Application of integral calculus in professional activities. Lesson summary "application of the integral"

"Omsk State Medical Academy"

Ministry of Health and Social Development of the Russian Federation

on the topic: application of the definite integral

in medicine

completed by a 1st year student

Department of General Medicine

group 102F

Glushneva N.A.

Introduction

The outstanding Italian physicist and astronomer, one of the founders of exact natural science, Galileo Galilei (1564-1642) said that “The Book of Nature is written in the language of mathematics.” Almost two hundred years later, the founder of German classical philosophy, Kant (1742-1804), argued that “In every science there is as much truth as there is mathematics in it.” Finally, almost another hundred and fifty years later, almost in our time, the German mathematician and logician David Hilbert (1862-1943) stated: “Mathematics is the basis of all exact natural science.”

Leonardo Da Vinci said: “Let no one who is not a mathematician read me in my fundamentals.” Trying to find a mathematical basis for the laws of nature, considering mathematics to be a powerful means of knowledge, he even applies it in such a science as anatomy.

Everyone needs mathematics. And doctors too. At least in order to correctly read a regular cardiogram. Without knowing the basics of mathematics, you cannot be proficient in computer technology or use the capabilities of computed tomography... After all, modern medicine cannot do without the most complex technology.

Today it is impossible to study hemodynamics - the movement of blood through vessels without using the integral.

For a long time, catheterization of the right parts of the heart was the only research method that made it possible to assess the conditions of the right parts of the heart, obtain characteristics of intracardiac blood flow, and determine the pressure in the right parts of the heart and pulmonary artery.
The main advantage of echocardiographic examination (EchoCG) is that it is possible to non-invasively, in real time, assess the size and movement of cardiac structures, obtain characteristics of intracardiac hemodynamics, and determine the pressure in the chambers of the heart and pulmonary artery. Good comparability of the results of echocardiography studies with data obtained during cardiac catheterization has been proven.
EchoCG examination allows not only to detect the presence of pulmonary hypertension, but also to exclude a number of diseases that cause secondary pulmonary hypertension: mitral valve defects, congenital heart defects, dilated cardiomyopathy, chronic myocarditis.

However, closer to practice. First, let's find the linear speed of blood flow

Change in the linear speed of blood flow in various vessels

This is the path traveled per unit time by a blood particle in a vessel. The linear velocity in vessels of different types is different (see figure) and depends on the volumetric velocity of blood flow and the cross-sectional area of ​​the vessels. In practical medicine, the linear speed of blood flow is measured using ultrasound and indicator methods; the time of complete blood circulation is often determined, which is 21-23 s.

To determine it, an indicator (erythrocytes labeled with a radioactive isotope, a solution of methylene blue, etc.) is injected into the cubital vein and the time of its first appearance in the venous blood of the same vessel in the other limb is noted.

To begin with, let us remember that an integral is a mathematical object that arose historically based on the need to solve various applied problems in physics and technology. These are also physical applications of the definite integral: calculating the path of a material point moving along a rectilinear or curved path according to the speed of its movement.

Those physical quantities that are determined using an integral are usually called integral, and those quantities through which integral quantities are expressed are called differential. For example, the speed of a body at a point is a differential characteristic of the body, and the mass of the body is an integral one.

Differential characteristics are determined by the value at a point and are usually different at different points in space.

Integral characteristics always express the properties of objects related to an entire region of space. For example, mass characterizes the entire body as an object occupying a region of space. The path traveled by a body is also an integral characteristic, since it characterizes an entire trajectory consisting of many points, and the speed is different at each point of the trajectory and characterizes each point separately.

The question arises - how to calculate the integral velocity for an entire vessel (artery or vein), knowing the linear velocity of blood flow. Very simple: you need

• divide the entire region of space into separate fairly small parts (for example, mutually perpendicular planes). In this case, we will get many small cubes inside the body, within which we conventionally consider the differential characteristic to be unchanged, constant.
• multiply the value of the differential characteristic inside each cube by the value of the volume of this cube and sum such products. At this stage we obtain the integral sum. The integral sum is not exactly equal to the integral, but can serve as an approximate value.
• go to the limit of the integral sum when the volume of the cubes of the partition of the body tends to zero. At this stage we obtain the exact value of the linear velocity integral.

Below are calculations of stroke volume (stroke volume of the heart (syn.: systolic blood volume, systolic heart volume, stroke volume of blood) - the volume of blood (in ml) ejected by the ventricle of the heart in one systole) - one of the main quantities in echocardiography, calculated when using the integral of the linear velocity of blood flow.

a - Schemes for calculating stroke volume, a - using the flow continuity equation, b - using the flow continuity equation in the presence of significant mitral regurgitation.

VTI = V cp ET,

where CSA is the cross-sectional area, VTI is the integral of the linear flow velocity, V cp is the average flow velocity in the outflow tract of the left ventricle, ET is the ejection time.

In the case when hemodynamically significant mitral regurgitation is present (more than 2nd degree), the total stroke volume of the left ventricle is calculated using the formula:

TSV = FSV + RSV,

[Linear velocity integral (FVI, or VTI)] = [Flow time (ET)] x [Average blood flow velocity (Vmean)];

Cardiac output can be determined by the integral of the linear velocity of the aortic and pulmonary flow.

In conclusion, I would like to add that my work is not intended for a mathematician who is thoroughly versed in integration, but for any person who has shown an interest in the use of integral in medicine. Therefore, I tried to make it as accessible as possible and interesting even for a child.

Bibliography:

1. Diseases of the heart and blood vessels http://old.consilium-medicum. com/media/bss/06_02/42.shtml
2. Hemodynamics http://ru.wikipedia.org/wiki/% D0%93%D0%B5%D0%BC%D0%BE%D0%B4% D0%B8%D0%BD%D0%B0%D0%BC% D0%B8% D0%BA%D0%B0
3. Integral sign http://ru.wikipedia.org/wiki/% C7%ED%E0%EA_%E8%ED%F2%E5%E3% F0%E0%EB%E0
4. Medical Consilium http://www.consilium-medicum. com/article/7144
5. Basic equations - Heart http://serdce.com.ua/osnovnye- uravneniya
6. Practical guide to ultrasound diagnostics http://euromedcompany.ru/ultrazvuk/prakticheskoe- rukovodstvo-po-ultrazvukovoj- diagnostike

Lesson motto: “Mathematics is the language spoken by all exact sciences” N.I. Lobachevsky

The purpose of the lesson: to summarize students’ knowledge on the topic “Integral”, “Application of the integral”; expand their horizons, knowledge about the possible application of the integral to the calculation of various quantities; consolidate skills in using integrals to solve applied problems; instill cognitive interest in mathematics, develop a culture of communication and a culture of mathematical speech; be able to learn to speak in front of students and teachers.

Lesson type: repetition-summarizing.

Type of lesson: lesson – defense of the project “Application of the Integral”.

Equipment: magnetic board, “Application of the Integral” posters, cards with formulas and tasks for independent work.

Lesson plan:

1. Project protection:

1. from the history of integral calculus;
2. properties of the integral;
3. application of integral in mathematics;
4. application of the integral in physics;

2. Solution of exercises.

During the classes

Teacher: A powerful research tool in mathematics, physics, mechanics and other disciplines is the definite integral - one of the basic concepts of mathematical analysis. The geometric meaning of the integral is the area of ​​a curvilinear trapezoid. The physical meaning of the integral is 1) the mass of an inhomogeneous rod with density, 2) the displacement of a point moving in a straight line with speed over a period of time.

Teacher: The guys in our class did a lot of work; they selected problems where a definite integral is used. They have the floor.

Student 2: Properties of the integral

Student 3: Application of the integral (table on the magnetic board).

Student 4: We are considering the use of integrals in mathematics to calculate the area of ​​figures.

The area of ​​any plane figure, considered in a rectangular coordinate system, can be composed of the areas of curvilinear trapezoids adjacent to the axis Oh and axles OU. Area of ​​a curved trapezoid bounded by a curve y = f(x), axis Oh and two straight lines x=a And x=b, Where a x b, f(x) 0 calculated by the formula cm. rice. If a curved trapezoid is adjacent to the axis OU, then its area is calculated by the formula , cm. rice. When calculating the areas of figures, the following cases may arise: a) The figure is located above the Ox axis and is limited by the Ox axis, the curve y = f (x) and two straight lines x = a and x = b. (See. rice.) The area of ​​this figure is found by formula 1 or 2. b) The figure is located under the Ox axis and is limited by the Ox axis, the curve y=f(x) and two straight lines x=a and x=b (see. rice.). The area is found by the formula . c) The figure is located above and below the Ox axis and is limited by the Ox axis, the curve y=f(x) and two straight lines x=a and x=b( rice.). d) The area is limited by two intersecting curves y = f (x) and y = (x) ( rice.)

5 student: Let's solve the problem

x-2y+4=0 and x+y-5+0 and y=0

Student 7: An integral, widely used in physics. Word to physicists.

1. CALCULATION OF THE PATH TAKEN BY A POINT

The path traveled by a point during uneven movement in a straight line with variable speed over the period of time from to is calculated by the formula.

Examples:

1. Speed ​​of point movement m/s. Find the path traveled by the point in 4 seconds.

Solution: according to the condition, . Hence,

2. Two bodies began to move simultaneously from one point in one direction in a straight line. The first body moves with speed m/s, the second - with speed v = (4t+5) m/s. How far apart will they be after 5 seconds?

Solution: it is obvious that the desired value is the difference in the distances covered by the first and second bodies in 5 s:

3. A body is thrown vertically upward from the surface of the earth with a speed u = (39.2-9.8^) m/s. Find the maximum height of the body lift.

Solution: the body will reach its maximum lifting height at a time t when v = 0, i.e. 39.2- 9.8t = 0, whence I= 4 s. Using formula (1) we find

2. CALCULATION OF FORCE WORK

Work done by variable force f(x) when moving along an axis Oh material point from x = A before x=b, is found by the formula When solving problems of calculating the work of force, Huck's law is often used: F=kx, (3) where F - force N; X- absolute elongation of the spring, m, caused by force F, A k- proportionality coefficient, N/m.

Example:

1. A spring at rest has a length of 0.2 m. A force of 50 N stretches the spring by 0.01 m. How much work must be done to stretch it from 0.22 to 0.32 m?

Solution: using equality (3), we have 50 = 0.01k, i.e. kK = 5000 N/m. We find the limits of integration: a = 0.22 - 0.2 = 0.02 (m), b=0.32- 0.2 = 0.12(m). Now, using formula (2), we get

3. CALCULATION OF WORK PERFORMED WHEN LIFTING A LOAD

Task. A cylindrical tank with a base radius of 0.5 m and a height of 2 m is filled with water. Calculate the work required to pump water out of the tank.

Solution: select a horizontal layer of height dх at depth x ( rice.). The work A that must be done to raise a layer of water weighing P to a height x is equal to Px.

A change in depth x by a small amount dx will cause a change in volume V by the amount dV = pr 2 dx and change in weight P by * dP = 9807 r 2 dx; in this case, the work A performed will change by the value dA = 9807пr 2 xdx. Integrating this equality as x changes from 0 to H, we obtain

4. CALCULATION OF FLUID PRESSURE FORCE

Strength value R the pressure of the liquid on the horizontal platform depends on the depth of immersion X of this area, i.e., from the distance of the area to the surface of the liquid.

The pressure force (N) on the horizontal platform is calculated by the formula P =9807Sx,

Where - liquid density, kg/m3; S - area of ​​the site, m2; X - immersion depth of the platform, m.

If the platform experiencing fluid pressure is not horizontal, then the pressure on it is different at different depths, therefore, the force of pressure on the platform is a function of the depth of its immersion P(x).

5. ARC LENGTH

Let the plane curve AB(rice.) given by the equation y =f(x) (axb), and f(x) And f?(x)- continuous functions in the interval [a,b]. Then the differential dl arc length AB expressed by the formula or , and arc length AB calculated by formula (4)

where a and b are the values ​​of the independent variable X at points A and B. If the curve is given by the equation x =(y) (with y)d), then the length of the arc AB is calculated by the formula (5) where With And d independent variable values at at points A and V.

6. CENTER OF MASS

When finding the center of mass, use the following rules:

1) x coordinate ? center of mass of a system of material points A 1, A 2,..., A n with masses m 1, m 2, ..., m n, located on a straight line at points with coordinates x 1, x 2, ..., x n , are found by the formula

(*); 2) When calculating the coordinates of the center of mass, you can replace any part of the figure with a material point, placing it at the center of mass of this part, and assign to it a mass equal to the mass of the part of the figure under consideration. Example. Let a mass of density (x) be distributed along the rod-segment [a;b] of the Ox axis, where (x) is a continuous function. Let's show that a) the total mass M of the rod is equal to; b) coordinate of the center of mass x " equal to .

Let's split the segment [a; b] into n equal parts with points a= x 0< х 1 < х 2 < ... <х n = b (rice.). On each of the n of these segments, the density can be considered constant for large n and approximately equal to (x k - 1) on the k-th segment (due to the continuity of (x). Then the mass of the k-th segment is approximately equal to and the mass of the entire rod is equal to

Considering each of n small segments as a material point of mass m k placed at point , we obtain from formula (*) that the coordinate of the center of mass is approximately found as follows

Now it remains to note that as n -> the numerator tends to the integral, and the denominator (expressing the mass of the entire rod) tends to the integral

To find the coordinates of the center of mass of a system of material points on a plane or in space, they also use the formula (*)

Teacher: You have a table and problems on your tables. Using the table, find: a) the amount of electricity; b) the mass of the rod based on its density.

Quantities

Derivative calculation

Calculation of the integral Option 1 Option 2 Lesson summary: We completed the topic “Integral”, learned to calculate antiderivatives, integrals, areas of figures, considered the application of the integral in practice, these problems may appear on the Unified State Exam, I think you can handle them.

Integral calculus arose in connection with the solution of problems of determining areas and volumes. 2000 BC the inhabitants of Egypt and Babylon already knew how to approximately determine the area of ​​a circle and knew the rule for calculating the volume of a truncated pyramid. The theoretical basis for the rules for calculating areas and volumes first appeared among the ancient Greeks. Materialist philosopher Democritus V century BC considers bodies as consisting of a large number of small particles. That is, the cone is a set of very thin cylindrical disks of different radii. The problem of squaring the circle played a huge role in the history of integral calculus.(squaring a circle - constructing a square whose area is equal to the area of ​​the given circle). Hippocrates found the exact quadrature of several curvilinear figures (middle 5th century).

The first known method for calculating the integral is the method of exhaustion of Eudoxus (ca. 370 BC). He tried to find areas and volumes by breaking them into an infinite number of parts for which the area or volume was already known. This method was taken up and developed by Archimedes and was used to calculate the areas of parabolas and approximate the area of ​​a circle.In his work “Quadrature of the Parabola,” Archimedes uses the exhaustion method to calculate the area of ​​the sector of the parabola. Those. Archimedes was the first to compile sums, which in our time are called integral sums. The first significant attempts to develop Archimedes' integration methods, which were crowned with success, were made in XVII century, when, on the one hand, significant advances were made in the field of algebra, and on the other hand, economics, technology, and natural science developed more and more intensively, and extensive and in-depth methods for studying and calculating quantities were required.

When calculating the area of ​​a curved trapezoid Newton and Leibniz come to the conceptantiderivative (or primitive) function for a given derivative functionf(X),WhereWITHcould be anything. Tacalled today formula Newton-Leibniz makes it possible to reduce the rather complex calculation of definite integrals, i.e. finding the limits of integral sums leads to a relatively simple operation of finding antiderivatives.Leibniz owns the differential symbol a p Later the integral symbol appearedDefinite integral symbolwas introduced by J. Fourier, and the term “integral” (from Latin integer - whole) was proposed by I. Bernoulli.

Work on the study of the foundations of differential and integral calculus begins in XIX century through the works of O. Cauchy and B. Bolzano. At the same time, Russian mathematicians M.V. made a significant contribution to the development of integral calculus. Ostrogradsky, V.Ya. Bunyakovsky, V.Ya. Chebyshev. This was the time when modern mathematical analysis was just being created. This was, perhaps, the only era of mathematical creativity in its intensity, and Euler united the extensive but scattered material of the new analysis into a solid science.

With time, man acquired more and more power over nature, but the dream of flying to the stars remained just as unrealizable. Science fiction writers mentioned rockets for space flight. However, these missiles were a technically unreasonable dream. The honor of opening people's way to the stars fell to our compatriot K. E. Tsiolkovsky. A whole galaxy of scientists, led by S.P., worked on the tasks of creating an artificial Earth satellite and calculating the trajectory of their entry into orbit. Korolev.

Particularly interesting are problems that are prototypes of problems for calculating the trajectories of spacecraft entering a given orbit, for finding the altitude and speed of ascent or descent of a body, and some other problems using integral calculus.

Problem 1. The speed of rectilinear motion of the body is given

equation Find the equation of path S if in time t = 2 sec the body has traveled 20 m.

Solution: where Let's integrate: where Using the data, we find C = 4. I.e. the equation of motion of the body has the form .

When flying into space, we must take into account all the factors of our environment, and in order to get where we need to go, we need to calculate the trajectory of movement using the initial data. All this must be done before the flight takes place.2016 marks the 55th anniversary of the flight into orbit of the first cosmonaut Yuri Alekseevich Gagarin. When calculating, it was necessary to solve such problems.

Problem 2. It is necessary to launch a rocket weighing P = 2·10 4 N(T) from the surface of the Earth to a heighth= 1500 km.Calculate the work required to run it.

Solution.f – the force of attraction of a body by the Earth is a function of its distance X to the center of the Earth: where On the surface of the Earth where the force of gravity is equal to the weight of the body R, A x = R- radius of the Earth, therefore When a rocket rises from the surface of the Earth to a height h variable X varies fromx = R before x= R+ h. We find the job we are looking for using the formula: Then we get: the work to launch a rocket is equal to

Problem 3. Strength in 10 N stretches the spring 2 cm. What kind of job is she

commits at the same time?

Solution . According to Hooke's law, force F , which stretches the spring, is proportional to the stretch of the spring, i.e.F =kh. From the problem statement

k= 10/0,02(N/m), That F= 500x. Job: .

Problem 4. From a mine deepl= 100 mit is necessary to lift the cage evenly with the weight P 1 = 10 4 N, which hangs on a rope wound on a drum. Calculate total work A full required to lift the cage if the weight of one linear meter of rope R 2= 20N.

Solution . The work of raising the cage: and the work of lifting the rope is proportional to the weight of the rope, i.e. Therefore, the complete work is complete:

Problem 5. The spring bends under the influence of a force of 1.5 10 4 N by 1 cm. How much work must be done to deform the spring by 3 cm? ( The deforming force is proportional to the deflection of the spring.)

Solution . F=kh,Where X- spring deflection. At x = 0.01m we have: . Then the work for deformation is:

Ascent into outer space is difficult and unsafe, but returning to Earth is no less difficult, when the spacecraft must land at a speed of no more than 2 m/s. Only in this case, the device, the instruments in it, and most importantly, the crew members, will not experience a sharp, hard impact. Konstantin Eduardovich Tsiolkovsky decided to use the braking of the spacecraft by the Earth's air envelope. Moving at a speed of 8 m/s, the spacecraft does not fall to Earth. The first stage of descent is turning on the braking motor for a short time. The speed decreases by 0.2 km/s, and the descent begins immediately. Let's consider an example of solving a problem of drawing up a law of motion under given conditions.

Problem 6. Find the law of motion of a freely falling body with constant acceleration g, if at the moment of movement the body was at rest.

Solution:It is known that the acceleration of a rectilinearly moving body is the second derivative of the path S with respect to time t , or derivative of speed with respect to time t: , but, therefore, from where. We integrate: , and From the condition: , from where we find the speed of movement: . Let's find the law of body motion: , or . Let's integrate: , . According to the initial conditions: , from where we find We have the equation of motion of a falling body: - this is a familiar formula of physics.

Problem 7. A body is thrown vertically upward with an initial velocity

Find the equation of motion of this body (neglect air resistance).

Solution:Let us take the vertical direction upward as positive, and the acceleration of gravity as directed downward as negative. We have: , from where . Let's integrate: then . Because and then C 1: and Velocity Equation: Find the law of body motion: because. and then where .We integrate: or When and we will find , and We have the equation of motion of the body: or .

The following example shows the calculation of the trajectory for discharging used sections, unnecessary devices, and materials. In this case, they are sent to Earth, calculating the orbit so that when passing through the atmospheric layers they burn up, and the unburned remains fall to Earth (most often into the ocean) without causing harm.

Problem 8. Write an equation for a curve passing through point M (2; -3) and having a tangent with an angle coefficient .

Solution:The problem statement states: or Integrating, we have: At x = 2 And y = -3, C = - 5, and the motion trajectory has the form: .

Builders sometimes have to solve problems of calculating the areas of unusual figures for which there are no generally known formulas. In this case, integrals come to the rescue again.

Problem 9. Calculate the area of ​​the figure bounded by the lines: and

Solution: Let's construct a drawing (Fig. 1), for which we solve a system of equations. Let's find the points of intersection of the lines: A(-2;4) And B(4;16). The required area is the difference of areas with the limits of integration, a = x 1 = -2 And b = x 2 = 4. Then we have the area:

.

Cosmonauts and scientists, working at the orbital station, solve and study many issues of astronomy, physics, chemistry, medicine, biology, etc., for the purity of the experiment. Let's accompany the following problem with a literary example. The famous science fiction novel by H.G. Wells, “The War of the Worlds,” describes the attack of the Martians on planet Earth, who decided to expand their overpopulated territories by capturing ours, because Earth's climatic conditions were suitable. The seizure of territory and the destruction of earthlings began, who received help from where they had not expected at all. Our “native” bacteria, which we have already learned to fight, having entered the body of the Martians with air, food, water, found in it a favorable environment for their development and reproduction, quickly adapted and, having destroyed the Martians, rid the Earth of the invaders. Let's consider the solution to a problem that gives an idea of ​​this.

Problem 10.The rate of reproduction of some bacteria is proportional to the number of bacteria present at a given time t. The number of bacteria tripled within 5 hours. Find the dependence of the number of bacteria on time.

Solution: Let x(t ) is the number of bacteria at time t, and at the initial moment then the rate of their reproduction. By condition we have: or the following: Let's find C: and the function It is known that i.e. or where the proportionality coefficient is equal to: and the function has the form: .

In the famous novel by A.N. Tolstoy’s “Hyperboloid of Engineer Garin” I would like to feel, feel, what is it - a hyperboloid? What are its dimensions, shape, surface, volume? The next task is about this.

Problem 11.Hyperbole, bounded by lines: y = 0, x = a, x = 2arotates around the OX axis. Find the volume of the resulting hyperboloid (Fig. 2).

Solution.We use the formula to calculate the volume of bodies of revolution around the OX axis using a certain integral:

UFO scientists are studying the facts cited by “eyewitnesses”, saying that they saw a flying spaceship in the form of a huge luminous disk (“plate”), approximately the same shape as in Figure 3. Let us consider the solution to the problem of determining the volume of such a “plate” "

Problem 12. Calculate the volume of a body formed by rotation around the OX axis of the area bounded by lines y = x 2 - 9 And y = 0.

Solution: When drawing a paraboloid (Fig. 3), we have integration limits from x = -3 before x = 3. Due to the symmetry of the figure relative to the op-amp axis, we replace the limits of integration by x = 0 And x = 3, and the result will be doubled. Therefore, the disk volume is equal to:

The economic meaning of a definite integral expresses the volume of production with a known function f(t ) - labor productivity at the moment t . Then the volume of output over the period is calculated using the formula Let's consider an example for an enterprise.

Problem 13. Find the volume of products produced in 4 years if the Cobb-Douglas function has the form

Solution. The volume of products produced by the enterprise is equal to:

To summarize, we can conclude that the use of the integral opens up great opportunities. When studying geometry, they consider calculating the areas of flat figures limited by straight segments (triangles, parallelograms, trapezoids, polygons), and the volumes of bodies obtained by rotating them. The definite integral allows you to calculate the areas of complex figures bounded by any curved lines, as well as find the volumes of bodies obtained by rotating curved trapezoids around any axis.

I would also like to note that the use of a definite integral is not limited only to the calculation of various geometric quantities, but is also used in solving problems from various fields of physics, aerodynamics, astronomy, chemistry and medicine, astronautics, as well as economic problems.

Bibliography:

1. Apanasov, P.T. Collection of problems in mathematics: textbook. allowance / P.T. Apanasov, M.I. Orlov. - M.: Higher School, 1987. - 303 p.
2. Bedenko, N.K. Lessons on algebra and principles of analysis: methodological manual / N.K. Bedenko, L.O. Denishcheva. - M.: Higher School, 1988. - 239 p.
3. Bogomolov, N.V. Practical classes in higher mathematics: textbook. allowance / N.V. Bogomolov. - M.: Higher School, 1973. - 348 p.
4. Higher mathematics for economists: textbook / ed. N.Sh. Kremer. – 3rd ed. – M.: UNITY-DANA, 2008.- 479 p.
5. Zaporozhets, G.I. Guide to solving problems in mathematical analysis: textbook. allowance / G.I. Zaporozhets. - M.: Higher School, 1966. - 460 p.

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"MR combined lesson for the teacher "Fundamentals of integral calculus. Definite integral."

STATE AUTONOMOUS EDUCATIONAL

INSTITUTION OF SECONDARY VOCATIONAL EDUCATION

NOVOSIBIRSK REGION

"BARABINSKY MEDICAL COLLEGE"

METHODOLOGICAL DEVELOPMENT

combined lesson for teacher

DISCIPLINE "MATHEMATICS"

Section 1.Mathematical analysis

Subject1.6. Fundamentals of integral calculus. Definite integral

Speciality

060101 General medicine

Well- first

Methodological sheet

Formation of State Standards requirements when studying the topic

« Fundamentals of integral calculus. Definite integral"

must know:

the importance of mathematics in professional activities and in mastering a professional educational program;

basic mathematical methods for solving applied problems;

fundamentals of integral and differential calculus.

As a result of studying the topic, the student should be able to:

solve applied problems in the field of professional activity;

Lesson objectives:

Educational goals: repeat and consolidate the skills of calculating the indefinite and definite integral, consider methods for calculating definite integrals, consolidate the skill of finding the definite integral

Educational goals: to promote the formation of a culture of communication, attention, interest in the subject, to promote the student’s understanding of the essence and social significance of his future profession, and the manifestation of sustainable interest in it.

Developmental goals:

contribute

developing the ability to use techniques of comparison, generalization, and highlighting the main thing;

development of mathematical horizons, thinking and speech, attention and memory.

Type of activity: combined lesson

Lesson duration: 90 minutes

Interdisciplinary connections: physics, geometry and all subjects where mathematics is used

Literature:

Gilyarova M.G. Mathematics for medical colleges. – Rostov n/d: Phoenix, 2011. – 410, p. - (Medicine)

Mathematics: textbook. allowance / V.S. Mikheev [and others]; edited by N.M. Demina. – Rostov n/d: Phoenix, 2009. – 896 p. – (Secondary vocational education).

Lesson equipment:

Handout

Progress of the lesson

№ p/p	Lesson stage	Time (min)	Guidelines
	Organizational part		Checking students' attendance and appearance. Communicate the topic, purpose and plan of the lesson.
	Motivation		The concept of integral is one of the basic ones in mathematics. By the end of the 17th century. Newton and Leibniz created the apparatus of differential and integral calculus, which forms the basis of mathematical analysis. The study of this topic completes the school course of mathematical analysis, introduces students to a new tool for understanding the world, and consideration in school of the application of integral calculus to the most important branches of physics shows students the importance and power of higher mathematics. The need to fully study the most important elements of integral calculus is associated with the enormous significance and importance of this material when mastering a professional educational program. In the future, knowledge of the definite integral will be useful to you when finding solutions to equations that determine the rate of radioactive decay, bacterial reproduction, muscle contraction, dissolution of a drug in a tablet and many other problems of differential calculus used in medical practice.
	Updating of reference knowledge		It is necessary to test computational skills and knowledge of the table of integrals (Annex 1)
	Presentation of new material		Presentation plan (Appendix 2) Definite integral Properties of the definite integral Newton-Leibniz formula Calculation of definite integrals by various methods Application of a definite integral to the calculation of various quantities. Calculating the area of ​​a flat figure
	Practical part Performing exercises to reinforce the topic material		(Appendix 3) Primary consolidation of acquired knowledge and skills Understanding the acquired knowledge and skills
	Summing up the lesson		Giving marks, commenting on mistakes made during the work
	Homework		Prepare theoretical material for the practical lesson and complete the tasks of the section “Self-control” (Appendix 4)

Annex 1

Updating of reference knowledge

Mathematical dictation

1 option

I.

II.

Option 2

I. Calculate indefinite integrals

II. Name the method for calculating integrals

Appendix 2

Information and reference material

Definite integral

The concept of an integral is related to the inverse problem of differentiating a function. It is convenient to consider the concept of a definite integral in solving the problem of calculating the area of ​​a curvilinear trapezoid.

To find the area of ​​a figure bounded on both sides by perpendiculars restored at points A And b, on top of a continuous curve y =f(X) and below the axis Oh, let's split the segment [A,b] for small sections:

a = x 0 x 1 x 2 ... x n -1 x n = b.

Let's restore perpendiculars from these points to the intersection with the curve y =f(X). Then the area of ​​the entire figure will be approximately equal to the sum of elementary rectangles having a base equal to  X i = x i -X i -1 , and the height is equal to the value of the function f(X) inside each rectangle. The smaller the value  X i, the more accurately the area of ​​the figure will be determined S . Hence:

Definition.If there is a limit to the integral sum that does not depend on the way the segment [a,b] and selecting points, then this limit is called the definite integral of the functionf(X) on the segment [a,b] and denote:

Wheref(x) is the integrand function, x is the integration variable, andb- limits of integration (read: definite integral ofado bef from x de x).

Thus, geometric meaning definite integral is associated with determining the area of ​​a curvilinear trapezoid bounded from above by the function y =f(X), bottom axis Oh, and on the sides - perpendiculars restored at the points A And b.

The process of calculating a definite integral is called integration. Numbers a andb are called accordingly lower and upper limits of integration.

Properties of a definite integral

If the limits of integration are equal, then the definite integral is equal to zero:



If you rearrange the limits of integration, the sign of the integral will change to the opposite:



The constant factor can be taken out of the sign of the definite integral:



Definite integral of the sum of a finite number of continuous functionsf 1 (x), f 2 (x)... f n (x), given on the interval [a,b], is equal to the sum of definite integrals of the summands of the functions:

The integration segment can be divided into parts:



If a function is always positive or always negative on the interval [a,b], then the definite integral is a number of the same sign as the function:

Newton-Leibniz formula

The Newton-Leibniz formula establishes the connection between the definite and indefinite integrals.

Theorem.The value of the definite integral of the functionf(X) on the segment [a,b] is equal to the increment of any of the antiderivatives for this function on a given segment:

From this theorem it follows that a definite integral is a number, while an indefinite integral is a set of antiderivative functions. Thus, according to the formula, to find a definite integral it is necessary:

1. Find the indefinite integral of this function by putting C = 0.

2. Substitute the antiderivative instead of the argument into the expression X upper limit first b, then lower limit A, and subtract the second from the first result.

Calculation of definite integrals by various methods

When calculating definite integrals, use the methods discussed for finding indefinite integrals.

Direct integration method

This method is based on the use of tabular integrals and the basic properties of the definite integral.

EXAMPLES:

1) Find

Solution:

2) Find

Solution:

3) Find

Solution:

Integration variable replacement method

EXAMPLE:

Solution. To find the integral, we use the change of variable method. Introduce a new variable

u=3 x ‑ 1 , Then du = 3 dx, dx = . When introducing a new variable, it is necessary to replace the limits of integration, since the new variable will have different limits of change. They are found using the variable replacement formula. So the upper limit will be equal to And b = 32 ‑ 1 = 5 , lower ‑ And A =31 ‑ 1 = 2 . Replacing the variable and limits of integration, we get:

Method of integration by parts

This method is based on the use of the integration by parts formula for a definite integral:

EXAMPLE:

1) Find

Solution:

Let u = ln x, dv = xdx, Then

Application of a definite integral to the calculation of various quantities.

Calculating the area of ​​a flat figure

It was previously shown that a definite integral can be used to calculate the area of ​​a figure enclosed between the graph of a function y =f(x), axis Oh and two straight lines X = a and x =b.

If the function y =f(x) is below the abscissa line, i.e. f(x)

If the function y =f(x) crosses the axis several times Oh, then it is necessary to separately find the areas for the plots when f(x) 0, and add them to the absolute values ​​of the areas when the function f(x)

EXAMPLE 1. Find the area of ​​a figure bounded by a function y = sinX and axis Oh Location on 0  X 2.

Solution. The area of ​​the figure will be equal to the sum of the areas:

S = S 1 + | S 2 |,

where S 1 - ; area at at0 ; S 2 - area at at 0.

S=2 + 2 = 4 sq. units.

EXAMPLE 2. Find the area of ​​the figure enclosed between the curve y = x 2 , axis Oh and straight x = 0, x = 2.

Solution. Let's build function graphs at= x 2 And x = 2.

The shaded area will be the desired area of ​​the figure. Because f(x) 0,then

Calculating the arc length of a plane curve

If the curve y =f(X) on the segment [A,b] has a continuous derivative, then the arc length of this curve is found by the formula:

EXAMPLE

Find the arc length of a curve y 2 = x 3 on the segment (y0)

Solution

The equation of the curve is y = x 3/2, then y’ = 1.5 x 1/2.

Making the replacement 1+we get:

Let's return to the original variable:

Calculation volume of a body of revolution

If a curved trapezoid bounded by a curve y =f(x) and straight x=a And x=b, rotates around an axis Oh, then the volume of rotation is calculated by the formula:

EXAMPLE

Find the volume of a body formed by rotation around an axis Oh half-wave sinusoid
y= sin x, at 0≤ x≤.

Solution

According to the formula we have:

To calculate this integral we will make the following transformations:

Appendix 3

Primary consolidation of the studied material

1. Calculation of definite integrals

2. Applications of the definite integral

Area of ​​the figure

Calculate the area of ​​the figure bounded by the lines:

The path traveled by a body (point) during rectilinear motion over a period of time fromt 1 beforet 2 (

v =3 t 2 +2 t -1 (tin c,vin m/s). Find the distance traveled by the body in 10 s from the start of the movement.

The speed of a point varies according to the law v =6 t 2 +4 (tin c,vin m/s). Find the path traveled by the point in 5s from the beginning of the movement.

Point movement speed v =12 t -3 t 2 (tin c,vin m/s). Find the path traveled by the point from the start of its movement to its stop.

Two bodies began to move simultaneously from one point in one direction in a straight line. The first body moves with speed v =6 t 2 +2 t(m/s), second
v =4 t+5 (m/s). At what distance from each other will they be after 5s?

Appendix 4

Self-monitoring on the topic

"Definite integral and its application"

1 option 1. Evaluate Integrals 2. y = - x 2 + x + 6 And y = 0 3. The speed of a point varies according to the law v =9 t 2 -8 t (tin c,vin m/s). Find the path traveled by the body in the fourth second from the start of its motion.

Option 2 1. Evaluate Integrals 2. Calculate the area of ​​a figure bounded by lines y = - x 2 + 2 x + 3 And y = 0 3. The speed of a point varies according to the law v = 8 t - 3 t 2 (tin c,vin m/s). Find the distance traveled by the body in five seconds from the start of the movement.

I. In physics

Work of force

(A=FScos, cos 1)

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

the increment in the kinetic energy of a particle over time dt is equal to the scalar product Fds, where ds is the movement of the particle over time dt. Magnitude

is called the work done by force F.

Let the point move along the OX axis under the influence of a force, the projection of which onto the OX axis is a function f(x) (f is a continuous function). Under the influence of force, the point moved from point S1(a) to S2(b). Divide the segment into n segments of equal length

The work done by the force will be equal to the sum of the work done by the force on the resulting segments. Because f(x) is continuous, then for small the work done by the force on this segment is equal to

Similarly, on the second segment f(x1)(x2-x1), on the nth segment --

f(xn-1)(b-xn-1).

Therefore the work is equal to:

And An = f(a)x +f(x1)x+...+f(xn-1)x= ((b-a)/n)(f(a)+f(x1)+...+f( xn-1))

The approximate equality becomes exact as n

A = lim [(b-a)/n] (f(a)+...+f(xn-1))= f(x)dx (by definition)

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

Ep = A= - (-F(s)) dx

From the course of mechanics it is known that

From here we find

Ep= - (-Cs)ds = CS2/2 | = C/2 l2/4

Answer: Cl2/8.

How much work must be done to stretch the spring by 4 cm, if it is known that under a load of 1 N it stretches by 1 cm?

According to Hooke's law, the force X N stretching the spring by x is equal to

We find the proportionality coefficient k from the condition: if x = 0.01 m, then X = 1 N, therefore, k = 1/0.01 = 100 and X = 100x. Then

Answer: A=0.08 J

Using a crane, a reinforced concrete hollow is removed from the bottom of a river 5 m deep. What work will be done if the hollow has the shape of a regular tetrahedron with an edge of 1 m? The density of reinforced concrete is 2500 kg/m3, the density of water is 1000 kg/m3.

Height of tetrahedron

tetrahedron volume

The weight of the hole in water, taking into account the action of the Archimedean force, is equal to

Now let's find the work Ai when removing the gouge from the water. Let the vertex of the tetrahedron reach a height of 5+y, then the volume of the small tetrahedron emerging from the water is equal, and the weight of the tetrahedron is:

Hence,

Hence A=A0+A1=7227.5 J + 2082.5 J = 9310 J = 9.31 kJ

Answer: A=9.31 (J).

What pressure force does a rectangular plate of length a and width b (a>b) experience if it is inclined to the horizontal surface of the liquid at an angle b and its larger side is at a depth h?

Center of mass coordinates

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

Let a material homogeneous plate o have the shape of a curvilinear trapezoid (x;y |axb; 0yf(x)) and the function

is continuous on , and the area of ​​this curved trapezoid is equal to S, then the coordinates of the center of mass of the plate o are found using the formulas:

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

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

Let's draw a semicircle in the OXY coordinate system.

y = (1/2S) (R2-x2)dx = (1/R2) (R2-x2)dx = (1/R2)(R2x-x3/3)|= 4R/3

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

Find the coordinates of the center of gravity of the figure bounded by the ellipse arc x=acost, y=bsint, located in the first quarter, and the coordinate axes.

In the first quarter, as x increases from 0 to a, the value of t decreases from p/2 to 0, therefore

Using the formula for the area of ​​an ellipse S=рab, we get

The path traveled by a material point

If a material point moves rectilinearly with speed =(t) and in time

T= t2-t1 (t2>t1)

passed the path S, then

In geometry

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

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 quantity.

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

Let's find a formula for calculating volume:

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

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

Let's introduce an auxiliary function S(x) that specifies the following correspondence: to each x from the segment we associate the cross-sectional area of ​​this figure with a plane passing through a given point x perpendicular to the OX axis.

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

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

and the volume of the part enclosed between two adjacent planes is equal to the volume of the cylinder Vc = SmainH.

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

where S(x) is the section of the plane passing through the selected point perpendicular to the OX axis.

To find the volume you need:

• 1) Select 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 this body with a plane perpendicular to the OX axis and passing through the corresponding point.
• 4) Express through known quantities a function expressing the area of ​​a given section.
• 5) Compose an integral.
• 6) After calculating the integral, find the volume.

Find the volume of a triaxial ellipse

Plane sections of an ellipsoid parallel to the xOz plane and spaced from it at a distance y=h represent an ellipse