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 in one case or another it will be more convenient to carry out integration? 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 whether our graphical solution coincides with the analytical one.

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 further solve using the Newton-Leibniz formula.

3.2. In the previous paragraph 3.1, we examined the case 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 the given function is not positive, and is also 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.

The article is not completed.

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

A)

Solution.

The first and most important point of the decision is the construction of the drawing.

Let's make the drawing:

The equation y=0 sets the “x” axis;

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

- y=x 2 +2 - a parabola, the branches of which are directed upward, with the 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 solving the corresponding quadratic equation, find the intersection with the axis Oh .

The vertex of a parabola can be found using the formulas:

You can also build lines point by point.

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

Answer: S =9 sq. units

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, “by eye” we count the number of cells in the drawing - 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.

What to do if the curved trapezoid is located under the 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 curved trapezoid completely located under the axis Oh , then its area can be found using 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 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.

With) Find the area of ​​a plane 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 area problems, we are most interested in the points of intersection of lines. Let's find the intersection points of the parabola and straight 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 .

You can 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).

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

On the segment , according to the corresponding formula:

Answer: S =4.5 sq. units

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 in the decision is the construction of the 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 from 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:

It is clear from the drawing 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 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, be able to construct a straight line and a hyperbola.

A curved trapezoid is a flat figure bounded by an 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 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.

From the point of view of geometry, the definite integral is AREA.

That is, a certain 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 the area of ​​the corresponding curvilinear trapezoid.

Example 1

This is a typical assignment statement. The first and most important point of the decision is the construction of the 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.

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

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

Answer:

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, “by eye” we count the number of cells in the drawing - 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 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 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 , .

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.” 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:

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 bounded 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:

Example 4

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

Solution: 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, a “glitch” often occurs that you need to find the area of ​​​​a figure that is shaded in 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: