AI Assistant
Transcript
00:34:390Paolo Guiotto: Good morning.
00:40:260Paolo Guiotto: We start from some of the exercises I left you, the five 7… 1… number… let's start.
00:51:20Paolo Guiotto: What we did, the one?
00:55:350Paolo Guiotto: Okay, let's start with number 2.
01:00:680Paolo Guiotto: So, we have integral of X over 2 plus Y in the XDY,
01:08:370Paolo Guiotto: Over this domain, which is described by these inequalities.
01:15:870Paolo Guiotto: 0 less or equal than Y, which is less or equal than 1 minus X squared.
01:26:70Paolo Guiotto: So, here we have this function.
01:29:190Paolo Guiotto: F, X, Y.
01:32:70Paolo Guiotto: which is, X divided 2 plus Y.
01:37:20Paolo Guiotto: Now, this is defined and continuous, provided Y is not equal to minus 2. So, it is a continuous function on R2, except to points where Y is equal to minus 2.
01:53:470Paolo Guiotto: If you want… You see this in plain XY.
02:01:360Paolo Guiotto: Y equal minus 2.
02:04:210Paolo Guiotto: Means that it is an original line down here.
02:09:320Paolo Guiotto: These points are not in the domain. Every… everything else is in the domain, and the function is continuous.
02:16:20Paolo Guiotto: Now, what is the integration domain? This is important because for the moment, to compute the integral, we're going to use, of course, the reduction formula that requires we know the function is integral, so…
02:30:720Paolo Guiotto: What is the domain? The domain is described by this,
02:34:950Paolo Guiotto: these conditions. So this is the set of points for which Y is greater or equal than zero. So we are above the x-axis, so we may discard all this part down here, so in particular that…
02:48:680Paolo Guiotto: The bad points, sir.
02:51:130Paolo Guiotto: And it is below 1 minus X squared. So, Y equal 1 minus X squared is a parabola, like that.
03:01:110Paolo Guiotto: This is Y equal 1 minus X squared. This means
03:06:110Paolo Guiotto: points on this parabola, they have the Y, which is equal to 1 minus X squared. So since I have to be below that parabola, I have to discard whatever is above the parabola.
03:18:500Paolo Guiotto: So, this part of the Cartesian plane here.
03:23:850Paolo Guiotto: And therefore, what remains is this,
03:28:830Paolo Guiotto: Region… well, let's maybe use another color.
03:34:540Paolo Guiotto: this region here.
03:40:380Paolo Guiotto: So, all these… region is the domain D where we have to compute integral.
03:48:920Paolo Guiotto: Now, the domain D… Which is the set of points XY in R2, Wow.
03:57:470Paolo Guiotto: Y is greater or equal than zero, and less or equal than 1 minus X squared.
04:03:860Paolo Guiotto: Is clearly closed, standard factor.
04:11:40Paolo Guiotto: And, you can see from the figure.
04:14:560Paolo Guiotto: And bounded.
04:19:760Paolo Guiotto: Okay?
04:21:160Paolo Guiotto: evident.
04:25:160Paolo Guiotto: from V… figure…
04:30:260Paolo Guiotto: But if we don't see the figure, we can, we can deduce that this set is bounded. Let's see how.
04:39:520Paolo Guiotto: without…
04:43:680Paolo Guiotto: the figure… We may notice.
04:53:380Paolo Guiotto: That.
04:54:890Paolo Guiotto: Well, if the point XY belongs to D,
04:59:40Paolo Guiotto: Well, this means that Y must be greater or equal than 0, less or equal than 1 minus X squared.
05:05:290Paolo Guiotto: So let's see what we can draw from this. First, we can draw that Y is greater or equal than zero, and that's the first fact.
05:14:850Paolo Guiotto: Then, from the other constraint, y less or equal than 1 minus X squared.
05:20:640Paolo Guiotto: Since this quantity will be less than 1, because it is 1 minus a positive number, so it is definitely less than 1, so together, these two gives Y between 0 and 1, so Y is bounded.
05:37:530Paolo Guiotto: And where do I get the bound for X?
05:44:860Paolo Guiotto: the exploratory.
05:47:480Paolo Guiotto: do easier.
05:49:960Paolo Guiotto: You see, 1 minus X squared must be greater than Y, which is greater than 0.
05:57:430Paolo Guiotto: So, in particular, from… still from the definition of the domain, I read that 1 minus X squared must be greater or equal than 0. This means that X squared must be less or equal than 1, this means that X is between minus 1 and 1.
06:15:230Paolo Guiotto: And that means that both X and Y X and Y are bundled.
06:23:550Paolo Guiotto: So… The point XY, is bound.
06:31:60Paolo Guiotto: Of course, it is more difficult to read this from the inequalities, because intuitively.
06:40:610Paolo Guiotto: If we can see a figure, at least if we see figures, because if we are blind, it becomes another story. So… but in any case, whenever we are blind, because we cannot draw figures, we have to learn how to draw these information from the constraints.
06:58:760Paolo Guiotto: Here I want to point you a warning.
07:02:770Paolo Guiotto: Because that's a very, very, very, very frequent error that then will become a big error when we describe the integration.
07:14:440Paolo Guiotto: Because apparently, from this, discussion, it seems that XY belongs to D, finally, if Y is between 0, 1, and that's between minus 1 and 1.
07:26:50Paolo Guiotto: So, one would… would say that, so, it seems… that, XY.
07:38:270Paolo Guiotto: belongs to the if and only if. Actually, I've not written if and only if, I've written only this arrow, okay? So that's a… that's the mistake.
07:49:160Paolo Guiotto: So, X is between minus 1 and 1.
07:53:960Paolo Guiotto: And Y is between 0 and 1.
07:57:90Paolo Guiotto: You understand that this is not correct, because if you draw this set, what is the set of points XY for which X is between minus 1 and 1, and Y is between 0, 1?
08:11:660Paolo Guiotto: No, it's not a circle.
08:15:640Paolo Guiotto: Yeah, almost. Rectangle. Yeah, because… because you have, from minus 1 to 1, for the X,
08:27:330Paolo Guiotto: and for the Y from 0 to 1.
08:30:540Paolo Guiotto: Now, points who have the abscessa between minus 1 and 1, and the ordinate between 0, 1, they are a rectangle.
08:39:590Paolo Guiotto: And that clearly is not the domain, because the domain is that region delimited above by the parabola y equals 1 minus square. So, this means that it is wrong.
08:50:820Paolo Guiotto: This is important, because then we will… when you will, compute this integral, and you have to identify where X varies, Y varies, the sections, no? If you… if you have in mind the wrong domain, you will have in mind
09:06:400Paolo Guiotto: You will write down the wrong parameterization for the integral.
09:12:420Paolo Guiotto: This is wrong.
09:20:70Paolo Guiotto: Okay? So be careful. In fact, from the logical point of view, it is clear what is going on. We say, if XY is in B, then
09:31:130Paolo Guiotto: X must be between minus 1 1 and Y between 0, 1, but I'm not saying that if X verifies this condition and why that one, I am in there. I'm not saying if and all if. You understand what is the point?
09:43:430Paolo Guiotto: Because the logical argument here is, I want to get that both X and Y are bound, right?
09:50:10Paolo Guiotto: I don't care. Of course, the domain is not a rectangle, it has a different shape. So I said, okay, let's use the inequalities to get that there is a bound for both X and Y, but this does not mean that I will have all these points, because my domain is, in fact, a little bit smaller. It's this thing.
10:10:90Paolo Guiotto: This is the domain, no?
10:13:400Paolo Guiotto: So you see there are points that have X between minus 1 and 1, and Y between 0, 1, like this point here. This point here is not in D.
10:26:170Paolo Guiotto: So, there are points who verify, these conditions, X between minus 1 and 1, Y between 0, 1, but they are not in D. So, they are not points of D.
10:38:540Paolo Guiotto: Okay, so in any case, now we know that the func… the domain D is closed and bounded, so F is continuous on D,
10:48:00Paolo Guiotto: Because we observed that the bad points, which are in red in that figure, are not in D, so F is continuous in D, and D is compact.
11:02:320Paolo Guiotto: So, under these conditions, we know that F is integral.
11:11:330Paolo Guiotto: on the… So, the integral on D.
11:16:710Paolo Guiotto: of F.
11:17:780Paolo Guiotto: Makes sense. And not only this, because last time we said.
11:23:970Paolo Guiotto: If the function is integral on D, then we can use the reduction formula. Otherwise, we are not authorized. This morning, I will show you that if it is not integral.
11:35:470Paolo Guiotto: the reduction formula could lead to a disaster.
11:39:660Paolo Guiotto: Okay? But let's stay on this. So, and… Let's activate that.
11:46:940Paolo Guiotto: And, reduction.
11:51:100Paolo Guiotto: formula.
11:54:600Paolo Guiotto: applause.
11:57:700Paolo Guiotto: Now, I told you, so last time we had to close the class because the time was over.
12:04:490Paolo Guiotto: I told you that there are two different formulas, and there is no one which is better than the other.
12:14:930Paolo Guiotto: One says,
12:17:40Paolo Guiotto: you integrate first in the variable X and then in the variable Y, and the other says you integrate first in Y and then in X.
12:24:830Paolo Guiotto: Which one is better?
12:27:120Paolo Guiotto: Well, this depends on the problem, because basically the problem is that you have, first of all, to describe these actions.
12:36:660Paolo Guiotto: So, if you, for example, are not able to describe these sections, because you are not… you are not able to say, for why fixed? What is the range for X? You do not arrive to do that.
12:51:930Paolo Guiotto: It's completely useless to write this formula, because you cannot write anything.
12:55:860Paolo Guiotto: Okay, so you have to move to the other. There are cases for which both possibilities are available, but maybe then you have troubles in computing the integral, because you know that we cannot compute any integrals. We are very limited in the possibilities.
13:13:240Paolo Guiotto: So, there are, many steps, and there is no recipe. You have to try.
13:18:760Paolo Guiotto: With a little bit of experience, you will learn what is better and what is not, no? For example, in this case.
13:28:100Paolo Guiotto: Something is already written here in the description of the domain.
13:33:150Paolo Guiotto: So, let's, for a moment, do graphically the description. Let's see what are the two formulas. So, that one that integrates first in X and then in Y, and that one that integrates first in Y, and the second in X. So, let's do this just for illustrative purposes.
13:51:00Paolo Guiotto: So, we can write that the integration on D of F is… so, to get used to apply correctly these formulas, I would say that the best thing to do is to write the two integrals.
14:07:960Paolo Guiotto: The function, it's always there, so in this case, the function is… what?
14:14:370Paolo Guiotto: X divided 2 plus Y.
14:19:90Paolo Guiotto: Now, let's say that this is the case where we integrate first in X, and then in Y.
14:27:120Paolo Guiotto: And the last step is to write the range, and the second, the formula will be the opposite. So the same function, X divided 2 plus Y. Here, we will integrate first in Y, and then in X.
14:40:470Paolo Guiotto: Now, let's take the domain, let's write here.
14:46:500Paolo Guiotto: The domain is, this, region.
14:51:730Paolo Guiotto: Here, so we have, well…
14:57:100Paolo Guiotto: The range for X is minus 1, and for Y is 01, but as I said, we do not take all possible X and Y with that range.
15:06:790Paolo Guiotto: So this is minus 1101.
15:10:280Paolo Guiotto: This is the parabola y equal 1 minus X squared.
15:14:150Paolo Guiotto: And the domain is this one.
15:16:900Paolo Guiotto: Now, let's do, maybe…
15:21:100Paolo Guiotto: the second one first, because what we have to write here is the so-called X section.
15:29:290Paolo Guiotto: And the second integration, the last one, which is not the first that you see. The first, apparently, that you see is this one, no? This seems to be the first. No. The first thing you do is this one, okay?
15:41:410Paolo Guiotto: So here we have to take DX such that the DX section is non-empty.
15:45:990Paolo Guiotto: And similar here, there is DY, and this is the integration on Y's for which the DY is non-
15:53:340Paolo Guiotto: So let's start with the second one.
15:55:710Paolo Guiotto: Because here I see that I have to choose an X on the x-axis, and take lines parallel to the Y axis, and cut the domain.
16:07:380Paolo Guiotto: So, you see that if I take this X here, there are no points of the domain on this blue line.
16:13:50Paolo Guiotto: So I would say that for this, this section is empty. There is no point, there is no section.
16:19:500Paolo Guiotto: So, as the common says, no Martin Lino Party is the same thing. So, BX here is…
16:27:650Paolo Guiotto: empty when X is where. So you see that when X is here, so at left and minus 1,
16:36:130Paolo Guiotto: Or when X is also here, at right of 1.
16:41:240Paolo Guiotto: Or, when X is between minus 1 and 1, I have something. I have an intersection.
16:46:820Paolo Guiotto: So, the intersection is this one, but this is not the section. The section is what? The DX section is a set of Ys. So, the section is a set of the other coordinates, such that
17:02:190Paolo Guiotto: the point, XY, is in the domain.
17:05:150Paolo Guiotto: So I have to think to what are the coordinates of these points in red, the ordinates, sorry, of these points in red. So, what can be said here? I know that if this is X, this point has a Y, which is 1 minus X squared.
17:25:300Paolo Guiotto: So the range will be… We'd be… Brother.
17:32:560Paolo Guiotto: 0 to 1 minus X squared, for which X? For X between minus 1 and 1.
17:42:880Paolo Guiotto: Okay, so now this means that this second parameterization now is at hand, because I can say that integral on D of the function F can be written as this double integration. The function
17:56:700Paolo Guiotto: is X divided 2 plus Y. This is first in Y, and then in X.
18:03:30Paolo Guiotto: Now, what are the X for which DDX is non-empty?
18:07:350Paolo Guiotto: Are those in minus 1-1.
18:09:980Paolo Guiotto: So, the X for which the X is non-empty, so this external integration is the interval minus 1 to 1.
18:19:350Paolo Guiotto: Then, for this X, what is the DX section? So, the set of Ys. Now, the set of Y is this one for DX between minus 1 and 1, you see? So, you have to indicate between 0 and 1 minus X squared.
18:34:410Paolo Guiotto: And now this is properly set.
18:38:310Paolo Guiotto: And I should continue. I will do later. Let's see the second way.
18:43:700Paolo Guiotto: So the second way I redo the figure.
18:46:670Paolo Guiotto: I will do the domain, minus 1, 1, the parabola…
18:52:900Paolo Guiotto: This is Y equal 1 minus X squared, the domain.
18:58:460Paolo Guiotto: In the second case, I'm using this formula. Now, I have to find the Y section.
19:05:750Paolo Guiotto: So you take a Y on the y-axis.
19:08:740Paolo Guiotto: So let's say that you see that there are…
19:11:340Paolo Guiotto: and you take lines with, parallel with respect to the x-axis. You see that there are also here, basically, two different situations.
19:23:930Paolo Guiotto: So, since this value is 1,
19:27:270Paolo Guiotto: When I am, in this case, Y is above 1. So, this section DYE, which is, I remind for your convenience, is a set of the other variables, so X, such that XY belongs to D.
19:43:20Paolo Guiotto: So it is the set of abuses of points with ordinate Y that are in B.
19:49:350Paolo Guiotto: So, we see that when Y is… is greater than 1.
19:58:560Paolo Guiotto: Or, also here, that's why less than zero.
20:04:350Paolo Guiotto: In these two cases, I get nothing.
20:08:60Paolo Guiotto: When Y is between 0 and 1, I get something. This is the case, and this something is this set.
20:17:50Paolo Guiotto: Which, again, is not the section, that's just the intersection between this blue line and the domain. This section is the set of the abyshisas of this point, so I have to go down here on the x-axis and read what is this domain here.
20:33:900Paolo Guiotto: Now, this is, a bit more complicated.
20:40:220Paolo Guiotto: And this would say, probably, that since it's more difficult, it would be better to use the other one, but…
20:48:860Paolo Guiotto: remind that there is a second step to do, which is the calculation of integral. Maybe something that is difficult either at the beginning becomes easier later. So, there is not a rule. But in any case, let's try to see what is the range here. So, if this point is Y,
21:06:10Paolo Guiotto: Now, I know that I have to find DX, you see that this is a symmetric interval, so I know that the DX that I have here, or there, it's related to Y through this equation. Y equals 1 minus X squared, this means X squared equals 1 minus Y.
21:25:80Paolo Guiotto: So I get X equal plus or minus the square root of 1 minus Y. So this guy here is clearly the root of 1 minus y, and this is minus the root of 1 minus Y.
21:42:60Paolo Guiotto: So, as you have seen, it's not impossible to determine the section.
21:46:540Paolo Guiotto: The section here is,
21:50:490Paolo Guiotto: For this case, the interval minus root of 1 minus y plus root of 1 minus y.
22:02:760Paolo Guiotto: Okay, so this means that now I have all the ingredients to do… to write down the reduction formula for this case. So I have, again, the double integration, my function X divided 2 plus Y,
22:17:330Paolo Guiotto: Now, this is integrating first in X, and then in Y.
22:22:520Paolo Guiotto: So, let's start from the external integration. I have to put the set of Ys for which this section, the Y, is non-empty.
22:34:40Paolo Guiotto: what is this range? So you see that for this Y section is empty, so discard those Ys, and the Y for which the section is non-empty is Y between 0, 1. So this will be integral between 0, 1.
22:48:640Paolo Guiotto: Then, when Y is between 0 and 1, dx must be between these two values, so minus…
22:55:780Paolo Guiotto: the root of 1 minus Y,
22:58:770Paolo Guiotto: Plus the root of 1 minus 1.
23:01:450Paolo Guiotto: And this is the second integration. So, as you can see, I have… I've been able to write two formulas, so let me just copy the first one down here, so to see the two together. So this is the integral minus 1, then we have the second integration from 0 to 1 minus 1.
23:20:470Paolo Guiotto: X squared. The function is the same, X over 2 plus Y.
23:25:650Paolo Guiotto: This one starts in Y and finishes in X.
23:31:540Paolo Guiotto: So, it seems that they are two entirely different things, because they look different, but they will yield the same output, okay?
23:41:570Paolo Guiotto: Now.
23:42:590Paolo Guiotto: To proceed, I do not need to continue with both. That's a completely stupid job, because if the scope is to compute the integral, I just need to find the best way, possibly the cheapest way to do that, okay?
23:58:60Paolo Guiotto: But at least I need to just carry out one way, not all the two ways, okay?
24:04:880Paolo Guiotto: And I don't even need to verify if they will yield the same value, because since the function is integral, the two for sure will give the same value.
24:14:530Paolo Guiotto: So there is no danger in this. Now, the unique argument to choose one of the two formulas is just a computational argument. Which one should be easier? Actually, I think that both are the same.
24:29:390Paolo Guiotto: But let's do just once, forever. I want two, and more this type of thing, because maybe it's too long also. So let's call them formula number one and this formula number two. Let's see what are the differences. So let's imagine that we continue with Formula 1.
24:48:60Paolo Guiotto: Well, in Formula 1, you see the first integration, which is the first thing you have to do, is now this one, now to the exterior.
24:57:560Paolo Guiotto: is an integration in X. That's the integration variable. So, what is independent of X?
25:06:740Paolo Guiotto: goes outside, if it is… if you can go by algebraic rules. For example, this factor here is a constant for that integral, so I can, well, write outside 1 over 2 plus Y, then I have to integrate from minus the root to plus the root, x dx, then do integration in Y.
25:27:110Paolo Guiotto: Now, this integration is easy because X is the derivative of X squared over 2, so X square over 2, to be integrated from X equals minus root 1 minus y to X equals root of 1 minus Y.
25:42:780Paolo Guiotto: So this value is, 1 half.
25:48:100Paolo Guiotto: Then I have to do the square. This counts as the root, and since there is the minus,
25:56:220Paolo Guiotto: You get, basically,
26:05:450Paolo Guiotto: You get zero, yeah.
26:08:560Paolo Guiotto: So it's much better, this one, because… because let's hit on the other one what happens. Okay, so this is zero, you see? Because you get, from evaluating a text equal root of 1 minus Y, you compute the X squared, you get 1 minus Y. Then you have to do minus…
26:28:530Paolo Guiotto: the value of X squared at the dissects, but the minus disappear in the square, and you get, again, 1 minus Y.
26:36:420Paolo Guiotto: So at the end, you get 0. So this integral is fortunately equal to 0, so we are integrating now from 0 to 1,
26:44:630Paolo Guiotto: 1 over 2 plus Y times 0, because that is a factor, this is a factor, DY, so of course, we are integrating 0, and the integral will be 0, so the integration is over.
26:58:380Paolo Guiotto: Let's see what happens if we do the other.
27:01:340Paolo Guiotto: formula.
27:02:970Paolo Guiotto: Which looks easier, no? Because, it was easier to be, written this one.
27:10:350Paolo Guiotto: So I would say integral from minus 1 to 1. Here, the first integration is now in Y. So, X, for example, is a constant factor I can carry outside that X.
27:24:770Paolo Guiotto: And they can write, X times integral from 0 to 1 minus X squared of X squared.
27:34:150Paolo Guiotto: 1 over 2 plus Y dy.
27:38:140Paolo Guiotto: 30 of the X.
27:41:610Paolo Guiotto: So this means integral between minus 1 and 1x, while this is a derivative in Y of log of
27:50:900Paolo Guiotto: Y plus 2, because when you do the derivative, you get one of them.
27:54:890Paolo Guiotto: Y plus 2 times the derivative of Y, which is 1. So we have to do the evaluation of log of Y plus 2,
28:04:140Paolo Guiotto: between Y equals 0 and y equals 1 minus X squared.
28:10:850Paolo Guiotto: And then…
28:12:200Paolo Guiotto: All this must be integrated next, so we have integral from minus 1 to 1, X times… when I plug y equals 1 minus X squared, plus 2 becomes log of 3 minus X squared, when I put y equals 0 minus log 2.
28:32:330Paolo Guiotto: Yes.
28:34:640Paolo Guiotto: So now I have a second integration to do.
28:38:280Paolo Guiotto: Here, I can do this integration in a smart way, which is you take this first integral, which is X log of 3 minus X squared.
28:49:450Paolo Guiotto: the Excel, we are decomposing into
28:52:830Paolo Guiotto: minus log 2 is a constant integral from minus 1 1 of X dx.
28:59:110Paolo Guiotto: Now, something which is sometimes useful to remind is that if you have a function of one variable, let's use a different letter, because, let's say F of U.
29:12:260Paolo Guiotto: is… Odd.
29:15:720Paolo Guiotto: So, F of minus U is equal to minus F of U.
29:20:870Paolo Guiotto: And you are integrating on an interval which is symmetric with respect to the origin, so something like minus A to A, F of U, the U,
29:32:480Paolo Guiotto: That integral will always be equal to zero.
29:36:100Paolo Guiotto: Because of this symmetry. The idea is that if the function is odd, it's like that, maybe.
29:42:760Paolo Guiotto: So, when you integrate from minus A to A, you are algebraically summing this area, but this area is the opposite of this one, the integral is 0. Okay?
29:54:240Paolo Guiotto: However, you can easily see this formally, but intuitively, this is the rule. That's useful here, because this is odd. If you change X with minus X, the logarithm part is the same, because depends on X squared, but the other one changed sine, so this is odd.
30:11:350Paolo Guiotto: And you are on a symmetric interval, minus 1, 1. So that integral, you don't need to compute. You can do by parts, etc, but it's useless. And the same here, also this one for the same reason is 0, and we reopen equals 0, okay? So let's say that this finishes the exercise.
30:32:280Paolo Guiotto: Before we leave the exercise, let me just stress, focus your attention on Other couple of facts.
30:44:00Paolo Guiotto: So the first factor, I will repeat many times this, because these are all very frequent errors that…
30:52:70Paolo Guiotto: Also, lots of you will do, because you are not different from your previous colleagues that passed through this course. So, that's a common disease. It's something that… it's endemic, impossible to be eliminated.
31:07:210Paolo Guiotto: So… As you can see from this.
31:12:320Paolo Guiotto: Once we, we write the reduction formula, okay.
31:19:790Paolo Guiotto: I want to… to… to foc… to focus your attention on two facts. Number one.
31:28:160Paolo Guiotto: You don't see any more diviables in this In this exteriorly, Okay?
31:36:570Paolo Guiotto: In particular, that cannot be this valuable.
31:41:280Paolo Guiotto: Because this variable here, so the variable…
31:44:580Paolo Guiotto: to integrate first disappear after the first integration. You see in the calculations, explicitly, look at this Formula 1,
31:54:460Paolo Guiotto: We computed the first integral… let's leave this one, because it is 0, maybe it's particular. Let's see the second one, no? Here, where we computed this integral in Y.
32:04:940Paolo Guiotto: Okay? We did the calculation, and that's the value that we calculated this is the value of that integral.
32:12:200Paolo Guiotto: You don't see any more than at the Y, and that's correct.
32:16:280Paolo Guiotto: It would be a bigger or bigger thing, wise to, yeah.
32:21:620Paolo Guiotto: This means you are not understanding what you are doing.
32:24:700Paolo Guiotto: That's why… This disappeared after this integration, so there cannot be any more than one.
32:32:140Paolo Guiotto: As well as you cannot have, for example, the acts of the mighty
32:37:640Paolo Guiotto: Because they shouldn't be able to find the internal battle.
32:41:710Paolo Guiotto: Okay, why you would normally have the variables into the innermost integrals, because probably there are conditions that depends on what is Y, so it's normal to have variables here.
32:56:280Paolo Guiotto: Not normal. Not correct.
33:00:280Paolo Guiotto: Okay.
33:01:400Paolo Guiotto: So be very, very, very careful on this, because these are very…
33:05:960Paolo Guiotto: Frequent errors that, unfortunately, we see along the exams.
33:12:410Paolo Guiotto: Now… Let's do the number 3. And this time, let's try to… Go a little bit faster.
33:22:450Paolo Guiotto: And let's try also to add something, which is we don't need to draw anything to do calculations. So here we have to compute this integral. The function is 1 over 1 plus Y,
33:37:220Paolo Guiotto: Even if it is a function of one variable, it doesn't mean that it is not a function of two variables.
33:43:620Paolo Guiotto: It's independent of X. Well, okay, it is well possible, no?
33:49:140Paolo Guiotto: Domain is modulus of y less or equal than 1 minus X pair.
33:55:240Paolo Guiotto: Salt.
33:56:270Paolo Guiotto: First of all, let's discuss the existence.
34:03:630Paolo Guiotto: of integral only of F.
34:07:20Paolo Guiotto: So, here… the function F,
34:10:860Paolo Guiotto: is always, since we are integrating in two variables, it's a function of two variables. If you are integrating in three variables, it will be a function of three variables, and so on. It's 1 over 1 plus Y, which is well-defined and continuous in R2, except at those points where Y is equal to minus 1.
34:39:620Paolo Guiotto: No, this one…
34:46:810Paolo Guiotto: we have to do later, because, here, I think that we can take Y equal,
34:52:630Paolo Guiotto: minus 1 in that domain. So, this is not continuous on the domain.
34:58:750Paolo Guiotto: So let's leave this later.
35:01:960Paolo Guiotto: Because I have to show you how to check this function here. If you look at the domain, so here, let's do a figure.
35:09:870Paolo Guiotto: So the bad points are Y equal minus 1 down here, so let's say we cannot touch these points. But if you look at the domain.
35:18:820Paolo Guiotto: Domain is modulus of Y less than or equal than 1 minus X squared, so it is Y less or equal than 1 minus X squared, and above minus that thing.
35:30:620Paolo Guiotto: So it is between this parabola and… so the parabola Y minus X squared is this one, 1 minus X squared.
35:39:780Paolo Guiotto: And the other one is the opposite, so as you can see, we touch this point, minus 1, and this is in the domain.
35:48:60Paolo Guiotto: So, unfortunately, we have a… we cannot say that the function is continuous on this domain, because there is a point where it is not even defined, the function.
35:58:120Paolo Guiotto: So, we don't, we don't know how for the moment of mathematics, it's, easier. So, we know that the function is not continuous.
36:11:860Paolo Guiotto: In the domain, but we will fix that later, so… We… do. The… Solution… Later.
36:25:820Paolo Guiotto: So because we don't know… since I don't know if the function is integral, I cannot use the reduction formula. Okay, so let's suspend this exercise. Let's do another one.
36:37:620Paolo Guiotto: Well, let's do the number 4, which is very easy.
36:40:830Paolo Guiotto: So lambda 4 is integral, the domain is this, 01 times 2, 4. I hope that these notations are easy for you.
36:54:670Paolo Guiotto: 1 over X minus Y squared, the X DY.
36:58:460Paolo Guiotto: So, you know what is a 01 Cartesian product 24? What does it mean?
37:04:750Paolo Guiotto: Yeah, the X is between 01, the Y is between 2 and 4. So, let's do the figure. In this case, again, we don't need, but however, X is between 01, so this, and Y between 2 and 4, the figure is not in scale, of course.
37:22:820Paolo Guiotto: So, we may say that the domain is something like this.
37:28:400Paolo Guiotto: is a rectangle. It seems a square, but it's not a square.
37:33:20Paolo Guiotto: Okay, now, this function is well-defined, a part where X is equal to Y. Now, there is a dangerous set of points, which, however, does not
37:43:80Paolo Guiotto: affect our domain, because this quarter here is 1, no, sorry, there is no intersection. So we can say that, here, the function f of xy, which is 1 over X minus Y squared.
38:03:210Paolo Guiotto: is continuous on the main. Domain D is clearly compact.
38:11:590Paolo Guiotto: Right? So there is no problem, so there exists the integral only of the function f. So now it's just a matter of computing the integral.
38:23:480Paolo Guiotto: Now, instead of doing the figure with sections, let's try to read the section directly from the domain.
38:32:270Paolo Guiotto: to compute, to compute.
38:39:230Paolo Guiotto: You need to go out.
38:41:180Paolo Guiotto: we apply… the reduction formula.
38:47:790Paolo Guiotto: Now, let's say integral on D of F is a double integral one variable integration.
38:55:770Paolo Guiotto: Let's say that we decide to integrate… in this case, should be perfectly indifferent, which is the first integration, which is the second. So let's say that we start with X and we finish with Y.
39:10:110Paolo Guiotto: Now, we have to write the range for these two variables.
39:15:480Paolo Guiotto: So, the fact is that the point XY belongs to D,
39:20:480Paolo Guiotto: If and only if the point XY belongs to the rectangle, so as we say, the X is between 0, 1,
39:29:520Paolo Guiotto: And Y is between 2 and 4.
39:32:310Paolo Guiotto: And as you can see, there is no relation, so we may say that when X is between 0, Y,
39:39:400Paolo Guiotto: 1, Y must be between 2 and 4. And when X is not in 0, 1, there is no Y.
39:46:30Paolo Guiotto: You see? Because X must be between 0, 1, and Y must be between 2 and 4. And vice versa, for Y, I could say, when Y is between 2 and 4, X, it must be between 0, 1.
39:59:650Paolo Guiotto: Okay, so in this case, since I'm getting first in X, I would say that when Y is between square 4 and y goes up there to the second integral.
40:12:300Paolo Guiotto: DX must be between 0 and 1.
40:15:610Paolo Guiotto: And that's now perfectly. Second, we compute the integral 2 to 4. So the first is integration in X.
40:24:140Paolo Guiotto: So that function, I should look 1 over X minus y squared as the relative respect to X of
40:37:680Paolo Guiotto: Huh?
40:42:680Paolo Guiotto: derivative with respect to X of what?
40:49:390Paolo Guiotto: No.
40:50:780Paolo Guiotto: I'm sorry, this is the derivative of…
40:57:260Paolo Guiotto: No. What is the derivative with respect to X of log of x minus y?
41:03:380Paolo Guiotto: It is 1 over X minus Y times the derivative of X, which is 1. That's not the derivative.
41:11:910Paolo Guiotto: To log, it's to this.
41:17:810Paolo Guiotto: When you derive a power, you get a power, okay? What? A power with an exponent which is diminished by 1. So when you do the opposite, you have to increase the exponent of 1 unit.
41:30:170Paolo Guiotto: So this is a power. X minus Y, but kids, you have to review integration, otherwise you cannot compute anything.
41:39:570Paolo Guiotto: I cannot redo integration in one variable. You have to review these things. So this is power X to minus 2. So this could come from differentiating with respect to X power X minus y to minus 1. More or less, maybe there is some factor to be adjusted.
41:56:390Paolo Guiotto: Let's see what happens if we do this derivative. This is minus 1 times X minus 1 to exponent minus 1, minus 1, right?
42:06:880Paolo Guiotto: Then there is the derivative of the argument of the power with respect to X, and that's 1, because derivative of X minus Y with respect to X is 1. So this is minus 1 over X minus Y to exponent 2. So basically, it's almost what we need, because if you put the minus here.
42:26:140Paolo Guiotto: Minus. This becomes plus. So this guy here is now the derivative of that.
42:33:810Paolo Guiotto: So, you know, what does it mean? 1 over X minus y squared is the derivative with respect to X of… of what? Of this function, okay? So this is, going back here. So…
42:47:70Paolo Guiotto: Well, let me write down.
42:49:270Paolo Guiotto: So this is equal to… Oh, integral 2 to 4.
42:55:740Paolo Guiotto: Then we add… well, let me just copy. Integral 0, 1, 1 over X minus y squared.
43:03:860Paolo Guiotto: DXDY, right? Now, we said that this is the derivative with respect to X of minus X minus y 2 exponent minus 1.
43:15:560Paolo Guiotto: So, the integral will be… there is still the integral to 4. This integral here is the evaluation of that thing, X minus y to X minus 1,
43:28:130Paolo Guiotto: between X equals 0 and X equals 1. And then, whatever comes out, I have to integrate in Y.
43:36:860Paolo Guiotto: Now, carry outside this minus, which is just a button, so we have integral 2 to 4,
43:44:90Paolo Guiotto: Then, for X equals 1, I get 1 minus Y to exponent minus 1, so 1 over 1 minus Y. Minus, for x equals 0, I get 1 over minus Y, right? DY.
43:58:440Paolo Guiotto: So, at the end, I get, if you want, we can restore the minus inside. This becomes 1 over Y minus 1.
44:08:360Paolo Guiotto: minus, minus, plus, with the minus outside, minus 1 over Y dY.
44:13:950Paolo Guiotto: Now, we do this second integration.
44:16:760Paolo Guiotto: This is the derivative with respect to Y of…
44:20:810Paolo Guiotto: log of Y minus 1, right? Well, actually, to be more precise, there should be the absolute value.
44:27:970Paolo Guiotto: This is the derivative with respect to Y of.
44:31:770Paolo Guiotto: log of Y, or absolute value of Y better. So we have… then these integrals will be the evaluation of log of modulus y minus 1 between y equal 2,
44:46:410Paolo Guiotto: Y equals 4. When I say de-evaluation, this means final value minus initial value.
44:52:360Paolo Guiotto: minus the same here, log of modulus y between y equals 2, Y equals 4.
45:00:930Paolo Guiotto: This yields. When Y is 4, we get log 3.
45:05:810Paolo Guiotto: Minus. When Y is 2, we get log 1, which is 0.
45:10:660Paolo Guiotto: minus this year, when Y is 4, we get log 4. When Y is 2, we get log 2.
45:18:580Paolo Guiotto: Log4 is, 2 log 2.
45:23:100Paolo Guiotto: And, forward this at the end of the log, too.
45:27:710Paolo Guiotto: So, the final… Output is log 3 minus log 2, so log 3 half.
45:33:850Paolo Guiotto: And that's the value of the integral.
45:37:140Paolo Guiotto: Okay?
45:41:30Paolo Guiotto: They were right in time to do the breath.
45:45:290Paolo Guiotto: 10 minutes today, I'm generous.
45:50:490Paolo Guiotto: You want 5 minutes?
46:06:450Paolo Guiotto: Let's see one more example. Here we have number 5. Well, these are very simple exercises.
46:14:120Paolo Guiotto: Of course, let's say that complexity here is computational complexity, so…
46:19:960Paolo Guiotto: it would be less easier to determine the description, the parameterization for the integral. However, let's see this one, which is still easy. Integral on XY, X between 1 and 2, and Y
46:36:60Paolo Guiotto: between, one of, X and, And, excellent.
46:44:510Paolo Guiotto: of X over Y, DXDY.
46:50:720Paolo Guiotto: So, again… the function FXY,
46:58:90Paolo Guiotto: is X over Y, which is well-defined and continues on R2, except when Y is equal to 0.
47:08:360Paolo Guiotto: Which is, however, not… In the domain, huh?
47:14:120Paolo Guiotto: Because if you look at these two constraints, you cannot get Y equals 0.
47:19:310Paolo Guiotto: Because Y must be greater than 1 over X. X is between 1 and 2.
47:24:250Paolo Guiotto: So, I noticed that,
47:26:390Paolo Guiotto: I don't want to do the figure, it's easy, but I don't want to do the figure. Here, I notice that the X…
47:35:250Paolo Guiotto: on the…
47:37:00Paolo Guiotto: D is the domain. On D, Y is greater or equal than 1 over X. And since X is between 1 and 2,
47:50:800Paolo Guiotto: So, what can be said about 1 over X?
47:59:490Paolo Guiotto: 1 over X.
48:02:270Paolo Guiotto: 1 half, you have to take this one, no? X less than 2 says 1 over X greater than 1 half.
48:09:770Paolo Guiotto: So, one half, so in particular, it's strictly positive, so there is no problem.
48:15:320Paolo Guiotto: So, our F… B is continuous on D.
48:20:890Paolo Guiotto: Clearly, D is closed.
48:25:750Paolo Guiotto: Usual argument, it is defined by large inequalities, and it is also bounded, though.
48:33:990Paolo Guiotto: And… bound it.
48:37:760Paolo Guiotto: Because, as we can see, X is already bounded, is even between 1 and 2. About Y, I can say it is larger than 1 over X. We say that 1 over X is always greater than 1 half, because X is between 1 and 2.
48:54:340Paolo Guiotto: and Y is less than X, X is less than 2, so you get at the end that Y is between 1 half and 2. So both X and Y are bounded, no?
49:05:880Paolo Guiotto: X bounded here, Y bounded here.
49:09:930Paolo Guiotto: So the domain is bound.
49:12:810Paolo Guiotto: And therefore, F is continuous. On B closed and bounded, compact, there exists the integral.
49:19:870Paolo Guiotto: of F on that domain. So, we can apply it, and… we can… Apply it.
49:31:670Paolo Guiotto: the… reduction formal.
49:36:170Paolo Guiotto: Now… It is better to integrate first in X and then in Y, or vice versa.
49:42:270Paolo Guiotto: Let's see, if you look at the function, the function is X over Y.
49:51:830Paolo Guiotto: So, integrating in X is easy, integrating in Y is easy. It's basically indifferent. So, there is no preference for the function.
50:02:420Paolo Guiotto: But for the domain, there is a, apparently, a preface, because the domain says when X is between 1 and 2,
50:09:680Paolo Guiotto: Y is between 1 over X and X.
50:13:40Paolo Guiotto: So, this description says that
50:17:480Paolo Guiotto: which kind of double integration? First in X and then in Y, or first in Y?
50:23:630Paolo Guiotto: First in Y and then in X, because this says that when X is between 1 and 2, you read there, in this case it is written… not… it won't be always like that, but let's see, X is between 1 and 2,
50:39:720Paolo Guiotto: Y is between 1 over X Index.
50:45:370Paolo Guiotto: Imagine you do the opposite.
50:47:590Paolo Guiotto: by using that description. Apparently, you could write all of this.
50:53:300Paolo Guiotto: So, imagine that you flip the order, you do first in X and then in Y. Can we say that this is… since X is between 1 and 2, I put 1 and 2 here, and Y is between 1 over X and X, I write this. What do you think about this?
51:11:810Paolo Guiotto: You see that there is something wrong here, but conceptually wrong.
51:17:00Paolo Guiotto: PX is here. So once I've done this integration, there is no more X. So how is it possible to see X here?
51:27:290Paolo Guiotto: If you do the calculation, you can do, but the final result, we contain X, Which sounds strange.
51:35:230Paolo Guiotto: Got it.
51:36:180Paolo Guiotto: And moreover, these sites cannot be here, because it's an integration variable. It should be inside here.
51:43:430Paolo Guiotto: So that's drama.
51:45:10Paolo Guiotto: And that's another typical error.
51:55:490Paolo Guiotto: Okay, so let's continue with the first one. Now, this is an integration in Y, so we carry outside the X, integral from 1 to 2 of X, integral from 1 over X to X of 1 over Y dy. Now, that's easy because it's logarithm, so this is the derivative with respect to Y of
52:15:380Paolo Guiotto: log of absolute value of Y. So, this means that we have to do the evaluation, integral 1 to 2, X,
52:25:470Paolo Guiotto: Evaluation logarithm of modulus of Y,
52:29:190Paolo Guiotto: 4Y equal 1 over X, and 4Y equal X, and then take the difference.
52:35:870Paolo Guiotto: As you can see, this guy here, this quantity here, is this int… oh, no, sorry, not that one.
52:44:60Paolo Guiotto: Is this integral here.
52:47:820Paolo Guiotto: At the end of the calculation, there is no more X, okay? X disappears. So we have,
52:57:430Paolo Guiotto: So if I… since I used the modulus, I should write log of modulus of X minus log of modulus 1 over X.
53:06:880Paolo Guiotto: then I remind that my X is between 1 and 2.
53:12:420Paolo Guiotto: So it's not everywhere, so X in particular is positive, so this is log of X minus log of 1 over X.
53:21:640Paolo Guiotto: I can also simplify a bit this by using properties of logarithms. This is log of X to minus 1, so when you carry down the minus 1, this becomes minus log of X. So with the minus here, it's plus, so I get 2 log of X.
53:39:100Paolo Guiotto: So, at the end, I get integral from 1 to 2 of X times 2 log of X. As you can see, there is no more Y.
53:51:310Paolo Guiotto: For Y is over, because we integrated Y. So if you see Y after you have integrated Y, there is an error. Stop immediately, and go back, because there is an error, 100%.
54:05:240Paolo Guiotto: Okay, maybe there is an error also here, but you don't see clearly that there is an error.
54:11:270Paolo Guiotto: But if you see again the Y after you integrated Y, there is an error. That's with the probability 1, 100%.
54:19:670Paolo Guiotto: Okay?
54:22:50Paolo Guiotto: Well, now we can do this integrating by parts.
54:26:300Paolo Guiotto: So, for example, I… I can see I give the 2 to X, yes, I see this as the derivative with respect to X of X squared, so this becomes, integrating by parts, I have the product X squared log X,
54:42:590Paolo Guiotto: Evaluated between X equals 1 and 2.
54:46:440Paolo Guiotto: minus integral between 1 and 2, the derivative moves through the log, so it gives 1 over X.
54:54:430Paolo Guiotto: Now, for X equals 2, I get, 4 log 2,
54:59:200Paolo Guiotto: minus, for X equals 1, the log is 0, so in the field, this simplifies this, I get integral of X, which is X square over 2, to be evaluated between x equals 1, x equal 2,
55:12:280Paolo Guiotto: So, at the end, I have a 4 log 2, Minus 1 half…
55:18:530Paolo Guiotto: Then for X equals 2, I get 4. For X equals 1, I get 1, so this is the final value, okay?
55:29:60Paolo Guiotto: Okay, so now, let's return back to some general
55:34:660Paolo Guiotto: Remark. Staying still on the deduction formula.
55:40:480Paolo Guiotto: Okay, now let's say that we, started, Working with this formula.
55:49:340Paolo Guiotto: And this formula says that So, let's, return that.
55:58:870Paolo Guiotto: On reduction formula.
56:02:110Paolo Guiotto: So, the property says that if the function f is
56:08:910Paolo Guiotto: Well, let's say it this way. If there exists the integral of F, DX, DY.
56:17:920Paolo Guiotto: Then, this integral, of F, the XDY, can be computed applying one of these two formulas.
56:27:920Paolo Guiotto: So you have a double integration, you do first… well, let's… of course, F would be F of XY, okay? But just to keep light notations, so let's say that I do first integration in X, and then in Y.
56:41:820Paolo Guiotto: In this case, what have I to write down here? The X sections or the Y sections?
56:50:100Paolo Guiotto: the Y sections.
56:51:980Paolo Guiotto: Because this is that of X, and here, Y such that the Y is known.
56:59:90Paolo Guiotto: And the alternative formula is that you do the first integration in Y, so this will be on the X section, and the second will be on X such that the X is non-empty.
57:12:370Paolo Guiotto: And this will be litigation units.
57:15:670Paolo Guiotto: Now, the problem, this formula requires that you know that the function is integral.
57:23:300Paolo Guiotto: Okay.
57:24:700Paolo Guiotto: So, if we have a function which is continuous on a closed unbounded domain, we know that this formula can be used. But what if, in general, I don't know if this happens?
57:36:290Paolo Guiotto: Now, the point is that,
57:39:570Paolo Guiotto: We may notice the following.
57:42:870Paolo Guiotto: So… If, F is integral.
57:50:190Paolo Guiotto: Then, also, modulus of F is integral, because by definition, you remind that
57:59:400Paolo Guiotto: integral on D of modulus of F is finite.
58:04:820Paolo Guiotto: If you go back to the definition, we did the… Couple of times ago, maybe.
58:13:400Paolo Guiotto: No.
58:14:480Paolo Guiotto: Yeah, so was last time, probably.
58:18:170Paolo Guiotto: So we say that, Somewhere…
58:30:250Paolo Guiotto: What is it?
58:34:50Paolo Guiotto: Yacht.
58:34:980Paolo Guiotto: F continuous, F integral if modulus of F is inter… is integral. So, let's say, let's use, let's start with this. So, if,
58:45:570Paolo Guiotto: F is interval, most of F is integral, but if we add a continuous, this becomes an if and all if, plus F continuous.
58:55:440Paolo Guiotto: If and only if… modulus of F is integral.
59:02:60Paolo Guiotto: Now… So… if, F is continuous.
59:12:890Paolo Guiotto: And… integral.
59:16:100Paolo Guiotto: then there exists the integral on D of modulus of F,
59:21:670Paolo Guiotto: And we may apply to this integral the reduction formula.
59:26:710Paolo Guiotto: So we write the same thing, but for the models.
59:30:710Paolo Guiotto: So we could have that… this nested integration.
59:35:780Paolo Guiotto: So let's write just BY different from empty, so we'll…
59:41:270Paolo Guiotto: Make it a slightly lighter notation.
59:44:480Paolo Guiotto: And also this other nested integration, they both are
59:49:920Paolo Guiotto: the integral of the modulus of F.
59:53:970Paolo Guiotto: Which is fine.
59:58:180Paolo Guiotto: So, if F is continuous and integrable, Then, huh?
00:04:270Paolo Guiotto: Both these two quantities that coincide with the value of the integral of the modulus are fined.
00:16:750Paolo Guiotto: Now, the point is that it is possible to prove that also the vice versa holds. So there is a theorem.
00:27:700Paolo Guiotto: Which is called the tonality.
00:31:510Paolo Guiotto: PRM?
00:32:970Paolo Guiotto: That says that if you have a function F, which is continuous on a certain domain B, it doesn't matter what is the nature of this domain.
00:42:130Paolo Guiotto: Such that… At least one of these two integrals is 5, such that at least One… Aw.
00:55:940Paolo Guiotto: or either the iterated integral of modulus of F integrating first in X, so along this Y section, the Y different from MT,
01:07:880Paolo Guiotto: Or the other one, the integral of modulus of F, first in Y,
01:16:30Paolo Guiotto: Along the X section, the X non-empty.
01:22:680Paolo Guiotto: So, if F is continuous, such that at least one of these two quantities is pi in it.
01:32:250Paolo Guiotto: Then… The function f is integral, on the…
01:39:570Paolo Guiotto: That is, there exists the integral.
01:43:60Paolo Guiotto: On the OVAP.
01:45:410Paolo Guiotto: So, what is the difference? This is a test to check integrability. It's an efficient test, because basically it says, if you compute one of these two iterated integrals, we now know what does it mean to compute iterated… well, here I miss DY, sorry.
02:05:190Paolo Guiotto: So there is this.
02:06:820Paolo Guiotto: We know what it means to compute an iterated integral, it's what we have done so far, and we discover that just one of the two is finite. It means that if the function is continuous, the function is integral. And then.
02:23:510Paolo Guiotto: At the end, applying the Fubini theorem, from this,
02:32:550Paolo Guiotto: applying.
02:37:430Paolo Guiotto: for being…
02:38:610Paolo Guiotto: theorem, which is the reduction formula, the integral on D of F will be computed to this alterated integration, either on the Y sections or on X sections.
02:57:190Paolo Guiotto: What does it mean? It means that combining together the two facts, we have this sort of algorithm. I have a continuous function and a domain. If the domain is compact, I know that it is integral. What if the domain is not compact, or the function,
03:16:570Paolo Guiotto: Yes, what if the domain is not conducted? In that case, I could prove that if one of these two is finite.
03:25:550Paolo Guiotto: then the function is integral, and if I needed to compute the integral, at that point, I could apply the reduction formula to have the value of the integral. So this is an algorithm, so in practice.
03:44:640Paolo Guiotto: we… But… the… following.
03:52:860Paolo Guiotto: Hi, good evening.
03:59:930Paolo Guiotto: 4… existence.
04:06:50Paolo Guiotto: And… value.
04:10:230Paolo Guiotto: off.
04:11:280Paolo Guiotto: an integral of a function of two variables, fxy dxdy.
04:19:360Paolo Guiotto: for a function F, which is continuous on a domain D, where I want to compute the integral. So I could say, no, so if F is continuous on D, so I should check this on the integration domain.
04:37:260Paolo Guiotto: Then… I should check if the domain D is, Compact. If D is compact.
04:49:570Paolo Guiotto: then I have that to exist the integral on V of F.
04:54:660Paolo Guiotto: And I can compute the… And… I can.
05:01:280Paolo Guiotto: Calculate.
05:03:410Paolo Guiotto: True.
05:08:410Paolo Guiotto: the reduction formula.
05:12:360Paolo Guiotto: Which is the type of way we followed so far. If D is not compact.
05:25:00Paolo Guiotto: Well, I cannot use this, but if I…
05:29:140Paolo Guiotto: Prove that one of these two, with the absolute value, is finite, And… If… One. Off.
05:41:100Paolo Guiotto: these two iterated integrals for the modulus of F, that one computing first in X, so there are the y-sections here.
05:53:870Paolo Guiotto: Or, the second one.
05:56:150Paolo Guiotto: This one, computing first in Y, and then in X. Here, there will be the X sections.
06:04:180Paolo Guiotto: So I start directly by the iterated integral for the modulus, is finite.
06:13:580Paolo Guiotto: Then, I again obtain that there exists the integral only of F,
06:19:60Paolo Guiotto: not of models of F, of F, and I can compute And the… I. Can.
06:28:680Paolo Guiotto: compute, With the reduction formula.
06:36:550Paolo Guiotto: In particular, this will happen in many cases, if we are in the cases, in this case, in particular.
06:48:380Paolo Guiotto: If I have a constant sine function, so, for example, a positive function.
06:53:730Paolo Guiotto: If F is positive and continuous, on B.
07:01:980Paolo Guiotto: Now, this, the wave would be, you know, and the domain D is not compact, and D.
07:10:980Paolo Guiotto: is not.
07:13:170Paolo Guiotto: Compacta.
07:17:60Paolo Guiotto: I should check.
07:20:290Paolo Guiotto: the way would be check that one of these two is fine, but the models of F would be F itself, in that case. So, this integral for one of these two iterated integrals, I don't want to rewrite everything in detail, so let's write this.
07:36:490Paolo Guiotto: So they iterated the integral first in X, then in Y. It's exactly the same of the iterated integral of F itself in X and Y, right?
07:47:390Paolo Guiotto: So, if this is finite, at once, I have that, number one, the function is integral, Okay?
07:55:650Paolo Guiotto: So this would say that there exists the integral of F, and this integral could be computed by using the reduction formula. That is, computing this quantity.
08:08:20Paolo Guiotto: So basically, I don't need to repeat the calculation, because I already done this.
08:12:250Paolo Guiotto: Okay? Now, I want to show you this, this,
08:17:310Paolo Guiotto: Situation on, some example here.
08:21:939Paolo Guiotto: So… Example, this is 535.
08:30:859Paolo Guiotto: So, the question is, Determine… if… the function FXY.
08:40:760Paolo Guiotto: Written as X cubed E minus YX squared.
08:45:850Paolo Guiotto: is integral.
08:51:990Paolo Guiotto: On this domain, huh?
08:55:149Paolo Guiotto: Which is, 0 plus infinity.
08:59:380Paolo Guiotto: Times.
09:00:710Paolo Guiotto: 1, 2… And… In this case…
09:11:630Paolo Guiotto: compute…
09:16:60Paolo Guiotto: the integral on B of F, if it is integral.
09:21:279Paolo Guiotto: So, let's see the situation. Let's do the figure, just to understand what is the domain.
09:27:340Paolo Guiotto: The domain here is in plain XY,
09:31:630Paolo Guiotto: 0 plus infinity, so X is positive, and Y is between 1 and 2, so it means that Y is between 1 and 2,
09:40:100Paolo Guiotto: And X is whatever positive. So what is this domain? The domain is an original infinite strip like that.
09:49:520Paolo Guiotto: That's the domain.
09:52:930Paolo Guiotto: It is clear that the domain is closed, because if you want, domain B could be written as X is between 1 and 2… no, sorry.
10:03:130Paolo Guiotto: Axis between is, greater equal than zero.
10:09:770Paolo Guiotto: And Y is between 1 and 2. So the domain is divided by large inequalities, so it is closed.
10:19:580Paolo Guiotto: But… I'm bounded.
10:24:530Paolo Guiotto: So, the function f is clearly a good function, no? It's continuous on everywhere, so in particular on the domain D. But since the domain is unbounded, I cannot say that the integral is, exists, okay? Because, I cannot use that.
10:43:210Paolo Guiotto: However, I also noticed that, in this case.
10:47:710Paolo Guiotto: Since, on the domain DX is positive.
10:51:130Paolo Guiotto: By the way, incidentally, I noticed that the function is positive only, and… So, notice that…
11:03:70Paolo Guiotto: that.
11:05:430Paolo Guiotto: F is positive on…
11:09:930Paolo Guiotto: So, to, to check if this function is integral, and as you will see at once, to compute the integral.
11:17:260Paolo Guiotto: what should I do is to check that one of the iterated integrals for the modulus is finite, okay? So, to check integrability.
11:30:230Paolo Guiotto: integrability.
11:32:420Paolo Guiotto: I check… First, though.
11:38:130Paolo Guiotto: beef.
11:39:450Paolo Guiotto: since F, F is continuous, if one of
11:44:410Paolo Guiotto: You don't have to check the two, but it's sufficient only one of these two. So, one of the two integrals. The first one is integrating in X and then in Y, and the second one…
11:56:180Paolo Guiotto: is the opposite, doing first the integration in Y, and then in X.
12:02:660Paolo Guiotto: is fired.
12:08:50Paolo Guiotto: Now, let's write down what are these two, because for the parametization of the integral, the two are more or less the same. But maybe for the calculation.
12:19:550Paolo Guiotto: something is different for the two, okay? So this first one is the integration of modulus of F, which, as I say, it's F, it's F, because F is positive.
12:32:390Paolo Guiotto: on the domain deeds. Not…
12:34:830Paolo Guiotto: on the full R3, because the quantity X cubed can be negative, no? But on the domain D, X is positive, so X cubed is positive, the exponential is, in any case, positive, so our function is positive.
12:48:390Paolo Guiotto: So this one, first in X, and then in Y, is… Precisely.
12:56:620Paolo Guiotto: So we say that this is F, so we write X cubed E minus,
13:01:940Paolo Guiotto: What is the function?
13:04:00Paolo Guiotto: Wa… No, YX squared… Yes, it's a square bit.
13:10:800Paolo Guiotto: minus YX squared.
13:13:580Paolo Guiotto: Now, this is first integration in X, and then in Y. So, we have to identify what is the range for X, what is the range for Y.
13:22:620Paolo Guiotto: look at the domain, and tell me what is the, in this case, for this integral. Let's…
13:31:830Paolo Guiotto: Shink a bit.
13:34:20Paolo Guiotto: What would your writer?
13:38:550Paolo Guiotto: We should start from this one, always, when we die.
13:44:810Paolo Guiotto: No, this is the length of Y.
13:48:740Paolo Guiotto: No.
13:50:450Paolo Guiotto: For why? Yeah, you have 1, 2, in this case, it's thousands. They are independent.
13:58:80Paolo Guiotto: Y must be dedicated between 1 and 2, and in that case, X must be followed.
14:03:590Paolo Guiotto: So, yeah, would we want to go to see local flexibility?
14:07:620Paolo Guiotto: So, we have, in this case, the appropriate description is 1 to 2 and 0 to plus infinity.
14:15:490Paolo Guiotto: This is the case we integrate first in X and then in Y. Let's take the second one, which is the same, but with the reverse order. So we first integrate in Y, and then in X, and then we choose which one should be better.
14:32:430Paolo Guiotto: So the function is the same, because it's a modulus of F, which is F on the domain, so X cubed E minus YX squared. Now we integrate in Y, and then in X.
14:43:730Paolo Guiotto: There is no difference, basically, because the conditions are the same. Y between 1 and 2, X between 0 plus infinity, so when X is between 0 plus infinity, Y is between 1 and 2.
14:54:620Paolo Guiotto: Now, by looking at these two.
14:57:380Paolo Guiotto: Which one would you choose to do the first stack?
15:04:830Paolo Guiotto: The second one, Yes, because,
15:08:770Paolo Guiotto: Yeah, it's not impossible to do the first one as well, but the second one as a function of Y, so emphasize what is as a function of Y. It is something like e to minus Y. Well, there is X squared, yes, but that's a constant. It's like if you have E minus 3Y.
15:27:710Paolo Guiotto: It's easy to integrate this thing.
15:30:690Paolo Guiotto: Yeah, there is also X cube, but X cube is a constant, it comes out, so it would be reasonable to try first with this, so I will continue with this. It is in the Gazzero price infinity, so first let's carry X cube outside.
15:44:960Paolo Guiotto: So we have integral from 1 to 2, E minus YX squared, DY, Now, DX.
15:53:60Paolo Guiotto: Okay, now we have to say that this is the derivative with respect to Y of something.
15:59:210Paolo Guiotto: So this is the derivative with respect to Y of what?
16:09:270Paolo Guiotto: Yeah, but okay, but if you don't know exactly, let's say that you know that when you differentiate the exponential, you get the exponential. So let's start with this, and let's see what happens. This is E minus YX squared.
16:24:10Paolo Guiotto: times the derivative of the exponent with respect to Y, so it is minus X squared.
16:30:800Paolo Guiotto: So, basically, as you can see, if I give an X squared here, so let's give X squared back here, and you also put a minus, you have to multiply 2 times by minus, you have that the quantity inside, now it's a derivative.
16:46:930Paolo Guiotto: You see? So I have minus integer 0 plus infinity, this is X,
16:53:620Paolo Guiotto: integral 1 to 2 of the derivative with respect to Y of E minus y X squared, right?
17:01:710Paolo Guiotto: DY, and that's DX. Now, the innermost integral is easy now, because it is the evaluation of E minus y x squared between y equal 1 and Y equal 2.
17:17:970Paolo Guiotto: So this is minus integral 0 plus infinity X times… when Y is 2, we get E minus 2x squared. When Y is 1, we get E minus X squared.
17:30:750Paolo Guiotto: As you can see once more, there is no more Y.
17:34:520Paolo Guiotto: Okay?
17:35:870Paolo Guiotto: Now, we have to do these integrals, but they are not particularly complicated, because, well, first, let's reorganize this, say that it is X minus X squared, DX,
17:49:380Paolo Guiotto: Minus integral 0 plus infinity Xe minus 2X squared dx.
17:58:170Paolo Guiotto: Now, you should recognize that these are more… yeah.
18:05:700Paolo Guiotto: AXA?
18:10:810Paolo Guiotto: No.
18:13:200Paolo Guiotto: Well,
18:18:970Paolo Guiotto: Yeah.
18:21:530Paolo Guiotto: This was XCUBE.
18:23:730Paolo Guiotto: We took X squared off from this and gave back to the inside to create the derivative.
18:30:950Paolo Guiotto: Okay? So what… this is what remains after this operation.
18:36:90Paolo Guiotto: Okay, now, do you see these are derivatives of…
18:42:270Paolo Guiotto: Well, you know, again, when you do the derivative of an exponential, you get the same exponential, so you have to start from e to minus X squared. So let's see what is the derivative of E minus X squared, just to…
18:54:670Paolo Guiotto: to take confidence with this function. It is E minus X squared times the derivative of the exponent, which is minus
19:02:960Paolo Guiotto: Two acts.
19:04:290Paolo Guiotto: So, we have almost this. So, for the first one, I need a minus 2. I do not have… I can do this. Multiply and divide by minus 2.
19:14:120Paolo Guiotto: For the second one, I have to adjust this, because if I do the derivative of E minus 2X squared, I get E minus 2X squared times the derivative, the exponent, which is minus 4x.
19:27:980Paolo Guiotto: So I need a minus 4, which I do not have. I divide and multiply by minus 4.
19:33:440Paolo Guiotto: Then I take out the factors I don't need inside these ones.
19:38:420Paolo Guiotto: And so I have.
19:40:540Paolo Guiotto: minus 1 half, the first integral is the integral of the derivative with respect to X of E minus X squared, right?
19:49:800Paolo Guiotto: Then the second one, when I put the minus 4 outside, is the denominator becomes plus 1 fourth integral 0 plus infinity of the derivative with respect to X of E minus 2x pair.
20:02:870Paolo Guiotto: the X.
20:04:870Paolo Guiotto: Okay, so now these are evaluations, so E minus X squared to be evaluated from 0 to plus infinity. Of course, plus infinity means a limit.
20:18:100Paolo Guiotto: plus 1 fourth evaluation E minus 2x squared from, again, X equals 0 to X equals plus infinity.
20:28:770Paolo Guiotto: At plus infinity, the two exponentials are 0, so I get 0, minus the value at 0 is e to 0, 1, here, and the same here.
20:40:60Paolo Guiotto: So at the end, I get, from the first one, minus 1 times minus 1 half plus one.
20:46:830Paolo Guiotto: Minus 1 fourth minus 1 minus 1 fourth.
20:52:760Paolo Guiotto: So, one thought.
20:54:860Paolo Guiotto: Now, look at the conclusion.
20:57:230Paolo Guiotto: So what… what have we computed?
20:59:880Paolo Guiotto: We computed the iterated integral for the models. We computed this guy here.
21:07:230Paolo Guiotto: Okay, one of the two iterated integrals for the models. It coincides with the iterated integral of F. That's just because F is positive, not for the same special reason. It's F positive on the domain.
21:20:400Paolo Guiotto: Okay?
21:21:650Paolo Guiotto: on the domain, modules F is the same of F. And what we obtain is that that integral is finite.
21:30:440Paolo Guiotto: Okay? So…
21:33:60Paolo Guiotto: The iterated integral of modulus of F, we did the first in… I do not remind… first in…
21:40:240Paolo Guiotto: Why? Okay. This one…
21:43:600Paolo Guiotto: On, so here there are the X sections, so DX for which DX is non-empty, etc. This DX is fine.
21:54:400Paolo Guiotto: the function F Jeez.
21:57:110Paolo Guiotto: integral.
21:59:440Paolo Guiotto: So, if you want now to compute the value of the integral of F, you can apply a reduction formula.
22:06:850Paolo Guiotto: And this is, again, why don't we use this one? This is the integration of first in Y, and then in X, along the X sections, and the X such that this is non-empty.
22:20:470Paolo Guiotto: But this is the same of this.
22:24:70Paolo Guiotto: So we already did this calculation.
22:26:820Paolo Guiotto: And this is the calculation we'll produce 1 fourth, so we don't need to redo anything, okay?
22:33:450Paolo Guiotto: This could be different if they function as really a variable sign.
22:38:170Paolo Guiotto: Because in that case, we would have not such an easy… confusion.
22:45:130Paolo Guiotto: Let's see… Second example… Still… 5… 3… 6, huh?
22:57:460Paolo Guiotto: So, discuss, if this function E2 minus X.
23:05:240Paolo Guiotto: is integral.
23:07:500Paolo Guiotto: on E… Which is… well, let's write down here.
23:12:850Paolo Guiotto: It's longer.
23:14:730Paolo Guiotto: B is the set of points, XY.
23:19:70Paolo Guiotto: Such that X is both Y is between 0 and X squared.
23:29:720Paolo Guiotto: If yes… compute.
23:37:40Paolo Guiotto: the integral only of the function n.
23:44:110Paolo Guiotto: So… The domain, just to have an idea, is this one. We are in plane XY.
23:52:890Paolo Guiotto: We have to take X positive, so we throw away the negative half plane.
23:59:400Paolo Guiotto: Y, you see here, is positive, so we draw also this half plane.
24:06:60Paolo Guiotto: And Y is less or equal than X squared. Y equal to X squared is this parabola.
24:13:830Paolo Guiotto: So, since it must be less or equal, you throw away this part.
24:18:790Paolo Guiotto: So, at the end, the domain is this plane… region.
24:25:140Paolo Guiotto: between the X, the positive direction of the x-axis, and the parabola.
24:32:60Paolo Guiotto: Clearly, it's an unbounded domain.
24:35:10Paolo Guiotto: So, Domain D is closed.
24:40:720Paolo Guiotto: bots.
24:42:640Paolo Guiotto: Unbounded.
24:47:150Paolo Guiotto: The function f is continuous on domain D, but unfortunately, we cannot say immediately that it is integral.
24:55:320Paolo Guiotto: It is positive, that's evident. It's E minus X.
25:01:310Paolo Guiotto: Okay, so let's keep in mind this. To check integrability.
25:10:420Paolo Guiotto: We… Check.
25:15:10Paolo Guiotto: if… One.
25:17:330Paolo Guiotto: off.
25:18:460Paolo Guiotto: D… iterated.
25:25:600Paolo Guiotto: Integrals.
25:29:980Paolo Guiotto: for the absolute value of F,
25:32:690Paolo Guiotto: which coincides with F, in this case, wherever. It doesn't matter, because the function is positive.
25:38:370Paolo Guiotto: is fine.
25:44:140Paolo Guiotto: Now… So we have to compute one iterated integral for F, which is the same of modulus of F,
25:52:910Paolo Guiotto: Deciding which one should be, done first, integration in X or in Y. If you look at the functions, we have to integrate sooner or later, this function. This function is easy for the integration, both in X and in Y, because in X,
26:10:400Paolo Guiotto: is not a particularly difficult function, it's just a plain exponential. In Y, it's even constant, no? So, it is…
26:18:990Paolo Guiotto: So, it doesn't… this means that it's not the function to decide which should be the first integration, because it's indifferent for the function.
26:27:690Paolo Guiotto: But if you look at the description.
26:30:650Paolo Guiotto: That description says that you should integrate first in.
26:37:990Paolo Guiotto: You see that the Y depends on X.
26:41:960Paolo Guiotto: Okay, so you should say, okay, let's first integrate in Y, then in X. So what are the range?
26:54:160Paolo Guiotto: 4X?
26:56:750Paolo Guiotto: 4X… 0 to positive infinity. For Y,
27:05:490Paolo Guiotto: 2X squared. Yes, in this case, we could also do the opposite there. So, the integration in first in X, and then in Y, are you able to see
27:16:600Paolo Guiotto: What is the integration?
27:23:680Paolo Guiotto: Sorry, Meiji, we have to start first from Y. What is the range for Y that you read here?
27:31:170Paolo Guiotto: Don't tell me 0x squared, because in this case, we are thinking that Y is the last integration variable, so it cannot depend on X.
27:41:10Paolo Guiotto: It's the contrary, it's X, that depends.
27:47:430Paolo Guiotto: No, tell me, why is from… No.
27:53:770Paolo Guiotto: I cannot write this.
27:56:810Paolo Guiotto: Because, you see, this X has disappeared after this, so that cannot be correct. That's wrong for… for sure. So, second possibility.
28:08:970Paolo Guiotto: Yes, it is from 0 to plus infinity, you can understand if you look at the figure, but do not look at the figure. Look at the inequalities.
28:17:880Paolo Guiotto: You now focus… Y is the last thing, so it cannot be depending on X. It would be the contrary. So, you see, here it's written Y is positive. Yes, it is also written Y less or equal than X squared, but that means that X must be greater since it is positive.
28:36:420Paolo Guiotto: must be greater than the root, so for X, I will have root of y to plus infinity. You see, because I have this combination, X positive and Y between 0 and X squared.
28:49:140Paolo Guiotto: I can transform this coop by this one, otherwise it's not correct.
28:55:580Paolo Guiotto: X squared greater than Y, it means that X is greater or equal than root of y. There would be another one-off, if you say X squared greater than Y, and Y is positive, in principle, we should say either X is greater than root of y, or X is less or equal than minus the root of Y, right?
29:17:100Paolo Guiotto: But the second one cannot be, because X is possible, so this is this card.
29:23:00Paolo Guiotto: Okay? Otherwise, you would have these two. And this says that if you now look at this, you can rewrite these things in this way. So, X is greater than root of Y,
29:34:90Paolo Guiotto: and y must be greater than 0. If you now look at this, you immediately recognize that Y is positive, so from 0 to plus infinity, and X is greater than root of y, so from root of y to plus infinity. This is the true parameterization. Let's take the first one. So we have integral 0 to plus infinity.
29:55:140Paolo Guiotto: integral 0 to X squared of… the function is e to minus X, this is the integration in dy, sorry, then we have dx, so that's the integral we have to compute.
30:07:590Paolo Guiotto: Let's go down here.
30:09:580Paolo Guiotto: So, as you can see, E minus X is not, depending on y, so I can carry outside as a constant. This becomes the integral from 0 to plus infinity, e to minus y, integral from 0 to X squared. What remains inside is not 0.
30:26:900Paolo Guiotto: is one.
30:28:900Paolo Guiotto: DY… And then I have the X. So now this integral is easy, it is equal to… X square.
30:38:620Paolo Guiotto: And therefore, we have to integrate from 0 to plus infinity, x squared e to minus X DX. And this can be made by a couple of integration by parts. Now, this is the derivative of E minus X.
30:55:200Paolo Guiotto: Minus, maybe.
30:56:980Paolo Guiotto: So this is, evaluation minus X squared E minus X between 0 and plus infinity.
31:06:80Paolo Guiotto: minus integral 0 to plus infinity, the derivative now goes on x squared, so 2X, times minus e to minus X, so plus e to minus X.
31:19:110Paolo Guiotto: Now, about the evaluation, when you go to plus infinity, the exponential kills the power, so the limit will be zero. In any case. At zero, you get zero, so all this kills zero.
31:31:920Paolo Guiotto: So we have two integrals from 0 to plus infinity, x e2 minus X. We repeat again the integration by parts, so this is the derivative with respect to X of minus e to minus X.
31:45:660Paolo Guiotto: So we have two.
31:48:100Paolo Guiotto: evaluation, minus Xe to minus X, between 0 and plus infinity, minus integral 0 to plus infinity. Derivative moves to X. We have 1,
32:03:30Paolo Guiotto: times minus E to minus X, DX.
32:09:130Paolo Guiotto: Now, the evaluation, again, at plus infinity, you get 0, because the exponential kills the power. At 0 gets 0, so nothing from this.
32:18:430Paolo Guiotto: So we have a minus here, and minus here becomes a plus, so at the end, a 2 integrals 0 to plus infinity, e to minus X.
32:28:470Paolo Guiotto: And now, once more, this is the derivative with respect to X of minus C to minus X. This is a 2, evaluation of minus C to minus X, again, between 0 and plus in 2.
32:41:950Paolo Guiotto: At plus infinity, you get 0. At zero, you get minus 1, so it is the difference, plus 1.
32:49:140Paolo Guiotto: So this is the result.
32:51:360Paolo Guiotto: So what does this say?
32:53:580Paolo Guiotto: This is the iterated integral of F, which is modulus of F for this function.
33:00:380Paolo Guiotto: So, since this is filed, we got that, F…
33:05:330Paolo Guiotto: is integral, so there exists the integral of F from that domain.
33:10:880Paolo Guiotto: And to compute, we would apply the reduction formula that would go down to this one, okay? So we get that this is 2, and we finished.
33:22:810Paolo Guiotto: Okay?
33:28:400Paolo Guiotto: Okay.
33:31:390Paolo Guiotto: Sold.
33:33:800Paolo Guiotto: No.
33:40:800Paolo Guiotto: We have only 3 minutes left, so let's say that, I just, leave you to do.
33:50:310Paolo Guiotto: So, do the, exercises, From 5, 7, 1… Number 6… 7… 8… And 9…
34:09:580Paolo Guiotto: And, this one, it's simple, but a little bit tricky, okay? So, I know that,
34:18:800Paolo Guiotto: You have to think about,
34:22:550Paolo Guiotto: Because the function is… there is a maximum between X squared and Y squared, you will see. Okay, so think about that.
34:29:900Paolo Guiotto: Now, the exercise, the next one is on triple integrals, we have not yet done.
34:37:660Paolo Guiotto: Maybe we will do it on Monday?
34:40:830Paolo Guiotto: This is on volumes.
34:44:210Paolo Guiotto: It is used now. Okay, let's stop here.
34:47:90Paolo Guiotto: Okay, that's it for today, and what's going on?
34:59:380Paolo Guiotto: Let's stop.