 Hi, I'm Zor. Welcome to New Zor education. I would like to illustrate certain concepts which we were discussing before in the previous lecture actually. We were discussing improper definite integrals and as illustration, I just have a few examples. Maybe I will have more, but at least these four examples for today which basically explain how these improper integrals are supposed to be handled with. Now this lecture is part of the course of advanced mathematics presented on unizor.com for students of high school teenagers in general. I do recommend you to watch this lecture from the website because it has very detailed notes plus many topics on this website have exams, so you can just check yourself how you how you study, how successful you're studying. And the site is completely free, so Alright, so few examples. Integral from 0 to 1, 1 over square root of x dx. Okay, first of all what's improper about this particular integral? Well, let's think about it. As x approaching 0 1 over square root of x obviously goes to infinity. So my function is not contiguous basically on the segment from 0 to 1. As a matter of fact, the graph is at 1, it's equal to 1 and as x is diminishing, since it's denominator, my function is increasing and it's increasing to infinity basically. So the graph is something like this. It's asymptotically goes to y-axis. Now we need to know this area, which seems to be definitely improper, so to speak, right? I'm not I'm not saying it's infinite because maybe it's not infinite since this thing is becoming narrower and narrower it's obviously infinite height, but infinitesimal widths, right? So in this particular case, I mean I know that this is actually the finite area, but we have to prove it somehow. So how do we calculate the value of this particular improper definite integral? Well, first of all, let me start from the definition. What is this? Considering 0 is a point of improperness. Well, this integral is basically a limit integral from a to 1, 1 over square root of x dx, as a goes to 0 if this limit exists, right? So first of all, let's just evaluate this particular integral. How? Well, obviously we have to take the indefinite integral and the derivative of this function and then use the muton label in its formula, right? Now this is basically x to the power of minus one-half, right? Minus because it's denominator and one-half because this is square root. Now indefinite integral of this is x to the power of minus one-half plus one and divided by minus one-half plus one, right? Obviously because derivative of this, when we are taking derivative from a power function, we diminish by one, it's exponent and then it's supposed to be multiplied by this particular power. So that's why I divided to get 0, to get 1, sorry, coefficient of 1. All right, so this is what I have. Well, you can actually say this is minus no, this is plus one-half, right? But we divide to one-half. So it's 2. All right, so it's 2 times x to which power? One-half. Well, plus c obviously, but we don't really care about plus c right now since we will have both limits of integration, they cancel out c. All right, so then we substitute 2 square root of x from 0 to 1. So we are using this formula Newton-Ladies formula. So for 1, we get 2 and for 0 minus we get 0, which is 2. Oh, sorry, to a not to 2. I have already went to a limit minus 2 square root of a, right? Now, as you see, whenever a goes to 0, this thing obviously converges to 2 because this is infinitesimal. So the answer is to... That's it. So what did we do? What's very important? Well, number one, we have expressed this as a proper definite integral by cutting off the point where our function is behaves, right? And then we basically calculated this integral as normal definite integral using indefinite integral and the derivative and the formula Newton-Labnitz. And then we actually went to a limit when the point which we are using to cut off our function goes really to the one which we need, which is 0. So that's basically a typical approach to evaluate these improper integrals. And that's what I will do maybe with a little bit less words for the next examples. Okay, now next example is from 0 to infinity e to the power of minus x dx. It's minus x. Okay. Now, what's improper about this guy? Well, obviously this infinity is improper. Now e to the power of minus x goes like this. This is 1, this is 0, and this is e to the power minus x. So again, we have misbehaving function. Well, because it's an infinity. So again, obviously the traditional approach is not working. And again, we will use this traditional definition of improper integral. So what do we have to cut in this particular case? Well, we have to cut infinity, obviously, to say that this is from 0 to b e to the power of minus x dx. And we have to take the limit of this as b goes to plus infinity. That's what this is by definition, right? So there is no other way to define this particular integral. We cannot use those Riemann sums when I divide the whole area of integration into a certain number of intervals, because it's infinite. We can divide it. All right. Now, this is easy. What is this? Well, let me just preserve this limit and say, okay, I will say what is my indefinite integral from e to the power of minus x? Well, that's, well, e to the power of something, it's usually e to the power of something, right? So in this case, minus x. But if I will differentiate this, I will have e to the power of minus x times derivative of the inner function, which is minus x, and derivative is minus 1. So it will be minus. So to nullify this minus, I have to put this. So this is my indefinite integral, which I have to take in these particular limits. Now the upper limit is minus e to the power of minus b, and the lower limit is minus e to the power of zero. Okay. Well, this is e to the power of zero is one, minus, it's minus one, and minus it's plus one, right? So this is one. Now, how about this piece? Well, as b goes to, sorry, not zero, infinity. As b goes to infinity, minus b goes to obviously minus infinity, which is one over e to very large number. Let's call it capital B. I don't know. Capital B is minus lowercase b. Well, no. Actually, I'm wrong. It's lowercase b. b is a large number. It's plus infinity. Minus b is minus infinity. So this is what we have here. Right? And obviously, if b goes to plus infinity, my denominator goes to infinity, so the whole fraction goes to zero. So this thing is goes to zero as b goes to plus infinity, and what remains is one. So this limit is equal to one, and this is the value of my integral. So again, what was important here? Important to realize where is impropriateness. In this case, that's the infinite right limit of integration. So we have to cut it with b and then take the limit as it goes to plus infinity. Next problem. They're all easy problems, as you see. The most important here is number one, to be able to take indefinite integrals, and obviously my examples are relatively easy, where indefinite integral is kind of obvious. And the second is to be able to take the limit and again, in our cases, the limit was very simple and obvious. So these are just, you know, simple examples. Of the approach rather than attempt to give you some hard time figuring out integrals or whatever. Next. Integral from zero to infinity. Well, when I'm saying infinity, it means plus infinity, obviously. One over one plus x square dx. Well, this is the same type of improper behavior as in the previous example. We have this infinity, which means we have to say that this is, by definition, integral from zero to b, as b goes to infinity of this particular function. So what do we do now? We take an indefinite integral from this function, right? So what's indefinite integral of one over one plus x square? Well, I just happen to remember that this is arc tangent. And we have to take it from zero to b. Okay, remember what arc tangent is? That's an angle. Well, in radians, by the way. Tangent of which is equal to x. So which is equal to limit b goes to infinity. It's arc tangent b minus arc tangent zero. Okay, what is Well, this is a constant. It's independent, right? Of b. So we can write the value of this. So what is the value of arc tangent of zero? It's the angle tangent of which is equal to zero. Well, it's zero, right? Remember what tangent is? Here's my function of the tangent. Minus pi over two and pi over two. So if my tangent is equal to zero, that's an angle zero. So I can just completely wipe it out. That's zero. Okay? Now, how about arc tangent of b if b goes to infinity? Well, let's just draw the graph of arc tangent. We did it many times before. Basically, it's the function which is inverse to this one. So the graph is also inverse. Here, the x have from minus pi over two to pi over two. That's the main angles where arc tangent is defined. So in this case, it would be on the y-axis from pi over two to minus pi over two. And the whole thing will be inverse. So now it's very easy to see where exactly is the angle tangent of which is equal to infinity. Obviously, that's it goes to pi over two. If b goes to infinity, my value arc tangent of b goes to pi over two. Now, you remember the tangent is from from the definition, tangent is of any angle. That's a xy. It's y over x, right? So as my angle is increasing from zero to pi over two, my y-coordinate is increasing to one and my x-coordinate is decreasing to zero, right? And at the pi over two, my x is equal to zero, so tangent is not defined. However, it really goes as the value to infinity as we see as we approach pi over two. So tangent is not defined, but it's limit goes to, well, I shouldn't say limit goes to infinity. It's unlimitedly going to the positive infinity, right? So that's why this particular limit is equal to pi over two. As function goes to, as argument b goes to infinity, value of arc tangent b goes to pi over two. And we are talking about limit in this case. So it's a regular limit. So this is equal to pi over two. And my last example is is more of an exercise for definite integrals than improper definite integrals. Because as we see, we always reduce our problem of taking improper definite integral to basically two things. Number one is to take indefinite integral and number two go to limit, right? So if you know how to do indefinite integrals and you know how to do the limits, that's basically a sufficient condition to understand how definite improper integrals are supposed to be handled. So my function is from one to infinity one over x squared times e to the power of one over x dx. Well, that looks complicated. I mean one over x in the exponent, it's difficult. Plus we have this thing which is messing around. However, we should be really I don't know how should I say it. It goes through all the problems sufficient number of times to recognize that this is one over x and the derivative of one over x is very close to this one, right? Remember? Why? Because again x to the power of minus one, that's what it is. We increase the power we decrease the power, we'll be taking a derivative, we decrease the power by one. So it goes to derivative of this goes to x to the power of minus two and then minus one would be our multiplier. So it's minus one over x squared, right? Now we have x squared. So it's almost minus integral one to infinity e to the power one over x one over x derivative dx. Okay, that's my first transformation. Now, why did they do it? Well, obviously because this is one over x and this is one over x. Now, what is one over x derivative times dx? Well, this is differential of d over one over x, right? Because what is differential of the function? It's a derivative of the function times differential of the argument, right? We basically covered that pretty well before. And why did I do this? Well, now I can very easily do the substitution. So let's do a substitution. t is equal to one over x. What happens? Well, obviously I can say minus integral e to the power of t dt. Now, what's the limit? If x is goes from one to infinity, one over x goes to, if it's one, then it's one. If it's infinity, one over infinity is zero, right? Well, what did I actually kind of cheat it a little bit? Well, I shortened the whole explanation because in theory I should have said that this is equal not exactly to this but to a limit of Let me just do it slightly more. It's b and limit b it goes to infinity, right? Same thing is here because now I have regular integral and I can do whatever I want with it, right? Including the substitution. Now this is better, right? But now you see I should probably put, instead of zero, I should put one over b as b goes to infinity, right? And one over b as b goes to infinity is equal to zero again, obviously. Okay, now what is this integral? Well, it's a very familiar thing. Indefinite integral of e to the power of t is equal to e to the power of t, right? Plus constant c. So basically we have to evaluate this thing which is minus, minus is retained, zero to the power of t, which is indefinite integral from e to the power of t from one to zero. Now, it's kind of unusual to have integration from one to zero and then the minus here. What's probably better is to have the plus here and then from zero to one, right? Remember when we change the order of integration? Same thing here. We change the sign. That's better, right? Remember the property of integral. Integral from a to b is equal to minus integral from b to a, right? Okay, and what is this? Well, e to the power of t when t is equal to one is e minus e to the power of zero, which is one. So this is my answer, e minus one. So it looks a little scary, but you have to recognize two things. Number one, we can simplify it because of this substitution. One over x is a great substitution in this particular case because we recognize it here and here is derivative of it and that's why we converted this integral into this. And everything else is exactly the same, with limits and integration using the indefinite integral and formula Newton-Leibnitz formula. Okay, so these are four very simple examples. Sometimes you have to get a little bit creative when you recognize that something can be done this way. I mean sometimes, yes, integral needs a little bit of creativity. Differentiation is simpler because there are certain known elementary functions and everything else, every other function can actually be represented, usually can be represented as a combination of elementary functions, composition, multiplication, product, etc. So if we know the rules of differentiation and we know what's the derivative of every elementary function, well, including maybe not so elementary like our tangent, for instance. But in any case, all these functions which we are talking about, they are known as far as their derivatives is concerned. So any combination of these functions give us a more complex function. But anyway, how do you define any complex function as some kind of a formula which involves simpler functions? So it's always decipherable in this case and we can always differentiate it. Integration is much more difficult because in integration not every function which can be constructed from elementary function can be integrated in terms of elementary functions. That's what I would like actually to point out. I can have something a little bit more complicated here. For instance, I know 1 plus x, if I will put even that, the function, it's already much more complex and I'm not sure really how to take this integral. You see this particular property of having 1 over x square here and 1 over x there immediately prompts me to use some substitution like this. If it's not like this, well, it needs some creativity, maybe, and even if you are very creative, there is still no guarantee that integral can be expressed in terms of composition of regular functions, right? So that's why there are some maybe problems. In certain cases, when you're trying to integrate some functions, although I'm pretty sure that the functions which you will be given during exams, for instance, to integrate, they are usually the ones which you can really integrate using certain simple methods, like substitution or by parts or something like this. All right, so basically that's it for today. Thank you very much. I do recommend you to go to the website and read all the notes and what probably would be very important for you is just take all these examples, do yourself this integration and see if you come up with the same answers. Very useful. Okay, that's it. Thanks very much and good luck.