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Transcript
00:34:980Paolo Guiotto: It does not work.
00:52:820Paolo Guiotto: Okay, now it's working.
00:54:820Paolo Guiotto: Okay, good morning. And, let's restart from…
01:01:600Paolo Guiotto: where we stopped yesterday. So, yesterday, we have seen a definition of integral for functions of two variables.
01:11:390Paolo Guiotto: positive functions of two variables. That definition can be, of course, in a basically straightforward way, can be extended to functions of any number of variables. So.
01:24:950Paolo Guiotto: the, construction…
01:33:850Paolo Guiotto: of… Instagram.
01:41:750Paolo Guiotto: can… be extended…
01:52:70Paolo Guiotto: Extended to… functions.
01:58:540Paolo Guiotto: Let's say F, function of a vector variable X,
02:03:730Paolo Guiotto: Where the vector variable X belongs to some space RM.
02:10:160Paolo Guiotto: Genetic.
02:12:10Paolo Guiotto: I do not repeat the construction, it's not the point here, it's just to know that,
02:19:200Paolo Guiotto: We have such a definition.
02:22:80Paolo Guiotto: F positive.
02:25:410Paolo Guiotto: And, bounded.
02:29:610Paolo Guiotto: So, we know that, F integral.
02:37:670Paolo Guiotto: F integral…
02:43:120Paolo Guiotto: means…
02:46:450Paolo Guiotto: that the integral of F on the domain interval on domain D means that this integral, that we denote with this symbol.
02:58:630Paolo Guiotto: is, well… defined We have seen that it is not always well-defined.
03:07:920Paolo Guiotto: If well-defined,
03:11:980Paolo Guiotto: we have this… this operation that, for the moment, we don't know how to compute. Well, an important fact to know is that,
03:23:590Paolo Guiotto: if, the function f is continuous.
03:30:590Paolo Guiotto: And the domain D is compact, so closed and bounded.
03:38:810Paolo Guiotto: Then… F is, integrable.
03:49:540Paolo Guiotto: on…
03:53:400Paolo Guiotto: So, in this case, and this reminds of a fact that you have seen in first-year calculus, this,
04:01:330Paolo Guiotto: is, similar.
04:04:20Paolo Guiotto: Actually, it is an expansion.
04:06:740Paolo Guiotto: to the… fact that…
04:11:270Paolo Guiotto: that if the function F is continuous on a closed and bounded interval AB, then the integral from A to B of F of X dx is well defined, okay?
04:25:810Paolo Guiotto: So, at least we have a certain number of cases for which we have the integral well-defined. So, it means that we do not have to check anything if we know that the function is continuous and the integration domain is compact.
04:43:340Paolo Guiotto: In other cases, we have to
04:45:460Paolo Guiotto: We have… we will see later how to decide if the function is integral or not.
04:53:90Paolo Guiotto: Now, this is the definition when the function is positive, and this, you remind, the basic idea was, to define,
05:05:600Paolo Guiotto: the area.
05:08:450Paolo Guiotto: in the case of functions of two variables, so the domain is a subset of the Cartesian plane XY, and the integral, geometrically, represents
05:21:390Paolo Guiotto: area.
05:23:280Paolo Guiotto: The limited… sorry, the volume, the limited above by the roof, which is the graph of the function F, and below by the domain, there is a sort of cylinder. So, this is for dimension 2.
05:39:640Paolo Guiotto: So, if the function is a function of, for example, 3 variables, we cannot draw anything, because imagine we have three axes for the three variables, so X, Y, Z.
05:53:890Paolo Guiotto: But then we should need another 4 taxes for the values of F, and we do not have such representation in an easy way.
06:01:700Paolo Guiotto: So, and therefore, what is the integral in three variables, F of XYZ, DX, DYDZ? What does this represent?
06:17:80Paolo Guiotto: It's a sort of volume of an object that has four dimensions. It's a sort of four-dimensional volume.
06:24:840Paolo Guiotto: Well, there is a name that, in mathematics, we use to make all these things with a unique terminology, which is the concept of measure.
06:37:720Paolo Guiotto: And in fact, we may notice also this. We introduced this definition.
06:45:310Paolo Guiotto: We already noticed this fact, that this is in the case of integration in two variables.
06:52:860Paolo Guiotto: If we have a genetic subset of the plane, X, Y.
06:59:430Paolo Guiotto: The integral of the function, which is constantly equal to 1 on the domain.
07:07:410Paolo Guiotto: It is the volume of a cylinder, let's say, with base D and H1, so…
07:14:740Paolo Guiotto: in borrowing the idea that the volume of that cylinder would be the product between the area of the base times the height, we define this quantity. So, if the
07:27:290Paolo Guiotto: subset of R2, let's start with R2. It's a subset of R2. We define
07:37:950Paolo Guiotto: the area of the… as the integral on D of the function, just constantly equal to 1.
07:51:90Paolo Guiotto: Provided this integral makes sense.
07:54:810Paolo Guiotto: Provided that…
07:59:560Paolo Guiotto: The… Integral.
08:02:700Paolo Guiotto: Makes sense.
08:07:280Paolo Guiotto: We have to specify this, because this is basically an integral, and it is not, granted that
08:16:500Paolo Guiotto: an integral is well-defined, even if the function is very simple, like this one. In particular, it can be proved, for example, that
08:27:940Paolo Guiotto: This is the case for common sets, like open sets, closed sets, for those sets, there is an area, okay? I will return on this in a moment. More in general, if I have a set
08:41:760Paolo Guiotto: D, which is contained in RM,
08:46:390Paolo Guiotto: So this, means that now we cannot,
08:50:330Paolo Guiotto: do a figure unless M is 3.
08:54:510Paolo Guiotto: So let's say that if M is 3,
08:58:430Paolo Guiotto: the domain is now generically a solid domain in our tree. So, like, I don't know, a potato, or a ginger food, I don't know. Imagine a kind of object like that. But in gen…
09:17:580Paolo Guiotto: Like what?
09:19:530Paolo Guiotto: Legacy. Okay.
09:21:480Paolo Guiotto: But in general, if it is in RM, so in R4, R5, I cannot see what is it. So the integral of 1 on that domain
09:35:120Paolo Guiotto: Provided it makes sense.
09:37:730Paolo Guiotto: We'll define what quantity. Well, by analogy, if D is in R2, the integral of 1 is the area of D,
09:46:230Paolo Guiotto: So, if D is in R3, I will expect that the integral of 1 will be the volume of D, and if B is in R4, it will be the 4-dimensional volume.
09:57:130Paolo Guiotto: Well, we call this just the measure, the M-dimensional measure of D. Well, we denote with this symbol, lambda M of D, and we call it the
10:09:290Paolo Guiotto: M, dimensional… measure…
10:19:140Paolo Guiotto: off… So, measure will stand for if M is equal to 2, measure means area.
10:31:280Paolo Guiotto: If M is equal to 3, measure means volume.
10:36:530Paolo Guiotto: and so on, and so on. There is not so on, because for every point, we do not have a word for the analogous of area and volume, and we could just measure
10:47:20Paolo Guiotto: M-dimensional measure. So the integral of one, of one on a certain sector is the measure of that sector.
10:57:890Paolo Guiotto: Okay, now, the next step is the extension to a variable sine function.
11:04:950Paolo Guiotto: So, it's a reminder of, our program, we…
11:09:820Paolo Guiotto: described yesterday, we said step one is the definition of
11:17:920Paolo Guiotto: the, integral for positive functions. Then, now, step two, definition of integral for real valued functions. Fortunately, this is not, complicated, it's very fast. Now, say, definition…
11:35:890Paolo Guiotto: Off.
11:37:510Paolo Guiotto: integral… for… R-value function.
11:50:820Paolo Guiotto: Now, we will help with the figure, the… the…
11:54:440Paolo Guiotto: The idea is the same of the integral for functions of one variable. So in this figure, I will plot. On the x-axis, actually, this stands for the domain. So I have a function, which is function of a vector variable.
12:10:880Paolo Guiotto: X, defined on a domain D of RM,
12:16:200Paolo Guiotto: real value. So, the horizontal axis represents Rn. I know, this is not R, so it should be a…
12:26:230Paolo Guiotto: And space, we've been…
12:28:560Paolo Guiotto: set the number of dimensions, M, but let's stay focused on the idea and not on the
12:34:140Paolo Guiotto: particular geometry. Now, let's say that the domain D would be represented as a subset of M, so, like, the green thing I'm doing here.
12:44:940Paolo Guiotto: And we have a function defined on this domain. Now, suppose that this function can be positive and negative, so somewhere it will be positive, and somewhere else will be negative.
12:55:770Paolo Guiotto: What is the idea that you already have in one variable calculus of the integral for this case?
13:03:400Paolo Guiotto: The integral, in this case, does not compute the area between the graph and the axis, so does not compute the geometrical area of this figure that I'm coloring in black.
13:14:810Paolo Guiotto: Here.
13:15:850Paolo Guiotto: But rather, it computes what is called the arithmetic area, or algebraic area. So it says, we take the area of the… which is between… included, between the graph of the function and the domain, when the function is positive.
13:34:540Paolo Guiotto: And we count negatively the area when the function is negative. So the idea is that the integral on this F is, let's say, this area
13:47:190Paolo Guiotto: let's call it A1. This is A2, the area. These are positive numbers. When we talk about areas, we mean geometical positive quantities. So the integral will be area 1 minus area 2, by definition. Now, the point is, how do we define this thing?
14:07:160Paolo Guiotto: Now, there is a simple trick that makes this definition easy.
14:11:100Paolo Guiotto: Now, if the black line is the function f, I define two auxiliary functions, which are this one. The first one is the red line I'm drawing, which is just F when F is positive, and I set 0 when F is negative.
14:26:90Paolo Guiotto: This thing is called the positive part of F. So, by definition, F plus is equal to F when F is positive.
14:36:180Paolo Guiotto: and 0 when F is negative.
14:38:510Paolo Guiotto: More precisely, I'm saying F plus is a function, so F plus of X is, by definition, F of X
14:47:460Paolo Guiotto: if F of X is positive.
14:52:100Paolo Guiotto: So X by X, you compute f of x, you see, if it is positive, you take f of x, otherwise you take 0, if f of x is negative.
15:03:870Paolo Guiotto: In this way, the idea is that F plus, as you can see from the figure, and from the analytical definition, F plus is a positive function, is always positive.
15:16:410Paolo Guiotto: And, as the intuition suggests, the integral of that positive function that has been defined, so now… at this… at this point, we know that there is an integral for positive functions, so we… this quantity makes sense.
15:34:120Paolo Guiotto: And the integral of F plus should be just the area A1. So, A1 should be the area
15:41:380Paolo Guiotto: of… so the integral of the positive part of F.
15:45:860Paolo Guiotto: Okay, so let's define a one piece.
15:48:890Paolo Guiotto: And similarly, to define it, I do this. I take a function which is 0 when F is positive, so this is going to be this blue line.
15:59:320Paolo Guiotto: And instead of taking F when F is negative, I take minus F, so this is something like a specula. So this is called the negative part.
16:10:190Paolo Guiotto: which is defined in this way. It is 0 when F is positive. It is minus F when F is negative.
16:19:530Paolo Guiotto: The reason why I take minus F and not F is to have a positive function, no?
16:25:420Paolo Guiotto: Now, also, this F- is positive.
16:29:430Paolo Guiotto: And you may notice that the integral of F minus, which is the region with the, say, the area for the moment, but here is not an area.
16:39:10Paolo Guiotto: Because I don't know what is the dimension. It can be an area only if M is 1, but if M is 2, that's a volume. If M is 3, it's a measure, etc. But, however, this thing I'm calling, which is the integral of F minus.
16:53:840Paolo Guiotto: should be geometrically the same of the area A2, because it's just reflected in respect to the x-axis. So, I define A2 as the integral on the of the negative part of it.
17:09:849Paolo Guiotto: And therefore, here we have the definition of integral.
17:14:339Paolo Guiotto: definition.
17:18:50Paolo Guiotto: Now, skip some line here, because I have to put something. So we put the integral of F on D, by definition, as the integral of the positive part.
17:31:330Paolo Guiotto: which is the, say, the area A1 minus the integral of the negative part, which is the area A2.
17:41:440Paolo Guiotto: Now, in order that this definition makes sense, this is the difference between two quantities that must be finite, otherwise I could have problems. So, we introduce this definition. We say.
17:56:760Paolo Guiotto: that.
17:58:450Paolo Guiotto: The function F is… integral.
18:02:600Paolo Guiotto: on.
18:03:680Paolo Guiotto: P… If… F plus.
18:08:290Paolo Guiotto: F minus, it's positive part, it's negative part, R… bold.
18:15:270Paolo Guiotto: integral.
18:20:960Paolo Guiotto: on D.
18:23:80Paolo Guiotto: And their integrals are finite, so the integral of the positive part
18:29:360Paolo Guiotto: It's find it, and the integral of the negative part is fine.
18:35:420Paolo Guiotto: In this,
18:37:320Paolo Guiotto: case, we can define, in a proper way, this thing. We can say that the integral is, by definition, the difference between the integral of the positive part and the integral of the negative part.
18:51:200Paolo Guiotto: I know…
18:52:280Paolo Guiotto: or what all of you are wondering. You see, this definition is just a very abstract definition, because it does not give any recipe how to compute. That's not yet the moment. We are doing the process of defining this quantity, then we will discuss later today.
19:11:20Paolo Guiotto: how to compute, okay? This is nothing different of what you have seen, for example, in first-year calculus.
19:19:180Paolo Guiotto: when you have seen the integral in one variable. There is a process saying that this is the definition, but then how to compute it's another story.
19:28:950Paolo Guiotto: And you don't use the definition to compute. But that's not a novelty. You don't use the definition to compute derivatives, usually. You use rules of calculus and known derivatives to derive new derivatives. You don't… you never, almost never use the definition of derivative.
19:45:410Paolo Guiotto: And that's normal mathematics, okay? So… Now, this is the definition, so…
19:56:430Paolo Guiotto: We may notice a nice factor that is this remark, If the two integrals,
20:06:540Paolo Guiotto: Of the positive path
20:09:50Paolo Guiotto: end of the negative part, I'll find it, I write plus-minus to say the 2.
20:14:710Paolo Guiotto: Okay, so if they are fined, in particular, also the sum of these two integrals.
20:21:880Paolo Guiotto: Not the difference, but the sum.
20:24:930Paolo Guiotto: will be finite, and also vice versa, because these two quantities are positive, no? So, if the sum is finite, you must have that, the two must be finite, because if one of them is infinite, the sum will never be finite.
20:40:950Paolo Guiotto: What is the sum of these two integrals? Well.
20:44:790Paolo Guiotto: The difference is what we call the integral of F, but what is the sum of the two? So let's redo the figure down here.
20:53:370Paolo Guiotto: So this is, again, RM.
20:56:590Paolo Guiotto: with the domain B, which is this thing.
21:00:880Paolo Guiotto: And we have a function that, for simplicity, take a very simple figure, because the point is to illustrate the property. Now we said that the positive part is this red function, it is F when F is positive, and 0 when F is negative.
21:18:30Paolo Guiotto: And the integral of the positive part is the area, basically, of this region. So this one is the integral only of F plus.
21:28:390Paolo Guiotto: For F minus, we take 0 when F is positive, so we have this blue line here, and minus F when F is negative, so we have this. This is F-, and the integral of F minus is basically this area.
21:45:30Paolo Guiotto: Yeah yeah, that's right.
21:49:890Paolo Guiotto: So, when you look at the two together.
21:53:110Paolo Guiotto: When you sum up, you see that the sum of the red plus the blue area is now the geometric area, which is between the domain D and the graph of a remarkable function that, let's say, followed in yellow here, which is this one.
22:11:770Paolo Guiotto: And this yellow function is the absolute value of F.
22:15:500Paolo Guiotto: Whatever is between the yellow line and the domain has area exactly the sum of the red plus the blue. So this quantity here.
22:25:00Paolo Guiotto: So, the integral of the positive part plus the integral of the negative part is nothing but the integral of the absolute value.
22:34:200Paolo Guiotto: So, in other words, we can say that That is, huh?
22:42:610Paolo Guiotto: The integral of positive-negative part is finite if and if the sum is finite, but the sum is the integral of the absolute value, so if and only if the integral of the absolute value is finite. So the integral of both positive and negative part is finite.
23:01:170Paolo Guiotto: I remind you that these are positive quantities, is true.
23:04:740Paolo Guiotto: If and only if the integral on the… of the absolute value is fined.
23:11:110Paolo Guiotto: So this is very important, because it is the condition we should verify, in principle, to discuss integrability. If the integral of the absolute value is finite, then the two are finite, and so on.
23:27:940Paolo Guiotto: And in fact, we have this,
23:33:970Paolo Guiotto: Well, actually, here there is a little, detail.
23:43:140Paolo Guiotto: Yeah, there is a tricky problem behind this remark.
23:50:410Paolo Guiotto: If you take this function, take this function, you remind, I, yesterday.
24:00:140Paolo Guiotto: We, when we refresh the idea of the integral in one variable, we…
24:08:420Paolo Guiotto: reminded of this example, which is an example of a function which is positive, well-defined on the interval 0, 1, but this function is not integral. So, to say that not every function is integral. So.
24:22:120Paolo Guiotto: In principle, you should always check integrability. So this is a problem, because the definition is not easy, so how do I check integrability? That's why, for example.
24:33:180Paolo Guiotto: A condition like this one, Is, sort of,
24:41:520Paolo Guiotto: is a good condition, because continuous functions is a quite large class of functions, and so saying that, continuous functions are integral, means that you don't have to check basically nothing, because continuity is much easier than checking with the definition that the function is integral.
25:00:350Paolo Guiotto: But this is an example, of course, of a function which is not continuous. It's continuously jumping up and down, so you can even see that this function is never continuous. There is no point where this function is continuous. Now, slightly modifying this example.
25:22:180Paolo Guiotto: Here, I want you just to tell you, and it doesn't matter if you want to catch exactly this thing, but I want you to remark this thing. If you take this function, let's stay in dimension 1 for simplicity, M equals 1,
25:40:340Paolo Guiotto: and take this function, f of x, which is a slight modification of the example of the Dirichlet function. I take a function which is equal to plus 1 when x is, I don't know, irrational.
25:56:230Paolo Guiotto: And in the interval 0, 1, as per the multiplayer function. Now, for the deploy function, I will put 0, but here I will put minus 1 for X, which is original, and 01.
26:09:250Paolo Guiotto: it's not such a different function in respect to the Dichler function, it's still the same music, because
26:16:430Paolo Guiotto: you have the interval 0, 1, you have at points where here, on the domain, if this is a rational number, so if this X is in Q, now the function here has value minus 1.
26:32:770Paolo Guiotto: If in another point, X, this is not in Q, the value, so it is irrational, the value of the function is plus 1. Now.
26:42:630Paolo Guiotto: At each point, you can be only rational or irrational, and everywhere you find rationals and irrationals. So this means that, again, this function is a continuous up and down jumping from rationals and irrationals.
26:59:60Paolo Guiotto: from minus 1 to 1. You see, it's basically the same function that you have here, only here you jump from value 0 to value 1. In this figure, you jump from value minus 1 to value 1.
27:13:290Paolo Guiotto: What is the point here?
27:15:650Paolo Guiotto: Now, the point here is that if you take the absolute value, so this function, F, is not integral.
27:25:130Paolo Guiotto: This is because it's basically the same example we have seen yesterday.
27:29:990Paolo Guiotto: But the absolute value of F is what?
27:33:900Paolo Guiotto: You can take the absolute value of the function is constant equal to 1, because it is either the modulus of 1 or the modulus of minus 1, so in any case, it would come
27:43:940Paolo Guiotto: constant equal to 1 on the interval 0, 1, so it's a constant on a closed and bounded interval, so modulus of F is integral.
27:56:690Paolo Guiotto: So this is to say that we may have that an F is not integral, but the modulus of F is integral. So this is to say that saying that modulus of F is integral is not sufficient to say that F is integral, because this example shows
28:15:730Paolo Guiotto: that. F… is not integral, but modulus of F is integral.
28:22:470Paolo Guiotto: So it's not true that F is integral if and if the modulus is integral, okay?
28:28:700Paolo Guiotto: F.
28:29:730Paolo Guiotto: integral, if and only if modus of F integral. This is false.
28:40:930Paolo Guiotto: However, if we add something, this becomes true.
28:45:70Paolo Guiotto: And this something is something that we usually have.
28:49:920Paolo Guiotto: So, if we have F, which is continuous.
28:54:470Paolo Guiotto: then this holds. Notice that here is false, because this function is never continuous. There is no point where the function is continuous. But if f is continuous, then F is integral if and only if the modulus of F is integral.
29:15:590Paolo Guiotto: So they are equivalent.
29:18:820Paolo Guiotto: Okay, so now, let's close this,
29:22:440Paolo Guiotto: this… this remark. Let's go back to the definition of it. Let's say that now we have an integral defined.
29:31:150Paolo Guiotto: So, once the integral is defined, we, have a number of elementary properties. So, the Integral.
29:43:750Paolo Guiotto: Verifies… number… Off.
29:54:10Paolo Guiotto: elementary… properties.
30:01:840Paolo Guiotto: at all natural, so we don't spend too much time, we lease these properties. So, the property one is so-called linearity.
30:15:320Paolo Guiotto: that says that if you have the two functions, F and G, which are integral.
30:24:940Paolo Guiotto: on a domain DE, Then… Any linear combination of the two, something like,
30:34:800Paolo Guiotto: alpha F plus beta G, where alpha and beta are constant numbers.
30:42:360Paolo Guiotto: This is integral, is integrable.
30:46:760Paolo Guiotto: on the… For every choice of these constants, alpha and beta.
30:54:100Paolo Guiotto: And, of course, the integral of the linear combination, alpha F plus beta G, will be the linear combination of the integrals. So, alpha integral of F plus beta integral of
31:10:70Paolo Guiotto: G.
31:11:460Paolo Guiotto: So this is the rule that says the integral of the sum is the sum of the integrals, and you can carry in and out factors, constant factors, if you need.
31:22:180Paolo Guiotto: Second, this is called the monotonicity.
31:26:680Paolo Guiotto: that says that you have functions FG that are integrable on D,
31:32:990Paolo Guiotto: And you know that F is smaller than G on D.
31:36:470Paolo Guiotto: When we say on D means F of X less or equal than G of X for every X in D.
31:45:750Paolo Guiotto: Then, what happens to the integrals?
31:48:510Paolo Guiotto: Well, the integral of F will be less or equal than the integral of G.
31:54:780Paolo Guiotto: So… That's quite natural.
31:58:240Paolo Guiotto: Third, this is called the triangular inequality.
32:08:160Paolo Guiotto: If F is integral on D,
32:14:590Paolo Guiotto: Then, we have that the absolute value of the integral
32:20:160Paolo Guiotto: Now, remind that the origin of this notation is that you are doing basically something like a sum of things. So, the modulus of a sum is less than the sum of the models. So, we have this, that you can carry the models inside
32:35:480Paolo Guiotto: And say that the modules of the integral is less lupal than the integrals of the models.
32:41:350Paolo Guiotto: Number 4, another property that we…
32:45:910Paolo Guiotto: We must know is the decomposition
32:53:640Paolo Guiotto: And it says that, if, F is, say, if F…
33:00:970Paolo Guiotto: is integral on two domains, so, say, D1 and D2,
33:07:720Paolo Guiotto: Suppose that they are disjoint, the one intersection D2, empty.
33:14:90Paolo Guiotto: So, F will be integral on the union, D1, union. D2,
33:23:280Paolo Guiotto: And the integral on the total domain, D1 in union D2, so you put together the two domains, this will be splitted into the sum of the integral on D1 plus the integral on D2 of F.
33:39:290Paolo Guiotto: Okay?
33:40:340Paolo Guiotto: So, if for some reason.
33:42:590Paolo Guiotto: It is convenient to split the domain, not the function, the domain in two parts. You can do, you can split the integration into sub-integrations into two subdomains.
33:57:410Paolo Guiotto: Hmm, what else?
34:01:810Paolo Guiotto: Have a good night.
34:03:420Paolo Guiotto: If there's anything else?
34:09:429Paolo Guiotto: Yeah, there is this property that sometimes is, convenient.
34:14:560Paolo Guiotto: Which is the property of the null sets.
34:20:420Paolo Guiotto: So, if, the, UF that F is integral on a domain D, RMD.
34:32:409Paolo Guiotto: And… E.
34:35:840Paolo Guiotto: Well, let's write it in this way. Is contended in the…
34:41:420Paolo Guiotto: And the measure, you remind that quantity we defined above, the measure of the difference between these two.
34:50:80Paolo Guiotto: is zero.
34:52:60Paolo Guiotto: So there is no difference in terms of… the area of the difference, the volume of the difference in zero.
34:58:680Paolo Guiotto: Then, computing the integral on D of F is the same than computing the integral on E of F. So, in other words, if two sets
35:09:510Paolo Guiotto: are different for a set which has measure zero, we will see easily, at least in an easy situation, how to determine this. The integral won't change the value. The integral does not fill
35:23:550Paolo Guiotto: the values, the domains with the measure equal to zero. That's what it is saying.
35:36:110Paolo Guiotto: No, no, I'm saying, okay, suppose, let's,
35:42:10Paolo Guiotto: Let's show an example, no? Suppose that, for example.
35:49:390Paolo Guiotto: Suppose I am integrating, I don't know, a function of two variables. It is not important which function I'm talking about, on a disk. X squared plus Y squared less or equal than 1.
36:04:130Paolo Guiotto: So this is the domain I'm considering here, and it is a circle centered, or a disk centered in the origin, with radius 1. It is all this.
36:16:80Paolo Guiotto: Okay?
36:17:470Paolo Guiotto: Now… Suppose that I consider a second integral, the same function, FXY, the X, DY.
36:28:830Paolo Guiotto: But on the disc, X squared plus Y squared, strictly less than 1.
36:34:940Paolo Guiotto: It's not the same. What is the second domain? The second domain is the disk, the same disk, but without the boundary. So let's say that in red, I have this.
36:46:560Paolo Guiotto: And the boundary is not included, so I put dashed point to say that these points of the circle are not in the
36:55:830Paolo Guiotto: So, the difference between this is in notations above, this is our E.
37:01:550Paolo Guiotto: What is the difference between D and E? There is something.
37:06:500Paolo Guiotto: Yeah, the difference is exactly that circle, this one. This set here, only the circle, not with the interior. Only the circle. This is D minus P. Now, this thing in plane, is a line.
37:21:940Paolo Guiotto: So, intuitively, it won't have an area.
37:26:320Paolo Guiotto: Because, it's, it's a set without area, as area zero. You take a segment.
37:33:750Paolo Guiotto: There is no area. You may think it's like a rectangle with a basis and height equals zero. So when you do the problem, you get zero. So that's the same for a line. So this means that doing these two integrals will yield the same value.
37:48:510Paolo Guiotto: the integral won't feel that difference between the two sets, because the difference is something that has no relevance for the integral. That's the idea.
37:58:770Paolo Guiotto: Behind.
37:59:790Paolo Guiotto: So, however, we will see in practice when needed this stage.
38:05:610Paolo Guiotto: Okay, let's say that now we have a definition of integral, we have the basic properties, yes.
38:27:130Paolo Guiotto: Well, this is delicate, because of…
38:30:290Paolo Guiotto: Well, let's see down here, no? For example, I am…
38:35:560Paolo Guiotto: I am integrating, let's say that this is, again, in two variables. Say that I am integrating on a domain which is made by D1 and D2 here.
38:46:680Paolo Guiotto: You see?
38:47:850Paolo Guiotto: So my function is defined on the domain, which is made by the union of these two sets. So you have a function, the part of the function here, and another part of function there.
39:00:300Paolo Guiotto: So what I'm saying is that the volume of the first cylinder plus the volume of the second cylinder is the total volume. So you can decompose the integral on the union
39:12:710Paolo Guiotto: Into the sum of the two integrals, okay?
39:17:310Paolo Guiotto: What if they overlap?
39:20:200Paolo Guiotto: well, I could say, in some cases, it would be wrong, it won't be true. For example, imagine that this is the situation. You have the two domains, and they have an overlapping like this.
39:36:860Paolo Guiotto: Now, if you compute the first integral, you are computing this volume here. So, let's say the volume of this cylinder. If you compute the second integral, you are doing the volume of this other cylinder. You see that there is a common part, you are counting twice.
39:52:920Paolo Guiotto: So that won't be the volume of the unified cylinder.
39:57:360Paolo Guiotto: Because there is an overlapping that you are counting once when you compute the integral on the 1 of F. Another time, you integral on the 2 of F.
40:09:370Paolo Guiotto: So, in that case, the sum won't be the integral on the union.
40:16:910Paolo Guiotto: Okay?
40:20:170Paolo Guiotto: Yeah, you have a true correct formula would be integral on D1, linear on D2. You do, the integral on D1,
40:29:350Paolo Guiotto: you do the integral on D2, but then, since you count the… you are counting two times the integral on the intersection, intersection is this one, D1 intersection D2, you should subtract that contribution.
40:44:780Paolo Guiotto: So that's the correct formula.
40:48:280Paolo Guiotto: So, in particular, when the two are disjoint, you boil down into the previous one, because if the two are disjointed, that set is empty, and the integral on nothing will be zero. That's the idea.
41:02:410Paolo Guiotto: Okay, now, good. Now, let's say that we have now a definition, and the major problem is now the second
41:13:420Paolo Guiotto: step of this, the second part of this program. How do we compute an integral, actually?
41:22:700Paolo Guiotto: And let's start the, focusing on this, so… computing…
41:33:380Paolo Guiotto: Multiple.
41:37:640Paolo Guiotto: integrals.
41:42:660Paolo Guiotto: As anticipated yesterday.
41:45:730Paolo Guiotto: There are two main techniques. One is the reduction formula, and the other is the change of variables that, of course, are not exclusive, so they can be combined, and in most of cases, they will be combined, okay?
41:59:220Paolo Guiotto: Now, since the second one, is something that you already know, you already know the idea of change of variable.
42:07:790Paolo Guiotto: And moreover, if you don't have the first one, even with the second one, you cannot do anything. Let's start with the first technique.
42:17:500Paolo Guiotto: which is actually a new technique, because you have never seen a multiple integral, so you don't have a… you have not seen something of this type. So, let's start talking about the reduction formula.
42:39:320Paolo Guiotto: So, I will start for simplicity.
42:42:300Paolo Guiotto: with the case of an integral for a function of two variables. Then we will extend this for a general case. So, suppose that we want to compute… we want…
42:55:670Paolo Guiotto: to compute.
42:59:430Paolo Guiotto: an integral on a domain D of a function of two variables, XY, say, the XCY.
43:07:270Paolo Guiotto: Now…
43:11:710Paolo Guiotto: So, well, here it's… we don't need for the moment to draw anything, but to think about this integral.
43:23:270Paolo Guiotto: Informally, of course, that's not a proof.
43:27:100Paolo Guiotto: As you know that this integral… well, yesterday, we have seen a bit of this construction, and you remind we started defining inferior sums, superior sums, and basically, these are sums of what?
43:44:860Paolo Guiotto: of these quantities, like DX, K, D, Y, J,
43:51:30Paolo Guiotto: HKJ, where this HKJ, these numbers, are the minimum or the maximum of F on this little rectangle, okay?
44:02:90Paolo Guiotto: So now, for simplicity, I will say that this discrete formula is the integral, and the value of the height is just the value of the function. So what they do is the following.
44:15:740Paolo Guiotto: So imagine that we take our domain.
44:20:250Paolo Guiotto: we divide the values of X and the values of Y in parts, so we divide these in intervals where these are the Xi and these are the YJ.
44:34:990Paolo Guiotto: So we have the domain… integration domain is a subset here.
44:40:290Paolo Guiotto: And let's say that,
44:45:150Paolo Guiotto: this division, of X and Y,
44:50:480Paolo Guiotto: Generates a division, a partition, of the domain into little rectangles.
44:57:710Paolo Guiotto: Let's call this the rectangle RIJ. This is the rectangle.
45:03:740Paolo Guiotto: In such a way that we can say, of course, I repeat, it's not a proof, but it gives the idea perfectly, that this is equal the sum of what? So, the values H are here, just the values of the function at points Xi and
45:22:870Paolo Guiotto: YJ, so at this point, I'm taking the value here.
45:27:310Paolo Guiotto: So I will have a quantity here, and a height here, so this is the value F.
45:33:730Paolo Guiotto: at point XIYJ. I use this as 8 of these parallel epipides here.
45:46:530Paolo Guiotto: And so, this is times D, Xi DYJ. So let's say that this is the integral.
45:54:860Paolo Guiotto: Well, properly, this is an approximation, so if I want to make this a precise argument, you should say, well, that integral will be approximated as much as I like with the sum of this type.
46:08:800Paolo Guiotto: So, it's not an exact identity, but since I'm not doing a proof, who cares, basically, no? So, this is some of all I and J.
46:17:740Paolo Guiotto: Now, if you look at this sum, this is a double sum of certain numbers. Let's call for a moment these numbers, AIJ.
46:27:610Paolo Guiotto: Okay? So we focus a second on this sum, sum of numbers AIJ, where I and J varies from certain index, let's say 0 to capital N.
46:39:770Paolo Guiotto: Now, what… how can… how can we look at this? Now, these numbers, AIJ, can be seen as the entries of a matrix. So, a matrix like A11, A12, A13, etc, A1N.
46:58:780Paolo Guiotto: A21, A22, A23, and so on, A2N.
47:05:820Paolo Guiotto: Until we have the end line AN1, AN2, and so on, ANN.
47:12:220Paolo Guiotto: So that quantity represents the sum of all the entries of this table.
47:17:870Paolo Guiotto: Now, I could decide, why not, to do the sum of all these entries.
47:23:660Paolo Guiotto: By, for example, summing first the first line.
47:29:400Paolo Guiotto: And this produces the value, sum of… you see that these elements are in common the first index, which is 1, so this is the sum of the A1, so we're keeping the same index as J, the sum of J. This is the sum
47:47:480Paolo Guiotto: of A to J.
47:50:680Paolo Guiotto: This is the sum of the… A N. J…
47:56:310Paolo Guiotto: And then, doing the sum of this, plus this, plus this, etc.
48:01:520Paolo Guiotto: The total, the sum of the sums, is the total sum of the numbers AIJ over all I and J.
48:12:360Paolo Guiotto: Or, in alternative, I could decide to sum by columns.
48:18:240Paolo Guiotto: the entries of this, so here I will produce a sum, like, sum of… you see that now this… the common index is the second one, so this is something like AI1,
48:30:340Paolo Guiotto: sum over i, this… Is now… this yields the sum
48:38:270Paolo Guiotto: a I2 of a I, and so on. This will give a sum of a i of the
48:45:930Paolo Guiotto: A, I, N, and then summing these quantities together, I will get what? The sum of all the A's, so again, this thing, yeah.
48:57:60Paolo Guiotto: Okay.
48:58:220Paolo Guiotto: So this means that I could write that the sum of the AIJ over all i and j can be written. Equivalently, in the first way, you see, this is
49:12:870Paolo Guiotto: the… when I do the sum in this way, I do the sum of, this.
49:19:700Paolo Guiotto: If you look at this, these are sum of J,
49:22:890Paolo Guiotto: over J of water. Wait, you see that there is A1J, then A2J, then A3J, then A and J. The generic will be AIJ for I equal 1, 2, 3, 4, and then you sum over the I's.
49:39:980Paolo Guiotto: Or, equivalently, you flip the order, and you get the sum over i, AIJ, and then sum over J.
49:49:660Paolo Guiotto: Okay, now let's apply this, which is just an algebraic,
49:55:790Paolo Guiotto: passage to this particular sum. So, I could say that my integral
50:02:220Paolo Guiotto: of the function f, which is the sum of all IJ of fxij.
50:10:560Paolo Guiotto: DXIDYJ could be… you see, let's take, for example, this one.
50:18:970Paolo Guiotto: This would be… so, I start… I say I start because the first impression you do is not the second… the… the first sum you see. You first do this sum.
50:30:710Paolo Guiotto: you have to compute this, and then you do the sum of the sums. So this is the first, and this is the last. So, let's say that we do first the sum of a J of these numbers, so F, Xi, YJ,
50:44:330Paolo Guiotto: D.
50:45:360Paolo Guiotto: Exile.
50:47:290Paolo Guiotto: DYJ, And then we will sum over I, this.
50:53:80Paolo Guiotto: Now, You notice that this factor here, for example, DXI,
50:59:280Paolo Guiotto: is independent of the summation index J. So it's like a common factor that you can factorize and say that that same factor will multiply the sum of FXIYJ
51:16:670Paolo Guiotto: DYJ, all this will be, at the end, multiplied by DXI, and then I will sum up this over I.
51:27:100Paolo Guiotto: Okay, now… Look at the sum that you have in the round parentheses.
51:33:380Paolo Guiotto: If you, for a second, would forget this,
51:37:170Paolo Guiotto: Exile, forget of… of… or cover.
51:40:630Paolo Guiotto: What's that?
51:41:740Paolo Guiotto: T-F-Y-J-V-A-J.
51:44:910Paolo Guiotto: Exactly for the same reason that this double sum comes from an interval up there, huh?
51:51:960Paolo Guiotto: So, that integral means you are doing sums of, these are volumes that are obtained by doing product 8 by the area of the base. Now, this is, again, a sum of values which are 8 times a length of the base.
52:09:110Paolo Guiotto: So that's an India.
52:11:190Paolo Guiotto: Integral of what?
52:14:570Paolo Guiotto: Well, this should be an integral… it's not an integral in the variable X, because
52:20:790Paolo Guiotto: the sum is done over the Y, so this is an integral in a variable.
52:29:10Paolo Guiotto: in the variable Y, of F, Xi Y.
52:36:560Paolo Guiotto: Okay?
52:39:660Paolo Guiotto: And then what I do is, now, for a second, call this integral. After you have done the integral, you don't see anymore the Y, but this is a quantity, let's call it a capital F, that depends on the Xi.
52:55:300Paolo Guiotto: So what you see here, this black box, you put inside the Xi, you get out a number, which is the integral of FXIY in the variable Y. So this is a sum of i of capital F, Xi.
53:11:190Paolo Guiotto: DXI, which is another integral.
53:14:320Paolo Guiotto: Because of the same philosophy, no? So this is the integral of F of X dx in the variable x.
53:22:430Paolo Guiotto: So, at the end, you would get that this is the integral of the function fx, but FX, now, if you want to see dx, just call this X.
53:33:30Paolo Guiotto: you see that fx is itself an integral, is the integral of little fxy in the variable y. So, you do this first integration.
53:43:940Paolo Guiotto: Once you've done the integration, you don't see anymore the Y. It's only a function of X. You integrate in X, and you see… you don't see anymore the variable X. You get what? In principle, you should get the integral of F.
54:00:150Paolo Guiotto: Now, there is a little point here to understand.
54:04:870Paolo Guiotto: Here we are talking about integras, and when we talk about integras, we need to specify what is the integration domain.
54:12:590Paolo Guiotto: So, what should we write here?
54:15:210Paolo Guiotto: for the range of X and Y.
54:23:360Paolo Guiotto: Now, to understand… to understand this, we may do this,
54:30:640Paolo Guiotto: Well, reminder, let's do… here we do a figure, let's finish this, then we do the bracket.
54:36:480Paolo Guiotto: Here we do… it is convenient to do a figure to understand what is going on. So now, this is a figure in the plane XY, and this, as usual, is the domain V.
54:47:80Paolo Guiotto: So now we are integrating on this domain, so the points we consider for this sum, these points are where are in the domain D, as you can see, no? So, these points, X, Y, are in
55:04:770Paolo Guiotto: Domain D.
55:06:560Paolo Guiotto: So the condition is that the point XY must be in the domain B, because we are integrating on D.
55:15:590Paolo Guiotto: But now, if you look at that integration, if you see this integration, the innermost integration, this is an integration in Y.
55:24:900Paolo Guiotto: X is freezer, it's… you… it's not variable, yeah. The variable is Y. So here, you have to look at this as an integral in Y for an X fixed.
55:36:750Paolo Guiotto: So, which one should I consider?
55:40:250Paolo Guiotto: Yeah. Now, the point… the condition is that this one.
55:44:200Paolo Guiotto: the point XY must be in D. And this, you see that if you take an X like this one.
55:53:100Paolo Guiotto: the set of the Ys for which this point… this point XY is in D, so the points XY that are in D are those on this, say.
56:03:710Paolo Guiotto: vertical segment, I'm coloring in green, huh?
56:07:490Paolo Guiotto: Okay, these are the points XY with WC psi X that are in B.
56:13:490Paolo Guiotto: So the set of the Ys, which is the ordinate, is this one.
56:20:490Paolo Guiotto: is the orthogonal projection on the y-axis. So this in blue, this is… the set of Y's.
56:31:270Paolo Guiotto: such that point XY belongs to D.
56:37:260Paolo Guiotto: with Abshesa X.
56:39:340Paolo Guiotto: That's not always the same set, because you change X, for example, take this X here.
56:46:610Paolo Guiotto: Now, you see that along this vertical line, the points XY are these ones.
56:54:280Paolo Guiotto: So the set of the ordinates of these points is different from the set of the ordinates of these points.
57:00:760Paolo Guiotto: So for each X, for each X, you have a different set of Y's for which the point XY is indeed.
57:10:700Paolo Guiotto: It's like if you have a potato, really, which is not a perfectly regular stuff, and you slice the potato. The slice is never the same, because maybe in some part of the potato is smaller, then you have a small slice, some other part is bigger, you have a big slice.
57:28:760Paolo Guiotto: So, yeah, you're doing exactly the same thing. You are taking your knife, and you slice the domain with, say, which slices parallel to the Y-axis.
57:39:330Paolo Guiotto: Each time you have a slice which is this green light.
57:42:940Paolo Guiotto: Now, here, the domain of integration must be the set of the ordinates, so the Ys of these points. This thing is called the X section of this. So this is…
57:57:110Paolo Guiotto: Let's give a name.
57:58:850Paolo Guiotto: DX is, by definition, the set of Y's, real.
58:05:680Paolo Guiotto: So the set of ordinates, such that D point XY belongs to D.
58:13:100Paolo Guiotto: So, in this figure, this thing here is the DY. I know that it's… maybe I should use the same color, so…
58:22:770Paolo Guiotto: The DX set is a set of Y's.
58:27:380Paolo Guiotto: real numbers, such that point XY belongs to D.
58:34:480Paolo Guiotto: And this blue set is here on the Y axis. So this thing is the BX section. You change X, this set changes, okay?
58:46:320Paolo Guiotto: So, the, solution of this puzzle is that in that innermost integral, you must put, so, integral only of F, X, Y,
58:58:580Paolo Guiotto: DXDY.
59:00:860Paolo Guiotto: Is equal to two integrals.
59:04:390Paolo Guiotto: FXY, this is DY, this is DX, huh?
59:08:740Paolo Guiotto: Well, in the innermost integral, you have to integrate on a set which is the X section of the domain. So, for each value of X, you will have a set DX.
59:23:330Paolo Guiotto: And now, for what is the range for X?
59:27:520Paolo Guiotto: Well, for X, let's do a second figure again of the same thing. So, this is the domain.
59:35:790Paolo Guiotto: Now, notice this phenomenon, that there are acts like this one.
59:40:450Paolo Guiotto: for which there is a slice. So, when…
59:44:800Paolo Guiotto: you cut with your knife, you get something. But there are the other Xs, for example, you take an X down here.
59:54:260Paolo Guiotto: If you take these sacs and use lice.
59:58:30Paolo Guiotto: Do you get any slice here? No, because there is no potato. If I cut, I don't find anything. So in this case, I could say that here, my… the X is this set down here.
00:12:40Paolo Guiotto: So this is the Z DX. But for this one, DX is just… empty. There is nothing.
00:21:230Paolo Guiotto: So, of course, this means that I am Outside of the domain.
00:26:750Paolo Guiotto: So the set of DX I should consider are those for which the X section is non-empty. So this is the condition. X such that DX is non-empty.
00:40:770Paolo Guiotto: Okay, so, let's just put in a form of theorem, then we do the back. Theorem.
00:49:290Paolo Guiotto: If the function f is integral.
00:54:860Paolo Guiotto: on the… then it happens that the integral on D of F, XY, the XDY,
01:06:70Paolo Guiotto: is equal to this double nested integration. Nested because, as you can see.
01:13:50Paolo Guiotto: It works exactly with a specific order.
01:26:40Paolo Guiotto: So, where… DX is the set of Y's.
01:31:630Paolo Guiotto: of the ordinance of points with abshesa X tetra in D.
01:37:970Paolo Guiotto: It seems complicated, but at the end, you will see.
01:40:890Paolo Guiotto: it will become rather mechanical. Now,
01:46:860Paolo Guiotto: In this formula, I said it is a nested integration, because you have to do first in this one.
01:53:380Paolo Guiotto: And then you do the second one. So there are two integrations.
01:58:450Paolo Guiotto: As you can see, each is on one single variable.
02:02:820Paolo Guiotto: So this… This is an ordinary integral.
02:07:240Paolo Guiotto: integral of type, you already know, but literally one variable right here, X is playing the role of the parameter, constant.
02:14:60Paolo Guiotto: Once you have done this round parenthesis, suppose that we are able to compute this integral, at the end, we will see something that is not pending anymore on Y, those bias do integrated, but it is dependent only on X, so it's a function of X that we now have to do another integration.
02:34:100Paolo Guiotto: A second integration.
02:35:860Paolo Guiotto: Once we have done this second disintegration, this theorem says that if the function is integral, this is exactly the value of the W.
02:44:900Paolo Guiotto: This theorem is called the Fabini theorem.
02:57:130Paolo Guiotto: Okay, let's go back and then…
03:01:580Paolo Guiotto: We have just, 5 minutes,
03:15:740Paolo Guiotto: Okay, so this is, called reduction formula.
03:33:960Paolo Guiotto: And there is another formula, which is for the inverse order, but we will see later. I want immediately to show you an example.
03:44:610Paolo Guiotto: to fix the… how it works, this. This is the example 532 from notes. Technically, Very easy.
03:56:30Paolo Guiotto: So, computer… the integral on domain D of cosine X plus Y, DXDY, what the domain D,
04:13:750Paolo Guiotto: is defined in this way. It's the set of points, XY in R2, Such that, huh?
04:22:590Paolo Guiotto: Y is positive, less or equal than X, which is less or equal than pi.
04:36:440Paolo Guiotto: So that's our first double integral with computers, so let's see how it works.
04:41:990Paolo Guiotto: Now, of course, we will apply this formula.
04:45:990Paolo Guiotto: Now, an important, very important fact that we will return a lot on this is that we need to know that f is integral to use this formula.
04:56:210Paolo Guiotto: Otherwise, we are not alone. I will show you not today.
05:00:540Paolo Guiotto: an example that if you are irresponsible in applying this, you could… you could get a disaster, okay? So, we must know if the function is integral. In this case, we are lucky, because if you look, the function f…
05:16:530Paolo Guiotto: XY is cosine of X plus Y.
05:20:950Paolo Guiotto: So, this is a function well-defined, continuous on the full plane at 2.
05:27:880Paolo Guiotto: So it will be continuous on whatever is the domain. So F is continuous also on the domain.
05:35:220Paolo Guiotto: And moreover, the domain is clearly closed, because it's defined by weak inequalities, so D is closed.
05:44:460Paolo Guiotto: And, clearly, also, if you just see the constraints, you see that the two coordinates, both X and Y, are positive and less than pi, so they are bounded.
05:55:40Paolo Guiotto: And so the domain is bounded.
05:59:990Paolo Guiotto: And therefore, this means that D is compact.
06:04:560Paolo Guiotto: Compacte.
06:06:100Paolo Guiotto: And, we said above that,
06:09:450Paolo Guiotto: If you have a continuous function on a compact domain, you are sure that the function is integral.
06:14:400Paolo Guiotto: So, F… Ease… Integrable.
06:19:100Paolo Guiotto: on D.
06:21:240Paolo Guiotto: And therefore, reduction formula
06:27:660Paolo Guiotto: applies.
06:29:580Paolo Guiotto: Okay, so now we can concretely compute the integral.
06:35:50Paolo Guiotto: So… integral on D of cosine X plus Y, the XDY.
06:44:230Paolo Guiotto: Now, since we have seen only one reduction formula, this one, but there is another one.
06:49:860Paolo Guiotto: a specular formula. This one starts computing integral in Y variable, then second integration in X variable. But there is a dual formula, where you start integrating in X, and you continue in Y.
07:04:480Paolo Guiotto: Okay? However, suppose that we know only this one, and let's see what happens if we apply. So let's write, for a second, just copy. So here we have the function cosine X plus Y, so the first integration here is y.
07:21:520Paolo Guiotto: So we have here the famous X section, let's see, what is it? And the second integration is in X for DX, such that DX is non-empty.
07:33:230Paolo Guiotto: Now, since this is the first time, and the set is not particularly complicated, to understand what are these DX sections.
07:42:680Paolo Guiotto: Let's do a figure of the domain.
07:45:410Paolo Guiotto: Later, we will learn how to do this without watching the domain, which is, of course, impossible if the domain is complicated, or is the domain in R3, and things like that. But in this case, it is easy, because this is the plane XY.
08:04:60Paolo Guiotto: Domain is made by all these inequalities. It says Y greater or equal than 0, less than or equal than X, which is less or equal than Y. What do they mean?
08:14:420Paolo Guiotto: So, first, Y must be greater or equal than zero. So, the Y
08:19:160Paolo Guiotto: We are above the x-axis, so we throw away this spark. This is Y negative.
08:25:740Paolo Guiotto: Then, it says that X must be less… greater than Y, or Y must be less than X. Y equal to X is along this line. This is the line Y equal X. Of course, the line continues down here, but who cares what happens, because we already discard.
08:44:60Paolo Guiotto: Now, why less than X means below that line, so we discard all part of the plane which is above that line.
08:54:810Paolo Guiotto: Then we have another condition here, X less or equal than pi. X equals pi, let's say it is here. X equal pi is a vertical line, by the way.
09:06:640Paolo Guiotto: points with absence equal pi, and we take X less so at left, so we discard this part here.
09:14:580Paolo Guiotto: Now, conditions are over, so what remains is the domain? The domain is this triangle here.
09:22:520Paolo Guiotto: So, whatever is inside. That's the integration domain.
09:26:819Paolo Guiotto: Okay?
09:29:450Paolo Guiotto: Okay, now we can… I clean up everything, and I will just keep all in the domains. So, I really do the… I redo the figure.
09:37:819Paolo Guiotto: By drawing only the domain.
09:42:130Paolo Guiotto: So we have this.
09:43:720Paolo Guiotto: Triangle. This is pi.
09:47:300Paolo Guiotto: By the way, this is Y equal X, so this will be also pi.
09:52:880Paolo Guiotto: And the domain is this one.
09:55:550Paolo Guiotto: Okay, now let's see what are the X sections. What are these X sections? I say that you get the X sections slicing the domain along lines parallel to the Y axis. So it means you fix an X.
10:12:280Paolo Guiotto: As you can see here, there are basically two cases. One is the case when I take an X here, or here.
10:20:760Paolo Guiotto: If I now slice the domain along this, lie, you know?
10:28:490Paolo Guiotto: Oh, this one.
10:31:860Paolo Guiotto: Do you see on these vertical lines points of the domain D? No. So what does it mean? That, in this case, the set DX, which I remind you, is the set of Ys, so of ordinates, of points with the abshesa X fixed, that are in D,
10:52:60Paolo Guiotto: So, for this axa, for the blue X, I choose there.
10:56:10Paolo Guiotto: There is no Y, so DX is empty. So the conclusion for this X is DX equal empty set when X is… here means x less than 0, or X greater than pi, you see?
11:12:430Paolo Guiotto: Now, what happens if I take an X between 0 and pi, like here?
11:18:400Paolo Guiotto: In this case, you see that there is something on this vertical line.
11:23:550Paolo Guiotto: And the slice is this one, okay?
11:28:200Paolo Guiotto: Now, the set DX is not the green set, because that's a set of points, so there are two coordinates. The set DX is rather the set of Ys, of the ordinates of these points. So I can say that for X between 0 and pi.
11:48:430Paolo Guiotto: My DX is this segment.
11:51:620Paolo Guiotto: You see?
11:55:10Paolo Guiotto: So, definitely, it starts with the value 0.
11:59:280Paolo Guiotto: And it ends where? What is this number?
12:03:300Paolo Guiotto: is X itself, because you are on the line Y equal X. So that's exactly X. So the X section is the interval 0 to X.
12:13:550Paolo Guiotto: You see?
12:14:860Paolo Guiotto: Okay? Now, let's go back to the formula.
12:18:780Paolo Guiotto: And this is the most of the complication with the multiple integrals, to do this job properly. The remaining part of the calculation may be tricky or not, but it's something… somehow elementary.
12:32:60Paolo Guiotto: But this part is fundamental, because if you do the wrong description of the domain.
12:39:230Paolo Guiotto: You do the wrong integral, and perhaps you end in something that cannot be computed, or…
12:45:590Paolo Guiotto: Whatever. So let's say integral of cosine X plus Y, DXDY,
12:52:140Paolo Guiotto: Now we have all the, all the informations needed, because it says, we have to integrate on the X for which the X is non-empty. What are these X?
13:05:00Paolo Guiotto: What are the X for which the X is non-empty?
13:09:610Paolo Guiotto: they are these DX for which DX is non-empty, because for these other X, the DX is empty. So the range for X is from 0 to pi.
13:21:720Paolo Guiotto: That's the last integral, the integral in X. Then we have an integral in Y.
13:26:740Paolo Guiotto: What is the range for Y? When x is between 0 and pi, the range for y is that one, is 0 to X.
13:36:690Paolo Guiotto: of what? The function cosine X plus Y.
13:40:940Paolo Guiotto: Now, this is a properly set integral we can now compute, okay?
13:46:680Paolo Guiotto: Do you see or not? Should I repeat?
13:52:580Paolo Guiotto: Okay, so… First, have you understood that the domain is that one, okay? Now…
13:59:780Paolo Guiotto: The formula is this one, and to apply this formula, I need to determine these sections, the exact, and for which acts they are dominated, because this determines the domain that they have to put here, and the domain for the exit here.
14:15:710Paolo Guiotto: This is the domain of Y, this is the domain of X.
14:19:30Paolo Guiotto: Now, the X sections are, what? A set of Y, which is written here as the definition.
14:27:730Paolo Guiotto: So, sets of ordinates.
14:30:440Paolo Guiotto: of points with the Abshysa X fixed that are in B. So it means that you take an abshesa X, and you cut your domain along, or you intersect, if you don't want to, a battery language, and you…
14:46:170Paolo Guiotto: You take a vertical line with upsis, so parallel to the y-axis. You do the intersection with the domain.
14:54:700Paolo Guiotto: In certain cases, you will get empty. For example, yeah, you see that style of this one?
15:01:460Paolo Guiotto: they have no common points with the red set. So this means that there is no Y of points that are in the red set here. So this means
15:13:580Paolo Guiotto: X section is empty, for which X for those X that are here, negative, or here, greater than 5?
15:23:310Paolo Guiotto: While for the X that are between 0 and pi, so we are here, you see that there is an intersection, that green section.
15:32:860Paolo Guiotto: Now, this set DX is the set of Ys of these points, DX in picture.
15:38:780Paolo Guiotto: It's X.
15:40:330Paolo Guiotto: Now, you see that from this figure, the set of Ys is the interval from 0 to X, because the last Y is this one, which is equal to X.
15:55:480Paolo Guiotto: So now we know that.
15:57:40Paolo Guiotto: The set of X for which the X section is non-empty is between 0 and pi.
16:04:520Paolo Guiotto: And this means that these, out exterior
16:10:680Paolo Guiotto: integration is on the range for X for which the X is not empty, so the range is X between 0 and pi, you see here, okay?
16:22:650Paolo Guiotto: The innermost integration is the integration on the X section, that for those x is the interval 0 to X. So, that's why you see 0 pi here and 0x here.
16:35:720Paolo Guiotto: Now, we completed the calculation, and as you will see now, it becomes a pair of one-variable integrations, so I have
16:46:300Paolo Guiotto: First operation to do is this one, okay? So, this, for the moment, remains integral 0 pi. Now, I have to look at this as a function of Y. X is just freeze, the parameter. So, how do you compute the integral from 0 to X of cosine x plus y in dy?
17:05:920Paolo Guiotto: So, as usual, you use tools of one variable, calculus. You try to see this as the derivative of something impossible. Now, this is the derivative with respect to what? The variable is y, so you, you say derivative with respect to Y of sine
17:23:770Paolo Guiotto: of X plus Y. In fact, if you do this derivative, you get cosine of X plus Y times the derivative of the argument with respect to Y at S1.
17:33:930Paolo Guiotto: So it's exactly there. And therefore, by the fundamental formula of integral calculus.
17:45:660Paolo Guiotto: Once you… the formula is integral from A to B of F prime T, dt equal FB minus FA,
17:56:140Paolo Guiotto: That's the formula. So, the, I have to take that function, sine, of X plus Y.
18:05:880Paolo Guiotto: Evaluate this function at Y equals 0, at Y equals X, and do the difference between these two values.
18:16:100Paolo Guiotto: What I get is, for Y equals X, this becomes sine of 2X.
18:23:370Paolo Guiotto: sine of 2x minus for Y equals 0, I get sine of X. So that's the value this
18:33:00Paolo Guiotto: is the value of these integral. As you can see, there is no more NEY.
18:40:190Paolo Guiotto: If you have a Y written here, it means it is wrong. 100%, no possibility.
18:46:740Paolo Guiotto: Okay?
18:48:30Paolo Guiotto: Because that's why it has been integrated and disappeared after the integration.
18:53:410Paolo Guiotto: Okay, so now this is the quantity we have to plug back here.
18:58:690Paolo Guiotto: In place of that integral. So we can say that, well, let's put the mark here.
19:06:430Paolo Guiotto: Star.
19:09:160Paolo Guiotto: So now, the second integration is integral 0 to pi of… we computed the innermost integral. We got sine of 2x minus sine x.
19:25:470Paolo Guiotto: DX. And now we do this and this other integration, which is
19:30:10Paolo Guiotto: Elementary, also this one, because sine 2x is more or less derivative of cosine 2X, maybe divided by 2, maybe with a minus. In fact, if I do the derivative of this, I get the minus, derivative of cos is minus sine.
19:46:430Paolo Guiotto: of 2X, then derivative of the argument is 2 divided by 2. As you can see, everything simplifies, it gets sine 2X. So the first will give the evaluation of minus cos 2x divided by 2.
20:00:420Paolo Guiotto: between x equals 0 and X equals pi minus. For the integral of sine, if you want, you keep a plus minus sign, which is the derivative
20:12:290Paolo Guiotto: of cosine.
20:15:390Paolo Guiotto: So we have a plus the evaluation of cos x between x equals 0 and X equals pi.
20:24:60Paolo Guiotto: So, this is minus 1 half times.
20:28:330Paolo Guiotto: Cosine of 2x at x equal pi. So cosine of 2 pi, which is 1, minus cosine of 0, which is 1.
20:37:990Paolo Guiotto: plus the second one is cosine of X at x equals phi, which is minus 1, minus cosine at x equals 0, which is 1. So, at the end, this is 0, and this is minus 2.
20:53:330Paolo Guiotto: So what is this minus 2?
20:56:660Paolo Guiotto: That's the value of the integral.
21:01:70Paolo Guiotto: Okay?
21:11:140Paolo Guiotto: Okay, so let's,
21:15:900Paolo Guiotto: let's add this now. So we have seen the reduction formula.
21:21:440Paolo Guiotto: That says, if you want to do this double integration, provided you know that the function is integral, you can do the double integration by doing two nested one-variable integrations, integrating first in Y and second in X.
21:35:910Paolo Guiotto: I told you there is an alternative formula that says you can do the same by doing integration first in X and second in Y. So, we have a second theorem. Actually, it's a unique theorem.
21:49:240Paolo Guiotto: So… Say, it is possible
21:57:330Paolo Guiotto: So… Dude.
22:02:260Paolo Guiotto: nested.
22:04:820Paolo Guiotto: Integration.
22:07:70Paolo Guiotto: Fast.
22:10:130Paolo Guiotto: in X, then… in Y. So we have this, say, TRM, still the same TRM.
22:21:170Paolo Guiotto: if F is… integral.
22:25:840Paolo Guiotto: on P.
22:27:830Paolo Guiotto: Then,
22:30:680Paolo Guiotto: the integral on D of FXY DXDY.
22:37:690Paolo Guiotto: Is equal to… so…
22:39:910Paolo Guiotto: to… you don't need to memorize the formula. It's a double operation, double integration of what? The function?
22:47:660Paolo Guiotto: Now, you decide which is the variable you do integration first, let's say X here.
22:53:770Paolo Guiotto: So, here, the domain is the domain of X. What is the domain of X? Let's say that the philosophy is the same. We look at the domain D in the plane XY,
23:09:30Paolo Guiotto: Now, since here it is Y to be fixed, so imagine that you fix a Y, and you look at points.
23:17:760Paolo Guiotto: that are in the domain D on that line, so you see that you get a segment for this figure.
23:24:160Paolo Guiotto: So, these are points of type XY, but Y is fixed. So, the set of X for which the point XY is in the domain is now a subset of the horizontal axis that we will call X section. Maybe we will use a different
23:41:870Paolo Guiotto: notation to do not confuse the two. DX is a set of X in real line, such that point XY belongs to D. It's similar, it's the same of the previous one. It's like now we slice along,
23:59:320Paolo Guiotto: with the knife parallel to the x-axis.
24:02:520Paolo Guiotto: So this is the stat, the X I have to put here.
24:08:510Paolo Guiotto: Of course.
24:11:900Paolo Guiotto: It's not the X3.
24:14:20Paolo Guiotto: I should kill you. Why?
24:18:790Paolo Guiotto: Who's…
24:19:750Paolo Guiotto: Okay, so this is the set DY, sorry, okay? Y is fixed, so DY is a set of X, DX is a set of Y. And now, the second integration will be on Y is such that DY is non-empty.
24:34:200Paolo Guiotto: DWOP.
24:38:100Paolo Guiotto: Okay, so it means that, basically, we have two reduction formulas.
24:42:420Paolo Guiotto: Does it mean that I have to use both? Not at all, because if you have to compute an integral, you just need to use one of the two, okay?
24:51:860Paolo Guiotto: But also means that there is a little danger behind this, because you don't know which one to use.
24:58:870Paolo Guiotto: Maybe one of the two is easier, or maybe one of the two is even impossible.
25:04:280Paolo Guiotto: So there is no way to… to know this, except and discover by your own which one works. But there are situations in which the two are not together possible.
25:16:690Paolo Guiotto: Since we have just few minutes, I would like to…
25:23:250Paolo Guiotto: Do another example, just to fix ideas, and leave you with something to do for the next time.
25:34:460Paolo Guiotto: Zit?
25:36:840Paolo Guiotto: Okay, let me check, so…
25:40:870Paolo Guiotto: I think… I think that, the exercise you should do is, 5-7-1…
25:58:10Paolo Guiotto: Let's say that, you could do… all the exercises, Maybe, except…
26:06:550Paolo Guiotto: two of them that requires some… some… some more consideration we have not yet done. So, do all… Accept.
26:19:430Paolo Guiotto: Number 7 and 9. So, for example, we do the number 1,
26:29:130Paolo Guiotto: It is integral for, well, let's write the domain. Y is between 0, 1,
26:36:10Paolo Guiotto: And X is between 0 and 1 minus Y squared.
26:42:520Paolo Guiotto: the function is Xe2Y, DXDY.
26:49:920Paolo Guiotto: Okay, so… First of all, the function F
26:54:630Paolo Guiotto: is the function we have to integrate is X into Y, and it is definitely a continuous function everywhere, so in particular, it will be on the integration domain, let's call it D.
27:08:520Paolo Guiotto: D is clearly closed because it is defined by weak inequalities that involve continuous functions, so it is closed.
27:19:250Paolo Guiotto: And, it is evident that it is also bounded, because the Y is between 0, 1, X is between 0 and 1 minus Y squared, that is less than 1, okay? So, D is also bounded.
27:34:300Paolo Guiotto: If you want, you can also see what is this domain, because we have… Y is between 0 and 1, and then X is between 0 and 1 minus Y squared. What does it mean?
27:48:590Paolo Guiotto: Well, this means that if I look at the y-axis as the horizontal axis, 1 minus Y squared is a parabola pointing downward, but in the y-axis, so it is like this.
28:04:840Paolo Guiotto: Okay, this is the parabola X equal 1 minus Y squared. Since you are taking X less, it means below that parabola, so at the left. So you discard all this part here.
28:19:790Paolo Guiotto: And also, you are taking Y between 0 and 1, so the Y is between 0 and 1, so…
28:28:50Paolo Guiotto: 0Y equals 0 is the axis, so Y greater than 0 means you discover this part. And Y below 1, well, exactly this point is Y equals 1, X equals 0.
28:42:70Paolo Guiotto: So, Y equals 1 is this one, and you discard the part.
28:46:510Paolo Guiotto: above this.
28:48:920Paolo Guiotto: X is also positive, I forgot there are lots of conditions here, so you don't have to miss anything. So at the end, you see that the domain is this thing.
29:01:30Paolo Guiotto: this, portion of parabola, okay?
29:05:860Paolo Guiotto: So, it is clear that it is boundary.
29:08:950Paolo Guiotto: So this means that F is integral.
29:13:590Paolo Guiotto: on… Dao.
29:17:260Paolo Guiotto: Let's start doing the integration, applying the reduction formula. So, the integral of F, because of reduction formula, now we have two options. Which one?
29:28:540Paolo Guiotto: Now, one of the ways to choice is,
29:32:630Paolo Guiotto: how is described the domain, because for this domain, you see that we could easily determine the X sections or the Y sections. You see that some have common points, some other we don't have. But forget the figure. Let's start looking only at the conditions.
29:52:670Paolo Guiotto: What this condition says, they say that you are in domain D when Y is between 0 and 1, and for those Y, X must be between 0 and 1 minus Y squared.
30:05:870Paolo Guiotto: You see? This is already saying that this is the description of these sections. You see the section, the Y sections? Because it is saying, for Y between 0, 1, you have this. For Y outside, you have nothing, because Y cannot be outside.
30:25:880Paolo Guiotto: So, this is saying that you should do first the integration in X of your function.
30:32:920Paolo Guiotto: E to X in Y, and then the integration in Y.
30:36:710Paolo Guiotto: Now, for X, when Y is between 0, 1, so that's Y between 0, 1, X is between 0 and 1 minus Y squared, between 0 and 1 minus Y squared. And here, we have…
30:52:850Paolo Guiotto: The… the nested integration properly set.
30:57:340Paolo Guiotto: Okay, now we compute this. So, as you can see, this is an integration in X, so E2Y is a constant for X, so I write it outside integry, then integral 0 to 1 minus y squared of X dx, then integrated in Y.
31:16:140Paolo Guiotto: Well, this is easy because this is the derivative of X squared divided 2.
31:21:640Paolo Guiotto: with respect to X. So this will be integral 0 to 1 e to y. Then we have the evaluation of X squared over 2 between x equals 0 and X equals 1 minus Y squared.
31:38:730Paolo Guiotto: And after that, we have to integrate in Y.
31:41:680Paolo Guiotto: Now, this can be written outside, it's a factor of 1 half, so for x equals 1 minus Y squared, we get 1 minus Y squared squared.
31:52:930Paolo Guiotto: For x equals 0, we get 0. So, after the first integration, we get 1 half integral 0, 1, e, 1 minus y squared squared.
32:07:00Paolo Guiotto: And this now is a more or less easy integral that you can compute by some application of integration by parts. So this is 1 minus 2Y squared plus Y to power 4,
32:20:670Paolo Guiotto: So you do integrals 0, 1, e to y, y to power 4.
32:26:150Paolo Guiotto: EY.
32:28:10Paolo Guiotto: I… I do this just to illustrate, but you should… you should finish by… by yourself. This, so we use this as derivative. This is the derivative of e to y.
32:39:640Paolo Guiotto: So this becomes the product e to y, y power 4, between y equals 0 and y equals 1, minus integral 0 to 1. Now, derivative moves from the exponential to the power, so 4Y cubed.
32:59:10Paolo Guiotto: DY This evaluation for Y equals 1, you get E,
33:05:70Paolo Guiotto: for y equals 0, you get 0. So minus 4, now we have the integral 0 to 1 of Y cubed e to y. We repeat again the integration by parts here. So we look at this as the derivative of…
33:20:370Paolo Guiotto: itself, so we have E minus 4, then parentheses, evaluation of Y cubed
33:29:970Paolo Guiotto: e to y between 0 and 1 minus integral 01, then derivative is 3Y square e toy.
33:43:780Paolo Guiotto: So this is, for y equals 1, is again E minus 3 integrals 0, 1, y square, e toy, etc. You can finish.
33:54:710Paolo Guiotto: Okay?
33:56:180Paolo Guiotto: It takes, two minutes to complete this calculation, okay?
34:02:930Paolo Guiotto: Good. Do the exercises, I left you, and
34:07:590Paolo Guiotto: on Friday, we will start solving some of these, okay?
34:12:400Paolo Guiotto: It's very important that you think about to the, profile.
34:17:270Paolo Guiotto: parameterization of the integral when you apply the reduction for it. That's the most important thing you have to think about. So, if you do not feel confident, use the figure, or if you feel confident, try to work directly with the inequalities that describe to me.
34:35:150Paolo Guiotto: Okay, let's see on Friday what we've done.
34:44:240Paolo Guiotto: What's going on?