero and this will again give you one. So your answer is going to be pi square by two minus two. Which option is that? Option A is the right answer. Option A becomes the right answer. So well done. I think the first one to answer this again was Srijan. Great. Let's have another one. So limit A tending to infinity one by A. Integral from zero to infinity x square plus A x plus one by one plus x to the power four. Plus tan inverse one into tan inverse one by x. This integral is pi square by K. K is a natural number. What is the value of K? Yes. Anybody? Okay. Santosh has given a response. I need one more response. Okay. So first of all, let's forget about one by A limit x A tending to infinity. We'll take this later on. Okay. First let us focus on the integral itself. Okay. So integral from zero to infinity x square plus A x plus one by one plus x to the power four tan inverse of one by x dx. Here, let us substitute one by x as t. So minus one by x square dx will become your dt. Limit of integration will become from infinity to zero. Okay. And let's make these substitutions in these terms. So x is one by t square A by t plus one by one plus one by t to the power four tan inverse t dx is going to be negative by x squared. x squared itself will be one by x squared itself would be t squared. Sorry, it's x squared dt. So it's one by t squared dt. So far so good. No problem. Now, if you simplify this, first of all, let us multiply throughout with t to the power four. Okay. So numerator will become one plus a t plus t square by one plus t to the power four tan inverse t dt. Okay. So what I did was I multiplied the numerator and denominator by t to the power four, but this t square will automatically cancel this off. So giving you the numerator term like this. Correct. Now, if I swap the position of this limit, I can get rid of this negative sign, right? Because there's a property which I'm going to talk about in couple of minutes from now. If you swap the position of the upper and the lower limit, your integral is going to become negative of the previous one. It's very obvious because let's say I'm integrating a function from a to b. My answer would be f of b minus f of a. Correct. If I integrate it from b to a, the same function, then my answer would be f of a minus f of b. And both are negatives of each other. Both are negatives of each other. Correct. Sir, we could use properties to solve this question, right? Sorry? We could use properties to solve the question, right? No, just one property is required, which is actually a very obvious property. Okay. Sir, I use another property. It's fine. It's fine. In the exam, they will not ask you whether you have used property or not. But this could be solved even without the use of the properties which are going to come, actually. Okay, sir. Okay. Now, nothing is there in the name, so I can put the name back as this tan inverse x dx. Correct. Now, this is your i. Okay. And initial was also your i. Okay. So, if you see this, this term was also your i. So, let me write that also simultaneously here. 0 to infinity x squared plus ax plus 1 by x to the power 4 plus 1. Now, tan inverse 1 by x, can I write it as cot inverse x? Because x is positive, when x is positive, remember tan inverse 1 by x is equal to cot inverse x. Okay. Now, when you add these two, let's add these two. Let's add these two. So, when you add these two, it gives you 2i is equal to 0 to infinity. Now, remember, these two terms are exactly the same. So, tan inverse and cot inverse will get added up. So, it will become x squared plus ax plus 1 by x to the power 4 plus 1 into tan inverse x plus cot inverse x, which is pi by 2. Okay. So, i is nothing. So, this is a property, right? Huh? So, this is a property, right? No, this is not a property. I'm just adding it. Okay. Okay, sir. I'm just adding it. It's not a property. Okay. Now, what I'm going to do is, I'm going to separate out x squared plus 1 by x to the power 4 plus 1 and I'm going to separate out x by x to the power 4 plus 1. Okay. I'm sorry. How did the tan inverse x become cot inverse x? I missed that part. No, no, no. It did not become tan inverse x. In the original question, there was tan inverse 1 by x. Correct? Yeah. Tan inverse 1 by x is cot inverse x. If x is positive. Oh, right. Sorry. That's what I did. And I added both the eyes. Okay. How do we integrate this? How do we integrate this part? You divide by x square throughout. You just have to follow your indefinite integrals. So, 1 by 1 plus x square. Here, if you take x square s t in the second integral, if you take your x square s t, then x dx will become dt by 2. Now remember, there will be no change in the limits of integration. It will still remain the same for both of them. Okay. Okay. So, I'm going to just take this up. So, in the first integral, we'll take x minus 1 by x to be your k. So, 1 plus 1 by x square dx will become your dk. Okay. So, this is going to be dk by k square plus 2. Okay. By the way, I'm going to evaluate it here itself tan inverse k by root 2. Okay. Now, what about the limits of integration? What will happen to the limit of integration? When you put infinity, it will remain infinity. Correct. But when you put a 0, what happens? It becomes minus infinity. Correct. Yes, sir. Okay. So, when you put up infinity, it becomes pi by 2 root 2. Okay. And minus infinity will become minus of minus pi by root 2. So, it will become plus this. Correct. Which is nothing but pi by 2 root 2 itself. So, the first integral evaluates as pi by 2 root 2. So, pi by 2 root 2. Integral is nothing but a by 2 tan inverse of t, which is again pi by 2. Okay. Remember, there was a pi by 4 factor outside. Don't forget that. So, finally, your result becomes pi square by a root 2 plus pi square a by 16. Correct. Now, the question actually was to evaluate the limit of 1 by limit of this multiplied with 1 by a. Okay. So, now let me finally use that. So, limit 1 by a of the result. The result was pi square by a root 2 plus pi square a by 16, a tending to infinity. So, that will become pi square by a root 2 a plus pi square by 16. Now, since a tends to infinity, this term will vanish, giving you the final answer as pi square by 16, which implies K is 16, which implies K is 16. So, option number C is correct. Let me check. Who all have given the right answer to this? Santosh was the only one who answered this and he was correct. Now, we are going to quickly talk about the properties. Properties of definite integrals. So, so far we learned that. One more important thing which I wanted to highlight. When you are putting the limits of integration. Okay. And you have already solved the problem by integration by parts. Okay. Be very, very careful. Let me ask you this question. Let's say if I integrate from 0 to pi by 2, x sin x. Okay. I have seen people, they apply integration by parts. Good. That's fine. But they misuse the limits of integration. How? They will do it like this. U into integration of this, that is going to be minus cos x. And in this minus cos x, they put the limit of integration only. X, they forget. Are you getting my point? This is wrong. Please do not do this. Okay. The basic that they do is minus integral of derivative of this, that is going to be one. Again, integral of this minus cos x. Again, here they put the limits of integration. Okay. And whatever they get, they try to integrate it again from 0 to pi by 2. Please note that these all operations are not acceptable. They are wrong operations. Okay. The right way to do it is if you are applying integration by parts. First, complete the problem thinking as if there is no limits of integration given to you. So first, integrate it normally without applying any limits of integration. So sin x integration is minus cos x. Okay. Just leave it like this. Minus 1 into minus of cos x. Okay. Finish off this agenda. So this becomes minus x cos x plus integration of cos x, which is going to be sin x. And finally, whatever result you get, there you put the limits of integration. This is the right way to solve the question. Are you getting it? Or if you want to put the limit of integration, you have to put the limit of integration on the whole result over here. Are you getting it? And you have to put the limit of integration on the whole integral of this. You can't put inside partially. Okay. What I suggest is do it after completing the integration by parts. Is that clear? So let me begin with the properties. The first property I've already discussed with you. There's nothing in the name. You can always change the name of the variable being used. So f of x integral from a to b is same as f of t integral from a to b. Second also I've discussed with you. If you change the position of the upper and the lower limit, it becomes negative of the previous integral. Third property is you can always break the limits of integration. Let's say from a to b, you can make it as a to c, then c to b. Okay. And the c can lie anywhere. If this function is defined for all real numbers, if let's say f of x is defined for all real numbers, that means it is from r to r. Okay. Then c can be anywhere. But if this function is defined only from a to b, then c has to be between a and b. Okay. So many books write that c has to be between a and b. This is to be only followed when your function is defined from a to b only. It is defined from or defined in the interval a to b. Okay. Else you can take c to be any real number. Okay. Here c could be any real number. So it can be only when f of x is defined in a to b. It can be when f of x exists at c. Yeah. That's what f of f of x is only defined between a and b, then c has to be between a and b. But if f of x is defined for all real numbers, then c can be anywhere. Okay. I've seen Adi Sharma writing this, but please note that this is only true when the function is defined from a to b. Now, this can be further generalized and you can break the limit of integration at any n number of points. So you can break it. Why is it only true like that? I mean, if it is in that case, if it is that case, then it can be written as minus of that, which is correct. Sorry. It can be written as a to c plus integral c to b would become negative. Why? Yeah, let it be. You're going less. Yeah. Let it be. So you need not be between a to b, right? But if your function is not defined beyond b, how will you go beyond c? How will you go beyond b? Yeah. Fine. Correct? That's what I said. If your function is defined for all real numbers, you can go anywhere you want. But if your function is only between a and b, it is not defined anywhere else. You can't choose a value beyond a and b. Okay. So this property says that you can keep breaking it at so many points, depending upon the requirement of the function. Okay. So let me finally break it from cn minus 1 to cn and finally from cn to b. Okay. Now on what basis do we break this function? Basically, there are certain criteria that we follow. First is if the function faces some kind of a discontinuity in between. Okay. Then we prefer breaking the function at the point of discontinuity. Okay. Because any interim substitution there on, if there is a discontinuity, it would influence your answer. So that is where we prefer breaking the limit of integration is where the function is changing its definition. We deal with a lot of special functions like mod function, gif function, piecewise defined function, inverse trigonometric function. These functions keep on changing their definition in different, different intervals of x. Okay. So there also it becomes important to break the limits of integration. Are you getting my point? These are the scenario under which we are going to break our limit of integration. Okay. If it is a special function, if it is a piecewise function, or if it is going to be an inverse trigonometric function, or if it is going to become discontinuous at some point, then it is advisable that you break the limit of integration. Now at the point of, at the critical points I would say, or at the point where there is a, there may be a problem arising. We'll take questions based on this first. Let's start with, let's take this question. The value of integral from 0 to 100 of gif. Now this is a gif function. Remember whenever there's a gif function, they'll mention it in the question itself. Unless until mentioned, don't treat any square bracket as gif. Two minutes to solve this. Okay. Srijan has given a response. Well done. Srijan you are correct. Now this type of problem is best solved by plotting the graph. If you plot the graph of tan inverse x. Okay. We all know the graph of tan inverse x looks like this. Okay. Where it dies at pi by 2. Pi by 2 is 1.57. Correct. This is minus 1.57. Okay. Now whenever you're plotting the graph of gif of any function, we normally make steps of size 1, 1 each. Correct. So this is your line y is equal to 1. In fact down also we can make a line y equal to minus 1. But that is not going to be of much use to us because we are integrating it from 0 to 100. Okay. So let's say 100 is some place over here. Okay. Now remember when you're taking the gif of this function, this part of the graph, this part of the graph is going to fall down on the floor like this. Okay. Leaving behind a hole over here and a solid dot over here. Correct. Now after this part that is now, which is this value actually if you see this is your 1. Correct. So when you're saying tan inverse x is your 1, that means x is tan of 1. So this value arises at tan of 1. Correct. Post this part, this entire part of the graph till infinity is just going to fall down on this red line. Okay. It's all going to fall down on this red line. So ultimately, ultimately the area that you have to find out, let me just show you that area. The area that you have to find out is the area under these two steps. Right. Now this step is not going to give you any area, but this step is going to give you an area of this rectangle. Okay. So dimension of this rectangle is with his 1 and length is going to be, this length is going to be 100 minus tan 1. So your answer will be 1 into 100 minus tan 1, which is option number C. Absolutely correct Gaurav and Shrijan. So try to approach this question. Try to approach this question from graphical point of view. Let's try another one. Okay. So integration of X to the power GIF of X square plus GIF of X square to the power of X from 1 to 2. Anyone? In one minute. Yeah. Yeah, one second, one second almost. Ashutosh came for a very brief time when he left. How is your preparation for tomorrow's exam? Good, manageable? No. What happened to Shrijan? He left so many days off, right? Yeah, but I can't really start until today morning, so messed up there. Okay. Some responses come. Okay. Not really. That's why Niranjan has muted himself because he's quietly opened his computer science book and he's studying it. Whenever a student says my phone is not working means don't expect any answer from me. Okay. Santosh says dash. I'll not say what he said. Okay. Let me tell you Santosh and Shrijan. Both of you have guessed it right. Or got it right. Mine wasn't getting the exact answer. I'm getting a root 3 in the answers because I'm getting the remaining part of the options. I'm getting a term extra. I don't know from where it's coming. You're getting a term extra? Yeah. Okay. Sure. Others? One second. Yeah. Is it B? Yes, it is B. See, since there is a JIF of X square, you know that this will change its value at root 2 and root 3 also, right? Okay. So critical points would be root 2 and root 3. So we have to break the limits of integration there. So 1, 2 root 2. Okay. 1, 2 root 2. You are integrating. Remember, you are integrating X to the power of 1. X to the power of 1. Plus this is going to be 1 to the power X, which is 1 itself. Okay. From root 2 to root 3. You'll be integrating X square. Plus 2 to the power X. Correct. And from root 3 to 1, you will be integrating X cube plus 3 to the power X. Okay. So remember here, there was a special function, which was JIF of X square. And hence in light of that, we had to break the limits of integration at root 2 and root 3. So this will just become X square by 2. X square by 2 plus X root 2 to 1. This will become X cube by 3 plus 2 to the power X by ln 2. From again root 2 to root 3. And this will become X to the power 4 by 4 plus 3 to the power X by ln 3. Again from root 3 to 1. Okay. So let's put these values. 2 square by 2 will give you 2 by 2, which is 1. So 1 plus root 2. Minus half plus 1. Yes, sir. Here I will get 3 root 3 by 3. 2 to the power root 3. Yeah. This will give you 3 square 9 by 4. Okay. Let's collect first the rational terms. So we'll have a 1, 1 gets cancelled over here. So we have a minus half. And this will become minus half plus 1 by 4 minus 9 by 4. Correct? Any other term which I'm missing out? Any other term which I'm missing out? Okay. So what does this give me? This gives me this is minus 1 by 4 minus 9 by 4. That gives you minus 10 by 4. Do we have such an answer? Oh, plus 5 by 4 is there. That means some term I'm missing out. 1 by 2 plus 1, 3 by 2. 3 root 3, this will not give me. 2 root 2 will also not give me. 1 by 4 is there. Minus. So how do we get the 1 by 4? One second, one second, one second. This limit of integration here goes till 2, right? So this is 2 over here. I wrote a 1 by mistake. I'm sorry. So this will become an 8 by 16 by 4. And this will become 3 to the power 2 by this. Sorry. Yeah. So this will give you. This will give you. This will give you 1 minus half minus 1 plus 4 minus 9 by 4. So I think this gives you take an LCM of that is 7 by 2 minus 9 by 4. That's going to be 14 by 4 minus 9 by 4 plus 5 by 4. So this is going to be plus 5 by 4. So I'm just going to write the answer down plus 5 by 4. Now, let us focus on the irrational terms. What do we get? Okay. But I realized my mistake. I got it. You got your mistake? Yes. Okay. So root 2 and minus 2 by root 2 will give you plus root 2 by 3. Okay. This should be there in your answer. Just check. This is going to be just root 3. Okay. Now let's figure out log 2 terms. Log 2 terms would be 1 by ln 2 2 to the power root 3 minus 2 to the power root 2. Is this somewhere in the answer? Yes, it is there. And we have to now see 1 by ln 3 terms. That is going to be 3 to the power root 3. In fact, it's minus 3 to the power root 3 and there would be a 3 squared term. So let me write that first. 3 square minus 3 to the power root 3. Okay. Is this there in the answer? I think, yeah. Option number B is correct. Question. I think after this you should be able to manage. After this it was just integration. Correct. I'm putting the limits of integration. Let's have another one. Let's take this question. This could be easily done through graphs. Hope you know how to deal with min function, max function. I already discussed this a couple of times in the class also. Draw both the graphs and deal with the bottom most part of the graph. Done. Okay. Shavan is given a response. Shavan, that is not correct. Try once again. Oh, happy birthday, Shavan. Thank you, sir. Where is the party? No party, sir. I've been studying computer science. So your birthday is very close to Narendra Modi's birthday also, right? Yes, sir. Great personality. Both of them. Yeah. So Rahul Gandhi wished Narendra Modi happy birthday and said where is the party? So Narendra Modi said I have a party in every state. Where is your party? Okay. So Santosh says D, Niranjan says C. Both of you are wrong. That means the only answer left is B. B is the right answer. So guys, I think you have not been able to identify this function. This function is your fractional part of X, isn't it? Okay. So let us try to draw fractional part of X. Fractional part of X, we all know that it's a line like this. 0, 1, 2, like that. Okay. This is nothing but fractional part of minus X. Let me just reflect the same thing about the Y axis. Right? Changing the sign of X means reflecting the graph about the Y axis. Correct? By the way, let me just draw more part of the graph. I should not stop here. Minus 1, like this. Okay. Now for the graph of negative X fraction part, you just have to mirror image it about the X axis. Oh, sorry, about the Y axis. Correct? So this will become something like this. So I'm drawing it in white. Okay. So this will become something like this. Okay. Something like this, something like this, something like this, and so on and so forth. Okay. Remember, my playing area is only from minus 2 to 2. So I don't bother about any part of the graph which is beyond this. Now, minimum means the lowest part of the white and the yellow graph. So as you can see, the lowest part, I'm just changing my color of my pen. Let me make it blue. The lowest part would be the one which I'm shading with blue. This is the lowest, most part of the graph. Correct? So the area that I'm looking out for is the area under these blue shaded graphs or blue boundary graphs. Correct? So that area is nothing but four triangles over here. Now, each triangle has a height of half and a base of, altitude is half, base is one. So each triangle, so one triangle area, area of each triangle is half base into height. That's actually one fourth. Okay. And there are four such triangles. So your answer is going to be four into one by four. That's nothing but one. Option number B is correct. It should be an easy one. Okay. Next question is, yeah. Next question is, find the integral of, find the integral of cot inverse of tan x from zero to pi. Find the integral of cot inverse of tan x from zero to pi. Oh, Santosh, bad luck. This time I'm not giving you the options. Zero? No, zero is not the right answer. Pi square by four, pi square by minus pi square by two. No, none of you are correct. Oh, Santosh said pi square by two. Yeah, Santosh, you are correct. Sorry, I didn't see that. Pi square by two. Right. Pi square by two is the right answer. How do we solve this question? How do we solve this question? Now, try to recall your definition of cot inverse of cot x. Yeah. If you recall the graph, I don't know how many of you recall the graph for this. The graph for this looked like this. Like this. Okay. So this end is pi. This end is zero. This is pi like that. Is that fine? Any doubt regarding the graph? No, sir. Now I have to find cot inverse of tan x. How do I do that? So can we substitute x as pi by two minus t? Absolutely correct. So let us substitute x as pi by two minus t. Okay. dx will become negative dt. Right. So the integral becomes cot inverse of tan pi by two minus t, which is cot t. Negative dt. What about the limits of integration? Limit of integration will now become. Pi by two to minus pi by two, right? Yes, I switch it over. But we switch it over and observe this negative sign. Okay. So I have to also write the definition of this from minus pi by two to this. So it will be like this. This is minus pi. Okay. Now, as I can see from pi by two to zero, the function definition is something else and zero to pi by two, the function definition is something else. So I have to accordingly. I have to change the definition according to the limits of integration over here. So from minus pi by two to zero, your function will behave as t plus pi, correct? And from zero to pi by two, the function will just behave as t. Correct? So this is the integration that we need to perform, which is quite easy. Okay. From minus pi by two to zero. And this will become a t square by two from zero to pi by two. So when you put a zero at zero, when you put a minus, this becomes pi square by eight. Okay. Minus pi square by two. And this becomes pi square by eight again. Minus zero. Okay. So pi square by eight and pi square by two will get cancelled, leaving you with pi square by two as your final answer. So be very, very careful while dealing with special functions and while dealing with inverse functions. Next property, which I'm going to talk about is the most important of all properties, which is called the King's property. Okay. It's property number four, I guess. This property says integral of a function, which is continuous from the interval a to b, can be written as integral of this. Right? It's very easy to prove this geometrically. If you look at the function graph, let's say this is your function graph from a to b. Okay. So you want to find out the area under the graph. Okay. Now this property is saying that even if you plot the graph of this function, the area is not going to change from a to b. How? Let us check it out. If I have to draw the graph of y is equal to f of a plus b minus x. Okay. First, what I'll do, I'll come from the graph of f of minus x first. Okay. So f of minus x graph is nothing but reflect this graph about the y axis. So it will be something like this. Okay. This point will become minus a, this point will become minus b. Now what I'm going to do is I'm going to change my x with x minus of a plus b. That means I'm going to shift this graph to the right by a plus b units. If I do that, this end will come at b, this end will come at a and the graph will start looking like this. So the property says that this is nothing but your this graph. So this property says that this area and this area are the same, which is actually correct. Because you just flipped the position of the corners of the graph, which doesn't change the area within the graph. Okay. So this end came over here. Okay. And this end came over here and your graph started looking somewhat like this. That doesn't mean the area is going to change. Now a very special case of this property is integration from, let me put a star here. A very special case. Take a special case. This integral from 0 to a f of x is integral of f of a minus x from 0 to a. All I've done is I've made my a as 0 b as a. Okay. And this property has come about. Let's take questions on this because this property is going to be used at so many places. So let's start solving few questions on them. Let's take this question. Done. Okay. Srijan is okay. Fine. No. That's not right, Srijan. B is also not right. No, that is also not right. No Santosh. Not right. No God of not right. Okay. Let me discuss this. Everybody has started. None of you have given the right answer. So let's apply. Let's call this integral is I. Let's apply King's property. So backings property. This will become 0 to pi. Pi minus x. Okay. Now the sin 2 pi minus 2x sin 2 pi minus 2x is minus sin 2x. Okay. And here I will get sin pi by 2 minus cos pi minus x, which is again minus pi by 2 cos x. Okay. Can I say this minus and this minus will take care of themselves because sin of minus theta is minus sin theta. So I'm just going to apply to the denominator. So yeah. Yeah, I'm sorry. In denominator also this will become. Yeah. So this becomes pi minus pi minus 2x. Correct. So I can write this as can I just make it as x minus pi here and make everything the same. Now let's add these two. Let's add these two. So 2i will be equal to 0 to pi 2x minus pi sin 2x sin pi by 2 cos x by 2x minus pi. This and this will get cancelled off. Correct. So I'll be left with 2y is equal to 0 to pi. Let's open this sin 2x as 2 sin x cos x. This 2 and this 2 gets cancelled. Let's take cos x as t. So sin x in fact minus sin x dx will become a dt. So I can write this as 1 to minus 1 instead of that I'll observe the minus sign. Can I write it as t sin pi by 2t. This is going to be right. Correct. Now on this I can apply integration by parts. I can take this as u. I can take this as v. Let's apply di method. So t sin pi by 2t. This will become derivative is 1. This will become a 0. This will become sin integral is minus cos pi by 2t divided by pi by 2. And again if you integrate it it becomes minus sin pi by 2t divided by pi square by 4. Plus minus plus. So your answer is going to become minus t cos pi t by 2 divided by pi by 2 plus sin pi t by 2 divided by pi square by 4. The limits of integration is from minus 1 to 1. Now when I put a 1 here I get a 0. When I put a 1 here I get 4 by pi square. Minus. When I put a minus 1 again I get a 0. And when I put a minus 1 here I get minus 4 by pi square. That means your result is 8 by pi square. Option number c is correct. None of you replied with c. I was mostly getting d as the answer. Is that fine? Got your mistake? Let's do the next question. Integrate. Integrate log of cot a plus tan x from 0 to a. Okay? A belongs to the interval 0 to pi by 2. So what was the previous answer? 8 by pi square. Option c. Yes sir. Thank you sir. Okay. Anyone? Sir one second. Can I say it? Sorry? Can I say the answer? Yeah tell me. Is it a by 2 log cos square a? a by 2 log cos square a. A log cos square a. Yeah I think it's correct. The answer that the book says is minus a log sin a. Minus a log sin a. Okay let me discuss this quickly. So first cot you can write it as cos a by sin a. This you can write it as sin x by cos x. Okay? 0 to a. If you take the LCM etc. Inside it becomes cos x cos a plus sin x sin a. By sin a cos x. Okay? So nothing but 0 to a log of cos x minus a. By sin a cos x. If you split this up as two separate terms. That is log of cos x minus a. And minus log sin a cos x. You can split this further as 0 to a log of sin a. And 0 to a log of cos x. Now on the first one over here you apply King's property. You apply King's property on this. So when you do that it becomes 0 to a log of cos. In case of x we'll put a minus x. So a minus x minus x will leave you with a minus x. Remember cos doesn't care about a minus over here. And as a result these two will get cancelled off. You don't have to actually evaluate them. So you boil down to integration of log of sin a. With respect to x which is nothing but minus x log sin a. Because it's a constant. Put your limit of integration it becomes minus a log sin a. That's your answer. Now I'm sure you would have come across a very special case of this in your NCRT textbook. Where they asked you to integrate 0 to pi by 4 log 1 plus tan x. How many of you have seen this question? Me. You have seen this question right? What was the answer for this? If you remember the answer the answer for this is going to be minus pi by 4 log 2 to 2. In fact minus pi by 8 log 2. Yeah log half. Or you can say like this one second. You can say pi by 8 log 2. In fact if you see this is the same answer. If you put your a as pi by 4 it becomes minus pi by 4 log of sin pi by 4. Which is nothing but minus pi by 4 log of 1 by root 2. That's nothing but pi by 8 log 2. So did I miss anything important? I just now joined. Like what has not been covered in school or anywhere else? Of course we are taking a variety of problems which may not have been asked in school. So can't really comment till you see the recordings. Don't worry. You can join in from here on. Where were you? Sir I had gone out. Okay. Let's try. I think this problem you had already done in school. Correct. Evaluate this or this. It's a one minute question. Pi by 4. But the first one will also be pi by 4. It's a super easy question. It's an NCRT question. So all you need to do is in this problem the first part you need to convert by substituting your x as a sin theta. So dx is going to be a cos theta d theta. So limit of integration will become 0 to pi by 2 a cos theta d theta divided by a sin theta plus a cos theta. Okay. In fact if you cancel out a pi by 2 it just becomes if you cancel out a from the numerator and denominator and divide by cos theta it becomes tan theta plus 1 which is actually the second problem. Okay. Now in order to deal with this at this stage I will apply in fact I can apply it at other stage also. I'll apply King's property. So this will become 0 to pi by 2 cos pi by 2 minus theta is sin theta sin pi by 2 minus theta is cos theta and this will also become a sin theta. Okay. Add these two up. If you add these two up 2i will become 0 to pi by 2. In fact we'll get a 1. Okay. And 2i will be equal to pi by 2. So i is equal to pi by 4. Easy question. Next. Let f of x be sin x by x. Then integral from 0 to pi by 2 f of x into f of pi by 2 minus x is equal to which of the following options. Okay. Sreejan you are correct. Anybody else? So once. Okay. Santosh you are correct. Omkar no you are not correct. Sukirt that's correct now. Omkar that's correct. Yes sir got my mistake. So much of a time you are taking. So 0 to pi by 2 f of x is nothing but sin x by x f of pi by 2 minus x is nothing but cos x by pi by 2 minus x. So it's 0 to pi by 2. Can I just multiply it through orbit 2 and make it sin 2x by x pi minus 2x dx. Now look at the options. Options all talk about 0 to pi 0 to pi 0 to pi. So the obvious substitution that comes into mind is let's put 2x as t. Right. So dx will become dt by 2. So it becomes 0 to pi sin of t dx is dt by 2. Okay. x here will become t by 2 pi minus t. Okay. This 2 and this 2 will get cancelled off. That means 0 to pi sin t by t pi minus t dt. Correct. Now what I can do next is I can split this up as sin t. Let me just work on this term only. Can I split this up as sin t 1 by t plus 1 by pi plus t. Correct. That's correct. Divided by a pi. Okay. So what I'm doing is I'm just splitting this as if I'm splitting up partial fractions. Okay. So this becomes integral of 0 to pi sin t by t pi minus t dt as integral of 0 to pi 1 by pi sin t by t plus 1 by pi sin t by pi minus t. Right. Now let me use King's property on the second integral. By the way, this is already my first integrals. I'm going to disturb it. Let me use King's property on this. Let's apply King's property on this. So this will give me 1 by pi is 0 to pi sin t by t dt plus 1 by pi 0 to pi sin pi minus t is going to be sin t. And this denominator is going to become a t. So it's the same thing written twice. 0 by pi 0 to pi f of x dx you can write or f of t dt you can write doesn't make a difference. There is nothing in the name as I already told you. So your answer number or option number A becomes the right option. Is that fine? Any questions? Let me move on to property number five now property number five is what I call as halving the limit property. This property says integral of a function which is continuous. Let's say a function my function is continuous in the interval 0 to 2a. You can write this as integral 0 to a f of x plus 0 to a f of 2a minus x. Now remember this is the parent property which many of us don't remember. We just remember the results of this parent property. Which I'm going to do a little later on. But first I want to I would like to prove this particular formula itself. Okay, both geometrically and non geometrically. So let me do it non geometrically first. If you start from left-hand side 0 to 2a f of x you can break this up as 0 to a f of x and a to 2a f of x dx correct. Now first integral I will not disturb. But in the second integral I'm going to replace my x with 2a minus t. That means my dx will become minus dt. So this will become this will be as such this will become now limit of integration when you put x as a t will become a. But when you put x as 2a t will become a 0. And instead of x I will put 2a minus t. Instead of dx I will put a minus of dt minus of dt. Okay, let me just write this also along with it. Now what we can do is we can observe the negative sign and switch the upper and the lower limit position. So 0 to a f of x dx is again 0 to a f of 2a minus x dx. Okay, in fact, I have changed the name of the variable from t to x back because there's nothing in the name and that's how the property comes up. We can prove this very easily by using geometry also geometrical proof. So if you are trying to find out the area under a function, let's say f of x was graph is like this from 0 to 2a. Okay, this particular property says that it is same as let's say I call this area as a1. Okay, the same as adding these two area. Now this area is nothing but the same graph the same graph but only from 0 to a. Okay, so this area. Okay, let me call this as a2. And the graph of this function and the graph of this function. So let me first draw the graph of f of 2a minus x. See f of 2a minus x first you need to go draw the graph of f of minus x. Okay, first draw the graph of f of minus x. You know the graph is going to be mirror image about the y-axis. So this is your graph of f of minus x. Correct. To get the graph of f of 2a minus x replace your x with x minus 2a. That means shift this graph to a units to the right. So if you shift this graph to a units to the right, it comes like this. Correct. In other words, if I draw it over here itself, your graph will appear to be like this. Now, 0 to a area would be this area. 0 to a area would be this area. Let me call this area as a3. Okay, now you would appreciate here that if you split a1, you would realize that this area was actually your a2 which you have shown over here. Okay, and this area is nothing but area a3. So a1 is made up of a2 plus a3, which is what this property says. So this is your a1. So this is your a2 plus a3. This is what this property says. So this is your a1. So this is your a1. This is your a2 and this is your a3, which is actually correct. Is that clear? Now, there are two important outcomes of this particular property. I call it as two important conclusions. First conclusion is if your function satisfies this functional equation, that is f of x is f of 2a minus x. Now, when does a function satisfy such kind of functional equation? Assuming function, not exactly. Symmetric about x is equal to a. Yeah, this will be satisfied when the function is symmetrical about the line x equal to a. Okay, if a function is symmetrical about x equal to a, let's say the graph of the function is like this. Okay, this is your x equal to a line. So if you're integrating this function from a to 2a, sorry, 0 to 2a, it is as good as saying, take this area and double it up. That means take the area from 0 to a and double this area up. Okay. Conclusion number two. If your function satisfies this functional equation, f of x is equal to negative of f of 2a minus x. When does the function satisfy this functional equation? When it is symmetrical about a point a comma 0. Okay, so as to say that the function must cross or look symmetrical about a comma 0. That means whatever is the area over here, remember the same area would be below the x axis also and they will cancel each other out. So what will happen? Your integral of f of x from 0 to 2a will collapse and become 0. Okay, so these are the two conclusions that we normally draw from this property. Please do not forget the parent property that is also very important. Many people just remember these conclusions which is not sufficient enough. Let's take questions on this. Let's start with this question. If u is defined as integral of cos of 2 pi by 3 sin square x from 0 to pi by 2 and v is defined as integral from 0 to pi by 2 cos of pi by 3 sin x, then what is the relationship between u and v? Okay, no problem. Anyone? Okay. Shrijan, that is correct. Santosh, that's also correct. Gaurav, that is correct. Okay, time to solve right now. So u is 0 to pi by 2 cos of 2 pi by 3 sin square x. So let's apply King's property. Let's apply King's property. So u will be 0 to pi by 2 cos of 2 pi by 3 cos square x, right? Let's add these two u's. Let us add these two u's. So if you add these two u's, you get 2 u is equal to 0 to pi by 2 cos 2 pi by 3 sin square x plus cos 2 pi by 3 cos square x. Let's use the transformation formula which is cos a plus cos b. Cos a plus cos b is 2 cos a plus b by 2 into cos a minus b by 2, okay? So 2 u will be equal to 0 to pi by 3 2 cos a plus b by 2, can I say it will become cos pi by 3? And cos a minus b by 2, can I say it becomes pi by 3 cos 2 x? Remember, cos doesn't care about the negative sign within it. It's an even function. Cos pi by 3 and 2 will get cancelled. So 2 u is equal to 0 to pi by 2 cos pi by 3 cos 2 x. In order to bring it to your expression of v, can I make a substitution over here? Can I call 2x as t? So dx will be equal to dt by 2. So 2 u equal to 0 to pi cos pi by 3 cos t. dx is dt by 2 which is half. Now here we have a function. Let's say I call it as f of t. Remember f of t is same as f of pi minus t. Yes or no? There will be no effect on the function because outside is a cos, okay? So can I half the limit here itself? Can I say it is 2 times 0 to pi by 2 cos of pi by 3 cos t? Yes or no? That's nothing but 0 to pi by 2 cos pi by 3 cos t. Are we there at the destination? Oh, there is a sign there. You have to use it again, sir. Yeah. Kings property, yeah. Now let's apply kings over here. Let's apply kings property over here. So if you apply kings property, it becomes 0 to pi by 2 cos of pi by 3 sin t. Now there is nothing in the name. You can put it back in terms of x. It doesn't matter. Okay. That's nothing but your v. So yes, 2u is equal to v is the desired result that I am getting. Which is nothing but option number a. Well done. Many of you replied with option number a. Very good. Correct. Next question. Let's take this up. If integral of x by 1 plus sin x whole square from 0 to pi is given as a, we have to find this integral in terms of a. No, that is not correct, Srijan. That's wrong. That is also wrong. Wrong, Srijan. Santosh, wrong. Sir, I seriously got b. That is also wrong. Yes sir, even I got b. Yeah, even I got b. That's wrong. Should I help you out? Okay, let's discuss this. First of all, let me call this integral as b. Let me call this as b. Okay. So let b be this integral 0 to pi. 2x square cos square x by 2. By 1 plus sin x the whole square. Okay. Let's do the operation b minus a. He's already given to us. So let's do b minus a. So b minus a will become remember since the denominator will be the same x square if you take common so x square. 1 plus sin x the whole square. You get 2 cos square x by 2 minus 1. Okay. Now 2 cos square x by 2 minus 1 is cos x. Okay. So b minus a is as good as integrating 0 to pi x square cos x by 1 plus sin x the whole square. Correct. Now what I'm going to do here is I'm going to take this as v. And I'm going to take this as u. Okay. Why I'm taking this as v is because it is easy to integrate. So if you want to integrate cos x by 1 plus sin x the whole square it is very easy to integrate because you can take 1 plus sin x to be your t. Okay. So cos x dx will become a dt. Okay. So it's just like integrating dt by t square which is nothing but minus 1 by t. So b minus a. Can I start integrating it? So the first term is going to be x square into integration of this is minus 1 plus sin x. Okay. You can put the limits of integration on the entire thing 0 to pi minus integral of derivative of this into integral of this will become 1 plus sin x from 0 to pi. By the way if you put the value of pi here b minus a becomes minus pi square by 1 plus 0 minus 0. And here we have 0 to pi x 1 plus sin x. Getting this. Now this is something which I'm going to evaluate separately. Let's do it separately. So I'm going to start with the problem 0 to pi x 1 plus sin x dx. First apply King's property. First apply King's property. So the same I will become 0 to pi. Pi minus x by 1 plus sin x. Okay. Add these two. So this implies 2i is equal to 2i is equal to 0 to pi. Pi dx by 1 plus sin x. Okay. So far so good. No doubt about this. Okay. Now since 1 by 1 plus sin x. Let's say I call this as my f of x function. Since f of x is equal to f of pi minus x I can half the limit. So I can write this as 2i is equal to 2 pi 0 to pi by 2 dx by 1 plus sin x. Okay. Now apply King's property. Once again it becomes 0 to pi by 2 dx by 1 plus cos x. Which is nothing but 2 pi 0 to pi by 2. This is nothing but integration of half secant square x by 2. Integration of half secant square x by 2. Correct. This two and this two will get cancelled. So it's pi integration of secant square x by 2. So it's tan x by 2 divided by half. So 2 will come up 0 to pi by 2. So it's going to become 2 pi. When you put a pi by 2 it becomes a 1 minus 0. So that's nothing but 2 pi itself. So how did you get secant square x by 2? This 1 plus cos x is 2 cos square x by 2. Correct. Yes sir. Okay. So 2 pi is 2 i. Remember I am finding here 2 i only. So I can replace this entire thing with... So b minus a is nothing but minus pi square plus 2 pi. That means b will be a plus 2 pi minus pi square. Does any option say that a plus 2 pi minus pi square a plus 2 pi minus pi square. Option a says that. So very good question. Is that fine everyone? Yes. Next is the property number 6. Which is called the even odd property. You'll find a lot of resemblance between property number 5 and property number 6. Which I am going to do right now. It says that if you are integrating a function from minus a to a. It becomes integration of f of x from 0 to a. It becomes integration of f of x from 0 to a. Plus integration of f of minus x from 0 to a. Let me prove this mathematically first. Yeah, yeah. I got your message. Okay. So let me prove this mathematically. Let me start with left hand side. So minus a to a f of x you can break it up as minus a to 0 and 0 to a. Let's not disturb this part. Let's work on this part. Okay. Now looking at the proof. Looking at the result only you'll get the idea how to prove it. You just have to put x is minus t. Correct. So this becomes dx becomes minus of dt. Now limits of integration will become a to 0 now. So f of minus t dx is minus dt. You can switch your position off the upper and the lower limit and observe the negative sign. By the way, right hand side will be just carried forward. So here just switch the upper and the lower limit and observe the minus sign. Finally, there's nothing in the name you can put your t back as x. So 0 to a f of minus x dx plus 0 to a f of x dx. This is what the property says. Not too important conclusions that we can draw from here. I'll write down the conclusions here. Conclusion number one. If your f of x is even function. If your f of x is an even function, that is to say that f of minus x is f of x. Also as to say that the graph is symmetrical about x equal to 0 line, which is nothing but your y axis. Then your integrant from minus a to a f of x dx is going to be 2 times 0 to a f of x. Guys remember I have not said this, but the function is continuous from minus a to a. Conclusion number two. Is this rotation symmetry than 0? Yes, if f of x is an odd function, that means f of minus x is negative of f of x. That means the graph is symmetrical about 0 comma 0. Then what will happen? Integral from minus a to a f of x dx will become 0. It's very obvious that if a function is an even function, what will happen? The graph will be exactly symmetrical about x equal to 0 line, which is your y axis. So instead of finding this area, you can just find out this area and double it up. So just this area you double it up, your job is going to be done. So this into 2 will be your total area. But when your function is symmetrical about origin, then that means there is a cancellation of area happening. Let's say I have a graph like this. This area and this area will exactly cancel each other out. So integral from minus a to a will vanish. So this area will be 0. This is a very important conclusion because we know that every function can be written as a sum of an even and an odd function. Correct? We have done this in our functions chapter. So whenever you see that your limits of integration are exactly opposite in sign, try to extract out as many odd functions as you can from your given function and just make it 0. The integral of that will be 0. Are you getting my point? And you can only work with the integral of the even function. That is going to save a lot of time for you. Let's take a quick question on this. Sir, is it 2? Okay. Shrijan says 2. Sir, no, sir, I'm 2. Sukhi also says 2. Sir, no, sir, I'm only... Oh, okay. Even he said 2. Privately he said 2. Okay. Let's check this out. This is quite easy. You just have to segregate the odd functions. Odd functions will be... Yes, sir, 2. Yeah. Odd functions will be everything except that 1. So that 1 you have to take separately and treat it as per the even function criteria. Okay. So this you don't have to evaluate at all because this is an odd function. Okay. So this is just nothing but twice 0 to pi by 4 integration of secant square x. That's nothing but twice tan x. 0 to pi by 4 answer is 2. Let's take the next one. Suppose a function g nx is given by this and it satisfies this equation integral minus 1 to this is equal to 0. For all linear function px plus q, then which of the following option is correct? Sir, if it satisfies all linear functions then won't g nx be 0, like 0 function. Why? Sir, because if you consider the areas then it should be like that, right? Because px plus q can be anything, it can be like any line. So the areas will be unequal. So if you want it to become 0 then I thought it should be equal. Give me a better reasoning. That's correct. Shrijan is correct. Anybody else? Gaurav, where are you? Shouldn't be a problem actually. Yes, sir. It's easy. It's very easy. That means this should be completely an odd function, correct? Now odd function, now what are the even parts here? Even parts I can see is px to the power 2n plus 2, okay? That's also p an x square and it is bn q, correct? These are the even, sorry, even part of the function. Yes or no? Odd part, I don't have to worry much about odd part because odd part anyway's integral is going to be giving me 0. So what they're saying that even the integral of these parts from minus 1 to 1 should give you 0, correct? One thing that I can say here for sure that this thing should be 0. That means bn has to be 0 for all q. Whichever option has bn equal to 0, only those option can qualify, okay? So a and b could be the possible options, c and d cannot be. Next, if you integrate this, you get px to the power 2n plus 3 by 2n plus 3 and this you get p an x to the power 3 by 3, okay? So the piece of this, I've already integrated it so I have to just put 0 to 1, okay? So it's p by 2n plus 3 is equal plus p an by 3, that's going to be 0, right? Now p need not be 0, that means if you take p common or if you directly drop off your p, remembering p is not 0, so your an will be negative 3 by 2n plus 3, negative 3 by 2n plus 3 which is option number b is what it is saying, b is the right option. Okay, so a bit of analysis has to go in. Next question, yeah. If your function is given by this determinant, then find the integral from minus pi by 2 to pi by 2 of x square plus 1 f of x plus f double dash x. Anyone? It's almost done, sir. Please also see the video lecture of the definite integral session taken with the other batch. So there are done more basic problems, here I am taking slightly more challenging ones, but also remember to see those videos once your exam is over, okay? No hurry, after the exam is over you can watch those videos. Let's see who has responded. So keep your correct. See guys, it's just a one minute problem, one minute, not more than that. If you look at the function, if you replace your x with minus x, okay? Remember nothing will change in the first column, nothing will change in the second column, but everything in the third column will become negative, isn't it? Which clearly implies this is negative of f of x, that means f of x is an odd function, correct? This clearly implies f of x is an odd function. Sir, how does it become one? Sorry. Okay, thank God. So even the double derivative will also be an odd function, correct? Which implies the sum of these two will also be an odd function, correct? Now remember x square plus one is an even function, x square plus one is an even function, okay? So x square plus one into f of x plus f double lash x is a product of even and odd, which will actually be odd. So if you're integrating an odd function from minus pi by two to pi by two, your answer has to be zero, which is none of these. Zero is the answer. So no need to evaluate the determinant at all. It was just based on your understanding of the basic property. They're actually evaluated with determinants. Oh my God. No sir, but then I cancelled off most of the things because it was odd. So this is a type of learning for you. Okay, hope you can read this question slightly blurred. Let me just scribble on it. So wait, how do you show that even the double derivative is also an odd function if at all f of x is odd? The derivative of an even function is odd, correct? And derivative of an odd function is even? Wait, I didn't know that. What? Function chapter we did this, no? Oh, yeah, yeah, correct, correct, correct. Oh yes, yeah, no way. Hope you can read this question now. This is normal brackets. So it's, it's nothing but integral of x plus one by x minus one square plus x minus one by x plus one square minus two. All to the power half from minus half to one. Zero. No, that's not zero. So one second. Sir. Two, let's check. Let's check. Okay. Now I'm sure all of you would have to. Okay, let's check. Most of you would have figured out that this is actually x plus one by x minus one minus x minus one by x plus one. Whole square under root, correct? Which actually is mod of x plus one by x minus one minus x minus one by x plus one. Okay. I'm not sure how many of you accounted for this mod thing. Many of us have this habit of writing square under root as the same thing, which is not correct. It's actually mod of the same thing. Form a mistake, sir. Okay. Yeah, Omkar seems to be correct. Let's just discuss it completely. So this will become minus half to half. If you take the LCM inside, it will become x square minus one. So four x will come up. Correct. Now, if you draw the wavey curve for four x by x square minus one, which is actually four x by x plus one x minus one. The wavey curve will give you the sign scheme like this. Correct. Now you're dealing with minus half to half minus half is here. And half is here. Correct. So can I say from, by the way, even even I don't need to do that. I can say that this function will always be an even function. This entire function is an even function. Why? Because if you change your x with minus x, it just becomes minus four x by x square minus one mod, which is same as mod four x by x square minus one. That means f of minus x is f of x. So I can first make it as two times zero to half mod four x by x square minus one. Now I don't even need this part. I can only deal with zero to half interval where the function is negative. So it becomes negative four x by x square minus one. I'm sure by this time you would have known how to integrate this. So if you take a minus two out, so minus four, zero to half, two x by x square minus one. This integral is minus four ln x square minus one. Okay. Put first half, half will give you ln three by four. Zero will give you zero. So that is just going to be minus four, sorry, plus four ln four by three. Okay. Sir, is it ln of minus or not defined? No, you're taking mod. Okay. Okay. Okay. So I think this is your answer. Sir, I didn't get it actually. I missed by a constant, I think. Yeah, I missed by a constant. Yeah. Yeah. Is that fine, guys? This chapter is pretty long. It's not a small chapter. We have a lot of things to cover. I think the next class, which is next class on Thursday, Friday is your English exam, right? Lol, yeah. So please study English before and don't bunk on that day. That's going to be an important class. Sorry, we won't bunk it but we won't study English before also. Exactly. I know last moment you guys are going to study. Anyways, we'll do a lot of important things the next class. So as of now, I'm just stopping the session over here and all the best for tomorrow's exam. Thank you, sir. Thank you. Thank you. Over and over. Thank you, sir. Bye-bye. Have a good day. Thank you, sir. Class over already. You joined late to the end. Four to seven. Thank you, sir.