Task No. 3. Make a drawing and calculate the area of ​​the figure, limited by lines

Application of the integral to the solution of 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 = f(x), the O x axis and the straight lines x = a and x = b. In accordance with this, the area formula is written as follows:

Let's look at some examples of calculating the areas of plane figures.

Task No. 1. Calculate the area bounded by the lines y = x 2 +1, y = 0, x = 0, x = 2.

Solution. Let's construct a figure whose area we will have to calculate.

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

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

Task No. 2. Calculate the area bounded by the lines y = x 2 – 1, y = 0 in the range from 0 to 1.

Solution. The graph of this function is a parabola of branches that are directed upward, and the parabola is shifted relative to the O y axis down by one unit (Figure 2).

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

Task No. 3. Make a drawing and calculate the area of ​​the figure bounded by the lines

y = 8 + 2x – x 2 and y = 2x – 4.

Solution. The first of these two lines is a parabola with its branches directed downward, since the coefficient of x 2 is negative, and the second line is a straight line intersecting both coordinate axes.

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

Now let’s find the intersection points of the parabola and the straight 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 = 2x – 4 or x 2 – 12 = 0, whence .

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

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

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

To construct a parabola, you can also use 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. Using Vieta’s theorem, it is easy to find its roots: x 1 = 2, x 2 = 4.

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 according to the formula .

In relation to this condition, we obtain the integral:

2 Calculation of the volume of a body of rotation

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

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

Task No. 4. Determine the volume of the body obtained from the rotation of a curved trapezoid bounded by straight lines x = 0 x = 3 and curve y = around the O x axis.

Solution. Let's draw a picture (Figure 4).

Figure 4. Graph of the function y =

The required volume is

Task No. 5. Calculate the volume of the body obtained from the rotation of a curved trapezoid bounded by the curve y = x 2 and straight lines y = 0 and y = 4 around the O y axis.

Solution. We have:

Review questions

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

Let's move on to consider applications of integral calculus. In this lesson we will analyze the typical and most common task – how to use a definite integral to calculate the area of ​​a plane figure. Finally searching for meaning in higher mathematics - may they find him. You never know. In real life, you will have to approximate a dacha plot using 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. You can establish warm friendly relations with certain integrals on the page Definite integral. Examples of solutions.

In fact, in order to find the area of ​​a figure, you don’t need that much knowledge of the indefinite and definite integral. The task “calculate the area using a definite integral” always involves constructing a drawing, so your knowledge and drawing skills will be a much more pressing issue. In this regard, it is useful to refresh your memory of the graphs of basic elementary functions, and, at a minimum, to be able to construct a straight line, parabola and hyperbola. This can be done (for many, it is necessary) using methodological material and articles on geometric transformations of graphs.

Actually, everyone is familiar with the task of finding the area using a definite integral since school, and we will not go much further from school curriculum. This article might not have existed at all, but the fact is that the problem occurs in 99 cases out of 100, when a student suffers from a hated school and enthusiastically masters 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 curved trapezoid.

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

Then the area of ​​a curvilinear trapezoid is numerically equal to a definite integral. Any definite integral (that exists) has a very good geometric meaning. At the lesson Definite integral. Examples of solutions 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 AREA.

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

Example 1

This is a typical assignment statement. The first and most important point in the decision is the construction of a drawing. Moreover, the drawing must be constructed RIGHT.

When constructing a drawing, I recommend the following order: at first it is better to construct all straight lines (if they exist) and only Then– parabolas, hyperbolas, graphs of other functions. It is more profitable to build graphs of functions point by point, the point-by-point construction technique can be found in reference material Graphs and properties of elementary functions. There you can also find very useful material for our lesson - how to quickly build a parabola.

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

I will not hatch a curved trapezoid, it is obvious here what the area is we're talking about. The solution continues like this:

On the segment, the graph of the function is located above the axis, That's why:

Answer:

Who has difficulties with calculating the definite integral and applying the Newton-Leibniz formula , refer to the lecture Definite integral. Examples of solutions.

After the task is completed, it is always useful to look at the drawing and figure out whether the answer is real. In this case, we count the number of cells in the drawing “by eye” - well, there will be about 9, it seems to be true. It is absolutely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.

Example 2

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

This is an example for you to solve on your own. Full solution and answer at the end of the lesson.

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

Example 3

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

Solution: Let's make a drawing:

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

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

1) If you are asked to solve simply 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 discussed.

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

Example 4

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

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

This means that the lower limit of integration is , the upper limit of integration is .
If possible, it is better not to use this method..

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

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

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

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

Here you no longer need to think about where the figure is located - above the axis or below the axis, and, roughly speaking, it matters which graph is HIGHER(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 completed solution might look like this:

The desired figure is limited by a parabola above and a straight line 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 special case formulas . Since the axis is specified by the equation, and the graph of the function is located not higher axes, then

And now a couple of examples for your own solution

Example 5

Example 6

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

When solving problems involving calculating area using a definite integral, a funny incident sometimes happens. The drawing was done correctly, the calculations were correct, but due to carelessness... the area of ​​the wrong figure was found, this is exactly how your humble servant screwed up several times. Here real case from life:

Example 7

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

Solution: First, let's make a drawing:

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

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

This example is also useful in that it calculates the area of ​​a figure using two definite integrals. Really:

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

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

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

Answer:

Let's move on to another meaningful task.

Example 8

Calculate the area of ​​a figure bounded by lines,
Let’s present the equations in “school” form and make a point-by-point drawing:

From the drawing it is clear that our upper limit is “good”: .
But what is the lower limit?! It is clear that this is not an integer, but what is it? May be ? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that... Or the root. What if we built the graph incorrectly?

In such cases, you have to spend additional time and clarify the limits of integration analytically.

Let's find the intersection points of a straight line and a 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 simplest.

On the segment , according to the corresponding formula:

Answer:

Well, to conclude the lesson, let’s look at two more difficult tasks.

Example 9

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

Solution: Let's depict this figure in the drawing.

Damn, I forgot to sign the schedule, and, sorry, I didn’t want to redo the picture. Not a drawing day, in short, today is the day =)

For point-by-point construction you need to know appearance sinusoids (and generally 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 possible to construct a schematic drawing, on which the graphs and limits of integration should be fundamentally correctly displayed.

There are no problems with the limits of integration here; they follow directly from the condition: “x” changes from zero to “pi”. Let's make a further decision:

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

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 ​​the plane figure (the region of integration). This is the simplest form of double integral, when the function of two variables is equal to one: .

Let's first consider the problem in general view. Now you will be quite surprised how simple everything 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 depict the area in the drawing:

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

Thus:

And immediately an important technical technique: iterated integrals can be calculated separately. First the inner integral, then the outer integral. This method I highly recommend it to beginners in the subject.

1) Let's calculate the internal integral, and the integration is carried out over the variable “y”:

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

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

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

The resulting formula is exactly the working formula for calculating the area of ​​a plane figure using the “ordinary” definite integral! Watch the lesson Calculating Area Using a Definite Integral, there she is at every step!

That is, problem of calculating area using double integral not much different from the problem of finding the area using a definite integral! In fact, it's the same thing!

Accordingly, no difficulties should arise! I won’t look at very many examples, since you, in fact, have repeatedly encountered this task.

Example 9

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

Let us choose the following order of traversal of the area:

Here and further I will not dwell on how to traverse the area, 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 stick to the same method:

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

2) The result obtained in the first step is substituted into the external integral:

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

Answer:

This is such a stupid and naive task.

An interesting example for an independent solution:

Example 10

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

An approximate example of a final solution at the end of the lesson.

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

But in some cases, the second method of traversing the area is more effective, and at the end of the young nerd’s course, let’s look at a couple more examples on this topic:

Example 11

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

Solution: We are looking forward to two parabolas with a quirk that lie on their sides. There is no need to smile; similar things occur quite often in multiple integrals.

What is the easiest way to make a drawing?

Let's imagine a parabola in the form of two functions:
– the upper branch and – the lower branch.

Similarly, imagine a parabola in the form of upper and lower branches.

Next, point-wise plotting of graphs rules, resulting in such a bizarre figure:

We calculate the area of ​​the figure using the double integral according to the formula:

What happens if we choose the first method of traversing the area? Firstly, this area will have to be divided into two parts. And secondly, we will observe this sad picture: . Integrals, of course, are not of a super-complicated level, but... there is an old mathematical saying: those who are close to their roots do not need a test.

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

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

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

Thus:

As they say, feel the difference.

1) We deal with the internal integral:

We substitute the result into the outer integral:

Integration over the variable “y” should not be confusing; if there were a letter “zy”, 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 integration according to the “Y” method.

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

What to add…. All!

Answer:

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

Example 12

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

This is an example for you to solve on your own. It is interesting to note that if you try to use the first method of traversing the area, the figure will no longer have to be divided into two, but into three parts! And, accordingly, we get three pairs of repeated integrals. Sometimes it happens.

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

I wish you success!

Solutions and answers:

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

Let us choose the following order of traversal of the area:

Thus:
Let's move on to inverse functions:


Thus:
Answer:

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


Let's make the drawing:

Let's change the order of traversing the area:

Answer:

Let's move on to consider applications of integral calculus. In this lesson we will analyze the typical and most common task calculating the area of ​​a plane figure using a definite integral. Finally, let all those who seek meaning in higher mathematics find it. You never know. In real life, you will have to approximate a dacha plot using elementary functions and find its area using a definite integral.

To successfully master the material, you must:

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

2) Be able to apply the Newton-Leibniz formula and calculate the definite integral. You can establish warm friendly relations with certain integrals on the page Definite integral. Examples of solutions. The task “calculate the area using a definite integral” always involves constructing a drawing, so your knowledge and drawing skills will also be a relevant issue. At a minimum, you need to be able to construct a straight line, parabola and hyperbola.

Let's start with a curved trapezoid. A curved trapezoid is a flat figure bounded by the graph of some function y = f(x), axis OX and lines x = a; x = b.

The area of ​​a curvilinear trapezoid is numerically equal to a definite integral

Any definite integral (that exists) has a very good geometric meaning. At the lesson Definite integral. Examples of solutions we 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 AREA. That is, the definite integral (if it exists) geometrically corresponds to the area of ​​a certain figure. Consider the definite integral

Integrand

defines a curve on the plane (it can be drawn if desired), and the definite integral itself is numerically equal to the area of ​​the corresponding curvilinear trapezoid.

Example 1

, , , .

This is a typical assignment statement. The most important point solutions - drawing. Moreover, the drawing must be constructed RIGHT.

When constructing a drawing, I recommend the following order: at first it is better to construct all straight lines (if they exist) and only Then– parabolas, hyperbolas, graphs of other functions. The point-by-point construction technique can be found in the reference material Graphs and properties of elementary functions. There you can also find very useful material for our lesson - how to quickly build a parabola.

In this problem, the solution might look like this.

Let's do the drawing (note that the equation y= 0 specifies the axis OX):

We will not shade the curved trapezoid; here it is obvious what area we are talking about. The solution continues like this:

On the segment [-2; 1] function graph y = x 2 + 2 located above the axisOX, That's why:

Answer: .

Who has difficulties with calculating the definite integral and applying the Newton-Leibniz formula

,

refer to lecture Definite integral. Examples of solutions. After the task is completed, it is always useful to look at the drawing and figure out whether the answer is real. In this case, we count the number of cells in the drawing “by eye” - well, there will be about 9, it seems to be true. It is absolutely clear that if we got, say, the answer: 20 square units, then it is obvious that a mistake was made somewhere - 20 cells obviously do not fit into the figure in question, at most a dozen. If the answer is negative, then the task was also solved incorrectly.

Example 2

Calculate the area of ​​a figure bounded by lines xy = 4, x = 2, x= 4 and axis OX.

This is an example for you to solve on your own. Full solution and answer at the end of the lesson.

What to do if the curved trapezoid is located under the axleOX?

Example 3

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 curved trapezoid completely located under the axis OX , then its area can be found using the formula:

In this case:

.

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

1) If you are asked to solve simply a definite integral without any geometric meaning, then it may 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 discussed.

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

Example 4

Find the area of ​​a plane figure bounded by lines y = 2x – x 2 , y = -x.

Solution: First you need to make a drawing. When constructing a drawing in area problems, we are most interested in the points of intersection of lines. Let's find the intersection points of the parabola y = 2x – x 2 and straight y = -x. This can be done in two ways. The first method is analytical. We solve the equation:

This means that the lower limit of integration a= 0, upper limit of integration b= 3. It is often more profitable and faster to construct lines point by point, and the limits of integration become clear “by themselves.” Nevertheless, the analytical method of finding limits still sometimes has to be used if, for example, the graph is large enough, or the detailed construction did not reveal the limits of integration (they can be fractional or irrational). Let's return to our task: it is more rational to first construct a straight line and only then a parabola. Let's make the drawing:

Let us repeat that when constructing pointwise, the limits of integration are most often determined “automatically”.

And now the working formula:

If on the segment [ 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 using the formula:

Here you no longer need to think about where the figure is located - above the axis or below the axis, but it matters which graph is HIGHER(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 from 2 x – x 2 must be subtracted – x.

The completed solution might look like this:

The desired figure is limited by a parabola y = 2x – x 2 on top and straight y = -x below.

On segment 2 x – x 2 ≥ -x. According to the corresponding formula:

Answer: .

In fact, the school formula for the area of ​​a curvilinear trapezoid in the lower half-plane (see example No. 3) is a special case of the formula

.

Because the axis OX given by the equation y= 0, and the graph of the function g(x) located below the axis OX, That

.

And now a couple of examples for your own solution

Example 5

Example 6

Find the area of ​​a figure bounded by lines

When solving problems involving calculating area using a definite integral, a funny incident sometimes happens. The drawing was done correctly, the calculations were correct, but due to carelessness... The area of ​​the wrong figure was found.

Example 7

First let's make a drawing:

The figure whose area we need to find is shaded blue(look carefully at the condition - how the figure is limited!). But in practice, due to inattention, people often decide that they need to find the area of ​​the figure that is shaded in green!

This example is also useful because it calculates the area of ​​a figure using two definite integrals. Really:

1) On the segment [-1; 1] above the axis OX the graph is located straight y = x+1;

2) On a segment above the axis OX the graph of a hyperbola is located y = (2/x).

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

Answer:

Example 8

Calculate the area of ​​a figure bounded by lines

Let’s present the equations in “school” form

and make a point-by-point drawing:

From the drawing it is clear that our upper limit is “good”: b = 1.

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

May be, a=(-1/3)? But where is the guarantee that the drawing is made with perfect accuracy, it may well turn out that a=(-1/4). What if we built the graph incorrectly?

In such cases, you have to spend additional time and clarify the limits of integration analytically.

Let's find the intersection points of the graphs

To do this, we solve the equation:

.

Hence, a=(-1/3).

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

, ,

according to the corresponding formula:

Answer:

To conclude the lesson, let's look at two more difficult tasks.

Example 9

Calculate the area of ​​a figure bounded by lines

Solution: Let's depict this figure in the drawing.

To construct a point-by-point drawing, you need to know the appearance of a sinusoid. In general, it is useful to know the graphs of all elementary functions, as well as some sine values. They can be found in the table of values trigonometric functions . In some cases (for example, in this case), it is possible to construct a schematic drawing, on which the graphs and limits of integration should be fundamentally correctly displayed.

There are no problems with the limits of integration here; they follow directly from the condition:

– “x” changes from zero to “pi”. Let's make a further decision:

On a segment, the graph of a function y= sin 3 x located above the axis OX, That's why:

(1) You can see how sines and cosines are integrated in odd powers in the lesson Integrals of trigonometric functions. We pinch off one sinus.

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

(3) Let's change the variable t=cos x, then: is located above the axis, therefore:

.

.

Note: note how the integral of the tangent cubed is taken; a corollary of the basic trigonometric identity is used here

.

In this article you will learn how to find the area of ​​a figure bounded by lines using integral calculations. For the first time we encounter the formulation of such a problem in high school, when we have just completed the study of definite integrals and it is time to begin the geometric interpretation of the acquired knowledge in practice.

So, what is required to successfully solve the problem of finding the area of ​​a figure using integrals:

• Ability to make competent drawings;
• Ability to solve a definite integral using the well-known Newton-Leibniz formula;
• The ability to “see” a more profitable solution option - i.e. understand how it will be more convenient to carry out integration in one case or another? Along the x-axis (OX) or the y-axis (OY)?
• Well, where would we be 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 are building a drawing. It is advisable to do this on a checkered piece of paper, on a large scale. We sign the name of this function with a pencil above each graph. Signing the graphs is done solely for the convenience of further calculations. Having received a graph of the desired figure, in most cases it will be immediately clear which limits of integration will be used. Thus, we solve the problem graphically. However, it 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 specified, then we find the points of intersection of the graphs with each other and see if our graphic solution with analytical.

3. Next, you need to analyze the drawing. Depending on how the function graphs are arranged, there are different approaches to finding the area of ​​a figure. Let's look at different examples of finding the area of ​​a figure using integrals.

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

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

What lines is the figure bounded by? 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 have positive values. Next, given straight lines x = 1 And x = 3, which run parallel to the axis OU, are the boundary lines of the figure on the left and right. Well y = 0, it is also the x-axis, which limits the figure from below. The resulting figure is shaded, as can be seen from the figure on the left. In this case, you can immediately begin solving the problem. Before us is a simple example of a curved trapezoid, which we then solve using the Newton-Leibniz formula.

3.2. IN previous paragraph 3.1 the case is analyzed when a curved 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 such a problem below.

Example 2 . Calculate the area of ​​a figure 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 the axis OH, straight x = -4, x = -1, y = 0. Here y = 0 limits the desired figure from above. Direct x = -4 And x = -1 these are the boundaries within which the 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 given function not positive, and still continuous on the interval [-4; -1] . What do you mean not positive? As can be seen from the figure, the figure that lies within the given x's 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.

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