 Hi, I'm Zor. Welcome to Unizor Education. We introduced the concept of indefinite integral, and now I would like to talk about the properties of this operation. This lecture is part of the advanced course of mathematics for teenagers and high school students. It's presented on unizor.com. I suggest you to watch this lecture from the website, because it has lots and lots of very detailed notes, and for registered students there is a possibility to take exams. Well, actually you can take any number of exams, any number of times each, until you will reach perfection, and the site is free, so basically use it at will. Alright, so properties of indefinite integral. Let me start from something which I said a couple of times before, that integration is an inverse operation to differentiation. So differentiating function we get its derivative, integrating derivative we get the function back. Well, plus constant, if you don't remember, right? Okay, so my first property is the following. Now, as I was saying, integration of derivative basically delivers you back your original function. This is derivative of the function, this is original function, plus constant C. Right, it can be expressed slightly differently, and maybe shorter actually. Now, we all remember that this constitutes differential of the function. It's basically like a definition, which means I can rewrite whatever I had on the top as differential of f of x is equal to f of x plus constant. This actually emphasizes even more the concept of differential to be kind of inverse of integral. So we first differentiate function, and then we integrate the function, and we get back the original function. That's the sense in which they are inverse to each other. Okay, fine. Now, this is basically my first property, if you wish, and it seems very trivial right now, but a little bit further when I will introduce another property, it will be obvious that this particular notation has its advantages. Okay, my second property is kind of expected. Integral of constant times function is equal to, oops, you can take the constant outside of integral. How can I prove this? Well, let's just assume that integral of f of x dx is equal to g of x plus C. What does it mean? Well, it means that the derivative of g of x is equal to f of x, right? Now, from this, obviously, we can multiply by a constant. Now, we know that derivative of the product of a function by a constant is equal to constant multiplied by derivative of the function. So this is actually the same as a times g of x derivative. So this is the same as this, right? Derivative of constant times function is equal to constant multiplied by derivative of the function, which means that this thing proves basically this, right? Because derivative of this thing is equal to, now, this is g of x, right? Well, plus constant, but doesn't matter when we differentiate it. So if I will take a derivative of this, it will be derivative of a times g of x, which is the same as a times derivative of g of x, and that's why we have this equality. So basically all these properties of the integrals are proven by the corresponding properties of the derivative. Since derivative of a product function by a constant is equal to constant multiplied by derivative, that's why we basically have this. Because this is, it means this is some kind of function where h of x, where h of x is equal to a times f of x, right? That's on the left side. And on the right side, I have this a times g of x. And derivative of this is also equal to a times f of x. So that's why they're equal. An absolute analogous is another theorem about the sum of two functions. Integral of f of x plus g of x, dx is equal to sum of integrals. How can I do it? Exactly the same way. I differentiate, differentiate of this sum. The derivative of the sum is equal to sum of derivative. The derivative of this thing is f of x, derivative of this thing is g of x, right? So derivative of this is this, which proves the same thing here, right? This means what? It means that there is some kind of a function, derivative of which is equal to, well, I should put, derivative of this is equal to this. So sum of derivatives is equal to this. And derivative of this is equal to exactly the same thing. So they are, but I did not really put plus c in these cases. Let me just say a couple of words about this. When I'm talking about indefinite integral, I'm actually talking about not just one particular function, which is the result of this indefinite integral. I'm talking about the family of functions, which are different, but they are really not very much different. They are different from each other only by a constant. So if this represents family of functions, derivative of which is equal to this, and this represents basically again family of functions, and these are not different more than by a constant and these no more by a constant, that actually makes this equal sign having some kind of a meaning. Okay, the function which is this is different from the representative of this result is different from representative of this result no more than by a constant. That's what basically it means. And that's why I'm kind of skipping the constant because it's obviously irrelevant for all these kind of things. What is relevant is that since the derivative of this is equal to whatever we have under integral there, that's why we can have this particular equality. And next, well, you know what before doing next, let me just give you a couple of examples. We started from this, right? What can be a good example of this? Look at it this way. What is this? Well, this is what is differential of sine. Well, it's a derivative of sine times dx, right? So it's cosine of x times dx. Now we are looking for a function derivative of which is equal to cosine. Now what is this function? Well, obviously it's a sine of x, n plus c. So you see this sine and this sine. This is kind of a demonstration that differentiation is inverse function to integration. Now, another simple case was when I was saying that I can take a multiplier out of the integral. Now, how can I prove this? Well, how can I exemplify this? Let me just give you an example. An example is like this, for instance. Now, on one hand, you can actually just remember, we were talking about the formula. Remember this formula? Why? Because differentiation of this is n plus 1 times x to the power reduced by 1, right? Which is n plus 1 minus 1, which is n. And n plus 1 would be multiplied here. That's why I put it here to get just x to the power of n, right? Now, if you multiply this both sides by n plus 1, you will see this. And you can recognize this as this for n is equal to 4, right? So you can actually very quickly come up with this result. This is x to the power of 5 plus c, right? Because derivative of x to the power of 5 is 5 times x to the power of 4. But let's do it slightly differently. Let's take 5 outside and have integral of just x to the power of 4 using the property which we have proven, right? Now, this thing, 5 times... Now, remember this formula, I was just demonstrating. This is x to the power of 5 divided by 4 plus 1, which is 5, right? Plus c. And this is exactly, again, x to the power of 5, right? So these are simple examples. And now, basically I would say the central example of whatever I was just saying before in today's lecture is integral of some kind of polynomial. And as a polynomial, I will take something like 4x to the power of 3, 3x to the 2, 2 to the x plus 1 dx. Now, since this is a sum, integral of sum is equal to sum of integrals, and each one of them has a multiplier, so I can take it outside of the integration. So it would be 4 integral of x to the power of 3 plus 3 integral of x to the power of 2 plus 2 integral of x dx plus 1 and integral of dx, right? Which is equal to... Well, this is x to the power of 3 and integral of that is x to the power of 4 divided by 4 and 4 and 4 will cancel out, so that would be x to the power of 4. Oops. x to the power of 4. Now, this would be obviously x to the power of 3 divided by 3 and multiplied by 3, so it would be x plus 3. And this would be x to the power of 2 divided by 2 and multiplied by 2, so that's 2. This would be an integral of 1, which is x plus constant c. So we have used right now all these basic properties, the linear properties, if you wish, which means you can multiply by a constant and add under integral expression and spread it over multiple integrals and take each one of them separately. Now, if it's a little bit more complicated function, but also which can be separated into pieces, something like this, you still can consider this as sum. Integral of this is x to the power of 4. Integral of logarithm, well, we don't know yet, but anyway it's easier than to integrate the whole expression. And later on in today's lecture I will explain how to find integral of logarithm. But in any case, that's easier than this. So this linearity of indefinite integral is a very, very basic fundamental property which is very, very important. Now we will do something a little bit more difficult. Well, you remember that derivative also is kind of a linear operator. The multiplier by constant can be brought out of the derivative. The derivative of sum of two functions can be separated into sum of derivatives. So these are linear properties. Now, what about multiplication? Hope you remember this function. If you want to derive, to have to take derivative of this, remember the formula? So that's not basically as easy as sum of two functions, right? The product of functions, of two functions, derivative of this is a little bit more complicated than the product of derivative. So it's not product of derivatives. Just remember this. Now, same thing with integrals. However, using this, we can find some very helpful formula for integration. And here is the formula. Let's integrate both of these. Now, integral of this, integral of derivative of the function, function being the product of these two, is the function itself. We talked about this as the first property. So integral of this is f of x times g of x, plus constant, okay? But let's just forget about this constant for a while. We assume that we know this. Now, on the right side, if we will integrate, that's the sum, right? So we can do it as a sum. Okay. Now, this is basically a straight consequence of the formula for derivative, right? What's really very important here, that integral of sum is equal to sum of two integrals. That's what I was using, because integral is a linear function of the function, linear operator, I should really say. And this integral of this is integral of derivative is equal to the original function from which we started, right? Because what is integral? It's a function derivative of which equal whatever you have here. And we already have a derivative, right? So that's what it is. Now, from this, we can derive a very helpful formula. Formula basically is very simple. We just resolve it for one of these. For instance, we have something like f of x times derivative of the function. Then what we can say is, we can say from this, it's equal to product of the functions minus integral, and we will reverse the integration here. Oh, sorry. GFX. Now, this is a formula which is called integration by parts. Well, I don't know why they called it this way, but anyway, that's the official name for this integration. What is important is to realize that this is not just any function. So we are not talking about integral of a product of two functions. We are talking about integral of the product of any function and derivative of any other function. Then in this case, we can actually change it into this. Now, why do we have to do this? This looks simpler than that. Well, it depends. It depends on what is the function f and what is the function g. And you will see in a couple of minutes that in some cases, it's much easier to deal with this than with this. So it all depends. If function f is kind of simpler, whatever, and especially it's simplified when we are getting a derivative from it, then it makes sense to do it this way. Now, which functions are simplified by taking derivative? Well, obviously, the power function, right, x to the power of n is some multiplier and then x to the power of n minus 1. So we are reducing the power by differentiation. So this might be a case when function f of x is some kind of a power of x and this allows us to reduce this power. So that's basically the reason. But you will see the example of this. Okay, now, let me just give you an example of this. Integral of x squared times e to the power of x dx. Well, obviously immediately you don't know what is the function derivative of which is this. So you can use this rule f being x squared, g being e to the power of x and derivative of g of x, derivative of e to the power of x is e to the power of x. We know that, right? So this e to the power of x is both g and g of x, which means that I can use this one, which is x squared times e to the power of x minus integral of this. Now, what is derivative of x squared? That's 2x, right? So it's 2 outside of integral x and the g is still e to the power of x. This is exactly what I was just talking about. This is simpler because this is the first power and this is the second power. First is simpler. Now, this is not the end of it. Now I don't know what this is, but I can do exactly the same thing again, right? So I have my x squared e to the power of x minus 2 and this integral I will also represent in this way x being f of x and e to the power of x is g of x. So it will be what? x times e to the power of x minus integral e to the power of x times derivative of x is 1 dx. Now, this is, again, simpler. So what's the result? The result is x squared e to the power of x minus 2x e to the power of x plus 2... Now, what is integral of e to the power of x? Well, e to the power of x is the function which has a derivative equal to itself. So that will be e to the power of x. Derivative of e to the power of x is e to the power of x. That's why integral is equal to exactly the same, plus c. And that's the answer. I mean, you obviously can factor out e to the power of x. x squared minus 2x plus 2 plus c. And that's the result. This is the result of integration. Now, I wiped out my main integration formula. So it was integral f of x g of x dx is equal to f of x g of x minus integral of f derivative g of x dx. Let me see if I can change my marker. Okay, now, let me slightly modify this. It will be a little bit shorter. You remember that g derivative times dx is basically a differential of f of x, right? Of g of x. So I can rewrite it as following. f of x dg of x is equal to f of x g of x minus integral. Now, this is g of x times differential of f, right? So this is basically another representation of exactly the same thing. I'm just combining g derivative times dx with differential of g and f derivative times dx as a differential of f, right? And that is a little bit more common representation of integration by parts. And let me just give you an example of this. For instance, I actually don't need this and I don't need this. It already played its role and I have more real estate. Now, for instance, we have integral of x square d sine of x. And again, remember this x square. If I will replace differential of sine with differential of x square, I will reduce the power by one and then I can do it another time. So this is a typical thing. So this is my f. This is my g. So it's x square times sine of x minus integral of sine of x d of x square, right? Now, what is differential of x square? Well, that's a derivative times dx, right? The derivative of x square is 2x times, so it's x square sine x minus 2 sine x times x times dx, right? And here I will do exactly the same thing. x square sine of x minus 2. I will combine sine of x and dx into differential of cosine, right? Actually, I need this minus because derivative of cosine is minus sine, which is minus sine and dx. So that's how I did it. Now, this, again, I will do exactly the same thing as before. I will integrate it by parts. So I have x square sine x plus 2. So it's f times g, x times cosine x minus integral cosine x dx, right? Cosine and differential of this function. So in this case, when I'm using differentials, it's kind of more symmetrical, equals 2. All I have to do is to find integral of cosine. Now, I know that the derivative of the sine is cosine, right? So this is basically my sine. And my result is plus 2x cosine x minus 2 sine x plus constant c. So that's the result of my integration, okay? So this function has integral, which is this. And as a consequence of this, I have a very simple case when I can integrate the logarithm, which I promised in the beginning. So how can I integrate logarithm? Well, I don't know any function, elementary function, which has a derivative equal to logarithm, right? But here's what I can do. Now, this would be my f of x and this would be my g of x. Plain x is also a function, right? g of x is equal to x and f of x is equal to logarithm x. So integral is equal to product of the functions, which is x times logarithm x minus. And now I will change whatever was a function becomes differential, whatever was differential becomes a function, right? So x times differential of logarithm x equals... Now, what is differential of logarithm? That's the derivative times gx, right? Derivative of logarithm is 1 over x, right? Well, obviously we are talking about logarithm, which means we are talking about only positive values of x. So we can reduce it and I will have just 1. And integral of 1 is what? It's just plain x, right? Because derivative of x is 1 and constant c. So this is integral of logarithm. And we have derived it using this integration by parts. Very important, actually. Obviously we have to know when to use it because sometimes we can actually use it and we get more complicated results. After applying this rule of integration by parts. But in some cases, like in this one or when this function is some kind of a polynomial, for instance, it helps. Because then when we differentiate it, it will reduce the power. And by the way, differentiation of logarithm is also kind of simplifying our story. Because logarithm is a little bit more complex function versus 1 over x, the derivative of x, which seems to be a little simpler, right? Alright, so these are basic properties. Now, they are linear properties, like constant multiplier or sum, which are really very simple and trivial. And this more complicated property of integration by parts, which is very much used in many, many different cases. I do suggest you to look at the notes to this lecture on unizord.com. The notes presents basically only the basic points of this lecture and exercises, which I have just done with you. Try to do these exercises yourself and see if you will receive the same results as I did. Alright, that's it. Thank you very much and good luck.