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 points of intersection with the 0x axis, that is, the roots of the equation 8 + 2x - x 2 = 0 or x 2 - 2x - 8 = 0. By 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 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
The task is a school one, but, despite the fact, almost 100% will meet in your course of higher mathematics. That's why in all seriousness we will treat ALL examples, and the first thing to do is to familiarize yourself with application Function Graphs to brush up on the technique of constructing elementary graphs. …Eat? Great! A typical task statement is as follows:
Example 10
.
AND first major step solutions consists just in building a drawing. That being said, I recommend the following order: at first it's better to build everything straight(if any) and only Then – parabolas, hyperbole, graphs of other functions.
In our task: straight defines the axis straight parallel to the axis and parabola is symmetrical about the axis , for it we find several reference points: 
It is desirable to hatch the desired figure: 
Second phase is to compose correctly And calculate correctly definite integral. On the segment, the graph of the function is located over axis, so the required area is:
Answer:
After the task is completed, it is useful to look at the blueprint
and see if the answer is realistic.
And we "by eye" count the number of shaded cells - well, about 9 will be typed, it seems to be true. It is quite clear that if we had, say, 20 square units, then, obviously, a mistake was made somewhere - 20 cells obviously do not fit into the constructed figure, at most a dozen. If the answer turned out to be negative, then the task was also solved incorrectly.
Example 11
Calculate the area of a figure bounded by lines 
and axis
We quickly warm up (necessarily!) And consider the “mirror” situation - when the curvilinear trapezoid is located under axle:
Example 12
Calculate the area of the figure bounded by lines and coordinate axes.
Solution: find several reference points for constructing the exponent:
and execute the drawing, getting a figure with an area of \u200b\u200babout two cells: 
If the curvilinear trapezoid is located not higher axis , then its area can be found by the formula: .
In this case: 
Answer: - well, very, very similar to the truth.
In practice, most often the figure is located in both the upper and lower half-planes, and therefore we move from the simplest school problems to more meaningful examples:
Example 13
Find area flat figure, bounded by lines , .
Solution: first you need to complete the drawing, while we are particularly interested in the intersection points of the parabola and the line, since there will be integration limits. You can find them in two ways. The first way is analytical. Let's make and solve the equation:
Thus:
Dignity analytical method consists in its accuracy, A flaw- V duration(and in this example we are still lucky). Therefore, in many problems it is more profitable to construct lines point by point, while the limits of integration are found out as if “by themselves”.
With a straight line, everything is clear, but to build a parabola it is convenient to find its vertex, for this we take the derivative and equate it to zero:
- this is the point where the top will be located. And, due to the symmetry of the parabola, we will find the remaining reference points according to the “left-right” principle: 
Let's make a drawing: 
And now the working formula: if on the interval some continuous function greater than or equal continuous functions, then the area of \u200b\u200bthe figure bounded by the graphs of these functions and line segments can be found by the formula: 
Here it is no longer necessary to think where the figure is located - above the axis or below the axis, but, roughly speaking, it matters which of the two graphs is ABOVE.
In our example, 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:
On the segment: , according to the corresponding formula:
Answer:
It should be noted that the simple formulas considered at the beginning of the paragraph are special cases of the formula 
. Since the axis is given by the equation, then one of the functions will be zero, and depending on whether the curvilinear trapezoid lies above or below, we get the formula either
And now a couple of typical tasks for independent decision
Example 14
Find the area of figures bounded by lines:
Solution with drawings and brief comments at the end of the book
In the course of solving the problem under consideration, a funny incident sometimes happens. The drawing was made correctly, the integral was solved correctly, but due to inattention ... found the area of the wrong figure, this is how your obedient servant was mistaken several times. Here is a real life case:
Example 15
Calculate the area of a figure bounded by lines 
Solution: let's make a simple drawing, 
the trick of which is that the required area is shaded in green(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 gray! A special insidiousness is that the line can be underdrawn to the axis, and then we will not see the desired figure at all.
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 there is a graph of a hyperbola.
It is quite clear that the areas can (and should) be added: 
Answer:
AND educational example for independent solution:
Example 16
Calculate the area of the figure bounded by lines , , and coordinate axes.
So, we systematize the important points of this task:
On the first step CAREFULLY study the condition - WHAT functions are given to us? Mistakes happen even here, in particular, arc to The tangent is often mistaken for the arc tangent. By the way, this also applies to other tasks where the arc tangent occurs.
Further the drawing must be done CORRECTLY. Better to build first straight(if any), then graphs of other functions (if any J). The latter are in many cases more profitable to build point by point- find several anchor points and carefully connect them with a line.
But here the following difficulties may lie in wait. First, it is not always clear from the drawing integration limits- this happens when they are fractional. On mathprofi.ru at relevant article I considered an example with a parabola and a straight line, where one of their intersection points is not clear from the drawing. In such cases, you should use analytical method, we make the equation:
and find its roots:
– lower limit of integration, – upper limit.
After the drawing is built, analyze the resulting figure - once again take a look at the proposed functions and double-check whether THIS is a figure. Then we analyze its shape and location, it happens that the area is quite complicated and then it should be divided into two or even three parts.
We compose a definite integral or several integrals according to the formula 
, we have analyzed all the main variations above.
We solve a definite integral(s). At the same time, it can turn out to be quite complicated, and then we apply a phased algorithm: 1) find the antiderivative and check it by differentiation, 2) We use the Newton-Leibniz formula.
The result is useful to check by using software / online services or just “estimate” according to the drawing by cells. But both are not always feasible, so we are extremely attentive to each stage of the solution!
A complete and up-to-date version of this course in pdf format,
as well as courses on other topics can be found.
You can also - simple, affordable, fun and free!
With best wishes, Alexander Emelin
Calculating the area of a figure This is perhaps one of the most difficult problems in area theory. In school geometry, they are taught to find the areas of the main geometric shapes such as, for example, a triangle, a rhombus, a rectangle, a trapezoid, a circle, etc. However, one often has to deal with the calculation of the areas of more complex figures. It is in solving such problems that it is very convenient to use integral calculus.
Definition.
Curvilinear trapezoid some figure G is called, 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 curvilinear trapezoid can be denoted by S(G).
The definite integral ʃ a b f(x)dx for the function f(x), which is continuous and non-negative on the segment [a; b], and is the area of the corresponding curvilinear trapezoid.
That is, to find the area of \u200b\u200bthe figure G, bounded by the lines y \u003d f (x), y \u003d 0, x \u003d a and x \u003d 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 the curvilinear trapezoid can be found by the formula S(G) = -ʃ a b f(x)dx.
Example 1
Calculate the area of \u200b\u200bthe figure bounded by the lines y \u003d x 3; y = 1; x = 2.
Solution.
The given lines form the figure ABC, which is shown by hatching on rice. 2.
The desired 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 \u003d x 3,
(y = 1.
Thus, we have x 1 \u003d 1 - the lower limit and x \u003d 2 - the upper limit.
So, S = S DACE - S DABE = ʃ 1 2 x 3 dx - 1 = x 4 /4| 1 2 - 1 \u003d (16 - 1) / 4 - 1 \u003d 11/4 (square units).
Answer: 11/4 sq. units
Example 2
Calculate the area of \u200b\u200bthe figure bounded by lines y \u003d √x; y = 2; x = 9.
Solution.
The given lines form the figure ABC, which is bounded from above by the graph of the function
y \u003d √x, and from below the graph of the function y \u003d 2. The resulting figure is shown by hatching on rice. 3.
The desired area is equal to S = ʃ a b (√x - 2). Let's find the limits of integration: b = 9, to find a, we solve the system of two equations:
(y = √x,
(y = 2.
Thus, we have that x = 4 = a is the lower limit.
So, S = ∫ 4 9 (√x – 2)dx = ∫ 4 9 √x dx –∫ 4 9 2dx = 2/3 x√x| 4 9 - 2x| 4 9 \u003d (18 - 16/3) - (18 - 8) \u003d 2 2/3 (square units).
Answer: S = 2 2/3 sq. units
Example 3
Calculate the area of \u200b\u200bthe figure bounded by the lines y \u003d x 3 - 4x; y = 0; x ≥ 0.
Solution.
Let's plot the function y \u003d x 3 - 4x for x ≥ 0. To do this, we find the derivative y ':
y’ = 3x 2 – 4, y’ = 0 at х = ±2/√3 ≈ 1.1 are critical points.
If we represent the critical points on numerical axis and arrange the signs of the derivative, then we get 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 is min = -16/(3√3) ≈ -3.
Let's determine the intersection points of the graph with the coordinate axes:
if x \u003d 0, then y \u003d 0, which means that A (0; 0) is the point of intersection with the Oy axis;
if y \u003d 0, then x 3 - 4x \u003d 0 or x (x 2 - 4) \u003d 0, or x (x - 2) (x + 2) \u003d 0, from where x 1 \u003d 0, x 2 \u003d 2, x 3 \u003d -2 (not suitable, because x ≥ 0).
Points A(0; 0) and B(2; 0) are the intersection points of the graph with the Ox axis.
The given lines form the OAB figure, which is shown by hatching on rice. 4.
Since the function y \u003d x 3 - 4x takes on (0; 2) a negative value, then
S = |ʃ 0 2 (x 3 – 4x)dx|.
We have: ʃ 0 2 (x 3 - 4x)dx = (x 4 /4 - 4x 2 /2)| 0 2 \u003d -4, from where S \u003d 4 square meters. units
Answer: S = 4 sq. units
Example 4
Find the area of the figure bounded by the parabola y \u003d 2x 2 - 2x + 1, the straight lines x \u003d 0, y \u003d 0 and the tangent to this parabola at the point with the abscissa x 0 \u003d 2.
Solution.
First, we compose the equation of the tangent to the parabola y \u003d 2x 2 - 2x + 1 at the point with the abscissa x₀ \u003d 2.
Since the derivative y' = 4x - 2, then for x 0 = 2 we get k = y'(2) = 6.
Find the ordinate of the touch point: y 0 = 2 2 2 – 2 2 + 1 = 5.
Therefore, the tangent equation has the form: y - 5 \u003d 6 (x - 2) or y \u003d 6x - 7.
Let's build a figure bounded by lines:
y \u003d 2x 2 - 2x + 1, y \u003d 0, x \u003d 0, y \u003d 6x - 7.
Г y \u003d 2x 2 - 2x + 1 - parabola. Points of intersection with the coordinate axes: A(0; 1) - with the Oy axis; with the Ox axis - there are no intersection points, because the equation 2x 2 - 2x + 1 = 0 has no solutions (D< 0). Найдем вершину параболы:
x b \u003d 2/4 \u003d 1/2;
y b \u003d 1/2, that is, the vertex of the parabola point B has coordinates B (1/2; 1/2).
So, the figure whose area is to be determined is shown by hatching on rice. 5.
We have: S O A B D \u003d S OABC - S ADBC.
Find the coordinates of point D from the condition:
6x - 7 = 0, i.e. x \u003d 7/6, then DC \u003d 2 - 7/6 \u003d 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 \u003d 10/3 (square units).
Finally we get: S O A B D \u003d S OABC - S ADBC \u003d 10/3 - 25/12 \u003d 5/4 \u003d 1 1/4 (sq. units).
Answer: S = 1 1/4 sq. units
We have reviewed examples finding the areas of figures bounded by given lines. To successfully solve such problems, you need to be able to build 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 and skills to calculate certain integrals.
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We figured out how to find the area of a curvilinear trapezoid G. Here are the resulting formulas:
for a continuous and non-negative function y=f(x) on the segment , 
for a continuous and non-positive function y=f(x) on the segment .
However, when solving problems of finding the area, one often has to deal with more complex figures.
In this article, we will talk about calculating the area of figures whose boundaries are explicitly specified by functions, that is, as y=f(x) or x=g(y) , and analyze in detail the solution of typical examples.
Page navigation.
Formula for calculating the area of a figure bounded by lines y=f(x) or x=g(y) .
Theorem.
Let the functions and be defined and continuous on the segment , and for any value x from . Then area of figure G, bounded by lines x=a , x=b , and is calculated by the formula 
.
A similar formula is valid for the area of \u200b\u200bthe figure bounded by the lines y \u003d c, y \u003d d, and: 
.
Proof.
Let us show the validity of the formula for three cases: 
In the first case, when both functions are non-negative, due to the additivity property of the area, the sum of the area of the original figure G and the curvilinear trapezoid is equal to the area of the figure. Hence, 
That's why, . The last transition is possible due to the third property of the definite integral.
Similarly, in the second case, the equality is true. Here is a graphic illustration: 
In the third case, when both functions are nonpositive, we have . Let's illustrate this: 
Now we can move on to the general case when the functions and cross the Ox axis.
Let's denote the intersection points. These points divide the segment into n parts , where . The figure G can be represented by the union of the figures 
. It is obvious that on its interval falls under one of the three cases considered earlier, therefore their areas are found as 
Hence, 
The last transition is valid due to the fifth property of the definite integral.
Graphic illustration of the general case.
Thus the formula proven.
It's time to move on to solving examples for finding the area of figures bounded by the lines y=f(x) and x=g(y) .
Examples of calculating the area of a figure bounded by lines y=f(x) or x=g(y) .
We will begin the solution of each problem by constructing a figure on a plane. This will allow us to represent a complex figure as a union of simpler figures. In case of difficulties with the construction, refer to the articles:; And .
Example.
Calculate the area of a figure bounded by a parabola 
and straight lines , x=1 , x=4 .
Solution.
Let's build these lines on the plane. 
Everywhere on the segment, the graph of a parabola above straight. Therefore, we apply the previously obtained formula for the area and calculate the definite integral using the Newton-Leibniz formula: 
Let's complicate the example a bit.
Example.
Calculate the area of the figure bounded by lines.
Solution.
How is this different from previous examples? Previously, we always had two straight lines parallel to the x-axis, and now only one x=7 . The question immediately arises: where to take the second limit of integration? Let's take a look at the drawing for this. 
It became clear that the lower limit of integration when finding the area of \u200b\u200bthe figure is the abscissa of the point of intersection of the graph of the straight line y \u003d x and the semi-parabola. We find this abscissa from the equality: 
Therefore, the abscissa of the intersection point is x=2 .
Note.
In our example and in the drawing, it can be seen that the lines and y=x intersect at the point (2;2) and the previous calculations seem redundant. But in other cases, things may not be so obvious. Therefore, we recommend that you always analytically calculate the abscissas and ordinates of the points of intersection of lines.
Obviously, the graph of the function y=x is located above the graph of the function on the interval . We apply the formula to calculate the area: 
Let's complicate the task even more.
Example.
Calculate the area of the figure bounded by the graphs of functions and 
.
Solution.
Let's build a graph inverse proportionality and parabolas .
Before applying the formula for finding the area of a figure, we need to decide on the limits of integration. To do this, we find the abscissas of the intersection points of the lines by equating the expressions and .
For values of x other than zero, the equality 
equivalent to third degree equation 
with integer coefficients. You can refer to the section to recall the algorithm for solving it.
It is easy to check that x=1 is the root of this equation: .
Dividing the expression 
to the binomial x-1 , we have:
Thus, the remaining roots are found from the equation 
:
Now from the drawing it became clear that the figure G is enclosed above the blue and below the red line in the interval 
. Thus, the required area will be equal to 
Let's look at another typical example.
Example.
Calculate the area of a figure bounded by curves 
and the abscissa axis.
Solution.
Let's make a drawing.
This is an ordinary power function with an exponent of one third, the plot of the function 
can be obtained from the graph by displaying it symmetrically about the x-axis and lifting it up by one. 
Find the intersection points of all lines.
The x-axis has the equation y=0 .
The graphs of the functions and y=0 intersect at the point (0;0) since x=0 is the only real root of the equation.
Function Graphs and y=0 intersect at (2;0) , since x=2 is the only root of the equation 
.
Function graphs and intersect at the point (1;1) since x=1 is the only root of the equation 
. This statement is not entirely obvious, but is a strictly increasing function, and - strictly decreasing, therefore, the equation has at most one root.
The only remark: in this case, to find the area, you will have to use a formula of the form 
. That is, the bounding lines must be represented as functions of the argument y , but with a black line . 
Let's define the points of intersection of the lines.
Let's start with graphs of functions and : 
Let's find the point of intersection of graphs of functions and : 
It remains to find the point of intersection of the lines and : 
As you can see, the values match.
Summarize.
We have analyzed all the most common cases of finding the area of a figure bounded by explicitly given lines. To do this, you need to be able to build lines on a plane, find the points of intersection of lines and apply the formula to find the area, which implies the ability to calculate certain integrals.
Let the function be non-negative and continuous on the interval . Then, according to geometric sense of a definite integral, the area of a curvilinear trapezoid bounded from above by the graph of this function, from below by the axis , from the left and right by straight lines and (see Fig. 2) is calculated by the formula
Example 9 Find the area of a figure bounded by a line 
and axis.
Solution. Function Graph 
is a parabola whose branches point downwards. Let's build it (Fig. 3). To determine the limits of integration, we find the points of intersection of the line (parabola) with the axis (straight line). To do this, we solve the system of equations
We get: 
, where , ; hence, , .
Rice. 3
The area of the figure is found by the formula (5):
If the function is non-positive and continuous on the segment , then the area of the curvilinear trapezoid, bounded from below by the graph of this function, from above by the axis, from the left and right by straight lines and , is calculated by the formula
. (6)
If the function is continuous on a segment and changes sign at a finite number of points, then the area of the shaded figure (Fig. 4) is equal to the algebraic sum of the corresponding definite integrals:
Rice. 4
Example 10 Calculate the area of \u200b\u200bthe figure bounded by the axis and the graph of the function for .
Rice. 5
Solution. Let's make a drawing (Fig. 5). The desired area is the sum of the areas and . Let's find each of these areas. First, we determine the limits of integration by solving the system 
We get , . Hence:
;
.
Thus, the area of the shaded figure is
(sq. units).
Rice. 6
Let, finally, the curvilinear trapezoid is bounded from above and below by the graphs of functions continuous on the segment and ,
and on the left and right - straight and (Fig. 6). Then its area is calculated by the formula
. (8)
Example 11. Find the area of the figure enclosed by the lines and .
Solution. This figure is shown in Fig. 7. We calculate its area using formula (8). Solving the system of equations, we find , ; hence, , . On the segment we have: . Hence, in formula (8) we take as x, and as - . We get:
(sq. units).
More challenging tasks the calculation of areas is solved by dividing the figure into non-intersecting parts and calculating the area of the whole figure as the sum of the areas of these parts.
Rice. 7
Example 12. Find the area of the figure bounded by the lines , , .
Solution. Let's make a drawing (Fig. 8). This figure can be considered as a curvilinear trapezoid bounded from below by the axis , from the left and right - by straight lines and , from above - by graphs of functions and . Since the figure is bounded from above by the graphs of two functions, to calculate its area, we divide this figure a straight line into two parts (1 is the abscissa of the point of intersection of the lines and). The area of each of these parts is found by the formula (4):
(sq. units); 
(sq. units). Hence:
(sq. units).
Rice. 8
|
Rice. 9
In conclusion, we note that if a curvilinear trapezoid is bounded by straight lines and , the axis and continuous on the curve (Fig. 9), then its area is found by the formula
Volume of a body of revolution
Let a curvilinear trapezoid bounded by a graph of a function continuous on a segment, an axis, straight lines and rotate around the axis (Fig. 10). Then the volume of the resulting body of revolution is calculated by the formula
. (9)
Example 13 Calculate the volume of a body obtained by rotating around the axis of a curvilinear trapezoid bounded by a hyperbola , straight lines , and the axis .
Solution. Let's make a drawing (Fig. 11).
It follows from the conditions of the problem that , . By formula (9) we obtain
.
Rice. 10
Rice. eleven
The volume of a body obtained by rotation around an axis OU curvilinear trapezoid bounded by straight lines y = c And y = d, axis OU and a graph of a function continuous on a segment (Fig. 12), is determined by the formula
. (10)
|
Rice. 12
Example 14. Calculate the volume of a body obtained by rotation around an axis OU curvilinear trapezoid bounded by lines X 2 = 4at, y= 4, x = 0 (Fig. 13).
Solution. In accordance with the condition of the problem, we find the limits of integration: , . By formula (10) we obtain:
Rice. 13
Arc length of a flat curve
Let the curve given by the equation , where , lie in a plane (Fig. 14).
Rice. 14
Definition. The length of an arc is understood as the limit to which the length of a polyline inscribed in this arc tends when the number of links of the polyline tends to infinity, and the length of the largest link tends to zero.
If the function and its derivative are continuous on the segment , then the arc length of the curve is calculated by the formula
. (11)
Example 15. Calculate the length of the arc of the curve enclosed between the points for which 
.
Solution. From the condition of the problem we have 
. By formula (11) we obtain:
.
4. Improper integrals
with infinite limits of integration
When introducing the concept definite integral it was assumed that the following two conditions are met:
a) limits of integration A and are finite;
b) the integrand is bounded on the segment .
If at least one of these conditions is not met, then the integral is called improper.
Let us first consider improper integrals with infinite limits of integration.
Definition. Let the function be defined and continuous on the interval , then and unbounded on the right (Fig. 15).
If the improper integral converges, then this area is finite; if the improper integral diverges, then this area is infinite.
Rice. 15
An improper integral with an infinite lower limit of integration is defined similarly:
. (13)
This integral converges if the limit on the right side of equality (13) exists and is finite; otherwise the integral is said to be divergent.
An improper integral with two infinite limits of integration is defined as follows:
, (14)
where с is any point of the interval . The integral converges only if both integrals converge on the right side of equality (14).
;G) 
= [select the full square in the denominator: ] = 
[replacement:
] = 
Hence, the improper integral converges and its value is equal to .