Task 1. (On the calculation of the area of \u200b\u200bthe curvilinear trapezium).

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

We break the segment [A; b] (the basis of the curvilinear trapezium) on n equal parts; This partition is carried out with the help of points x 1, x 2, ... x k, ... x n-1. We will spend directly direct, parallel axes. Then the specified curvilinear trapezium breaks on N parts, on N narrow columns. The area of \u200b\u200bthe entire trapezium is equal to the sum of the area of \u200b\u200bthe columns.

Consider a separate K-b color, i.e. A curvilinear trapezium, the base of which serves a segment. Replace it with a rectangle with the same base and a height of F (x k) (see Figure). The area of \u200b\u200bthe rectangle is equal to \\ (f (x_k) \\ cdot \\ delta x_k \\), where \\ (\\ delta x_k \\) is the length of the segment; Naturally consider the composed of the work with the approximate value of the area of \u200b\u200bthe K-th column.

If you now do the same with all the other columns, we will come to the following result: the area S of a given curved trapezion is approximately equal to the area S N stepped figure composed of n rectangles (see figure):
\\ (S_n \u003d 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 the designation, we believe that a \u003d x 0, b \u003d x n; \\ (\\ Delta x_0 \\) - length of the segment, \\ (\\ delta x_1 \\) - length length, etc.; At the same time, as we agreed above, \\ (\\ Delta X_0 \u003d \\ DOTS \u003d \\ Delta X_ (N-1) \\)

So, \\ (s \\ approx s_n \\), and this is an approximate equality, the more accurately, the more n.
By definition, it is believed that the desired area of \u200b\u200bthe curvilinear trapezium is equal to the sequence limit (s n):
$$ S \u003d \\ Lim_ (n \\ to \\ infty) s_n $$

Task 2. (about moving point)
The material point is moving on the straight. The dependence of the speed on time is expressed by the formula V \u003d V (T). Find the movement of the point over the time interval [A; b].
Decision. If the movement was uniform, then the task would be very simple: S \u003d VT, i.e. S \u003d V (B - A). For uneven traffic, you have to use the same ideas on which the decision of the previous task was based.
1) we divide the time interval [A; b] on n equal parts.
2) Consider the time interval and we assume that during this period of time the speed was constant, such as at the time of T k. So, we believe that V \u003d V (T k).
3) Find the approximate value of the movement of the point over the time interval, this is an approximate value indicate S K
\\ (S_K \u003d V (T_K) \\ Delta T_K \\)
4) Find the approximate movement of s:
\\ (s \\ approx s_n \\) where
\\ (S_n \u003d s_0 + \\ dots + s_ (n-1) \u003d V (T_0) \\ Delta T_0 + \\ DOTS + V (T_ (N - 1)) \\ Delta T_ (N-1) \\)
5) the desired movement is equal to the sequence limit (s):
$$ S \u003d \\ Lim_ (n \\ to \\ infty) s_n $$

Let's summarize. Solutions of various tasks have been drove to the same mathematical model. Many challenges from various areas of science and technology lead in the process of solving the same model. So this mathematical model must be specifically learned.

The concept of a specific integral

We give a mathematical description of the model that was constructed in the three considered tasks for the function y \u003d f (x), continuous (but not necessarily nonnegative, as it was assumed in the considered tasks) on the segment [A; b]:
1) divide the segment [A; b] on n equal parts;
2) We make an amount $$ s_n \u003d f (x_0) \\ Delta X_0 + F (x_1) \\ Delta X_1 + \\ DOTS + F (X_ (N-1)) \\ Delta X_ (N-1) $$
3) Calculate $$ \\ LIM_ (N \\ To \\ Infty) s_n $$

In the course of mathematical analysis, it is proved that this limit in the case of a continuous (or piecewise continuous) function exists. He's called a specific integral from the function y \u003d f (x) by segment [A; b] And denote:
\\ (\\ int \\ limits_a ^ b f (x) dx \\)
The numbers a and b are called the limits of integration (respectively by the lower and upper).

Let us return to the above tasks. The definition of an area given in task 1 can now rewrite as follows:
\\ (S \u003d \\ int \\ limits_a ^ b f (x) dx \\)
Here S is the area of \u200b\u200bthe curvilinear trapezoid depicted in the figure above. This is consisting the geometric meaning of a specific integral.

Determining the movement s point moving in a straight line with a speed V \u003d V (T) during the period of time from T \u003d A to T \u003d B, given in Task 2, you can rewrite it:

Newton's Formula - Leibnia

To begin with, they will answer the question: what is the relationship between a specific integral and primitive?

The answer can be found in problem 2. On the one hand, the movement s point moving in a straight line with a speed V \u003d V (T) during the period of time from T \u003d A to T \u003d B and is calculated by the formula
\\ (S \u003d \\ int \\ limits_a ^ b V (t) dt \\)

On the other hand, the coordinate of the moving point is a primitive for the speed - denote its S (T); It means that the movement S is expressed by the formula S \u003d S (B) - S (a). As a result, we get:
\\ (S \u003d \\ int \\ limits_a ^ b V (t) dt \u003d s (b) -s (a) \\)
where S (T) is primitive for V (T).

The following theorem is proved in the course of mathematical analysis.
Theorem. If the function y \u003d f (x) is continuous on the segment [A; b], then the formula is valid
\\ (S \u003d \\ int \\ limits_a ^ b f (x) dx \u003d f (b) -f (a) \\)
where f (x) is primitive for f (x).

The resulting formula is usually called newton Formula - Leibnia In honor of the English Physics of Isaac Newton (1643-1727) and the German philosopher of Gottfried Leibnitsa (1646-1716), which received it independently from each other and almost simultaneously.

In practice, instead of recording F (B) - F (a), they use the record \\ (\\ left. F (x) \\ right | _a ^ b \\) (it is sometimes called it double substitution) And, accordingly, rewrite Newton's formula - Leibnitsa in this form:
\\ (S \u003d \\ int \\ limits_a ^ b f (x) dx \u003d \\ left. F (x) \\ right | _a ^ b \\)

Computing certain integral, first find the primitive, and then carry out a double substitution.

Relying on Newton's Formula - Leibnitsa, you can get two properties of a specific integral.

Property 1. The integral from the amount of functions is equal to the sum of the integrals:
\\ (\\ int \\ limits_a ^ b (f (x) + g (x)) dx \u003d \\ int \\ limits_a ^ b f (x) dx + \\ int \\ limits_a ^ b g (x) dx \\)

Property 2. A permanent multiplier can be reached by the integral sign:
\\ (\\ int \\ limits_a ^ b kf (x) dx \u003d k \\ int \\ limits_a ^ b f (x) dx \\)

Calculation of flat features using a specific integral

With the help of the integral, you can calculate the area not only curvilinear trapeats, but also flat figures more complex view, for example, this presented in the figure. Figure p is limited to straight x \u003d a, x \u003d b and graphs of continuous functions y \u003d f (x), y \u003d g (x), and on the segment [A; b] The inequality \\ (G (x) \\ LEQ F (X) \\) is performed. To calculate the square s of such a figure, we will act as follows:
\\ (S \u003d S_ (ABCD) \u003d S_ (ADCB) - S_ (aabb) \u003d \\ int \\ limits_a ^ b f (x) dx - \\ int \\ limits_a ^ b g (x) dx \u003d \\)
\\ (\u003d \\ int \\ limits_a ^ b (f (x) -g (x)) dx \\)

So, the area S is a figure bounded by straight x \u003d a, x \u003d b and graphs of functions y \u003d f (x), y \u003d g (x), continuous on the segment and such as for any x from the segment [A; b] The inequality \\ (G (x) \\ LEQ F (X) \\) is performed, calculated by the formula
\\ (S \u003d \\ int \\ limits_a ^ b (f (x) -g (x)) dx \\)

Table of indefinite integrals (primitive) some functions

$$ \\ int 0 \\ cdot dx \u003d c $$$$ \\ int 1 \\ cdot dx \u003d x + c $$$$ \\ int x ^ n dx \u003d \\ FRAC (x ^ (n + 1)) (n + 1 ) + C \\; \\; (N \\ NEQ -1) $$$$ \\ int \\ FRAC (1) (X) DX \u003d \\ ln | x | + C $$$$ \\ int e ^ x dx \u003d e ^ x + c $$$$ \\ int a ^ x dx \u003d \\ FRAC (a ^ x) (\\ ln a) + c \\; \\; (A\u003e 0, \\; \\; A \\ NEQ 1) $$$$ \\ int \\ cos x dx \u003d \\ sin x + c $$$$ \\ int \\ sin x dx \u003d - \\ cos x + C $$$ $ \\ int \\ FRAC (DX) (\\ cos ^ 2 x) \u003d \\ Text (TG) x + C $$$$ \\ int \\ FRAC (DX) (\\ Sin ^ 2 x) \u003d - \\ Text (CTG) x + C $$$$ \\ int \\ FRAC (DX) (\\ SQRT (1-x ^ 2)) \u003d \\ Text (ArcSin) x + C $$$$ \\ int \\ FRAC (DX) (1 + x ^ 2 ) \u003d \\ Text (ArCTG) X + C $$$$ \\ int \\ text (CH) x DX \u003d \\ Text (SH) x + C $$$$ \\ INT \\ TEXT (SH) x DX \u003d \\ TEXT (CH ) x + C $$

Certain integral. How to calculate the area of \u200b\u200bthe figure

Go to consideration of integral application applications. In this lesson, we will analyze the typical and most common task. - How to calculate the plane shape with a specific integral. At last seeing meaning In the highest mathematics - let him find it. Little. We'll have to bring closer in life country cottage area Elementary functions and find its area using a specific integral.

For successful material development, it is necessary:

1) understand uncertain integral at least at the average level. Thus, teapotes should be familiar with the lesson Not.

2) To be able to apply the Newton labnic formula and calculate a specific integral. To establish warm friendly relations With certain integrals, you can on the page Certain integral. Examples of solutions.

In fact, in order to find the area of \u200b\u200bthe figure, there is no such knowledge of the uncertain and defined integral. The task "Calculate the area with the help of a specific integral" always implies the construction of the drawing, so much more actual question There will be your knowledge and skills to build drawings. In this regard, it is useful to refresh in the memory of graphics of the main elementary functions, and at least, be able to build a straight, parabola and hyperbola. It can be done (many - needed) with methodical material and articles on geometric transformations of graphs.

Actually, with the task of finding an area with the help of a specific integral, everyone is familiar from school, and we will eat little forward from school program. This article could not even be, but the fact is that the task is found in 99 cases out of 100, when the student suffers from a hateful tower with enthusiasm departing the course of higher mathematics.

Materials of this workshop are presented simply, in detail and with a minimum of theory.

Let's start with a curvilinear trapezium.

Curvilinear trapezium A flat figure is called a limited axis, straight, and a continuous schedule on a segment of a function that does not change the sign on this interval. Let this figure be located not less The abscissa axis:

Then the area of \u200b\u200bthe curvilinear trapezium is numerically equal to a specific integral. Any particular integral (which exists) has a very good geometric meaning. At the lesson Certain integral. Examples of solutions I said that a certain integral is a number. And now it's time to state one more useful fact. From the point of view of geometry, a certain integral is an area.

I.e, a specific integral (if it exists) geometrically corresponds to the area of \u200b\u200bsome figure. For example, consider a specific integral. The integrand function sets the curve on the plane, located above the axis (which wishes can do the drawing), and the determined integral itself is numerically equal to Square The corresponding curvilinear trapezium.

Example 1.

This is a typical task formulation. First I. the most important moment Solutions - Building drawing. And the drawing must be built RIGHT.

When building a drawing, I recommend next order: first it is better to build all straight (if they are) and only later - Parabolas, hyperbolas, schedules of other functions. Function graphs are more profitable to build potochoe, with the technique of check-in construction can be found in reference material Charts and properties of elementary functions. There you can also find a very useful material in relation to our lesson the material - how to quickly build a parabola.

In this task, the decision may look like that.
Perform the drawing (note that the equation sets the axis):

I will not stroke a curvilinear trapeze, here it is obvious about which area this is speech. The decision continues like this:

On the segment schedule a function is located over the axis, so:

Answer:

Who has difficulties with the calculation of a certain integral and the use of Newton-Leibnia formula , refer to the lecture Certain integral. Examples of solutions.

After the task is completed, it is always useful to look at the drawing and estimate, the real one turned out. In this case, "on the eyes" we count the number of cells in the drawing - well, approximately 9 will be flown, it seems to the truth. It is quite clear that if we had, say, answer: 20 square units, it is obvious that an error is made somewhere - in the figure of 20 cells, it is clearly not fitted, from the strength of a dozen. If the answer turned out negative, the task is also decided incorrectly.

Example 2.

Calculate the area of \u200b\u200bthe shape, limited lines, and axis

This is an example for self-decide. Complete solution And the answer at the end of the lesson.

What to do if the curvilinear trapezium is located under the axis?

Example 3.

Calculate the area of \u200b\u200bthe shape, limited lines, and the coordinate axes.

Decision: Perform drawing:

If the curvilinear trapezium is located under the axis (or at least not higher This axis), then its area can be found by the formula:
In this case:

Attention! Do not confuse two types of tasks:

1) If you are invited to solve a simple integral without any geometric meaning, then it may be negative.

2) If you are invited to find the figure of the figure using a specific integral, then the area is always positive! That is why in just the considered formula appears minus.

In practice, the figure is most often located in the upper and lower half plane, and therefore, from the simplest school charts, go to more meaningful examples.

Example 4.

Find the area of \u200b\u200ba flat figure, limited lines ,.

Decision: First you need to draw a drawing. Generally speaking, when building a drawing in tasks to the area, we are most interested in the intersection points of the lines. Find points of intersection of parabola and direct. This can be done in two ways. The first method is analytical. We solve the equation:

So, the lower integration limit, the upper limit of integration.
This way is better, if possible, do not use.

It is much more profitable and faster to build the lines of the line, while the integration limits are clarified as if "by themselves". The technique of the cessation for various graphs is considered in detail in the help Charts and properties of elementary functions . However, an analytical way to find the limits after all, it is sometimes necessary to apply if, for example, the schedule is large enough, or a trained construction did not reveal the integration limits (they can be fractional or irrational). And such an example, we also consider.

We return to our task: more rational first build a straight line and only then Parabola. Perform drawing:

I repeat that in the current construction, the integration limits are most often found out by the "automatic".

And now the working formula: If on the segment some continuous function more or equal Some continuous function, the area of \u200b\u200bthe figure, limited by graphs of these functions and direct, can be found by the formula:

Here it is no longer necessary to think where the figure is located - over the axis or under the axis, and, roughly speaking, important what is the graph above(relative to another schedule) and what - below.

In this example, it is obvious that on the segment of Parabola is located above straight, and therefore it is necessary to subtract

Completion of the solution may look like this:

The desired figure is limited to parabola from above and direct bottom.
On the segment, according to the corresponding formula:

Answer:

In fact, the school formula for the area of \u200b\u200bthe curvilinear trapezium in the lower half-plane (see simple example No. 3) - private case Formulas . Since the axis is defined by the equation, and the function graph is located not higher Axis, T.

And now a couple of examples for an independent decision

Example 5.

Example 6.

Find the area of \u200b\u200bthe figure limited lines ,.

In the course of solving tasks for calculating the area with a specific integral, a funny case occurs sometimes. The drawing is completed correctly, calculations - right, but intensified ... found the area is not the figureThat this is how your humble servant was packed. Here is a real case from life:

Example 7.

Calculate the area of \u200b\u200bthe shape, limited lines ,,,.

Decision: First do the drawing:

... oh, the drawing of Khrenovynsky came out, but everything seems to be picking up.

Figure whose area we need to find is shaded in blue (Look carefully on the condition - than the figure is limited!). But in practice, "glitch" often arises in mindfulness, which you need to find an area of \u200b\u200bthe figure, which is shaded green!

This example is still useful and the fact that in it the area of \u200b\u200bthe figure is considered using two specific integrals. Really:

1) A straight schedule is located on the segment over the axis;

2) On the segment over the axis there is a graph of hyperboles.

It is clear that the square can (and need) to decompose, so:

Answer:

Go to another substantive task.

Example 8.

Calculate the area of \u200b\u200bthe shape, limited lines,
Imagine the equation in the "school" form, and perform the current drawing:

From the drawing it is clear that the upper limit we have "good" :.
But what is the lower limit?! It is clear that this is not an integer, but what? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well be that. Or root. And if we generally improperly built a schedule?

In such cases, you have to spend extra time and specify the integration limits analytically.

Find the intersection points of the direct and parabola.
To do this, solve the equation:


,

Indeed.

Further solution is trivial, the main thing is not to get confused in substitutions and signs, the calculations here are not the simplest.

On cut According to the corresponding formula:

Answer:

Well, and in the conclusion of the lesson, consider two tasks more difficult.

Example 9.

Calculate the area of \u200b\u200bthe shape, limited lines ,,

Decision: Show this shape in the drawing.

Damn, forgot the schedule to sign, but to redo the picture, sorry, not a hotz. Not inherited, shorter, day today \u003d)

For check-in construction you need to know appearance sinusoids (and it is generally helpful to know graphs of all elementary functions), as well as some sinus values, they can be found in trigonometric table. In some cases (as in this), it is allowed to build a schematic drawing on which the graphs and integration limits must be reflected in principle.

With the limits of integration, there are no problems here, they follow directly from the condition: - "X" varies from zero to "pi". We draw up a further solution:

On the segment, the function graph is located above the axis, so:

Suppose that the function is non-negative and continuous on the segment. Then, according to the geometric meaning of a specific integral, the area of \u200b\u200ba curvilinear trapezium, limited from the top of the graph of this function, from the bottom - axis, on the left and right - direct and (see Fig. 2) is calculated by the formula

Example 9. Find the area of \u200b\u200bthe figure, limited line and axis.

Decision. The graph of the function is parabola, whose branches are directed down. We construct it (Fig. 3). To determine the integration limits, find the line intersection (parabola) with axis (straight). To do this, solve the system of equations

We get:, from where,; hence, .

Fig. 3

Figure area Formula (5):

If the function is non-positive and continuous on the segment, then the area of \u200b\u200bthe curvilinear trapezium, limited to the bottom with a graph of this function, top-axis, on the left and right - direct and, calculated by the formula

If the function is continuous on the segment and changes the sign in the end number of points, then the area of \u200b\u200bthe shaded figure (Fig. 4) is equal to the algebraic sum of the corresponding specific integrals:

Fig. 4

Example 10. Calculate the area of \u200b\u200bthe figure, limited by the axis and the graph of the function at.

Fig. 5

Decision. Let's draw a drawing (Fig. 5). The desired area is the sum of the square and. We will find each of these areas. Initially, we define the integration limits, solving the system we get ,. Hence:

Thus, the area of \u200b\u200bthe shaded figure is equal to

Fig. 6

Let, finally, the curvilinear trapezium is limited from above and below the charts of continuous functions on the segment and, and on the left and right - straight and (Fig. 6). Then its area is calculated by the formula

Example 11. Find the area of \u200b\u200bthe figure limited lines and.

Decision. This figure is depicted in fig. 7. The area is calculated by formula (8). Solving the system of equations we find ,; hence, . On the segment we have :. It means that in formula (8) as we take X, and as -. We get:

More complex tasks for calculating areas are solved by splitting the figure on the inverse parts and calculating the area of \u200b\u200bthe entire figure as the sum of the areas of these parts.

Example1 . Calculate the area of \u200b\u200bthe figure, limited lines: X + 2U - 4 \u003d 0, y \u003d 0, x \u003d -3, and x \u003d 2

We will execute the construction of the figure (see Fig.) We build a straight X + 2U - 4 \u003d 0 by two points A (4; 0) and in (0; 2). Expressing y through x, we obtain y \u003d -0.5x + 2. by formula (1), where f (x) \u003d -0.5x + 2, and \u003d -3, B \u003d 2, we find

S \u003d \u003d [-0.25 \u003d 11.25 kV. elf

Example 2. Calculate the area of \u200b\u200bthe figure, limited lines: x - 2au + 4 \u003d 0, x + y - 5 \u003d 0 and y \u003d 0.

Decision. Perform the construction of the figure.

We construct a straight x - 2au + 4 \u003d 0: y \u003d 0, x \u003d - 4, a (-4; 0); x \u003d 0, y \u003d 2, in (0; 2).

We construct the straight x + y - 5 \u003d 0: y \u003d 0, x \u003d 5, c (5; 0), x \u003d 0, y \u003d 5, d (0; 5).

We will find the point of intersection of direct, solving the system of equations:

x \u003d 2, y \u003d 3; M (2; 3).

To calculate the desired area, we break the AMS triangle on two triangles of the AMN and NMS, since with a change in x from a to n, the area is limited to direct, and when x from n to C - direct

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

For the triangle NMS, we have: y \u003d - x + 5, i.e. f (x) \u003d - x + 5, a \u003d 2, b \u003d 5.

By calculating the area of \u200b\u200beach of the triangles and folding the results, finding:

sq. units.

sq. units.

9 + 4, 5 \u003d 13.5 square meters. units. Check: \u003d 0.5as \u003d 0.5 kV. units.

Example 3. Calculate the area of \u200b\u200bthe figure, limited lines: Y \u003d x 2 , y \u003d 0, x \u003d 2, x \u003d 3.

In this case, it is required to calculate the area of \u200b\u200bthe curvilinear trapezium, limited by parabola y \u003d x 2 , straight x \u003d 2 and x \u003d 3 and oh (see Fig.) By formula (1) we find the area of \u200b\u200bthe curvilinear trapezium

\u003d \u003d 6kv. units.

Example 4. Calculate the area of \u200b\u200bthe figure, limited lines: y \u003d - x 2 + 4 and y \u003d 0

Perform the construction of the figure. The desired area is concluded between parabola y \u003d - x 2 + 4 and the axis oh.

Find points of intersection of parabola with the axis Oh. Believing y \u003d 0, we find x \u003d since this figure is symmetric with respect to the OU axis, then we calculate the area of \u200b\u200bthe figure located on the right of the OU axis, and the resulting result will be doubly: \u003d + 4x] kV. units. 2 \u003d 2 kV. units.

Example 5. Calculate the area of \u200b\u200bthe figure, limited lines: y 2 \u003d x, yx \u003d 1, x \u003d 4

It requires calculating the area of \u200b\u200bthe curvilinear trapezium, limited to the top branch of the parabolia 2 \u003d x, axis oh and straight x \u003d 1 and \u003d 4 (see Fig.)

By formula (1), where f (x) \u003d a \u003d 1 and b \u003d 4 we have \u003d (\u003d sq.

Example 6. . Calculate the area of \u200b\u200bthe figure, limited lines: y \u003d sinx, y \u003d 0, x \u003d 0, x \u003d.

The desired area is limited to the half-wave sinusoid and the axis oh (see Fig.).

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

Example 7. Calculate the area of \u200b\u200bthe figure, limited lines: y \u003d - 6x, y \u003d 0 and x \u003d 4.

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

Consequently, its area is found by formula (3)

= =

Example 8. Calculate the area of \u200b\u200bthe figure, limited lines: Y \u003d and x \u003d 2. The curve y \u003d construct by points (see Fig.). Thus, the area of \u200b\u200bthe figures are found by formula (4)

Example 9. .

h. 2 + U. 2 \u003d R. 2 .

It requires calculating the area, limited circle x 2 + U. 2 \u003d R. 2 , i.e., the area of \u200b\u200bthe radius circle R with the center at the beginning of the coordinates. We will find the fourth part of this area, taking the integration limits from 0

doR; We have: 1 = = [

Hence, 1 =

Example 10. Calculate the area of \u200b\u200bthe figure, limited lines: y \u003d x 2 and y \u003d 2x

This figure is limited to parabola y \u003d x 2 and direct y \u003d 2x (see Fig.) To determine the intersection points of the specified lines by solving the system of equations: x 2 - 2x \u003d 0 x \u003d 0 and x \u003d 2

Using the area for finding the area formula (5), we get

\u003d The area of \u200b\u200bthe curvilinear trapezium formed by the function f,equal to the increment of a primitive function:

Exercise 1:

Find the area of \u200b\u200bthe curvilinear trapezoid, limited by the graph of the function: f (x) \u003d x 2 And straight y \u003d 0, x \u003d 1, x \u003d 2.

Decision: ( according to the algorithm slide 3)

Draw a function schedule and straight

Find one of valid functions f (x) \u003d x 2 :

Self-test on the slide

Integral

Consider a curvilinear trapezion specified by the function f. On the segment [ a; B.]. Discuss this segment into several parts. The area of \u200b\u200bthe entire trapezium will break up the amount of the squares of smaller curvilinear trapezes. ( slide 5). Every such a trapezium can be approximately considered a rectangle. The amount of the area of \u200b\u200bthese rectangles gives an approximate idea of \u200b\u200bthe entire area of \u200b\u200bthe curvilinear trapezium. The smaller we break the segment [ a; B.], the more accurately calculate the area.

We write these arguments in the formula formulas.

We divide the segment [ a; B.] on n parts points x 0 \u003d a, x1, ..., xn \u003d b. Length k-gO denote by xk \u003d xk - xk-1. Let's make up

Geometrically, this amount is an area of \u200b\u200bshape, shaded in the figure ( sch.m..)

The sum of the species is called integral sums for the function. f.. (Sch.m.)

Integral sums give an approximate value of the area. The exact value is obtained using the limit transition. Imagine that we crush the splitting of the segment [ a; B.] So that the lengths of all small segments tend to zero. Then the area of \u200b\u200bthe composed figure will approach the area of \u200b\u200bthe curvilinear trapezium. It can be said that the area of \u200b\u200bthe curvilinear trapezium is equal to the limit of integral sums, SK.T. (Sch.m.)or integral, i.e.,

Definition:

Integral function f (x) from a. before b. called the limit of integrated amounts

= (Sch.m.)

Formula Newton Labitsa.

Remember that the limit of integrated amounts is equal to the area of \u200b\u200bthe curvilinear trapezion, it means that you can write:

SK.T. \u003d. (Sch.m.)

On the other hand, the cryvilinear trapezium area is calculated by the formula

S k. T. (Sch.m.)

Compare these formulas, we get:

= (Sch.m.)

This equality is called Newton Labits Formula.

For the convenience of calculations, the formula is written in the form:

= = (Sch.m.)

Tasks: (Shch.m.)

1. Calculate the integral according to the Newton Labits formula: ( checking the slide 5)

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

3. Find the area of \u200b\u200bthe figure limited lines: y \u003d x 3, y \u003d 0, x \u003d 1, x \u003d 2. ( Slide 7.)

Finding the squares of flat figures ( slide 8.)

How to find the square of the figures that are not curvilinear trapezes?

Let two functions be given, the graphs of which you see on the slide . (Sch.m.) It is necessary to find the area of \u200b\u200bthe painted figure . (Sch.m.). The figure about which is speaking is a curvilinear trapeze? And how can I find its area using the property of the additivity of the area? Consider two curvilinear trapeats and from the square of one of them to subtract the area of \u200b\u200banother ( sch.M.)

We will make an algorithm for finding an animation area on a slide:

1. Build graphs of functions
2. Sprogit the points of intersection of graphs on the axis of the abscissa
3. Sharp the figure obtained when crossing the graphs
4. Find curvilinear trapeats, intersection or combining which is a given figure.
5. Calculate the area of \u200b\u200beach of them
6. Find a difference or amount of space

Oral task: how to get the area of \u200b\u200bthe shaded figure (tell us with the help of animation, slide 8 and 9)

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

Bibliography

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3. Bashmakov M.I. Mathematics: Tutorial for institutions of the beginning. and media. prof. Education / M.I. Shoes. - M: Academy, 2010.
4. Kolmogorov A.N. Algebra and start analysis: Tutorial for 10-11 cells. Education institutions / A.N. Kolmogorov. - M: Enlightenment, 2010.
5. Ostrovsky S.L. How to make a presentation to the lesson? / C.L. Ostrovsky. - M.: The first of September, 2010.