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Keywords: integral, curvilinear trapezoid, area of ​​figures bounded by lilies

Equipment: whiteboard, computer, multimedia projector

Lesson type: lesson-lecture

Lesson Objectives:

  • educational: to form a culture of mental work, to create a situation of success for each student, to form a positive motivation for learning; develop the ability to speak and listen to others.
  • developing: the formation of the independence of the student's thinking in the application of knowledge in various situations, the ability to analyze and draw conclusions, the development of logic, the development of the ability to correctly pose questions and find answers to them. Improving the formation of computational, calculating skills, developing the thinking of students in the course of performing the proposed tasks, developing an algorithmic culture.
  • educational: to form concepts about a curvilinear trapezoid, about an integral, to master the skills of calculating the areas of flat figures

Teaching method: explanatory and illustrative.

During the classes

In the previous classes, we learned how to calculate the areas of figures whose boundaries are broken lines. In mathematics, there are methods that allow you to calculate the area of ​​\u200b\u200bfigures bounded by curves. Such figures are called curvilinear trapezoids, and their area is calculated using antiderivatives.

Curvilinear trapezoid ( slide 1)

A curvilinear trapezoid is a figure bounded by the function graph, ( w.m.), straight x = a And x = b and abscissa

Various types of curvilinear trapezoids ( slide 2)

We consider various types of curvilinear trapezoids and notice: one of the lines is degenerate into a point, the role of the limiting function is played by the line

Area of ​​a curvilinear trapezoid (slide 3)

Fix the left end of the interval A, and right X we will change, i.e., we move the right wall of the curvilinear trapezoid and get a changing figure. The area of ​​a variable curvilinear trapezoid bounded by the function graph is the antiderivative F for function f

And on the segment [ a; b] the area of ​​the curvilinear trapezoid formed by the function f, is equal to the increment of the antiderivative of this function:

Exercise 1:

Find the area of ​​a curvilinear trapezoid bounded by the graph of a function: f(x) = x 2 and direct y=0, x=1, x=2.

Solution: ( according to the slide 3 algorithm)

Draw a graph of the function and lines

Find one of the antiderivatives of the function f(x) = x 2 :

Slide Self-Check

Integral

Consider a curvilinear trapezoid given by the function f on the segment [ a; b]. Let's break this segment into several parts. The area of ​​the entire trapezoid will be divided into the sum of the areas of smaller curvilinear trapezoids. ( slide 5). Each such trapezoid can be approximately considered a rectangle. The sum of the areas of these rectangles gives an approximate idea of ​​the entire area of ​​the curvilinear trapezoid. The smaller we break the segment [ a; b], the more accurately we calculate the area.

We write these considerations in the form of formulas.

Divide the segment [ a; b] into n parts with dots x 0 \u003d a, x1, ..., xn \u003d b. Length k- th denote by xk = xk - xk-1. Let's sum up

Geometrically, this sum is the area of ​​the figure shaded in the figure ( sh.m.)

Sums of the form are called integral sums for the function f. (sch.m.)

Integral sums give an approximate value of the area. The exact value is obtained by passing to the limit. Imagine that we refine the partition of the segment [ a; b] so that the lengths of all small segments tend to zero. Then the area of ​​the composed figure will approach the area of ​​the curvilinear trapezoid. We can say that the area of ​​a curvilinear trapezoid is equal to the limit of integral sums, Sk.t. (sch.m.) or integral, i.e.,

Definition:

function integral f(x) from a before b is called the limit of integral sums

= (sch.m.)

Newton-Leibniz formula.

Remember that the limit of integral sums is equal to the area of ​​a curvilinear trapezoid, so we can write:

Sk.t. = (sch.m.)

On the other hand, the area of ​​a curvilinear trapezoid is calculated by the formula

S to. t. (sch.m.)

Comparing these formulas, we get:

= (sch.m.)

This equality is called the Newton-Leibniz formula.

For the convenience of calculations, the formula is written as:

= = (sch.m.)

Tasks: (sch.m.)

1. Calculate the integral using the Newton-Leibniz formula: ( check slide 5)

2. Compile integrals according to the drawing ( check on slide 6)

3. Find the area of ​​​​a figure bounded by lines: y \u003d x 3, y \u003d 0, x \u003d 1, x \u003d 2. ( Slide 7)

Finding the areas of plane figures ( slide 8)

How to find the area of ​​figures that are not curvilinear trapezoids?

Let two functions be given, the graphs of which you see on the slide . (sch.m.) Find the area of ​​the shaded figure . (sch.m.). Is the figure in question a curvilinear trapezoid? And how can you find its area, using the additivity property of the area? Consider two curvilinear trapezoids and subtract the area of ​​the other from the area of ​​one of them ( w.m.)

Let's make an algorithm for finding the area from the animation on the slide:

  1. Plot Functions
  2. Project the intersection points of the graphs onto the x-axis
  3. Shade the figure obtained by crossing the graphs
  4. Find curvilinear trapezoids whose intersection or union is the given figure.
  5. Calculate the area of ​​each
  6. Find difference or sum of areas

Oral task: How to get the area of ​​a shaded figure (tell using animation, slide 8 and 9)

Homework: Work out the abstract, No. 353 (a), No. 364 (a).

Bibliography

  1. Algebra and the beginning of analysis: a textbook for grades 9-11 of the evening (shift) school / ed. G.D. Glazer. - M: Enlightenment, 1983.
  2. Bashmakov M.I. Algebra and the beginning of analysis: a textbook for grades 10-11 of middle school / Bashmakov M.I. - M: Enlightenment, 1991.
  3. Bashmakov M.I. Mathematics: a textbook for institutions beginning. and avg. prof. education / M.I. Bashmakov. - M: Academy, 2010.
  4. Kolmogorov A.N. Algebra and the beginning of analysis: a textbook for 10-11 cells. educational institutions / A.N. Kolmogorov. - M: Enlightenment, 2010.
  5. Ostrovsky S.L. How to make a presentation for the lesson? / S.L. Ostrovsky. – M.: First of September, 2010.

    This term has other meanings, see Trapezium (meanings). Trapeze (from other Greek τραπέζιον "table"; ... Wikipedia

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Task 1(on the calculation of the area of ​​a curvilinear trapezoid).

In the Cartesian rectangular coordinate system xOy, a figure is given (see figure), bounded by the x axis, straight lines x \u003d a, x \u003d b (a curvilinear trapezoid. It is required to calculate the area of ​​\u200b\u200bthe curvilinear trapezoid.
Solution. Geometry gives us recipes for calculating the areas of polygons and some parts of a circle (sector, segment). Using geometric considerations, we will be able to find only an approximate value of the required area, arguing as follows.

Let's split the segment [a; b] (base of a curvilinear trapezoid) into n equal parts; this partition is feasible with the help of points x 1 , x 2 , ... x k , ... x n-1 . Let us draw lines through these points parallel to the y-axis. Then the given curvilinear trapezoid will be divided into n parts, into n narrow columns. The area of ​​the entire trapezoid is equal to the sum of the areas of the columns.

Consider separately the k-th column, i.e. curvilinear trapezoid, the base of which is a segment. Let's replace it with a rectangle with the same base and height equal to f(x k) (see figure). The area of ​​the rectangle is \(f(x_k) \cdot \Delta x_k \), where \(\Delta x_k \) is the length of the segment; it is natural to consider the compiled product as an approximate value of the area of ​​the kth column.

If we now do the same with all the other columns, then we arrive at the following result: the area S of a given curvilinear trapezoid is approximately equal to the area S n of a stepped figure made up of n rectangles (see figure):
\(S_n = f(x_0)\Delta x_0 + \dots + f(x_k)\Delta x_k + \dots + f(x_(n-1))\Delta x_(n-1) \)
Here, for the sake of uniformity of notation, we consider that a \u003d x 0, b \u003d x n; \(\Delta x_0 \) - segment length , \(\Delta x_1 \) - segment length , etc; while, as we agreed above, \(\Delta x_0 = \dots = \Delta x_(n-1) \)

So, \(S \approx S_n \), and this approximate equality is the more accurate, the larger n.
By definition, it is assumed that the desired area of ​​the curvilinear trapezoid is equal to the limit of the sequence (S n):
$$ S = \lim_(n \to \infty) S_n $$

Task 2(about moving a point)
A material point moves in a straight line. The dependence of speed on time is expressed by the formula v = v(t). Find the displacement of a point over the time interval [a; b].
Solution. If the motion were uniform, then the problem would be solved very simply: s = vt, i.e. s = v(b-a). For uneven motion, one has to use the same ideas on which the solution of the previous problem was based.
1) Divide the time interval [a; b] into n equal parts.
2) Consider a time interval and assume that during this time interval the speed was constant, such as at time t k . So, we assume that v = v(t k).
3) Find the approximate value of the point displacement over the time interval , this approximate value will be denoted by s k
\(s_k = v(t_k) \Delta t_k \)
4) Find the approximate value of the displacement s:
\(s \approx S_n \) where
\(S_n = s_0 + \dots + s_(n-1) = v(t_0)\Delta t_0 + \dots + v(t_(n-1)) \Delta t_(n-1) \)
5) The required displacement is equal to the limit of the sequence (S n):
$$ s = \lim_(n \to \infty) S_n $$

Let's summarize. The solutions of various problems were reduced to the same mathematical model. Many problems from various fields of science and technology lead to the same model in the process of solution. So, this mathematical model should be specially studied.

The concept of a definite integral

Let us give a mathematical description of the model that was built in the three considered problems for the function y = f(x), which is continuous (but not necessarily non-negative, as was assumed in the considered problems) on the segment [a; b]:
1) split the segment [a; b] into n equal parts;
2) sum $$ S_n = f(x_0)\Delta x_0 + f(x_1)\Delta x_1 + \dots + f(x_(n-1))\Delta x_(n-1) $$
3) compute $$ \lim_(n \to \infty) S_n $$

In the course of mathematical analysis, it was proved that this limit exists in the case of a continuous (or piecewise continuous) function. He is called a definite integral of the function y = f(x) over the segment [a; b] and are denoted like this:
\(\int\limits_a^b f(x) dx \)
The numbers a and b are called the limits of integration (lower and upper, respectively).

Let's return to the tasks discussed above. The definition of area given in problem 1 can now be rewritten as follows:
\(S = \int\limits_a^b f(x) dx \)
here S is the area of ​​the curvilinear trapezoid shown in the figure above. This is what geometric meaning of the definite integral.

The definition of the displacement s of a point moving in a straight line with a speed v = v(t) over the time interval from t = a to t = b, given in Problem 2, can be rewritten as follows:

Newton - Leibniz formula

To begin with, let's answer the question: what is the relationship between a definite integral and an antiderivative?

The answer can be found in problem 2. On the one hand, the displacement s of a point moving along a straight line with a speed v = v(t) over a time interval from t = a to t = b and is calculated by the formula
\(S = \int\limits_a^b v(t) dt \)

On the other hand, the coordinate of the moving point is the antiderivative for the speed - let's denote it s(t); hence the displacement s is expressed by the formula s = s(b) - s(a). As a result, we get:
\(S = \int\limits_a^b v(t) dt = s(b)-s(a) \)
where s(t) is the antiderivative for v(t).

The following theorem was proved in the course of mathematical analysis.
Theorem. If the function y = f(x) is continuous on the segment [a; b], then the formula
\(S = \int\limits_a^b f(x) dx = F(b)-F(a) \)
where F(x) is the antiderivative for f(x).

This formula is usually called Newton-Leibniz formula in honor of the English physicist Isaac Newton (1643-1727) and the German philosopher Gottfried Leibniz (1646-1716), who received it independently of each other and almost simultaneously.

In practice, instead of writing F(b) - F(a), they use the notation \(\left. F(x)\right|_a^b \) (it is sometimes called double substitution) and, accordingly, rewrite the Newton-Leibniz formula in this form:
\(S = \int\limits_a^b f(x) dx = \left. F(x)\right|_a^b \)

Calculating a definite integral, first find the antiderivative, and then carry out a double substitution.

Based on the Newton-Leibniz formula, one can obtain two properties of a definite integral.

Property 1. The integral of the sum of functions is equal to the sum of the integrals:
\(\int\limits_a^b (f(x) + g(x))dx = \int\limits_a^b f(x)dx + \int\limits_a^b g(x)dx \)

Property 2. The constant factor can be taken out of the integral sign:
\(\int\limits_a^b kf(x)dx = k \int\limits_a^b f(x)dx \)

Calculating the areas of plane figures using a definite integral

Using the integral, you can calculate the area not only of curvilinear trapezoids, but also of plane figures of a more complex type, such as the one shown in the figure. The figure P is bounded by straight lines x = a, x = b and graphs of continuous functions y = f(x), y = g(x), and on the segment [a; b] the inequality \(g(x) \leq f(x) \) holds. To calculate the area S of such a figure, we will proceed as follows:
\(S = S_(ABCD) = S_(aDCb) - S_(aABb) = \int\limits_a^b f(x) dx - \int\limits_a^b g(x) dx = \)
\(= \int\limits_a^b (f(x)-g(x))dx \)

So, the area S of the figure bounded by the straight lines x = a, x = b and the graphs of functions y = f(x), y = g(x), continuous on the segment and such that for any x from the segment [a; b] the inequality \(g(x) \leq f(x) \) is satisfied, is calculated by the formula
\(S = \int\limits_a^b (f(x)-g(x))dx \)

Table of indefinite integrals (antiderivatives) of some functions

$$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n+1))(n+1 ) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch )x+C $$

Calculating the area of ​​a figure This is perhaps one of the most difficult problems in area theory. In school geometry, they are taught to find the areas of basic geometric shapes such as, for example, a triangle, a rhombus, a rectangle, a trapezoid, a circle, etc. However, one often has to deal with the calculation of the areas of more complex figures. It is in solving such problems that it is very convenient to use integral calculus.

Definition.

Curvilinear trapezoid some figure G is called, bounded by the lines y = f(x), y = 0, x = a and x = b, and the function f(x) is continuous on the segment [a; b] and does not change its sign on it (Fig. 1). The area of ​​a curvilinear trapezoid can be denoted by S(G).

The definite integral ʃ a b f(x)dx for the function f(x), which is continuous and non-negative on the segment [a; b], and is the area of ​​the corresponding curvilinear trapezoid.

That is, to find the area of ​​\u200b\u200bthe figure G, bounded by the lines y \u003d f (x), y \u003d 0, x \u003d a and x \u003d b, it is necessary to calculate the definite integral ʃ a b f (x) dx.

Thus, S(G) = ʃ a b f(x)dx.

If the function y = f(x) is not positive on [a; b], then the area of ​​the curvilinear trapezoid can be found by the formula S(G) = -ʃ a b f(x)dx.

Example 1

Calculate the area of ​​\u200b\u200bthe figure bounded by the lines y \u003d x 3; y = 1; x = 2.

Solution.

The given lines form the figure ABC, which is shown by hatching on rice. 2.

The desired area is equal to the difference between the areas of the curvilinear trapezoid DACE and the square DABE.

Using the formula S = ʃ a b f(x)dx = S(b) – S(a), we find the limits of integration. To do this, we solve a system of two equations:

(y \u003d x 3,
(y = 1.

Thus, we have x 1 \u003d 1 - the lower limit and x \u003d 2 - the upper limit.

So, S = S DACE - S DABE = ʃ 1 2 x 3 dx - 1 = x 4 /4| 1 2 - 1 \u003d (16 - 1) / 4 - 1 \u003d 11/4 (square units).

Answer: 11/4 sq. units

Example 2

Calculate the area of ​​\u200b\u200bthe figure bounded by lines y \u003d √x; y = 2; x = 9.

Solution.

The given lines form the figure ABC, which is bounded from above by the graph of the function

y \u003d √x, and from below the graph of the function y \u003d 2. The resulting figure is shown by hatching on rice. 3.

The desired area is equal to S = ʃ a b (√x - 2). Let's find the limits of integration: b = 9, to find a, we solve the system of two equations:

(y = √x,
(y = 2.

Thus, we have that x = 4 = a is the lower limit.

So, S = ∫ 4 9 (√x – 2)dx = ∫ 4 9 √x dx –∫ 4 9 2dx = 2/3 x√x| 4 9 - 2x| 4 9 \u003d (18 - 16/3) - (18 - 8) \u003d 2 2/3 (square units).

Answer: S = 2 2/3 sq. units

Example 3

Calculate the area of ​​\u200b\u200bthe figure bounded by the lines y \u003d x 3 - 4x; y = 0; x ≥ 0.

Solution.

Let's plot the function y \u003d x 3 - 4x for x ≥ 0. To do this, we find the derivative y ':

y’ = 3x 2 – 4, y’ = 0 at х = ±2/√3 ≈ 1.1 are critical points.

If we draw the critical points on the real axis and place the signs of the derivative, we get that the function decreases from zero to 2/√3 and increases from 2/√3 to plus infinity. Then x = 2/√3 is the minimum point, the minimum value of the function y is min = -16/(3√3) ≈ -3.

Let's determine the intersection points of the graph with the coordinate axes:

if x \u003d 0, then y \u003d 0, which means that A (0; 0) is the point of intersection with the Oy axis;

if y \u003d 0, then x 3 - 4x \u003d 0 or x (x 2 - 4) \u003d 0, or x (x - 2) (x + 2) \u003d 0, from where x 1 \u003d 0, x 2 \u003d 2, x 3 \u003d -2 (not suitable, because x ≥ 0).

Points A(0; 0) and B(2; 0) are the intersection points of the graph with the Ox axis.

The given lines form the OAB figure, which is shown by hatching on rice. 4.

Since the function y \u003d x 3 - 4x takes on (0; 2) a negative value, then

S = |ʃ 0 2 (x 3 – 4x)dx|.

We have: ʃ 0 2 (x 3 - 4x)dx = (x 4 /4 - 4x 2 /2)| 0 2 \u003d -4, from where S \u003d 4 square meters. units

Answer: S = 4 sq. units

Example 4

Find the area of ​​the figure bounded by the parabola y \u003d 2x 2 - 2x + 1, the straight lines x \u003d 0, y \u003d 0 and the tangent to this parabola at the point with the abscissa x 0 \u003d 2.

Solution.

First, we compose the equation of the tangent to the parabola y \u003d 2x 2 - 2x + 1 at the point with the abscissa x₀ \u003d 2.

Since the derivative y' = 4x - 2, then for x 0 = 2 we get k = y'(2) = 6.

Find the ordinate of the touch point: y 0 = 2 2 2 – 2 2 + 1 = 5.

Therefore, the tangent equation has the form: y - 5 \u003d 6 (x - 2) or y \u003d 6x - 7.

Let's build a figure bounded by lines:

y \u003d 2x 2 - 2x + 1, y \u003d 0, x \u003d 0, y \u003d 6x - 7.

Г y \u003d 2x 2 - 2x + 1 - parabola. Points of intersection with the coordinate axes: A(0; 1) - with the Oy axis; with the Ox axis - there are no intersection points, because the equation 2x 2 - 2x + 1 = 0 has no solutions (D< 0). Найдем вершину параболы:

x b \u003d 2/4 \u003d 1/2;

y b \u003d 1/2, that is, the vertex of the parabola point B has coordinates B (1/2; 1/2).

So, the figure whose area is to be determined is shown by hatching on rice. 5.

We have: S O A B D \u003d S OABC - S ADBC.

Find the coordinates of point D from the condition:

6x - 7 = 0, i.e. x \u003d 7/6, then DC \u003d 2 - 7/6 \u003d 5/6.

We find the area of ​​triangle DBC using the formula S ADBC ​​= 1/2 · DC · BC. Thus,

S ADBC ​​= 1/2 5/6 5 = 25/12 sq. units

S OABC = ʃ 0 2 (2x 2 - 2x + 1)dx = (2x 3 /3 - 2x 2 /2 + x)| 0 2 \u003d 10/3 (square units).

Finally we get: S O A B D \u003d S OABC - S ADBC ​​\u003d 10/3 - 25/12 \u003d 5/4 \u003d 1 1/4 (sq. units).

Answer: S = 1 1/4 sq. units

We have reviewed examples finding the areas of figures bounded by given lines. To successfully solve such problems, you need to be able to build lines and graphs of functions on a plane, find the points of intersection of lines, apply a formula to find the area, which implies the ability and skills to calculate certain integrals.

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