Area curvilinear trapezium Numerically equal a certain integral

Any particular integral (which exists) has a very good geometric meaning. At the lesson, 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 some curve on the plane (it can always be drawn if 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. 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.

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 this.
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.

Solution: Perform drawing:

If a curvilinear trapezium fully located under the 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 ,.

Solution: 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 corresponding figure 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 below the axis,

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 ,,,.

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 on inattention, it is often that it is necessary 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) 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:

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:

Hence, .

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 ,,

Solution: Show this shape in the drawing.

For checking the drawing, 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:

(1) How to integrate sinuses and cosines in odd degrees can be viewed at the lesson Integrals OT. trigonometric functions . This is a typical reception, pressing one sinus.

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

(3) We will replace the variable, then:

New alteration integration:

Who has very bad things with replacements, please go to the lesson Replacement method in an indefinite integral. Who is not very clear to the replacement algorithm in a specific integral, visit the page Certain integral. Examples of solutions.

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 (curvilinear trapezium base) 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 \\)

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, 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 $$$$ \\ int \\ FRAC (DX) (\\ Sin ^ 2 x) \u003d - \\ Text (CTG) x + C $$$$ \\ int \\ FRAC (DX) (\\ SQRT (1-x ^ 2)) \u003d \\ Text (ArcSin) x + C $$$$ \\ int \\ FRAC (DX) (1 + x ^ 2 ) \u003d \\ Text (ArCTG) X + C $$$$ \\ int \\ text (CH) x DX \u003d \\ Text (SH) x + C $$$$ \\ INT \\ TEXT (SH) x DX \u003d \\ TEXT (CH ) x + C $$

The figure, limited by a chart of continuous non-negative on a segment $$ Functions $ F (x) $ and direct $ y \u003d 0, \\ x \u003d a $ and $ x \u003d b $, is called a curvilinear trapezium.

The area of \u200b\u200bthe corresponding curvilinear trapezium is calculated by the formula:

$ S \u003d \\ int \\ limits_ (a) ^ (b) (F (x) DX). $ (*)

Tasks for finding the area of \u200b\u200bcurvilinear trapezium We will be conditionally divided by $ 4 $ type. Consider each type of Read more.

I Type: The curvilinear trapezium is clearly set. Then immediately apply the formula (*).

For example, find the area of \u200b\u200bcurvilinear trapezium, limited by the graph of the function $ y \u003d 4- (x-2) ^ (2) $, and direct $ y \u003d 0, \\ x \u003d 1 $ and $ x \u003d $ 3.

Draw this curvilinear trapezium.

Using the formula (*), we will find the area of \u200b\u200bthis curvilinear trapezium.

$ S \u003d \\ int \\ limits_ (1) ^ (3) (\\ left (4- (x-2) ^ (2) \\ Right) dx) \u003d \\ int \\ limits_ (1) ^ (3) (4DX) - \\ int \\ limits_ (1) ^ (3) ((x-2) ^ (2) dx) \u003d 4x | _ (1) ^ (3) - \\ left. \\ FRAC ((X-2) ^ (3) ) (3) \\ RIGHT | _ (1) ^ (3) \u003d $

$ \u003d 4 (3-1) - \\ FRAC (1) (3) \\ left ((3-2) ^ (3) - (1-2) ^ (3) \\ RIGHT) \u003d 4 \\ CDOT 2 - \\ FRAC (1) (3) \\ left ((1) ^ (3) - (- 1) ^ (3) \\ RIGHT) \u003d 8 - \\ FRAC (1) (3) (1 + 1) \u003d $

$ \u003d 8- \\ FRAC (2) (3) \u003d 7 \\ FRAC (1) (3) $ (unit $ ^ (2) $).

II Type: The curvilinear trapezoid is defined implicitly. This case usually does not specify or are specified partially straight $ x \u003d a, \\ x \u003d b $. In this case, you need to find the intersection points of the functions $ y \u003d f (x) $ and $ y \u003d 0 $. These points will be points $ a $ and $ b $.

For example, find the area of \u200b\u200bthe figure limited by the graphs of the functions $ y \u003d 1-x ^ (2) $ and $ y \u003d 0 $.

Find the intersection points. To do this, equate the right parts of functions.

Thus, $ a \u003d -1 $, and $ b \u003d 1 $. Draw this curvilinear trapezium.

Find the area of \u200b\u200bthis curvilinear trapezium.

$ S \u003d \\ int \\ limits _ (- 1) ^ (1) (\\ left (1-x ^ (2) \\ Right) dx) \u003d \\ int \\ limits _ (- 1) ^ (1) (1dx) - \\ int \\ Limits _ (- 1) ^ (1) (x ^ (2) dx) \u003d x | _ (- 1) ^ (1) - \\ left. \\ FRAC (X ^ (3)) (3) \\ Right | _ (-1) ^ (1) \u003d $

$ \u003d (1 - (- 1)) - \\ FRAC (1) (3) \\ left (1 ^ (3) - (- 1) ^ (3) \\ RIGHT) \u003d 2 - \\ FRAC (1) (3) \\ left (1 + 1 \\ RIGHT) \u003d 2 - \\ FRAC (2) (3) \u003d 1 \\ FRAC (1) (3) $ (units $ ^ (2) $).

III Type: Figure area, limited by the intersection of two continuous non-negative functions. This figure will not be a curvilinear trapezium, and therefore, with the help of formula (*), its area does not calculate. How to be?It turns out that the area of \u200b\u200bthis figure can be found as the difference in the areas of curvilinear trapezes bounded by the upper function and $ Y \u003d 0 $ ($ s_ (UF) $), and the lower function and $ Y \u003d 0 $ ($ s_ (LF) $), where In the role of $ x \u003d a, \\ x \u003d b $, coordinate are coordinates of $ X $ of the intersection of these functions, i.e.

$ S \u003d S_ (UF) -S_ (LF) $. (**)

The most important thing when calculating such areas is not "miss" with the choice of upper and lower function.

For example, find the area of \u200b\u200bthe figure limited by the functions $ y \u003d x ^ (2) $ and $ y \u003d x + $ 6.

Find the intersection points of these graphs:

On the Vieta theorem,

$ x_ (1) \u003d - 2, \\ x_ (2) \u003d 3. $

That is, $ a \u003d -2, \\ b \u003d $ 3. I'll show the figure:

Thus, the upper function is $ y \u003d x + $ 6, and the lower is $ y \u003d x ^ (2) $. Next, we find $ s_ (UF) $ and $ s_ (LF) $ by formula (*).

$ S_ (UF) \u003d \\ int \\ Limits _ (- 2) ^ (3) ((x + 6) dx) \u003d \\ int \\ limits _ (- 2) ^ (3) (xDX) + \\ int \\ limits _ (- 2 ) ^ (3) (6DX) \u003d \\ left. \\ FRAC (X ^ (2)) (2) \\ Right | _ (- 2) ^ (3) + 6x | _ (- 2) ^ (3) \u003d 32 , $ 5 (unit $ ^ (2) $).

$ S_ (LF) \u003d \\ int \\ limits _ (- 2) ^ (3) (x ^ (2) dx) \u003d \\ left. \\ FRAC (X ^ (3)) (3) \\ Right | _ (- 2) ^ (3) \u003d \\ FRAC (35) (3) $ (unit $ ^ (2) $).

We substitute found in (**) and we get:

$ S \u003d 32.5- \\ FRAC (35) (3) \u003d \\ FRAC (125) (6) $ (unit $ ^ (2) $).

IV Type: Figure area, limited by function (s), not satisfying (s) condition of non-negativity. In order to find the area of \u200b\u200bsuch a figure you need symmetrically relative to the $ OX $ axis ( in other words, Put "minuses" before functions) to display the area and using the methods set out in the types I - III, find the area of \u200b\u200bthe displayed area. This area will be the desired area. Previously, you may have to find the points of intersection of graphs of functions.

For example, find the area of \u200b\u200bthe figure limited by the graphs of the functions $ y \u003d x ^ (2) -1 $ and $ y \u003d 0 $.

Find the points of intersection of graphs of functions:

those. $ a \u003d -1 $, and $ b \u003d 1 $. Hatch the area.

Symmetrically display the area:

$ y \u003d 0 \\ \\ rightarrow \\ y \u003d -0 \u003d 0 $

$ y \u003d x ^ (2) -1 \\ \\ rightarrow \\ y \u003d - (x ^ (2) -1) \u003d 1-x ^ (2) $.

It turns out a curvilinear trapezion, limited by a graph of the function $ y \u003d 1-x ^ (2) $ and $ y \u003d 0 $. This is the task of finding a curvilinear trapezium of the second type. We have already solved it. The answer was such: $ s \u003d 1 \\ FRAC (1) (3) $ (unit $ ^ (2) $). It means that the area of \u200b\u200bthe desired curvilinear trapezium is equal to:

$ S \u003d 1 \\ FRAC (1) (3) $ (unit $ ^ (2) $).

It is required to calculate the area of \u200b\u200bthe curvilinear trapezion bounded by straight,
,
both curve
.

We break the cut
spoken elementary segments, length
th segment
. Restore perpendicular from the separation points of the segment to intersection with the curve
, let be
. As a result, we get elementary trapeziums, the sum of their area is obviously equal to the sum of a given curvilinear trapezium.

We define the greatest and smallest values \u200b\u200bof the function on each elementary interval, on the first interval.
, on the second
etc. Calculate sums

The first sum is the area of \u200b\u200ball described, the second - there is an area of \u200b\u200ball inscribed in the curvilinear trapezion of rectangles.

It is clear that the first sum gives the approximate value of the area of \u200b\u200bthe trapezoid "with an excess", the second - "with the disadvantage". The first amount is called the upper amount of Darboux, the second - according to the lower amount of Darboua. Thus, the area of \u200b\u200bthe curvilinear trapezium satisfies inequality
. Find out how the amounts of Darboux behave with increasing the number of segmentation points
. Let the number of partition points increased by one, and it is in the middle of the interval
. Now the number is like

inspected and described rectangles increased by one. Consider how the lower amount of Darbu has changed. Instead of Square
a fisher rectangle equal
we get the amount of the areas of two rectangles
Since Length
can not be less
the smallest values \u200b\u200bof the function on
. On the other hand,
, insofar as
can not be more
the greatest value of the function on the interval
. So, adding new segment breaking points increases the value of the bottom amount of Darboux and reduces the upper amount of Darbou. At the same time, the lower amount of Darboux with any increase in the number of partition points cannot exceed the values \u200b\u200bof any upper amount, since the sum of the areas of the described rectangles is always more sum Squares inscribed in a curvilinear trapezium of rectangles.

Thus, the sequence of the lower amounts of Darboux increases with increasing the number of points of splitting the segment and is limited from above, according to the well-known theorem, it has a limit. This limit is the area of \u200b\u200ba given curvilinear trapezium.

Similarly, the sequence of the upper amounts of Darboux decreases with an increase in the number of interval separation points and is limited to from the bottom of any lower amount of Darboux, it means that it also has a limit, and it is also equal to the area of \u200b\u200bthe curvilinear trapezium.

Therefore, to calculate the area of \u200b\u200bthe curvilinear trapezium is enough for splitting the interval to determine either the bottom or the upper amount of Darboua and then calculate
, or
.

However, such a solution to the task assumes with any, arbitrarily large number of partitions
, Finding on each elementary interval of the greatest or smallest value of the function, which is a very time-consuming task.

A simpler solution is obtained by the integral amount of Riemann, which is

where
some point of each elementary interval, that is
. Consequently, the integral amount of Riemann is the sum of the areas of all sorts of rectangles, and
. As shown above, the limits of the upper and lower amount of Darboux are the same and equal to the area of \u200b\u200bthe curvilinear trapezium. Using one of the properties of the limit of the function (the rule of two policemen), we get that with any splitting of the segment
and choosing points the area of \u200b\u200bthe curvilinear trapezium can be calculated using the formula
.