 Hi, I'm Zor. Welcome to Inizor Education. Subject of today's lecture is integration. This is my first lecture on this subject and on the integration. So I will probably divide this into indefinite integrals and definite integrals. So this is the first one about indefinite integrals. And obviously we have to start from the definition. So this is basically about what is indefinite integral. Now this lecture is part of the course of advanced mathematics presented on Unizor.com for teenagers and high school students. I do recommend you to watch this lecture from the Unizor.com website because every lecture including this one obviously has detailed notes. Plus if you are registered, then you can take exams for instance and basically participate in the educational process under supervision for instance of somebody like teacher or parent or your own supervision. Okay, and the site is free by the way. Alright, so let's talk about indefinite integral. Well, first of all, let's recall what we did before. Before we were talking about differentiation. So what is differentiation? I mean we took some kind of a function and applied the procedure of differentiation to find its derivative. So assuming that we are dealing with smooth functions, it's basically an operation. This is an operation on all the different functions which, well, can be differentiated. Let's call it smooth functions. And the result is basically for every function from this set we have found the corresponding derivative of this function. So basically it's an operation, right? So for any function, let's say you have a function y equals to let's say logarithm x, okay? We applied the procedure of differentiation and come out with another function which is 1 over x. It's a different function, but this function is derivative of this function. So that's why differentiation is an operation on the set of functions. From one function we can get another function. Okay, so it's an unary operation by the way because there is only one particular argument of the differentiation which is the function which we are trying to take the derivative of. And as with any operation we are interested in properties. And we have already examined the properties of derivatives, something like derivative of a sum is equal to sum of derivatives, etc. Now what is always very important with operations is their inversibility, if you wish. So if we can get from a to b, can we get from b to a? For instance, if you are applying a procedure and operation of adding 1 to a number. So from 5 we get 6, from 27 we get 28. Is there an inverse operation? Yes, it's a subtraction of 1. From 6 we can get to 5, from 28 we can get to 27. So there is always an operation and there is an inverse operation. I mean we would like it to have this kind of property of inversibility. Does it always exist this inversibility? Well, not necessarily even among numbers. Let's just recall an operation of raising a number to the power of 2. So 2 to the power of 2 is equal to 4. 27 to the power of 2 is equal to something else, but it's some kind of a concrete number. But now let's look at another thing. Minus 2 also square gives you 4. So what kind of inverse operation to raising to the power of 2? Squaring, let's call it. If we are talking about inverse operation, then from 4 we can get either 2 or minus 2, right? So that's not exactly an operation. If we are getting from 2, we are getting to 4. That's a concrete result. So the squaring is actually a very determined and defined operation. Square root, which is we are thinking about square root to be an inverse of raising to the power of 2, is not exactly of this type because square root of 4 can give you either 2 or minus 2, right? So what do we do? Well, first of all, in a classical meaning of the function square root among real numbers is not exactly a function because it has always at least 2 values, sometimes 0 actually if it's a negative, right? So within the realm of real numbers, square root is not a unique operation, if you wish. And that's why we cannot say that raising to the power of 2 is only an operation and it has inverse, which is square root. No, that's not true actually. However, we would like to use square root as basically some kind of an inverse to a squaring and we would like to specify all possible values of this operation. So what do we do? Well, we have invented a special notation for this. We are saying this is plus minus 2, right? So this plus minus indicates that there are two different numbers as a result of this operation plus 2 and minus 2. Well, now let's go back to integration. Now, differentiation as we have found out is an unary operation among the functions which are sufficiently smooth to do this operation. But now, can we talk about a reverse operation? Well, obviously we can because if you will take some function which happened to be a derivative of something then if we will be able to find original function from which we have started then that will be an operation, right? So this operation is called integration. So for instance, from 1 over x we can get to logarithm x, right? Now, can I say that this is a unique operation, the unique answer to this procedure of integration? Well, no, because obviously we know this particular property of derivative. If you have f of x, I will use this differentiation by x, I will use this particular notation. If this is equal to, let's use different function, let's say g of x. So the derivative of f of x is g of x. Now, derivative of f of x plus some kind of constant, as we know derivative of sum is equal to sum of derivatives. Now, what's the derivative of constant? Well, we were talking about this a few times, it's 0. So this is exactly the same as derivative of f of x which is g of x. Now, what does it mean? It means that if I start from the g of x, not only I can get f of x as a result of integration, it to inverse my process of differentiation I'm integrating, right? So not only I'm getting f of x, but also any other function which differs from f of x by a constant. So, similarly, from 1 over x I get not only the logarithm x, but also any function which looks like logarithm x plus c. Well, we should actually relate to this inversibility but not exactly in the pure sense of this word as we were talking about square root. Square root is also not exactly an inverse operation to raising to the power of 2 because it delivers two different answers. In this case, it's not 2, it's infinite number of answers, but at least we are suspecting that they are very much lookalike and differ only by addition of constant. Okay, so let's talk about syntax now. So, with square roots we actually came to this plus minus notation to indicate that it's not exactly pure result, but at least a reasonable set, in this case set of two elements as a result. In this case we will do relatively the same. So, let's consider that you have this situation. So, I would like to say that if I will apply this procedure of integration to this function, I should get original f of x, right? So, notation is the following. This is the sign of an integral. Okay, I'll leave empty space here and this is f of x, right? Now, as I was saying, not only f of x is the result but also all other functions which differ by constant. So, I have the right to have it like this where c any constant, right? Plus, there is another little detail. You see, I'm usually using this index here to specify differentiating by x. So, here I also have to specify that I'm integrating by x and the syntactical approaches here. I will put dx differential of x. Let's not think about this as anything which has certain meaning. I mean, it does have a meaning but we will talk about this much later. So far, just accept this as a syntax. So, if you have a function and you would like to integrate it, you have the symbol of integral which is like stretched up and down letter s and you put this dx where x is an argument actually. And this is the syntax for operation and integration applied to the function gfx and if we can find the function f of x derivative of which is equal to gfx then this is an answer in plus constant c. Okay, now, let me just give a couple of examples before I will prove that this is actually the only possible result. Okay, couple of examples. Now, let's talk about differentiation. Now, differentiation sometimes, you know, we are using differentiation as d by dx, right? f of x is equal gfx. That's the same thing. By the way, this also dx has certain meaning. But again, let's not pay attention to the meaning. Let's just talk about this as just a symbol, symbol of differentiation. So, what do we know? For instance, d by dx of x to the power of n is equal to n, x to the power of n minus 1, right? That's the derivative of power function. Now, as a result, what can I say about integral of power function? Let's just think about this. Integral x to the power of n dx is equal to... So, we need to find out the function derivative of which is x to the power of n. Well, derivative actually diminishing the power by 1, right? So, I should start with the power of n plus 1, then my derivative would deliver x to the power of this minus 1, which is exactly n. But now there is also this multiplier, right? So, if I will derive, if I will make a derivative from this, it will be n plus 1 times x to the power of n. So, I have to neutralize this n plus 1, and that would be the correct answer, right? Because if you will take a derivative of this, it will be n plus 1, and it will cancel with this one, and x to the power of n plus 1 minus 1, which is n. And plus constant c, of course, if we are talking about integral. So, here is my first example of integration of the power function. So, I know how to integrate any power function. And from the power function, after we will talk about the properties of integral, especially additive properties, like integral over sum is equal to sum of integrals, then you can probably derive from here any kind of polynomial function. But let's talk about this later on. That's not a subject today. Okay, some other examples. Okay, what's the derivative of e to the power of x? Well, we all know that derivative is e to the power of x. What does it mean? It means the integral of e to the power of x. This is called, by the way, indefinite integral. I started from explaining that we're talking about indefinite integrals, because there are definite, again, unrelated to today's lecture. So, if I'm saying integral right now, I mean indefinite integral. So, this is, again, we have to find the function derivative of which is equal to e to the power of x. Well, we know the derivative of e to the power of x is e to the power of x. So, that's e to the power of x. But we have to add the constant. So, not only e to the power of x is an integral of this, but also any other function which differs by constant, by any constant. So, that's my second example, right? Next. Okay, if I will have function sin of x, and if I will do a derivative, that's equal to cos sin of x, right? I'm trying to use as many different notation for derivative as possible. So, this is one, d by dx is another, this prime symbol is another symbol. So, derivative has different notations. Integral, by the way, has only one notation, which is this one, right? So, what can I say as a consequence of this one? That integral of cos sin of x dx is equal to sin of x, right? If derivative of sin is cos sin, then integral of cos sin is sin plus c, plus any constant. And my last example is this. Actually, I already spoke about this. Last example, d by dx logarithm x is equal to 1 over x, and therefore integral of 1 over x dx is equal to logarithm x plus constant c. There is one little detail which I would like you to pay attention to. Remember, integral of x to the power of n dx, I said that's what it is, right? Plus c. Well, this is not exactly always true for any n. It's true for n not equal to minus 1, right? Otherwise, this would be zero in denominator. So, for all n not equal to minus 1, this is a true formula. And what about n equal to minus 1? Well, x to the power of minus 1, which is 1 over x, that's exactly this case. So, in this case, we have a completely different situation. You see? Just a small detail. Okay. So, these are a few examples of, if we know the derivative, then integral of that derivative is the original function we started from, right? So, these are basically the ways to approach this, because first you have to kind of think about what is a function derivative of which is equal to my original function I would like to integrate. And then you basically derive it. Okay. Now, what I have not actually addressed is the following. I said this, that for instance, I know this. So, I have one particular function, and I know that derivative of this function is g of x. Now, I would like to take integral of g of x, g of x. And now I'm saying that obviously from this, I know that this is the result. So, these are all the different functions derivative of which is equal to g of x. I know about this, and therefore it's obviously the derivative of this is equal to g of x. But I said something which I have not proven yet. I said it's all the functions derivative of which is equal to g of x. Is that a true statement? That's something which I really have to prove. That this plus c actually allows me to specify all the possible functions, and there are no other functions. So, there is no function which differs from f of x, not just by a constant, but by something else, maybe. And derivative of that other function would also be g of x. Now, this is not the case, and they can prove it. So, this actually is complete set of functions derivative of which is equal to g of x. How can I prove it? Well, let's just think about this way. For instance, derivative of function f1 is equal to g of x. And derivative of function f2 is also equal to g of x. I would like to prove that the difference between these is constant. Which means it allows me actually to take only one of them and say that plus c completely covers all the possible functions which have a derivative equal to g of x. How can I prove that these two functions differ only by a constant? Well, let's just think about it. Let's subtract. I have zero on the left, on the right. Now, on the left, I have derivative of one minus derivative of another. Now, derivative is additive function, so the difference of derivatives is derivative of difference. So, I have dx of f1 of x minus f2 of x. Now, what's interesting is, it's for all possible values of argument. So, if argument is defined, let's say, on some segment from a to b, then for every argument, derivative of this function is equal to zero. Now, if you go back to a lecture, it's among the main theorems of derivatives. There is a lecture which was dedicated to constants. And one of the things which was actually proven in that lecture was that if a derivative of a function is equal to zero everywhere where this function is defined, let's say in this segment in particular, then the function is constant. It's very easy to prove. The proof was based on the theorem of Lagrange. And the theorem of Lagrange, by the way, let me just remind you a little bit, might be just as a refreshing thing. If you have a function, then there is a point somewhere in the middle where tangential line is parallel to this one, right? Now, this is basically Lagrange's theorem. And now what we see is that if this line is parallel to this one, and they know that this line is always horizontal because the derivative is equal to zero, right? That actually means that these two are supposed to be the same, right? So it's horizontal line. So that's what's very easy to prove. Which means that f1 of x minus f2 of x is constant. It can be any different constant, and that's why I'm specifying that integral of g of x dx can be equal to, let's say in this case, f1 of x plus c where c any real constant. It actually describes all the possible functions because all the functions which have the same derivative, g of x, are different from each other only by a constant. So that constant actually allows me to specify all the different possibilities of this particular result of integration. By the way, this result of integration sometimes it's called anti-derivative because it is really kind of an inverse thing. And the meaning of this plus c should be understood in exactly the same way as you understand that square root of 4 is equal to plus minus 2. This is syntactical notation which allows you to specify that there are different answers to this problem. Same thing here. This is the problem and these are different answers. In this case there are only two answers. In this case there are infinite number of answers and they all differ from each other by a constant c. So this actually completes this lecture. I have proven that this is a really complete, completely exhaustive description of all the possibilities of this particular operation of integration. So we can get just one particular function derivative of which is equal to g of x and then saying that this function plus any constant is a complete answer to the operation of integration. So try not to forget to add c or whatever, any letter which you prefer. c is where customers think. So whenever you have this particular result of integration, whenever you find anti-derivative, you always have to add constants to specify all the possible functions derivative of which equals to whatever is under integral. Now, here is an interesting consideration. There are certain rules to use when you are calculating the derivative. I mean, if it's, let's say, polynomial, well, you basically divide it by every term and then you have a derivative of every term, factor out multiplier, then you have some kind of a power function, you know how to deal with this, etc. So no matter how complex your function is, using the rules of derivation, you can always find its derivative. With integration, it's not exactly the same way. No matter how complex your function is, you can always find the derivative using the rules. But the complexity in case of integrals is significantly different because you see with derivatives, you can use this chain rule whenever you have a function of function, etc. With integrals, it's not exactly that simple. So if taking derivative is actually a skill, taking integrals is a little bit of an art. All right, that would be a nice point to finish this particular lecture. I have introduced you to the art of integration. Well, primarily the purpose of this lecture was just to define what is integral and how to specify it whenever you're dealing with it. I do suggest you to read notes for this lecture on unizor.com, so you go to calculus, indefinite integrals and definition. And other than that, that's it. Thank you very much and good luck.