 Can you move on to the next question then, next question is find the area bounded by find the area bounded by y is equal to mod of ln of mod x and y equal to 0, find the area bounded by mod of ln of mod x and y equal to 0, needless to tell you y equal to 0 is your x axis actually, so remember step number one plot the graph, so the biggest prerequisite of this chapter is graphs, we cannot proceed without knowing the graph, so first let us ok you just take one minute and plot this graph and then I will tell you how to deal with this graph, are you done with the graph, ok first of all let us talk about y is equal to ln x, forget about any mod, how does this graph look like, this graph actually looks like this and if you are modding x what happens, let me go step by step, if you are modding x what happens, you just have to make the same graph on the other side as well like this, ok, so this is the graph of, this entire thing is the graph of ln of mod x, now when you mod this what happens, let us go to the step number 4, let me choose the blue color, what do you do with the graph, the part which is below the x axis will now go above the x axis that means this part will now go like this, yes or no, ok, similarly this part will also go above like this, yes or no and these two parts will get erased off, these two parts will get erased off, so this is your final graph of mod of ln of mod x and let me show this to you on the algebra as well, so y is equal to mod of, do you see this, now these two arms will come very close to each other but they will never actually touch each other, are you getting this point, yes, now how would I get this area, how would I get the area which is between these two that means how do I get this entire area, that is the question. As I said sir before we have to consider both vertical and horizontal strips, so we can consider horizontal strips in the blue area, can we do that, yes of course this is the final graph so there is no blue and green now, so we have to find this area right now, ok, so what I am trying to say here is that if you want to find this area, can we first focus on finding the area of this part, yeah, can I say doubling up this area would give me the answer, yeah because it is symmetrical, right, so area would be nothing but twice the area of this part, let me call it as A1, ok, now first focus on getting the equation of this blue line, what is the equation of this blue line because of so many mods we tend to lose track that which curve represents what or what is the equation of this arm of the curve, so can somebody tell me what is this blue line which do you see the cursor moving on the screen, my cursor or let me just show you this part, this part, ok, what is the equation of that part I want to know, you realize that it is y equal to ln x mirrored image about x axis, this line is a part of y is equal to ln x it is a part of y is equal to ln x mirrored about the x axis, yes or no, yes, so if a curve is mirrored image about x axis what is going to be its equation minus y equal to ln x, correct, yeah, yeah, and my dear students if I am going to take horizontal strips for it, I would need this as x equal to e to the power minus y, yes or no, yes, so can I say A1 area I am calling this area as the A1 area and A is my final answer, so how would I get A1 area can I say A1 area would be integral of minus e to the power minus y dy from 0 to infinity, yes or no, yeah, and this is nothing but gamma 1 which is known to us as 1, so for the students who are not aware about gamma function just a quick highlight on what is the gamma function gamma n is basically integral from 0 to infinity e to the power minus x x to the power n minus 1 dx, right, you can also write it by changing the variable as e to the power minus y, y to the power n minus 1 dy does not make a difference and if you put your n as 1 this term is going to disappear because y to the power 1 minus 1 is y to the power 0 which is going to be 1, anyhow if you do not want to get into improper integrals you can still solve this, so let me just clear this up, you can still solve this very easily, you just have to integrate it from 0 to b e to the power minus y dy and take the limit on b tending to infinity which is just going to be minus e to the power minus y from b to infinity, so it is minus e to the power minus b minus of minus 1 and as b tends to infinity this will become 0, so your answer will be 1, so a 1 is going to be 1, any doubt about this, so a is going to be 2 a 1 which is nothing but 2 into 1, so your answer will be 2 square units, so when we are writing the exam and we get such a type of an integral, right, it is safer to remember the improper integrals. Yeah, you can just verify your answer by using the improper integrals, okay, after all these improper integrals have been derived from our basic understanding of definite integrals, right? Yeah. Great, can we take a question which is very similar to this, I am clearing this screen now, okay. Next question is, find the area enclosed between, find the area enclosed between y is equal to ln x plus e and x is equal to ln 1 by y and the x axis and the x axis, solve this, right, so now, don't look at this screen, I will be just drawing my curves, yeah, now you can just check, did you get this figure, yeah, so and we need this area, please note that this area is open from this side, right, so even if it is open the sum is going to be finite because e to the power minus x is basically a convergent term, so it is going to just die down on the x axis, however we will never touch it, we say that x axis is going to be an asymptote to this function. Now, how many different types of strips will be required here, very good, two strips, one is be like this and the other one would be like this, yes or no, right, now for this strip, let me draw the other one with a different color so that we can easily identify it and this be red, yeah, so for the blue strip what expression should I write, anyone, which is the upper curve for this strip, first tell me the equation of that curve, tell me the y for that curve, why I am taking y for that curve because I am using vertical strips, then y for the lower part of the strip, tell me those, these two things, x plus e minus 0 and I am integrating this from which value of x to which value of x, 1 minus e to 0, correct, so this definition will last only from this point till this point, correct, plus, now for this red strip, tell me the y of the upper part of the strip, you will say e to the power minus x, lower part of the strip again 0, see I am just writing the 0 for clarity, in the normal case you may not choose to write that also, okay, and integrate this from 0 to which point, does it stop anywhere? Infinity, no, yeah, it goes all the way to infinity, correct, so now we have to evaluate this integral, now while evaluating this integral you may choose to substitute x plus e s t, your dx will still remain your dt, so this integral, this integral the double ticked 1 will convert as ln t dt, okay, what will be the lower limit, when you put x as 1 minus c lower limit will be 1 and when you put x as 0 upper limit will be e and this answer is very well known to us, it is gamma 1, this answer is 1, what is ln t integration, t ln t minus 1, upper limit e lower limit 1, so when you put e you get a 0 and when you put a 1 you get minus of minus 1, so your answer will finally be 1 plus 1 which is nothing but 2 square units again, is that clear?