 So today I'm going to begin with property number four, sorry, property number five, four we had already done in the last session, property number five, property five. Okay, property five normally I call it half the or having the limit property. Okay. Now we'll come to know why we call it having the limit property. So let us say you have a function f of x, okay, which is our function continuous in the interval zero to two way. That means you can apply FTC to it because the function is continuous in the interval zero to two way, then you can write this function as zero to a f of x zero to a f of two a minus x dx. Okay. And as you can see the upper limit has been halved in both of these sub integrals which have come over here. That is why the name of the property is given like that half the limit property. Let us prove this property first of all and we will see the application of this property through some questions. So we're doing it in both ways, non geometrically as well as geometrically through our algebraic some concept. So I'll first start with the proof. I'll first start with the proof are not very similar we'll see when we prove it. So let's say zero two to a, I break this interval as zero to a. Okay, so our splitting of the limit, limit property basically allows us to split the limit at any point between a and b also. So I've used that property to begin with. Now as you can see the first integral has already appeared over here. Okay, so I will not disturb that term. Maybe I will work on the second integral to show that it is equal to the second part of it. So I have to show the equality of these two integrals. Okay, so let us do that. So for the second interval I'll write it separately a to two a. Okay. Now here what I'm going to do I'm going to use a substitution. What is the substitution I'm going to substitute x as two a minus t. Okay, so DX becomes negative dt. So let us write it down so f of two a minus t DX becomes negative dt. Now what will happen to our limits on t as X is a as X is a T is also a please check it out. Okay, so let me write it here as X is a T will also become a and as X is two a T will become a zero. Okay, that means we are integrating this from a to zero. Of course it has a negative sign. Now with respect to negative sign I discussed it yesterday that if you want to observe the negative sign just stop the position of the upper and the lower limit so it becomes zero to a f of two a minus T dt. And now the very first property which is very important. There is nothing in the name, call it T call it X back again doesn't make any difference. So it is back to zero to a f of two a minus X DX. Okay, so in short, basically this integral could be replaced by this integral could be replaced by this. And hence, and hence we can say integral of f of x from zero to two a could be written as integral from zero to a f of X, and integral of zero to a f of two a minus X DX. Now we'll be proving this property will be proving this property geometrically as well. So we'll see how the algebraic area under the function and the curve can be also expressed in both the ways. Okay, so let's try to do it. Let's try to do it geometrically as well. First note, note this prove down if you want to. And if you have any questions to let me know. Okay, no questions anybody. So those are joined in first of all good morning, we were doing the property number five, which is called halving the limit property. Yes, I'll explain everything once again. So to prove this property. So this is the property that you have over it. So to prove this property I first use the splitting of the limit property. So zero to a and then a to two a. The first limit I did not, sorry, first integral I did not disturb. Second integral, that is this integral. Let me write it here. Right, what I did, I made this substitution X is equal to two a minus T. Right, so your X became two a minus T DX becomes negative DT. When your lower limit is a lower limit of T will become a when your upper limit is to a then the upper limit of T will become zero. So this integral becomes minus integral from a to zero of f of two a minus T DT. Then I switch the position of I interchange the position of the upper and the lower limit minus sign got absorbed. And now there's nothing in the name. So I wrote my T back in terms of X. Okay, so once I wrote it back, this whole thing the second integral I replaced it back with zero to a f of two a minus X DX and this becomes your final property. Is this fine? Any questions? Any concerns? Okay, now we'll try to understand it from geometrical perspective. So geometrically, how do you see this property? See, it's very important to analyze these properties geometrically because that will leave a long lasting impact. Or you can say a retention in your mind with respect to these properties as what will happen you'll keep forgetting them. Okay, so let us say I have a function like this, which you are integrating from zero to two a. Okay, so you want to find out the algebraic area between the function and between the two vertical lines X equal to zero and X equal to two a. So this algebraic area is given by this expression. Let's call it as a. Okay, please note that a has got a sign of its own. It is not an absolute area. It is an algebraic area. Now see, this property says that you could find out the same area first of all. Okay, by finding the area under the curve from zero to a, which is definitely correct. Okay, half the area can be obtained from here. Let me call it as a one. Okay. And then what do you do is you make some small graphical transformation. So what is the transformation. So first, change your X with a negative X. So when you change your X with a negative X what happens to this graph, this graph gets reflected. Am I audible to everybody because it's showing my internet is unstable. Am I audible. Can you hear me on is my voice. Okay, okay. Thanks. Thanks. So once you change your X with a minus X the graph gets reflected about the y axis. So it becomes like this. And then what do you do, you change your X with a X minus two a. So when you do that you are shifting the graph. By the way, this will this end will become minus two a. Here it was at two a. Okay. So here you're doing what you are shifting this graph to a to the right. So when you do that, just like what happened in the King's property, the graph will start looking like this. Okay, the graph will start looking like this. So now this graph which is f of two a minus X. If you see if you look for this area, which I'm shading in white. This white area is let's say a to area that a to area will be the same as as the half of this area on this site. So this is a to area and this is a one area. Okay, so what is happening, you are obtaining half the area from the actual graph. And then you're flipping it, you're not taking it to the original position and then the next half of the area will now come between zero to a again. Is it fine. Voice is breaking up. Is it is it now breaking up. Now it is fine. Okay, so this is the geometrical way of looking at the property. Now there are conclusions that we are going to draw from this property. Two important conclusions by the way in school. Many of the students just remember these conclusions. But what is important is knowing the parent property. Okay, this is the parent property and the conclusions are derived from the parent property. Okay. So, the conclusions are number one, if your function satisfies this property that means f of x is equal to f of two a minus x. Okay, then what will happen the integral zero to two a f of x dx can be written as twice of zero to a f of x. Now can somebody tell me that if a function is satisfying this functional equation. Is there any specialty in the graph of that function. When do you think a function will satisfy a property like this f of x is equal to f of two a minus x. Right. This property will be satisfied when the function is symmetrical about x equal to a line. Okay, that's very easy to figure out. See, if I make a graph out of it, let me make a graph over here. See, what is trying to say, let's say f zero, f zero and f two a are same, let's say f zero is here, and this is your two a point. So, f zero and f two a are same value. f one and f two a minus one is same value. Let's say f one is there and f two a minus one is there. Okay, f three and f two a minus three has same value f three and f two a minus three has same value. So basically what are you trying to say the whatever is the value of the function at one the same is the value at two a minus one. Whatever is the value at two same is that two a minus two I'm just taking some integral shifts, but it is continuously following the property here. What is going to happen f a and f a will definitely be the same the moment you put x as a f a and f a would definitely be the same. So basically it tries to show that the function is of such a nature that it is following a symmetric nature about which line about the vertical line x equal to a. So when such a thing happens, and we are trying to find out the area of this entire function between zero to two a. Then is it not, you know, easier to figure out the area between zero to a and double it up. So just find out this area. Let me share it in orange color. Find out this area, which is your this term and double it up. Okay, that is what this property says. Similarly, another conclusion is if your f of x is negative of f of two a minus x, then what will happen this area is going to become a this area is going to become a zero. Can somebody tell me what is special about just like in this case you told me it will be symmetrical about about x equal to a line. Okay, what will be happening in this case. It will be symmetrical about a comma zero point right. So in this case the graph is showing symmetrical nature about a comma zero point. Okay, so you can see very well from, you know, a rough analysis of it. So I'll also make a graph for it. See what are you trying to say. This function equation is trying to say what value the function has at zero. It has negative of that value at two a. Okay, so let us say the function value at zero is here, then at two a it will have a negative of the same value. Correct. Yes. Now remember the function is continuous. Don't forget that whatever it has at one the same it will have at two a minus one negative of that. So let's say one value is here, then at two a minus one it will have negative of that value. So if you start doing it, you'll realize that when you put x as a, when you put x as a it will become f a is equal to negative effect. That means to f a is equal to zero that means the graph will cut the x axis at a comma zero so basically it will be following a nature like this. So you can see that and you can see that this part would be algebraically opposite in sign as compared to this part so they will each cancel out. So let's say this is a positive area, then this is a negative area, they will cancel each other out to give you a algebraic sum as a zero. Okay, so as rightly said this function will be showing a symmetrical nature about a point a comma zero. That's fine. So these are the conclusions that normally in school, we need but overall you should remember this parent property as well very very important. So please make a note of everything whatever we have discussed so far. Okay, now the same integral which is f of x from zero to two a is also presented in an alternative way which I will give you as a question. Whether you are fine in proving these properties. Okay. So let's take a question first question is prove that if f of x is a continuous function between zero to two a integral of f of x from zero to two a can also be written as zero to a f of a minus x dx plus zero to a zero to a f of a minus x dx. Okay, many books will claim this to be a separate property itself but I mean, this is just another version of the same property. So please try to prove it and do let me know if you're done with the proof. So tomorrow morning we'll be continuing with our pair of straight lines topic. And Monday we will have I think Monday you don't have a session right YPR and PSYP and Monday is there any. Oh, in the evening you have physics Monday evening I'm not talking about evening Monday morning same time. See today like Rajaji Nagar have physics class right so they're still coming in the morning session right so you'll have two sessions. Some days. Okay, so I was planning to have a session with you on Monday morning. Can we again keep it at 1130 to 130. Then 12 to 222 becomes a lunchtime know. So what is your timing for Monday 1130 no 825 to 1115. We'll have to follow the same time actually. Okay, are we done with this done. So I decided to prove this so our effort would be to basically you know, see this is already zero to a can I use King's property on this. So on this if you use KP, you'll get zero to a f of a minus a minus x dx. Okay, which is basically zero to a f of x dx. So my aim will be to convert this from a to 2 a f of x dx. So see what I will do for it. See you should know what is your end goal mind goal is to create zero to 2 a f of x correct so f of x is what I need everywhere. And preferably if the first one is zero to a f of x the second one I want it to be a to 2 a so that I can apply the reverse of splitting of limit property to get the left inside. So in this what I plan to do please note here carefully. So in this I plan to substitute a plus x as a t. Okay, so dx and dt will be same as per the substitution. So f of t dt put your lower limit of x as zero then lower limit of t will be a put your upper limit of x as a then the upper limit of t will be doing. Okay, and then there is nothing in the name. So there is nothing in the name so you can write it back as an x. Okay, so the same as a to 2 a f of x so if you combine these two, then from the reverse of the splitting of limit property you can say it's zero to a, so zero to 2 a f of x dx. Okay, is this fine any questions any concerns. Now let us take some problems. Let us take some problems. The first problem that I would like all of you to try out here is the one which is also asked very commonly in school exams. And I normally request students to remember this result because it is not only there in your NCRT but also a very commonly seen result or use result in definite integrals. So that is zero to pi by two LN sin x dx prove that this is equal to zero to pi by two LN cos x dx is equal to pi by two LN half. Okay, and let me just put a note here please remember this result because it is very useful. So let's solve this question prove that integral of LN sin x that is log of sin x to the basic from zero to pi by two is same as LN cos x integral from zero to pi by two, and this result is pi by two LN half. So yes one book again sorry to disturb. So you asked a lot of questions and yeah, till when is this event of 8 to 11 will go in school. Any any idea or is it like ad hoc, just a Monday. Okay, so please wait for my next message. If Monday is the last thing then we can do something about it. Yes. So if you don't have any idea how to begin with at least the first part of the proof is easy. How do you prove this. KP. So let us say I is integral of LN sin x from zero to pi by two. So by King's property I can say I is also equal to zero to pi by two LN of sin pi by two minus x. And that is nothing but zero to pi by two LN of LN of cos x. Okay. Now just add these two eyes. So these two eyes you add them. Fine. So when you add them you get to I is equal to zero to pi by two. See LN sin x plus LN cos x will give you LN sin x cos x. I hope lot properties is clear to everybody. Okay, then what I'm going to do. I'm going to put a two here and two here. She doesn't good. So this problem normally if you have not solved it before it becomes slightly difficult for people to crack it. So this is nothing but LN sin 2x by two. Correct. Which you can write it as zero to pi by two LN sin 2x dx and minus LN 2. I will take it as a separate integral. Okay. Please let me know if you have any problem, any concerns, any questions till this stage of the solution. Fine. Process is clear. What did I do? First, I use my King's property. I got the second integral. Both the integrals are same. So I added them. I got to I. So when I added them, LN sin x plus LN cos x gives LN sin x into cos x. I introduced the two inside here. Okay. And I made it like this. Okay. No problem so far. Now, the second integral that you see over here is just an integral of a constant. So LN 2 is a constant. Right. So this integral will be simply similar assembly LN of x. Sorry. LN of 2 into x. And you just put the upper and the lower limit apply your FTC to and just figure it out. So you end up getting 2 y is equal to 0 to pi by 2 LN sin. LN sin 2 x. And this will become minus pi by 2 LN 2 minus pi by 2 LN 2. Okay. No problem so far. Now, in the first integral here, you will put 2 x is equal to T. So DX will be equal to half DT. Correct. So when you do that, you'll end up getting LN sin T DX is half DT. Okay. And watch out for the limits. This is very important to your students because people in the, you know, hurry, they forget to change the limits. I've seen many times the student has made a change in the function DX function. Everything is changed limit. He has forgotten to change. And because of that, everything goes for a toss. So your limit of integration will now be zero to pi. Correct me if I'm wrong. Okay. Now, see my effort is to get back to the LN sin x function once again, whether you get it as LN sin T, LN sin U, LN sin x doesn't matter. Name of the variable doesn't matter. Now here, the only problem is my limit is from zero to pi, but my original I was from zero to pi by two. Right. So what can I do? I can make quickly make a check. So let us say I have a function LN sin T. Okay. Does this function change if I put T as a pi minus T? You say, no, it doesn't change. It still becomes LN sin T. So indirect, indirectly, what we are trying to say is that treat this as two A in your mind. So you're trying to say f of T is equal to f of two A minus T. Okay. So basically, let me go back to the previous slide. Basically, you're trying to say that not this one, even before this one. Yeah. So basically, you're trying to say that your function is satisfying this first conclusion. If f of x is equal to f of two A minus x, then zero to two A, which is in our case zero to pi f of x can be broken down into two times zero to pi by two LN sin T. Isn't it? So I will use this first conclusion to get my job done there. So let me go back to the slide. Yeah. So this fellow, what I'm going to do, I'm going to break it up. First of all, let me take a snapshot because I think I have reached the end of the slide. Yeah, so let me go to the top. So can I say, can I say this term I can write it as two times zero to pi by two LN sin T because my first property was basically fulfilled. Okay. And this is back to your eye. Isn't it? This whole thing is back to your eye because there's nothing in the name, right? So two to gets canceled and you get two I is equal to I minus pi by two LN two. Take this I to the left side. So two I minus I will give you an eye. So it gives you minus pi by two LN two, which is as good as pi by two LN half. Very important result. If you remember it, you will save a lot of your time. I know in questions which basically uses this integral. So we'll take some questions where you'll find that if you remember the result of this particular integration, you can save a bit of your time. So please note this down. This is there in your NCRT also and it asks it comes in school exams also quite a lot. So if you want to copy from anywhere, let me know I can drag the screen to that point. So if you repeat the second step here, second step aware in this part, this part in the right. Okay, see what happened is this integral, which is that I'm treating as so I'm just writing it separately. I'm treating as zero to two a just the one which I've shown with the curly bracket forget the half part. Okay, so I'm treating it like this. Okay, and I just showed you that f of t and f of two a minus t are same. So I can break it up as two times zero to a f of t dt. Now our to a is pi here. So half of pi is pi by two and this two here comes over here. Good. And then I realize that oh this is back to my eye. So this is back to why so after that simplification is quite obvious. Any questions anybody else. Okay, so let's take another set of question here. Integrate integrate zero to pi by two. x cot x integrate zero to pi by two. x cot x. If you want, please let me know your response in the chat box. Oh yeah, minus correct. Correct. Okay, anybody else. Right. Okay, so if you see, we can apply integration by parts on this taking this as my first function and taking this as my second function. Okay. So integration my part says it will be you into integral of cortex integral of cortex is Ellen sign x. Okay, now in this whole thing you have to put your upper and lower limit. Fine. Minus derivative of x is going to be one integral of very good for the chair and why does the sign come in and I couldn't get that. Okay, never mind, we'll check whether it is right or wrong. So integral of one into Ellen sign x from. Okay, zero to pi by two. But this is that we just now saw. Right, as I told you, if you know happen to remember this is that you are going to save a bit of your time, you don't have to do everything right from scratch. So this problem would have become a double doubly long problem where you have to separate to evaluate Ellen sign x from zero to pi by two. So this result is already pi by two Ellen half pi by two Ellen half. Okay. Now what about this. So when you put a pi by two, you'll get a zero. Okay. But when you put a zero, it actually becomes an indeterminate form. So here in such cases, what do we do, we take a limiting case over here. We're going to take zero minus limit extending to zero of x Ellen sign x. Okay, so when you put a pi by two, it gives you a zero, no doubt, which is this zero. And when you put a zero, you have to take a limiting case here. Okay, now this thing you can evaluate easily because all of you are very well versed in your evaluation of limits you can use Lopital. So right now it is tending to zero into infinity form so we can convert it to infinity by infinity form also. So Ellen sign x upon one by x. So now it is infinity by infinity form you can apply Lopital. Apply your LH rule. So I'm applying LH rule this becomes limit extending to zero. Derivative of Ellen sign x is one by sign x into cos x derivative of this will be minus one by x square. Okay. So this becomes if I'm not mistaken, Ellen of negative x square by tan x. Okay, and this anyways is going to be zero because you can only compensate one of the x and one of the x will be, you know, free to make everything zero. But this guy will be the game changer. It will make it zero. So everything here becomes zero. So this limit is also a zero. So overall your answer is minus pi by two Ellen half or you can say pi by two Ellen. This is your answer. Is it fine? Any questions are HL because an element become yes. Ellen will become undefined. Correct. Could you explain the second step in the limit? This step to this step. See, right now, however, this is ending to zero into infinity form. We cannot evaluate. We cannot apply Lopital on this. Is there any way to evaluate the limit of this or above this one? This part or this part. This part. See, this part when you put a pi by two, what do you get? You get pi by two Ellen one. And when you put a zero, please note that you can't put it directly zero because Ellen zero is going to be undefined. So in those case, we take a limiting case. Are you getting my point? So that is why the second term is a limiting term extending to zero X. Ellen X. Oh, sorry. Excellent. Sign X. Kinshuk, does it work? We could have X as we had. Does it work? Okay, fine. Then you could use it. No, in this case, your right hand limit exists because you're going to put a value between zero to pi by two. So you're going to put a value between zero to pi by two. So before zero, you're not interested. So only one sided limit is good enough. If a function doesn't exist to the left of something, we are fine with one sided limit only, right? That is what we had done in our case of limits. Next question. Let's move on to another one. If an integral zero to pi by two cost of two pi by three sine square X. Okay, is you. And another integral zero to pi by two cost of two pi by three sine X DX is V. Then which of the following option is correct? Two U is equal to V. Two U is equal to three U is equal to V. U is equal to two V. I'm launching the poll for this. If you're done, you can put your response on the poll as well. Somebody is fast enough to solve this in 30 seconds. It happens when somebody wants to unblock this and unblock the view because this poll, they come in front of your phone actually. Okay, so three people have responded. And three people, three different options. Okay, four people for different options now. Okay, last 30 seconds. 10 of you have already responded. Last 30 seconds for those who want to respond. Five, three, two, one. Now the people who are making a guesswork, they'll make option B grow. Okay, so most of you have given vote to either option A or B. Okay, B I think it's more because when you don't know how to solve it, put B as your safe as seven, but I can see most of the time it becomes wrong actually. Now let us start with you actually and okay, so you, if I apply King's property on this, so if I apply KP on this, so King's property make you as zero to pi by two costs, two pi by three cost square X. Okay. So I'm not writing the intermediate steps like writing X as pi by two minus X. So it is very, very much understood that's a sine square pi by two minus X is cost square X. I'm directly writing it. Now add the two us. Add this you and this you. Okay, so these two you add. So when you add it, you get zero to pi by two costs of two pi by three costs square X plus cost of two pi by three cost square X. Okay. You can use your transformation formula cost A plus cost B over A. So this becomes zero to pi by two to cost this plus this by two. Now see this plus this will give you two pi by three because sine square plus cost square will become one by two will give you a pi by three correct me if I'm wrong. Okay. Then cost this minus this by two will give you a pi by three cost to X. In fact, ideally it should give you minus pi by three cost to X. But as you know, cost is unaffected by the negativity of the, you know, argument of the angle to it. Okay. Now to cost pi by three will be one. So just leave it aside. So now you have zero to pi by two cost of pi by three cost to X. Now is it any close to our V? So we at least have a cost pi by three. Now there's a sign X and the limit is limit is limit is pi by two. Okay. So what I can do here is yeah, can I do one thing? Can I put my two X as a T? So DX is half DT. So this will become zero to pi. Correct me if I'm wrong. Cos pi by three cost T. And half DT. Is it fine? Any questions? Now let us see whether halving the limit property can help us over here because eventually my V is in terms of zero to pi by two. So let us see. So let's say I call this function as f of T for the time being is f of T and f of pi minus T give you the same result. Give you the same result. Yes, because cost pi minus T is minus cost T but there's a cost outside. So that negative sign will not matter. So this is satisfied. So this is satisfied. I can write two times zero to pi by two cost of pi by three cost T. Two to goes off. Okay. Now again, let us apply KP because I need, I need sign inside not costs. So if I use KP again, this becomes zero to pi by two cost pi by three sign T. And this is what is your V actually. Isn't it? So your two U is equal to V. That is option number A is the right option. What happened to you? You got it right? Good. Good. Is it fine? Any questions? Any questions? Any concerns? Okay. So we can go on and on with the, you know, more problems. But now we are, I'm going to begin with the next property, property number six. That is more or less inspired from this property itself. Could you explain the first step? First step from here to here. Kings property. I added these two, you know. So what is cost? Let's say I call this angle as A and this angle is B. What is cost? A plus cost B. Two cost A plus B by two cost A minus B by two. Okay. Okay. Okay. So now I'll be moving on to the next set of properties. The next set of property. I call it as the even odd property. So this is property number six for us. Property number. Six for us. Okay. So this property name is even odd property. Why I'm giving some names to it so that, you know, when I talk about it. With the name, you will be able to connect to it. Even odd property. So this is property number six for us. Property number six for us. You can connect to it. Even odd property. This property says that. If you are evaluating an integral. Which is from minus A to A. F of X DX. You can write it as zero to A. F of X DX. And zero to A. F of minus X DX. Okay. So what is this property number six for us. So if you are evaluating what is even an odd in this. We'll talk about it. We'll talk about it. It has to do with the conclusions of this property. So this is the parent property. First of all, I would request everybody to prove this property. I want everybody. So instead of me moving this property. If it comes from you, I will be more than happy. And we'll also try to prove it geometrically as well. Once done, please. Let me know on the chat box. It is done good. You can choose is also done. Why are you raising your hand. All right. Okay. Let's discuss it. Most of you have done it. So I think let's start doing it. Okay. So first of all, again, splitting the limit property. So if I have to start solving this problem against, please remember the function is considered to be continuous or minus to it. Okay. Then only you are basically able to write such kind of a Yeah, so minus a to a, let's split the limit as minus a to zero and zero to A. Okay. Now the second term I'm not going to disturb because the second term, okay, do not disturb this, do not disturb because this is same as this term. Okay. So why do disturb it? Right? All I need to do is I have to show that this guy, I have to show that this guy is equal to this guy. This is what I need to show. Okay. So how do I show this? So very simple, we'll take this term separately. Minus a to zero f of x. Okay. Now, see, how do you get a minus x over a and how the limits become, you know, opposite in sign? The answer to those questions is very obvious when you basically substitute x as a negative t. So dx becomes negative dt. Okay. So this whole expression becomes f of negative t, f of negative t, dx will become negative dt. Okay. And the limits of integration will now become if x is minus a, t will become a and if x is zero, t will become a zero. So all we need to do is absorb the negative sign and then there is nothing in the name. So absorb the negative sign by flipping your upper and the lower limits positions. And now there is nothing in the name, so you can call it as zero to a f of minus x dx. Okay. So if you see the first property is actually important, it seemed to be a trivial one, but it saves us at so many places. So here the second, the first term becomes your second term. So hence we can say, please remember this parent property, integral of minus a to a f of x is same as zero to a f of x and zero to a f of minus x. Okay. Geometrically speaking, this proof is very, very simple to understand. So you're trying to, you're trying to, let's say I have a function like this. Okay. You're trying to find out the algebraic area between minus a to a. Okay. And let me do one thing. Let me call this half as a one, this half as a two. So your answer to this limit that is minus a to a answer to this integral is a one plus a two. Okay. So the first part says this part is directly your a two. No problem. Right. Right. So zero to a f of x is your a two part. Okay. So the left hand side is a one plus a two and the very first integral has already given you a two. Now let's check what does the second integral give us. So for that, we need to sketch f of minus x f of minus x means again, taking the mirror image of this graph about the y axis. So it'll be something like this. So now zero to a will give you this area, which actually matches with this area. Okay. So this area will be your a one area. So this is how your left and right hand side become equal to each other. Is this fine? Any questions? Any questions with non geometrical and geometrical proof? Do let me know. Now what is important and what is mostly used is the two conclusions that we draw from this property. So we'll take those conclusions on the next slide. But if you want to copy down anything, do let me, please do so so that we can go to the next slide. Done. Copy. So from this property minus a to a f of x dx is equal to zero to a f of x dx plus zero to a f of minus x dx. Two important conclusions we draw. The first conclusion is if your function is an even function, okay, that is to say f of minus x is same as f of x. Okay. Then what happens? This integral minus a to a f of x dx will yield twice of zero to a f of x dx. Of course, because your second integral will also be the same as the first one. And geometrically speaking, it happens because for an even function, the graph is already known to be symmetrical about the y axis. Isn't it? So when you're integrating and trying to find out the algebraic area between minus a to a. So this says that if you realize your function is even, then just find out this area and double it up. Okay. So double of this area will be your entire answer. Okay. So double of this area will give you integral from minus a to a f of x dx. Okay. Very obvious. And if your function happens to be an odd function, if your function is an odd function. Okay. That means f of negative x is negative of f of x. Okay. Then what will happen? Your integral from minus a to a f of x dx will give you a zero area because as we know, the function has to be continuous. And if the function is continuous, and it has to, it has to be an odd function, it has to pass through origin like this. Let me make a, I mean, any graph you can make like this. Okay. It could be other way around also. Okay. I'm just drawing a dummy graph here. So this part, this part will be a positive area as per this diagram, and this part will be a negative area as per this diagram, and they will cancel each other out giving you a zero answer. No, no, no, no. Not necessarily. An odd function. An odd function is just symmetrical about the origin. Right? But if it is continuous, it will pass through the origin. Is this fine? Any questions? Any concerns? Anybody has? See, if you remember the graph of, no, I think that would not be a good example because the domain doesn't contain that. The graph of cosec inverse x, right? It was an odd function, but it was not passing through the origin, maybe because origin was not a part of your domain. So it could be symmetric about origin without passing through it. Maybe y equal to 1 by x could be an example as well. Okay. All right. Now here, these properties will be using in order to solve many questions. So what we do in light of this property, wherever we find there is a function given to us, we first try to analyze whether the function is completely even or completely odd, or it could be broken down into even and odd combination so that my odd part doesn't take much of my time. Okay. So odd part, if you remember, if I had given a question in the determinant also, so the moment you realize your function is an odd function and you're integrating it from minus a to a, let's not waste even a second to write it as a zero. So that is basically a time saver for us. Okay. So we'll see some questions and we'll try to break up a function into even odd components. We know that all functions can be written down as a sum of an even and odd. And wherever the odd part is there, you should not waste any time writing that part as a zero, that integral as a zero and save your time. So let's take a few questions based on the same. So let me begin with, let me begin with this question. Integral from minus pi by four to pi by four x to the power nine minus three x to the power five plus seven x cube minus x plus one upon cos square x. Simple question. Please think and then solve. Very good, Sutej, Aditya. Nice. Good one. Anybody else other than Sutej and Aditya? Gayatri, very good. Now, I think in this session there will be two batteries. One is with TH, other is without an H. Yes, should we discuss it? I don't think so. This is a difficult problem. See, if you have a good watch on this entire problem, this is an even function. Okay. This is an even function and all this is an odd function. Okay. So let me write over it. Even, even odd. Okay. So what I'm going to do, I'm going to use the property that odd by even is an odd. Okay. So I'm going to first break this integrand as this. Hariharan, very good. Okay. And this. Correct me if I'm wrong. Okay. So even by even is even but odd by even or even by odd or same goes with product also. They will be odd. So this guy will be completely odd if you have paid attention to the functions chapter. So we have done this property. So this will be completely odd. So here you don't have to waste even a single second. Write this as a zero. So the problem boils down to integrating minus pi by 4 to pi by 4 secant square x. Secant square x integration. Everybody knows. Okay. It's going to be, by the way, before, before you solve it, you can do one more step. So since this is an even function, you can write this as zero to pi by 4. I mean, even if you don't do it, it's not change your answer. So zero to pi by 4 twice of it. So it will be two times tan of x from zero to pi by 4. So when you put a pi by 4, you get a one. When you put a zero, you get a zero answer simple as that. Any questions? Anybody? Any questions? Any concerns? Next. Can I go to the next slide? Okay. Try this out integral from minus half to half x plus one by x minus one whole square x minus one x plus one the whole square minus two whole raise to the power of half whole raise to the power of half. This is a normal bracket. Okay. Don't read this as a gif unless until stated. This is a normal bracket. So integral from minus half to half x plus one by x minus one whole square x minus one by x plus one whole square minus two whole raise to the power of half whole raise to the power of half. Okay. Okay. Okay. It has. It has. I think, please check your results. They don't seem to be right. Anybody else apart from two of you? Okay. Very good. Okay. So first of all, let us try to discuss it. Now, possibly one mistake that you might have made is writing this whole thing as x plus one x minus one minus x minus one x plus one. But it is actually mod of this. Okay. We will check. Okay. If you would have missed out a mod here, things will go for in a different direction all together. Okay. So minus half to half. If you just take the simplification over here, it's x plus one the whole square minus x minus one the whole square upon x square minus one mod of the whole thing. So this gives you this gives you minus half to half numerator will become a four x and denominator will become x square minus one more of the whole thing. Is it correct? Okay. Now, if you look at the wavy curve. So since there is a mod, we need to see how does this mod change its definition in the in our interval of integration. So let us focus on this function f of x, which is x square, sorry, four x my x square minus one. Let's try to see the wavy curve for this. Okay, wavy curve is very simple. All these terms are, you know, sweet and simple factors. So zero minus one plus one. So this is plus minus plus minus. Okay. My area of interest is only between minus half to half. I'm not interested beyond it. Okay. So I'm only interested in this zone minus half to half. Okay. I don't want to go beyond though that interval. So see, so minus half to zero minus half to zero, the function is going to behave as minus half to zero, the function is going to behave as four x by x square minus one because it is already a positive function and zero to half the function is going to behave as a negative four x by x square minus one. Yes or no. Now, if you want to do this, you can do this, but you can actually bypass this by using the fact of your even odd property behavior. So if you see as a function, as a complete function, this is going to be an even function. Why? Because if you put your x as a minus x, let's say I call this as my g of x function, then if you put x as a minus x, it actually doesn't change the function. Isn't it? So even if you have a negative on the top, the mod takes care of it, isn't it? So you could either do it like this or alternately, you could write this as zero to half mod of four x by x square minus one into two by using our even odd property. And now zero to half, zero to half, you know the function is going to be a negative one. So it is going to be zero to half negative of this guy, which is minus four x by x square minus one. This is easier to integrate than integrating these two. Okay, both of them will give the same answer, but why to go for a longer route when we have a shorter option available. So you are integrating zero to half twice of four x by one minus x square. Now, I think everybody can integrate this. This is a simple game for us. So take one minus x square as a t, correct? So negative two x dx is a dt. Okay, so I can write this as correct me if I'm wrong. I can write it as four negative dt by t, correct? And when your x is zero, this guy is going to be a one and when x is half, this guy is going to be three by four, correct? So you can do one thing. You can switch the positions of your upper and lower limit and absorb the negative sign. So it is four times lnt from three by four to one. Now, when you put a one, you get a zero. When you put at this, you get ln of three by four. Okay, so the answer is minus four ln three by four, minus four ln three by four. And I do not think anybody got it right. However, Vavav Manigar and Siddhish were quite close to it, but messed up in the sign. Yeah, so answer is minus four, or you can say four ln four by three. That would have been another way to write it. Is this fine? Any questions? Any concerns? Anybody? So sorry. Okay, let's try. So in the first step, we'll take the LCM inside the square root. It doesn't matter. I mean, even if you do it, you will end up getting a perfect square only, right? Does it make a difference? Aditya, does it make a difference doing it by LCM method or only in the numerator? Are you? Send me your working. I'll go through it. Okay, let's take one more because I think people are not very confident. Any questions? Class? Definitely. Okay, we'll take this question. This is a question which basically comes in school exam also. And why I have chosen this question and why I had chosen this question because it had come in, I think NPSHSR exam. And many people made mistake here, school exam. And when people make mistake in school questions, I feel really, very, very bad. So that's why I had picked up this question. So please evaluate this integral minus pi to pi cos square x upon one plus a to the power x tx. A is positive. Okay, Aditya. Okay. Now see here, many people start looking out for that even odd nature. Okay, rather, you could basically focus on the parent property, which we had discussed. So this was the parent property. And the two children of this parent property was that, you know, when the function is even and the function is odd. So don't forget the parent property. Most of us when we are solving the problem, we only try to figure out whether it's even or odd. Okay, so use this property over it. Okay. So here I can write this as negative pi to pi cos square x. Okay. By the way, you could also use your King's property as well here. Okay, so King's property will also work fine. So I can write it as zero to pi cos square x one plus a to the power x dx. And again, zero to pi cos square x one plus a to the power minus x. Is it fine? So now what I'm going to do is I'm going to multiply in the second one with a to the power x in both numerator and denominator. So this leads to zero to pi cos square x one plus a to the power x dx. And again, zero to pi a to the power x cos square x again, one plus a to the power x. Correct. So let's, you know, club the two integrants because they are under the same limits of integration. So you can take cos square x one plus a to the power x common, you'll have one plus a to the power x and that will happily get cancelled with this fellow. So overall, you're integrating cos square x from zero to pi. Now, so many ways to do it, you can integrate it normally also, you can integrate it by the use of property also that is your call how you want to do it. Okay, so here, let us use some property. Can we first use half the limit property because cos pi minus x whole square is same as cos square x. So I like it like this. Okay. And now let's say call this as I one. Okay, so I one is integral of zero to pi by two cos square x use KP over here. So KP will give you I one as zero to pi by two sine square x dx. So two I one is going to be zero to pi by two of one. So that is going to be pi by two. So we actually need to I one, isn't it? So we actually need to I one only. So to I one, the answer is going to be pi by two one. Why you change your answer from pi by two to pi? Any special reason why you did that? It's pi by two only. Okay. Okay. Is it fine? Any questions? KP can also be used. Okay, you can also use KP to do it. Here A plus B will be zero actually. So you have to replace your X with a minus X. Okay. So automatically you'll get the same result from there as well. Is it fine? Any questions, any concerns? Next question. Next question is actually a property which I want you all to figure it out. But I'll give you this as a question. If f of t is an odd function. Okay. And you make a function phi x which uses the function f of t integration from A to X. Okay. A could be any real number over here. Okay. Then prove that then prove that phi X is an even function. Then prove that phi X is an even function. But integration is a different thing because you don't get the same function every time when you integrate it just because of a yes. Anybody who has done this can please write down done on the chat box. If you could do this please write it down on the chat box. Okay. Pradhyan is done. So admit cards for KVPY is released. So what are the timings for the exam? Exactly. You are in the morning slot or in the afternoon slot? Okay. Afternoon slot. Morning slot may be like KVPY SA people will be there. Check it out if it has come I think. Seventh of November. Okay. So let's figure it out. So the only way we can figure out the parity for phi of X is let's try to see what does phi of minus X gives us. If phi of minus X gives us phi of X back then it is an even function. Okay. So let's see what happens. So if you change X with the minus X all you need to do is just change the upper limit sign. Right. Now let's try to do some manipulation with it. The first thing that I will be doing here is I'll be writing this function or writing this integral as my A to minus A. Okay. And then minus A to minus X. Okay. Now since my F of t is an odd function this whole thing will go for a zero. This whole thing will go for a zero since F of t is an odd function. Okay. So I'll be left with minus A to minus X F of minus t. Sorry, F of t dt. Okay. So here what I can do I can make a replacement of t with a minus you can say some other variable let's say minus K so that the negativities of the upper and the lower limit are absorbed and it starts assembling my phi X. So I'll change my t with a let's say minus K. So your dt will become a minus dk. So your entire expression will now become this entire expression will now become F of minus K minus dk and limit of integration will now become A to X. Why? Because when t is minus A K will be A and when t is minus X K will be X. Correct. Now being an odd function F of minus K will give you minus F of K. So minus minus will become plus and now there is nothing in the name I can again change it back to X. Right. Thereby giving you Phi of X back. So we started with Phi of minus X and doing all these manipulations I ended up getting this. So basically since Phi of minus X is equal to Phi of X it implies that it implies that your function your implies that your Phi of X function is even. So this is even function. Remember this many a times it helps you to solve few tricky questions also. I've seen some questions which are based on the use of this property. Even though I gave this as a question, please treat this as a property itself. Now let's see the other way around what happens if your F of t is even function then what happens? Anything that you would like to copy here please do so. Sure Shraddha. Let me know once you're done. Done. Okay. Thank you. The next property in this I will normally I'll just call it as another question. This question is slightly tricky. If F of t is an even function. Okay. Then integral from A to X F of t dt will be odd if either A is 0 or integral from 0 to A F of t dt is 0. So if you want your this function to be an odd function then it will only happen either when A is 0 or the integral of the function from 0 to A is 0. In fact, if A is 0 this will automatically be true. Okay. That is to say that for a non-zero A if this area is not 0 it will not be an odd function. Are you getting my point? So for a non-zero A if this area is not 0 then this will not be an odd function. Is the property understood? See the property says if you want this to be odd either A is 0 then it will be odd. Okay. Or if A is non-zero then the area under the function F of t from 0 to A should be 0 for it to be odd. Let's try to prove it. In fact, I will assist you in proving it because we don't have much time. So let's try to prove it. By the way, let me name it as a function phi of X. Okay. So now you're trying to claim that I want to make this function an odd function. So if you want to have this as an odd function then this criteria must be satisfied as per our rules of odd function. Phi of minus X should be negative of phi X. Okay. Now what is phi of minus X? Phi of minus X will be integral from A to minus X F of t dt. Right. Now what can we do here? Any future step, any further step anybody? Again, same thing A to minus A and minus A to minus X. Now F being an even function, do you all agree that this part I can write it as negative 2 times 0 to A F of t dt. First of all, I swapped the position of the upper and the lower limit that created a negative sign over it and then I twice it up from 0 to A because F of t is known to be an even function. And in this part, you just substitute, I'll just do it separately here. In this part, you just substitute t as a negative k. So your dt will become negative dk. Your limit of integration will become A to X and let me copy this as it is. Is this fine? Any questions here so far? Now F of minus k is as good as F of k and there's nothing in the name. So I can change it back to X itself. So far so good. Okay. And you can see that this is already a negative phi of X. Okay. So if you want this function to be negative phi of X, it is very obvious that this integral, that means if you cancel this off from both the places, then this integral that is negative 2 from 0 to A F of t dt must be 0. That is to say that 0 to A F of t dt must be 0. Okay. So if A is 0, no problem at all because this will anyways be 0. So if A is 0, no problem at all, it will behave as an odd function. But if A is non-zero and you still want it to be an odd function, then this area which is between F of t from 0 to A, that area should be 0. Okay. So if A is 0, then anyways it is 0. Then it is true. Then it is definitely true. Okay. Otherwise it should be, you know, if A is a non-zero value, then this area must definitely be true. So it basically proves this particular property or result you can call it. Is it fine? Any questions? No, no, no, no. See, if F of X is odd, then phi of X will be even. That's what we proved in the first problem. But if F of X is even, then A to X F of t dt need not be odd. It will be odd only when either A is 0 or this area under the function from 0 to A is 0. Got the point. Okay. So only when either A is 0 or this property is satisfied, then we can say that this function is odd. Else we cannot say it is an odd function. Yes. If A is 0, then phi will definitely be odd. If F of t is, if the function F is an even function, we'll prove it separately. Let's prove that separately. So if A is 0, then phi of X will definitely be odd. Ruchita, are you done with the copying part or is it? Let me know once. Okay. Thank you, Ruchita. Now see, if F of t is even, okay, then phi of X, which is from 0 to X, F of t dt, this will definitely be odd. Let's check. So let's call phi of X. Okay, phi of X I've already written. Let me not waste time. 0 to minus X, F of t dt. Okay. Here you just change your t with a minus k. Okay. So dt is minus dk. So this will become 0 to X, F of minus k minus dk. Since F of t is an even function, it is as good as 0 to X, F of k dk. Okay. Yes or no? Okay. dk, not dx t. Now there's nothing in the name, so you can call it as 0 to X, F of t dt back. So this is nothing but negative phi of X. So you started with phi of negative X and you got negative phi of X. Right? So it has to be odd. So if A is non-zero, then for it to be odd, this extra condition must be fulfilled. That is what I'm trying to say. This extra condition must be fulfilled if A is non-zero. Let's take a small question. I mean, not a very difficult one. If your function F of X is integral of ln 1 minus t by 1 plus t dt from 0 to X. Okay. Then F of X is option A, even function, odd function, neither even nor odd. Data is insufficient. One second, Shraddha. I'll just show you the last part. But let people copy this question. Then I'll go to the last part. So we have a lot of properties left. We have got scaling property. We have got periodic function properties. We have got Leveny's rule. We have got inequality property. Then we have definite integral, a limit of a sum as definite integral. So I would need at least one class for sure. So Monday we'll have to check, because if Monday is an issue, we'll have to do something about it. Oh, sorry. Okay. So I'm going to the previous slide for Shraddha. She wants to copy something. Shraddha, you want to copy this part? The one which I wrote on the right side. Let me know once you're done. Do we answer on the chat? Yes. No problem, Shraddha. Or should I put the poll? Okay. I can put the poll. No issues. Yeah. Please give me a response on the poll as well. Nice. Nice. Nice. So I can, I'm getting one of the options as the most chosen one. Somebody has also put data is insufficient. Before we question, before we should question the data of the question, we should question our own data. Is my learning data proper? Okay. So see here, there is a function I'll call it as g of t. Okay. I cannot use f of t anymore because f of t is already, my phi of x is now playing the role of f of x. Sorry, our f of x is playing the role of phi of x. So f of t I'm calling as g of t now. So is g of t an odd function? Yes. Because if you change your t with a minus t, it just becomes negative of ln 1 minus t by 1 plus t. So this is an odd function. Okay. So this gives you negative of g of t. So it's an odd function. So odd function integral from a to x, a could be zero also. No problem. The problem comes when your g of t was even. Then for the odd part, your a should be either zero or the integral of zero to a g of t should be zero. But for odd function, it doesn't matter whether this is a or zero, any real number. That will always be giving you an even function. Okay. So please remember, this function in that case will always be even. So option number a is correct. Okay. Is this right? Any questions, any concerns? So I will not start a new property. We just have 12, 13 minutes more. We'll take a few more questions. Okay. The question says, if f of x and g of x are two continuous functions being even an odd respectively, so f is even g is odd. a is a non-zero number. b is a positive real number not equal to one. So this integral is independent of f, independent of g, independent of both f and g, none of these. So I'll put the poll on for you all. In case you have solved it, please answer on the poll. Oh my God, it is weighing heavily on my side. Same in Kormangala house. Kormangala, are you in HSR? Oh, very close to my house since yesterday night. Sure. Sure. Let's meet up in HSR also. What year all the shareavocation plan will be washed off because of this rain? Okay. So did that one year ago? In fact, that all the plans are washed since your class 10th days, right? 10th, our pains won't allow us to do anything, board year, board year. And then 11th, 12th preparation year, 12th again, this only despite we get is in our first year of engineering. No, first year you'll definitely get a respite, okay? Because if at all colleges start and you're in the campus, there's not much of a pressure in the first year. Unless and until you want to get a department change. See, what happens in most of the ITs, you have an option to change your department. If you get exceptionally good score in the first year, okay, then I think 1% of the, the entire strength is allowed to change their branch. 1% or 5% I'm not very sure about the numbers. No, no, no, this goes like this. See, if you are BTEC, okay, you have an option to go to an MTEC, okay? Let's say I'm BTEC Electrical and I want to switch my branch to MTEC, let's say computer science. I can go, I can choose for that. Many people do it because they want dual degree, they realize later on that I want dual degree. But if you are an MSC, then you'll only get MTEC. You cannot be a BTEC from an MSC. Let's say you got MSC Integrated Physics in IIT Kharagpur. So it's a five-year course, okay? So you can only change to a five-year course. That is, you can be an MTEC in some other branch. Now, there's a commercial angle to it because from five years, you cannot reduce it to four years because one year of fees, who will pay? So there's a loss of business for IIT, right? They also want money to run. That's why you can increase the number of years you want to stay in IIT because that's good for them, but you cannot decrease it. Now, it doesn't happen to any subject. It depends upon your CGPA. For example, let us say your CGPA, that is cumulative grade point average. There's something called SGPA, semester grade point average and CGPA. CGPA is one year combined, average of the two SGPAs. If that is, let's say, you know, top 10 in the institute, then you can go from any branch to any branch. Even if you have gone into IIT, let's say geophysics, you can go directly to computer science, okay? But let's say it is not very great. Let's say you are like high eight point something, then maybe you can go from a geophysics to a civil engineering branch, maybe like that. But don't rely on this because it is very rare because if you are working hard, there are students who are also working hard there and they're all IIT and don't forget it. So you are competing against the people who are already good in the competition field. At a different branch, no shortage. Why will you, why will you like to stay in the same branch? Let's say you've got civil engineering and you're not happy. Sir, I don't want to study civil engineering. Okay. So do well in first year, get like 9.5 or something and then choose your branch. You say, okay, I want to go to electronics, computer science, mechanical, they will definitely not take you there. Same college. Yes, same college. You can't change the college. You can't change the college. It happened in BITS also. We call it as DIPC, department change. That's a lingo which we use, DIPC. We say, Bandhani DIPC, DIPC means he has changed his department. Yes, guys, most of you have voted for option B, should we discuss it now? Okay. So yeah, B and a bit of A and a C was, so let's discuss it out. So this, you can apply King's property first of all. So if you apply King's property or if you use your, you can say the parent property, which we had discussed a little ago, minus A to A, F of X by B to the power G of X plus 1. You can write it as 0 to A, F of X, B of G of X plus 1. And 0 to A, F of negative X, B to the power G of negative X plus 1. And you know your function is, F is even. Okay. And G is odd. So because of this, because of this, you end up getting, this is unaffected. This will become 0 to A, F of X. And this is negative G of X. So you can do one thing. You can multiply with B to the power G of X and write it as 1 plus B to the power G of X. Okay. On adding, you will automatically see that B to the power G of X will get cancelled off. Integration from 0 to A, DX. So it is independent of G. So option number B. Okay. So this guy becomes independent of G. Any questions? Any questions, any concerns with respect to this? Last question for the day today, then we'll meet again tomorrow. F of X is a determinant. Could you scroll down? Okay. F of X is a determinant, which is made up of cos X, X square 1 as the first column, e to the power X square, ckx2 as the second column, 2x cos square X by 2, sin X plus X cube and X plus tan X as the third column. Okay. Question is integral from minus pi by 2 to pi by 2, X square plus 1 times F of X plus F double dash X is equal to which of the following? 1 minus 1 to no tau. Easy question. You should be able to do it quickly with respect to X, of course. Sorry about that. All is on. But even if you do that, you will not be wrong actually. Done. Easy. I'll give you one more minute. It's not a difficult question. It's just an odd even game. Okay. Four people, four different options. Good. Okay. How was the math exam? NPS, R and R? I forgot to ask. How was your paper? Easy. Went well. Okay. You got your scores also? I mean, your teachers must have got the scores. Have you shared your scores with you? No. Okay. Every question was from some textbook called the MCQs. Which one? Elder source. Center books. Okay. Chalo, we'll do this question. I don't know. It would have definitely come from Oswal. See, it's a new pattern. So we don't have much options, but to follow whatever these big publishers provide. Okay. Let's discuss it. So end of poll. Anybody wants to vote? Anybody wants to vote? Nobody wants to vote. Okay. So most of you gave option A as the right answer, but of source that is not the right answer. Answer is actually, we'll discuss. See, is it clear that this is an odd function? Is that part clear to everybody? Because if you change x with a minus x, nothing will happen to column one. Nothing will happen to column two, but column three elements will all become negative, negative, negative. That means f of minus x would have been negative of f of x. So it's an odd function. Now any odd function, if you differentiate, it'll give you even. Again, if you differentiate, it'll give you odd. Now see, this is odd. Double derivative is also odd. So overall, this is odd and this is even. So even into odd is again an odd. So ultimately you are doing what? You are integrating some odd function. Odd function integral from minus a to a is zero. So option D is correct. None of these. Okay. Anyways, so as I told you, there are a lot of properties to be left. I at least need five hours of class. So let's try to finish this chapter during the Deshera only so that after Deshera, I can start differential equations and I have to at least end 10 days before your board exams also so that you can have enough time to prepare.