a)

Solution.

First and crucial moment solutions - building a drawing.

Let's make a drawing:

The equation y=0 sets the x-axis;

- x=-2 and x=1 - straight, parallel to the axis OU;

- y \u003d x 2 +2 - a parabola whose branches are directed upwards, with a vertex 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 a parabola can be found using the formulas:

You can draw lines and point by point.

On the interval [-2;1] the graph of the function y=x 2 +2 located over axis Ox , that's why:

Answer: S \u003d 9 square units

After the task is completed, it is always useful to look at the drawing and figure out 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 seems to be true. It is quite clear that if we had, say, the answer: 20 square units, then, obviously, a mistake was made somewhere - 20 cells clearly do not fit into the figure in question, at most a dozen. If the answer turned out to be negative, then the task was also solved incorrectly.

What to do if the curvilinear trapezoid is located under axle Oh?

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

Solution.

Let's make a drawing.

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

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

Attention! Don't confuse the two types of tasks:

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 the 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 plane figure bounded by lines y \u003d 2x-x 2, y \u003d -x.

Solution.

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

We solve the equation:

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

We build the given lines: 1. Parabola - vertex at the point (1;1); axis intersection Oh - points(0;0) and (0;2). 2. Straight line - the bisector of the 2nd and 4th coordinate angles. And now Attention! If on the interval [ a;b] some continuous function f(x) 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 does not 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 one 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 lines point by point, while the limits of integration are found out as if "by themselves". Nevertheless, the analytical method of finding the limits still sometimes has to be used if, for example, the graph is large enough, or the threaded construction did not reveal the limits of integration (they can be fractional or irrational).

The desired figure is limited by a parabola from above and a straight line from below.

On the segment , according to the corresponding formula:

Answer: S \u003d 4.5 sq. units

The figure bounded by the graph of a continuous non-negative function $f(x)$ on the interval $$ and the lines $y=0, \ x=a$ and $x=b$ is called a curvilinear trapezoid.

The area of ​​the corresponding curvilinear trapezoid calculated by the formula:

$S=\int\limits_(a)^(b)(f(x)dx).$ (*)

The problems of finding the area of ​​a curvilinear trapezoid we will conditionally divide into $4$ types. Let's consider each type in more detail.

Type I: a curvilinear trapezoid is given explicitly. Then immediately apply the formula (*).

For example, find the area of ​​a curvilinear trapezoid bounded by the graph of the function $y=4-(x-2)^(2)$ and the lines $y=0, \ x=1$ and $x=3$.

Let's draw this curvilinear trapezoid.

Applying the formula (*), we find the area of ​​this curvilinear trapezoid.

$S=\int\limits_(1)^(3)(\left(4-(x-2)^(2)\right)dx)=\int\limits_(1)^(3)(4dx)- \int\limits_(1)^(3)((x-2)^(2)dx)=4x|_(1)^(3) – \left.\frac((x-2)^(3) )(3)\right|_(1)^(3)=$

$=4(3-1)-\frac(1)(3)\left((3-2)^(3)-(1-2)^(3)\right)=4 \cdot 2 - \frac (1)(3)\left((1)^(3)-(-1)^(3)\right) = 8 – \frac(1)(3)(1+1) =$

$=8-\frac(2)(3)=7\frac(1)(3)$ (unit$^(2)$).

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

For example, find the area of ​​the figure bounded by the graphs of the functions $y=1-x^(2)$ and $y=0$.

Let's find the intersection points. To do this, we equate the right parts of the functions.

So $a=-1$ and $b=1$. Let's draw this curvilinear trapezoid.

Find the area of ​​this curvilinear trapezoid.

$S=\int\limits_(-1)^(1)(\left(1-x^(2)\right)dx)=\int\limits_(-1)^(1)(1dx)-\int \limits_(-1)^(1)(x^(2)dx)=x|_(-1)^(1) – \left.\frac(x^(3))(3)\right|_ (-1)^(1)=$

$=(1-(-1))-\frac(1)(3)\left(1^(3)-(-1)^(3)\right)=2 – \frac(1)(3) \left(1+1\right) = 2 – \frac(2)(3) = 1\frac(1)(3)$ (unit$^(2)$).

Type III: the area of ​​a figure bounded by the intersection of two continuous non-negative functions. This figure will not be a curvilinear trapezoid, which means that using the formula (*) you cannot calculate its area. How to be? It turns out that the area of ​​this figure can be found as the difference between the areas of curvilinear trapezoids bounded by the upper function and $y=0$ ($S_(uf)$) and the lower function and $y=0$ ($S_(lf)$), where the role of $x=a, \ x=b$ is played by the $x$ coordinates of the intersection points of these functions, i.e.

$S=S_(uf)-S_(lf)$. (**)

The most important thing when calculating such areas is not to “miss” with the choice of the upper and lower functions.

For example, find the area of ​​a figure bounded by the functions $y=x^(2)$ and $y=x+6$.

Let's find the intersection points of these graphs:

According to Vieta's theorem,

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

That is, $a=-2, \ b=3$. Let's draw a figure:

So the top function is $y=x+6$ and the bottom one is $y=x^(2)$. Next, find $S_(uf)$ and $S_(lf)$ using the formula (*).

$S_(uf)=\int\limits_(-2)^(3)((x+6)dx)=\int\limits_(-2)^(3)(xdx)+\int\limits_(-2 )^(3)(6dx)=\left.\frac(x^(2))(2)\right|_(-2)^(3) + 6x|_(-2)^(3)= 32 ,5$ (unit $^(2)$).

$S_(lf)=\int\limits_(-2)^(3)(x^(2)dx)=\left.\frac(x^(3))(3)\right|_(-2) ^(3) = \frac(35)(3)$ (unit$^(2)$).

Substitute found in (**) and get:

$S=32,5-\frac(35)(3)= \frac(125)(6)$ (unit $^(2)$).

Type IV: the area of ​​a figure bounded by a function(s) that does not satisfy the non-negativity condition. In order to find the area of ​​such a figure, you need to be symmetrical about the $Ox$ axis ( in other words, put “minuses” in front of the functions) display the area and, using the methods described in types I - III, find the area of ​​the displayed area. This area will be the required area. First, you may have to find the intersection points of the function graphs.

For example, find the area of ​​the figure bounded by the graphs of the functions $y=x^(2)-1$ and $y=0$.

Let's find the intersection points of the function graphs:

those. $a=-1$ and $b=1$. Let's draw the area.

Let's display the area symmetrically:

$y=0 \ \Rightarrow \ y=-0=0$

$y=x^(2)-1 \ \Rightarrow \ y= -(x^(2)-1) = 1-x^(2)$.

You get a curvilinear trapezoid bounded by the graph of the function $y=1-x^(2)$ and $y=0$. This is a problem of finding a curvilinear trapezoid of the second type. We already solved it. The answer was: $S= 1\frac(1)(3)$ (units $^(2)$). So, the area of ​​the desired curvilinear trapezoid is equal to:

$S=1\frac(1)(3)$ (unit$^(2)$).

Definite integral. How to calculate the area of ​​a figure

We now turn to the consideration of applications of the integral calculus. In this lesson, we will analyze a typical and most common task. How to use a definite integral to calculate the area of ​​a plane figure. Finally, those who seek meaning in higher mathematics - may they find it. You never know. We'll have to get closer in life country cottage area elementary functions and find its area using a definite integral.

To successfully master the material, you must:

1) understand indefinite integral at least at an average level. Thus, dummies should first read the lesson Not.

2) Be able to apply the Newton-Leibniz formula and calculate the definite integral. Forge warm friendly relations with definite integrals can be found on the page Definite integral. Solution examples.

In fact, in order to find the area of ​​\u200b\u200ba figure, you do not need so much knowledge of the indefinite and definite integral. The task "calculate the area using a definite integral" always involves the construction of a drawing, so much more topical issue will be your knowledge and drawing skills. In this regard, it is useful to refresh the memory of the graphs of the main elementary functions, and, at a minimum, to be able to build a straight line, a parabola and a hyperbola. This can be done (many need) with the help of methodological material and articles on geometric transformations of graphs.

Actually, everyone is familiar with the problem of finding the area using a definite integral since school, and we will go a little ahead of the school curriculum. This article might not exist at all, but the fact is that the problem occurs in 99 cases out of 100, when a student is tormented by a hated tower with enthusiasm mastering a course in higher mathematics.

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

Let's start with a curvilinear trapezoid.

Curvilinear trapezoid called a flat figure bounded by the axis , straight lines , and the graph of a function continuous on a segment that does not change sign on this interval. Let this figure be located not less abscissa:

Then the area of ​​a curvilinear trapezoid is numerically equal to a certain integral. Any definite integral (that exists) has a very good geometric meaning. On the lesson Definite integral. Solution examples I said that a definite integral is a number. And now it's time to state another useful fact. From the point of view of geometry, the definite integral is the AREA.

That is, the definite integral (if it exists) geometrically corresponds to the area of ​​some figure. For example, consider the definite integral . The integrand defines a curve on the plane that is located above the axis (those who wish can complete the drawing), and the definite integral itself is numerically equal to the area of ​​the corresponding curvilinear trapezoid.

Example 1

This is a typical task statement. The first and most important moment of the decision is the construction of a drawing. Moreover, the drawing must be built RIGHT.

When building a blueprint, I recommend the following order: first it is better to construct all lines (if any) and only after- parabolas, hyperbolas, graphs of other functions. Function graphs are more profitable to build point by point, the technique of pointwise construction can be found in reference material Graphs and properties of elementary functions. There you can also find material that is very useful in relation to our lesson - how to quickly build a parabola.

In this problem, the solution might look like this.
Let's make a drawing (note that the equation defines the axis):

I will not hatch a curvilinear trapezoid, it is obvious what area we are talking about here. The solution continues like this:

On the segment, the graph of the function is located over axis, that's why:

Answer:

Who has difficulty calculating the definite integral and applying the Newton-Leibniz formula , refer to the lecture Definite integral. Solution examples.

After the task is completed, it is always useful to look at the drawing and figure out 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 seems to be true. It is quite clear that if we had, say, the answer: 20 square units, then, obviously, a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer turned out to be negative, then the task was also solved incorrectly.

Example 2

Calculate the area of ​​the figure bounded by the lines , , and the axis

This is a do-it-yourself example. Full solution and answer at the end of the lesson.

What to do if the curvilinear trapezoid is located under axle?

Example 3

Calculate the area of ​​the figure bounded by lines and coordinate axes.

Solution: Let's make a drawing:

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

Attention! Don't confuse the two types of tasks:

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 the minus appears in the formula just considered.

In practice, most often the figure is located in both the upper and lower half-planes, and therefore, from the simplest school problems, we move on to more meaningful examples.

Example 4

Find the area of ​​a flat figure bounded by lines , .

Solution: First you need to complete the drawing. Generally speaking, when constructing a drawing in area problems, we are most interested in the intersection points of lines. Let's find the points of intersection of the parabola and the line. This can be done in two ways. The first way is analytical. We solve the equation:

Hence, the lower limit of integration , the upper limit of integration .
It is best not to use this method if possible..

It is much more profitable and faster to build the lines point by point, while the limits of integration are found out as if “by themselves”. The point-by-point construction technique for various charts is discussed in detail in the help Graphs and properties of elementary functions. Nevertheless, the analytical method of finding the limits still sometimes has to be used if, for example, the graph is large enough, or the threaded construction did not reveal the limits of integration (they can be fractional or irrational). And we will also consider such an example.

We return to our task: it is more rational to first construct a straight line and only then a parabola. Let's make a drawing:

I repeat that with pointwise construction, the limits of integration are most often found out “automatically”.

And now the working formula: If there is some continuous function on the interval greater than or equal some continuous function, then the area of ​​the figure bounded by the graphs of these functions and straight lines, can be found by the formula:

Here it is no longer necessary to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it matters which chart is ABOVE(relative to another graph), and which one 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

The completion of the solution might look like this:

The desired figure is limited by a parabola from above and a straight line from below.
On the segment , according to the corresponding formula:

Answer:

In fact, the school formula for the area of ​​a curvilinear trapezoid in the lower half-plane (see simple example No. 3) is a special case of the formula . Since the axis is given by the equation , and the graph of the function is located not higher axes, then

And now a couple of examples for an independent decision

Example 5

Example 6

Find the area of ​​the figure enclosed by the lines , .

In the course of solving problems for calculating the area using a certain integral, a funny incident sometimes happens. The drawing was made correctly, the calculations were correct, but due to inattention ... found the area of ​​the wrong figure, that's how your obedient servant screwed up several times. Here real case from life:

Example 7

Calculate the area of ​​the figure bounded by the lines , , , .

Solution: Let's make a drawing first:

…Eh, the drawing came out crap, but everything seems to be legible.

The figure whose area we need to find is shaded in blue.(carefully look at the condition - how the figure is limited!). But in practice, due to inattention, a “glitch” often occurs, that you need to find the area of ​​\u200b\u200bthe figure that is shaded in green!

This example is also useful in that in it the area of ​​\u200b\u200bthe figure is calculated using two definite integrals. Really:

1) On the segment above the axis there is a straight line graph;

2) On the segment above the axis is a hyperbola graph.

It is quite obvious that the areas can (and should) be added, therefore:

Answer:

Let's move on to one more meaningful task.

Example 8

Calculate the area of ​​a figure bounded by lines,
Let's present the equations in a "school" form, and perform a point-by-point drawing:

It can be seen from the drawing that our upper limit is “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 turn out that. Or root. What if we didn't get the graph right at all?

In such cases, one has to spend additional time and refine the limits of integration analytically.

Let's find the points of intersection of the line and the parabola.
To do this, we solve the equation:


,

Really, .

The further solution is trivial, the main thing is not to get confused in substitutions and signs, the calculations here are not the easiest.

On the segment , according to the corresponding formula:

Answer:

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

Example 9

Calculate the area of ​​the figure bounded by lines , ,

Solution: Draw this figure in the drawing.

Damn, I forgot to sign the schedule, and redoing the picture, sorry, not hotz. Not a drawing, in short, today is a day =)

For pointwise construction, you need to know appearance sinusoids (and in general it is useful to know graphs of all elementary functions), as well as some sine values, they can be found in trigonometric table. In some cases (as in this case), it is allowed to construct a schematic drawing, on which graphs and integration limits must be displayed in principle correctly.

There are no problems with the integration limits here, they follow directly from the condition: - "x" changes from zero to "pi". We make a further decision:

On the segment, the graph of the function is located above the axis, therefore:

Task number 3. Make a drawing and calculate the area of ​​\u200b\u200bthe figure bounded by lines

Application of the integral to solving applied problems

Area calculation

The definite integral of a continuous non-negative function f(x) is numerically equal to the area of ​​a curvilinear trapezoid bounded by the curve y \u003d f (x), the O x axis and the straight lines x \u003d a and x \u003d b. Accordingly, the area formula is written as follows:

Consider some examples of calculating the areas of plane figures.

Task number 1. Calculate the area bounded by the lines y \u003d x 2 +1, y \u003d 0, x \u003d 0, x \u003d 2.

Solution. Let's build a figure, the area of ​​​​which we will have to calculate.

y \u003d x 2 + 1 is a parabola whose branches are directed upwards, and the parabola is shifted upwards by one unit relative to the O y axis (Figure 1).

Figure 1. Graph of the function y = x 2 + 1

Task number 2. Calculate the area bounded by the lines y \u003d x 2 - 1, y \u003d 0 in the range from 0 to 1.

Solution. The graph of this function is the parabola of the branch, which is directed upwards, and the parabola is shifted down by one unit relative to the O y axis (Figure 2).

Figure 2. Graph of the function y \u003d x 2 - 1

Task number 3. Make a drawing and calculate the area of ​​\u200b\u200bthe figure bounded by lines

y = 8 + 2x - x 2 and y = 2x - 4.

Solution. The first of these two lines is a parabola with branches pointing downwards, since the coefficient at x 2 is negative, and the second line is a straight line crossing both coordinate axes.

To construct a parabola, let's find the coordinates of its vertex: y'=2 – 2x; 2 – 2x = 0, x = 1 – vertex abscissa; y(1) = 8 + 2∙1 – 1 2 = 9 is its ordinate, N(1;9) is its vertex.

Now we find the points of intersection of the parabola and the line by solving the system of equations:

Equating the right sides of an equation whose left sides are equal.

We get 8 + 2x - x 2 \u003d 2x - 4 or x 2 - 12 \u003d 0, from where .

So, the points are the points of intersection of the parabola and the straight line (Figure 1).

Figure 3 Graphs of functions y = 8 + 2x – x 2 and y = 2x – 4

Let's build a straight line y = 2x - 4. It passes through the points (0;-4), (2; 0) on the coordinate axes.

To build a parabola, you can also have its intersection points with the 0x axis, that is, the roots of the equation 8 + 2x - x 2 = 0 or x 2 - 2x - 8 = 0. By the Vieta theorem, it is easy to find its roots: x 1 = 2, x 2 = four.

Figure 3 shows a figure (parabolic segment M 1 N M 2) bounded by these lines.

The second part of the problem is to find the area of ​​this figure. Its area can be found using a definite integral using the formula .

With regard to this condition, we obtain the integral:

2 Calculation of the volume of a body of revolution

The volume of the body obtained from the rotation of the curve y \u003d f (x) around the O x axis is calculated by the formula:

When rotating around the O y axis, the formula looks like:

Task number 4. Determine the volume of the body obtained from the rotation of a curvilinear trapezoid bounded by straight lines x \u003d 0 x \u003d 3 and a curve y \u003d around the O x axis.

Solution. Let's build a drawing (Figure 4).

Figure 4. Graph of the function y =

The desired volume is equal to

Task number 5. Calculate the volume of the body obtained from the rotation of a curvilinear trapezoid bounded by a curve y = x 2 and straight lines y = 0 and y = 4 around the axis O y .

Solution. We have:

Review questions

How to insert mathematical formulas on the site?

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 universal method will help improve the visibility of the site in search engines. It has been working for a long time (and I think it will work forever), but it is morally outdated.

If you are constantly using math formulas on your site, then I recommend 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) using a simple code, you can quickly connect a MathJax script to your site, which will be automatically loaded from a remote server at the right time (list of servers); (2) upload the MathJax script from a remote server to your server and connect it to all pages of your site. The second method is more complex and time consuming and will allow you to speed up the loading of the pages of your site, and if the parent MathJax server becomes temporarily unavailable for some reason, 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 within 5 minutes you will be able to use all the features of MathJax on your website.

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

One of these code options needs to 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 paste the second code, then 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 the site control panel, add a widget designed to insert third-party JavaScript code, copy the first or second version of the load code above into it, and place the widget closer to the beginning of the template (by the way, this is not necessary at all , since 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 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 a 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 cubes adjacent to it along the faces are removed from it. It turns out a set consisting of 20 remaining smaller cubes. Doing the same with each of these cubes, we get a set consisting of 400 smaller cubes. Continuing this process indefinitely, we get the Menger sponge.