Zavdannya, what do you understand underwire integral.

It is acceptable that the function of the elements is assigned and write down the amount

as it is called integral.

Q: Under the sing integral (o.i.) as a function and as a choice

Designation:

Numbers are called integrated (according to Rіman) on .

T. іsnuvannya: For your mind, sho.

Vіdpovіdno until the appointment of o.i. it is significant that the integral may be deposited in the form

Vіdpovіdno up to cl.17.1.1 and 17.1.2 and o. we write down the formulas for the area of ​​the curvilinear trapezoid: , robotic forces

on the:

Understanding the sub-integral, integral sums.

The basis of the underwire integral, so that the interintegral sum is obvious for the given, so that the boundary gives the volume of a cylindrical body. However, tse mirkuvannya not є suvorim. In higher courses of hardening, it is necessary to bring up and call the theorem on the basis of the subvertical integral.

Theorem іsnuvannya. For whether it be a function, without interruption in an enclosed area, where I can measure a, there is a sub-integral, so that there is a boundary between the integral sums with an unrestricted increase in the number of small maidanchiks for the mind, that the skin from them is drawn into a speck. Tsya boundary can not be deposited in the way of breaking up the region, but on the part, not in the choice of points.

Nadalі mi vzglyaditimemo no function, without interruption in the sphere of integration.

From the foundation theorem, we can, for example, divide the region into small rectangles with straight sides parallel to the coordinate axes (Fig. 230). When tsimu. Vibirayuchi potim at the skin small rectum by points we can write, zgіdno z denominations of the subvertical integral

In order to substantiate that the underlying integral can be gained, as between the sum of the mind, the name of the sign of vicorist is also the sign of

Viraz is called an element of a square in Cartesian coordinates and an additional square of a rectangle with sides parallel to the coordinate axes.

It is important that, when the integral bag is folded, the maidans, which adjoin to the inter-region, do not form rectangles. However, it can be brought that the pardon for replacing such maidanchiks with squares between squares will be reduced to zero.

The power of the subordinate integrals

The dominance of the sub-integral (that yogo visnovok) is analogous to the dominance of the one-time singing integral.

1°. Additiveness. What is the function f(x, y) integrated in the region D and as an area D for help curve G the zero area is divided into two links and does not muffle the high internal points of the region D 1 ta D 2 , then the function f(x, y) integrated into skin areas D 1 ta D 2 , moreover

2°. Linear power . What functions f(x, y) that g(x, y) integration in the area D, a α і β - be it speech numbers, then the function [ α · f(x, y) + β · g(x, y)] is also integrated in the region D, moreover

3°. What functions f(x, y) that g(x, y) integration in the area D, then additional functions of these functions are integrated into D.

4°. What functions f(x, y) that g(x, y) offensive integration in the region D and everywhere in my gallery f(x, y) ≤ g(x, y), then

5°. What is the function f(x, y) integrated in the region D, those function | f(x, y)| integrated in the region D, moreover

(Of course, with integration | f(x, y)| in D not showing integration f(x, y) in D.)

6°. The mean value theorem. What an insulting function f(x, y) that g(x, y) integration in the area D, function g(x, y) is invisible (non-positive) everywhere in this gallery, Mі m- exact upper and lower bounds of the function f(x, y) in the region D, then there is a number μ that satisfies the nervousness m ≤ μ ≤ M and so that the formula is valid

Zocrema, as a function f(x, y) is uninterrupted in D, and the area D zv'yazkova, then in this gallery there is such a point ( ξ , η ), what μ = f(ξ , η ), and formula (11) looks like

MOVING INTEGRALS

lecture 1

Sustained integrals.The purpose of the undercurrent integral is that of power. Repeated integrations. Links of lower integrals to repeated ones. Placement between integration. Calculation of the underlying integrals of the Cartesian coordinate system.

The sub-integral is a deepening of the understanding of the sing integral in different functions of two variables. In this way, the vice-versa of integration will be present as a flat figure.

Come on D- Dejaka is a closed, bordered area, and f(x,y) - a sufficient function, it was marked out by this gallery. Let's assume that between regions D are added up from the last number of curves, assignments by peers mind y=f(x) or x=g( y), de f(x) that g(y) are uninterrupted functions.

Rozib'ёmo region D decent rank on n part. area i-ї delyanki is meaningful by the symbol D s i. On the skin dilyantsi, quite a vibe is a point Pi, and let it out in be-yak_y fixing the cartesian system maє coordinates ( x i ,y i). Sklademo integral sum for the function f(x,y) by region D, for which value of the function is known at all points Pi, multiplying їх by the area of ​​the double plots Ds i And we assume all the results are taken away:

Nazvemo diam(G) areas G the largest distance between the boundary points of the region.

Integral functions f(x,y) in the area D is called the boundary, to what extent the sequence of integral sums (1.1) with an unrestricted increase in the number of breaks n (at whom). Write down like this

It is worth noting that, apparently, the integral sum for a given function and a given area of ​​integration is to fall in the way of expanding the area D ta select point Pi. Prote yakshcho podviyny іsnuє іsnuє, tse means, that between the vіdpovіdny іntegrаlny sums it is not possible to lie between the appointed chinnikіv. In order(or, as it seems, general function f(x,y) is integrated in the domain D), it is sufficient that the integral function of the bool uninterrupted at the task gallery integration.

Come on function f(x,y) integrated in the region D. Shards between the cumulative sums for such functions cannot be accumulated by the method of splitting the area of ​​integration, splitting can be carried out for the help of vertical and horizontal lines. Todі more businessmen of the region D matime straight-cut looking, the area of ​​​​such a dorivnyu D s i=D x i D y i. Therefore, the area differential can be written as ds=dxdy. Otzhe, in the Cartesian coordinate system under the integrals you can write down at the sight

Respect. Like the integrand function f(x,y)º1, then the under-integral of the area of ​​the region of integration is:

Significantly, that underlined integrations can be the same power, as well as singly integrated. The deeds of them are significant.

Power of the subordinate integrals.

1 0 .Linear power. Integral of the sum of func- tions of the other sum of the integrals:

and constant multiplier can be blamed for the sign of the integral:

2 0 .Additive power. Since the area of ​​integration D is divided into two parts, then the subintegral is more complete than the sum of the integrations over the skin part:

3 0 .The mean theorem. What is the function f( x,y)is continuous in region D, then there is such a point in the gallery(x, h) , what:

Further post nutrition: how are the sub-integrals calculated? Yogo can be virahuvati approximately, with this method it is broken effective methods folded sums of cumulative sums, which are then calculated numerically with additional EOM. With an analytical calculation of the sub-integrals, they are reduced to two sing-integrals.

Suspended integrals for teapots

This lesson introduces the great topic of multiple integrals, from which the students will sound out another course. Move that by triple integrals you can zalyakate the upholsterer not higher, lower differential equals to that vіdrazu, let's take a look at the food: well, what is it? Zvichayno, deakim will be coherent, and, honestly, I’ve slandered a little bit with naming stats - in order to learn how to virishuvati podvіynі іntegrali, it’s necessary mother deyakі novice. First, if you go about integrations, then, obviously, you will have to integrate. Logic. Otzhe, for the development of applications, you need to know insignificant integrals that count linear integrals hot bi on the middle level. A good novelty is in the fact that by powerful forces they have integrated in greater vipadkiv to do it simply.

Who gets hard? I understood on the right. Tim, who drank a lot of beer during the first semesters. However, normal students can be overwhelmed - on the site there are all materials to fill in the gaps or incomprehensibly. You just happen to spend more than an hour. Possilannya on those, yakі slid vyvchiti or repeat, dodavatimutsya pіd hour statti.

At the introductory lesson, step by step, the following basic moments will be discussed:

– Understanding the undercurrent integral

– Area of ​​integration. The order of bypassing the region of integration. How to change the order of bypass?

After that, as you understand all Asia WELL, you can go to the article How to calculate the underestimated integral? Apply solution. In addition, there is a wider task of calculation of the running integral in polar coordinates that typical supplement about significance to the center of gravity of a flat circumscribed figure.

Let's start with impudent food - what is it?

Understanding the undercurrent integral

The subvariant integral of the ignorant person is written as follows:

Let's take a look at the terms and definitions:
- The icon of the vertical integral;
- Area of ​​integration (flat figure);
- Integral function of two variables, often it is simple;
- Badges of differentials.

What does it mean to calculate the underestimated integral?

Calculate the underlying integral - tse means know KILO. Simplest number:

І vkray bazhano know yogo correctly =)

The result (number) can be negative. And zero can easily be entered. Specially zupinivsya for a given moment, oskolki chimalo students were restless, if you come out "something else marvellous."

Bagato who remembers that "superb" value integral- Tezh number. It's just like that here. The underwire integral has іsnuє i vіdminny geometric zmist, Ale about tse pіznіshe, usumu svіy hour.

How to calculate the underestimated integral?

In order to calculate the underlying integral, it is necessary to call it up to the following we repeat the integrals. Zrobiti tse possible two ways. The largest extension of the offensive method:

Deputy food supply is necessary to set up between integration. Moreover, the nutritional signs of the outer integral are alone - tse numbers, and the subwires of the nutritional signs of the inner integral are functions odnієї zminnoї , scho to deposit in the form of "iks".

Take the stars between integration? The stench is to lie in a v_d tsgogo, as for the mind of the head of the given area. The area is a splendid flat figure, with which you repeatedly stuck together, for example, when numbered areas of flat figures or calculation of the volume of body wrapping. Nezabar you know how to correctly place the boundaries of integration.

After that, as the transition to the repeated integrals is decided, the following calculation is carried out without intermediary: the inner integral is taken first, and then the outer one. One after another. Zvіdsi name - iterative integrals.

Roughly seeming, the task is to count up to the calculation of two singing integrals. How everything is not so coherent and scary, and how did you run into the “singular” sing integral, what do you need to be separated from two integrals ?!

Another way to pass to iterated integrals is simpler:

What has changed? The order of integration has changed: now the inner integral is taken for "ix", and the outer one - for "iplayer". Between integration, marked with stars - be different! One by one the stars of the outer integral - tse numbers, and the underlying stars of the inner integral are turning functions, scho to lie in the form of "igrok".

Which bi mi did not choose the way of transition to repeated integrations, residual sound of obov'yazkovo viide that itself:

Please, remember the importance of power, as it is possible to vikoristovuvati, zokrema, for re-verification of the decision.

Algorithm for solving the sub-integral:

We systematize information: in what order should you look at the task?

1) It is necessary to visconate the chair. Without an armchair, the task is not to break. More precisely, you won’t win, but it will be similar to a game in checks. On the armchair, depict the area as a flat figure. Most of the time, the figure is simple and surrounded by lines, parabolas, hyperbolas, etc. Competent that shvidku tekhnіku pobudovі koslene can be mastered in the lessons Graphs and basic powers of elementary functions, geometric transformation of graphics. Otzhe, the first stage is the vikonati of the armchair.

2) Set up between integrations and go to repeated integrations.

3) Take the inner integral

4) Take the current integral and subtract the difference (number).

Area of ​​integration. The order of bypassing the region of integration.
How to change the order of bypass?

In this paragraph, we can see the most important food - how to go to repeated integrations and correctly place between integrations. As it was said more, you can do it like this:

I like this:

In practice, learning the most awkward tasks leads to the most difficulties, and students often get lost in the process of integrating. Let's look at a specific example:

butt 1

Solution: Let's visualize the area of ​​integration on the armchair:

Zvichayna flat figure and nothing special.

Now I'll see a skin care tool for you - a digging stick, a laser pointer. The task is to scan the skin dot of the shaded area:

Promin the laser to pass the area of ​​integration strictly downhill, then you must finish your order lower flat figures. Promin to enter the region through the entire abscissa, as you are asked by equals and exit the region through a parabola (red arrow). To enlighten the whole region, you need sharply angry to the right draw a vkazіvka vzdovzh osі vіd 0 to 1 (green arrow).

Otzhe, what happened:
"Іgrek" changes from 0 to ;
"ix" changes from 0 to 1.

For tasks, the above is said to be written in case of inconsistencies:

Name the data of inconsistency bypassing the gallery of integration or just order of integration

After we have sorted out the bypass order, we can go from the undercurrent integral to the iterated integrals:

Half of the task is completed. Now it is necessary to pass to iterated integrals in a different way. For whom should you know the pivotal functions. Who is familiar with another paragraph of the lesson Volume body wrap that will be easier. Looking at the functions that define the area . It is even simpler, then go to the return functions, which means to say "iksi" through "igreeks". Single function, de є i "iks" and "iplayer", є.

Yakscho, then, moreover:
the reversal function sets the right side of the parabola;
The reversal function sets the left corner of the parabola.

It is not uncommon to blame sumniv, the axis, for example, the function of signing the left or the right of the parabola? Summing up the differences is even simpler: take a parabolic point, for example, (from the right needle) and set the coordinates in a straight line, for example, the same line:

Correct equality is taken away, henceforth, the function is assigned to the right hand of the parabola, but not to the left.

More than that tsyu reverb(thoughts abo in black) spend the evening after that, as you have passed to the return functions. I’ll borrow nothing for an hour, but I’ll sing a pardon on the coast!

The region of integration is bypassed in another way:

Now we can trim the laser pointer levoruch view gallery integration. Promin laser pass area sharply angry to the right. At to this particular type vіn enter the region through the parabolic needle and exit from the region through a straight line, as set by the equals (red arrow). To scan the entire area with a laser, it is necessary to carry out a vzdovzh axis strictly downhill enter 0 to 1 (green arrow).

In this manner:
"iks" changes from up to 1;
"Іgrek" changes from 0 to 1.

The procedure for bypassing the area next to record as irregularities:

Then, the transition to repeated integrations is as follows:

Vidpovid can be written like this:

Once again, I will guess that the residual result cannot be calculated, depending on which order of bypassing the regions was chosen (equality is established for this). However, it is still far from the final result, now our task is no longer to properly place the interintegration.

butt 2

Denmark's sub-integral with the region of integration. Go to repeated integrations and expand between integrations in two ways.

This is an example of an independent solution. Competently wake up the chair take a straight line on the detour(stars and where to shine with a laser pointer). Zrazok fine design like a lesson.

Most typical tasks trochs are shriveled up in a different formula:

butt 3

Induce the area of ​​integration

Solution: For the mind, the first way to bypass the region was given. Decisions start again from the chair. Here the region does not lie on a silver platter with a blaky oblyamіvkoy, but to encourage you not to create special difficulties. On the back of the hand "knіmaєmo" functions from interintegration: , . The function, arbitrarily, sets a straight line, but what does the function set? Let's її troch change:
- near the center on the cob of coordinates of radius 2. The function w sets the upper line (do not forget, if you sum it up, then you can always set a point that lies on the upper or lower line).

We wonder between the outer integral: "ix" changes from -2 to 0.

Vikonaemo armchairs:

To be precise, I have indicated with arrows the first way to bypass the region, which in turn tells the iterative integrals of the mind: .

Now it is necessary to change the order of bypassing the region, for which we will go to the turning functions (viz. “iksi” through “іgreki”):

Not long ago, we rearranged the function to the level of the stake, let’s say “iks”:
As a result, two pivotal functions are required:
- Signing the right pivkolo;
- Significant leva pіvkolo.
I know the same, as if blaming sumniv, take a point of stake and say, de left, and de right.

Change the order of bypassing the area:

Zgidno with another way around, laser promin enter to the region levoruch through the lion pivkolo i go right-handed through a straight line (red arrow). At the same hour laser insertion be carried out on the axis of ordinates uphill enter 0 to 2 (green arrow).

In this order, the order to bypass the area:

Zagal can be written hint:

butt 4

This is an example of an independent solution. The butt is no longer foldable, but respect that the procedure for bypassing a handful of tasks in a different way! Why work in such vipadkah? First, blaming the difficulties of the chairs, the shards of the chair make the schedule of the hemorrhagic function suddenly insinuate me myself. I recommend the next order dіy: first of all, we take the “primary” function (we say “gravets” through “iks”). We will give you a schedule for the “extreme” function (you can always indulge if you want to krapkovo). Similarly, we can use a larger simple linear function: we can “grave” and conduct a straight line.

We analyze the inter-integration gaps: we enter the middle region through i and exit through . When you do everything right, you can go to the “Greek” smoothness from -1 to 0. Since you have designated an integration area on the chair, change the order of going around in the warehouse of special difficulties. Zrazok designed solution for a lesson.

A similar butt I will sort out a little more later.

Navіt yakscho u have understood everything well, be kind, don't hurry, go straight ahead until the calculation of the running integral. The order of the detour is rich, and it’s important to get your hands on some of the heads, it’s more, I haven’t looked at everything yet!

In the front chotyroh butts, the area of ​​integration was seen as a whole in the 1st, 2nd, 3rd and 4th coordinate quarters. Why wait like that? No, obviously.

butt 5

Change the order of integration

Solution: It’s like a chair, with which the graph of the function is actually a cubic parabola, just won’t “lie on the side”:

The order of bypassing the region, which instructs the iterated integrals , marked with arrows. It should be revered that in the course of the vikonnanny of the chair, one more outlined figure was painted (levoruch in the axis of ordinates). Therefore, we should respect the integration of the region - for the region you can pardon the wrong figure.

Let's move on to the return functions:
- we need the right head of a parabola;

Let's change the order of bypassing the region. As you remember, with another way of bypassing the area, it is necessary to scan the area with a laser switch to the right. Ale here posterіgaєtsya cіkava rіch:

How to repair in similar situations? In such cases, divide the area of ​​integration into two parts and for the skin part, add up your repeated integration:

1) If the "gravity" changes from -1 to 0 (green arrow), then please enter the region through a cubic parabola and exit through a straight line (red arrow). Therefore, the order to bypass the area will be as follows:

2) If the “gravity” changes from 0 to 1 (brown arrow), then promptly enter the area through the parabolic needle and exit through that very straight line (crimson arrow). Then, the order to bypass the area will be as follows:

I repeated repeated integrals:

Singing and multiple integrals have a certain power additivity so that they can be folded, which in a given way and next to grow:
- And the axis and our walk around the region in a different way, looking at the sum of two integrals.

Vidpovid write like this:

What order of circumvention is the most obvious? Zvichayno the one, which letter is given in the mind of a task - you will count twice less!

butt 6

Change the order of integration

This is an example of an independent solution. At the new one there is a pivkol, the analysis of which was reportedly reviewed in Appendix 3. The solution was drawn up like a lesson.

And at once, obіtsyane zavdannya, if a list of tasks is another way to bypass the region:

butt 7

Change the order of integration

Solution: If the procedure for bypassing tasks in a different way, before the wake-up chair, it is necessary to go to the “superior” functions. At whom butt, there are two patients for transformation: i.
With a linear function, everything is simple:

The graph of the function is a parabola with a claim to canonicity.

Virazimo "iplayer" through "ix":

We take two pins of the parabola: i. Yaku from them vibrati? The simplest thing is to see the viconate of the armchair. If you have forgotten the material of analytical geometry about a parabola, then all one insult can be made to krapkovo:

Once again, I pay respect to the fact that on this armchair there was a sprinkling of flat figures and it’s important to choose a figure! At the choice of figures, who are joking around, they will help you between the integration of the other integrals:
, at which point do not forget what the reversal function sets all parabola.

Arrows, which indicate the detour of the figures, exactly indicate the boundaries of the integration of integrations .

Dosit shvidko you learn to carry out such an analysis in your mind and know the need for an area of ​​integration.

If the figure is found, the final part of the solution is even simpler, we change the order of bypassing the area:

The return functions are already known, and the necessary procedure for bypassing the area:

Suggestion:

The final example of a paragraph for independent development:

butt 8

Change the order of integration

Outwardly, the solution is that it is similar to the lesson.

Let's start by looking at the process of calculating the subvariant integral and getting to know it with geometrical complexity.

The sub-integral is numerically superior to the area of ​​a flat figure (regions of integration). The simplest form of the under-integral, if the function of two changeable ones is: .

We can look at the back of the head at the infamous look. At the same time, you are moving, everything is really simple! Calculate the area of ​​flat figures, surrounded by lines. For singing, it’s important that you have a vіdrіzku. The areas of the figures are numerically more advanced:

We depict the area on the armchair:

Choose the first way to bypass the area:

In this manner:

І once important technical trick: repeated integrals can be entered into. First of all, the inner integral, then the outer integral. I strongly recommend Danish way to those teapots.

1) Calculate the internal integral, with which integration is carried out for the change "gravet":

Non-significance integral here is the simplest, and farther victorious, the banal formula of Newton-Leibnitz, with the same difference, that between integrations are not numbers, but functions. The back of the head was put into the "igrok" (original function) upper boundary, then the lower boundary

2) The result, subtracting from the first point, must be added to the current integral:

A larger compact record of the whole solution looks like this:

Otriman formula - this is exactly a working formula for calculating the area of ​​a flat figure for the help of the "extreme" singing integral! Marvel at the lesson Calculation of the area for the help of the sing integral, There it is on the skin croci!

Tobto, the task of calculating the area for the help of the underwire integral few people care for the help of the singing integral! In fact, it's the same!

Clearly, it is not to blame for any difficulties! I'll take a look at the small butts, the shards of you, in fact, more than once stuck together with these tasks.

butt 9

For the help of the underwire integral, calculate the area of ​​the flat figure, surrounded by lines,

Solution: We depict the area on the armchair:

The area of ​​the figure is calculated for the help of the subvertical integral for the formula:

Choose the next order to bypass the area:

Here and further, I do not dwell on how to look around the region, the shards in the first paragraph were brought up even more clearly.

In this manner:

As I have already designated, it is easier for the citizens to calculate the repeated integrals more quickly, which I will add to the method:

1) Row behind the back of the Newton-Leibniz formulas are analyzed with the internal integral:

2) The result, subtracting the first croc, is substituted for the existing integral:

Point 2 - in fact, rebuffing the area of ​​the flat position with the help of the sing integral.

Suggestion:

The axis is so bad and naїvne zavdannya.

Cicavi butt for independent cherry:

butt 10

For the help of the underwire integral, calculate the area of ​​the flat figure, surrounded by lines,

A glimpse of a final design solution for a lesson.

In Prikladakh 9-10, it is significantly more important to win the first way to bypass the region, drink more reading, before speech, you can change the order of bypass and calculate the area in another way. If you don’t pardon, then, naturally, you will see your own significant area.

1.1 Determination of the vertical integral

1.2 Dominance of the sub-integral

The dominance of the sub-integral (that yogo visnovok) is analogous to the dominance of the one-time singing integral.

1°. Additiveness. If the function f(x, y) is integrated in the domain D and if the domain D beyond the additional curve Г the area zero is divided into two links and cannot have joint internal points of the domain D 1 and D 2, then the function f(x, y) is integrated in skin from areas D 1 and D 2, moreover

2°. Linear power. How are the functions f(x, y) and g(x, y) integrable in the space D, huh? i? - be it speech numbers, then the function [? f (x, y) + ? g (x, y)] is also integrated in the domain D, moreover

3°. Since the functions f(x, y) and g(x, y) are integrable in the domain D, then the additional functions of these functions are integrable in D.

4°. How can the functions f(x, y) and g(x, y) be integrated in the domain D and cross f(x, y)? g(x, y), then

5°. Since the function f(x, y) is integrated by the domain D, then the th function |f(x, y)| integrated in region D, moreover

(Obviously, the integration of | f (x, y) | D does not show the integration of f (x, y) in D.)

6°. The mean value theorem. Although the offensive functions f(x, y) and g(x, y) are integrated in the domain D, the function g(x, y) is invisible (non-positive) everywhere in this region, M and m are the exact upper and lower bounds of the function f( x, y) in the region D, then there is a number? that satisfies the unevenness of m? ? ? M i so that the formula is valid

Sokrema, since the function f(x, y) is continuous D, and the domain D is connected, then in this domain there is such a point (?, ?), What? = f(?, ?), and the formula looks like

7°. Important geometric power. living area area D

Let the body T (Fig. 2.1) be given to the space, below the area D, to the beast - a graph of an uninterrupted and invisible function) z \u003d f (x, y,) as it is assigned to the space D, from the sides - a cylindrical surface, a direct one є between the area D, and are parallel to the Oz axis. A body of this type is called a cylindrical body.

1.3 Geometrical interpretation of the vertical integral

1.4 Understanding the vertical integral of a rectangle

Let a sufficient function f(x, y) be assigned everywhere to the rectangle R = ? (div. Fig. 1).

Rosemary segment a? x? b by n partial segments beyond the auxiliary point a = x 0< x 1 < x 2 < ... < x n = b, а сегмент c ? y ? d на p частичных сегментов при помощи точек c = y 0 < y 1 < y 2 < ... < y p = d.

Tsomu razbittya for the help of straight lines, parallel axes Ox і Oy splits the rectangle R into n · p partial rectangles R kl = ? (k = 1, 2, ..., n; l = 1, 2, ..., p). Indicated by the division of the rectangle R, it is significant by the symbol T. We gave it a division under the term "rectangle" to understand the rectangle with sides parallel to the coordinate axes.

On the skin chastkovy rectangle Rkl, we choose a full point (?k,?l). Having put ?x k = x k - x k-1, ?y l = y l - y l-1, it is significant through ?R kl of the area of ​​the rectangle R kl . Obviously, ?R kl = ?x k ?y l .

is called the integral sum of the function f(x, y), which gives a given distribution T of a rectangle R and a given choice of intermediate points (? k, l) on partial rectangles of a distribution T.

The diagonal is called the diameter of the rectangle R kl . A symbol? Significantly the largest of the diameters of all common rectocuts R kl .

The number I is called the boundary of the integral sums (1) at? > 0, how can it be any positive number? can you say so date?, What at?< ? независимо от выбора точек (? k , ? l) на частичных прямоугольниках R выполняется равенство

| ? - I |< ?.

The function f(x, y) is called integrated (according to Riemann) on the rectangle R, because there is a final boundary between the I integral sums of the function at? >0.

The designated boundary I is called the subintegral of the function f(x, y) by the rectangle R and is denoted by one of the following symbols:

Respect. So, just like for a one-time sing integral, it is established that the function f(x, y) is integrated on the rectangle R and is circumscribed on this rectangle.

This gives the opportunity to look farther away from the boundary of the functions f(x, y).

The under-integral has the power, analogous to the power of the sing integral. Significantly less than the main ones:

1. What are the functions
integration in the region
, then integration into them is the amount and difference, moreover

2. The constant multiplier can be blamed for the sign of the subvariant integral:

3. Yakscho
integrated in the region
, and this area is divided into two areas that do not overlap і
, then

.

4. Yakscho
і
integration in the region
, in yakіy

, then

.

5. What's in the area
function
satisfied with the inconsistencies


de
і
acts dіysnі numbers, then

,

de – area of ​​the region
.

The proofs of these powers are analogous to the proof of the second theorems for the simple integral.

Calculation of the vertical integral in rectangular Cartesian coordinates

Let it be necessary to calculate the underlying integral
, de area - Rectangular, which is characterized by irregularities ,.

Let's assume that
is uninterrupted in the same rectangle and swells in the new unknown value, even though the integral of the volume of the body with the basis , fringed with the beast on top
, from sides - flats
,
,
,
:

.

From the other side, such a figure can be calculated for the help of the sing integral:

,

de
- the area of ​​\u200b\u200bcrossing this body with a plane that passes through a point and perpendicular to the axis
. Shards of analysis crossed with a curvilinear trapezium
, surrounded by the beast with a function graph
, de fixed, and , then

.

Z tsikh triokh equalities vyplivaє, scho

.

Henceforth, the calculation of the base integral was the calculation of the two sing integrals; when calculating the "internal integral" (written in arches) to be immutable.

Respect. Can you explain that the rest of the formula is correct when
, as well as at a glance, if the function
change the sign of the indicated rectangle.

The rights of the part of the formula are called the iterated integral and are designated as follows:

.

Similarly, it can be shown that

.

Above what has been said, you whine

.

Remaining equality means that the result of integration should fall in the order of integration.

To take a look at the deepest slope, let's introduce the understanding of the standard area. The standard (or correct) area directly given to the axis is called such an area, for which it should be straight, parallel to the center of the axis, interspersed between the area no more, lower at two points. Otherwise, it seems, overturning the region itself, that її cordon is only one breeze straight.

It is acceptable that the region is surrounded

that is surrounded by the beast with a function graph
, at the bottom - graph of the function
. Come on R ( ,) - minimal rectangle, in which the region is laid
.

Go to the area
that uninterrupted function is assigned
. Let's introduce a new function:

,

similar to the powers of the underwire integral

.

I, later,

.

Oskіlki vіdrіzok
to cover the area
then, later,
at


, but lie in a vіdrіzkom position, then
.

With fixed we can write:

.

The first and third integrals on the right side add up to zero, then

.

Otzhe,

.

Why is it necessary to use the formula for calculating the running integral over the area of ​​the standard axis
by way of the link to the repeated integral:

.

Yakscho region
є standard y straight axis
she shows up as inconsistencies ,

, similarly, one can prove that

.

Respect. For the region
, standard y straight axes
і
, there will be vicons

For this formula, there is a change in the order of integration and the hour of the calculation of the sub-linear integral.

Respect. As soon as the area of ​​integration ceases to be standard (correct) at both coordinate axes, її split at the sum of standard areas and represent the integral as the sum of the integrations in these areas.

butt. Calculate the running integral
by region
, surrounded by lines:
,
,
.

Solution.

Tsya area є standard yak schodo axis
, so i
.

We calculate the integral, taking into account the area of ​​the standard axis
.

.

Respect. How to calculate the integral, taking into account the area of ​​the standard axis
, we take the same result:

.

butt. Calculate the running integral
by region
, surrounded by lines:
,
,
.

Solution. Representably, the region of integration is given to the little one.

Tsya area є standard schodo axis
.

.

butt. Change the order of integration for the repeated integration:

Solution. Let's imagine the region of integration.

From the inter-integration lines, we know the lines that enclose the area of ​​integration: ,
,
,
. To change the order of integration, we can as functions in and we know the crossing points:

,
,
.

So, on one of the intervals, the function is expressed by two analytical virases, then the area of ​​integration must be divided into two areas, and the repeated integral of the tax is the sum of two integrations.

.