Consider a curvilinear trapezoid bounded by the Ox axis, a curve y \u003d f (x) and two straight lines: x \u003d a and x \u003d b (Fig. 85). Take an arbitrary value of x (only not a and not b). Let us give it an increment h = dx and consider a strip bounded by straight lines AB and CD, by the Ox axis, and by an arc BD belonging to the curve under consideration. This strip will be called the elementary strip. The area of ​​an elementary strip differs from the area of ​​the rectangle ACQB by a curvilinear triangle BQD, and the area of ​​the latter is less than the area of ​​the rectangle BQDM with sides BQ = =h=dx) QD=Ay and area equal to hAy = Ay dx. As the side h decreases, the side Du also decreases and, simultaneously with h, tends to zero. Therefore, the area of ​​BQDM is infinitesimal of the second order. The area of ​​the elementary strip is the area increment, and the area of ​​the rectangle ACQB, equal to AB-AC==/(x) dx> is the area differential. Therefore, we find the area itself by integrating its differential. Within the limits of the figure under consideration, the independent variable l: changes from a to b, so the required area 5 will be equal to 5= \f (x) dx. (I) Example 1. Calculate the area bounded by the parabola y - 1 -x *, the straight lines X \u003d - Fj-, x \u003d 1 and the axis O * (Fig. 86). at Fig. 87. Fig. 86. 1 Here f(x) = 1 - l?, the limits of integration a = - and t = 1, therefore 3) |_ 2 3V 2 / J 3 24 24* Example 2. Calculate the area bounded by the sinusoid y = sinXy, the Ox axis and the straight line (Fig. 87). Applying formula (I), we obtain L 2 S \u003d J sinxdx \u003d [-cos x] Q \u003d 0 - (-1) \u003d lf Example 3. Calculate the area bounded by the arc of the sinusoid ^y \u003d sin jc enclosed between two adjacent intersection points with the Ox axis (for example, between the origin and the point with the abscissa i). Note that from geometric considerations it is clear that this area will be twice the area of ​​the previous example. However, let's do the calculations: i 5= | s \ nxdx \u003d [ - cosx) * - - cos i- (- cos 0) \u003d 1 + 1 \u003d 2. o Indeed, our assumption turned out to be fair. Example 4. Calculate the area bounded by the sinusoid and the ^ axis Ox on one period (Fig. 88). Preliminary ras-figure judgments suggest that the area will turn out to be four times larger than in pr. 2. However, after doing the calculations, we get “i G, * i S - \ sin x dx \u003d [- cos x] 0 = = - cos 2n - (-cos 0) \u003d - 1 + 1 \u003d 0. This result requires clarification. To clarify the essence of the matter, we also calculate the area bounded by the same sinusoid y \u003d sin l: and the Ox axis ranging from l to 2n. Applying formula (I), we obtain Thus, we see that this area turned out to be negative. Comparing it with the area calculated in Ex. 3, we find that their absolute values ​​are the same, but the signs are different. If we apply property V (see Ch. XI, § 4), then we get by accident. Always the area below the x-axis, provided that the independent variable changes from left to right, is obtained by calculating using integrals negative. In this course, we will always consider unsigned areas. Therefore, the answer in the example just analyzed will be as follows: the desired area is 2 + |-2| = 4. Example 5. Let's calculate the area of ​​the BAB shown in Fig. 89. This area is limited by the axis Ox, the parabola y = - xr and the straight line y - = -x + \. Area of ​​a curvilinear trapezoid The required area of ​​OAB consists of two parts: OAM and MAB. Since point A is the point of intersection of the parabola and the straight line, we will find its coordinates by solving the system of equations 3 2 Y \u003d mx. (we only need to find the abscissa of point A). Solving the system, we find l; =~. Therefore, the area has to be calculated in parts, first pl. OAM, and then pl. MAV: .... G 3 2, 3 G xP 3 1/2 Y 2. QAM-^x (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].
Decision. 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 constructed 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's 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 $$

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Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.

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.

Decision: ( 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.

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Example1 . Calculate the area of ​​the figure bounded by lines: x + 2y - 4 = 0, y = 0, x = -3, and x = 2

Let's build a figure (see Fig.) We build a straight line x + 2y - 4 \u003d 0 along two points A (4; 0) and B (0; 2). Expressing y in terms of x, we get y \u003d -0.5x + 2. According to formula (1), where f (x) \u003d -0.5x + 2, a \u003d -3, b \u003d 2, we find

S \u003d \u003d [-0.25 \u003d 11.25 sq. units

Example 2 Calculate the area of ​​​​the figure bounded by lines: x - 2y + 4 \u003d 0, x + y - 5 \u003d 0 and y \u003d 0.

Decision. Let's build a figure.

Let's build a straight line x - 2y + 4 = 0: y = 0, x = - 4, A (-4; 0); x = 0, y = 2, B(0; 2).

Let's construct a straight line x + y - 5 = 0: y = 0, x = 5, С(5; 0), x = 0, y = 5, D(0; 5).

Find the point of intersection of the lines by solving the system of equations:

x = 2, y = 3; M(2; 3).

To calculate the required area, we divide the AMC triangle into two triangles AMN and NMC, since when x changes from A to N, the area is limited by a straight line, and when x changes from N to C, it is a straight line

For triangle AMN we have: ; y \u003d 0.5x + 2, i.e. f (x) \u003d 0.5x + 2, a \u003d - 4, b \u003d 2.

For the NMC triangle we have: y = - x + 5, i.e. f(x) = - x + 5, a = 2, b = 5.

Calculating the area of ​​each of the triangles and adding the results, we find:

sq. units

sq. units

9 + 4, 5 = 13.5 sq. units Check: = 0.5AC = 0.5 sq. units

Example 3 Calculate the area of ​​a figure bounded by lines: y = x 2 , y = 0, x = 2, x = 3.

AT this case it is required to calculate the area of ​​a curvilinear trapezoid bounded by a parabola y = x 2 , straight lines x \u003d 2 and x \u003d 3 and the Ox axis (see Fig.) According to formula (1), we find the area of ​​\u200b\u200ba curvilinear trapezoid

= = 6kv. units

Example 4 Calculate the area of ​​​​a figure bounded by lines: y \u003d - x 2 + 4 and y = 0

Let's build a figure. The desired area is enclosed between the parabola y \u003d - x 2 + 4 and axis Oh.

Find the points of intersection of the parabola with the x-axis. Assuming y \u003d 0, we find x \u003d Since this figure is symmetrical about the Oy axis, we calculate the area of ​​\u200b\u200bthe figure located to the right of the Oy axis, and double the result: \u003d + 4x] sq. units 2 = 2 sq. units

Example 5 Calculate the area of ​​a figure bounded by lines: y 2 = x, yx = 1, x = 4

Here it is required to calculate the area of ​​the curvilinear trapezoid bounded by the upper branch of the parabola y 2 \u003d x, the Ox axis and straight lines x \u003d 1x \u003d 4 (see Fig.)

According to formula (1), where f(x) = a = 1 and b = 4, we have = (= sq. units

Example 6 . Calculate the area of ​​the figure bounded by lines: y = sinx, y = 0, x = 0, x= .

The desired area is limited by a half-wave sinusoid and the Ox axis (see Fig.).

We have - cosx \u003d - cos \u003d 1 + 1 \u003d 2 square meters. units

Example 7 Calculate the area of ​​the figure bounded by lines: y \u003d - 6x, y \u003d 0 and x \u003d 4.

The figure is located under the Ox axis (see Fig.).

Therefore, its area is found by the formula (3)

= =

Example 8 Calculate the area of ​​\u200b\u200bthe figure bounded by the lines: y \u003d and x \u003d 2. We will build the curve y \u003d by points (see figure). Thus, the area of ​​\u200b\u200bthe figure is found by the formula (4)

Example 9 .

X 2 + y 2 = r 2 .

Here you need to calculate the area bounded by the circle x 2 + y 2 = r 2 , i.e. the area of ​​a circle of radius r centered at the origin. Let's find the fourth part of this area, taking the limits of integration from 0

dor; we have: 1 = = [

Hence, 1 =

Example 10 Calculate the area of ​​\u200b\u200bthe figure bounded by lines: y \u003d x 2 and y = 2x

This figure is limited by the parabola y \u003d x 2 and straight line y \u003d 2x (see Fig.) To determine the intersection points of the given lines, we solve the system of equations: x 2 – 2x = 0 x = 0 and x = 2

Using formula (5) to find the area, we obtain

= }