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] (base 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

Dadim mathematical description 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 \\)

Calculating a specific 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, it is possible to calculate the area of \u200b\u200bnot only curvilinear trapezes, but also flat figures of a more complex species, 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 $$$$ \\ int \\ FRAC (DX) (\\ Sin ^ 2 x) \u003d - \\ Text (CTG) x + C $$$$ \\ int \\ FRAC (DX) (\\ SQRT (1-x ^ 2)) \u003d \\ Text (ArcSin) x + C $$$$ \\ int \\ FRAC (DX) (1 + x ^ 2 ) \u003d \\ Text (ArCTG) X + C $$$$ \\ int \\ text (CH) x DX \u003d \\ Text (SH) x + C $$$$ \\ INT \\ TEXT (SH) x DX \u003d \\ TEXT (CH ) x + C $$

Go to consideration of integral application applications. In this lesson, we will analyze the typical and most common task. calculations of a flat figure with a specific integral. Finally, all the meaning of the meaning in the highest mathematics - will be found him. 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 friendships with certain integrals on the page Certain integral. Examples of solutions. The task "Calculate the area with the help of a specific integral" always implies the construction of the drawingTherefore, your knowledge and skills of building drawings will also be an urgent issue. At a minimum, you need to be able to build a straight, parabola and hyperbola.

Let's start with a curvilinear trapezium. A curvilinear trapezium is a flat figure limited by a graph of some feature. y. = f.(x.), axis OX. and lines x. = a.; x. = b..

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 solutionswe said that a certain integral is a number. And now it's time to state another 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. Consider a specific integral

Integrand

specifies the curve on the plane (it can be drawn when desired), and a certain integral itself is numerically equal to the area of \u200b\u200bthe corresponding curvilinear trapezium.

Example 1.

, , , .

This is a typical task formulation. The most important point of the decision is to build a drawing. And the drawing must be built RIGHT.

When building a drawing, I recommend the following order: first it is better to build all straight (if they are) and only later - Parabolas, hyperbolas, schedules of other functions. 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 drawing (Note that the equation y. \u003d 0 sets the axis OX.):

Strike a curvilinear trapezion will not, it is obvious here about which area there is a speech. The decision continues like this:

On the segment [-2; 1] Function Schedule y. = x. 2 + 2 is located over the axisOX., 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 xY. = 4, x. = 2, x. \u003d 4 and axis OX..

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

What to do if the curvilinear trapezium is located under the axisOX.?

Example 3.

Calculate the area of \u200b\u200bthe shape, limited lines y. = e - X., x. \u003d 1 and coordinate axes.

Solution: Perform drawing:

If a curvilinear trapezium fully located under the axis OX. , 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 flat shape limited lines y. = 2x. – x. 2 , y. = -x..

Solution: First you need to draw a drawing. 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 y. = 2x. – x. 2 and direct y. = -x.. This can be done in two ways. The first method is analytical. We solve the equation:

So, the lower limit of integration a. = 0, upper limit integration b. \u003d 3. Often it is often more profitable and faster to build the lines of the flow, while the integration limits are clarified as if "by themselves". 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). We return to our task: more rational first build a straight line and only then Parabola. Perform drawing:

Repeat that in the current construction, the integration limits most often find out "automatically".

And now the working formula:

If on the segment [ a.; b.] Some continuous function f.(x.) more or equal Some continuous function g.(x.), then the area of \u200b\u200bthe corresponding figure can be found by the formula:

Here, no need to think where the figure is located over the axis or under the axis, and 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 out of 2 x. – x. 2 need to subtract x..

Completion of the solution may look like this:

The desired figure is limited to parabola y. = 2x. – x. 2 top and straight y. = -x. bottom.

On the segment 2. x. – x. 2 ≥ -x.. 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 example number 3) - private case Formulas

.

Since the axis OX. Valid by equation y. \u003d 0, and a function schedule g.(x.) Located below the axis OX.T.

.

And now a couple of examples for an independent decision

Example 5.

Example 6.

Find an area of \u200b\u200bfigures 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, by inattention, ... found area is not the figure.

Example 7.

First execute the drawing:

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

This example is also useful in that it is considered to be in it the size of two specific integrals. Really:

1) on the segment [-1; 1] over the axis OX. Located schedule direct y. = x.+1;

2) on the segment above the axis OX. Located a chart of hyperbola y. = (2/x.).

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

Answer:

Example 8.

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

Imagine the equation in the "school" form

and do the current drawing:

From the drawing it is clear that the upper limit we have "good": b. = 1.

But what is the lower limit?! It is clear that this is not an integer, but what?

May be, a.\u003d (- 1/3)? But where is the guarantee that the drawing is made with ideal accuracy, it may well be that a.\u003d (- 1/4). 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 of graphs

To do this, solve the equation:

.

Hence, a.=(-1/3).

Further solution is trivial. The main thing is not to get confused in substitutions and signs. Calculations are not the simplest. On cut

, ,

according to the corresponding formula:

Answer:

In the conclusion of the lesson, consider two tasks more difficult.

Example 9.

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

Solution: Show this shape in the drawing.

For checking the drawing, you need to know appearance sinusoids. In general, it is useful to know the graphs of all elementary functions, as well as some sinus values. They can be found in the table of values trigonometric functions. In some cases (for example, 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 cut graph function y. \u003d SIN 3. x. Located above the axis OX., so:

(1) How to integrate sines and cosines in odd degrees, you can look at the lesson Integrals from trigonometric functions. Plug off one sine.

(2) We use the main trigonometric identity in the form of

(3) We will replace the variable t. \u003d COS. x., then: Located above the axis, so:

.

.

Note: Please note how the integral from Tangent in Cuba is taken, a consequence of the main trigonometric identity

.

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 drawingTherefore, a much more relevant issue will be your knowledge and skills of building drawings. In this regard, it is useful to refresh in the memory of graphics of basic elementary functions, and at least be able to build a straight, and hyperbola.

The curvilinear trapezion is called a flat figure, limited to the 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.

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 shape. For example, consider a specific integral. The integrand function sets a curve on the plane, located above the axis (which wishes can draw the drawing), and the specific integral itself is numerically equal to the area of \u200b\u200bthe corresponding curvilinear trapezium.

Example 1.

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

When building a drawing, I recommend the following 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 potion.

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

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

Answer:

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 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". 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:

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:

Example 4.

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

Decision: First do the drawing:

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 with 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:

From this article, you will learn how to find an area of \u200b\u200bfigures limited by lines using calculations using integrals. For the first time, we encounter such a task in high school, when we just passed the study of certain integrals and it is time to begin the geometric interpretation of the knowledge gained in practice.

So, what will be required to successfully solve the problem of searching for an area of \u200b\u200bthe figure with the help of integrals:

• Skill competently build drawings;
• The ability to solve a specific integral with the help of the well-known Newton-Leibnic formula;
• The ability to "see" a more profitable solution solution - i.e. Understand how in such an case it will be more convenient to carry out integration? Along the X axis (ox) or the axis of the game (OY)?
• Well, where without correct computing?) This includes an understanding how to solve that other type of integrals and the correct numerical calculations.

The algorithm for solving the task of calculating the area of \u200b\u200bthe figure, limited lines:

1. Build a drawing. It is advisable to do it on a piece in a cage, with a large scale. We subscribe a pencil over each chart the name of this function. The signature of the graphs is made exclusively for the convenience of further computing. After receiving a graph of the desired figure, in most cases it will be seen immediately which integration limits will be used. Thus, we solve the task with a graphic method. However, it happens that the values \u200b\u200bof the limits are fractional or irrational. Therefore, you can make additional calculations, go to step two.

2. If the integration limits are clearly not specified, we find the intersection points of the graphs with each other, and we look at whether our graphic solution with analytical is coincided.

3. Next, it is necessary to analyze the drawing. Depending on how graphics of functions are located, there are different approaches to finding the area of \u200b\u200bthe figure. Consider different examples To find the area of \u200b\u200bthe figure with the help of integrals.

3.1. The most classic and simple task option is when you need to find the area of \u200b\u200bthe curvilinear trapezium. What is a curvilinear trapeze? This is a flat figure limited to X axis (y \u003d 0)straight x \u003d a, x \u003d b and any curve continuous on the interval from a. before b.. At the same time, this figure is non-negative and is not lower than the abscissa axis. In this case, the area of \u200b\u200bthe curvilinear trapezium is numerically equal to a specific integral calculated by the Newton Labender formula:

Example 1. y \u003d x2 - 3x + 3, x \u003d 1, x \u003d 3, y \u003d 0.

What lines is the figure limited? We have parabola y \u003d x2 - 3x + 3which is located above the axis OH, it is non-negative, because All points of this parabola are positive. Next, the direct x \u003d 1. and x \u003d 3.who run parallel to the axis OUare the restrictive lines of the figure on the left and right. Well y \u003d 0.She is an X axis that limits the figure below. The resulting figure is shaded, as can be seen from the drawing on the left. In this case, you can immediately start solving the problem. We have a simple example of a curvilinear trapezium, which is further solving with the help of Newton-Leibnic formula.

3.2. In the previous paragraph 3.1, the case is disassembled when the curvilinear trapezium is located above the X axis. Now consider the case when the conditions of the task are the same, except that the function runs under the X axis. The standard Newton-Labender formula is added minus. How to solve such a task Consider further.

Example 2. . Calculate the area of \u200b\u200bthe shape, limited lines y \u003d x2 + 6x + 2, x \u003d -4, x \u003d -1, y \u003d 0.

In this example, we have a parabola y \u003d x2 + 6x + 2which originates from the axis OH, straight x \u003d -4, x \u003d -1, y \u003d 0. Here y \u003d 0. Limits the desired figure from above. Straight x \u003d -4. and x \u003d -1. These are borders within which a specific integral will be calculated. The principle of solving the problem of finding an area of \u200b\u200bthe figure almost completely coincides with the example number 1. The only difference is that the specified function is not positive, and everything is also continuous on the interval [-4; -1] . What does not mean positive? As can be seen from the figure, the figure, which lies within the specified ICs, has exclusively "negative" coordinates, which we need to see and remember when solving the problem. The area of \u200b\u200bthe figure is looking for Newton Labitsa formula, only with a minus sign in the beginning.

The article is not completed.

In the previous section dedicated to the analysis of the geometric meaning of a certain integral, we received a number of formulas for calculating the area of \u200b\u200bthe curvilinear trapezium:

Yandex.rtb R-A-339285-1

S (g) \u003d ∫ a b f (x) d x for continuous and non-negative functions y \u003d f (x) on the segment [a; b]

S (G) \u003d - ∫ a b f (x) d x for continuous and non-positive function y \u003d f (x) on the segment [a; b].

These formulas are applicable to solve relatively simple tasks. In fact, we will most often have to work with more complex figures. In this regard, this section we devote to the analysis of the algorithms for calculating the area of \u200b\u200bfigures, which are limited by the functions explicitly, i.e. As y \u003d f (x) or x \u003d g (y).

Theorem

Let the functions y \u003d F 1 (x) and y \u003d f 2 (x) are determined and continuous on the interface [A; b], with F 1 (x) ≤ F 2 (x) for any value x from [a; b]. Then the formula for calculating the area of \u200b\u200bthe figure G, bounded by the lines x \u003d a, x \u003d b, y \u003d f 1 (x) and y \u003d f 2 (x) will be viewed s (G) \u003d ∫ ABF 2 (X) - F 1 (x) DX.

A similar formula will be applicable to the area of \u200b\u200bthe figure, limited by the lines y \u003d c, y \u003d d, x \u003d g 1 (y) and x \u003d g 2 (y): s (g) \u003d ∫ Cd (G 2 (y) - G 1 (y) DY.

Evidence

We will analyze the three cases for which the formula will be fair.

In the first case, given the property of the additivity of the area, the sum of the area of \u200b\u200bthe original figure G and the curvilinear trapezium G 1 is equal to the area of \u200b\u200bthe figure G 2. It means that

Therefore, S (G) \u003d S (G 2) - S (G 1) \u003d ∫ ABF 2 (X) DX - ∫ ABF 1 (X) DX \u003d ∫ AB (F 2 (X) - F 1 (x)) DX.

Perform the last transition We can using the third property of a specific integral.

In the second case, equality is true: S (G) \u003d S (G 2) + S (G 1) \u003d ∫ ABF 2 (X) DX + - ∫ ABF 1 (X) DX \u003d ∫ AB (F 2 (X) - F 1 (x)) dx

Graphic illustration will look:

If both functions are non-positive, we obtain: S (G) \u003d S (G 2) - S (G 1) \u003d - ∫ ABF 2 (X) DX - - ∫ ABF 1 (X) DX \u003d ∫ AB (F 2 (X) - F 1 (x)) DX. Graphic illustration will look:

Let us turn to the consideration of the general case when Y \u003d F 1 (x) and Y \u003d F 2 (x) crosses the O x axis.

The intersection points we denote as x i, i \u003d 1, 2 ,. . . , N - 1. These points break the segment [a; b] on n parts x i - 1; x i, i \u003d 1, 2 ,. . . , n, where α \u003d x 0< x 1 < x 2 < . . . < x n - 1 < x n = b . Фигуру G можно представить объединением фигур G i , i = 1 , 2 , . . . , n . Очевидно, что на своем интервале G i попадает под один из трех рассмотренных ранее случаев, поэтому их площади находятся как S (G i) = ∫ x i - 1 x i (f 2 (x) - f 1 (x)) d x , i = 1 , 2 , . . . , n

Hence,

S (G) \u003d Σ i \u003d 1 n s (G i) \u003d Σ i \u003d 1 n ∫ xixif 2 (x) - F 1 (x)) dx \u003d \u003d ∫ x 0 xn (F 2 (x) - F ( x)) dx \u003d ∫ ABF 2 (X) - F 1 (X) DX

We can implement the last transition using the fifth properties of a specific integral.

We illustrate in the chart a general case.

The formula S (G) \u003d ∫ A B F 2 (x) - F 1 (x) D x can be considered proven.

And now we proceed to the analysis of examples of calculating the area of \u200b\u200bfigures, which are limited to the lines y \u003d f (x) and x \u003d g (y).

Consideration of any of the examples we will begin with the construction of the schedule. The image will allow us to represent complex figures as associations more simple figures. If the construction of graphs and figures make it difficult for them, you can explore the section on basic elementary functions, geometric conversion of graphs of functions, as well as building graphs during function research.

Example 1.

It is necessary to determine the area of \u200b\u200bthe figure, which is limited to the parabola y \u003d - x 2 + 6 x - 5 and the straight lines y \u003d - 1 3 x - 1 2, x \u003d 1, x \u003d 4.

Decision

Show lines on the graph in the Cartesian coordinate system.

On the segment [1; 4] The chart of parabola y \u003d - x 2 + 6 x - 5 is located above straight y \u003d - 1 3 x - 1 2. In this regard, to obtain an answer, we use the formula earlier, as well as a method for calculating a specific integral according to Newton-Leibnitsa formula:

S (G) \u003d ∫ 1 4 - x 2 + 6 x - 5 - - 1 3 x - 1 2 dx \u003d \u003d ∫ 1 4 - x 2 + 19 3 x - 9 2 dx \u003d - 1 3 x 3 + 19 6 x 2 - 9 2 x 1 4 \u003d \u003d - 1 3 · 4 3 + 19 6 · 4 2 - 9 2 · 4 - - 1 3 · 1 3 + 19 6 · 1 2 - 9 2 · 1 \u003d - 64 3 + 152 3 - 18 + 1 3 - 19 6 + 9 2 \u003d 13

Answer: S (G) \u003d 13

Consider a more complex example.

Example 2.

It is necessary to calculate the area of \u200b\u200bthe figure, which is limited to the lines y \u003d x + 2, y \u003d x, x \u003d 7.

Decision

In this case, we have only one straight line located parallel to the abscissa axis. This is x \u003d 7. It requires us to find the second integration limit on your own.

We construct a schedule and bring lines on it, data on the task condition.

Having a chart in front of the eyes, we can easily determine that the lower limit of the integration will be the abscissa of the intersection point of the schedule y \u003d x and the floor of parabola y \u003d x + 2. To find the abscissa, use equality:

y \u003d x + 2 o d z: x ≥ - 2 x 2 \u003d x + 2 2 x 2 - x - 2 \u003d 0 d \u003d (- 1) 2 - 4 · 1 · (- 2) \u003d 9 x 1 \u003d 1 + 9 2 \u003d 2 ∈ O d z x 2 \u003d 1 - 9 2 \u003d - 1 ∉ o d z

It turns out that the abscissa of the intersection point is x \u003d 2.

We draw your attention to the fact that general example On the drawing of the line y \u003d x + 2, y \u003d x intersect at the point (2; 2), so such detailed calculations may seem unnecessary. We brought here detailed solution just because in more complex cases The decision may not be so obvious. This means that the coordinates of the intersection of lines are better to always calculate analytically.

On the interval [2; 7] The graph of the function y \u003d x is located above the graph of the function y \u003d x + 2. Apply the formula for calculating Square:

S (G) \u003d ∫ 2 7 (x - x + 2) dx \u003d x 2 2 - 2 3 · (x + 2) 3 2 2 7 \u003d 7 2 2 - 2 3 · (7 + 2) 3 2 - 2 2 2 - 2 3 · 2 + 2 3 2 \u003d 49 2 - 18 - 2 + 16 3 \u003d 59 6

Answer: S (G) \u003d 59 6

Example 3.

It is necessary to calculate the area of \u200b\u200bthe figure, which is limited by graphs of functions y \u003d 1 x and y \u003d - x 2 + 4 x - 2.

Decision

Apply the lines on the schedule.

Determine with the limits of integration. To do this, we define the coordinates of the intersection points of the lines, equating expressions 1 x and - x 2 + 4 x - 2. Provided that x is not zero, the equality 1 x \u003d x 2 + 4 x - 2 becomes an equivalent equation of the third degree - x 3 + 4 x 2 - 2 x - 1 \u003d 0 with integer coefficients. To refresh the algorithm in memory by solving such equations, we can, contacting the section "Solution of Cubic Equations".

The root of this equation is x \u003d 1: - 1 3 + 4 · 1 2 - 2 · 1 - 1 \u003d 0.

Dividing the expression - x 3 + 4 x 2 - 2 x - 1 per bounce x - 1, we obtain: - x 3 + 4 x 2 - 2 x - 1 ⇔ - (x - 1) (x 2 - 3 x - 1) \u003d 0.

The remaining roots we can find from equation x 2 - 3 x - 1 \u003d 0:

x 2 - 3 x - 1 \u003d 0 d \u003d (- 3) 2 - 4 · 1 · (- 1) \u003d 13 x 1 \u003d 3 + 13 2 ≈ 3. 3; x 2 \u003d 3 - 13 2 ≈ - 0. 3.

We found an interval x ∈ 1; 3 + 13 2, on which the figure G is concluded above the blue and below the red line. It helps us to determine the area of \u200b\u200bthe figure:

S (G) \u003d ∫ 1 3 + 13 2 - x 2 + 4 x - 2 - 1 xdx \u003d - x 3 3 + 2 x 2 - 2 x - Ln x 1 3 + 13 2 \u003d - 3 + 13 2 3 3 + 2 · 3 + 13 2 2 - 2 · 3 + 13 2 - Ln 3 + 13 2 - - - 1 3 3 + 2 · 1 2 - 2 · 1 - Ln 1 \u003d 7 + 13 3 - Ln 3 + 13 2.

Answer: S (G) \u003d 7 + 13 3 - Ln 3 + 13 2

Example 4.

It is necessary to calculate the area of \u200b\u200bthe figure, which is limited to the curves y \u003d x 3, y \u003d - log 2 x + 1 and the axis of the abscissa.

Decision

We will apply all lines on the schedule. We can get a function of the function y \u003d - log 2 x + 1 from the graph y \u003d log 2 x, if we place it symmetrically relative to the abscissa axis and raise one unit upwards. The abscissa axis equation y \u003d 0.

Denote the intersection points of the lines.

As can be seen from the figure, the graphs of the functions y \u003d x 3 and y \u003d 0 intersect at the point (0; 0). This is obtained because x \u003d 0 is the only valid root of the equation x 3 \u003d 0.

x \u003d 2 is the only root of the equation - log 2 x + 1 \u003d 0, so the graphs of the functions y \u003d - log 2 x + 1 and y \u003d 0 intersect at the point (2; 0).

x \u003d 1 is the only root of the equation x 3 \u003d - log 2 x + 1. In this connection, the graphs of the functions y \u003d x 3 and y \u003d - log 2 x + 1 intersect at the point (1; 1). The last statement may be unclear, but equation x 3 \u003d - log 2 x + 1 cannot have more than one root, since the function y \u003d x 3 is strictly increasing, and the function y \u003d - log 2 x + 1 is strictly decreasing.

A further solution involves several options.

Option number 1

Figure G we can imagine as the sum of two curvilinear trapezes located above the abscissa axis, the first of which is located below the midline on the segment x ∈ 0; 1, and the second below the red line on the segment x ∈ 1; 2. This means that the area will be equal to S (G) \u003d ∫ 0 1 x 3 D x + ∫ 1 2 (- log 2 x + 1) d x.

Option number 2.

Figure G can be represented as a difference of two figures, the first of which is located above the abscissa axis and below the blue line on the segment x ∈ 0; 2, and the second between the red and blue lines on the segment x ∈ 1; 2. This allows us to find the area as follows:

S (G) \u003d ∫ 0 2 x 3 D x - ∫ 1 2 x 3 - (- Log 2 x + 1) D x

In this case, to find the area will have to use the formula of the form S (G) \u003d ∫ C D (G 2 (Y) - G 1 (Y)) D y. In fact, the lines that limit the figure can be represented as functions from the argument y.

Allowed the equations y \u003d x 3 and - log 2 x + 1 relative to x:

y \u003d x 3 ⇒ x \u003d y 3 y \u003d - log 2 x + 1 ⇒ log 2 x \u003d 1 - y ⇒ x \u003d 2 1 - y

We get the desired area:

S (G) \u003d ∫ 0 1 (2 1 - y - y 3) dy \u003d - 2 1 - y ln 2 - y 4 4 0 1 \u003d - 2 1 - 1 Ln 2 - 1 4 4 - - 2 1 - 0 ln 2 - 0 4 4 \u003d 1 ln 2 - 1 4 + 2 ln 2 \u003d 1 ln 2 - 1 4

Answer: S (G) \u003d 1 LN 2 - 1 4

Example 5.

It is necessary to calculate the area of \u200b\u200bthe figure, which is limited to the lines y \u003d x, y \u003d 2 3 x - 3, y \u003d - 1 2 x + 4.

Decision

Red line will apply on the schedule line, specified function y \u003d x. Blue with a line y \u003d - 1 2 x + 4, in black, we denote the line y \u003d 2 3 x - 3.

Note the intersection points.

Find points of intersection of graphs of functions y \u003d x and y \u003d - 1 2 x + 4:

x \u003d - 1 2 x + 4 o d z: x ≥ 0 x \u003d - 1 2 x + 4 2 ⇒ x \u003d 1 4 x 2 - 4 x + 16 ⇔ x 2 - 20 x + 64 \u003d 0 d \u003d (- 20) 2 - 4 · 1 · 64 \u003d 144 x 1 \u003d 20 + 144 2 \u003d 16; x 2 \u003d 20 - 144 2 \u003d 4 p r o in e p k a: x 1 \u003d 16 \u003d 4, - 1 2 x 1 + 4 \u003d - 1 2 · 16 + 4 \u003d - 4 ⇒ x 1 \u003d 16 n I'm in l i eat x p and in n and i x 2 \u003d 4 \u003d 2, - 1 2 x 2 + 4 \u003d - 1 2 · 4 + 4 \u003d 2 ⇒ x 2 \u003d 4 I'm in l i e t s i r e n e m u r a in n and n i ⇒ (4; 2) t o h k a p e r e s e n and I y \u003d x and y \u003d - 1 2 x + 4

We will find the point of intersection of graphs of functions y \u003d x and y \u003d 2 3 x - 3:

x \u003d 2 3 x - 3 o d z: x ≥ 0 x \u003d 2 3 x - 3 2 ⇔ x \u003d 4 9 x 2 - 4 x + 9 ⇔ 4 x 2 - 45 x + 81 \u003d 0 d \u003d (- 45 ) 2 - 4 · 4 · 81 \u003d 729 x 1 \u003d 45 + 729 8 \u003d 9, x 2 45 - 729 8 \u003d 9 4 П Р О Е Р K a: x 1 \u003d 9 \u003d 3, 2 3 x 1 - 3 \u003d 2 3 · 9 - 3 \u003d 3 ⇒ x 1 \u003d 9 I'm in l i e m i r a n e n i ⇒ (9; 3) t o h to A P E R E C E C H I Y \u003d X and Y \u003d 2 3 x - 3 x 2 \u003d 9 4 \u003d 3 2, 2 3 x 1 - 3 \u003d 2 3 · 9 4 - 3 \u003d - 3 2 ⇒ x 2 \u003d 9 4 n e i l i e t with i r e n e m u r a in n e n i

We will find the point of intersection of the lines y \u003d - 1 2 x + 4 and y \u003d 2 3 x - 3:

1 2 x + 4 \u003d 2 3 x - 3 ⇔ - 3 x + 24 \u003d 4 x - 18 ⇔ 7 x \u003d 42 ⇔ x \u003d 6 - 1 2 · 6 + 4 \u003d 2 3 · 6 - 3 \u003d 1 ⇒ (6 ; 1) T about h k a n e r e c e n i y \u003d - 1 2 x + 4 and y \u003d 2 3 x - 3

Method number 1.

Imagine the area of \u200b\u200bthe desired figure as the sum of the areas of individual figures.

Then the figure of the figure is equal to:

S (G) \u003d ∫ 4 6 x - - 1 2 x + 4 dx + ∫ 6 9 x - 2 3 x - 3 dx \u003d 2 3 x 3 2 + x 2 4 - 4 x 4 6 + 2 3 x 3 2 - x 2 3 + 3 x 6 9 \u003d 2 3 · 6 3 2 + 6 2 4 - 4 · 6 - 2 3 · 4 3 2 + 4 2 4 - 4 · 4 + + 2 3 · 9 3 2 - 9 2 3 + 3 · 9 - 2 3 · 6 3 2 - 6 2 3 + 3 · 6 \u003d - 25 3 + 4 6 + - 4 6 + 12 \u003d 11 3

Method number 2.

The area of \u200b\u200bthe original figure can be represented as the sum of the two other figures.

Then we solve the equation of the line relative to X, and only after that we apply the formula for calculating the figure of the figure.

y \u003d x ⇒ x \u003d y 2 to r and with n and i l and i y \u003d 2 3 x - 3 ⇒ x \u003d 3 2 y + 9 2 h e r n i l and i y \u003d - 1 2 x + 4 ⇒ x \u003d - 2 y + 8 s and n i l and n i

Thus, the area is equal:

S (G) \u003d ∫ 1 2 3 2 y + 9 2 - - 2 y + 8 DY + ∫ 2 3 3 2 y + 9 2 - y 2 dy \u003d ∫ 1 2 7 2 y - 7 2 DY + ∫ 2 3 3 2 y + 9 2 - y 2 dy \u003d 7 4 y 2 - 7 4 y 1 2 + - y 3 3 + 3 y 2 4 + 9 2 y 2 3 \u003d 7 4 · 2 2 - 7 4 · 2 - 7 4 · 1 2 - 7 4 · 1 + + - 3 3 3 + 3 · 3 2 4 + 9 2 · 3 - - 2 3 3 + 3 · 2 2 4 + 9 2 · 2 \u003d 7 4 + 23 12 \u003d 11 3

As you can see, the values \u200b\u200bcoincide.

Answer: S (G) \u003d 11 3

RESULTS

To find the area of \u200b\u200bthe figure, which is limited to the specified lines, we need to build lines on the plane, find points of their intersection, apply the formula for finding the area. In this section, we considered the most common tasks options.

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