 Okay, so let's start. Definite integrals. Definite integrals. A very interesting chapter almost 70 to 80% of the question which is related to integration asked in cognitive exams are based on definite integrals because of course the options are plain and simple So as the name itself suggests definite integrals that means the answer is the definite quantity Okay, now first of all we need to understand the first formula that we are going to talk about is basically the fundamental principle or fundamental theorem of integral calculus which says that if you have a function whose derivative is f of x Okay, so as to say that integral of f of x is capital f of x plus c. Okay, remember capital F is called the primitive or antiderivative whereas small x is called the derivative So the fundamental theorem of integral calculus says this result is going to be f of b minus f of a And since this is a very definite result, it's a numeric value. That's why such kind of integrals are called definite integrals Okay, quickly talking about the geometric interpretation when I say I'm integrating f of x from a to b Okay, and let's say this is my graph of the function and I'm integrating from a to b So if this is your function f of x and I'm integrating it from a to b It actually gives you the algebraic area between the function and the x-axis What do I mean by algebraic area? That means let's say this area which is above the x-axis This is called a positive area Now the question will arise in your mind. Why do we call this area as positive area? Very simple. If you look at the expression you are actually integrating this So at a distance x if you choose a very thin rectangular strip of dx So if you choose at a distance of x a very thin rectangular strip of the thickness dx Of course the height of this will be f of x itself Then f of x basically represents the height d of x represents the width So this gives you the differential area or the infinitesimally small area of this strip When you integrate this basically you are adding all these small areas Now remember since height and width both are positive here Now height is positive I can see because you are measuring in the positive direction up How do you know width is positive? So we are going from left to right? Yes if you are going from left to right means your x is increasing If your x is increasing dx will be positive dx is positive means x is increasing So in this way the entire area that comes out in this part let's say I call this part as c over here From a to c the area is a positive area Whereas when you find this area Here f of x itself Sorry this is from c But if you find the area from c to b your f of x is a negative quantity Because the height here is negative So this becomes a negative quantity while this is still positive Thereby giving you a negative answer over here So this entire thing is positive whereas this entire thing is negative So this area comes out to be negative So integral of the function from a to b gives you the algebraic sum of the area between the function and the x axis Many people say it gives you the area under that curve That is not completely correct actually It doesn't give you the area but however it helps you to find the area In order to find the area you have to cleverly choose your points You have to cleverly choose your signs of that particular part of the curve When you look at this formula it actually reminds you of the fact that You first have to find out the primitive through antiderivative or use of indefinite integrals And then put the limits of integration like this Remember all the methods that you have learned for finding indefinite integrals will work fine over here Your method of substitution etc everything will work Integration by parts that is also going to be working in this case So what I am going to do is before I introduce you to Further parts of definite integrals which is basically a property driven topic This is a property driven topic I am going to ask you few questions just on the basis of how much you understand When you say integration of f of x from a to b is f of b minus f of a So I will begin with a question Let me start with this question This is a very strange question It says evaluate this directly as well as by substitution of x is equal to 1 by t And examine as to why the results do not tally Okay if you look at this question Would you all like to try it first? Yes sir Sir can we answer? Yes sir Sir because when we substitute x equal to 1 by t the limits also change From minus 1 by 2 and 1 by 2 and when we solve the integral we get For t we get the integral as half tan inverse 2t and for x we get 2 times tan inverse of x by 2 So when you solve and since the limits are different for both We will get a different answer Limit if it is different it doesn't mean the answer would be different Sir when we substitute x is 1 by t and x goes from minus 2 to 2 But when we do that t will also become 0 at one point which can't happen Exactly the answer is there is a point 0 somewhere in between So when you say t is equal to 1 by x Okay and you can see that your limits of integration go from minus 2 to 2 That means there is a point where x becomes 0 Okay there your t will become undefined That means the function will suffer discontinuity At a point between minus 2 to 2 if you follow this substitution In such a case we cannot use this substitution Getting the point So a very important learning from here is that whenever we are using a substitution Make sure that nowhere between that interval of x The newly substituted variable becomes discontinuous Getting the point So if you solve it directly Minus 2 to 2 dx by x square plus 2 square Your answer is going to be half tan inverse x by 2 Okay if you put your limits of integration This is how we put the limits of integration When you put a 2 it becomes half Tan inverse 1 is pi by 4 Minus of minus pi by 4 Okay so I think this gives you pi by 4 itself as the answer Right But the moment you put x is equal to 1 by t dx is equal to minus 1 by t square dt Okay what will happen? Limit of integration of course will also change minus half to half Okay this will become minus 1 by t square dt 4 plus 1 by t square Okay which is nothing but minus dt by t square plus 4 Minus half to half Okay so this gives you the answer as half tan inverse of t by 2 Okay yes or no? So it's 4t square plus 1 Oh sorry 4t square This is 2t So it's 2 tan inverse 2t And it's a minus also What happened? Nothing sir Okay so when you put a half it becomes a minus pi by 4 Correct? And again when you put a minus half it again becomes a minus of half Minus of minus pi by 4 So it becomes minus pi by 4 That means your answer becomes negative which is impossible why? It is impossible because We know that this function is a positive function Correct? 1 by 4 plus x square is always a positive function That means it's always above the x axis So how can the area under that become negative? Okay now this discrepancy has arisen because of a faulty substitution over here Because my function becomes discontinuous at some point between minus 2 and 2 which is at zero point So if you have any questions please be careful when you are using substitutions to solve the questions in definite integrals By the way just to clear that myth If substitution changes that doesn't mean your answer is going to definitely change In fact substitution will change the limits of integration will change every time you substitute x with a new variable Okay so in definite integral we have to change the limits of integration according to the new substitution Okay let's take another question Hope you can read the question properly Find the value of integral of 0 to e to the power x square plus 2x minus 1 by 2 by x plus 1 Plus integral from 1 to e x ln x e to the power x square minus 2 by 2 Just 2 minutes to answer this Just typing your response so you can also speak it out Okay any idea anyone? One second one second Okay time up so what I will do here is I will not disturb the second integral Okay let the second integral be integral from 1 to e x log x e to the power x square minus 2 by 2 Okay While in the first integral what I will do is I will substitute x plus 1 st correct So dx will be equal to dt Okay limit of integration will change from 1 to e hope everybody knows how to change the limit of integration It's very simple when you put x as 0 t becomes 1 so lower limit will become 1 When you substitute x as e minus 1 t will become e so upper limit will become e So e to the power this term here is nothing but x plus 1 the whole square This is x plus 1 the whole square minus 2 correct So it's t square minus 2 by t okay dt Is that fine Now So divided by 2 Yeah now if you see the limits of integration for both of them has become the same Okay and let me tell you in a definite integral it doesn't matter the name of the variable Okay so you can call again t as x Now this is very surprising that you can I'm again calling t as x Because there's actually nothing in the name Ultimately have to integrate that function and put the limits of integration That's why the name of the variable at any instant can be changed to anything you want Okay so instead of this t here I will put everywhere x doesn't matter So what I'm going to do is I'm going to just change them to x again So wherever there was a t I put x Okay So but this shouldn't be fair right I mean Because in the beginning of substituting something as x Function of x is t But this is giving out a value See Shrijan what is happening is treat this as a new problem itself don't have any history to it Okay once you have made the substitution you have changed the limits Treat it as the beginning of the problem See if you want you can keep on keeping different different names But it's just going to make your problem more complicated I'm just using the same name just in order to bring it back to the same old function Okay ultimately you are going to put the limits of integration So the variable name should not matter at all Getting my point So when you're talking about conversion of t back to x Think as if there was no previous relationship between t and x Start with a new problem per se Okay So this becomes the same function Okay so you can combine these two functions and write it as single function So e to the power x square minus 2 by 2 1 by x plus x log x Okay integration from 1 to e Now how do you evaluate this? How do you evaluate this? If you look at this very carefully You would realize that this is the exact differential of log x into e to the power x square minus 2 by 2 Okay if you differentiate this You would realize you will end up getting the integrand over here Try it out If you keep this as intact and differentiate log x you get 1 by x If you keep log x and this Derivate differentiate this you will get log x e to the power x square minus 2 by 2 into 2x by 2 which is actually x Okay so this is an exact differential over here So the answer for this will be just log x into e to the power x square minus 2 by 2 You just have to put the limits of integration now So when you put an e you get 1 e to the power e square minus 2 by 2 Okay and when you put a 1 it becomes a 0 This is as we were saying root e to the power of e square minus 2 I think such an option is present In option number D Is this clear? So one thing that we learned over here something which was surprising That there is nothing in the name of the variable being used So if you say integration of f of x from a to b Or integration of f of t dt from a to b Integration of f of z dz from a to b They are all the same that is nothing but Capital f of b minus capital f of a So in the primitive of the function you have to substitute upper limit And substitute lower limit and take the difference Let's take another question Find the value of integral of Big pi x plus r That is the product of x plus r r from 1 to n times Summation of 1 by x plus k k from 1 to n And this product you are integrating from 0 to 1 2 minutes to solve this again In fact 2 minutes is also a big time for this Should solve it within 1 minute I think Omkar needs to mute his mic Done? Yes sir Okay D Fine Okay So let's discuss this See here When you look at this symbol Big pi of x plus r Times summation of 1 by x plus k Okay If you expand it it will be nothing but x plus 1, x plus 2, x plus 3 Duh-duh-duh-duh-duh-duh till you reach x plus n Times 1 by x plus 1, 1 by x plus 2 Till you reach 1 by x plus n Okay I don't know how many of you are able to Recognize that this is actually the derivative of x plus 1, x plus 2 All the way till x plus n How many of you are able to identify this? Remember when you are differentiating this That means you are using product rule You differentiate one at a time Correct? So let's say you differentiate the first term And keep others as such That means you are going to get the same Product with the first term x plus 1 will automatically get cancelled off And hence x plus 1 will automatically disappear Right? So when you differentiate it You just first differentiate the first term Keep others as such, correct? Okay, then you keep the first Differentiate the second which is again 1 And then keep the others as such Okay, and keep on doing so And the last term would be x plus 1, x plus 2 All the way till x plus n minus 1, correct? Now if you take x plus 1, x plus 2, etc Common from all of them The first term is just going to be 1 by x plus 1 Second term is just going to be 1 by x plus 2, etc Till 1 by x plus n, okay? So this entire expression Is just the integral of An exact differential, correct? Okay, so this dx Integration you are doing from 0 to 1 Okay, so basically you are integrating d of something So answer is that something itself So this is going to be your answer And you just have to put The upper and the lower limit in that So when you put a 1 You get 2 plus 3 all the way till n plus 1 Which is n plus 1 factorial And when you put a 0 You get a n factorial So this, okay? If you take n factorial as common It will be n plus 1 minus 1 That's nothing but n into n factorial And absolutely correct, Shijian? Answer is d, clear? Everyone?