 All right, AUC area under curves, area under curves, we call it as AUC in short form. So area under curves was introduced to you while you were having geometrical interpretation of definite integrals. So when we had a function and we integrated this function from A to B, when you found the integral of this function f of x from A to B, what did it give us? It gave us the algebraic area which is between the function and the x-axis. Now, if you interpret this in a much closer manner, if you see f of x into d of x, f of x into d of x, I'll write it as f of x minus 0 into d of x. So what is this actually? f of x minus 0 is basically, let's say at a distance x. You have chosen a strip dx, a differential change in x, a differential element x dx. And from there, if you let's say, go all the way up to the curve, go all the way up to the curve, this height, the height of this rectangular differential strip, this height is going to be the difference of the y-coordinate here and the difference of the y-coordinate here. Remember, y-coordinate on the x-axis is 0. So when you write f of x dx, it is to be interpreted like you have written f of x minus 0 into dx. And since f of x is at a higher level, that means it is positive as compared to this 0. That means f of x is more than 0. This term comes out to be automatically a positive quantity. dx will be positive if you're integrating from a lower value to a higher value. So if you're integrating from a lower value to a higher value, that means if a is less than b, then even your d of x is positive. So as a result, what you see here is a differential area of the strip. So let's say this strip has an area of da. So f of x minus 0 shows the height and it is positive height because f of x is more than 0 as seen in the graph, above the x-axis, that's why f of x is greater. And f of x minus 0 is the height into dx. dx will be positive if you're integrating from a lower value to a higher value. Now while you're finding the area, what are you doing? You are integrating this, correct? You're integrating this from a to b. So in fact, you are performing a computation on the area between the function and the x-axis. But mind you, in this case, your answer is going to come out to be positive because your f of x is above the x-axis. If your f of x is below the x-axis and if you use the same formula, then what is going to happen? So let's say this is my f of x. This is my f of x. And if I start integrating from, let me just show you a to b also in this diagram because I've not shown a to b. Let's say this is a, this is b. So this whole area is obtained by, yeah. Now if your function is below the x-axis, then what happens? At a distance x, if you take a differential x and construct a rectangle over here. Now if the same function, the same expression is written f of x dx from a to b. Now remember, this f of x is treated like as if you have written f of x minus zero. So this term is treated like this into dx. Now again, I'm assuming a is lesser than b. This is a and this is b. Now what will happen in this case? In this case, you realize that at the bottom of the strip, y is zero. In fact, you can treat this part to be your head now this time and this part to be your bottom. So you can see that your y f of x will be negative quantity here. That means it will be lesser than zero. When you do that, you end up getting this quantity as negative. This quantity becomes a negative quantity. Let me write it just a simple negative thing. Okay, and if you're integrating from a lower value to a higher value, this quantity becomes positive, right? So as a result, when you're integrating this whole thing, so let's say this is your DA, and you're integrating this whole thing from A to B. You will end up getting a negative answer out of this whole process, right? But when we are trying to find the area, let's say the question setter says, find the area between the function and the x-axis. He doesn't mean negative area. He only means positive area. So many of us, what do we do? In such cases, we take a modulus, right? But instead of taking a modulus, I have a better approach. And the approach is, why are we using f of x minus zero dx from A to B? Instead, we can use zero minus f of x dx from A to B. So because we get a negative area over here in order to do away with the effort of taking a modulus, which anyways we can do. See, I'm not denying that modulus should not be used. You can anyways use modulus to find the positive area. But the area always that we require or desire, or when the question setter says he wants the area, he always means the positive area. Until he uses the word algebraic area, okay? So there is a small amount of misconception in students that integration gives us area. No, the area that we need actually integration helps us to get the area. But in integration, we always obtain algebraic area along with the sign part of it. Okay, so integration is used, the word here is used. Integration is not equal to understand the difference. So we will be using integration to find the area. So please use it very carefully because the sign issues can arise, okay? So in order to do away with this, we normally recommend that if you want to, if you don't want, but area is always positive, but area is positive. So if you want your area to be positive, instead of doing this, you will have to do this. Integral of zero minus f of x because zero is at a higher level from a to p. This will give you the actual area which is going to be positive as well, okay? So my indication here or my hint here is whenever you're finding the area, you should at least know the relative position of the graph. With respect to any other graph. For example, in this case, there's an x-axis and there's a function. So you should know how is your function oriented or positioned or drawn with respect to the x-axis. So that you don't have to take a mod anytime. Now, many people think, sir, what is wrong with using mod? See, many times what will happen, your area will fluctuate. I mean, some area will go positive, negative, positive, negative, positive, negative. And somebody asked you the area. He actually means to have a positive values of all those small, small, small, small, small area which is created. So you'll have to keep taking mod wherever it is going down. So instead of taking a mod, why don't we use the fact that we should always do upper minus lower? Always upper minus lower. So this is upper because your zero is on the upper part of the strip. This is the lower part of the strip. This is the lower part of the strip. So this is the lower part. This is the upper part. Okay. If you do always upper minus lower, your height of that differential strip, or you can say that thing, infinitesimal rectangle, that will always be positive. And if you're integrating from lower value to a higher value, that dx will always be positive. So overall, that expression dA will become positive. So there is no requirement for you to take mod anywhere whatsoever. Is it fine? Any questions? Okay. So now this was a very special case where you had to find the area between the function and the x-axis. The same thing could be generalized. If you want to find the area between any two curves, if you want to find the area between any two curves. Okay. So let's move on to that. Anything that you would like to ask or note down over here, please do so. Please be very, very careful about the sign part. Just don't blindly integrate. If you blindly integrate, you may end up either getting negative part or you may end up doing a, you can say, a compensation of negative and positive part, thereby reducing the magnitude of your answer. Anything can happen. Anything can happen. So you have to be very, very careful. How do we decide the upper and the lower part? That's a very, very good question that Aditya asked. So Aditya, long, long back, there was a topic taught to you during the bridge course, which was graph and its transformation. So mostly you will have to be good in sketching your graph. That's what, that is something which I was about to tell you in some time. So when you're trying to solve questions on area under curves, you will have to be at least, you know, roughly knowing the position of the two functions which you're dealing with. Can you explain the negative part again? Yes, why not? See, in this case, if you just do this, your answer is going to come out to be a negative value. Why? Because f of x dx is treated as f of x minus zero dx. Now f of x is at a lower level, correct? And zero is at a higher level. So something is below the ground thing like that. Okay. So if you do f of x minus zero, this answer is going to be a negative value, correct? And if you integrate from lower value to a higher value, your dx is anyways a positive. Remember, dx depends upon from where to where you are integrating. If you're integrating from lower to higher, dx is positive because x is increasing as you are moving on, right? Change in x, dx is what? Change in x, differential in x. So change in x is positive if you are integrating from lower to a higher value, correct? So as a result, this component that you see here, which you are trying to integrate, integrate means sum up, okay? Summation is, integration is what? It's a kind of a summation, right? So this whole thing is going to be a negative quantity. If you're summing up negative quantities, you're going to get negative answers. But remember when the question setter says, I want the area, I want this area. It doesn't mean a negative area, right? He means always a positive area, right? No matter how the function is oriented, for him, he needs a positive area. I want this area. So two things you can do, either you can take a modulus of your answer, but that has got a lot of complications. You will soon realize why there is a lot of complications in the modulus. Other approach would be you instead of writing f of x minus 0 dx or f of x dx, for that matter, you write 0 minus f of x or minus of f of x dx. But for that, as I was talking to Aditya, you should know how the graph is oriented with respect to your x-axis in this case. Is it fine? By the way, your name is appearing to be Pratima. Who is this, by the way? Okay. Anyways, change your. Okay. Okay. All right. So now we are going to talk about area between two functions in general. Now, I'm going to spread this topic as two subtopics, finding the area between. I can write it down. Area between, between any two curves. Okay. Area between any two curves. And we are basically going to use two types of strips. One is the use of vertical strip. Okay. Vertical strips. Now, vertical strip is something which I was already using in the previous example also. Okay. However, we could also use horizontal strips to find the area, but it depends upon situation to situation. So when do we use vertical strips to find the area between any two given curves? Okay. So let's say the two curves given to you are like this. Okay. One is like this. Another is, you know, like this. Right. And we want to find out the area between the two curves. Let me name them. This is your f of x curve and this is your g of x curve. Fine. And we want to find out the area between the two curves from x equal to a. This is x equal to a. And let's say this part x equal to b. Well, let me just extend it down and this part is x equal to b. Fine. We are looking for finding this area. Okay. Yes, sir. Now, in such cases, we prefer using vertical strips to find the area under this curve. Okay. So what do we do again at a distance of x? You make a differential strip. Let me make it white and make it in gray color. Yeah. Make a differential strip like this. Okay. So this strip that you are seeing over here. Okay. Now ask yourself if you want to know what is the height, positive height, what are you going to do? See, at the top of the strip, your y is your f of x. At the bottom of your strip, your y is your g of x. So if you want to know a positive height of the strip, you have to always use upper minus lower. If you do lower minus upper or you are not, you know, taking into account which is upper, which is lower and doing any difference, it may lead to faulty results. Please be careful. So if you want to know the positive height of the strip, you should always use this. So this is anyway is going to be positive for sure. If you're doing upper minus lower for that, you should know which is the, what is the relative position of the two functions or the two graphs with respect to each other. That means a bare minimum idea of sketching the graph should be there. Okay. So one of the prerequisite is you must be good with your graph sketching. Pre-requisite that you should have a good hand at sketching the graph. Okay. Don't try to solve any area under curve concept or problem without sketching a graph, at least a rough graph is required. You should know where they're cutting each other. What is happening? Are they cutting x-axis somewhere? When they're going down, when they're going up with respect to each other. All those things should be, you know, roughly, you know, clear in your mind. I've seen people, they just take the function and start integrating or just, they just take the difference of the function and start integrating. It doesn't work that way. Okay. Now, once you have taken f of x minus g of x to be upper minus lower, this will always yield to be a positive answer. And since you're integrating from a lower value to a higher value while you're using A to B, okay, this dx will also be positive. So overall, this answer is going to be a positive answer and that answer is going to be the area. Okay. Whatever area is required. Now, many people ask me, sir, when do you use what horizontal strips? We will come to that case in some time, but let me finish the agenda over here. So here, please note that if you are trying to solve this question, okay, by, you know, any other route, I'm not denying that it cannot be solved by any other route, it would become negative, it will become complicated. For example, let's say many of you would say, sir, can I find out this area first, as you can see on my screen, I'm just showing you some path. Okay. Can I find this area first? Okay. Of course, if you want to find that area, you can integrate your function from A to B, correct? But this part is not desired, right? This part was not desired. So now you have to subtract the area under g of x function from this part. Let's say this is C part to B. Okay. Then you have to also find out this area. Okay. And that is the area between zero minus g of x dx, that is the area between g of x and dx axis, but you have to take zero minus g of x from A to C. See, everything here would lead to multiple, you know, expressions that you're going to write, and it is going to become extremely complicated. Okay. So there are some people who say, you know, I, you know, I want to, you know, find out the area by that primitive method of what we had discussed in the previous slide. Okay. So using x axis always as a reference part, please do not do so because it will lead to faulty results. Okay. Another misconception people have is that sir, when you're finding this area, won't this part be a negative part and won't it reduce your answer? No. That has already been taken care of. I'm using f of x minus g of x because I don't want anything to become any strip to become a negative height strip. All the strips here will be of positive height. That's the whole and sole purpose of doing this. So there'll be no such case that this part will become negative and it'll reduce the magnitude of the overall area. Those kind of issues are not going to arise if you're going to use this. So this is highly recommended that whenever you are finding the area, you should be using this part. Okay. Yes. Of course, a lot of things can happen. Aditya, so Aditya is saying, what if the functions cut each other? See, when you're, when you know the graph, then the life is, you know, super easy for you. Okay. So let's say your functions are like this. So one function is, you know, like this. Okay. I'm just making one function. Another function is like this. Just making a graph like this. Okay. And you're trying to find out the area. Okay. You're trying to find out this area. Let's say area from this part to area from this part. Okay. Let's say you are looking for this area. Okay. And let me just, you know, write down a few things. Let's say this is your x equal to a, this is your x equal to b. And this is your x equal to c. Yellow curve is your function. White curve is your g of x function. So yellow curve is your f of x function. White curve is your g of x function. Okay. Let me give a dummy scenario like this. Tell me what is the expression for the area? Now again, area means a positive value. So this whole thing positive, both added. What is the answer going to be? You tell me what expressions you're going to write. Anybody can unmute himself or herself and say. So here you would see that your definition of that differential area is going to change in both the loops. Okay. So in this slope, you would realize that in this slope, if you realize that if you take a strip, okay, if you take a strip, you realize that your strip area would be positive. If you do f of x in minus g of x into dx, and you're integrating from lower value a to a higher value c. Correct. But for the other part, let's say I show this part with a blue strip. So if I take any differential area in this part, okay, in this part, what is going to happen? You're integrating g of x minus f of x, because as you can see at the head of the strip, this is g of x. At the bottom of the strip, you have f of x. Correct. So here you will be doing g of x minus f of x, and you will be integrating from c to b. You'll be integrating it from c to b. So thereby accounting for the fact that both of these will be positive, positive, and you'll end up getting your area from it. Okay. So you should know where they're intersecting. You should know the coordinates. You should know the, how the definition of the strip is going to change in the different, different segments depending upon the relative positioning of the curves. So here my strip was having the differential area was having the definition f of x minus g of x into dx, whereas in the second part, in the blue part, it became g of x minus f of x into dx. Clear? Okay. So as I already told you, this is the thing that you need to be good at, sketching off curves. Without this, it will be very, very difficult to ace in this chapter. Okay. See concept is easy, but if you are not backed up by good interpretation of your positioning of the graphs, you may end up getting quality results. Okay. Now coming to the second part. So see moral of the story is very simple. I mean, if I just summarize it in simple words, sketch both the graphs or whatever the other graphs involved, in fact, let me not write both sketch the graphs involved. Okay. Second. Okay. See what kind of, you know, definition you have to adopt while you are framing a differential strip depending upon the position. Okay. So follow always upper minus lower. Okay. And integrate it from a lesser to a higher value. If you do this, you will always get a positive answer. Nobody is going to ask you to do a mod anywhere if you're doing this. Okay. I have seen some people have this habit of always relying on mod. They will do anything and they realize, oh, my answer is coming negative. Right. So if you're, if you're oblivious of how the graphs are behaving with respect to each other, even though the total answer will come out to be positive, but it may be a, you know, cancellation of some negative with the positive part. So be careful. So next, the second case that we are going to talk about is in some cases we need horizontal strips to find the area. So when do we need horizontal strips? In fact, you know, there could be a mix and match of both the type of steps, depending upon questions also. So don't be like, I always have to use one type of a strip to do a particular question. No, there could be mix and matches also in some cases I have seen. So horizontal strips are used when your curves are of this nature. Let's say there is a curve like this. Let's say there's a function like this. You know, okay, so let's say this is your function. Now, in such kind of a function, you would realize the function could be written x in terms of y. That means instead of y in terms of x, like we normally see, here normally x is written in terms of y. Okay. Even if it is not given, you have to write x in terms of y because that is what is going to be utilized. Okay. And let's say there's another function like this. I'm just making some dummy graph. Okay. Let's say something like this. Okay. And you want to find out the area from y equal to a till let's say y equal to b. Okay. You're looking for this area. Okay. And this function, sorry, I have not written the definition. This is let's say x equal to g of y. Again, I'm telling you the question setter need not mention x in terms of y. He may write y in terms of x only for both. Right. But when you are using vertical steps to find the area, it is your duty to convert or make x the subject of the formula. He is not going to give you. If he's given you, he's kind enough, but don't expect that kindness from the question setter. If you are going to use, if you, if you analyze the situation and you think that, oh, I'm going to use horizontal steps over here, because I feel that is going to be less complicated. So first of all, how do you make a choice that which kind of a strip you have to use? Or maybe you have to use a mix and match is depending upon the situation. Is there any complexity in finding the area? Just by the use of vertical steps, or is it a quite a complex situation to, you know, solve the question by using only vertical or using the vertical, vertical steps. There you have to understand that, okay, no, I need horizontal steps over here. This will be easier for me to find the area. And if it is easier for you to find the area, it is your duty to convert the given functions or make x in terms of five. Don't worry. Your functions will be of such a nature that you will be able to convert x in terms of five. Okay. Let me not call them as functions, by the way, let me call them as curves. Okay, so please now come out of the word function. Function is not a right word to be used anymore. So let's use the word equation of the curves are given to you. Make sense? Raghav. Okay. Fine. So now here at a distance y, at a distance y, let's say I make a horizontal strip like this. Okay. So let's say I make a horizontal strip like this. Okay. Now tell me in this case, what is the width of the strip? The width of the strip will be the difference of the right arm. So whatever is the x value over here, remember x value here is f of y minus the x value here. X value here is g of y. So if you want the width of the strip, if you want this width of the strip, and you want it to be a positive width, you should always do right minus left, the x on the right minus x on the left, because normally the x on the right is higher in value than the x on the left. Isn't it? As we move right, x increases, isn't it? As we move left, x decreases, just like as we move up, y increases, as we move down, y decreases. So I'm not saying both, one of them is positive, another is negative. Don't, don't get me wrong. Maybe I've drawn the curve that it looks in this way. But even if let's say the whole graph was towards the negative side of x-axis, the right side x will be at a higher value, maybe negative only, but at a higher value as compared to the left side x. So f y will be higher than g y. So this difference, f y minus g y, this will become a positive value. Okay. So if you multiply it with the thickness of this strip, this thickness is d y, you'll end up getting. Okay. And of course you should be integrating from lower to a higher value unless until you integrate lower to a higher value, d y will not be positive. So as a result, you have ensured that you have formed a differential strip whose area is a positive value because you have taken the width to be positive and you're integrating from lower to a higher value. So d y is also positive. So as a result, this whole thing on summation will give you a positive area. No need of a mod, et cetera. Anyway, no need of a mod, et cetera. Any. Is this like any questions, any concerns? So here the moral of the story is you have to integrate right minus left. The previous one was upper minus lower when you were taking vertical strips. Okay. So if you do right minus left and you integrated from lower value to a higher value. Okay. Let's say A is lesser than B, then this area will already come out to be positive. Is it okay? Any questions? Don't worry through our problems. We are going to understand which strip will be helpful in what type of question and what kind of complexities can come if we use something otherwise, let's say if it requires horizontal and you're using vertical. See, answer can be found out. It's not a hard and fast that you have to use one of the strips to solve it. But you have to, as a problem solver, you have to see which is a lesser time order effort, which is less complicated effort. Okay. So for that, you have to first analyze the question before you decide what you're going to do. Okay. So both the opportunities, both the methods are there at your disposal, but you have to see which situation fits what occasion. In fact, you can use a combination of the two. If at all it is required. Okay. So let's, let's jump into problem solving. Okay. We have done enough theory part. Let's jump into problem solving. I'll start with a very simple question, very, very simple question. Very, very simple question. Find the area bounded by the parabola y is equal to x square plus one and the straight line x plus y equal to three. So everybody, please sketch this graph. First of all, so what a coincidence. The first chapter that we had started our mathematical journey of our J preparation or competitive exam preparation. Okay. That chapter is helping us toward the end of our journey. Okay. That's how things are linked. So vertical shifting, horizontal shifting, etc, etc, whatever you had learned reflection about x axis, y axis, mod, they will all be very, very helpful in this chapter, extremely helpful. Okay. So I'll be also pitching in while solving this question. So first of all, y equal to x square plus one is a parabola y is equal to x square, which has been shifted one unit up. Okay. So let's say this is, this is your y is equal to x square plus one parabola. Okay. And this is the line y is equal to three minus x. So negatively slope line, cutting the y axis at zero comma three. So let me make use of another pen. Okay. Maybe I'll make it like this so that we don't go much higher. Now again, it's a rough sketch. Okay. It's a rough sketch. You don't have to be, you know, very, very accurate, but yes, make no mistake in knowing what all points they intersect. Okay. So as per the question center, you want the area bounded between the function y equal to x square plus one and the straight line. That means this is the area that is required. Okay. Now ask yourself first, from where to where I have to integrate, do I need to use vertical steps? Do I need to use horizontal steps? All those questions should start coming in your mind. Okay. Now here, if at all, let's say I'm giving you an option, if at all you want to use horizontal steps to integrate it, what all do you need to know? If you want to use horizontal steps to integrate it horizontal, not vertical, what all you need to know? Now you need to know, yes, very good. I'll just explain you what you need to know. You need to know, you need to know, what is the y coordinate at this position? Okay. So please note, please note, you need to know, let's say I call it as y equal to, this is this point I'm calling it y equal to a, y equal to b and this point I'm calling as y equal to c, y equal to c. Now, all of you, please pay attention. When you're making a horizontal strip to sum up this or to find this area, please note that your strip will be having the line on its right side. Okay. And it will be having the parabola on the left side, but only from c to b. Are you getting my point? Only from c to b, your right side part of this strip will have a line and the left side will have the parabola, but from a to b, sorry, a to c, if you make a horizontal strip, your horizontal strip will have a parabola to the right also parabola to the left also. Now, let me write down the expression. Let me write down the expression that will keep that'll give you. Okay. Okay. People have got the answer good, but I'm not interested in knowing the answer right now. I'm interested in, you know, making the concept clear. So right now I'm using, let us say, horizontal strips to find the area. So what I'm going to do, I'm going to figure out this point, coordinate first of all, can somebody tell me what are the coordinates of this point? This point, let me call it as a, this point. I know this is zero comma one. Okay. Please give me coordinates of a and b. B is over here. So for that, you're going to solve it. So you're going to solve x plus y equal to three along with this. Okay. So what I'm going to do is I'm going to write my y as three minus x. Okay. So I'm going to simultaneously solve these two. So when I do that, I end up getting a quadratic, correct me if I'm wrong. Okay. So this is factorizable as x plus two times x minus one. So x is equal to one, x is equal to minus two. So this point is one comma two. And this point is minus two comma five. Correct me if I'm wrong. Right. Yes, sir. So now from y equal to one till y equal to two. Remember this, this position y is two. This position y is one. So from y equal to one to y equal to two. What is the x on the right side of the strip? Who will tell me? What is the x here under root of y minus one? See, guys, please be careful here. Very, very important. This is equivalent to saying x square is y minus one. Isn't it? So x is plus minus under root y minus one. Right. Yes or no? Now remember, since this part of the strip, this part of your, you know, right side of your strip lies in the positive side, you're going to take positive under root y minus one. Correct. So now see, this is a very, very interesting thing that you're going to learn today. So right side will be under root y minus one minus. What is the x on the left side of the strip? Negative under root y minus one. How do you get it? So it'll become negative under root y minus one. Now, why did I tell you this approach is because many people say, sir, you're going from the same curve to the same curve. Won't that be zero? No. No, it will not be zero because two parts of the curve, they don't have the same functionality of x when represented in y. That's a mistake which many people do. In fact, I was, I was seeing some of the lectures of one of the institutes. The teacher was saying, has a very, very well known institute in online field. The teacher was saying, since you can't use a horizontal strip because you're going from the same curve to the same curve, so it'll become a zero. That'll give you a wrong result. No, it doesn't give you zero. You just have to understand that x here in terms of y is differently represented as compared to x on this side of the strip. So on one side, it is under root y minus one. On the other side, it is negative under root y minus one. That's where you need to understand that the CMC, many a times we see a circle. Let's say you want to find out the area of this circle by using horizontal strips. So let's say this circle is x square plus y square is equal to one. Now, what is the x in terms of y here? x in terms of y here will be under root of one minus y square. So this is your x in terms of y on the right side of the strip. On the left side of the strip, your x would be negative under root one minus y square. Understand it. Same goes when you're taking some cases of a vertical strip. So vertical strip, let's say you want to use this strip. So let's say this is our age, our good old friend y square is equal to x. So here, y is root x. And here, y is negative root x. So even though you're going from same curve to same curve, the top and the bottom of the strips do not follow the same definition or do not follow the same functionality of x as they will get cancelled. Thereby giving you a false impression that this, that SIP was having a zero height. The SIP was not having a zero height. Are you getting my point? What to take care of when you're trying to solve the question? Okay, so these are some misconceptions that people have while solving questions. Anyways, we'll come back to the question. So now from one to two, you are integrating this part of the strip. What about from two to five? So from two to five, your x of the strip here will be three minus y and this part will be again negative under root y minus 1. So you're going to do right minus left. Sorry, I wrote a dx minus take, it's a dy. From two to five. Now, if you see this overall process looks complicated, isn't it? So this will have a higher time order, even though it was going to give you the answer, don't worry about the answer being different. Answer is going to come out in this way also, but don't you think making two definitions is going to make this process little longer? Is going to make the process little longer? So it's better that you make a wiseful decision that what kind of SIP you are going to use to make your process faster. So here I would like to use vertical strips. Okay, so let me show you the second method. So I would like to use the vertical strips to do it. Okay, please don't mention this in the examination hall. Let's say even if it is your second semester topic, don't have to mention this. It is for your internal understanding. Okay, so let me sketch the diagram once again. I'll just quickly do it. So here if you see, if you use vertical strips, the best part of the vertical strip is, you don't have to take multiple definitions as what you did in the previous approach. You have to take two different definitions, isn't it? You have to frame two different integrals to find the answer because your strip definition changed, isn't it? In this case, the best part is, you don't have to change the strip definition. You always start with the top on the line and bottom on the parabola, always wherever you go. Even if you go here, even if you go to this extreme, okay, your head is always on the line and the bottom is always on the parabola. Head is always on the line, bottom is always on the parabola. Okay, so you don't have to change the definition of the strip as you're finding the area. Now, we already know this point. This is 1, 2, and this was 2, sorry, minus 2, 5. Okay, so what is y here? y here will be 3 minus x. What is y here? y here will be x square plus 1. So top minus bottom or upper minus lower, as I have used the word while I was discussing the theory, upper minus lower. This will always be positive and you're integrating from a lower value to a higher value. Simple, one simple expression and you're done with your answer. So let's calculate the result. I think this comes out to be 2 minus x minus x square integral from minus 2 to 1. That's 2x minus x square by 2 minus x cube by 3 from minus 2 to 1. Let's put the upper limit. 2 minus half minus one-third minus minus 4 minus 2 plus 8 by 3. Let's check how much it comes out to be. I hope I've written everything correctly. Just please verify. So this is going to be 6 plus 2 and minus half and this is going to be minus 3 if I'm not mistaken. So how much it comes out to be? 9 by 2. Please check the calculation. More importantly, is the process clear? Is the process clear to everybody? That is more important. So we'll take more questions and we'll try to explore which approach is going to be of a lesser time order, of a lesser complexity. And we're not going to just do problems. Let's take this problem. Yes, Gayathri, but you have to take a call when to use what? Sometimes vertical strips are easier. They will give you the result faster. Sometimes horizontal strips are faster. Sometimes you can see, I mean, in some rare cases, sometimes you'll see a mixed match approach will be faster. So okay, from here to here, I will use vertical. From here to here, I'll use horizontal because I feel I can take, you know, find the area without much change of the definition if I use horizontal. So you have to take a call. You mean to say an oblique strip? Is that what you want to say, Kinshuk? Oblique strip will come with a lot of complexities. I mean, I have not seen it being used, but I don't deny the use of it as well. See, in oblique strip, both x and y are changing, right? In vertical and horizontal strip, only one of the variables are changing. That complexity I'm talking about. Okay, let's take this question. Find the area lying above the x-axis and include it between the circle x square plus y square is equal to 8x and the parabola y square is equal to 4x. Try this out. Yeah, but when you talk about that area of that rectangle, it is going to have an oblique sides. So calculation of those oblique sides will be more complicated. Hariharan, that's what I was talking. Why not? It has to be done by variable substitution, of course, but the complexity will be there. If at all, you're not able to find out a suitable replacement variable. Okay, Pranav. So we had started with this question. Find the area lying above the x-axis and include it in between the circle and the parabola. Anybody has any success? Okay, take your time. Take your time. No issues. So if you see this circle, I'm just trying to sketch one. So this is, as you were saying, x minus 4 the whole square plus y square is equal to 4 square, isn't it? So it's a circle which is going to touch the y-axis, having center at 4 comma 0. So it'll look like this. Y square is equal to 4x. Now I'm just making a rough estimation because the question says that there will be an area between the parabola and the circle. So it has to be, the parabola has to cut past the circle like this. I'm just making a rough estimate here. Okay, so something of this nature. Okay, so this is your parabola. So they're asking us for this area. I'm showing it in blue color. That is the area which is required. Okay, Aditya, anybody else? Oh, sorry, sorry, sorry. One second, one second. Aditya, you may be correct because the question says, find the area above the x-axis and included between the circle and the parabola. I think this area only, right? Because that area is the area. Okay, some of you are saying, why not this area between the circle? That's what initially I was, okay, many of you are thinking it could be this area, right? Yeah, it could be, it could be, yeah. Because if you read the question, they're saying lying above the x-axis and included between the circle and the parabola. So both of them are basically making that, again, there is a bit of ambiguity in the question, but I'll assume that, let's say we are finding this white area. Okay, that will be between x-axis circle and parabola. So that will be lying above the x-axis and included between the circle and the parabola. So why this area will be wrong, Anusha? Is this, you're saying that this will be a wrong area? I mean, this will be a wrong interpretation of the question. No, it is satisfying all the criteria above the x-axis, included between the circle also and between the parabola also. No, no, no, no. Anusha, that's, this is correct actually. In fact, this is the area that we should be eyeing for. Okay, anyways, it could be both. I think there is a bit of ambiguity in the question. They should have written that it is above the x-axis, bounded by the x-axis. That part is missing. Anyways, we'll talk about this. We'll talk about this only. Okay, let's not worry about, in the examination room, don't worry, the question will be very clear cut mentioned. Okay. Now, see here, while you're solving this question, you would realize that from this point, whatever is this point, till this point, your strip will be such that it will start from the parabola and end on the x-axis. Yes or no? This part, the strip will be starting from the circle and end on the x-axis. So as you can see here, there is a change in the definition of the strip, even though you're using vertical strips only for both the cases. So that clarity has to be there, where you are changing the definition of the strip. Are you getting my point? So first of all, we need, what is the x-coordinate of intersection of the parabola and the circle? By the way, here, x is 0. And if you put y-square is 4x. So that will become x-square is equal to 4x. So x is equal to 4. So this point, x is has to be 4. Okay. And this point, we already know x is going to be 8. Okay. Now here, many people will say, sir, then we don't need to know the area of this strip through integration because this is quarter of a circle. Isn't it? This is quarter of a circle. So 1 fourth pi r square. Yes or no? Okay. So all you need to figure out what is the area of this part. And for that, you need to use integration. So upper minus lower dx. Now remember, very, very important, as I already told you in the previous slide, this amounts to two definition, y is equal to two under root x and y is equal to negative two under root x. Okay. This, the down part is the down part of the parabola. The bottom arm of the parabola is y is equal to negative two root x. The upper part is y is equal to two root x. Okay. So the upper part definition, y here will be two under root x. So lower part is zero. This part is zero. Y is equal to zero. So upper minus lower will be two root x minus zero, integrated from zero to four. Okay. So that's going to be simple. Half is going to be three by two divided by three by two from zero to four. So that's four by three x to the power of three by two from zero to four. If I'm not mistaken, that's going to be 32 by three. So your total area will be your area of quarter of a circle, which happens to be four pi and this 32 pi by three. Sorry, 32 by three. This is going to be area. Now, again, it depends on how you are interpreting the question. But again, as I told you, I'm not very interested in knowing the answer. I'm interested in whether you are fine with the process of integration. Okay, if you have, if you're assuming that you want this area, that is the area shown by the blue, this thing. So for this blue area, what are you going to do? You're going to do zero to four. Upper part will be, now see, Y on the upper part would be under root of, under root of, see here, under root of eight x minus x square. Okay. And Y on the bottom part will be two root x. So this area you are going to integrate from zero to four. If let's say you're finding the blue part. Okay. So for a circle, the upper half of the circle, that is the semi-circle on the upper part of the x-axis, the definition is Y is equal to positive under root, whatever it is coming. Below part, the definition would be Y equal to negative under root because Y is negative in the lower part. Okay. So please be careful. I'll, I'll draw it again over here and show you. See, this was our circle. This was our circle. So this part of the circle, which I'm showing by, you know, making a spring around it. This part of the circle, your definition will be Y is equal to under root eight x minus x square. Okay. And this part of the circle, which I'm showing with the blue, wiggly, wiggly thing, that will be Y is equal to negative under root eight x minus x square. Okay. So now when you're making this, when you're making a differential area over here, let's say I make an area over here, the top part of this strip will be on the circle, which is Y equal to under root of eight x minus x square. Bottom part will be on the parabola, which is Y equal to two under root x part. Got it, Pradap? Now, can you integrate this part also? Are you all aware how to integrate this? Yes, Pradap? Yes. Pradap, any other question? Complete this square and that is a standard integral. Right, Harian? Okay. So you can use, I mean, when I write it, actually it is four minus x minus four the whole square. So you can use your under root of a square minus x square. Is it fine? Any questions? So we'll move on to the next question. Find the area of the figure bounded by the parabola, or parabolas, both are fine. Given by the equation, x is equal to negative two y square and x is equal to one minus three y square. Please make the graphs properly. If you're making incorrect graphs, things will go for a toss. All right, so this graph will be, the graph of y square is equal to negative x by three shifted. Which way? Towards right or shifted towards left? Shifted towards right. Okay, absolutely. So this will be having vertex at this point. So somewhat like this. So I'm just making them intersect because the question setter himself has mentioned that bounded by, so they have to intersect somewhere. So he is basically expecting us to get this area. So this boomerang, this boomerang is formed, right? That boomerang area we want. Okay. Now, before you start solving this, ask yourself, what kind of strips is going to help you out here? Vertical or horizontal? I mean, which is going to be faster? Right, horizontal strips. Go for it now. I think you are on the right track. Go for it. Let's solve it. So you need to know what is y here, what is y here, and what is y here. And at any distance, you take, let's say, I make a dummy, I mean, just an exemplary strip over here. The right side will always be a white parabola. Left side will always be yellow. No matter where you make it, you want to make it here, you can make it, no issues. Right side will always be the white parabola. Left side will always be the yellow parabola. If you make it here also, right will always be the white parabola. Left will always be the yellow parabola. Okay. So thankfully, if I take a horizontal strips, I don't have to change the definition of my strips. That means I don't have to break my area into sub areas and try to find it out. Okay. So, and another interesting factor is that they have already mentioned x in terms of y. So that's a blessing in disguise for us. Okay. And let's find out their y coordinate. Why am I getting different answers? Hariharan has a different answer. Inchik has a different answer. Yeah. So let's try to solve it together. So minus 2y square is 1 minus 3y square. So from here, y square is equal to 1. So y is equal to plus minus 1. Okay. So this y is going to be plus 1. This guy is going to be minus 1. Now, right minus left, right y, so right x minus left x. This you are going to integrate right from minus 1 to 1. Automatically, you'll realize that you are integrating an even function. So if you write it down, you'll be integrating 1 minus y square dy. So since it is an even function, you can also use a bit of your properties of definite integrals. So it is 2 times 0 to 1. As you can see, area is also symmetrically placed about the x axis. So it's y minus y cube by 3. Twice of this from 0 to 1. Twice of this from 0 to 1. So that's nothing but 1 minus 1 by 3. That's going to be 4 upon 3. Is this fine? Now, again, there would be an element of question in your mind. What if I had used vertical strips? See the complexity with vertical strips. Again, you can do it. I'm not denying it. So let me make those strips once again. This was your yellow parabola. This was your white parabola. Yeah. Now see, if I had taken vertical strips, then what will happen? Your vertical strip definition will be from white parabola to yellow till you reach this red line. Okay. So let me just put some value here. Let's say x is equal to a here. x is equal to c here and x is equal to b here. So from a to c, you're going to follow this strip definition. That means from the white parabola to the yellow parabola. You can double it up probably because the area will be the same. And from c to b, you have to start from white and end on white. But remember, for the white parabola, your definition was y is equal to, sorry, y square is, what was it? x is equal to 1 minus 3 y square. So this part, you will write 3 y square is equal to 1 minus x. So y is equal to 1 minus x by 3. Now there will be plus minus under root of this. So this part will have a definition under root 1 minus x by 3. And this part will have a definition negative under root 1 minus x by 3. Okay. So you have to take the difference so that will all automatically twice up and you have to integrate that from c to b part. Okay. So unnecessarily, you have to change the definition of the strip twice while solving the question. But still it will give you the same answer you can write out that also you can do. You can find the area of, you can find the area of the white under white parabola and subtract the area under yellow parabola. Yes, why not? All those things, you know, that is up to you to figure it out from the question. You can definitely take any approach which you feel it is more convenient to you. Now these questions were just the, you can say the basic ones which you will also get in your school exams, etc. So now we'll start taking up questions which will come in your competitive exams. We'll come in the competitive exams. Let's take this question. I hope you can read this. Area of the region, regions enclosed by the curve y is equal to x square and y is equal to under root mod x options are there in front of you one by three two by three one by six and one. Let me put a poll for this. Okay. One of the options is leading. Why have I integrated from zero to one? It was integrated from minus one to one, but I use the property which you have learned in your definite integrals that if you have an even function to integrate from minus a to a, you can write it as two times zero to a. See, I'll just go back. See, integration is from minus one to one only, but I know one minus y square is an even function, isn't it? Even function in y. So minus a to a becomes twice zero to a. Remember even odd property. Okay. So let's close the poll now in another 30 seconds. Five, four, three, two, one, go. Okay. So I'll end the poll. Let me ask in front of which option did you choose? Give me a response. Okay, Pranav. So Pranav chose a. Let's see whether a was the right answer or not. By the way, most of the people also say a. Let's check. See, area includes between y is equal to x square. So first of all, let's draw y equal to x square. That's an upward opening parabola, which everybody knows since class 11th days. And of course, we did our parabola conic sections in 12th also. Okay. Now this is the graph of x square y equal to x square. Let me use the yellow pen on there. How would you draw the graph of y equal to under root mod x? So first of all, y equal to under root x. Everybody knows that this graph looks like this. But since you are modding it, since you're modding it, it will also be like this. Okay. So it will look like a butterfly kind of a thing. Yes, absolutely right. Can check. Okay. So we need to find out the area. Now we know that from the symmetry of the figure, if I just find this area and double it up, I should be able to find the entire area. Okay. So let's do one thing. Let's make a differential strip over here. Okay. So top of the strip, top of the strip, your y is under root x bottom of the strip, your y is x square. Now please be careful. Okay. Don't don't mess this up. This is root x and the bottom of the strip, it is x square. Okay. Why is root x on the top? Why is x square on the bottom? So this will give you a positive height because you have taken upper minus lower dx integrate from lower value to a higher value. Now, if I'm not mistaken, this x value is going to be one for sure because x square is equal to root x. Okay. At zero and one. So this is from zero to one and you just have to double this answer to get your desired area. So this is going to be x to the power three by two divided by three by two minus x cube by three from zero to one and twice of this entire area. So that's going to be, if I'm not mistaken, two by three minus one by three and twice of that, that's going to be two by three. Okay. For the examiners, your answer is a wrong answer. I forgot I did a silly mistake. Those words should go off your dictionary. Say I am wrong. Say to yourself five times. I solved it incorrectly. I listened to you feel that pain that there was a mistake which was not supposed to be done. You will not realize the importance of it. And I'm sure you will be the first person who will come and say, sir, I lost 20 marks in the exam because I forgot this. I did this. So don't say that word for it any more further. Say I made a mistake and I am regretful about it that I should not be doing such mistakes because you marked the wrong option at the end of the day. Examiner doesn't know what you have done. He just sees that you have marked the wrong answer. And this is applicable to each one of you who hide under this excuse that I knew it. I forgot. I knew I forgot. I knew I did this mistake. Does it get registered anywhere that you have done a silly mistake or a mistake or forgetting and you get some bonus marks? No, right. So why do you give that excuse to yourself? Okay. So as a serious J aspirant, this problem is wrong. I could not solve this problem. That is the end goal. Right? Excuse me, sir. Yes, Kintrick. Tell me. So why didn't you take the bottom half as well, the negative half as well into the question for the second function? Y equal to root of X can go to any. Why can... Hello? Yes, sir. Like Y will be like minus one. Why will Y be minus one? Under root of anything can give you a negative answer. Kintrick? Yes, sir. Minus one and one satisfy that. Why? Y equal to minus one and one? One comma minus one. One comma minus one satisfies it. Okay. No, sir. No, sir. Nothing. Under root of anything, output is always positive. Okay. Any even through answer is positive. See, guys, you are nearing the phase where you are going to write you off the most important exams of your life. And if this is the... I mean, forgetting and all these excuses if you're giving yourself, you'll do mistakes there also. Okay. So be slightly critical towards your preparation. Okay. I should not be doing this mistake. Anyways, let's take this question. The area bounded by this curve Y is equal to modulus of cos inverse sin X minus sin inverse cos X. And the X axis from three pi by two to two pi is which of the following? I hope this is legible. This is pi square square units. This is pi square by four, pi square by two and no time. All options are on your screen. Okay. Two people have responded. Two minutes. So you'll have two more classes the coming week, regular classes. That is Thursday and Friday. Post that, we will keep some revision and problem practice for your 7-1. Okay. So in these two classes which are coming next week, I'll be trying to cover up the vectors and 3D part as much as I can. If anything is left, we are going to take up post your semester one. I don't think so. We'll be having... I would try to complete vectors at least, scalar product, dot product, STP, VTP, vector equation, etc. And maybe in one class we can start a bit of your basics of 3D geometry. Okay. Next 30 seconds. Yeah, yeah. If you sketch your curve properly and you know what is your function, you don't even have to use calculus. I can sense that even though I'm not... I'm going to solve it now. Okay, 5, 4, 3, 2, 1, go. Okay. So let me ask Aditya, what answer you got? What was your response? Okay. Aditya got a C and out of 11 people who voted, actually there was a confusion between B and C. But by one vote, C has got, I don't know, more number of response. Okay. Anyways, we'll discuss it out. So see, first of all, if you look at this curve, y is equal to cos inverse sin x. Can I write it as pi by 2 minus sin inverse sin x? The first part. And same, the second guy can write pi by 2 minus cos inverse cos x. Okay. Now please note your area of integration or your interval of integration is from 3 pi by 2 to 2 pi. In this interval, try to figure out what does sin inverse sin x and cos inverse cos x actually give you. And for this, I always rely on my graph. I only know the graph of sin inverse sin x, which looks like this mountains, isn't it? Yes or no? So this is 0. This is pi by 2. This is pi. This is 3 pi by 2. Now this is 3 pi by 2. Okay. And this is going to be 2 pi. Correct me if I'm wrong. Okay. So from 3 pi by 2 to 2 pi, my required expression would be the equation of this line, which is actually y is equal to x minus 2 pi. Isn't it? Yes or no? So in short, in short, the first expression that you have written over here, it will become mod of pi by 2 minus x minus 2 pi. Correct me if I'm wrong. Correct. Now, the other one for cos inverse cos x from 3 pi by 2 to 2 pi. Let me sketch that graph of that as well. So this reminds me or this helps me to recall my all-inverse trig concepts, which I had already learned long back. Okay. And like that. Okay. So this is your 2 pi zone. This is your pi zone. So 3 pi by 2 is somewhere going to be here. Okay. So 3 pi by 2 to 2 pi, you have to find the equation of this part of the line, which is y is equal to 2 pi minus x. So this is going to be this minus 2 pi minus x. So let's write it down now properly. So it's mod of 5 pi by 2 minus x. And this is going to be x minus 3 pi by 2. Correct. Now, see x is a value, x is a value which is between 3 pi by 2 to 2 pi. Yes or no? So it's a value between 3 pi by 2 to 2 pi. So this expression 5 pi by 2 minus x. Is it a positive expression or a negative expression? Write a response on the chat box. If I take any value between 3 pi by 2 to 2 pi, 5 pi by 2 minus x will be a positive or a negative expression. Positive expression. So will mod do anything to it? No, because it is only positive. So you can do away with the mod. Right. Same. If I take any value between 3 pi by 2 to 2 pi, x minus 3 pi by 2 will be a positive expression or a negative expression. That is also going to be positive. So you do away with that modulus. So when you open the bracket, if I'm not mistaken, that's going to give you 4 pi minus x. Oh, sorry, 4 pi minus 2 x. Yes or no? In short, this whole function which was given to you in a complicated version was actually a line. Y is equal to 4 pi minus x. So the question setter was basically giving you a line like this. So it is a line like this. Of course, slope will be minus 2. So he wants you to find the area between this line and the x-axis from 3 pi by 2 to 2 pi. Now this position is 2 pi for sure. This position is 3 pi by 2. So he wants you to find this area. Now as somebody was, I think Aditya was rightly pointing out, you don't need calculus for this. It's just the area of this triangle, half base. Base is going to be pi by 2. What's going to be the height? Height is going to be pi. Correct me if I'm wrong. Correct. So half base into height, your answer is going to be pi square by 4 units. Pi square by 4 square units. So which option is right? Which option is right? Option number B is right. Yes or no? Okay, so Aditya, what mistake you made? Some mistake somewhere happened? Anything that you missed out? Okay, so that's a mistake now. Okay, so I remember one. I think you're super senior, not super senior, senior. No, no, super senior, Bharat. He used to be very frustrated with himself when he used to do some mistakes. And I really liked the way he was frustrated because with time he started doing lesser and lesser mistakes because he was very much careful that such mistakes happens with him very frequently. So he was very, very sensitive towards those mistakes. And with due course of time, he started getting, his accuracy level went up very high. I think by the time he wrote, I think he started scoring around 230s to 40s out of 300, which is a good score on full length paper. Okay, let's take this question now. Consider a function f of x and g of x both defined from r to r. Your function f of x is 2x minus x square, g of x is x to the power n, n being a natural number. Area between the function and g of x is half. Find the value of n. Find the value of n. I'll relaunch the poll. Easy. I don't think so. This is difficult. Three of you have responded. Sorry, five of you have responded so far. One more minute I can give you. Okay, five, four, three, two, one, two. Okay, end of poll. Now let me hear it out from one of you. Gurman, which option did you go for? Okay, so Gurman says he went for option number b. Okay, let's check. By the way, b has got substantial amount of wood. But a is leading the tally, actually. Most of you have voted for a. Let's check whether a was correct or Gurman's answer was correct. Let's find it out. See the first curve is a parabola, which is going to be opening downwards. So something like this. Okay, having its maxima at one. So it touches the x-axis or cuts the x-axis at zero and two, isn't it? Now, when you talk about x to the power n-curve, okay, x to the power n-curve will either be, let's say if n is, you can say odd number, the curve will look like this. I mean, I'm just drawing a rough estimate. Okay, something like this. Okay. Or if the n value is an even curve, the curve will go like this. I'm just using a different color. Okay, something like this. I mean, depends upon the, okay. So x to the power four, x to the power, sorry, x to the power six, x to the power eight, et cetera. If you look at the options, there are two even powers and they're two odd powers or two even values of n into odd values of n. Okay. So anyways, we need to figure out where does the curve x to the power n cut 2x minus x square. Now that is very obvious that 2x minus x square and x to the power n will be, you know, in the interval zero to two, they will be cutting at x equal to one. This value is going to satisfy this equation. 2x minus x square is x to the power n. Okay. Or any n, which is a natural number. Okay. In short, whether it is a, whether it is a even power of n or odd power of n, you will have to find out this area. Okay. Or you can say this area, whatever it depends. Okay. In short, if you make a differential strip, it will always going to start from our parabola, download opening parabola and end on the x to the power n curve. So this is going to be a positive height and DX. If you're integrating from zero to one, that will also be positive. This area is given to us as a half. Okay. So let's figure it out. What's the answer coming out to me? So this is going to be x square. This is going to be x cube by three. This is going to be x to the power n plus one by n plus one, zero to one. This is half. So if you put a one, you end up getting one minus one by three, which is two by three, minus one by n plus one, that is equal to half. Correct me if I'm wrong. Okay. So one by n plus one is equal to two by three minus half. That's going to be one by six. So one by n plus one is one by six. So n has to be a five. So if n is five, option number A is correct. Option number A is correct. Gurman, what is happening? What excuse you have? There should be no excuse at all, Gurman. For that matter, anybody wrong is wrong. So people who gave B as the answer, please analyze where you went wrong. Shatij, if you have a value like x to the power of five, let's try to check it out. See, let me make a Jujibra too. If I have x to the power of five. Okay. So basically you are looking for the area, consider both the functions. And if you're considering this area to be, this will become yeah, yeah, yeah. The other area varies with n. Okay. Let's say if I put a seven, it is going to be even more steeper, isn't it? So Pranav Shatij's point is why aren't we considering the area in the third quadrant? I believe we should have taken this area plus this area and then add it up to make it equal to half. Right. That could have been a possibility, Shatij, because nowhere in the question they have, yeah, the area between the function and this gx, the function and gx. But finding this point of intersection would have been a trickier one because, yeah, that point will come out, yeah, that will make the problem actually complicated because then from this point, which will come out in terms of n, till zero, okay, you'll have one definition. You'll find that area plus from zero to one, you'll find that add it up. That answer will come to be a complicated expression in n. Yes, but that was quite possible. That scenario could have been used in solving the question. No, the question is not, the question is not saying it has to be defined area between the function and g of x, right? So this area is also between the function and the g of x, isn't it? Both the functions. This is also an area between them only, no, let it lie outside it, but the question never said it is inside the parabola, right? Question never said it is area inside the parabola, inside this curve. It says area between the curve, between what is mentioned. Okay, so Siddhish, yes, you are absolutely right. So Siddhish has a point here and we must admit that Siddhish has brought out an interesting thing here. So if you are able to find out this point of intersection, then clearly we'll be able to figure that out. But in the absence of that point, it's very difficult to calculate it. But nevertheless, that would be a complicated question to solve. Okay, anyways, for making it a simpler question, we'll assume that they were only asking the area in the first quadrant. So I would have assumed that it's the first quadrant area that we are assuming here. Okay, just to make the question correct. Else solving this will be a Herculean task. Solving this will be a Herculean task, right, Siddhish? Okay, let's move on to the next question. Next question, let's take this one. Again, in many of the questions, I will just try to scare you with the functions. Integration part is not challenging. So I'll be scaring you with the functions here, mostly. So that you get over the fear of, you know, that okay, how will this function be sketched and everything. So I have taken a function here which is a piecewise function which has gi, sorry, which has got fractional part also into it. So as you can see, the question center has mentioned this is the fractional part and signal function is also there. Okay, we have to find the area bounded by this function and the x-axis. Let me launch the poll. Good if you don't need integration. I'll be happier to see that. Okay, I got responses from three of you so far. Okay, four of you have responded. Five of you and all of you have given the same answer. Yes, this is mod of, mod of signal mix. Okay, last 45 seconds. After this problem will take a five, six minutes break. Okay, just to take a breather, have some water, maybe have some snacks and come back. The class will be till 11 o'clock. Okay, we'll do as many problems as we can. Ten minutes, okay, fine, ten minutes break. Okay, five, two, no, there's not much of a theory. It's all about, you already learned the theory, it's just about fine-tuning it to suit your area requirement. That's why it's a one-class topic, not more than that. Okay, end of poll. I can see most of you have voted for option number. I will not say that. Let me ask somebody. Parvati, which option did you go for? Parvati, can you hear me? Okay, Parvati, if you have to choose an option, which option would you choose? Pick up an option. Okay, Harshini, which option did you go for? Why are you responding, Charan? I'm asking Harshini. Sinchan, okay, Sinchan says A. Okay, let's check. Only one person voted for A, by the way. Most of you have voted for B, B for Bangalore. Let's check. See, a gi fractional part graph, we all know fractional part graph looks like this. Now, between minus two to, between minus two to minus one, you just have to use this part of the graph. Okay, so I'll just demarcate minus two here, minus two, minus one. So use this part of the graph. Okay, please note that its top will stop at one. I mean, it did not include one. Okay, now between minus one to one, your signum function changes its definition. So minus one to zero, zero exclusive, it will be minus one. At zero, it will be zero. And from zero to one, it'll be one. But modulus will make everything positive. So it will be a line like this. Okay, and there would be a hole over here and there will be a solid dot here. Am I right? However, those are not going to impact our result. Next, fractional part of negative X. Fractional part of negative X means you're just reflecting the graph. You're just taking this graph and reflecting it about Y axis. So if you reflect this graph about Y axis, your graph is going to appear like this. Okay, so this is going to be like this. Now you're asking the graph to be between one to two, one to two, this part of the graph, you just pluck it up and put it over here, between one to two. Sorry for that. Make it properly. This is also going to be a straight. Okay. Now here, we have to be very, very careful. This minus one is included over here. So this hole is not going to be coming. It's going to be just a straight line like this. And as you can see here, this creates a kind of a trapezium structure for you. And somebody rightly pointed out, you don't need the integration process over here if it is a trapezium. Okay. So trapezium area, we already know it's half into height of the trapezium. Height of the trapezium is one into some of the parallel sites. Okay. So as you can see, this side is of length two. Okay. And this side is of length four. So it's two plus four. Right. So this answer is going to be conveniently three square units. Option number B is right. So it's all about sketching the graph in this case. Nothing difficult. Doesn't matter. That whole area is, that area is as good as a zero. Doesn't influence anything. Okay. So there is a hole over here. It doesn't make any difference. Anyway, if you see at that hole, the value of the function is zero. Most of the question in JEE and JEE main in advance is going to be almost like this where you will be observing that the exterior part of the question will be difficult, but the interior part of the question will be very, very easy. So once you break through the question, get your function properly, sketch it, you realize that, oh, this is a simple area to figure out. As you can see, the question center has tried to test you on your understanding of fractional part, sketching graph, signum function. That's it. Area part, area component is very less. We'll take a break now. Right now, the time is almost 10 a.m. We'll meet at 10, 10 a.m. Okay. On the other side for 50 minutes, we'll take some good questions. So see you in 10 minutes time. Enjoy your break. All right. So I hope people are back. All right. So let's take this question up. Area bounded by the function x square minus one by x square plus one and the line y equal to one. Let me put the poll on. Sorry, I forgot that. Let me put the poll on for this. Almost going to be two minutes now. Four of you have responded so far. Five of you now. Not same options. Almost three options have been touched upon. Okay, Kinshukh. I can give one more minute from here on. So those who are solving it, please conclude this in another one minute. Okay, Sanchin. Okay, five, four, three, one. Okay, let me end the poll. Let me ask one of you, Atharv. Atharv, what was your pick? Or if you have not solved it, what would have been your pick? Okay, Atharv says he would have gone for option A. Okay, let's check the response first. Most of you have said option B, by the way. Okay, we'll check it out whether B is right or wrong and whether Atharv's answer was right or wrong. See, first of all, sketching this graph. Couple of things kept to be kept in mind while you're sketching this. Where does it cut the x-axis? Where does it cut the x-axis? Plus minus one. Correct. So the graph will cut the x-axis at plus minus one. Great. Where does it cut the y-axis? Minus one. So it cut the y-axis at minus one. Okay. And since this function has only even powers of x, that means this function is even function. Even function means symmetrical about y-axis. And not only that, if you find the limiting value of y as x tends to a very large value, either plus infinity or minus infinity, you would realize that the limiting value would be, I mean, divided by x square. And whether x tends to plus infinity or minus infinity, these guys will be zero, zero. So that means the limiting value is one. That means at y equal to one, the function is going to stagnate in both the directions. Now, in light of this, if I want to sketch this graph, so what are the factors which I'm accounting for? Number one, the function is cutting the x-axis at plus minus one. The function is cutting the y-axis at zero comma minus one. The function is going to have a horizontal asymptote at y equal to one. In light of all this, do you realize that the graph of the function would appear to be like this? Okay. This is your y equal to one line. Anybody has any objection with respect to this graph? All right. Now, the question setter is asking us the area between this function and y equal to one line. So y equal to one line and this function, the area would be this. By the way, this is an unbounded area because the line y equal to one is never going to cut or test this curve. But having said that, the area is going to be a converging one. It is not going to be infinitely big. Okay. So what I'm going to do, I'm first going to find out this area, which I'm showing by blue shading. Okay. And I'm going to double this up. I'm going to double this up. Okay. So for this, I'm going to choose vertical strips starting from one and ending on the curve. So height of the strip will be one minus the given function. Correct. And I'm integrating it from zero to infinity because, okay, because let me tell you, this area is never going to cut the y equal to one line. Or you can say that it meets y equal to one line at infinity. And I'm going to twice this up. Can I say this is going to be my area of the given curve? Yes, no, maybe agreed. Fine. So if this is the case, the numerator part, if I take the LCM as x square plus one, the numerator part is going to be two and there's already a two waiting outside. So let me pull that two, two out and make it a four. So it's the integral of the function one by x square plus one from zero to infinity, four times that. This is as good as four times tan inverse x from zero to infinity. So that's nothing but two pi square units. So option number B is correct. Not A at half B is correct. Is it fine? Any questions? By the way, many of you said B. Well done. So again, here, what is tested? So what is tested is the art of sketching the curve. Okay. And of course, limits, et cetera, were required because that y can never become a one. So y equal to one would be a half asymptote to this curve. See, again, it depends on question to question, Aditya. You should try to analyze everything possible that will help you to sketch the curve. Just by knowing it is cutting x-axis, y-axis at this place and you're knowing it's symmetrical, it could give so many answers. Why won't the curve be like this in that case? That also is required. It is never going to cross y equal to one. So for this case, I'm going to need that information. Or anyways, if you're going to equate this to one, you are not going to get any solution because you're going to get minus one equal to one. So that indicates you that somewhere the curve is not meeting y equal to one line. So all these aspects have to be taken into consideration. Many times you have to see whether it's concave down and increasing concave up and decreasing. All those concepts may be tested in these kinds of questions. Is this how the graph actually looks? Okay, you want me to test the graph against the tool? Okay, let's test it. So Pranav wants to see the graph. So y is equal to x-square minus one upon x-square plus one. So this is how the graph appears to me, to the graph as well. I think we had done a decent job. Almost the same. Very close. Oh, there are two Pranavs. No, I can see only one Pranav. Okay, we'll resolve that. Let's take another question. Let's take this question. As you can see, the question talks about some max min. So it's mostly about you handling functions very well. The graph of the functions very well. The question says the area bounded by min of mod x, mod y equal to one and max of mod x, mod y equal to two is switch off the following. I'll put the poll on. Take around four, four and a half minutes. Only five people have voted so far. I can give one more minute. If I give one more minute, it'll be almost four and a half minutes for this question, which is the upper bound, not one option. That's a rare, rare occasion. Even if I give a very easy question, there will be somebody who will mark more than one option. Last 15 seconds, five, two, one. Okay, okay, Athar, fine. So I'll stop the poll now and let me pick one of you. Akash Kumar, Akash, which option did you go for? Akash, you're there? Arjun, guess it. If you are not solid, guess it. Tell me, tell me, tell me. All everybody is waiting for you. Okay, Arjun is saying B. Okay, most of you, by the way, have voted for option A, which is four. Let's check it out whether Arjun is correct or others are correct. Now, see here, first of all, let's try to analyze this function. Can I say this function will take the shape of mod x equal to 1 if mod x is less than mod y? Because then this guy will be the minimum of the two. And it will take the shape of this if mod x is greater than mod y, because mod y will now be the minimum of the two. Correct. Now, this is equivalent to saying x is equal to plus minus 1. This is also equivalent to saying y is equal to plus minus 1. Okay, now try to sketch a graph of x equal to plus minus 1. If your mod x is less than mod y, okay. So how would this graph look like? So let me just make one here. Let me take minus one here. Okay. So when mod x is greater than mod y. Now, remember, everybody, please pay attention here. This is a point where x and y both are equal in mod. Okay. So 1 comma 1 minus 1 comma 1 minus 1 comma minus 1 and 1 comma minus 1 mod x and mod y are equal. Okay. Equality can be taken in both the cases because it doesn't matter. Now, if you want your mod y to be greater than mod x, in those situations, you have to travel along mod x equal to one line. Yes or no. Yes or no. Okay. So you have to walk on x equal to one line when mod x is greater than, sorry, mod x is less than mod y. How would that graph appear? How would that appear? Okay, x equal to one line is this. This is x equal to one line. Okay, let me make it. And this is x equal to minus one line. Okay. So I have to walk on this line. If I go up, if I go up, mod y will become more than mod x. Okay. So can I say I will have to trace this part of the line which I'm showing with a solid white? And can I say I have to go down this line also? Agreed or not? This part will not be there. I'm raising this part in between so that nobody is getting confused because of that. Okay. Same with this fellow also. I'll have to go up and I'll have to go down because I have to travel on x equal to one and x equal to minus one, thereby also maintaining that mod y is more than mod x. Do you all agree that this is going to be my sketch for the first function? No, not completely first function. I'll have to take the other one also. So y equal to plus minus one when mod x is greater than one. Now here I will have to sketch something like this. So can I say the graph of minimum of mod x, mod y equal to one would become a kind of a street like this or crossroads like this? Yes or no? Any questions? Okay. So this is the graph of the first one. Now for the second one, I have to do a similar activity. So for the second one, my mod x will become two if my mod x is greater than mod y. In fact, I can include equal to sign also. So plus minus two will be somewhere over here. Okay. Now you have to travel on y, sorry, you have to travel on x equal to plus minus two whenever your mod x is greater than mod y. Okay. Now two comma two is a point where both of them are equal, right? So you have to either go up or down this line. Let me let me just demarcate the four points which I'm basically interested in. Yes. So when mod x is more than mod y, that means you are going to come down like this. Or let me make it in my, okay, this will be the part of the line. Am I right? So x is equal to plus minus two whenever mod x is greater than mod y. Please note that if I come down, if I come at these positions, then only my mod x will be greater than mod y because mod y will, y will start decreasing in value, right? X will still remain at two. Okay. Same, can I say mod y equal to two means y is plus minus two if your mod x is less than mod y. So can I say it will be covering up this part which I'm showing here. So that would be a square. Am I right? Now what is the area bounded by this two curve? The question is saying area bounded by this two curve. So area bounded by only contains these four square boxes which you can see on the border. Okay. And there will be just squares of area one, one, one each. So what's the answer to this question? Answer to this question is four square units which is option number A. As you can see, area calculation was almost negligible because you got a regular figure which is a square. What was important is how to sketch those functions. That's what j, j advance type of questions. The question per se will be so rough in exterior, will so tough in exterior that you will not be able to break inside to get to the area what? And that's where people make mistakes. Is it fine? Okay, be careful. No mistakes going forward. Try to reduce the number of mistakes. Let's take this question. Question says consider the region given by some set of points x comma y such that x square plus y square is less than equal to 100. And there's another set of points x comma y such that mod of x plus y is greater than zero in the pick. Find the area of A intersection B. Why can't we say that area inside is not bounded by the function in the previous question? Why it is bounded? See, you have to, there is a curve like this. There's a curve like this. There's a curve like this. Okay. And there's a square like this. So what is the area bounded by the two? This is the area bounded by the two. Why can't we say that? Achit. I mean, if you have some other thinking, please unmute and talk. Okay. All right. So let me run the poll for this as well. Four, four and a half minutes for this question also. Okay. Two minutes gone. I have got almost six people responding to this question. Good. Last minute. Don't be worried about pre-boards. Be worried about boards. Worrying is not going to solve anything, Prakul. So think as if pre-board is giving you an opportunity to make mistakes as much as you can so that you learn from them. So main is the board exam for us, which is going to be in, I think for you also it is shifted to, exactly it has come out, Prakul, for ISC. Where is your first exam? Twenty second memory. I got, so worry about that, that exam. Okay. So we'll stop the poll now. Five, four, three, two, one, go. Okay. So let me call upon somebody. Let's ask Bhumika. Bhumika, what answer did you pick up? Which, which option did you pick up? Okay. Bhumika says she went for C. But Bhumika, most of the people say D, D for Delhi. Let's see whether people are right or your answer is correct. See, first of all, X square plus Y square less than equal to 100 is basically going to be a circle. Okay. Inside of a circle. Okay. Whose radius is going to be 10. So let's say this is 10. So all the set of points which are within the circle is going to be your A, within or on the circle because it is less than equal to. Now sign X plus Y is greater than zero. This is going to be those values of X plus Y which are in the interval zero to pi or two pi to three pi. See, remember, try to remember your sign X graph. Sign X graph is like this, isn't it? Yes or no? So your X plus Y either should be zero to pi. Then it will be positive or two pi to three pi. Then it will all be positive or let's say four pi to, I mean, it goes three pi to four pi is negative, four pi onwards. I mean, I don't require this because four pi would have conveniently crossed this guy. Okay. But anyways, we'll write it down. Okay. So four pi will be here. It will be going in the negative side as well, I believe, because negative side, it will be positive from minus pi to minus two pi. Okay. So let me write it down. Union minus two pi to minus pi. Okay. Union not minus two to three will be so minus four pi to minus three pi. Okay. And so on. Okay. So your X plus Y should be in this zone. Okay. Now let us first talk about X plus Y between X plus Y between zero to pi. Let's take this part first. So X plus Y between zero to pi. What area will it represent? What area will it represent? Anybody. So let me just draw a line here. Let's draw X plus Y equal to X plus Y equal to zero. And X plus Y equal to pi. So let me draw these two lines over here. So this is X plus Y equal to zero. Okay. And this is X plus Y equal to pi. Let's say I'm just taking a rough estimate. Okay. Now, as per this question, as per this inequality, X plus Y should be greater than zero. That means you should lie above this line. But it should be lying below this line. Right? Yes or no? In short, the area that you would be looking out for will be sandwiched between these two parallel lines. And of course, you have to be within the circle because they want the intersection of the two areas. Is it correct or not? Everybody agrees that the area that we are looking for or the region that we are looking for, which is satisfying X plus Y between zero to pi, should be this shaded area. Of course, I have also restricted myself to be within the circle. Okay. Else it will be like the infinite areas. Can I say the same trend is going to continue? Let me just make some lines here. Now, this area will be the one which is between 2 pi to 3 pi. Agreed? This one? Correct? And this area will be the one which is between 4 pi to 5 pi within the circle. It may go beyond it if I make a line, but I have to be within the circle. Similarly, I also need this area. That is between minus 2 pi to pi, minus pi. And this part is between minus 4 pi to minus 5 pi. Sorry, minus 3 pi to minus 4 pi. This is your minus 3 pi. Okay. Now, which line? This line, the last line. Why won't it cut the circle? I mean, on the top side, okay, on the top side, it will be, let's say, it is slightly, I mean, it's a rough estimate actually. So it would be like this. Now, fine. However, it will be completely outside the circle. Why it will be completely outside the circle? See, 4x plus y is equal to 4 pi line. Okay. If you find the perpendicular distance of the origin from here, will it exceed 10? That's what we are trying to figure out. So the distance will be 4 pi by root 2. 4 pi by root 2. Let's try to figure it out. The value of 4 pi by root 2. I know I'm just using my calculator for it, just to be very accurate in my value. This divided by root 2. That gives out to be 8.88. It has to be within the circle. Okay. Now, look at the entire scenario very, very carefully. You would realize that the shaded area that you need, it is actually half of the answer or half of the area of the circle. Now, why C? This part that you have included, that part here, you have excluded. Correct? The part where it is excluded here, that part is included here. So what is happening is, you're exactly covering half the area of the circle in this particular question, which is half pi r square, which is 50 pi. Agreed? Any question related to why the area that I have shaded will ultimately be half the area of the circle. So symmetrically placed, and one is included, other is excluded. So basically you're covering whatever part you're covering from, you can say this side of the diameter. So let's say this is your diameter line. This is your diameter line. Okay? So you are covering half the area over here and the same half you're leaving out from it. Correct? So if you combine the area together, you are basically covering up half the area of the entire circle. Any questions? Beautiful question. This was a very good question to solve. Let's take another one. The previous question you're talking about, Harir, that was there in one of the mocks. Okay. All right. So we'll take this up as well and I'll run the poll first of all. S is the region of points which satisfy a y square less than 16x, x less than four, and they've given an inequality. Two minutes gone. Two people have responded so far. This is the last question for the day. In fact, last question for the topic as well. And as I told you, and I would remind this again to you, Thursday and Friday again, combined sessions would happen on vectors and 3D. Last minute for the question. One moment. Okay. Five, four, three, one. Okay. You want me to wait for how many seconds? 30 more seconds is fine, Kinshuk. 30 seconds. Okay. Let the poll go on for 30 more seconds. Okay. Should I now close the poll? Five, four, three, two, one, go. Okay. Before I share the result, let me name a person. Himanshu. Himanshu, what was your pick? Which option did you choose? Okay. Himanshu went for... What option, Himanshu? So, Himanshu went for A. Okay. Let's see. Most of the people said B, by the way. Okay. Let's see whether Janta was correct or Himanshu was correct. Let's check it out. See, first of all, Y square is less than 16X. Basically points to all the set of points within the parabola, Y square is equal to 16X. So, you are supposed to be within the parabola. X less than four. X less than four is, let's say, this line. So, you are supposed to be on the left side. That means your... Whatever points you're looking at, that has to be within this zone. Okay. Now, again, one more inequality has to be satisfied because you have to take that also as one of the overlapping cases. So, those cases where this inequality is satisfied, that is also to be accounted for. So, within this parabola, left to the line, and also satisfying this inequality. Now, what I'm going to do is, I'm going to break this into two cases. As you can see, there is a Y term sitting over here. So, if I say Y is positive, then the other term must also be positive for the inequality to be satisfied. This is one case. Or Y could be negative, and the other term also could be negative. So, I have to take the union of such X comma Ys or I have to take an addition of two zones which are given by this condition and this condition. So, I have to add the two zones. These two zones will be added. Not overlapped, added, because there is an odd condition sitting over it. So, within these two, you have to take the overlap. Within these two, you have to take the overlap and add the two zones. Add the two zones means, you will collectively take both the regions into your picture. Now, let's take the first case itself where Y is positive, and Y is positive means above the X axis and even this condition is satisfied. Now, this condition is, I think, as good as a wavey curve. This is factorizable. So, you have a zero, you have a one, two, three, four. Positive, negative, positive, negative, positive, negative. Okay. So, Y should be greater than zero and you want your X to be in the interval, zero to one, union two to three. Four to infinity anyways is not a part of our zone. So, Y is positive and X should be zero to one. Okay. Now, let's check. This is, let's say, one, two, three. So, Y should be positive and X should be between zero to one. So, can I say this zone? This is one, zero, two, three, and this zone. Am I right? So, the zones where Y is positive and X is between zero to one and two to three is this zone only. And of course, we have to be within the parabola. Everybody agrees with this first part? Okay. Second part, this one says Y should be less than zero and again, the wavey curve here will be almost the same. So, plus, minus, plus, minus, plus, minus. You want this guy to be negative as well, right? So, if you want this guy to be negative, you have to be between one to two. X should be between one to two or three to four. So, Y negative means below the X axis and X should be between one to two, means I am looking at this zone. Okay. Or three to four. Or three to four. Okay. So, do you realize that the area that we need is actually this area? Correct? Now, if you see this missing part is sitting over here, this missing part is sitting over here, this missing part is sitting over here. So, let's say you fill the missing part with the other part which basically means half the area of the parabola. So, the required area is half the area of the, you know, this whole thing. Okay. Area of the parabola. Between the parabola and X equal to four line. So, you can do this thing. Half the area of the parabola is as good as saying, zero to four, zero to four. And this parabola equation was Y square is equal to 16X. So, you can take four root X minus zero dx. So, it's eight by three, four to the power three by two is eight. So, our answer is 64 by three. Option number B is correct. So, Himanshu, either ways your answer became wrong. Is this fine? Any question? Any concerns? Okay. So, with this, we close this chapter. All the best for your coming week, three board exams. Okay. Never mind. So, we'll meet again on Thursday and Friday. Till then, take care. Bye-bye. All the best for your pre-boots.