Determine the number of solutions from the antiderivative graph. Antiderivative and integrals
Municipal Autonomous General Educational Institution
"Secondary school No. 56 with in-depth study of mathematics" of the city of Magnitogorsk
Methodological development of the lesson
mathematics
Antiderivatives and a definite integral on the exam. Overview of USE assignments on the topic "Antiprimitive")
for 11th grade students
(summary lesson)
Filimonova Tatyana Mikhailovna
Magnitogorsk 2018
annotation
The lesson is intended for 11th grade students. The topic of the lesson is “Anti-derivative and definite integral on the exam.Overview of USE assignments on the topic "Antiprimitive". The stage of training on this topic is the final one. The motivation for studying this topic is provided through the use of ICT, the use of various types of tasks, the involvement of FIPI tasks and the tasks of the Reshu Unified State Exam website. The priority goal in the lesson is the application of the acquired knowledge, the development of skills, the solution of problems with the exam.
Explanatory note
Methodological development is the development of a specific lesson in mathematics using ICT tools. The relevance of the development lies in the fact that students solve the problem of finding the area of \u200b\u200ba figure using different methods. Different ways of solving one problem, visibility, historical information and the presence of interdisciplinary connections contribute to the development of cognitive interest in mathematics, awareness of the importance of mathematics in everyday life.
During the test, students repeat theoretical information about the antiderivative and the integral, which will help them systematize the theory on this topic and prepare for the upcoming exam.
Lesson summary
Lesson type: summary lesson.
Goals:
Educational:
Formation of educational, cognitive and informational competencies, through generalization, systematization of knowledge on the topic "Primitive.Integral".
Educational:
Formation of informational, general cultural competencies through the development of cognitive activity, interest in the subject, creative abilities of students, broadening their horizons, developing mathematical speech.
Educational:
Formation of communicative competence and competence of personal self-improvement, through work on communication skills, the ability to work in cooperation, on the development of such personal qualities as organization, discipline.
Equipment:PC, projector, screen.
During the classes
I. Organizational moment:
Hello guys! I am glad to welcome you to the lesson.Cthe purpose of our lesson is to generalize, systematize knowledge on the topic “Antiprimitive. Integral”, to prepare for the upcoming exam.
II . Checking homework:
Find the area of a figure bounded by linesy= x2 , y=. The solution is on the slide.
On the board, a task has been prepared in advance to derive the formula for the volume of a ball.
2 people take turns coming to the board and briefly explaining the solution that
The rest are checking.
I II . Warm up.
Each student is given a test.
Collect completed tests.
Parsing of tasks is carried out frontally according to the displayed tasks on the screen.
I V . Mathematical relay.
Now on the road! The ascent to the "Peak of Knowledge" will not be easy, there may be blockages, collapses, and drifts. But there are also halts, where not only tasks are waiting for you. To move forward, you need to show knowledge.
Students for each desk receive sheets with tasks on the topic "Antiprimitive".
1. The value of the antiderivativeF( x) functionsf( x)=11 x+5 at 0 equals 6. FindF(-3).
2. The value of the antiderivativeF( x) functionsf( x)=8 cosxat the point -π is equal to 13. FindF( π /6).
3. The value of the antiderivative functionF( x) functionsf( x)=6 at point 0 is equal to -18. FindF(ln3).
4. The figure shows a graph of the antiderivativey= F( x) functionsf( x) and eight points on the x-axis: , , …, . At how many of these points does the functionf( x) is positive?
5. The figure shows a graph of the antiderivative y \u003dF( x) functionsf( x) and eight points on the x-axis: , , , …,. At how many of these points does the functionf(x)negative?
V . Halt.
“A happy accident falls only to the prepared minds” (Louis Pasteur).
Information from the history of integral calculus is read out. Newspapers prepared by students on the history of integral calculus are shown. Newspapers are dedicated to Newton and Leibniz.
VI. The most difficult climb.
The next task is supposed to be done in writing, so students work in notebooks.
Task. In how many ways can you find the area of a figure bounded by lines (slide)
Who has suggestions? (the figure consists of two curvilinear trapezoids and a rectangle) (choose a solution method, slide)
After discussing this problem, a record appears on the slide
1 way: S=S1 +S2 +S
2 way: S=S1 +SABCD-SOCD
Two students decide at the blackboard followed by an explanation of the solution, the rest of the students work in notebooks, choosing one of the solutions.
Conclusion (students do): we found two ways to solve this problem, getting the same result. Discuss which way is easier.
Everyone is very tired, but the closer to the goal, the tasks become easier and easier.
VSH. Lesson summary (slides)
“Thinking begins with surprise,” Aristotle observed 2,500 years ago. Our compatriot Sukhomlinsky believed that “the feeling of surprise is a powerful source of desire to know; from surprise to knowledge - one step. And mathematics is a wonderful subject for surprise.
Integrals are used for:
solving problems from the field of physics;
solving economic problems (for optimizing the work of a company in a competitive environment, calculating the profitability of a consumer loan);
solving socio-demographic problems (mathematical model of the Earth's population, etc.).
IX . Homework. (slide)
The assignment compiled by the teacher on the site "I will solve the exam".
X . Putting marks.
Bibliography
Vilenkin N.Ya. and etc. Algebra and beginning of mathematical analysis. Grade 11. V. Ch.2. (profile level). - M.: Mnemosyne, 2009. - 264 p.
Aleksandrova L.A. Algebra and beginning of mathematical analysis. Grade 11. Independent work. - M.: Mnemozina, 2009. - 100 p.
3. Shipova T.A. Algebra and the beginnings of analysis: Derivative. Definite integral. Tests. - M.: School-Press, 1996. - 64 p.
4. Website metaschool.ru for the development of lessons.
5. Website I will solve the Unified State Examination, catalog of tasks, primitive.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight line segments). Using the figure, compute F(9)-F(5), where F(x) is one of the antiderivatives of f(x).
Show SolutionSolution
According to the Newton-Leibniz formula, the difference F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of the curvilinear trapezoid bounded by the graph of the function y=f(x), straight lines y=0 , x=9 and x=5. According to the graph, we determine that the specified curvilinear trapezoid is a trapezoid with bases equal to 4 and 3 and a height of 3.
Its area is equal to \frac(4+3)(2)\cdot 3=10.5.
Answer
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of the function y=F(x) - one of the antiderivatives of some function f(x) defined on the interval (-5; 5). Using the figure, determine the number of solutions to the equation f(x)=0 on the segment [-3; 4].
Solution
According to the definition of the antiderivative, the equality holds: F "(x) \u003d f (x). Therefore, the equation f (x) \u003d 0 can be written as F "(x) \u003d 0. Since the figure shows the graph of the function y=F(x), we need to find those interval points [-3; 4], in which the derivative of the function F(x) is equal to zero. It can be seen from the figure that these will be the abscissas of the extreme points (maximum or minimum) of the F(x) graph. There are exactly 7 of them on the indicated interval (four minimum points and three maximum points).
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight line segments). Using the figure, compute F(5)-F(0), where F(x) is one of the antiderivatives of f(x).
Solution
According to the Newton-Leibniz formula, the difference F(5)-F(0), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of the curvilinear trapezoid bounded by the graph of the function y=f(x), straight lines y=0 , x=5 and x=0. According to the graph, we determine that the specified curvilinear trapezoid is a trapezoid with bases equal to 5 and 3 and a height of 3.
Its area is equal to \frac(5+3)(2)\cdot 3=12.
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of the function y=F(x) — one of the antiderivatives of some function f(x), defined on the interval (-5; 4). Using the figure, determine the number of solutions to the equation f (x) = 0 on the segment (-3; 3].
Solution
According to the definition of the antiderivative, the equality holds: F "(x) \u003d f (x). Therefore, the equation f (x) \u003d 0 can be written as F "(x) \u003d 0. Since the figure shows the graph of the function y=F(x), we need to find those interval points [-3; 3], in which the derivative of the function F(x) is equal to zero.
It can be seen from the figure that these will be the abscissas of the extreme points (maximum or minimum) of the F(x) graph. There are exactly 5 of them on the specified interval (two minimum points and three maximum points).
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of some function y=f(x). The function F(x)=-x^3+4.5x^2-7 is one of the antiderivatives of the function f(x).
Find the area of the shaded figure.
Solution
The shaded figure is a curvilinear trapezoid bounded from above by the graph of the function y=f(x), the straight lines y=0, x=1 and x=3. According to the Newton-Leibniz formula, its area S is equal to the difference F(3)-F(1), where F(x) is the antiderivative of the function f(x) specified in the condition. That's why S= F(3)-F(1)= -3^3 +(4,5)\cdot 3^2 -7-(-1^3 +(4,5)\cdot 1^2 -7)= 6,5-(-3,5)= 10.
Answer
Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Job type: 7
Topic: An antiderivative of a function
Condition
The figure shows a graph of some function y=f(x). The function F(x)=x^3+6x^2+13x-5 is one of the antiderivatives of the function f(x). Find the area of the shaded figure.
One of the operations of differentiation is finding the derivative (differential) and applying it to the study of functions.
Equally important is the inverse problem. If the behavior of a function is known in the vicinity of each point of its definition, then how to restore the function as a whole, i.e. over the entire range of its definition. This problem is the subject of study of the so-called integral calculus.
Integration is the reverse action of differentiation. Or the restoration of the function f(x) from the given derivative f`(x). The Latin word “integro” means restoration.
Example #1.
Let (f(x))' = 3x 2 . Find f(x).
Solution:
Based on the differentiation rule, it is easy to guess that f (x) \u003d x 3, because
(x 3) ' = 3x 2 However, it is easy to see that f (x) is found ambiguously. As f (x) you can take f (x) \u003d x 3 +1 f (x) \u003d x 3 +2 f (x) \u003d x 3 -3, etc.
Because the derivative of each of them is 3x2. (The derivative of the constant is 0). All these functions differ from each other by a constant term. Therefore, the general solution of the problem can be written as f(x)= x 3 +C, where C is any constant real number.
Any of the found functions f(x) is called primitive for the function F`(x) = 3x 2
Definition.
The function F(x) is called antiderivative for the function f(x) on a given interval J, if for all x from this interval F`(x) = f(x). So the function F (x) \u003d x 3 is antiderivative for f (x) \u003d 3x 2 on (- ∞ ; ∞). Since, for all x ~ R, the equality is true: F`(x)=(x 3)`=3x 2
As we have already noticed, this function has an infinite set of antiderivatives.
Example #2.
The function is antiderivative for all on the interval (0; +∞), because for all h from this interval, the equality holds.
The task of integration is to find all of its antiderivatives for a given function. The following assertion plays an important role in solving this problem:
A sign of the constancy of a function. If F "(x) \u003d 0 on some interval I, then the function F is a constant on this interval.
Proof.
Let us fix some x 0 from the interval I. Then for any number x from such an interval, by virtue of the Lagrange formula, one can specify such a number c between x and x 0 that
F (x) - F (x 0) \u003d F "(c) (x-x 0).
By condition, F’ (c) = 0, since c ∈1, therefore,
F(x) - F(x 0) = 0.
So, for all x from the interval I
i.e. the function F remains constant.
All antiderivative functions f can be written using one formula, which is called general form of antiderivatives for the function f. The following theorem is true ( basic property of primitives):
Theorem. Any antiderivative for the function f on the interval I can be written as
F(x) + C, (1) where F(x) is one of the antiderivatives for the function f(x) on the interval I, and C is an arbitrary constant.
Let us explain this statement, in which two properties of the antiderivative are briefly formulated:
- whatever number we put in expression (1) instead of C, we get the antiderivative for f on the interval I;
- whichever antiderivative Ф for f on the interval I is taken, one can choose such a number C that for all x from the interval I the equality will be satisfied
Proof.
- By condition, the function F is the antiderivative for f on the interval I. Therefore, F "(x) \u003d f (x) for any x∈1, therefore (F (x) + C)" \u003d F "(x) + C" \u003d f(x)+0=f(x), i.e. F(x) + C is the antiderivative for the function f.
- Let Ф (х) be one of the antiderivatives for the function f on the same interval I, i.e. Ф "(x) = f (х) for all x∈I.
Then (Ф (x) - F (x)) "= Ф" (x) - F '(x) = f (x) - f (x) \u003d 0.
It follows from here. due to the sign of the constancy of the function, that the difference Ф (х) - F (х) is a function that takes some constant value C on the interval I.
Thus, for all x from the interval I, the equality Ф(х) - F(x)=С is true, which was to be proved. The main property of the antiderivative can be given a geometric meaning: graphs of any two antiderivatives for the function f are obtained from each other by parallel translation along the y-axis
Questions for abstracts
The function F(x) is an antiderivative for the function f(x). Find F(1) if f(x)=9x2 - 6x + 1 and F(-1) = 2.
Find all antiderivatives for a function
For the function (x) = cos2 * sin2x, find the antiderivative F(x) if F(0) = 0.
For a function, find the antiderivative whose graph passes through the point
Function F(x ) called primitive for function f(x) on a given interval, if for all x from this interval the equality
F"(x ) = f(x ) .
For example, the function F(x) = x 2 f(x ) = 2X , because
F "(x) \u003d (x 2 )" = 2x = f(x). ◄
The main property of the antiderivative
If F(x) is the antiderivative for the function f(x) on a given interval, then the function f(x) has infinitely many antiderivatives, and all these antiderivatives can be written as F(x) + C, Where WITH is an arbitrary constant.
For example. Function F(x) = x 2 + 1 is the antiderivative for the function f(x ) = 2X , because F "(x) \u003d (x 2 + 1 )" = 2 x = f(x); function F(x) = x 2 - 1 is the antiderivative for the function f(x ) = 2X , because F "(x) \u003d (x 2 - 1)" = 2x = f(x) ; function F(x) = x 2 - 3 is the antiderivative for the function f(x) = 2X , because F "(x) \u003d (x 2 - 3)" = 2 x = f(x); any function F(x) = x 2 + WITH , Where WITH is an arbitrary constant, and only such a function is antiderivative for the function f(x) = 2X . ◄ |
Rules for computing antiderivatives
- If F(x) - original for f(x) , A G(x) - original for g(x) , That F(x) + G(x) - original for f(x) + g(x) . In other words, the antiderivative of the sum is equal to the sum of the antiderivatives .
- If F(x) - original for f(x) , And k is constant, then k · F(x) - original for k · f(x) . In other words, the constant factor can be taken out of the sign of the derivative .
- If F(x) - original for f(x) , And k,b- permanent, and k ≠ 0 , That 1 / k F( k x + b ) - original for f(k x + b) .
Indefinite integral
Indefinite integral from function f(x) called expression F(x) + C, that is, the set of all antiderivatives of the given function f(x) . The indefinite integral is denoted as follows:
∫ f(x) dx = F(x) + C ,
f(x)- called integrand ;
f(x) dx- called integrand ;
x - called integration variable ;
F(x) is one of the antiderivatives of the function f(x) ;
WITH is an arbitrary constant.
For example, ∫ 2 x dx =X 2 + WITH , ∫ cosx dx = sin X + WITH and so on. ◄
The word "integral" comes from the Latin word integer , which means "restored". Considering the indefinite integral of 2 x, we sort of restore the function X 2 , whose derivative is 2 x. Restoring a function from its derivative, or, what is the same, finding an indefinite integral over a given integrand, is called integration this function. Integration is the inverse operation of differentiation. In order to check whether the integration is correct, it suffices to differentiate the result and obtain the integrand.
Basic properties of the indefinite integral
- The derivative of the indefinite integral is equal to the integrand:
- The constant factor of the integrand can be taken out of the integral sign:
- The integral of the sum (difference) of functions is equal to the sum (difference) of the integrals of these functions:
- If k,b- permanent, and k ≠ 0 , That
(∫ f(x) dx )" = f(x) .
∫ k · f(x) dx = k · ∫ f(x) dx .
∫ ( f(x) ± g(x ) ) dx = ∫ f(x) dx ± ∫ g(x ) dx .
∫ f( k x + b) dx = 1 / k F( k x + b ) + C .
Table of antiderivative and indefinite integrals
| f(x)
| F(x) + C
| ∫
f(x) dx = F(x) + C
|
|
| I. | $$0$$ | $$C$$ | $$\int 0dx=C$$ |
| II. | $$k$$ | $$kx+C$$ | $$\int kdx=kx+C$$ |
| III. | $$x^n~(n\neq-1)$$ | $$\frac(x^(n+1))(n+1)+C$$ | $$\int x^ndx=\frac(x^(n+1))(n+1)+C$$ |
| IV. | $$\frac(1)(x)$$ | $$\ln |x|+C$$ | $$\int\frac(dx)(x)=\ln |x|+C$$ |
| v. | $$\sin x$$ | $$-\cos x+C$$ | $$\int\sin x~dx=-\cos x+C$$ |
| VI. | $$\cos x$$ | $$\sin x+C$$ | $$\int\cos x~dx=\sin x+C$$ |
| VII. | $$\frac(1)(\cos^2x)$$ | $$\textrm(tg) ~x+C$$ | $$\int\frac(dx)(\cos^2x)=\textrm(tg) ~x+C$$ |
| VIII. | $$\frac(1)(\sin^2x)$$ | $$-\textrm(ctg) ~x+C$$ | $$\int\frac(dx)(\sin^2x)=-\textrm(ctg) ~x+C$$ |
| IX. | $$e^x$$ | $$e^x+C$$ | $$\int e^xdx=e^x+C$$ |
| x. | $$a^x$$ | $$\frac(a^x)(\ln a)+C$$ | $$\int a^xdx=\frac(a^x)(\ln a)+C$$ |
| XI. | $$\frac(1)(\sqrt(1-x^2))$$ | $$\arcsin x +C$$ | $$\int\frac(dx)(\sqrt(1-x^2))=\arcsin x +C$$ |
| XII. | $$\frac(1)(\sqrt(a^2-x^2))$$ | $$\arcsin \frac(x)(a)+C$$ | $$\int\frac(dx)(\sqrt(a^2-x^2))=\arcsin \frac(x)(a)+C$$ |
| XIII. | $$\frac(1)(1+x^2)$$ | $$\textrm(arctg) ~x+C$$ | $$\int \frac(dx)(1+x^2)=\textrm(arctg) ~x+C$$ |
| XIV. | $$\frac(1)(a^2+x^2)$$ | $$\frac(1)(a)\textrm(arctg) ~\frac(x)(a)+C$$ | $$\int \frac(dx)(a^2+x^2)=\frac(1)(a)\textrm(arctg) ~\frac(x)(a)+C$$ |
| XV. | $$\frac(1)(\sqrt(a^2+x^2))$$ | $$\ln|x+\sqrt(a^2+x^2)|+C$$ | $$\int\frac(dx)(\sqrt(a^2+x^2))=\ln|x+\sqrt(a^2+x^2)|+C$$ |
| XVI. | $$\frac(1)(x^2-a^2)~(a\neq0)$$ | $$\frac(1)(2a)\ln \begin(vmatrix)\frac(x-a)(x+a)\end(vmatrix)+C$$ | $$\int\frac(dx)(x^2-a^2)=\frac(1)(2a)\ln \begin(vmatrix)\frac(x-a)(x+a)\end(vmatrix)+ C$$ |
| XVII. | $$\textrm(tg) ~x$$ | $$-\ln |\cos x|+C$$ | $$\int \textrm(tg) ~x ~dx=-\ln |\cos x|+C$$ |
| XVIII. | $$\textrm(ctg) ~x$$ | $$\ln |\sin x|+C$$ | $$\int \textrm(ctg) ~x ~dx=\ln |\sin x|+C$$ |
| XIX. | $$ \frac(1)(\sin x) $$ | $$\ln \begin(vmatrix)\textrm(tg) ~\frac(x)(2)\end(vmatrix)+C $$ | $$\int \frac(dx)(\sin x)=\ln \begin(vmatrix)\textrm(tg) ~\frac(x)(2)\end(vmatrix)+C $$ |
| XX. | $$ \frac(1)(\cos x) $$ | $$\ln \begin(vmatrix)\textrm(tg)\left (\frac(x)(2)+\frac(\pi )(4) \right) \end(vmatrix)+C $$ | $$\int \frac(dx)(\cos x)=\ln \begin(vmatrix)\textrm(tg)\left (\frac(x)(2)+\frac(\pi )(4) \right ) \end(vmatrix)+C $$ |
| The primitive and indefinite integrals given in this table are usually called tabular primitives
And table integrals
. |
|||
Definite integral
Let in between [a; b] given a continuous function y = f(x) , Then definite integral from a to b functions f(x) is called the increment of the primitive F(x) this function, that is
$$\int_(a)^(b)f(x)dx=F(x)|(_a^b) = ~~F(a)-F(b).$$
Numbers a And b are called respectively lower And top integration limits.
Basic rules for calculating the definite integral
1. \(\int_(a)^(a)f(x)dx=0\);
2. \(\int_(a)^(b)f(x)dx=- \int_(b)^(a)f(x)dx\);
3. \(\int_(a)^(b)kf(x)dx=k\int_(a)^(b)f(x)dx,\) where k - constant;
4. \(\int_(a)^(b)(f(x) ± g(x))dx=\int_(a)^(b)f(x) dx±\int_(a)^(b) g(x) dx \);
5. \(\int_(a)^(b)f(x)dx=\int_(a)^(c)f(x)dx+\int_(c)^(b)f(x)dx\);
6. \(\int_(-a)^(a)f(x)dx=2\int_(0)^(a)f(x)dx\), where f(x) is an even function;
7. \(\int_(-a)^(a)f(x)dx=0\), where f(x) is an odd function.
Comment . In all cases, it is assumed that the integrands are integrable on numerical intervals whose boundaries are the limits of integration.
Geometric and physical meaning of the definite integral
| geometric sense definite integral | physical meaning
definite integral |
 |  |
Square S curvilinear trapezoid (a figure bounded by a graph of continuous positive on the interval [a; b] functions f(x) , axis Ox and direct x=a , x=b ) is calculated by the formula $$S=\int_(a)^(b)f(x)dx.$$ | Path s, which the material point has overcome, moving in a straight line with a speed that changes according to the law v(t)
, for a time interval a ;
b], then the area of the figure bounded by the graphs of these functions and straight lines x = a
, x = b
, is calculated by the formula $$S=\int_(a)^(b)(f(x)-g(x))dx.$$ |
 | For example. Calculate the area of a figure bounded by lines y=x 2 And y= 2- x . We will schematically depict the graphs of these functions and highlight the figure whose area needs to be found in a different color. To find the limits of integration, we solve the equation: x 2 = 2- x ; x 2 + x- 2 = 0 ; x 1 = -2, x 2 = 1 . $$S=\int_(-2)^(1)((2-x)-x^2)dx=$$ |
$$=\int_(-2)^(1)(2-x-x^2)dx=\left (2x-\frac(x^2)(2)-\frac(x^3)(2) \right )\bigm|(_(-2)^(~1))=4\frac(1)(2). $$ ◄ |
|
Volume of the body of revolution
 | If the body is obtained as a result of rotation about the axis Ox curvilinear trapezoid bounded by a graph of continuous and non-negative on the interval [a; b] functions y = f(x) and direct x = a And x = b , then it is called body of revolution . The volume of a body of revolution is calculated by the formula $$V=\pi\int_(a)^(b)f^2(x)dx.$$ If the body of revolution is obtained as a result of the rotation of a figure bounded above and below by function graphs y = f(x) And y = g(x) , respectively, then $$V=\pi\int_(a)^(b)(f^2(x)-g^2(x))dx.$$ |
 | For example. Calculate the volume of a cone with a radius r
and height h
. Let us place the cone in a rectangular coordinate system so that its axis coincides with the axis Ox
, and the center of the base was located at the origin of coordinates. Generator rotation AB defines a cone. Since the equation AB $$\frac(x)(h)+\frac(y)(r)=1,$$ $$y=r-\frac(rx)(h)$$ |
| and for the volume of the cone we have $$V=\pi\int_(0)^(h)(r-\frac(rx)(h))^2dx=\pi r^2\int_(0)^(h)(1-\frac( x)(h))^2dx=-\pi r^2h\cdot \frac((1-\frac(x)(h))^3)(3)|(_0^h)=-\pi r^ 2h\left (0-\frac(1)(3) \right)=\frac(\pi r^2h)(3).$$ ◄ |
|
antiderivative function f(x) in between (a;b) such a function is called F(x), that equality holds for any X from the given interval.
If we take into account the fact that the derivative of the constant WITH equals zero, then the equality holds. So the function f(x) has many prototypes F(x)+C, for an arbitrary constant WITH, and these antiderivatives differ from each other by an arbitrary constant value.
Definition of the indefinite integral.
The whole set of antiderivatives of a function f(x) is called the indefinite integral of this function and is denoted 
.
The expression is called integrand, A f(x) – integrand. The integrand is the differential of the function f(x).
The action of finding an unknown function by its given differential is called uncertain integration, because the result of integration is more than one function F(x), and the set of its primitives F(x)+C.
The geometric meaning of the indefinite integral. The graph of the antiderivative D(x) is called the integral curve. In the x0y coordinate system, the graphs of all antiderivatives of a given function represent a family of curves that depend on the value of the constant C and are obtained one from the other by a parallel shift along the 0y axis. For the example above, we have:
J 2 x^x = x2 + C.
The family of antiderivatives (x + C) is interpreted geometrically as a set of parabolas.
If you need to find one from the family of antiderivatives, then additional conditions are set that allow you to determine the constant C. Usually, for this purpose, initial conditions are set: for the value of the argument x = x0, the function has the value D(x0) = y0.
Example. It is required to find that of the antiderivatives of the function y \u003d 2 x, which takes the value 3 at x0 \u003d 1.
The desired antiderivative: D(x) = x2 + 2.
Solution. ^2x^x = x2 + C; 12 + C = 3; C = 2.
2. Basic properties of the indefinite integral
1. The derivative of the indefinite integral is equal to the integrand:
2. The differential of the indefinite integral is equal to the integrand:
3. The indefinite integral of the differential of some function is equal to the sum of this function itself and an arbitrary constant:
4. A constant factor can be taken out of the integral sign:
5. The integral of the sum (difference) is equal to the sum (difference) of the integrals:
6. The property is a combination of properties 4 and 5:
7. The invariance property of the indefinite integral:
If 
, That
8. Property:
If 
, That
In fact, this property is a special case of integration using the change of variable method, which is discussed in more detail in the next section.
Consider an example:
3. integration method, in which the given integral is reduced to one or more table integrals by identical transformations of the integrand (or expression) and applying the properties of the indefinite integral, is called direct integration. When reducing this integral to a tabular one, the following transformations of the differential are often used (the operation " bringing under the sign of the differential»):
At all, f'(u)du = d(f(u)). this (formula is very often used in the calculation of integrals.
Find the integral
Solution. We use the properties of the integral and reduce this integral to several tabular ones.
4. Integration by the substitution method.
The essence of the method is that we introduce a new variable, express the integrand in terms of this variable, and as a result we arrive at a tabular (or simpler) form of the integral.
Very often, the substitution method helps out when integrating trigonometric functions and functions with radicals.
Example.
Find the indefinite integral 
.
Solution.
Let's introduce a new variable . Express X through z:
We perform the substitution of the obtained expressions into the original integral: 
From the table of antiderivatives we have 
.
It remains to return to the original variable X:
Answer:
