Presentation on the topic "integral and its application". Application of the integral Application of the definite integral in biology

Vladimir 2002

Vladimir State University, Department of General and Applied Physics

Introduction

The symbol of the integral has been introduced since 1675, and the issues of integral calculus have been dealt with since 1696. Although the integral is studied mainly by mathematicians, physicists have also contributed to this science. Almost no formula of physics is complete without differential and integral calculus. Therefore, I decided to explore the integral and its application.

History of integral calculus

The history of the concept of an integral is closely connected with the problems of finding quadratures. Mathematicians of Ancient Greece and Rome called the tasks of squaring one or another flat figure as tasks for calculating areas. The Latin word quadratura translates as "squaring". The need for a special term is explained by the fact that in ancient times (and later, up to the 18th century), ideas about real numbers were not yet sufficiently developed. Mathematicians operated with their geometric counterparts, or scalars that cannot be multiplied. Therefore, the tasks for finding areas had to be formulated, for example, as follows: "Construct a square that is equal in size to a given circle." (This classical problem “of squaring a circle” cannot, as is well known, be solved with a compass and straightedge.)

The symbol ò was introduced by Leibniz (1675). This sign is a change of the Latin letter S (the first letter of the word summ a). The very word integral was coined by J. Bernulli (1690). It probably comes from the Latin integro, which translates as bringing back to its previous state, restoring. (Indeed, the operation of integration "restores" the function whose differentiation yields the integrand.) Perhaps the origin of the term integral is different: the word integer means whole.

During the correspondence, I. Bernoulli and G. Leibniz agreed with J. Bernoulli's proposal. Then, in 1696, the name of a new branch of mathematics appeared - integral calculus (calculus integralis), which was introduced by I. Bernoulli.

Other well-known terms related to the integral calculus appeared much later. The name antiderivative function now in use replaced the earlier "primitive function" introduced by Lagrange (1797). The Latin word primitivus is translated as "initial": F(x) = ò f(x)dx - initial (or initial, or antiderivative) for f (x), which is obtained from F(x) by differentiation.

In modern literature, the set of all antiderivatives for the function f(x) is also called the indefinite integral. This concept was singled out by Leibniz, who noticed that all antiderivative functions differ by an arbitrary constant. b

is called a definite integral (the designation was introduced by K. Fourier (1768-1830), but Euler already indicated the limits of integration).

Many significant achievements of the mathematicians of ancient Greece in solving problems of finding quadratures (i.e., calculating areas) of flat figures, as well as cubature (calculating volumes) of bodies, are associated with the use of the exhaustion method proposed by Eudoxus of Cnidus (c. 408 - c. 355 BC). .e.). Using this method, Eudoxus proved, for example, that the areas of two circles are related as the squares of their diameters, and that the volume of a cone is equal to 1/3 of the volume of a cylinder having the same base and height.

Eudoxus' method was improved by Archimedes. The main stages characterizing the method of Archimedes: 1) it is proved that the area of a circle is less than the area of any regular polygon described around it, but greater than the area of any inscribed one; 2) it is proved that with an unlimited doubling of the number of sides, the difference in the areas of these polygons tends to zero; 3) to calculate the area of a circle, it remains to find the value to which the ratio of the area of a regular polygon tends with an unlimited doubling of the number of its sides.

With the help of the exhaustion method and a number of other witty considerations (including involving models of mechanics), Archimedes solved many problems. He gave an estimate for p (3.10/71

Archimedes anticipated many ideas of integral calculus. (Let us add that in practice the first limit theorems were proved by him.) But it took more than a thousand and a half years before these ideas found a clear expression and were brought to the level of calculus.

The mathematicians of the 17th century, who obtained many new results, learned from the works of Archimedes. Another method was also actively used - the method of indivisibles, which also originated in Ancient Greece (it is associated primarily with the atomistic views of Democritus). For example, they imagined a curvilinear trapezoid (Fig. 1, a) composed of vertical segments of length f(x), to which, nevertheless, they attributed an area equal to an infinitely small value f(x)dx. In accordance with this understanding, the required area was considered equal to the sum

an infinite number of infinitely small areas. Sometimes it was even emphasized that the individual terms in this sum are zeros, but zeros of a special kind, which, added up in an infinite number, give a well-defined positive sum.

On such a now seeming at least dubious basis I. Kepler (1571-1630) in his writings “New Astronomy”.

(1609) and "Stereometry of Wine Barrels" (1615) correctly calculated a number of areas (for example, the area of a figure bounded by an ellipse) and volumes (a body was cut into 6c finitely thin plates). These studies were continued by the Italian mathematicians B. Cavalieri (1598-1647) and E. Torricelli (1608-1647). The principle formulated by B. Cavalieri, introduced by him under some additional assumptions, retains its significance in our time.

Let it be required to find the area of the figure shown in Figure 1,b, where the curves bounding the figure from above and below have the equations y = f(x) and y=f(x)+c.

Representing a figure composed of "indivisible", in the terminology of Cavalieri, infinitely thin columns, we notice that they all have a common length c. Moving them in a vertical direction, we can make a rectangle out of them with base b-a and height c. Therefore, the required area is equal to the area of the resulting rectangle, i.e.

S \u003d S1 \u003d c (b - a).

The general Cavalieri principle for the areas of plane figures is formulated as follows: Let the lines of a certain bundle of parallels intersect the figures F1 and F2 along segments of equal length (Fig. 1, c). Then the areas of figures Ф1 and Ф2 are equal.

A similar principle operates in stereometry and is useful in finding volumes.

In the 17th century many discoveries related to integral calculus have been made. So, already in 1629, P. Fermat, the problem of squaring any curve y \u003d xn, where n is an integer (that is, he essentially derived the formula ò xndx \u003d (1 / n + 1) xn + 1), and on this basis decided a number of tasks to find the centers of gravity. I. Kepler, in deriving his famous laws of planetary motion, actually relied on the idea of approximate integration. I. Barrow (1630-1677), Newton's teacher, came close to understanding the connection between integration and differentiation. Works on the representation of functions in the form of power series were of great importance.

However, for all the significance of the results obtained by many extremely inventive mathematicians of the 17th century, calculus did not yet exist. It was necessary to highlight the general ideas underlying the solution of many particular problems, as well as to establish a connection between the operations of differentiation and integration, which gives a fairly general algorithm. This was done by Newton and Leibniz, who independently discovered a fact known as the Newton-Leibniz formula. Thus, the general method finally took shape. We still had to learn how to find the antiderivatives of many functions, give logical new calculus, etc. But the main thing had already been done: differential and integral calculus had been created.

The methods of mathematical analysis were actively developed in the next century (first of all, the names of L. Euler, who completed the systematic study of the integration of elementary functions, and I. Bernoulli should be mentioned). Russian mathematicians M.V. Ostrogradsky (1801-1862), V.Ya. Bunyakovsky (1804-1889), P.L. Chebyshev (1821-1894) took part in the development of integral calculus. Of fundamental importance were, in particular, the results of Chebyshev, who proved that there are integrals that cannot be expressed in terms of elementary functions.

A rigorous exposition of the theory of the integral appeared only in the last century. The solution of this problem is associated with the names of O. Cauchy, one of the greatest mathematicians, the German scientist B. Riemann (1826-1866), the French mathematician G. Darboux (1842-1917).

Answers to many questions related to the existence of areas and volumes of figures were obtained with the creation of measure theory by K. Jordan (1838-1922).

Various generalizations of the concept of the integral were already proposed at the beginning of this century by the French mathematicians A. Lebesgue (1875-1941) and A. Denjoy (1884-1974), and the Soviet mathematician A. Ya. Khinchinchin (1894-1959).

Definition and properties of the integral

If F(x) is one of the antiderivatives of the function f(x) on the interval J, then the antiderivative on this interval has the form F(x)+C, where CнR.

Definition. The set of all antiderivatives of the function f(x) on the interval J is called the definite integral of the function f(x) on this interval and is denoted by ò f(x)dx.

ò f(x)dx = F(x)+C, where F(x) is some antiderivative on the interval J.

f is the integrand, f(x) is the integrand, x is the integration variable, C is the integration constant.

Properties of the indefinite integral.

(ò f(x)dx) ¢ = ò f(x)dx ,

ò f(x)dx = F(x)+C, where F ¢(x) = f(x)

(ò f(x)dx) ¢= (F(x)+C) ¢= f(x)

ò f ¢(x)dx = f(x)+C – from the definition.

ò k f (x)dx = k ò f¢(x)dx

if k is a constant and F ¢(x)=f(x),

ò k f (x)dx = k F(x)dx = k(F(x)dx+C1)= k ò f¢(x)dx

ò (f(x)+g(x)+...+h(x))dx = ò f(x)dx + ò g(x)dx +...+ ò h(x)dx

ò (f(x)+g(x)+...+h(x))dx = ò dx =

= ò ¢dx = F(x)+G(x)+...+H(x)+C=

= ò f(x)dx + ò g(x)dx +...+ ò h(x)dx, where C=C1+C2+C3+...+Cn.

Integration

tabular way.

Substitution method.

If the integrand is not a table integral, then it is possible (not always) to apply this method. For this you need:

split the integrand into two factors;

designate one of the multipliers of the new variable;

express the second factor in terms of a new variable;

write the integral, find its value and perform back substitution.

Note: for the new variable, it is better to designate the function that is associated with the remaining expression.

1. ò xÖ(3x2–1)dx;

Let 3x2–1=t (t³0), take the derivative of both parts:

ó dt 1 1 ó 1 1 t 2 2 1 ---Ø

ô- t 2 = - ô t 2dt = – --– + C = -Ö 3x2–1 +C

ò sin x cos 3x dx = ò – t3dt = – – + C

Let cos x = t

Method for converting an integrand to a sum or difference:

ò sin 3x cos x dx = 1/2 ò (sin 4x + sin 2x) dx = 1/8 cos 4x – ¼ cos 2x + C

ó x4+3x2+1 ó 1 1

ô---- dx = ô(x2+2 – --–) dx = - x2 + 2x – arctg x + C

õ x2+1 õ x2+1 3

Note: when solving this example, it is good to make polynomials "angle".

In parts

If it is impossible to take the integral in a given form, and at the same time, it is very easy to find the antiderivative of one factor and the derivative of another, then you can use the formula.

(u(x)v(x))^=u^(x)v(x)+u(x)v(x)

u^(x)v(x)=(u(x)v(x)+u(x)v^(x)

We integrate both parts

ò u^(x)v(x)dx=ò (u(x)v(x))^dx – ò u(x)v^(x)dx

ò u^(x)v(x)dx=u(x)v(x)dx – ò u(x)v^(x)dx

ò x cos (x) dx = ò x dsin x = x sin x – ò sin x dx = x sin x + cos x + C

Curvilinear trapezoid

Definition. The figure bounded by the graph of a continuous, sign-constant function f(x), the abscissa axis and the straight lines x=a, x=b, is called a curvilinear trapezoid.

Ways to find the area of a curvilinear trapezoid

Theorem. If f(x) is a continuous and non-negative function on the interval , then the area of the corresponding curvilinear trapezoid is equal to the increment of antiderivatives.

Given: f(x) is a continuous indef. function, xО.

Prove: S = F(b) – F(a), where F(x) is the antiderivative of f(x).

Proof:

Let us prove that S(a) is the antiderivative of f(x).

D(f) = D(S) =

S^(x0)= lim(S(x0+Dx) – S(x0) / Dx), for Dx®0 DS is a rectangle

Dx®0 with sides Dx and f(x0)

S^(x0) = lim(Dx f(x0) /Dx) = lim f(x0)=f(x0): since x0 is a point, then S(x) is

Dx®0 Dx®0 antiderivative f(x).

Therefore, by the theorem on the general form of the antiderivative, S(x)=F(x)+C.

Because S(a)=0, then S(a) = F(a)+C

S = S(b)=F(b)+C = F(b)–F(a)

The limit of this sum is called a definite integral.

The sum under the limit is called the integral sum.

The definite integral is the limit of the integral sum on the segment as n®¥. The integral sum is obtained as the limit of the sum of products of the length of the segment obtained by splitting the domain of the function at any point of this interval.

a - lower limit of integration;

b - top.

Newton-Leibniz formula.

Comparing the formulas for the area of a curvilinear trapezoid, we conclude:

if F is the antiderivative of b on , then

ò f(x)dx = F(b)–F(a)

ò f(x)dx = F(x) ô = F(b) – F(a)

Properties of a definite integral.

ò f(x)dx = ò f(z)dz

ò f(x)dx = F(a) – F(a) = 0

ò f(x)dx = – ò f(x)dx

ò f(x)dx = F(a) – F(b) ò f(x)dx = F(b) – F(a) = – (F(a) – F(b))

If a, b and c are any points of the interval I on which the continuous function f(x) has an antiderivative, then

ò f(x)dx = ò f(x)dx + ò f(x)dx

F(b) - F(a) = F(c) - F(a) + F(b) - F(c) = F(b) - F(a)

(this is the additivity property of a definite integral)

If l and m are constants, then

ò (lf(x) + m j(x))dx = l ò f(x)dx + m òj(x))dx –

is the linearity property of a definite integral.

ò (f(x)+g(x)+...+h(x))dx = ò f(x)dx+ ò g(x)dx+...+ ò h(x)dx

ò (f(x)+g(x)+...+h(x))dx = (F(b) + G(b) +...+ H(b)) –

– (F(a) + G(a) +...+ H(a)) + C =

F(b)–F(a)+C1 +G(b)–G(a)+C2+...+H(b)–H(a)+Cn=

= ò f(x)dx+ ò g(x)dx+...+ ò h(x)dx

Set of standard pictures

S=ò f(x)dx + ò g(x)dx

Application of the integral

I. In physics.

Force work (A=FScosa, cosa ¹ 1)

If a force F acts on a particle, the kinetic energy does not remain constant. In this case, according to

the increment of the particle's kinetic energy in time dt is equal to the scalar product Fds, where ds is the displacement of the particle in time dt. Value

is called the work done by the force F.

Let a point move along the OX axis under the action of a force whose projection onto the OX axis is a function f(x) (f is a continuous function). Under the action of the force, the point moved from the point S1(a) to S2(b). Let's divide the segment into n segments of the same length Dx = (b - a)/n. The work of the force will be equal to the sum of the work of the force on the resulting segments. Because f(x) is continuous, then for small, the work of the force on this segment is equal to f(a)(x1–a). Similarly, on the second segment f(x1)(x2–x1), on the nth segment - f(xn–1)(b–xn–1). Therefore, work on is equal to:

А » An = f(a)Dx +f(x1)Dx+...+f(xn–1)Dx=

= ((b–a)/n)(f(a)+f(x1)+...+f(xn–1))

Approximate equality turns into exact one for n®¥

А = lim [(b–a)/n] (f(a)+...+f(xn–1))= ò f(x)dx (by definition)

Let a spring of stiffness C and length l be compressed by half its length. Determine the value of the potential energy Ep is equal to the work A performed by the force –F (s) the elasticity of the spring when it is compressed, then

Ep = A= – ò (–F(s)) dx

It is known from the course of mechanics that F(s)= –Cs.

From here we find

En= – ò (–Cs)ds = CS2/2 | = C/2 l2/4

Answer: Cl2/8.

Center of mass coordinates

The center of mass is the point through which the resultant of gravity passes for any spatial arrangement of the body.

Let the material homogeneous plate o have the shape of a curvilinear trapezoid (x; y |a £ x £ b; 0 £ y £ f (x)) and the function y \u003d f (x) is continuous on , and the area of this curvilinear trapezoid is equal to S, then the coordinates of the center The plate mass o is found by the formulas:

x0 = (1/S) ò x f(x) dx; y0 = (1/2S) ò f 2(x) dx;

Center of Mass.

Find the center of mass of a homogeneous semicircle of radius R.

Draw a semicircle in the OXY coordinate system.

y = (1/2S) òÖ(R2–x2)dx = (1/pR2) òÖ(R2–x2)dx =

= (1/pR2)(R2x–x3/3)|= 4R/3p

Answer: M(0; 4R/3p)

The path traveled by a material point

If the material point moves rectilinearly with the speed u=u(t) and for the time T= t2–t1 (t2>t1) has passed the path S, then

In geometry

Volume is a quantitative characteristic of a spatial body. A cube with an edge of 1mm (1di, 1m, etc.) is taken as a unit of volume.

The number of cubes of a unit volume placed in a given body is the volume of the body.

Axioms of volume:

Volume is a non-negative value.

The volume of a body is equal to the sum of the volumes of the bodies that make it up.

Let's find the formula for calculating the volume:

choose the OX axis in the direction of the location of this body;

determine the boundaries of the location of the body relative to OX;

Let's introduce an auxiliary function S(x) that defines the following correspondence: to each x from the segment we put in correspondence the sectional area of the given figure by the plane passing through the given point x perpendicular to the OX axis.

let's divide the segment into n equal parts and draw a plane perpendicular to the OX axis through each point of the division, while our body will be divided into parts. According to the axiom

V=V1+V2+...+Vn=lim(S(x1)Dx +S(x2)Dx+...+S(xn)Dx

Dx®0, and Sk®Sk+1, and the volume of the part enclosed between two adjacent planes is equal to the volume of the cylinder Vц=SonH.

We have the sum of the products of the function values at the partition points by the partition step, i.e. integral amount. By the definition of a definite integral, the limit of this sum as n®¥ is called the integral a

V= ò S(x)dx, where S(x) is the section of the plane passing through

b selected point perpendicular to the OX axis.

To find the volume you need:

one). Choose the OX axis in a convenient way.

2). Determine the boundaries of the location of this body relative to the axis.

3). Construct a section of a given body by a plane perpendicular to the OX axis and passing through the corresponding point.

4). Express in terms of known quantities a function that expresses the area of a given section.

5). Make an integral.

6). After calculating the integral, find the volume.

Volume of figures of rotation

The body obtained as a result of the rotation of a flat figure about some axis is called the figure of rotation.

The function S(x) of the rotation figure has a circle.

Ssec(x)=p f 2(x)

Arc length of a flat curve

Let the function y = f(x) have a continuous derivative y^ = f^(x) on the interval. In this case, the arc length l of the “piece” of the graph of the function y = f(x), xн can be found by the formula

l = ò r(1+f^(x)2)dx

Bibliography

M.Ya.Vilenkin, O.S.Ivashev–Musatov, S.I.Shvartsburd, “Algebra and Mathematical Analysis”, Moscow, 1993

“Collection of problems in mathematical analysis”, Moscow, 1996.

I.V. Savelyev, “Course of General Physics”, Volume 1, Moscow, 1982

For the preparation of this work, materials from the site http://referatovbank.ru/ were used.

Solving integrals is an easy task, but only for the elite. This article is for those who want to learn to understand integrals, but know little or nothing about them. Integral... Why is it needed? How to calculate it? What are definite and indefinite integrals? If the only use of the integral you know is to get something useful from hard-to-reach places with a hook in the shape of an integral icon, then welcome! Learn how to solve integrals and why you can't do without it.

We study the concept of "integral"

Integration was known in ancient Egypt. Of course, not in a modern form, but still. Since then, mathematicians have written a great many books on the subject. Particularly distinguished Newton and Leibniz but the essence of things has not changed. How to understand integrals from scratch? No way! To understand this topic, you will still need a basic knowledge of the basics of mathematical analysis. Information about , which is also necessary for understanding integrals, is already in our blog.

Indefinite integral

Let's have some function f(x) .

The indefinite integral of the function f(x) such a function is called F(x) , whose derivative is equal to the function f(x) .

In other words, an integral is a reverse derivative or antiderivative. By the way, about how to read in our article.

An antiderivative exists for all continuous functions. Also, a constant sign is often added to the antiderivative, since the derivatives of functions that differ by a constant coincide. The process of finding an integral is called integration.

Simple example:

In order not to constantly calculate the antiderivatives of elementary functions, it is convenient to bring them into a table and use ready-made values.

Complete table of integrals for students

Definite integral

When dealing with the concept of an integral, we are dealing with infinitesimal quantities. The integral will help calculate the area of the figure, the mass of an inhomogeneous body, the path traveled during uneven movement, and much more. It should be remembered that the integral is the sum of an infinitely large number of infinitely small terms.

As an example, imagine a graph of some function. How to find the area of a figure bounded by a graph of a function?

With the help of an integral! Let's break the curvilinear trapezoid, bounded by the coordinate axes and the graph of the function, into infinitesimal segments. Thus, the figure will be divided into thin columns. The sum of the areas of the columns will be the area of the trapezoid. But remember that such a calculation will give an approximate result. However, the smaller and narrower the segments, the more accurate the calculation will be. If we reduce them to such an extent that the length tends to zero, then the sum of the areas of the segments will tend to the area of the figure. This is the definite integral, which is written as follows:

The points a and b are called the limits of integration.

Bari Alibasov and the group "Integral"

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Rules for Calculating Integrals for Dummies

Properties of the indefinite integral

How to solve indefinite integral? Here we will consider the properties of the indefinite integral, which will be useful in solving examples.

The derivative of the integral is equal to the integrand:

The constant can be taken out from under the integral sign:

The integral of the sum is equal to the sum of the integrals. Also true for the difference:

Properties of the Definite Integral

Linearity:

The sign of the integral changes if the limits of integration are reversed:

At any points a, b and with:

We have already found out that the definite integral is the limit of the sum. But how to get a specific value when solving an example? For this, there is the Newton-Leibniz formula:

Examples of solving integrals

Below we consider several examples of finding indefinite integrals. We offer you to independently understand the intricacies of the solution, and if something is not clear, ask questions in the comments.

To consolidate the material, watch a video on how integrals are solved in practice. Do not despair if the integral is not given immediately. Turn to a professional student service, and any triple or curvilinear integral over a closed surface will be within your power.

Department of education of the Gomel city

Executive Committee

State educational institution

"Gymnasium No. 71 in Gomel"

Competitive work

"Application of differential and integral calculus to solving physical and geometric problems in MATLab"

Artist: Orekhova Ksenia Ivanovna,

9B class student

Head: Gorsky Sergey Mikhailovich,

IT-teacher

State educational institution

"Gymnasium No. 71 in Gomel"

Introduction

1. History of integral and differential calculus

2. Differential in physics

3. Applications of the definite integral to the solution of some problems in mechanics and physics

4. Differential equations

5. Examples of solving problems in matlab

List of sources used

Introduction

The optional course "Application of differential and integral calculus to solving physical and geometric problems" aims to study the course of mathematical analysis based on practical coverage of the material, based on the use of methods of this section of mathematics for solving problems of geometry and physics; as well as the implementation of these tasks on a computer (using the MATLAB package).

As a result, we can say that such a voluminous, non-specific formulation of the topic and purpose of the optional course makes it possible to implement it at school. In the school course of algebra and began analysis, the course "Application of differential and integral calculus to solving physical and geometric problems" is aimed at studying the definite integral.

The place of the topic in the school mathematics course .

The optional course "Application of integral calculus to solving physical and geometric problems" deepens the material of the algebra course and began analysis in the eleventh grade and reveals the possibilities for practical consolidation of the material on topics included in the school mathematics course. These are the topics "Derivative of a function", "Definite integral" in algebra, and some topics in geometry and physics. As a result, this optional course implements the interdisciplinary connection of algebra and mathematical analysis with geometry, computer science and physics.

The development of students' correct ideas about the nature of the reflection by algebra of the main elements in geometry and physics, the role of mathematical modeling in scientific knowledge is facilitated by their acquaintance with the solution and visualization of various mathematical problems on a computer. The presentation of the elective course is based on the main features of version 6.1 of the MATLAB mathematical and engineering computing package, which has now become the standard tool for supporting the study of higher mathematics, numerical analysis and other training courses at many universities. Students are introduced to the main features of numerical and symbolic calculations, programming and visualization of results provided by the core of the MATLAB system and its extension package SymbolicMathToolbox.

Basic concepts of an optional course: definite integral, curve length, area, surface of revolution, cylindrical surface, body volume, etc.

Objectives of the elective course.

1. Educational: conduct a practical consolidation on the topic "Definite Integral", introduce students to the MATLAB 6.1 package of mathematical and engineering calculations, illustrate the implementation of the interdisciplinary connection of mathematical analysis with geometry, computer science and physics.

2. Educators: creating conditions for successful professional self-determination of students by solving difficult problems using a computer, educating a worldview and a number of personal qualities, by means of in-depth study of mathematics.

3. Developing: expanding the horizons of students, developing mathematical thinking, forming an active cognitive interest in the subject, developing students' professional interests, developing skills for independent and research activities, developing students' reflection (awareness of their inclinations and abilities necessary for future professional activities).

Program:

Building an astroid

t=-2*pi:pi/20:2*pi;

h=300; figure("Units","Pixels","position",

xlabel("x"); ylabel("y");

axis([-3, 3, -3, 3]);

% Surface of rotation

t=-2*pi:pi/20:2*pi;

meshgrid(t,v);

set(hFigure, "Color",);

set(hAxes, "Color",);

xlabel("x"); ylabel("y"); zlabel("z");

hPlot=plot(X,Y);

set(hPlot,"LineWidth",5)

set(hPlot, "Color")

Task 5. Construct the Bernoulli lemniscate in polar coordinates: .

Program:

for p=0:pi/60:2*pi

if 2*a^2*cos(2*p)>=0

set(hFigure, "Color",);

hP=polar(phi,r);

set(hP,"LineWidth",2);

Result (Fig. 17):

Task 6. Using numerical and symbolic calculations in MATLAB find: a) a definite integral; b) double integral; c) surface integral (of the first kind).

a) The classical problem of numerical analysis is the problem of calculating definite integrals. Of all the methods for calculating definite integrals, the trapezoid method is the simplest, but at the same time quite successfully used. MATLAB provides a function for this method: trapz(x,y) (the edit trapz command allows you to display the text of this function). One-dimensional array x (vector) contains discrete values of the arguments of the integrand. The values of the integrand at these points are concentrated in a one-dimensional array y. Most often, a uniform grid is chosen for integration, that is, the values of the elements of the array x are separated from each other by the same value - the integration step. The accuracy of calculating the integral depends on the value of the integration step: the smaller this step, the greater the accuracy.

Task 7. Calculate the integral using the trapezoid method with different integration steps (to observe 14 decimal digits after the decimal point, you must first enter and execute the formatlong command).

Program: Result:

function=trap(dx)

y=sin(x).*exp(-x);

t=traps(x,y); >> format long

ans = 0.42255394026468

>> trap(0.1)

ans = 0.50144886299125

>> trap(0.01)

ans = 0.50226667654901

>> trap(0.001)

ans = 0.50227485744814

The trapezoidal method is a very versatile method and is well suited for integrating functions that are not too smooth. If the function under the integral sign is smooth (several first derivatives exist and are continuous), then it is better to use integration methods of higher orders of accuracy. With the same integration step, methods of higher orders of accuracy achieve more accurate results.

In the MATLAB system, integration methods of higher orders of accuracy are implemented by the functions quad (Simpson's method) and quad8 (Newton-Cotes method of the 8th order of accuracy). Both of these methods are also adaptive. The latter means that the user does not need to control the achieved accuracy of the result by comparing successive values corresponding to different integration steps. All these functions perform these functions independently.

The quad8 function has a higher order of accuracy compared to the quad function, which is very good for smooth functions, since it provides a higher accuracy of the result with a larger integration step (less deductions). However, the quad function can have no less, but even greater performance for functions that are not too smooth (the second or third derivatives are discontinuous or large in absolute value). In any case, both of these functions provide the same relative precision of 0.001 by default.

Like many other functions of the MATLAB system, the quad and quad8 functions can take a different number of parameters. The minimum format for calling these functions includes three parameters: the name of the integrand, the lower limit of integration, and the upper limit of integration. If the fourth parameter is used, then it is the required relative accuracy of the calculation result. By the way, if both of these adaptive functions cannot provide the required accuracy (divergent or close to this integral), then they return the symbolic infinity Inf.

To calculate definite integrals by symbolic methods, two solutions can be used: directly or in stages (with substitution of symbolic numbers).

Task 8. Calculate a definite integral.

Program: Result:

a1=sym("0"); b1=sym("2");

% 1 way: work with substitution of symbolic numbers

symbol=int(w,"t",a,b)

symbol2a=subs(symbol,,)

number=vpa(symbol2a)

% 2 way: working with symbolic numbers

symbol2b=int(w,"t",a1,b1) symbol =

2.6666666666666666667

Task 9. Calculate the surface area obtained by rotating the astroid around the axis Ox : . (surface rendered in task 2) .

Program: Result:

t1=sym("0"); t2=sym("pi/2"); a=sym("1");

x=a*cos(t)^3; y=a*sin(t)^3;

f=y.*sqrt(diff(x)^2+diff(y)^2);

symbol=simplify(int(4*pi*f,"t",t1,t2))

number=vpa(symbol) symbol =

b) Double integrals are reduced to the calculation of iterated definite integrals, one of which is internal and the other external. The inner integral is the integrand for the outer integral. For numerical calculations, it would be possible to write some chain of calculations in which multiple calculations of the integrand would be reduced to multiple calls to the quad function. However, there is no need to do this yourself, since MATLAB has a special function dblquad for this.

Task 8. Calculate the integral , where .

Program:

Result:

function z=fof(x,y)

z=x.*sin(y)+y.*sin(x); >> format long

>> dblquad("fof",0,1,1,2)

1.16777110966887

Problem 9. Using symbolic calculations, obtain the following integrals , , , , , where .

Program:

z=sym("x*sin(y)+y*sin(x)");

i2=int(z,"x",0,1)

i3=int(int(z,"x"),"y")

i4=int(int(z,"x",1,2),"y",0,1)

i5=int(int(x+y,"y",x,1),"x",0,1) i1 =

1/2*x^2*sin(y)-y*cos(x)

1/2*sin(y)-y*cos(1)+y

1/2*x^2*cos(y)-1/2*y^2*cos(x)

1/2*cos(2)-cos(1)+3/2

Since symbolic calculations do not give an error in the calculation method and are more accurate in themselves, it can be seen that the dblquad function gives an accurate result up to 7 decimal places.

c) It is known from higher mathematics that many other types of integrals can be reduced to definite and double integrals, for example, the surface integral of the first kind. Since when it is found, differentiation under the integral sign is used, it is incorrect to use numerical calculations.

Problem 10. Calculate the surface integral of the 1st kind: , where S is the part of the plane lying in the first octant (by Theorem 2).

Program: Result:

fun=subs(f2,z,f1)

d=1+diff(f1,x)^2+diff(f1,y)^2

syms x1 x2 y1 y2

intpov1=int(int(fun*sqrt(d),"y",y1,y2),"x",x1,x2)

number=vpa(intpov1) fun =

Problem 11. Calculate the surface integral of the 1st kind , where S- sphere (by Theorem 3).

First, let's create a function that describes the surface over which the integration takes place:

function=pov;

syms x y z u v a

x=a*sin(u)*cos(v);

y=a*sin(u)*sin(v);

Program:

syms x y z u v a

f=sym("x^2+y^2");

E=diff(x0,"u")^2+diff(y0,"u")^2+diff(z0,"u")^2;

G=diff(x0,"v")^2+diff(y0,"v")^2+diff(z0,"v")^2;

F=diff(x0,"u")*diff(x0,"v")+diff(y0,"u")*

diff(y0,"v")+diff(z0,"u")*diff(z0,"v");

W=sqrt(E*G-F^2); f2=W*subs(f,,);

syms u1 u2 v1 v2

intpov=p*int(int(f2,"v",v1,v2),"u",u1,u2)

intpov2=simplify(intpov)

number=vpa(intpov2)

int=subs(intpov2,a,b) intpov =

4/3*a^2*pi*(a^4)^(1/2)*4^(1/2)

8/3*a^4*pi*csgn(a^2)

8.377580412*a^4*csgn(a^2)

Note. The csgn function is MATLAB specific. It cannot be entered by the user and occurs only when operating with the simplify function (simplification of symbolic expressions). For example:

>> syms a t

>> t=csgn(a^2)*a^2

Undefined function or variable "csgn".

>>simplify((a^4)^(1/2))

>>simplify((a^8)^(1/4))

>>simplify((a^9)^(1/3))

1. Anufriev, I.E. MatLab 5.3/6.x Tutorial / I.E. Anufriev. - St. Petersburg: BHV-Petersburg, 2002. - 736 p.

2. Berman, G.N. Collection of problems on the course of mathematical analysis / G.N. Berman, I.G. Aramanovich, A.F. Bermant and others - M.: Nauka, 1966. - 456 p.

3. Bermant, A.F. A short course of mathematical analysis for technical colleges / A.F. Bermant, I.G. Aramanovich. - M.: Nauka, 1966. - 736 p.

4. Gultyaev, A. Visual modeling in the MatLab environment / A. Gultyaev. - St. Petersburg: Peter, 2001. - 553 p.

5. Demidovich, B.P. Tasks and exercises in mathematical analysis for higher educational institutions / B.P. Demidovich, G.S. Baranenkov, V.A. Efimenko and others - M .: Nauka, 1966. - 472 p.

6. Lazarev, Yu.F. MatLab 5.x / Yu.F. Lazarev. - Kyiv: BHV, 2000. - 388 p.

7. Martynov, N.N. Matlab 5.x: calculations, visualization, programming / N.N. Martynov, A.P. Ivanov. - M.: KUDITS-OBRAZ, 2000. - 336 p.

8. Kurinnoy, G.Ch. Mathematics: A Handbook / G.Ch. Chicken. - Kharkiv: Folio; Rostov-on-Don: Phoenix, 1997. - 463 p.

9. Piskunov, N.S. Differential and integral calculus for technical colleges in 2 volumes / N.S. Piskunov. - M.: Nauka, 1966. - 2 vol. - 312 p.

10. Fikhtengolts, G.M. Course of differential and integral calculus in 3 volumes / G.M. Fikhtengolts. - M.: State publishing house of physical and mathematical literature, 1959. - v. 1-3.

11. Sites http://www/informika.ru, htt://www.softline.ru, http://matlab.ru.

And integral calculus for solving physical problems” aims to study the course of physics based on mathematical analysis.

This course deepens the material of the algebra course and the beginning of analysis in the tenth and eleventh grades and reveals the possibilities for practical consolidation of the material on topics included in the school physics course. These are the topics "Mechanics", "Electrostatics", "Thermodynamics" in physics, and some topics of algebra and beginnings of analysis. As a result, this elective course implements the interdisciplinary connection of algebra and mathematical analysis with physics.

Objectives of the elective course.

1. Teaching: conduct a practical consolidation on the topics "Mechanics", "Electrostatics", "Thermodynamics", illustrate the implementation of the interdisciplinary connection between mathematical analysis and physics.

2. Educators: creating conditions for successful professional self-determination of students by solving difficult problems, educating a worldview and a number of personal qualities, by means of an in-depth study of physics.

Examples of solving problems in physics by means of a mathematical apparatus.

differential application calculus to the solution of some problems of mechanics.

1. Job. Find the work done by a given force F when moving along a segment of the axis X. If strength F constant, then work BUT is equal to the product F for the path length. If the force changes, then it can be considered as a function of X:F = F(x). work increment BUT on the segment [X,x+ dx] cannot be calculated exactly as a product F(x) dx, as the force changes, on this segment. However, at small dx we can assume that the force changes slightly and the product represents the main part, i.e., is the differential of work ( dA = = F(x) dx). Thus, force can be considered a derivative of the work of displacement.

2. Charge. Let be q - the charge carried by electric current through the cross section of the conductor during the time t. If the current strength / is constant, then in time dt current will carry a charge equal to IDt. With a current strength that changes with time according to the law / \u003d / (/), the product I(t) dt gives the main part of the charge increment over a short time interval [ t, t+- dt], i.e. - is the charge differential: dq = I(t) dt. Therefore, the current strength is the derivative of the charge with respect to time.

3. Mass of a thin rod. Let there be an inhomogeneous thin rod. If you enter the coordinates as shown in Fig. 130, then the function t= t(1)- mass of a piece of rod from a point O to the point /. The inhomogeneity of the rod means that its linear density is not constant, but depends on the position of the point / according to some law p = p(/). If we assume on a small segment of the rod that the density is constant and equal to p(/), then the product p(/)d/ gives the mass differential dm. So linear density is the derivative of mass with respect to length.

4. Heat. Consider the process of heating a substance and calculate the amount of heat Q{ T), required to heat 1 kg of a substance from 0°C to T. Addiction Q= Q(T) very complex and determined experimentally. If the heat capacity with of this substance did not depend on temperature, then the product cdT would give a change in the amount of heat. Counting on a small interval [ T, T+ dT] the heat capacity is constant, we obtain the differential of the amount of heat dQ = c(T) dT. Therefore, heat capacity is the derivative of heat with respect to temperature.

5. Work again. Consider work as a function of time. We know the characteristic of work that determines its speed over time - this is power. When working with constant power N work for time dt is equal to Ndt. This expression represents the differential of work, i.e. dA = N(t) dt, and power is the derivative of work with respect to time.

All the examples given were built according to one and the same familiar to us from the course of physics: work, displacement, force; charge, time, current strength; mass, length, linear density; and so on. Each time one of these quantities acted as a coefficient of proportionality between the differentials of the other two, i.e., each time a relation of the form dy = k(x) dx. This ratio can be viewed as a way to determine the value k(x). Then k(x) is found (or defined) as a derivative at on X. We fixed this conclusion in each example. The reverse statement of the question is also possible: how to find the dependence at from X from a given relation between their differentials.

Applications of a definite integral to the solution of some problems in mechanics.

1. Moments and centers of mass of plane curves. If the arc of the curve is given by the equation y= f(x), a≤ x≤ b, and has a density = (x) , then the static moments of this arc Mx and My relative to the coordinate axes Ox and O y are equal

https://pandia.ru/text/80/201/images/image004_89.gif" width="215" height="101 src="> and the coordinates of the center of mass and - according to the formulas where l- mass of the arc, i.e.

2. Physical problems. Some applications of the definite integral in solving physical problems are illustrated below in the examples.

The speed of the rectilinear movement of the body expressed by the formula (m/s). Find the path traveled by the body in 5 seconds from the start of the movement.

Since the path traveled by the body with speed ( t) for a period of time , is expressed by the integral, then we have:

The equation of mechanical motion. Let the material point of the mass t moving under the influence of force F along the axis X. Denote t the time of her movement, and- speed, a- acceleration. Newton's second law, am = F takes the form of a differential equation, if we write the acceleration, a as the second derivative: a= x’’.

Integral calculus is a branch of mathematical analysis that studies integrals, their properties, methods of calculation and applications. Together with differential calculus, it forms the basis of the apparatus of mathematical analysis.

Dates of occurrence of some mathematical signs

	Meaning		When the sign is entered, year
Object signs
	infinity	J. Wallis
	ratio of circumference to diameter
	square root of
	unknowns or variables	R. Descartes

Operations signs
	addition	German mathematicians	end of the 15th century
	subtraction
	multiplication	W. Outred
	multiplication	G. Leibniz
		G. Leibniz
		R. Descartes
		X. Rudolf
	logarithm	I. Kepler
		B. Cavalieri


	arcsine	J. Lagrange
	differential	G. Leibniz
	integral	G. Leibniz
	derivative	G. Leibniz
	definite integral

	factorial
		W. Hamilton many mathematicians
		I. Bernoulli
relationship signs
	equality	R. Record
		T. Harriot
	comparability
	parallelism	W. Outred
	perpendicularity	P. Erigon

Integral calculus arose from the consideration of a large number of problems in natural science and mathematics. The most important of them is the physical problem of determining the path traveled in a given time along a known, but perhaps variable, speed of movement and a much older problem of calculating the areas and volumes of geometric figures (see Geometric problems for extremum).

The central concept in integral calculus is the concept of an integral, which, however, has two different interpretations, leading, respectively, to the concepts of indefinite and definite integrals.

In differential calculus, the operation of differentiation of functions was introduced. Considered in integral calculus, the mathematical operation inverse to differentiation is called integration or, more precisely, indefinite integration.

What is this inverse operation and what is its indeterminacy?

The differentiation operation compares a given function with its derivative. Suppose that we want, based on a given function , to find a function whose derivative is the function , i.e. . Such a function is called an antiderivative function.

This means that the operation inverse to differentiation - indefinite integration - consists in finding the antiderivative of a given function.

Note that, along with the function antiderivative for the function , there will obviously also be any function , which differs from the constant term : after all .

Thus, unlike differentiation, which compared functions with a single other function - the derivative of the first one, indefinite integration leads not to one specific function, but to a whole set of functions, and this is its uncertainty.

However, the degree of this uncertainty is not so great. Recall that if the derivative of some function is equal to zero at all points of some interval, then this function is constant on the interval under consideration (on intervals where the rate of change of the variable is zero everywhere, it does not change). This means that if on some interval , then the function is constant on this interval, since its derivative is equal to zero at all points of the interval.

So, two antiderivatives of the same function can differ on the interval only by a constant term.

Antiderivative functions are denoted by the symbol

where the sign is read: integral. This is the so-called indefinite integral. According to what has been proved, the indefinite integral depicts on the interval under consideration not one specific function, but any function of the form

, (1)

where is some antiderivative of a function on a given interval, and is an arbitrary constant.

For example, on the whole number axis

; ; .

Here we have specially marked the arguments of the integrands with different symbols: , in order to draw attention to the independence of the antiderivative as a function of the choice of the letter used to denote its argument.

The written equalities are checked by simple differentiation of their right parts, which results in the functions , standing in the left parts under the integral sign, respectively.

It is also useful to keep in mind the following obvious relations, which follow directly from the definitions of the antiderivative, derivative, differential, and from relation (1) for the indefinite integral:

, , , .

Finding the antiderivative is often facilitated by some general properties of the indefinite integral:

(taking out a constant multiplier);

(sum integration); if

(change of variable).

These relationships are also verified directly using the appropriate differentiation rules.

Let us find the law of motion of a freely falling body in a vacuum, based on the single fact that in the absence of air, the acceleration of free fall near the Earth's surface is constant and does not depend on the features of the falling body. We fix the vertical coordinate axis; we choose the direction on the axis towards the Earth. Let be the coordinate of our body at the moment . We know, therefore, that and is a constant. It is required to find a function - the law of motion.

Since , where , then, integrating successively, we find

So we have found that

, (3)

where and are some constants. But the falling body is still subject to one specific law of motion, in which there is no longer any arbitrariness. This means that there are some other conditions that we have not yet used; they allow among all the "competing" laws (3) to choose the one that corresponds to a particular movement. These conditions are easy to specify if you understand the physical meaning of the constants and . If we compare the extreme terms of relation (2) at , then it turns out that , and from (3) at , it turns out that . Thus, mathematics itself reminded us that the desired law of motion

is completely determined if you specify the initial position and initial velocity of the body. In particular, if and , we obtain .

We now note that between the operation of finding the derivative (differentiation) and the operation of finding the antiderivative (indefinite integration), there are, in addition to the above, a number of fundamental differences. In particular, one should keep in mind that if the derivative of any combination of elementary functions is itself expressed in terms of elementary functions, i.e. is an elementary function, then the antiderivative of an elementary function is no longer always a function of an elementary one. For example, primitive

elementary function (called the integral sine and denoted by the special symbol ) can be shown not to be expressed in elementary functions. Thus, the fundamental mathematical question about the existence of an antiderivative y given in advance should not be confused with the not always solvable problem of finding this antiderivative among elementary functions. Integration is often a source of introduction of important and widely used special functions, which are studied no worse than such "school" functions as or , although they are not included in the list of elementary functions.

Finally, we note that finding the antiderivative, even when it is expressed in elementary functions, is more like an art than a canonical computational algorithm like the differentiation algorithm. For this reason, the found antiderivatives of the most frequently occurring functions are collected in the form of reference tables of indefinite integrals. The following microtable of this kind is obviously equivalent to a microtable of derivatives of the corresponding basic elementary functions:

While we were talking about the inversion of the operation of differentiation, we came in this connection to the concepts of antiderivative, indefinite integral and gave the initial definition of these concepts.

Now we point out a different, much more ancient approach to the integral, which served as the main original source of the integral calculus and led to the concept of a definite integral or integral in the proper sense of the word. This approach can be clearly seen already in the ancient Greek mathematician and astronomer Eudoxus of Cnidus (approximately 408-355 BC) and Archimedes, i.e. it arose long before the advent of differential calculus and the operation of differentiation.

The question that was considered by Eudoxus and Archimedes, who created the “exhaustion method” in solving it, which anticipated the concept of an integral, is the question of calculating the area of a curvilinear figure. Below we will consider this issue, but for now we will put, following I. Newton, the following task: using the speed of the body known at any moment from the time interval, find the amount of movement of the body over this period of time.

If the law of motion were known, i.e. the dependence of the coordinate of the body on time, then the answer, obviously, would be expressed by the difference . Moreover, if we knew any antiderivative function on the interval , then, since , where is a constant, we could find the desired displacement value in the form of a difference , which coincides with the difference . This is a very useful observation, but if the antiderivative of a given function cannot be found, then one has to proceed in a completely different way.

We will argue as follows.

If the interval by separate moments , such that , is divided into very small time intervals , , then on each of these short intervals the speed of the body does not have time to noticeably change. Having fixed the moment arbitrarily, we can thus approximately assume that the motion occurs at a constant speed over the time interval. In this case, for the value of the path traveled during the time interval , we obtain an approximate value , where . Adding these quantities, we obtain an approximate value

for the entire displacement on the interval .

The found approximate value is the more accurate, the finer the division of the interval we make, i.e. the smaller the value of the largest of the intervals into which the interval is divided.

Hence, the amount of displacement we are looking for is the limit

(5)

sums of the form (4) when the value tends to zero.

Sums of a special form (4) are called integral sums for a function on the interval , and their limit (5), obtained by infinitely smaller partitions, is called the integral (or definite integral) of the function on the interval . The integral is denoted by the symbol

in which the numbers are called the limits of integration, and - the lower, a - the upper limit of integration; the function under the integral sign is called the integrand; - integrand; - integration variable.

So, by definition,

. (6)

This means that the desired value of the body displacement over a time interval at a known speed of movement is expressed by the integral (6) of the function over the interval .

Comparing this result with the one indicated in the antiderivative language at the beginning of this example, we arrive at the famous relation:

if . Equality (7) is called the Newton-Leibniz formula. In its left part there is an integral understood as a limit (6), and in its right part there is a difference of values (at the ends and an interval of integration) of the function , an antiderivative integrand function . Thus, the Newton-Leibniz formula connects the integral (6) and the antiderivative. This formula can, therefore, be used in two opposite directions: to calculate the integral by finding the antiderivative, or to obtain an increment of the antiderivative by finding the integral from relation (6). We will see below that both of these uses of the Newton-Leibniz formula are very important.

Integral (6) and formula (7) in principle solve the problem posed in our example. So, if (as is the case in the case of free fall, starting from a state of rest, i.e. from), then, having found the antiderivative functions according to formula (7), we obtain the value

movement in the time elapsed from moment to moment .

On the basis of the physical problem that has just been analyzed, which led us to the integral and the Newton-Leibniz formula, generalizing the observations made, we can now say that if a function is given on a certain interval, then, dividing the interval with points, composing the integral sums

where , , and passing to the limit at , where , we obtain by definition the integral

(6")

from a function over the interval . If, in addition, on , i.e., is the antiderivative of the function on the interval , then the Newton-Leibniz formula holds:

. (7)

LEONARD EULER
(1707-1783)

Euler, the greatest mathematician of the 18th century, was born in Switzerland. In 1727, at the invitation of the St. Petersburg Academy of Sciences, he came to Russia. In St. Petersburg, Euler fell into the circle of outstanding scientists: mathematicians, physicists, astronomers, received great opportunities to create and publish his works. He worked with enthusiasm and soon became, according to the unanimous recognition of his contemporaries, the first mathematician in the world.

Euler's scientific legacy is striking in its volume and versatility. The list of his works includes more than 800 titles. The complete works of the scientist occupies 72 volumes. Among his works are the first textbooks on differential and integral calculus.

In number theory, Euler continued the work of the French mathematician P. Fermat and proved a number of statements: Fermat's little theorem, Fermat's great theorem for exponents 3 and 4 (see Fermat's great theorem). He formulated problems that determined the horizons of number theory for decades.

Euler proposed to apply the means of mathematical analysis in number theory and took the first steps along this path. He understood that, moving further, it is possible to estimate the number of primes not exceeding , and he outlined a statement that would then be proved in the 19th century. mathematicians P. L. Chebyshev and J. Hadamard.

Euler does a lot of work in the field of mathematical analysis. Here he constantly uses complex numbers. The formula is named after him. , which establishes the connection between trigonometric and exponential functions that arises when using complex numbers.

The scientist was the first to develop the general doctrine of the logarithmic function, according to which all complex numbers, except zero, have logarithms, and each number corresponds to an infinite number of logarithm values.

In geometry, Euler laid the foundation for a completely new field of research, which later grew into an independent science - topology.

Euler's name is given to the formula relating the number of vertices (B), edges (P) and faces (G) of a convex polyhedron: .

Even the main results of Euler's scientific activity are difficult to enumerate. Here is the geometry of curves and surfaces, and the first exposition of the calculus of variations with numerous new concrete results. He had works on hydraulics, shipbuilding, artillery, geometric optics, and even music theory. For the first time he gives an analytical presentation of mechanics instead of a geometric presentation of Newton, builds the mechanics of a fixed point or a solid plate.

One of Euler's most remarkable achievements is related to astronomy and celestial mechanics. He built an exact theory of the motion of the Moon, taking into account the attraction not only of the Earth, but also of the Sun. This is an example of solving a very difficult problem.

The last 17 years of Euler's life were marred by an almost complete loss of sight. But he continued to create as intensely as in his younger years. Only now he no longer wrote himself, but dictated to his students, who carried out the most cumbersome calculations for him.

For many generations of mathematicians, Euler was a teacher. According to his mathematical manuals, books on mechanics and physics, several generations studied. The main content of these books is included in modern textbooks.

So, the most important concepts of the integral calculus are defined and the Newton-Leibniz formula is obtained, which connects integration and differentiation.

Just as in differential calculus, not only the problem of determining the instantaneous speed of movement led to the concept of a derivative, but also the problem of drawing a tangent, so in integral calculus, not only the physical problem of determining the distance traveled at a given speed of movement, but also many other problems lead to the concept of an integral. and among them the ancient geometric problems of calculating areas and volumes.

Let it be required to find the area shown in Fig. 1 figure (called a curvilinear trapezoid), the upper "side" of which is a graph of a function given on a segment. By points we divide the segment into small segments , in each of which we fix some point . The area of a narrow curvilinear trapezoid lying above the segment is replaced approximately by the area of the corresponding rectangle with base and height . In this case, the approximate value of the area of the whole figure will be given by the familiar integral sum, and the exact value of the desired area will be obtained as the limit of such sums when the length of the largest segment of the partition tends to zero. Thus, we get:

Let us now try, following Archimedes, to find out in what respect the parabola divides the area shown in Fig. 2 unit squares. To do this, simply calculate, based on formula (8), the area of the lower parabolic triangle. In our case and . We know the antiderivative of the function, which means that we can use the Newton-Leibniz formula (7") and easily obtain

Therefore, the parabola divides the area of the square in a ratio of 2:1.

When dealing with integrals, especially using the Newton-Leibniz formula, you can use the general properties of the indefinite integral, which are named at the beginning of the article. In particular, the rule for the change of variable in the indefinite integral, provided that , , and taking into account the Newton-Leibniz formula, allows us to conclude that

and thus, a very useful formula for the change of variable in a definite integral is obtained:

. (9)

With the help of integrals, the volumes of bodies are also calculated. If shown in Fig. 1 rotate a curvilinear trapezoid around the axis, then you get a body of revolution, which can approximately be considered composed of narrow cylinders (Fig. 3) obtained by rotating the corresponding rectangles. Keeping the previous notation, we write the volume of each of these cylinders in the form (the product of the area of \u200b\u200bthe base and the height). The sum gives an approximate value of the volume of the considered body of revolution. The exact value will be obtained as the limit of such sums at . Means,

. (10)

In particular, to calculate the volume shown in Fig. 4 cones, it is enough to put in the formula (10) , and , where is the slope of the rotated line. Finding the antiderivative of the function and using the Newton-Leibniz formula, we obtain

where is the area of the circle at the base of the cone.

In the examples analyzed, we exhausted the geometric figure with such figures, the areas or volumes of which we could calculate, and then we made the passage to the limit. This technique, coming from Eudoxus and developed by Archimedes, is called the method of exhaustion. This is the most common method of reasoning in most applications of the integral.

"Since the barrels are associated with a circle, a cone and a cylinder - the figures are regular, they are thus amenable to geometric changes." I. Kepler

Meaning is where the snakes of the integral are. Between numbers and letters, between and! V. Ya. Bryusov

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