Let the position of any point on the plane be uniquely determined by two numbers, where

.

Let
non-negative, continuous on the segment
function,
.

Consider a set of points

which can be interpreted as a curvilinear triangle

To calculate the area of ​​a curvilinear triangle, we divide this triangle into elementary curvilinear triangles.

Let us replace elementary curvilinear triangles with right triangles.

Let the heights of these triangles be equal,

and the bases, respectively, are .

Square the th elementary triangle will obviously be equal to

.

Area of ​​a curvilinear triangle will be approximately equal

. (1)

Expression (1) can be considered as an integral sum for the function
on the segment
.

Let us introduce the notation
.- this is petty

partitions
.

Then the area of ​​a curvilinear triangle

we obtain when passing in expression (1) to the limit at

=
. (2)

So, the area of ​​a plane figure in the polar coordinate system is equal to

.

EXAMPLE Calculate the area of ​​a figure enclosed by a curve (cardioid)

.

Solution. Let's draw a graph of a cardioid

As we can see, the cardioid is a line symmetrical about the axis
.

P 15. Calculation of curve length

Let the curve specified parametrically

,
.

Let's split the segment
on parts with dots.

Let us denote by
corresponding points on the curve . Let's connect these points with straight lines.

The resulting broken
called a broken line inscribed in a curve .

Elementary link length
equal to

Line length
in this case it will be equal

. (1)

Let us denote by
. Then the length of the curve we obtain by passing in expression (1) to the limit at

. (2)

So the length of the curve according to expression (2) is determined by the formula

. (3)

Spatial curve length , specified parametrically

,
,

will be equal

.

If the plane curve is given explicitly

,
,

then the parametric equations of the curve

can in this case be represented in the form

,
,
.

As a result, expression (3) is obtained in the form

.

EXAMPLE Find the length of a curve given parametrically.

Solution. Let's plot a given curve

Since the curve is symmetrical about the coordinate axes, it is enough to find .

Therefore, the length of the curve will be equal to

.

P 16. Improper integral of the first kind. Cauchy criterion. Signs of comparison.

The student was detained while trying to take

improper integral. The owner of the integral is being clarified.

The definition of the Riemann integral introduced earlier is not applicable if the function f(x) is unbounded on the interval or the integration interval is infinite. In these cases, the concept of a definite integral can be generalized and the concept of an improper integral can be introduced.

Let the function f(x) be defined on an infinite half-interval V x≥a. Then we have the function F(x) defined by the integral

(1)

with a variable upper limit.

Let us go to the limit in (1) as x→+∞ and formally introduce the following notation

F(x)=
(2)

Symbol
called an improper integral of the first kind. Moreover, if limit (2) exists, then the improper integral is called convergent. If the limit does not exist or is equal to ∞, then the improper integral is called divergent.

Improper integrals of the first kind on (-∞, b] and

,

(3)

Note that in (3) a and b tend to infinity independently of each other.

Note also that if the function f(x) is continuous on . That is, lines such as the cut of a mushroom are not taken into account, the stem of which fits well into this segment, and the cap is much wider.

Side segments can degenerate into points . If you see such a figure in the drawing, this should not confuse you, since this point always has its value on the “x” axis. This means that everything is in order with the limits of integration.

Now you can move on to formulas and calculations. So the area s curved trapezoid can be calculated using the formula

If f(x) ≤ 0 (the graph of the function is located below the axis Ox), That area of ​​a curved trapezoid can be calculated using the formula

There are also cases when both the upper and lower boundaries of the figure are functions, respectively y = f(x) And y = φ (x) , then the area of ​​such a figure is calculated by the formula

. (3)

Solving problems together

Let's start with cases where the area of ​​a figure can be calculated using formula (1).

Example 1.Ox) and straight x = 1 , x = 3 .

Solution. Because y = 1/x> 0 on the segment , then the area of ​​the curvilinear trapezoid is found using formula (1):

.

Example 2. Find the area of ​​the figure bounded by the graph of the function, line x= 1 and x-axis ( Ox ).

Solution. The result of applying formula (1):

If then s= 1/2 ; if then s= 1/3, etc.

Example 3. Find the area of ​​the figure bounded by the graph of the function, the abscissa axis ( Ox) and straight x = 4 .

Solution. The figure corresponding to the conditions of the problem is a curvilinear trapezoid in which the left segment has degenerated into a point. The limits of integration are 0 and 4. Since , using formula (1) we find the area of ​​the curvilinear trapezoid:

.

Example 4. Find the area of ​​the figure bounded by the lines , , and located in the 1st quarter.

Solution. To use formula (1), let’s imagine the area of ​​the figure given by the conditions of the example as the sum of the areas of the triangle OAB and curved trapezoid ABC. When calculating the area of ​​a triangle OAB the limits of integration are the abscissas of the points O And A, and for the figure ABC- abscissas of points A And C (A is the intersection point of the line O.A. and parabolas, and C- the point of intersection of the parabola with the axis Ox). Solving jointly (as a system) the equations of a straight line and a parabola, we obtain (the abscissa of the point A) and (the abscissa of another point of intersection of the line and the parabola, which is not needed for the solution). Similarly we obtain , (abscissas of points C And D). Now we have everything we need to find the area of ​​a figure. We find:

Example 5. Find the area of ​​a curved trapezoid ACDB, if the equation of the curve CD and abscissas A And B 1 and 2 respectively.

Solution. Let us express this equation of the curve through the game: The area of ​​the curvilinear trapezoid is found using formula (1):

.

Let's move on to cases where the area of ​​a figure can be calculated using formula (2).

Example 6. Find the area of ​​the figure bounded by the parabola and the x-axis ( Ox ).

Solution. This figure is located below the x-axis. Therefore, to calculate its area, we will use formula (2). The limits of integration are the abscissa and the points of intersection of the parabola with the axis Ox. Hence,

Example 7. Find the area enclosed between the abscissa axis ( Ox) and two adjacent sine waves.

Solution. The area of ​​this figure can be found using formula (2):

.

Let's find each term separately:

.

.

Finally we find the area:

.

Example 8. Find the area of ​​the figure enclosed between the parabola and the curve.

Solution. Let's express the equations of lines through the game:

The area according to formula (2) is obtained as

,

Where a And b- abscissas of points A And B. Let's find them by solving the equations together:

Finally we find the area:

And finally, cases when the area of ​​a figure can be calculated using formula (3).

Example 9. Find the area of ​​the figure enclosed between the parabolas And .

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, those who are looking for meaning in higher mathematics - may they 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.

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) with the help of methodological material and an article on geometric transformations of graphs.

Actually, everyone has been familiar with the task of finding the area using a definite integral since school, and we will not go much further than the 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 lower 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. In class 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 the area of ​​the corresponding curvilinear 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 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 complete the drawing (note that the equation defines the axis):

I will not shade the curved trapezoid; it is obvious here what area we are 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 no 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.” 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 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:

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 specified by the equation, and the graph of the function is located no 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 is a real life case:

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 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:

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, it is necessary to know the appearance of a sinusoid (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 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:

Calculating the area of ​​a figure- This is perhaps one of the most difficult problems in area theory. In school geometry, they teach you to find the areas of basic geometric shapes such as, for example, a triangle, rhombus, rectangle, trapezoid, circle, etc. However, you often have to deal with calculating the areas of more complex figures. It is when solving such problems that it is very convenient to use integral calculus.

Definition.

Curvilinear trapezoid call some figure G bounded by the lines y = f(x), y = 0, x = a and x = b, and the function f(x) is continuous on the segment [a; b] and does not change its sign on it (Fig. 1). The area of ​​a curved trapezoid can be denoted by S(G).

A definite integral ʃ a b f(x)dx for the function f(x), which is continuous and non-negative on the interval [a; b], and is the area of ​​the corresponding curved trapezoid.

That is, to find the area of ​​a figure G bounded by the lines y = f(x), y = 0, x = a and x = b, it is necessary to calculate the definite integral ʃ a b f(x)dx.

Thus, S(G) = ʃ a b f(x)dx.

If the function y = f(x) is not positive on [a; b], then the area of ​​a curvilinear trapezoid can be found using the formula S(G) = -ʃ a b f(x)dx.

Example 1.

Calculate the area of ​​the figure bounded by the lines y = x 3; y = 1; x = 2.

Solution.

The given lines form the figure ABC, which is shown by hatching in rice. 2.

The required area is equal to the difference between the areas of the curvilinear trapezoid DACE and the square DABE.

Using the formula S = ʃ a b f(x)dx = S(b) – S(a), we find the limits of integration. To do this, we solve a system of two equations:

(y = x 3,
(y = 1.

Thus, we have x 1 = 1 – the lower limit and x = 2 – the upper limit.

So, S = S DACE – S DABE = ʃ 1 2 x 3 dx – 1 = x 4 /4| 1 2 – 1 = (16 – 1)/4 – 1 = 11/4 (sq. units).

Answer: 11/4 sq. units

Example 2.

Calculate the area of ​​the figure bounded by the lines y = √x; y = 2; x = 9.

Solution.

The given lines form the ABC figure, which is limited above by the graph of the function

y = √x, and below is a graph of the function y = 2. The resulting figure is shown by hatching in rice. 3.

The required area is S = ʃ a b (√x – 2). Let's find the limits of integration: b = 9, to find a, we solve a system of two equations:

(y = √x,
(y = 2.

Thus, we have that x = 4 = a - this is the lower limit.

So, S = ∫ 4 9 (√x – 2)dx = ∫ 4 9 √x dx –∫ 4 9 2dx = 2/3 x√x| 4 9 – 2х| 4 9 = (18 – 16/3) – (18 – 8) = 2 2/3 (sq. units).

Answer: S = 2 2/3 sq. units

Example 3.

Calculate the area of ​​the figure bounded by the lines y = x 3 – 4x; y = 0; x ≥ 0.

Solution.

Let’s plot the function y = x 3 – 4x for x ≥ 0. To do this, find the derivative y’:

y’ = 3x 2 – 4, y’ = 0 at x = ±2/√3 ≈ 1.1 – critical points.

If we plot the critical points on the number line and arrange the signs of the derivative, we find that the function decreases from zero to 2/√3 and increases from 2/√3 to plus infinity. Then x = 2/√3 is the minimum point, the minimum value of the function y min = -16/(3√3) ≈ -3.

Let's determine the intersection points of the graph with the coordinate axes:

if x = 0, then y = 0, which means A(0; 0) is the point of intersection with the Oy axis;

if y = 0, then x 3 – 4x = 0 or x(x 2 – 4) = 0, or x(x – 2)(x + 2) = 0, whence x 1 = 0, x 2 = 2, x 3 = -2 (not suitable, because x ≥ 0).

Points A(0; 0) and B(2; 0) are the points of intersection of the graph with the Ox axis.

The given lines form the OAB figure, which is shown by hatching in rice. 4.

Since the function y = x 3 – 4x takes a negative value on (0; 2), then

S = |ʃ 0 2 (x 3 – 4x)dx|.

We have: ʃ 0 2 (x 3 – 4х)dx =(x 4 /4 – 4х 2 /2)| 0 2 = -4, whence S = 4 sq. units

Answer: S = 4 sq. units

Example 4.

Find the area of ​​the figure bounded by the parabola y = 2x 2 – 2x + 1, the lines x = 0, y = 0 and the tangent to this parabola at the point with the abscissa x 0 = 2.

Solution.

First, let’s create an equation for the tangent to the parabola y = 2x 2 – 2x + 1 at the point with the abscissa x₀ = 2.

Since the derivative y’ = 4x – 2, then for x 0 = 2 we get k = y’(2) = 6.

Let's find the ordinate of the tangent point: y 0 = 2 2 2 – 2 2 + 1 = 5.

Therefore, the tangent equation has the form: y – 5 = 6(x ​​– 2) or y = 6x – 7.

Let's build a figure bounded by lines:

y = 2x 2 – 2x + 1, y = 0, x = 0, y = 6x – 7.

Г у = 2х 2 – 2х + 1 – parabola. Points of intersection with the coordinate axes: A(0; 1) – with the Oy axis; with the Ox axis - there are no points of intersection, because the equation 2x 2 – 2x + 1 = 0 has no solutions (D< 0). Найдем вершину параболы:

x b = 2/4 = 1/2;

y b = 1/2, that is, the vertex of the parabola point B has coordinates B(1/2; 1/2).

So, the figure whose area needs to be determined is shown by hatching on rice. 5.

We have: S O A B D = S OABC – S ADBC.

Let's find the coordinates of point D from the condition:

6x – 7 = 0, i.e. x = 7/6, which means DC = 2 – 7/6 = 5/6.

We find the area of ​​triangle DBC using the formula S ADBC ​​= 1/2 · DC · BC. Thus,

S ADBC ​​= 1/2 · 5/6 · 5 = 25/12 sq. units

S OABC = ʃ 0 2 (2x 2 – 2x + 1)dx = (2x 3 /3 – 2x 2 /2 + x)| 0 2 = 10/3 (sq. units).

We finally get: S O A B D = S OABC – S ADBC ​​= 10/3 – 25/12 = 5/4 = 1 1/4 (sq. units).

Answer: S = 1 1/4 sq. units

We've looked at examples finding the areas of figures bounded by given lines. To successfully solve such problems, you need to be able to draw lines and graphs of functions on a plane, find the points of intersection of lines, apply a formula to find the area, which implies the ability to calculate certain integrals.

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