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 underwire 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
uninterrupted in the same rectangle and nabuvay 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 point:

,
,
.

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.

.

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 G, the area of ​​zero is divided into two links, and if there are no 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 circle, 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.

Why splitting for the help of straight lines, parallel to the axes Ox і Oy, is splitting the rectangle R into n · p partial rectangles R kl = ? (k = 1, 2, ..., n; l = 1, 2, ..., p). In the indication of the division of the rectangle R, it is significant by the symbol T. We have given the division under the term "rectangle" to mean 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 finite 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 a rectangle R and is circumscribed on that rectangle.

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

Power of the subordinate integrals.

A part of the power of the sub-integrals without a middle blazes out from the meaning of which understanding that power of the integral sums, but itself:

1. What is the function f(x, y) integrated into D, then kf(x, y) tezh it is integrated in this galusi, moreover (24.4)

2. What's in the area D integration functions f(x, y)і g(x, y), then those functions are integrated into this gallery f(x, y) ± g(x, y), i at

3. How to integrate in the region D functions f(x, y)і g(x, y) nerіvnіst f(x, y)≤g(x, y), then

(24.6)

Let's add more power to the under-integral:

4. Yakscho area D divided into two regions D 1 ta D 2 without glowing internal dots and function f(x, y) uninterrupted in the region D, then

(24.7) Bringing . Integral sum by region D you can see at a glance:

de division of the region D carried out in such a way that between D 1 ta D 2 is built up between parts of the battle. Passing sweat to the border, while taking away equality (24.7).

5. At the time of integration on D functions f(x, y) this function is integrated in my galus | f(x, y) |, and maє mistse nerіvnіst

(24.8)

Bringing.

stars for help at the border crossing in case of possessed nervousness (24.8)

6. de S D– area of ​​the region D. The proof of which assertion is taken away, substituting the integral sum f(x, y)≡ 0.

7. Still integrated in the region D function f(x, y) satisfies the nervousness

m ≤ f(x, y) ≤ M,

then (24.9)

Bringing.

The proof is carried out by a boundary transition from obvious unevenness

Consequence.

How to subdue all parts of the nervousness (24.9) on D, you can take the so-called mean value theorem:

Zokrema, for the mind of the uninterrupted function f in D there is such a point in the region ( x 0, y 0), in yakіy f(x 0, y 0) = μ , then

-

Another formulation of the mean value theorem.

Geometric zmist lower integral.

Let's see the body V, surrounded by a partial surface, what is asked by equals z = f(x, y), projection D tsієї surface per plane hu a tabular cylindrical surface, cut off from the vertical ones, which connects the points between the surfaces with their projections.

z = f(x, y)

V

y P i D Fig.2.

Shukatimemo the volume of the body as between the sum of the volumes of the cylinders, the bases of which are the parts Δ Si regions D, and by heights - vіdrіzki zavdovka f(Pi), de points Pi lie Δ Si. Passing to the border with, otrimaemo, scho

(24.11)

that is under the influence of the integral of the so-called cylinder, surrounded by the beast on the surface z = f(x, y), and below - the area D.

Calculation of the underline integral by the path of the yoga link to the second one.

Perspective area D, bordered by lines x=a, x=b(a< b ), de φ 1 ( X) and φ 2 ( X) without a break on [ a, b]. Then be straight, parallel to the coordinate axis at and pass through the inner point of the area D, crossing the cordon of the region at two points: N 1 ta N 2 (Fig. 1). Let's call this area correct in on-

at correct axle O at. Similarly, it is

y=φ 2 (x) there is an area that is right in a straight line

N 2 axes O X. Region, correct in direct-

Nії both coordinate axes, we will

D just call it right. For example,

The correct area is shown in Fig.1.

y=φ 1 (x) N 1

O a b x

Come on function f(x, y) uninterrupted in the region D. Look at Viraz

, (24.12)

rank dvorazovym integral type of function f(x, y) by region D. Let's calculate the internal integral (standing at the arms) by changing at, despite X postiynim. As a result, we see uninterrupted function view X:

Otrimanu function is integrable for X in between a before b. As a result, we take the number

We bring the important power of the yard-wise integral.

Theorem 1. Yakscho region D, correct straight ahead at, divided into two areas D 1 ta D 2 straight, parallel axis Pro at abo axis O X, then the dvorazovy integral over the region D more sums of the same integrals by regions D 1 ta D 2:

Bringing.

a) Go straight ahead x = c breaks D on the D 1 ta D 2, straight ahead at. Todi

+

+

b) Go straight ahead y=h breaks D on the right straight ahead at regions D 1 ta D 2 (Fig. 2). Significantly through M 1 (a 1 , h) that M 2 (b 1 , h) points of the cross line of the straight line y=h from the cordon L regions D.

y Region D 1 surrounded by uninterrupted lines

y=φ 2 (x) 1) y=φ 1 (x);

D 2 2) curve BUT 1 M 1 M 2 At, equal to what we write down

h M 1 M 2 y=φ 1 *(x), de φ 1 *(X) = φ 2 (X) at a ≤ x ≤ a 1 ta

A 1 D 1 Bb 1 ≤ x ≤ b, φ 1 *(X) = h at a 1 ≤ x ≤ b 1 ;

3) straight x = a, x = b.

Region D 2 surrounded by lines y=φ 1 *(x),

A y= φ 2 (X),a 1 ≤ x ≤ b 1 .

y=φ 1 (x) We can prove to the inner integral the theorem about

breaking through the integration:

O a a 1 b 1 b

+

Let's give another s otrimanih іntegraіv v vyglyadі sumi:

+ + .

Oskilki φ 1 *(X) = φ 2 (X) at a ≤ x ≤ a 1 ta b 1 ≤ x ≤ b, The first and the third ones take away the integrals and equalize to zero. Otzhe,

I D = , then .

The main power 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. 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

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 summed up from the final number of curves, given by the equals of the 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

Dearly, scho, vzagali seeming, the integral sum for set functions and the given area of ​​integration 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.