Definite integrals online on the site for students and schoolchildren to consolidate the material they have covered. And training your practical skills. A complete solution of definite integrals online for you in a matter of moments will help you determine all stages of the process. Online integrals - definite integral online. Certain integrals online on the site for students and schoolchildren to fully consolidate the material they have covered and train their practical skills. A complete solution of definite integrals online for you in a matter of moments will help you determine all stages of the process. Online integrals - definite integral online. For us, taking a definite integral online does not seem to be something super natural, having studied this topic from a book by outstanding authors. We thank them very much and express our respect to these individuals. An online service for calculating such problems will help you determine a certain integral in no time. Just provide the correct information and everything will be Good! Any definite integral as a solution to a problem will improve students' literacy. Every lazy person dreams of this, and we are no exception, we admit it honestly. If you still manage to calculate a definite integral online with a solution for free, then please write the website address to everyone who wants to use it. As they say, if you share a useful link, good people will reward you with a gift. The question of analyzing a problem in which a certain integral will be solved by the calculator on its own, and not by wasting your precious time, will be very interesting. That's why they are machines, to work for people. However, solving certain integrals online is not something every site can handle, and this is easy to check, namely, just take a complex example and try to solve it using each such service. You will feel the difference firsthand. Often, finding a definite integral online without any effort will become quite difficult and your answer will look ridiculous against the background of the overall picture of the result. It would be better to first take a course for a young fighter. Any solution to improper integrals online is reduced first to calculating the indefinite, and then using the theory of limits to calculate, as a rule, one-sided limits from the resulting expressions with substituted boundaries A and B. Having examined the definite integral you indicated online with a detailed solution, we concluded that you were mistaken on the fifth step, namely when using the Chebyshev variable replacement formula. Be very careful in your further decision. If the online calculator couldn’t take your specific integral the first time, then first of all you should double-check the written data in the appropriate forms on the website. Make sure everything is in order and go, Go-Go! 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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 us depict in the drawing the graph of the integrand function. A typical graph and curved 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. It may be so. In this case they 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 curved trapezoid is located below the axis?.gif" width="217" height="51 src=">.
: 
.
Example 1
The integrand function https://pandia.ru/text/80/057/images/image017_0.gif" width="43" height="23">, which means everything is fine and the improper integral can be calculated using the “standard” 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 curved trapezoid is equal to infinity.
When solving improper integrals, it is very important to know what the graphs of basic elementary functions look like!
Example 2
Calculate the improper integral or establish its divergence.
Let's make the drawing: 
First, we note 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 sign beyond the limit sign so that it does not get in the way in further calculations.
(2) We substitute the upper and lower limits using the Newton-Leibniz formula.
(3) We point out that https://pandia.ru/text/80/057/images/image024.gif" width="56" height="19 src="> (Gentlemen, this needs to be understood a long time ago) and simplify answer.
Here the area of an infinite curved trapezoid is a finite number! Unbelievable but true.
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).
Which of the table integrals is the integrand similar to? It reminds me of an arctangent: 
. These considerations suggest that it would be nice to have a square in the denominator. This is done by replacement.
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 with further calculations.
(2) We substitute the upper and lower limits in accordance with the Newton-Leibniz formula..gif" width="56" height="19 src=">? See the arctangent graph in the already repeatedly recommended article.
(3) We get the final answer. A fact that is useful to know by heart.
Advanced students may not find the indefinite integral separately and not use the replacement method, but rather use the method of substituting the function under the differential sign and solving the improper integral “immediately.” In this case, the solution should look something like this:
“
The integrand function 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 found very often. Work it out well! The antiderivative function is found here using the method of isolating a complete square.
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 making a change of variable. Or you can solve it “immediately” - by subsuming the function under the differential sign..
Improper integrals of unbounded functions
Sometimes such improper integrals are called improper integrals of the second kind. Improper integrals of the second kind are insidiously “encrypted” under the usual definite integral and look exactly the same: ..gif" width="39" height="15 src=">, 2) or at point , 3) or at both points at once, 4) or even on the integration segment.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 goes to zero, that is, the integrand simply does not exist at this point!
In general, when analyzing an improper integral you always need to substitute both integration limits into the integrand..jpg" alt="Improper integral, discontinuity point at the lower limit of integration" width="323" height="380">!}
Here everything is almost the same as in the integral of the first kind.
Our integral is numerically equal to the area of the shaded curved 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!).
All that remains is to modify the Newton-Leibniz formula. It is also modified with the help of a 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 has an infinite discontinuity at a point (don’t forget to check verbally or on a draft that everything is fine with the upper limit!)
First, let's calculate the indefinite integral: 
Replacement: 
Let's calculate the improper integral:
(1) What's new here? There is practically nothing in terms of solution technology. The only thing that has changed is the entry under the limit icon: . The addition means that we are striving for the value on the right (which is logical - see the graph). Such a limit in the theory of limits is called a one-sided limit. In this case we have a right-handed limit.
(2) We substitute the upper and lower limits using the Newton-Leibniz formula.
(3) Let's understand 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 , 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 curved trapezoid for such an improper integral fundamentally looks like this:
Here everything is absolutely the same, except that our limit tends to to valuehttps://pandia.ru/text/80/057/images/image052.gif" width="28" height="19"> we must approach infinitely close to the breaking point left.
Definite integral
\[ I=\int_a^bf(x)dx \]
was constructed under the assumption that the numbers $a,\,b$ are finite and $f(x)$ is a continuous function. If one of these assumptions is violated, we speak of improper integrals.
10.1 Improper integrals of the 1st kind
An improper integral of the 1st kind occurs when at least one of the numbers $a,\,b$ is infinite.
10.1.1 Definition and basic properties
Let us first consider the situation when the lower limit of integration is finite and the upper limit is equal to $+\infty$; we will discuss other options a little later. For $f(x)$, continuous for all $x$ of interest to us, consider the integral
\begin(equation) I=\int _a^(+\infty)f(x)dx. \quad(19) \label(inf1) \end(equation)
First of all, we need to establish the meaning of this expression. To do this, we introduce the function
\[ I(N)=\int _a^(N)f(x)dx \]
and consider its behavior for $N\rightarrow +\infty$.
Definition. Let there be a finite limit
\[ A=\lim_(N \rightarrow +\infty)I(N)=\lim_(N \rightarrow +\infty)\int _a^(N)f(x)dx. \]
Then we say that the improper integral of the 1st kind (19) is convergent and the value $A$ is assigned to it; the function itself is called integrable on the interval $\left[ a, \, +\infty \right)$. If the specified limit does not exist or it is equal to $\pm \infty$, then the integral (19) is said to diverge.
Consider the integral
\[ I=\int _0^(+\infty) \frac(dx)(1+x^2). \]
\[ I(N)=\int _0^(N) \frac(dx)(1+x^2). \]
In this case, the antiderivative of the integrand function is known, so
\[ I(N)=\int _0^(N) \frac(dx)(1+x^2)=arctgx|_0^(N)=arctgN. \]
It is known that $arctg N \rightarrow \pi /2 $ for $N \rightarrow +\infty$. Thus, $I(N)$ has a finite limit, our improper integral converges and is equal to $\pi /2$.
Convergent improper integrals of the 1st kind have all the standard properties of ordinary definite integrals.
1. If $f(x)$, $g(x)$ are integrable on the interval $\left[ a, \, +\infty \right)$, then their sum $f(x)+g(x)$ is also is integrable on this interval, and \[ \int _a^(+\infty)\left(f(x)+g(x)\right)dx=\int _a^(+\infty)f(x)dx+\int _a^(+\infty)g(x)dx. \] 2. If $f(x)$ is integrable on the interval $\left[ a, \, +\infty \right)$, then for any constant $C$ the function $C\cdot f(x)$ is also integrable on this interval, and \[ \int _a^(+\infty)C\cdot f(x)dx=C \cdot \int _a^(+\infty)f(x)dx. \] 3. If $f(x)$ is integrable on the interval $\left[ a, \, +\infty \right)$, and on this interval $f(x)>0$, then \[ \int _a^ (+\infty) f(x)dx\,>\,0. \] 4. If $f(x)$ is integrable on the interval $\left[ a, \, +\infty \right)$, then for any $b>a$ the integral \[ \int _b^(+\infty) f(x)dx \] converges, and \[ \int _a^(+\infty)f(x)dx=\int _a^(b) f(x)dx+\int _b^(+\infty) f( x)dx \] (additivity of the integral over the interval).
The formulas for change of variable, integration by parts, etc. are also valid. (with natural reservations).
Consider the integral
\begin(equation) I=\int _1^(+\infty)\frac(1)(x^k)\,dx. \quad (20) \label(mod) \end(equation)
Let's introduce the function
\[ I(N)=\int _1^(N)\frac(1)(x^k)\,dx. \]
In this case the antiderivative is known, so
\[ I(N)=\int _1^(N)\frac(1)(x^k)\,dx\,=\frac(x^(1-k))(1-k)|_1^N = \frac(N^(1-k))(1-k)-\frac(1)(1-k) \]
for $k \neq 1$,
\[ I(N)=\int _1^(N)\frac(1)(x)\,dx\,=lnx|_1^N= lnN \]
for $k = 1$. Considering the behavior for $N \rightarrow +\infty$, we come to the conclusion that integral (20) converges for $k>1$, and diverges for $k \leq 1$.
Let us now consider the option when the lower limit of integration is equal to $-\infty$, and the upper one is finite, i.e. let's look at the integrals
\[ I=\int _(-\infty)^af(x)dx. \]
However, this option can be reduced to the previous one if we make a change of variables $x=-s$ and then change the limits of integration in places, so that
\[ I=\int _(-a)^(+\infty)g(s)ds, \]
$g(s)=f(-s)$. Let us now consider the case when there are two infinite limits, i.e. integral
\begin(equation) I=\int _(-\infty)^(+\infty)f(x)dx, \quad (21) \label(intr) \end(equation)
and $f(x)$ is continuous for all $x \in \mathbb(R)$. Let's split the interval into two parts: take $c \in \mathbb(R)$, and consider two integrals,
\[ I_1=\int _(-\infty)^(c)f(x)dx, \quad I_2=\int _(c)^(+\infty)f(x)dx. \]
Definition. If both integrals $I_1$, $I_2$ converge, then integral (21) is called convergent and is assigned the value $I=I_1+I_2$ (in accordance with additivity over the interval). If at least one of the integrals $I_1$, $I_2$ diverges, integral (21) is called divergent.
It can be proven that the convergence of integral (21) does not depend on the choice of point $c$.
Improper integrals of the 1st kind with integration intervals $\left(-\infty, \, c \right]$ or $(-\infty, \, +\infty)$ also have all the standard properties of definite integrals (with the corresponding reformulation taking into account the choice integration interval).
10.1.2 Tests for the convergence of improper integrals of the 1st kind
Theorem (the first sign of comparison). Let $f(x)$, $g(x)$ be continuous for $x>a$, and $0 a$. Then
1. If the integral \[ \int _a^(+\infty)g(x)dx \] converges, then the integral \[ \int _a^(+\infty)f(x)dx converges. \] 2. If the integral \[ \int _a^(+\infty)f(x)dx \] diverges, then the integral \[ \int _a^(+\infty)g(x)dx diverges. \]
Theorem (second comparison criterion). Let $f(x)$, $g(x)$ be continuous and positive for $x>a$, and let there be a finite limit
\[ \theta = \lim_(x \rightarrow +\infty) \frac(f(x))(g(x)), \quad \theta \neq 0, \, +\infty. \]
Then the integrals
\[ \int _a^(+\infty)f(x)dx, \quad \int _a^(+\infty)g(x)dx \]
converge or diverge simultaneously.
Consider the integral
\[ I=\int _1^(+\infty)\frac(1)(x+\sin x)\,dx. \]
The integrand expression is a positive function on the integration interval. Further, for $x \rightarrow +\infty$ we have:
$\sin x$ is a "small" correction to the denominator. More precisely, if we take $f(x)=1/(x+\sin x)$, \, $g(x)=1/x$, then
\[ \lim _(x \rightarrow +\infty)\frac(f(x))(g(x))=\lim _(x \rightarrow +\infty)\frac(x)(x+\sin x) =1. \]
Applying the second comparison criterion, we come to the conclusion that our integral converges or diverges simultaneously with the integral
\[ \int _1^(+\infty)\frac(1)(x)\,dx . \]
As was shown in the previous example, this integral diverges ($k=1$). Consequently, the original integral diverges.
Calculate the improper integral or establish its convergence (divergence).
1. \[ \int _(0)^(+\infty)e^(-ax)\,dx. \] 2. \[ \int _(0)^(+\infty)xe^(-x^2)\,dx. \] 3. \[ \int _(-\infty)^(+\infty)\frac(2xdx)(x^2+1). \] 4. \[ \int _(0)^(+\infty)\frac(xdx)((x+2)^3). \] 5. \[ \int _(-\infty)^(+\infty)\frac(dx)(x^2+2x+2). \] 6. \[ \int _(1)^(+\infty)\frac(lnx)(x^2)\,dx. \] 7. \[ \int _(1)^(+\infty)\frac(dx)((1+x)\sqrt(x)). \] 8. \[ \int _(0)^(+\infty)e^(-\sqrt(x))\,dx. \] 9. \[ \int _(0)^(+\infty)e^(-ax)\cos x\,dx. \] 10. \[ \int _(0)^(+\infty)\frac(xdx)(x^3+1). \]
Are you here now? =) No, I wasn’t trying to intimidate anyone, it’s just that the topic of improper integrals is a very good illustration of how important it is not to neglect higher mathematics and other exact sciences. Everything you need to learn the lesson is on the website - in a detailed and accessible form, if you wish...
So, let's start with. Figuratively speaking, an improper integral is an “advanced” definite integral, and in fact there are not so many difficulties with them, and besides, the improper integral has a very good geometric meaning.
What does it mean to evaluate an improper integral?
Calculate improper integral - this means finding the NUMBER(exactly the same as in the definite integral), or prove that it diverges(that is, you end up with infinity instead of a number).
There are two types of improper integrals.
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? At the upper limit. It is endless: .
Less common are integrals with an infinite lower limit or with two infinite limits: , and we will look at them later - when you get the hang of it :)
Well, now let’s look at 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! Because if there are gaps, then there are additional nuances. For definiteness, let us assume that even then the typical curved trapezoid will look like this:
Note that it is infinite (not bounded on the right), and improper integral numerically equal to its area. The following options are possible:
1) The first thought that comes to mind: “since the figure is infinite, then 
", in other words, the area is also infinite. It may be so. In this case they 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 this be true? Easily. In the second case, the improper integral converges.
3) About the third option a little later.
In what cases does an improper integral diverge and in what cases does it converge? This depends on the integrand, and we'll look at specific examples very soon.
What happens if an infinite curved trapezoid is located below the axis? In this case, the improper integral 
(diverges) or is equal to a finite negative number.
Thus, improper integral can be negative.
Important! When you are given ANY improper integral to solve, then, generally speaking, there is no talk about any area and there is no need to build a drawing. I explained 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, let us remember the Newton-Leibniz formula: 
. In fact, the formula is also applicable to improper integrals, only it needs to be slightly modified. What is the difference? At the infinite upper limit of integration: . Probably, many guessed that this already smacks of the application of the theory of limits, and the formula will be written like this: 
.
What is the difference from a definite integral? Nothing special! As in the definite integral, you need to be able to find the antiderivative function (indefinite integral), and be able to apply the Newton-Leibniz formula. The only thing that has been added is the calculation of the limit. Whoever has a bad time with them, learn a lesson Function limits. Examples of solutions, because it’s better late than in the army.
Let's look at two classic examples:
Example 1
For clarity, I will draw a drawing, although, I emphasize once again, on practice There is no need to build drawings in this task.
The integrand function is continuous on the half-interval, which means that everything is fine and the improper integral can be calculated by the “standard” method.
Application of our formula 
and the solution to the problem looks like this:
That is, the improper integral diverges, and the area of the shaded curved trapezoid is equal to infinity.
In the example considered, we have the simplest table integral and the same technique for applying the Newton-Leibniz formula as in the definite integral. But this formula will be applied under the sign of the limit. Instead of the usual letter of a “dynamic” variable, the letter “be” appears. This should not confuse or baffle, because any letter is no worse than the standard “X”.
If you do not understand why at , then this is very bad, either you do not understand the simplest limits (and generally do not understand what a limit is), or you do not know what the graph of a logarithmic function looks like. In the second case, attend a lesson Graphs and properties of elementary functions.
When solving improper integrals, it is very important to know what the graphs of basic elementary functions look like!
The finished task should look something like this:
“
! When preparing an example, we always interrupt the solution and indicate what happens to the integrand – is it continuous on the interval of integration or not?. With this we identify the type of improper integral and justify further actions.
Example 2
Calculate the improper integral or establish its divergence.
Let's make the drawing: 
First, we note the following: the integrand is continuous on the half-interval. Hood. We solve using the formula 
:
(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 sign beyond the limit sign so that it does not get in the way in further calculations.
(2) We substitute the upper and lower limits using the Newton-Leibniz formula.
(3) We indicate that at (Gentlemen, this should have been understood a long time ago) and simplify the answer.
Here the area of an infinite curved trapezoid is a finite number! Unbelievable but true.
The finished example should look something like this:
“
The integrand function is continuous on
“
What to do if you come across an integral like - with break point on the integration interval? This means there is a typo in the example. (Most likely), or about an advanced level of training. In the latter case, due to additivity properties, we should consider two improper integrals on intervals and then deal with the sum.
Sometimes, due to a typo or intent, an improper integral may not exist at all, so, for example, if you put the square root of “x” in the denominator of the above integral, then part of the integration interval will not be included in the domain of definition of the integrand at all.
Moreover, the 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 appropriate limit.
And such examples, although rare, do occur in practice! Thus, in addition to convergence and divergence, there is also a third outcome of the solution with a valid answer: “there is no improper integral.”
It should also be noted that the strict definition of an 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 be able to solve the improper integral either.
Which of the table integrals is the integrand similar to? It reminds me of an arctangent: 
. These considerations suggest that it would be nice to have a square in the denominator. This is done by replacement.
Let's replace:
The indefinite integral has been found; in this case, it makes no sense to add a constant.
It is always useful to check the draft, that is, differentiate the result obtained:
The original integrand has been obtained, which means that the indefinite integral has been 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 with further calculations.
(2) We substitute the upper and lower limits in accordance with the Newton-Leibniz formula. Why 
at ? See the arctangent graph in the already recommended article.
(3) We get the final answer. A fact that is useful to know by heart.
Advanced students may not find the indefinite integral separately and not use the replacement method, but rather use the method of substituting the function under the differential sign and solving 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 found very often. Work it out well! The antiderivative function here is found using the method of selecting a complete square; more details on the method can be found in the lesson Integrating 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 making a change of variable. Or you can solve it “immediately” - by subsuming the function under the differential sign. Who has any mathematical training?
Complete solutions and answers at the end of the lesson.
Examples of solutions to improper integrals with an infinite lower limit of integration can be found on the page Efficient methods for solving improper integrals. There we also analyzed the case when both limits of integration are infinite.
Improper integrals of unbounded functions
Or improper integrals of the second kind. Improper integrals of the second kind are insidiously “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) or at the point , 3) or at both points at once, 4) or even on the integration segment. We will look at 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 our denominator goes to zero, that is, the integrand simply does not exist at this point!
In general, when analyzing an improper integral you always need to substitute both integration limits into the integrand. In this regard, let's check the upper limit: 
. Everything is fine here.
The curvilinear trapezoid for the type of improper integral under consideration fundamentally looks like this:
Here everything is almost the same as in the integral of the first kind.
Our integral is numerically equal to the area of the shaded curved 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!).
* by default we usually assume that the improper integral exists
All that remains is to modify the Newton-Leibniz formula. It is also modified with the help of a limit, but the limit no longer tends to infinity, but to the value on the right. It is easy to follow from 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 has an infinite discontinuity at a point (don’t forget to check verbally or on a draft that everything is fine with the upper limit!)
First, let's calculate the indefinite integral: 
Replacement: 
If you have any difficulties with replacement, please refer to the lesson Substitution method in indefinite integral.
Let's calculate the improper integral:
(1) What's new here? There is practically nothing in terms of solution technology. The only thing that has changed is the entry under the limit icon: . The addition means that we are striving for the value on the right (which is logical - see the graph). Such a limit in the theory of limits is called one-sided limit. In this case we have right-hand limit.
(2) We substitute the upper and lower limits using the Newton-Leibniz formula.
(3) Let's deal with at . How to determine where an expression is going? Roughly speaking, you just need to substitute the value into it, substitute three quarters and indicate that . Let's comb the answer.
In this case, the improper integral is equal to a negative number. There is no crime in this, just the corresponding curved trapezoid is located under the axis.
And now two examples for independent solutions.
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 curved trapezoid for such an improper integral fundamentally looks like this.
Improper integrals of the first kind: extension of the concept of a definite integral to cases of integrals with infinite upper or lower limits of integration, or both limits of integration are infinite.
Improper integrals of the second kind: extension of the concept of a definite integral to the cases of integrals of unbounded functions; the integrand does not exist at a finite number of points of a finite segment of integration, turning to infinity.
For comparison. When introducing the concept of a definite integral, it was assumed that the function f(x) is continuous on the interval [ a, b], and the integration segment is finite, that is, it is limited by numbers, and not by infinity. Some tasks lead to the need to abandon these restrictions. This is how improper integrals appear.
Geometric meaning of the improper integral It turns out quite simply. In the case when the graph of a function y = f(x) is above the axis Ox, the definite integral expresses the area of a curvilinear trapezoid bounded by a curve y = f(x) , x-axis and ordinates x = a , x = b. In turn, the improper integral expresses the area of an unlimited (infinite) curvilinear trapezoid enclosed between the lines y = f(x) (in the picture below - red), x = a and the abscissa axis.
Improper integrals are defined similarly for other infinite intervals:
The area of an infinite curved trapezoid can be a finite number, in which case the improper integral is called convergent. The area can also be infinity, and in this case the improper integral is called divergent.
Using the limit of an integral instead of the improper integral itself. In order to evaluate the improper integral, you need to use the limit of the definite integral. If this limit exists and is finite (not equal to infinity), then the improper integral is called convergent, and otherwise - divergent. What a variable tends to under the limit sign depends on whether we are dealing with an improper integral of the first kind or of the second kind. Let's find out about this now.
Improper integrals of the first kind - with infinite limits and their convergence
Improper integrals with infinite upper limit
So, writing an improper integral differs from the usual definite integral in that the upper limit of integration is infinite.
Definition. An improper integral with an infinite upper limit of integration of a continuous function f(x) in the interval from a before ∞ the limit of the integral of this function with the upper limit of integration is called b and the lower limit of integration a provided that the upper limit of integration grows without limit, i.e.
.
If this limit exists and is equal to some number rather than infinity, then an improper integral is called convergent, and the number to which the limit is equal is taken as its value. Otherwise an improper integral is called divergent and no meaning is attributed to it.
Example 1. Calculate improper integral(if it converges).
Solution. Based on the definition of the improper integral, we find
Since the limit exists and is equal to 1, then this improper integral converges and is equal to 1.
In the following example, the integrand is almost the same as in example 1, only the degree x is not two, but the letter alpha, and the task is to study the improper integral for convergence. That is, the question remains to be answered: at what values of alpha does this improper integral converge, and at what values does it diverge?
Example 2. Examine the improper integral for convergence(the lower limit of integration is greater than zero).
Solution. Let us first assume that , then
In the resulting expression, we move to the limit at:
It is easy to see that the limit on the right side exists and is equal to zero when , that is , and does not exist when , that is .
In the first case, that is, when . If , then 
and doesn't exist.
The conclusion of our study is as follows: this improper integral converges at and diverges at .
Applying the Newton-Leibniz formula to the type of improper integral being studied 
, you can derive the following formula, which is very similar to it:
.
This is a generalized Newton-Leibniz formula.
Example 3. Calculate improper integral(if it converges).
The limit of this integral exists:
The second integral, making up the sum expressing the original integral:
The limit of this integral also exists:
.
We find the sum of two integrals, which is also the value of the original improper integral with two infinite limits:
Improper integrals of the second kind - from unbounded functions and their convergence
Let the function f(x) given on the segment from a before b and is unlimited on it. Suppose that the function goes to infinity at the point b , while at all other points of the segment it is continuous.
Definition. An improper integral of a function f(x) on the segment from a before b the limit of the integral of this function with the upper limit of integration is called c , if when striving c To b the function increases without limit, and at the point x = b function not defined, i.e.
.
If this limit exists, then the improper integral of the second kind is called convergent, otherwise it is called divergent.
Using the Newton-Leibniz formula, we derive.