home Social science An improper integral with an infinite integration limit is defined. A method for solving an improper integral with an infinite lower limit. Improper integrals with an infinite upper limit

An improper integral with an infinite integration limit is defined. A method for solving an improper integral with an infinite lower limit. Improper integrals with an infinite upper limit

Are you here now? =) No, I wasn't trying to scare anyone, it's just that the topic of improper integrals is a very good illustration of how important it is not to run higher mathematics and others exact sciences. To master the lesson on the site, everything is there - in a detailed and accessible form, there would be a desire ....

So, let's start. Figuratively speaking, the improper integral is an “advanced” definite integral, and in fact there are not so many difficulties with them, moreover, the improper integral has a very good geometric sense.

What does it mean to calculate an improper integral?

Calculate Improper Integral - it means to find a NUMBER(exactly the same as in the definite integral), or prove that it diverges(that is, end up with infinity instead of a number).

Improper integrals are of two types.

Improper integral with infinite limit(s) of integration

Sometimes such an improper integral is called improper integral of the first kind. In general, an improper integral with an infinite limit most often looks like this: . How is it different from a definite integral? In the upper limit. It is endless:

Less common are integrals with an infinite lower limit or with two infinite limits: , and we will consider them later - when you get a taste :)

Well, now let's analyze the most popular case. In the vast majority of examples, the integrand function continuous in between and this one important fact should be checked first! For if there are gaps, then there are additional nuances. For definiteness, we assume that even then the typical curvilinear trapezoid will look like this:

Note that it is infinite (not bounded on the right), and improper integral numerically equal to its area. In this case, the following options are possible:

1) The first thought that comes to mind is: “since the figure is infinite, then ”, in other words, the area is also infinite. So it may be. In this case, we say that the improper integral diverges.

2) But. As paradoxical as it may sound, the area of an infinite figure can be equal to ... a finite number! For example: . Could it be? Easy. In the second case, the improper integral converges.

3) About the third option a little later.

When does an improper integral diverge and when does it converge? It depends on the integrand, and concrete examples we will review very soon.

But what happens if an infinite curvilinear trapezoid is located below the axis? In this case, the improper integral (diverges) or is equal to a finite negative number.

In this way, improper integral can be negative.

Important! When ANY improper integral is offered to you to solve, then, generally speaking, there is no talk of any area and there is no need to build a drawing. I told the geometric meaning of the improper integral only to make it easier to understand the material.

Since the improper integral is very similar to the definite integral, then we recall the Newton-Leibniz formula: . In fact, the formula is also applicable to improper integrals, only it needs to be slightly modified. What's the Difference? In the infinite upper limit of integration: . Probably, many have guessed that this already smacks of applying the theory of limits, and the formula will be written as follows: .

How is it different from a definite integral? Yes, nothing special! As in the definite integral, one must be able to find the antiderivative function ( indefinite integral), be able to apply the Newton-Leibniz formula. The only thing that has been added is the calculation of the limit. Who's bad with them, learn a lesson Limits of functions. Solution examples because better late than in the army.

Consider two classic examples:

Example 1

For clarity, I will build a drawing, although, I emphasize once again, on practice it is not necessary to build drawings in this task.

The integrand is continuous on the half-interval , which means that everything is fine and the improper integral can be calculated using the “regular” method.

Application of our formula and the solution looks like this:

That is, the improper integral diverges, and the area of the shaded curvilinear trapezoid is equal to infinity.

In the considered example, we have the simplest tabular integral and the same technique for applying the Newton-Leibniz formula as in the definite integral. But this formula is applied under the sign of the limit. Instead of the usual letter of the "dynamic" variable, the letter "be" appears. This should not confuse or confuse, because any letter is no worse than the standard "X".

If you don’t understand why when , then this is very bad, either you don’t understand the simplest limits (and don’t understand what a limit is at all), or you don’t know what the graph of a logarithmic function looks like. In the second case, visit the lesson Graphs and properties of elementary functions.

When solving improper integrals, it is very important to know how the graphs of the main elementary functions!

A clean job design should look something like this:

“

! When designing an example, we always interrupt the solution and indicate what happens to the integrand – is it continuous on the interval of integration or not. By this we identify the type of improper integral and substantiate further actions.

Example 2

Calculate the improper integral or establish its divergence.

Let's make a drawing:

First, we notice the following: the integrand is continuous on the half-interval . Good. Solving with formula :

(1) We take the simplest integral of power function(this particular case is found in many tables). It is better to immediately move the minus beyond the limit sign so that it does not get underfoot in further calculations.

(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.

(3) We indicate that when (Gentlemen, this has long been understood) and simplify the answer.

Here, the area of an infinite curvilinear trapezoid is equal to a finite number! Unbelievable, but it is a fact.

The clean design of the example should look something like this:

“

The integrand is continuous on

“

What to do if you come across an integral like - with breaking point on the interval of integration? This means that there is a typo in the example (Most likely) or an advanced level of education. In the latter case, due to additivity properties, one should consider two improper integrals on the intervals and and then deal with the sum.

Sometimes, due to a typo or the intention of an improper integral, it can not exist at all, so, for example, if we put in the denominator of the above integral Square root from "x", then part of the integration interval will not enter the domain of definition of the integrand at all.

Moreover, an improper integral may not exist even with all the "apparent well-being". Classic example: . Despite the definiteness and continuity of the cosine, such an improper integral does not exist! Why? It's very simple because:
- does not exist corresponding limit.

And such examples, although rare, are found in practice! Thus, in addition to convergence and divergence, there is also a third outcome of the solution with a full answer: "there is no improper integral."

It should also be noted that the strict definition of the improper integral is given precisely through the limit, and those who wish can familiarize themselves with it in the educational literature. Well, we continue the practical lesson and move on to more meaningful tasks:

Example 3

Calculate the improper integral or establish its divergence.

First, let's try to find the antiderivative function (indefinite integral). If we fail to do this, then naturally we will not solve the improper integral either.

Which of the table integrals does the integrand look like? It reminds me of the arc tangent: . From these considerations, the thought suggests itself that it would be nice to get a square in the denominator. This is done by substitution.

Let's replace:

The indefinite integral has been found, it makes no sense to add a constant in this case.

On a draft, it is always useful to perform a check, that is, to differentiate the result:

The original integrand was obtained, which means that the indefinite integral was found correctly.

Now we find the improper integral:

(1) We write the solution in accordance with the formula . It is better to immediately move the constant beyond the limit sign so that it does not interfere in further calculations.

(2) We substitute the upper and lower limits in accordance with the Newton-Leibniz formula. Why at ? See the arc tangent graph in the already repeatedly recommended article.

(3) We get the final answer. The fact that it is useful to know by heart.

Advanced students may not find the indefinite integral separately, and not use the replacement method, but use the method of summing the function under the differential sign and solve the improper integral "immediately". In this case, the solution should look something like this:

“

The integrand is continuous on .

“

Example 4

Calculate the improper integral or establish its divergence.

! This is a typical example, and similar integrals are very common. Work it out well! The antiderivative function is found here by the method of selecting a full square, more details about the method can be found in the lesson Integration of some fractions.

Example 5

Calculate the improper integral or establish its divergence.

This integral can be solved in detail, that is, first find the indefinite integral by changing the variable. And you can solve it "immediately" - by summing up the function under the sign of the differential. Who has some mathematical background.

Complete Solutions and answers at the end of the lesson.

Examples of solutions of improper integrals with an infinite lower limit of integration can be found on the page Efficient Methods for Solving Improper Integrals. The case where both integration limits are infinite is also considered there.

Improper integrals of unbounded functions

Or improper integrals of the second kind. Improper integrals of the second kind are cunningly "ciphered" under the usual definite integral and look exactly the same: But, unlike the definite integral, the integrand suffers an infinite discontinuity (does not exist): 1) at the point, 2) or at the point, 3) or at both points at once, 4) or even on the interval of integration. We will consider the first two cases, for cases 3-4 at the end of the article there is a link to an additional lesson.

Just an example to make it clear:. It seems to be a definite integral. But in fact, this is an improper integral of the second kind, if we substitute the value of the lower limit into the integrand, then the denominator vanishes, that is, the integrand simply does not exist at this point!

In general, when analyzing the improper integral it is always necessary to substitute both integration limits into the integrand. In this regard, we also check the upper limit: . Everything is good here.

The curvilinear trapezoid for the considered variety of the improper integral fundamentally looks like this:

Here, almost everything is the same as in the integral of the first kind.

Our integral is numerically equal to area a hatched curvilinear trapezoid that is not bounded from above. In this case, there can be two options *: the improper integral diverges (the area is infinite) or the improper integral is equal to a finite number (that is, the area of an infinite figure is finite!).

* by default, we habitually assume that the improper integral exists

It remains only to modify the Newton-Leibniz formula. It is also modified with the help of the limit, but the limit no longer tends to infinity, but to the value on the right. It is easy to follow along the drawing: along the axis, we must approach the breaking point infinitely close on right.

Let's see how this is implemented in practice.

Example 6

Calculate the improper integral or establish its divergence.

The integrand suffers an infinite break at a point (do not forget to check verbally or on a draft if everything is fine with the upper limit!)

First, we calculate the indefinite integral:

Replacement:

For those who have difficulty with the replacement, refer to the lesson Replacement method in indefinite integral.

We calculate the improper integral:

(1) What's new here? Practically nothing in terms of technique. The only thing that has changed is the entry under the limit icon: . The addition means that we are aiming for the value on the right (which is logical - see graph). Such a limit in the theory of limits is called unilateral limit. In this case we have right-hand limit.

(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.

(3) Dealing with at . How do you determine where an expression is headed? Roughly speaking, you just need to substitute the value into it, substitute three quarters and indicate that . Combing the answer.

In this case, the improper integral is equal to a negative number. There is no crime in this, just the corresponding curvilinear trapezoid is located under the axis.

And now two examples for an independent solution.

Example 7

Calculate the improper integral or establish its divergence.

Example 8

Calculate the improper integral or establish its divergence.

If the integrand does not exist at the point

An infinite curvilinear trapezoid for such an improper integral basically looks like this.

Consider two types of improper intervals:

1. Improper integrals of the first kind with infinite limits of integration;
2. Improper integrals of the second kind of functions with infinite discontinuities.

Improper integrals of the first kind with infinite limits of integration

Definition: Integrals of the form: are called improper integrals of the first kind with infinite limits, which are defined using the limits:

Definition Improper integrals are called convergent if there are finite limits by which these integrals are determined.

Improper integrals are said to be divergent if these limits do not exist or are infinite.

Indeed, let the function f(x) be defined and continuous for any value x=v from a semi-infinite segment of functions we have:

It converges to 1. Then, according to Theorem 1, the improper integral of the smaller function: also converges and its value is less than 1.

Theorem 2. If for positive-sign functions for which the inequality 0?g(x)?f(x) holds, for any x? a, the improper integral of the smaller function diverges, then the improper integral of the larger function also diverges.

Example. Investigate the convergence of the integral:

Solution. Let's compare the integrand with the function. For sign-positive on the interval . Improper integrals of the second kind are defined differently, depending on the location of the discontinuity points on the interval [ a; b].

1) Suppose the function f(x) has an infinite discontinuity at some interior point of the integration domain ( cÎ( a; b)) At other points of the segment [ a; b] the function is assumed to be continuous.

Then, if the limits and exist and are finite, then we say that the integral converges and is equal to

. (8.22)
2) Let the only discontinuity point of the function f(x) coincides with the point a

. (8.23)
3) Let the only discontinuity point of the function f(x) coincides with the point b. Then, if the limit exists and is finite, then we say that the integral converges and is equal to

. (8.24)
It is assumed everywhere that e > 0 and d > 0.

Problem 8.12. Calculate the improper integral.

Solution. x= 2. Therefore,

Problem 8.13. Calculate the improper integral.

Solution. The integrand has a discontinuity of the second kind at the point x= 0 (inside the region of integration). Consequently,

The first limit exists and is finite, but the second limit is equal to infinity ( at ). Therefore, this integral diverges.

Chapter 9

§9.1. Definition n-dimensional Euclidean space R n .

Before proceeding to the study of functions of several variables, it is useful to introduce the concept n-dimensional space for any n = 1, 2, 3,… .

2 point x n-dimensional space (vector) is an ordered collection n real numbers.

The number is called i-th coordinate of the vector .

2 Distance between two points n-dimensional space and is determined by the formula:

Distance from point to point x is called the modulus of the vector x and is denoted. From formula (9.1) it follows that .

AT n-dimensional space, the notion of scalar product is naturally introduced:

Angle between vectors x and y can be determined by the formula:

As before, vectors x and y are perpendicular if and only if they scalar product equals zero.

2Collection of all points n-dimensional space in which the distance is defined according to the formula (9.1) and the scalar product is called n-dimensional Euclidean vector space and is denoted by .

When n= 1 the space coincides with the line, in the case n= 2 - with a plane, and in the case n= 3 - with space.

2 Let and . The set of all points such that , is called n-dimensional ball centered at a point x or e-neighborhood of a point x in space and is denoted by .

In coordinate form, this definition looks like this:

In the case of a straight line, i.e. at n= 1, the neighborhood of the point is an interval centered at the point of radius e. In the case of a plane, i.e. at n= 2, the neighborhood of the point is an open circle centered at the point of radius e. In the case of space, i.e. at n= 3 the neighborhood of the point is an open ball centered at the point of radius e.

§9.2. The scope of a function of several variables. Continuity

2 function n variables is called such a rule (law), according to which each set consisting of n variables taken from some area D n-dimensional space , is put in correspondence singular z. In the simplest case .

2 A function of 2 variables is a rule (law) according to which each point M(x; y) belonging to some area D plane xOy, is assigned a single number z.

A set of points in space with coordinates form a certain surface (Fig. 9.1), rising above the area D(geometric meaning of a function of two variables).

2 Region D, for which the above correspondence is constructed, is called the domain of the function .

Problem 9.1. Find the scope of a function

Solution. The desired domain of definition is a set of points on the plane xOy, satisfying the system of inequalities . The inequalities and change their sign to the opposite (respectively) at the intersection of the following lines: x = y and x = 0, y= 0. These lines break the plane xOy for 6 regions. Sequentially, substituting arbitrary points from each region into the system , we make sure that the union of regions (1) and (3) is the domain of definition of the original function. Moreover, the direct x = y, except for the point (0; 0), is included in the domain of definition, and the lines x= 0, and y= 0 - not included (Fig. 9.2).

2 The closure of an area is a set of points in space , in any neighborhood of each of which there are points of the area D.

Let, for example, D– some open (the boundary is not included) area on the plane xOy. Then the closure of the region will be obtained if to the region D attach her border G .

2 Let in some area D plane xOy given a function , and let be some closure point of the region D(). Number BUT is called the limit of the function at the point M 0 if for any number e> 0 there is such a number δ > 0, which for all points other than the point M 0 and remote from it less than δ , the inequality is satisfied.

2 A function is called continuous at a point if it is defined at this point () and the equality takes place.

§9.3. Level lines of a function of two variables

2 Lines on the plane xOy, given by the equations , where FROM is an arbitrary constant, are called the level lines of the function .

Level lines are lines of intersection of a surface, a given function, and a plane z = C, parallel to the plane xOy. With the help of level lines, you can study the shape of the surface given by the function.

Example 9.2. Find level lines and determine the shape of the surface given by the equation.

The equations of the level lines in this case have the form . At C< 0 уравнение дает пустое множество решений (следовательно, вся поверхность расположена выше плоскости xOy). At C= 0 only one point satisfies the equation of the level line x = 0, y= 0 (with plane xOy surface intersects only at the origin). At C> 0 level lines are ellipses with semi-axes and . Level lines corresponding to different values FROM, shown in Fig. 9.3. Surface, given by the equation, is called an elliptical paraboloid (Fig. 9.4).

§9.4. Partial derivatives of the first order

Let in some area D plane xOy the function is given, and is some point of the area D.

, (9.2)

2 Partial derivative of a function at a point with respect to a variable y(denoted or ) is called

, (9.3)
if the given limit exists and is finite.

2 Partial derivative of a function n variables at point by variable x i called

, (9.4)
if the given limit exists and is finite.

As can be seen from formulas (9.2) - (9.4), partial derivatives are determined in the same way as the derivative of a function of one variable was determined. When calculating the limit, only one of the variables is incremented, the other variables are not incremented and remain constant. Therefore, partial derivatives can be calculated according to the same rules as ordinary derivatives, treating all free variables (except the one with respect to which differentiation is performed) as constants.

Problem 9.3. Find Partial Derivatives of Functions

Solution. .

Problem 9.4. Find partial derivatives of a function.

Solution. When differentiating this function with respect to the variable x we use the rule of differentiation of a power function, and when finding a partial derivative with respect to a variable y- the rule of differentiation exponential function:

Problem 9.5. Calculate the partial derivatives of the function at the point .

Solution. Applying the rule of differentiation complex function, find partial derivatives

Substituting into partial derivatives the coordinates of the point M, we get

§9.5. Gradient function of several variables.
Directional derivative

2 The gradient of a function at a point is a vector composed of partial derivatives of a given function calculated at a given point:

2 The derivative of a function at a point in the direction of the vector is the projection of the gradient vector of this function, calculated at the point M 0 , to this direction

Calculating the projection of a vector onto a vector in accordance with formula (2.6), we obtain

. (9.7)
Noticing that where a is the angle that the vector makes with the axis OX, we obtain another formula for calculating the derivative with respect to the direction of the vector

Problem 9.6. Find the gradient of a function at a point M 0 (4; 2) and the derivative in the direction of the vector

Solution. Let's find partial derivatives

Calculate the values of partial derivatives at the point M 0:

Function Gradient at a Point M 0 we find by the formula (9.5):

Problem 9.7. At the point M 0 (0; 1) calculate the derivative of the function in the direction of the bisector of the second coordinate angle.

Solution. Let's find the partial derivatives of the function:

Calculate the values of partial derivatives and the gradient of the function at a point M 0:

Derivative of a function at a point M 0 in the direction of the bisector of the second coordinate angle (this direction is with the axis OX corner a= 135°) we find by formula (9.8):

§9.6. Differential of a function of several variables
and its application to approximate calculations

1 If at the point the function has continuous partial derivatives and , then its total increment when passing from the point M 0 to a point can be represented as:

, (9.9)
where at , .

2 The expression is called the total differential of the function at the point .

From formula (9.9) it follows that the differential of the function is the main linear part of the total increment of the function . For sufficiently small D x and D y the expression is much smaller than the differential and can be neglected. Thus, we arrive at the following approximate formula:

. (9.10)
Comment. Formula (9.10) can be used for approximate calculation of the values of functions only at points sufficiently close to the point . How less value, the more accurate is the value found by formula (9.9).

Example 9.8. Calculate approximately using the differential.

Let's consider a function. It is required to calculate the value z 1 of this function at the point ( x 1 ; y 1) = (0.09; 6.95). Let us use the approximate formula (9.9), choosing the point (0; 7) as a point. Then D x = x 1 – x 0 = 0.09 - 0 = 0.09, D y =y 1 – y 0 = 6,95 – 7 = – 0,05.

Consequently,

§9.7. Partial derivatives of higher orders

Let in the area D a function is given that has continuous partial derivatives and in this region. Thus, in the area D we got two new continuous functions two variables and . If at some point in the region D functions and have partial derivatives both with respect to the variable x, and by change y, then these derivatives are called second-order derivatives of the function . They are designated as follows:

1 If at some point in the region D function has continuous mixed derivatives and , then at the point these derivatives are equal: . D , the following conditions must be met: D = 32 – 9 = 23.

Since the discriminant is greater than zero, then at the point M the function has an extremum. Namely, a local minimum, since BUT and FROM Above zero. Wherein

Sometimes such improper integrals are called improper integrals of the second kind. Improper integrals of the second kind are cunningly "encrypted" under the usual definite integral and look exactly the same: .

But, unlike the definite integral, the integrand suffers an infinite discontinuity (does not exist):

1) at the point ,

2) point,

3) at both points at once,

4) or even on the interval of integration.

We will consider the first two cases, for cases 3-4 at the end of the article there is a link to an additional lesson.

Let's look at an example to make it clear:

It seems to be a definite integral. But in fact, this is an improper integral of the second kind, since if we substitute into the integrand, then the value of the lower limit

then the denominator vanishes, that is, the integrand simply does not exist at this point!

When analyzing the improper integral it is always necessary to substitute both integration limits into the integrand. In this regard, we also check the upper limit:

Everything is good here. The curvilinear trapezoid for the considered variety of the improper integral fundamentally looks like this:

Here, almost everything is the same as in the integral of the first kind. Our integral is numerically equal to the area of the shaded curvilinear trapezoid, which is not bounded from above. In this case, there can be two options: the improper integral diverges (the area is infinite), or the improper integral is equal to a finite number (when the area of an infinite figure is finite!).

It remains only to modify the Newton-Leibniz formula. It is also modified with the help of the limit, but the limit no longer tends to infinity, but to value on right. It is easy to follow the drawing, which is along the axis OX on right.

Let's see how this is implemented in practice.

Example 6

(do not forget to check verbally or on a draft if everything is fine with the upper limit!). First, we calculate the indefinite integral:

For those who have difficulty with the replacement, refer to the lesson Replacement method in indefinite integral.

We calculate the improper integral:

(1) What's new here? Practically nothing in terms of technique. The only thing that has changed is the entry under the limit icon:

Adding +0 means that we are aiming for the value ¾ on the right, which is logical (see graph). Such a limit in the theory of limits is called unilateral limit. In this case we have right-hand limit.

(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.

In this case, the improper integral is equal to a negative number. There is no crime in this, just the corresponding curvilinear trapezoid is located under the axis OX. And now examples for independent decision.

Example 7

Calculate the improper integral or establish its divergence.

Example 8

Calculate the improper integral or establish its divergence.

If the integrand does not exist at the point

An infinite curvilinear trapezoid for such an improper integral fundamentally looks like this:

Here we absolutely do everything the same, except that the limit tends to to value bleft. Axis OX we must get infinitely close to the breaking point left.

Example 9

Calculate the improper integral or establish its divergence.

The integrand suffers an infinite discontinuity at a point b = 3 (we verbally check that everything is fine with a different integration limit!).

For a change, we will solve this limit immediately - by bringing the function under the sign of the differential. Those who find it difficult can first find the indefinite integral according to the scheme already considered.

The addition (-0) means that we have a limit left-sided, and to the point b = 3 we are approaching the axis OX left.

We understand why the fraction

(this is best done orally or in draft form).

We substitute the limit value under the root b = 3 - 0.

Finally:

The improper integral diverges.

The minus sign means that the corresponding curvilinear trapezoid is located under the axis OX. Be very careful with the signs.

Yes, of course, the improper integral diverges, but they are different things, different genres, and if you overlook the signs, then, strictly speaking, make a serious mistake.

And the final two examples for self-consideration:

Example 10

Calculate the improper integral or establish its divergence.

Example 11

Calculate the improper integral or establish its divergence.

An analysis of the situation when both integration limits are “bad”, or the discontinuity point is contained directly on the integration segment, can be found in the article Effective Methods solutions of definite and improper integrals.

Solutions and answers:

Example 4: Solution:

Example 5: Solution:

The integrand is continuous on .

Example 7: Solution:

The integrand suffers an infinite discontinuity at a point

The improper integral diverges.

Note: with expression limit

Improper integral with infinite integration limit

Sometimes such an improper integral is also called an improper integral of the first kind..gif" width="49" height="19 src=">.

Less common are integrals with an infinite lower limit or with two infinite limits: .

We will consider the most popular case https://pandia.ru/text/80/057/images/image005_1.gif" width="63" height="51"> ? No not always. Integrandhttps://pandia.ru/text/80/057/images/image007_0.gif" width="47" height="23 src=">

Let's depict the graph of the integrand in the drawing. A typical graph and a curvilinear trapezoid for this case looks like this:

Improper integralhttps://pandia.ru/text/80/057/images/image009_0.gif" width="100" height="51">", in other words, the area is also infinite. So it may be. In this case, we say that the improper integral diverges.

2) But. As paradoxical as it may sound, the area of an infinite figure can be equal to ... a finite number! For example: .. In the second case, the improper integral converges.

What happens if an infinite curvilinear trapezoid is located below the axis?.gif" width="217" height="51 src=">.

: .

Example 1

The integrand https://pandia.ru/text/80/057/images/image017_0.gif" width="43" height="23">, which means that everything is fine and the improper integral can be calculated using the "regular" method.

Application of our formula https://pandia.ru/text/80/057/images/image018_0.gif" width="356" height="49">

That is, the improper integral diverges, and the area of the shaded curvilinear trapezoid is equal to infinity.

When solving improper integrals, it is very important to know how the graphs of the main elementary functions look like!

Example 2

Calculate the improper integral or establish its divergence.

Let's make a drawing:

First, we notice the following: the integrand is continuous on the half-interval . Good..gif" width="327" height="53">

(1) We take the simplest integral of a power function (this special case is in many tables). It is better to immediately move the minus beyond the limit sign so that it does not get underfoot in further calculations.

(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.

(3) We indicate that https://pandia.ru/text/80/057/images/image024.gif" width="56" height="19 src="> (Gentlemen, this has long been understood) and simplify answer.

Here, the area of an infinite curvilinear trapezoid is equal to a finite number! Unbelievable, but it is a fact.

Example 3

Calculate the improper integral or establish its divergence.

The integrand is continuous on .

First, let's try to find the antiderivative function (indefinite integral).

Let's replace:

It is always useful to perform a check, that is, to differentiate the result obtained:

Now we find the improper integral:

(1) We write the solution in accordance with the formula . It is better to immediately move the constant beyond the limit sign so that it does not interfere in further calculations.

(2) We substitute the upper and lower limits in accordance with the Newton-Leibniz formula..gif" width="56" height="19 src=">?

(3) We get the final answer. The fact that it is useful to know by heart.

“

The integrand is continuous at https://pandia.ru/text/80/057/images/image041.gif" width="337" height="104">
“

Example 4

Calculate the improper integral or establish its divergence.

! This is a typical example, and similar integrals are very common. Work it out well! The antiderivative function is found here by the full square selection method.

Example 5

Calculate the improper integral or establish its divergence.

This integral can be solved in detail, that is, first find the indefinite integral by changing the variable. And you can solve it "immediately" - by summing the function under the sign of the differential ..

Improper integrals of unbounded functions

Sometimes such improper integrals are called improper integrals of the second kind. Improper integrals of the second kind are cunningly "encrypted" under the usual definite integral and look exactly the same: ..gif" width="39" height="15 src=">, 2) or at the point , 3) or at both points at once, 4) or even on the interval of integration.We will consider the first two cases, for cases 3-4 at the end of the article there is a link to an additional lesson.

Just an example to make it clear: https://pandia.ru/text/80/057/images/image048.gif" width="65 height=41" height="41">, then our denominator turns to zero, that is, the integrand simply does not exist at this point!

In general, when analyzing the improper integral it is always necessary to substitute both integration limits into the integrand..jpg" alt="(!LANG:Improper integral, discontinuity point in the lower limit of integration" width="323" height="380">!}

Here, almost everything is the same as in the integral of the first kind.
Our integral is numerically equal to the area of the shaded curvilinear trapezoid, which is not bounded from above. In this case, there can be two options: the improper integral diverges (the area is infinite) or the improper integral is equal to a finite number (that is, the area of an infinite figure is finite!).

It remains only to modify the Newton-Leibniz formula. It is also modified with the help of the limit, but the limit no longer tends to infinity, but to valuehttps://pandia.ru/text/80/057/images/image052.gif" width="28" height="19"> on right.

Example 6

Calculate the improper integral or establish its divergence.

The integrand suffers an infinite break at a point (do not forget to check verbally or on a draft if everything is fine with the upper limit!)

First, we calculate the indefinite integral:

Replacement:

We calculate the improper integral:

(1) What's new here? Practically nothing in terms of technique. The only thing that has changed is the entry under the limit icon: . The addition means that we are aiming for the value on the right (which is logical - see graph). Such a limit in the theory of limits is called a one-sided limit. In this case, we have a right-hand limit.

(2) We substitute the upper and lower limits according to the Newton-Leibniz formula.

(3) Understanding https://pandia.ru/text/80/057/images/image058.gif" width="69" height="41 src=">. How to determine where the expression should go? Roughly speaking, in you just need to substitute the value for it, substitute three quarters and indicate that... We comb the answer.

In this case, the improper integral is equal to a negative number.

Example 7

Calculate the improper integral or establish its divergence.

Example 8

Calculate the improper integral or establish its divergence.

If the integrand does not exist at the point

An infinite curvilinear trapezoid for such an improper integral fundamentally looks like this:

Everything is exactly the same here, except that the limit tends to to valuehttps://pandia.ru/text/80/057/images/image052.gif" width="28" height="19"> we must get infinitely close to the breaking point left.