The area of ​​the shaded figure bounded by a parabola and a straight line. Examples of

but)

Solution.

First and the most important moment solutions - drawing building.

Let's execute the drawing:

The equation y = 0 sets the x-axis;

- x = -2 and x = 1 - straight lines parallel to the axes OU;

- y = x 2 +2 - a parabola, the branches of which are directed upwards, with apex at the point (0; 2).

Comment. To construct a parabola, it is enough to find the points of its intersection with the coordinate axes, i.e. putting x = 0 find the intersection with the axis OU and deciding the appropriate quadratic equation, find the intersection with the axis Oh .

The vertex of the parabola can be found by the formulas:

You can draw lines and point by point.

On the segment [-2; 1] the graph of the function y = x 2 +2 situated above the axis Ox , so:

Answer: S = 9 square units

After the task is completed, it is always helpful to look at the blueprint and estimate if the answer is real. In this case, "by eye" we count the number of cells in the drawing - well, about 9 will be typed, it looks like the truth. It is quite clear that if we got, say, the answer: 20 square units, then, obviously, a mistake has been made somewhere - the figure under consideration clearly does not fit 20 cells, at most ten. If the answer is negative, then the task was also solved incorrectly.

What to do if the curved trapezoid is located under the axis Oh?

b) Calculate the area of ​​a shape, limited by lines y = -e x , x = 1 and coordinate axes.

Solution.

Let's complete the drawing.

If the curved trapezoid completely located under the axle Oh , then its area can be found by the formula:

Answer: S = (e-1) sq. units "1.72 sq. units.

Attention! The two types of tasks should not be confused:

1) If you are asked to solve just a definite integral without any geometric meaning, then it can be negative.

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

In practice, most often the figure is located in both the upper and lower half-planes.

with) Find the area of ​​a flat figure bounded by lines y = 2x-x 2, y = -x.

Solution.

First you need to complete the drawing. Generally speaking, when constructing a drawing in problems on an area, we are most interested in the points of intersection of lines. Find the intersection points of the parabola and straight This can be done in two ways. The first way is analytical.

We solve the equation:

Hence, the lower limit of integration a = 0 , the upper limit of integration b = 3 .

We build the given lines: 1. Parabola - the vertex at the point (1; 1); axis intersection Oh - points (0; 0) and (0; 2). 2. Straight line - bisector of the 2nd and 4th coordinate angles. Now Attention! If on the segment [ a; b] some continuous function f (x) is greater than or equal to some continuous function g (x), then the area of ​​the corresponding figure can be found by the formula: . And it doesn't matter where the figure is located - above the axis or below the axis, but it is important which chart is HIGHER (relative to another chart) and which is BELOW. In the example under consideration, it is obvious that on the segment the parabola is located above the straight line, and therefore it is necessary to subtract from

It is possible to construct the lines point by point, while the limits of integration become clear, as it were, "by themselves." Nevertheless, the analytical method of finding the limits still has to be used sometimes if, for example, the graph is large enough, or the precise construction did not reveal the limits of integration (they can be fractional or irrational).

The required figure is bounded by a parabola at the top and a straight line at the bottom.

On the segment , according to the corresponding formula:

Answer: S = 4.5 square units

How to insert mathematical formulas into a website?

If you ever need to add one or two mathematical formulas to a web page, then the easiest way to do this is as described in the article: mathematical formulas are easily inserted into the site in the form of pictures that Wolfram Alpha automatically generates. In addition to simplicity, this versatile method will help improve your site's visibility in search engines. It has been working for a long time (and I think it will work forever), but it is already morally outdated.

If you regularly use math formulas on your site, then I recommend that you use MathJax, a special JavaScript library that displays math notation in web browsers using MathML, LaTeX, or ASCIIMathML markup.

There are two ways to start using MathJax: (1) with a simple code, you can quickly connect a MathJax script to your site, which will be in the right moment automatically download from a remote server (server list); (2) upload the MathJax script from a remote server to your server and connect it to all pages of your site. The second method, which is more complicated and time-consuming, will speed up the loading of your site's pages, and if the parent MathJax server for some reason becomes temporarily unavailable, this will not affect your own site in any way. Despite these advantages, I chose the first method, as it is simpler, faster and does not require technical skills. Follow my example, and in 5 minutes you will be able to use all the MathJax features on your website.

You can connect the script of the MathJax library from a remote server using two versions of the code taken from the main MathJax site or from the documentation page:

One of these code variants must be copied and pasted into the code of your web page, preferably between the tags and or right after the tag ... According to the first option, MathJax loads faster and slows down the page less. But the second option automatically tracks and loads the latest versions of MathJax. If you insert the first code, then it will need to be updated periodically. If you insert the second code, the pages will load more slowly, but you will not need to constantly monitor MathJax updates.

The easiest way to connect MathJax is in Blogger or WordPress: in your site's dashboard, add a widget for inserting third-party JavaScript code, copy the first or second version of the loading code presented above into it, and place the widget closer to the beginning of the template (by the way, this is not necessary at all because the MathJax script is loaded asynchronously). That's all. Now, learn the MathML, LaTeX, and ASCIIMathML markup syntax, and you're ready to embed math formulas into your website's web pages.

Any fractal is built according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.

The iterative algorithm for constructing the Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 adjacent cubes are removed from it. The result is a set consisting of the remaining 20 smaller cubes. Doing the same with each of these cubes, we get a set, already consisting of 400 smaller cubes. Continuing this process endlessly, we get a Menger sponge.

We begin to consider the actual process of calculating the double integral and get acquainted with its geometric meaning.

The double integral is numerically equal to the area of ​​a flat figure (region of integration). This is simplest view double integral when the function of two variables is equal to one:.

First, consider the problem in general view... Now you will be surprised how simple it really is! Let's calculate the area of ​​a flat figure bounded by lines. For definiteness, we assume that on the segment. The area of ​​this figure is numerically equal to:

Let's draw the area in the drawing:

Let's choose the first way to traverse the area:

Thus:

And immediately an important technical trick: iterated integrals can be considered separately... First the inner integral, then the outer integral. This method I highly recommend to beginners in the topic teapots.

1) We calculate the internal integral, while the integration is carried out over the variable "game":

The indefinite integral is the simplest here, and then the banal Newton-Leibniz formula is used, with the only difference that the limits of integration are not numbers, but functions... First, the upper limit was substituted into the "game" (antiderivative function), then - the lower limit

2) The result obtained in the first paragraph must be substituted into the external integral:

A more compact record of the entire solution looks like this:

The resulting formula Is exactly the working formula for calculating the area of ​​a flat figure using the "ordinary" definite integral! Watch the lesson Calculating area using a definite integral, there she is at every turn!

I.e, area calculation problem using double integral not much different from the problem of finding the area using a definite integral! In fact, they are the same thing!

Accordingly, no difficulties should arise! I will consider not very many examples, since you, in fact, have repeatedly encountered this task.

Example 9

Solution: Let's draw the area in the drawing:

Let's choose the following order of traversing the region:

Hereinafter, I will not dwell on how to perform area traversal, since very detailed explanations were given in the first paragraph.

Thus:

As I already noted, it is better for beginners to calculate iterated integrals separately, and I will follow the same method:

1) First, using the Newton-Leibniz formula, we deal with the internal integral:

2) The result obtained at the first step is substituted into the outer integral:

Point 2 is actually finding the area of ​​a flat figure using a definite integral.

Answer:

Here is such a stupid and naive task.

An interesting example for independent decision:

Example 10

Using a double integral, calculate the area of ​​a flat figure bounded by lines,

An approximate sample of the final design of the solution at the end of the lesson.

In Examples 9-10, it is much more profitable to use the first way to traverse the area; curious readers, by the way, can change the order of the traversal and calculate the areas in the second way. If you do not make a mistake, then, naturally, the same values ​​of the areas will turn out.

But in a number of cases, the second method of bypassing the area is more effective, and at the end of the course of a young nerd, consider a couple more examples on this topic:

Example 11

Using the double integral, calculate the area of ​​a flat figure bounded by lines,

Solution: we are looking forward to two parabolas with a quirk, which lie on one side. You don't need to smile, similar things in multiple integrals are common.

What is the easiest way to make a drawing?

We represent the parabola in the form of two functions:
- upper branch and - lower branch.

Similarly, we represent the parabola in the form of an upper and a lower branches.

Further, point-by-point charting rules, as a result of which such a bizarre figure is obtained:

We calculate the area of ​​the figure using a double integral by the formula:

What happens if we choose the first way to traverse the area? First, this area will have to be divided into two parts. And secondly, we will observe this very sad picture: ... Integrals, of course, are not of a super-complicated level, but ... there is an old mathematical adage: those who are friendly with roots do not need a test.

Therefore, from a misunderstanding given in the condition, we express the inverse functions:

Inverse functions in this example have the advantage that they set the entire parabola at once without any leaves, acorns, branches and roots.

According to the second method, the traversal of the area will be as follows:

Thus:

Feel the difference, as they say.

1) Deal with the internal integral:

Substitute the result into the outer integral:

Integration with respect to the variable "igrek" should not be embarrassing, if there were the letter "siu", it would be great to integrate over it. Although who read the second paragraph of the lesson How to calculate the volume of a body of revolution, he no longer experiences the slightest awkwardness with the integration according to the "game".

Also pay attention to the first step: the integrand is even, and the integration segment is symmetric about zero. Therefore, the segment can be halved, and the result can be doubled. This technique is commented out in detail in the lesson. Effective methods calculating a definite integral.

What to add…. Everything!

Answer:

To test your integration technique, you can try calculating ... The answer should be exactly the same.

Example 12

Using the double integral, calculate the area of ​​a flat figure bounded by lines

This is an example for a do-it-yourself solution. It is interesting to note that if you try to use the first method of traversing the area, then the figure will have to be divided not into two, but into three parts! And, accordingly, you get three pairs of repeated integrals. Sometimes it happens.

The master class has come to an end, and it's time to move to the grandmaster level - How do I calculate the double integral? Examples of solutions... I will try not to be so maniac in the second article =)

Wish you luck!

Solutions and Answers:

Example 2:Solution: Let's draw the area on the drawing:

Let's choose the following order of traversing the region:

Thus:
Let's move on to the inverse functions:


Thus:
Answer:

Example 4:Solution: Let's move on to direct functions:


Let's execute the drawing:

Let's change the order of traversing the area:

Answer:

Problem 1(on calculating the area of ​​a curved trapezoid).

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

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

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

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

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

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

Let's summarize. Solutions different tasks reduced to the same mathematical model. Many problems from various fields of science and technology lead in the process of solving to the same model. This means that this mathematical model must be specially studied.

Definitive integral concept

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

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

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

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

Formula of Newton - Leibniz

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

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

On the other hand, the coordinate of the moving point is the antiderivative for the velocity - let us denote it by s (t); hence, the displacement s is expressed by the formula s = s (b) - s (a). As a result, we get:
\ (S = \ int \ limits_a ^ b v (t) dt = s (b) -s (a) \)
where s (t) is the antiderivative for v (t).

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

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

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

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

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

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

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

Calculating the areas of planar figures using a definite integral

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

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

Table of indefinite integrals (antiderivatives) of some functions

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

This article will show you how to find the area of ​​a shape bounded by lines using integral calculations. For the first time, we come across the formulation of such a problem in high school, when the study of definite integrals has just passed and it is time to start a geometric interpretation of the knowledge gained in practice.

So what is required for successful solution problems of finding the area of ​​a figure using integrals:

• Ability to competently build drawings;
• The ability to solve a definite integral using famous formula Newton-Leibniz;
• The ability to “see” a more advantageous solution - that is, to understand how in this or that case it will be more convenient to carry out the integration? Along the x-axis (OX) or the y-axis (OY)?
• Well, where without correct calculations?) This includes understanding how to solve that other type of integrals and correct numerical calculations.

Algorithm for solving the problem of calculating the area of ​​a figure bounded by lines:

1. We build a drawing. It is advisable to do this on a piece of paper in a cage, with a large scale. We sign the name of this function with a pencil above each graph. The signature of the graphs is done solely for the convenience of further calculations. Having received the graph of the desired figure, in most cases it will be immediately visible which limits of integration will be used. Thus, we solve the problem graphically. However, it so happens that the values ​​of the limits are fractional or irrational. Therefore, you can make additional calculations, go to step two.

2. If the limits of integration are not explicitly set, then we find the points of intersection of the graphs with each other, and see if our graphical solution coincides with the analytical one.

3. Next, you need to analyze the drawing. Depending on how the function graphs are located, there are different approaches to finding the area of ​​a figure. Consider different examples to find the area of ​​a figure using integrals.

3.1. The most classic and simple version of the problem is when you need to find the area of ​​a curved trapezoid. What is a curved trapezoid? It is a flat figure bounded by the x-axis. (y = 0), straight x = a, x = b and any curve continuous from a before b... Moreover, this figure is non-negative and is located not below the abscissa axis. In this case, the area of ​​a curvilinear trapezoid is numerically equal to a definite integral calculated by the Newton-Leibniz formula:

Example 1 y = x2 - 3x + 3, x = 1, x = 3, y = 0.

What are the lines bounding the figure? We have a parabola y = x2 - 3x + 3 which is located above the axis OH, it is non-negative, because all points of this parabola are positive. Further, the straight lines x = 1 and x = 3 that run parallel to the axis OU, are the bounding lines of the shape on the left and right. Well y = 0, it is the x-axis, which limits the figure from below. The resulting shape is shaded as seen in the picture on the left. In this case, you can immediately start solving the problem. We have before us a simple example of a curvilinear trapezoid, which we further solve using the Newton-Leibniz formula.

3.2. In the previous paragraph 3.1, we analyzed the case when the curvilinear trapezoid is located above the x-axis. Now consider the case when the conditions of the problem are the same, except that the function lies under the x-axis. A minus is added to the standard Newton-Leibniz formula. We will consider how to solve a similar problem further.

Example 2 ... Calculate the area of ​​a shape bounded by lines y = x2 + 6x + 2, x = -4, x = -1, y = 0.

In this example, we have a parabola y = x2 + 6x + 2 which originates from under the axis OH, straight x = -4, x = -1, y = 0... Here y = 0 bounds the desired shape from above. Direct x = -4 and x = -1 these are the boundaries within which a definite integral will be calculated. The principle of solving the problem of finding the area of ​​a figure almost completely coincides with example number 1. The only difference is that preset function not positive, and everything is also continuous on the interval [-4; -1] ... What does not mean positive? As you can see from the figure, the figure, which is within the given Xs, has exclusively "negative" coordinates, which is what we need to see and remember when solving the problem. We look for the area of ​​the figure using the Newton-Leibniz formula, only with a minus sign at the beginning.

The article is incomplete.