Integrals for dummies: how to solve, calculation rules, explanation. Method of integration by parts: explanation, solution of examples
By a definite integral from a continuous function f(x) on the final segment [ a, b] (where ) is the increment of some of its antiderivatives on this segment. (In general, understanding will be noticeably easier if you repeat the topic of the indefinite integral) In this case, the notation is used
As can be seen in the graphs below (the increment of the antiderivative function is indicated by ), a definite integral can be either a positive or a negative number(It is calculated as the difference between the value of the antiderivative in the upper limit and its value in the lower limit, i.e. as F(b) - F(a)).
Numbers a And b are called the lower and upper limits of integration, respectively, and the segment [ a, b] – segment of integration.
Thus, if F(x) – some antiderivative function for f(x), then, according to the definition,
(38)
Equality (38) is called Newton-Leibniz formula . Difference F(b) – F(a) is briefly written as follows:
Therefore, we will write the Newton-Leibniz formula like this:
(39)
Let us prove that the definite integral does not depend on which antiderivative of the integrand is taken when calculating it. Let F(x) and F( X) are arbitrary antiderivatives of the integrand. Since these are antiderivatives of the same function, they differ by a constant term: Ф( X) = F(x) + C. That's why
This establishes that on the segment [ a, b] increments of all antiderivatives of the function f(x) match up.
Thus, to calculate a definite integral, it is necessary to find any antiderivative of the integrand, i.e. First you need to find the indefinite integral. Constant WITH excluded from subsequent calculations. Then the Newton-Leibniz formula is applied: in antiderivative function the value of the upper limit is substituted b , further - the value of the lower limit a and the difference is calculated F(b) - F(a) . The resulting number will be a definite integral..
At a = b by definition accepted
Example 1.
Solution. First, let's find the indefinite integral:
Applying the Newton-Leibniz formula to the antiderivative
(at WITH= 0), we get
However, when calculating a definite integral, it is better not to find the antiderivative separately, but to immediately write the integral in the form (39).
Example 2. Calculate definite integral
Solution. Using formula
Properties of the definite integral
Theorem 2.The value of the definite integral does not depend on the designation of the integration variable, i.e.
(40)
Let F(x) – antiderivative for f(x). For f(t) the antiderivative is the same function F(t), in which the independent variable is only designated differently. Hence,
Based on formula (39), the last equality means the equality of the integrals
Theorem 3.The constant factor can be taken out of the sign of the definite integral, i.e.
(41)
Theorem 4.The definite integral of the algebraic sum of a finite number of functions is equal to algebraic sum definite integrals of these functions, i.e.
(42)
Theorem 5.If a segment of integration is divided into parts, then the definite integral over the entire segment is equal to the sum of definite integrals over its parts, i.e. If
(43)
Theorem 6.When rearranging the limits of integration, the absolute value of the definite integral does not change, but only its sign changes, i.e.
(44)
Theorem 7(mean value theorem). A definite integral is equal to the product of the length of the integration segment and the value of the integrand at some point inside it, i.e.
(45)
Theorem 8.If the upper limit of integration is greater than the lower one and the integrand is non-negative (positive), then the definite integral is also non-negative (positive), i.e. If
Theorem 9.If the upper limit of integration is greater than the lower one and the functions and are continuous, then the inequality
can be integrated term by term, i.e.
(46)
The properties of the definite integral allow us to simplify direct calculation integrals.
Example 5. Calculate definite integral
Using Theorems 4 and 3, and when finding antiderivatives - table integrals (7) and (6), we obtain
Definite integral with variable upper limit
Let f(x) – continuous on the segment [ a, b] function, and F(x) is its antiderivative. Consider the definite integral
(47)
and through t the integration variable is designated so as not to confuse it with the upper bound. When it changes X the definite integral (47) also changes, i.e. it is a function of the upper limit of integration X, which we denote by F(X), i.e.
(48)
Let us prove that the function F(X) is an antiderivative for f(x) = f(t). Indeed, differentiating F(X), we get
because F(x) – antiderivative for f(x), A F(a) is a constant value.
Function F(X) – one of the infinite number of antiderivatives for f(x), namely the one that x = a goes to zero. This statement is obtained if in equality (48) we put x = a and use Theorem 1 of the previous paragraph.
Calculation of definite integrals by the method of integration by parts and the method of change of variable
where, by definition, F(x) – antiderivative for f(x). If we change the variable in the integrand
then, in accordance with formula (16), we can write
In this expression
antiderivative function for
In fact, its derivative, according to rule of differentiation of complex functions, is equal
Let α and β be the values of the variable t, for which the function
takes values accordingly a And b, i.e.
But, according to the Newton-Leibniz formula, the difference F(b) – F(a) There is
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What is integration by parts? To master this type of integration, let's first remember the derivative of a product:
$((\left(f\cdot g \right))^(\prime ))=(f)"\cdot g+f\cdot (g)"$
The question arises: what do integrals have to do with it? Now let's integrate both sides of this equation. So let's write it down:
$\int(((\left(f\cdot g \right))^(\prime ))\text(d)x=)\int((f)"\cdot g\,\text(d)x+\ int(f\cdot (g)"\,\text(d)x))$
But what is an antiderivative of a stroke? It's just the function itself, which is inside the stroke. So let's write it down:
$f\cdot g=\int((f)"\cdot g\,\text(d)x+\int(f\cdot (g)"\,\text(d)x))$
IN given equation I propose to express the term. We have:
$\int((f)"\cdot g\,\text(d)x=f\cdot g-\int(f\cdot (g)"\,\text(d)x))$
That's what it is integration by parts formula. Thus, we are essentially interchanging the derivative and the function. If initially we had an integral of a stroke multiplied by something, then we get an integral of a new something multiplied by a stroke. That's all the rule. At first glance, this formula may seem complicated and meaningless, but in fact, it can greatly simplify calculations. Let's see.
Examples of integral calculations
Problem 1. Calculate:
\[\int(\ln x\,\text(d)x)\]\[\]
Let's rewrite the expression by adding 1 before the logarithm:
\[\int(\ln x\,\text(d)x)=\int(1\cdot \ln x\,\text(d)x)\]
We have the right to do this because neither the number nor the function will change. Now let's compare this expression with what is written in our formula. The role of $(f)"$ is 1, so we write:
$\begin(align)& (f)"=1\Rightarrow f=x \\& g=\ln x\Rightarrow (g)"=\frac(1)(x) \\\end(align)$
All these functions are in the tables. Now that we have described all the elements that are included in our expression, we will rewrite this integral using the formula for integration by parts:
\[\begin(align)& \int(1\cdot \ln x\,\text(d)x)=x\ln x-\int(x\cdot \frac(1)(x)\text(d )x)=x\ln x-\int(\text(d)x)= \\& =x\ln x-x+C=x\left(\ln x-1 \right)+C \\\ end(align)\]
That's it, the integral has been found.
Problem 2. Calculate:
$\int(x((\text(e))^(-x))\,\text(d)x=\int(x\cdot ((e)^(-x))\,\text(d )x))$
If we take $x$ as the derivative, from which we now need to find the antiderivative, we will get $((x)^(2))$, and the final expression will contain $((x)^(2))( (\text(e))^(-x))$.
Obviously, the problem is not simplified, so we swap the factors under the integral sign:
$\int(x\cdot ((\text(e))^(-x))\,\text(d)x)=\int(((\text(e))^(-x))\cdot x\,\text(d)x)$
Now let’s introduce the notation:
$(f)"=((\text(e))^(-x))\Rightarrow f=\int(((\text(e))^(-x))\,\text(d)x) =-((\text(e))^(-x))$
Let's differentiate $((\text(e))^(-x))$:
$((\left(((\text(e))^(-x)) \right))^(\prime ))=((\text(e))^(-x))\cdot ((\ left(-x \right))^(\prime ))=-((\text(e))^(-x))$
In other words, the minus is added first and then both sides are integrated:
\[\begin(align)& ((\left(((\text(e))^(-x)) \right))^(\prime ))=-((\text(e))^(- x))\Rightarrow ((\text(e))^(-x))=-((\left(((\text(e))^(-x)) \right))^(\prime )) \\& \int(((\text(e))^(-x))\,\text(d)x)=-\int(((\left(((\text(e))^(- x)) \right))^(\prime ))\text(d)x)=-((\text(e))^(-x))+C \\\end(align)\]
Now let's look at the $g$ function:
$g=x\Rightarrow (g)"=1$
We calculate the integral:
$\begin(align)& \int(((\text(e))^(-x))\cdot x\,\text(d)x)=x\cdot \left(-((\text(e ))^(-x)) \right)-\int(\left(-((\text(e))^(-x)) \right)\cdot 1\cdot \text(d)x)= \ \& =-x((\text(e))^(-x))+\int(((\text(e))^(-x))\,\text(d)x)=-x( (\text(e))^(-x))-((\text(e))^(-x))+C=-((\text(e))^(-x))\left(x +1 \right)+C \\\end(align)$
So, we have performed the second integration by parts.
Problem 3. Calculate:
$\int(x\cos 3x\,\text(d)x)$
In this case, what should we take for $(f)"$ and what for $g$? If $x$ acts as a derivative, then during integration we will get $\frac(((x)^(2)))(2 )$, and our first factor will not disappear anywhere - it will be $\frac(((x)^(2)))(2)\cdot \cos 3x$. Therefore, let’s swap the factors again:
$\begin(align)& \int(x\cos 3x\,\text(d)x)=\int(\cos 3x\cdot x\,\text(d)x) \\& (f)"= \cos 3x\Rightarrow f=\int(\cos 3x\,\text(d)x)=\frac(\sin 3x)(3) \\& g=x\Rightarrow (g)"=1 \\\ end(align)$
We rewrite our original expression and expand it according to the integration formula by parts:
\[\begin(align)& \int(\cos 3x\cdot x\ \text(d)x)=\frac(\sin 3x)(3)\cdot x-\int(\frac(\sin 3x) (3)\text(d)x)= \\& =\frac(x\sin 3x)(3)-\frac(1)(3)\int(\sin 3x\,\text(d)x) =\frac(x\sin 3x)(3)+\frac(\cos 3x)(9)+C \\\end(align)\]
That's it, the third problem is solved.
In conclusion, let's take another look at integration by parts formula. How do we choose which of the factors will be derivative and which will be this function? There is only one criterion here: the element that we will differentiate must either give a “beautiful” expression, which will then be reduced, or disappear altogether during differentiation. This concludes the lesson.
Complex integrals
This article concludes the topic of indefinite integrals, and includes integrals that I find quite complex. The lesson was created at the repeated requests of visitors who expressed their wish that more difficult examples be analyzed on the site.
It is assumed that the reader of this text is well prepared and knows how to apply basic integration techniques. Dummies and people who are not very confident in integrals should refer to the very first lesson - Indefinite integral. Examples of solutions, where you can master the topic almost from scratch. More experienced students can become familiar with techniques and methods of integration that have not yet been encountered in my articles.
What integrals will be considered?
First we will consider integrals with roots, for the solution of which we successively use variable replacement And integration by parts. That is, in one example two techniques are combined at once. And even more.
Then we will get acquainted with interesting and original method of reducing the integral to itself. Quite a few integrals are solved this way.
The third issue of the program will be integrals of complex fractions, which flew past the cash desk in previous articles.
Fourthly, additional integrals from trigonometric functions will be analyzed. In particular, there are methods that avoid time-consuming universal trigonometric substitution.
(2) In the integrand function, we divide the numerator by the denominator term by term.
(3) We use the linearity property of the indefinite integral. In the last integral immediately put the function under the differential sign.
(4) We take the remaining integrals. Note that in a logarithm you can use parentheses rather than a modulus, since .
(5) We carry out a reverse replacement, expressing “te” from the direct replacement:
Masochistic students can differentiate the answer and get the original integrand, as I just did. No, no, I did the check in the right sense =)
As you can see, during the solution we had to use even more than two solution methods, so to deal with such integrals you need confident integration skills and quite a bit of experience.
In practice, of course, the square root is more common; here are three examples for solving it yourself:
Example 2
Can't find definite integral
Example 3
Find indefinite integral
Example 4
Find the indefinite integral
These examples are of the same type, so the complete solution at the end of the article will only be for Example 2; Examples 3-4 have the same answers. Which replacement to use at the beginning of decisions, I think, is obvious. Why did I choose examples of the same type? Often found in their role. More often, perhaps, just something like 
.
But not always, when under the arctangent, sine, cosine, exponential and other functions there is a root of linear function, you have to use several methods at once. In a number of cases, it is possible to “get off easy,” that is, immediately after the replacement, a simple integral is obtained, which can be easily taken. The easiest of the tasks proposed above is Example 4, in which, after replacement, a relatively simple integral is obtained.
By reducing the integral to itself
A witty and beautiful method. Let's take a look at the classics of the genre:
Example 5
Find the indefinite integral
Under the root is a quadratic binomial, and when trying to integrate this example the kettle can suffer for hours. Such an integral is taken in parts and reduced to itself. In principle, it’s not difficult. If you know how.
Let us denote the integral under consideration by a Latin letter and begin the solution: 
Let's integrate by parts: 
(1) Prepare the integrand function for term-by-term division.
(2) We divide the integrand function term by term. It may not be clear to everyone, but I’ll describe it in more detail:
(3) We use the linearity property of the indefinite integral.
(4) Take the last integral (“long” logarithm).
Now let's look at the very beginning of the solution: 
And at the end: 
What happened? As a result of our manipulations, the integral was reduced to itself!
Let's equate the beginning and the end: 
Move to the left side with a change of sign: 
And we take the deuce to right side. As a result: 
The constant, strictly speaking, should have been added earlier, but I added it at the end. I strongly recommend reading what the rigor is here:
Note:
More strictly, the final stage of the solution looks like this:
Thus:
The constant can be redesignated by . Why can it be redesignated? Because he still accepts it any values, and in this sense there is no difference between constants and.
As a result:
A similar trick with constant renotation is widely used in differential equations. And there I will be strict. And here I allow such freedom only in order not to confuse you with unnecessary things and to focus attention precisely on the integration method itself.
Example 6
Find the indefinite integral
Another typical integral for independent solution. Full solution and answer at the end of the lesson. There will be a difference with the answer in the previous example!
If under square root is a quadratic trinomial, then the solution in any case reduces to two analyzed examples.
For example, consider the integral 
. All you need to do is first select a complete square:
.
Next, a linear replacement is carried out, which does “without any consequences”:
, resulting in the integral . Something familiar, right?
Or this example, with a quadratic binomial:
Select a complete square:
And, after linear replacement, we obtain the integral, which is also solved using the algorithm already discussed.
Let's look at two more typical examples of how to reduce an integral to itself:
– integral of the exponential multiplied by sine;
– integral of the exponential multiplied by the cosine.
In the listed integrals by parts you will have to integrate twice:
Example 7
Find the indefinite integral
The integrand is the exponential multiplied by the sine.
We integrate by parts twice and reduce the integral to itself: 
As a result of double integration by parts, the integral was reduced to itself. We equate the beginning and end of the solution:
We move it to the left side with a change of sign and express our integral: 
Ready. At the same time, it is advisable to comb the right side, i.e. take the exponent out of brackets, and place the sine and cosine in brackets in a “beautiful” order.
Now let's go back to the beginning of the example, or more precisely, to integration by parts: 
We designated the exponent as. The question arises: is it the exponent that should always be denoted by ? Not necessary. In fact, in the considered integral fundamentally doesn't matter, what do we mean by , we could have gone the other way: 
Why is this possible? Because the exponential turns into itself (both during differentiation and integration), sine and cosine mutually turn into each other (again, both during differentiation and integration).
That is, we can also denote a trigonometric function. But, in the example considered, this is less rational, since fractions will appear. If you wish, you can try to solve this example using the second method; the answers must match.
Example 8
Find the indefinite integral
This is an example for you to solve on your own. Before you decide, think about what is more profitable in in this case denote by , exponential or trigonometric function? Full solution and answer at the end of the lesson.
And, of course, do not forget that most of the answers in this lesson are quite easy to check by differentiation!
The examples considered were not the most complex. In practice, integrals are more common where the constant is both in the exponent and in the argument of the trigonometric function, for example: . Many people will get confused in such an integral, and I often get confused myself. The fact is that there is a high probability of fractions appearing in the solution, and it is very easy to lose something through carelessness. In addition, there is a high probability of an error in the signs; note that the exponent has a minus sign, and this introduces additional difficulty.
At the final stage, the result is often something like this:
Even at the end of the solution, you should be extremely careful and correctly understand the fractions: 
Integrating Complex Fractions
We are slowly approaching the equator of the lesson and begin to consider integrals of fractions. Again, not all of them are super complex, it’s just that for one reason or another the examples were a little “off topic” in other articles.
Continuing the theme of roots
Example 9
Find the indefinite integral
In the denominator under the root there is a quadratic trinomial plus an “appendage” in the form of an “X” outside the root. An integral of this type can be solved using a standard substitution.
We decide: 
The replacement here is simple: 
Let's look at life after replacement: 
(1) After substitution, we reduce the terms under the root to a common denominator.
(2) We take it out from under the root.
(3) The numerator and denominator are reduced by . At the same time, under the root, I rearranged the terms in a convenient order. With some experience, steps (1), (2) can be skipped by performing the commented actions orally.
(4) The resulting integral, as you remember from the lesson Integrating Some Fractions, is being decided complete square extraction method. Select a complete square.
(5) By integration we obtain an ordinary “long” logarithm.
(6) We carry out the reverse replacement. If initially , then back: .
(7) The final action is aimed at straightening the result: under the root we again bring the terms to a common denominator and take them out from under the root.
Example 10
Find the indefinite integral 
This is an example for you to solve on your own. Here a constant is added to the lone “X”, and the replacement is almost the same: 
The only thing you need to do additionally is to express the “x” from the replacement being carried out: 
Full solution and answer at the end of the lesson.
Sometimes in such an integral there may be a quadratic binomial under the root, this does not change the method of solution, it will be even simpler. Feel the difference:
Example 11
Find the indefinite integral
Example 12
Find the indefinite integral
Brief solutions and answers at the end of the lesson. It should be noted that Example 11 is exactly binomial integral, the solution method of which was discussed in class Integrals of irrational functions.
Integral of an indecomposable polynomial of the 2nd degree to the power
(polynomial in denominator)
A more rare type of integral, but nevertheless encountered in practical examples.
Example 13
Find the indefinite integral
But let’s return to the example with lucky number 13 (honestly, I didn’t guess correctly). This integral is also one of those that can be quite frustrating if you don’t know how to solve.
The solution starts with an artificial transformation: 
I think everyone already understands how to divide the numerator by the denominator term by term.
The resulting integral is taken in parts: 
For an integral of the form ( – natural number) withdrawn recurrent reduction formula:
, Where 
– integral of a degree lower.
Let us verify the validity of this formula for the solved integral.
In this case: , , we use the formula: 
As you can see, the answers are the same.
Example 14
Find the indefinite integral
This is an example for you to solve on your own. The sample solution uses the above formula twice in succession.
If under the degree is indivisible square trinomial, then the solution is reduced to a binomial by isolating the perfect square, for example:
What if there is an additional polynomial in the numerator? In this case, the method of indefinite coefficients is used, and the integrand function is expanded into a sum of fractions. But in my practice there is such an example never met, so I missed this case in the article Integrals of fractional-rational functions, I'll skip it now. If you still encounter such an integral, look at the textbook - everything is simple there. I don’t think it’s advisable to include material (even simple ones), the probability of encountering which tends to zero.
Integrating complex trigonometric functions
The adjective “complex” for most examples is again largely conditional. Let's start with tangents and cotangents in high degrees. From the point of view of the solving methods used, tangent and cotangent are almost the same thing, so I will talk more about tangent, implying that the demonstrated method for solving the integral is valid for cotangent too.
In the above lesson we looked at universal trigonometric substitution for solutions certain type integrals of trigonometric functions. The disadvantage of universal trigonometric substitution is that its use often results in cumbersome integrals with difficult calculations. And in some cases, universal trigonometric substitution can be avoided!
Let's consider another canonical example, the integral of one divided by sine:
Example 17
Find the indefinite integral
Here you can use universal trigonometric substitution and get the answer, but there is a more rational way. I will provide the complete solution with comments for each step:
(1) We use the trigonometric formula for the sine of a double angle.
(2) We carry out an artificial transformation: Divide in the denominator and multiply by .
(3) Using the well-known formula in the denominator, we transform the fraction into a tangent.
(4) We bring the function under the differential sign.
(5) Take the integral.
Pair simple examples for independent solution:
Example 18
Find the indefinite integral
Note: The very first step should be to use the reduction formula 
and carefully carry out actions similar to the previous example.
Example 19
Find the indefinite integral
Well, this is a very simple example.
Complete solutions and answers at the end of the lesson.
I think now no one will have problems with integrals: 
and so on.
What is the idea of the method? The idea is that, using transformations, trigonometric formulas organize only tangents and the derivative of tangent in the integrand. That is, we're talking about about replacement: 
. In Examples 17-19 we actually used this replacement, but the integrals were so simple that we got by with an equivalent action - subsuming the function under the differential sign.
Similar reasoning, as I already mentioned, can be carried out for the cotangent.
There is also a formal prerequisite for applying the above replacement:
The sum of the powers of cosine and sine is a negative integer EVEN number, For example:
for the integral – a negative integer EVEN number.
! Note : if the integrand contains ONLY a sine or ONLY a cosine, then the integral is also taken for a negative odd degree (the simplest cases are in Examples No. 17, 18).
Let's look at a couple of more meaningful tasks based on this rule:
Example 20
Find the indefinite integral
The sum of the powers of sine and cosine: 2 – 6 = –4 is a negative integer EVEN number, which means that the integral can be reduced to tangents and its derivative: 
(1) Let's transform the denominator.
(2) Using the well-known formula, we obtain .
(3) Let's transform the denominator.
(4) We use the formula 
.
(5) We bring the function under the differential sign.
(6) We carry out replacement. More experienced students may not carry out the replacement, but it is still better to replace the tangent with one letter - there is less risk of getting confused.
Example 21
Find the indefinite integral
This is an example for you to solve on your own.
Hang in there, the championship rounds are about to begin =)
Often the integrand contains a “hodgepodge”:
Example 22
Find the indefinite integral 
This integral initially contains a tangent, which immediately leads to an already familiar thought:
I will leave the artificial transformation at the very beginning and the remaining steps without comment, since everything has already been discussed above.
A couple of creative examples for your own solution:
Example 23
Find the indefinite integral 
Example 24
Find the indefinite integral 
Yes, in them, of course, you can lower the powers of sine and cosine, and use a universal trigonometric substitution, but the solution will be much more efficient and shorter if it is carried out through tangents. Full solution and answers at the end of the lesson
Definite integral. Examples of solutions
Hello again. In this lesson we will examine in detail such a wonderful thing as a definite integral. This time the introduction will be short. All. Because snowstorm outside the window.
In order to learn how to solve definite integrals you need to:
1) Be able to find indefinite integrals.
2) Be able to calculate definite integral.
As you can see, in order to master a definite integral, you need to have a fairly good understanding of “ordinary” indefinite integrals. Therefore, if you are just starting to dive into integral calculus, and the kettle has not yet boiled at all, then it is better to start with the lesson Indefinite integral. Examples of solutions. In addition, there are pdf courses for ultra-fast preparation- if you literally have a day, half a day left.
IN general view the definite integral is written as follows:
What is added compared to the indefinite integral? More limits of integration.
Lower limit of integration
Upper limit of integration is standardly denoted by the letter .
The segment is called segment of integration.
Before we get to practical examples, a small faq on the definite integral.
What does it mean to solve a definite integral? Solving a definite integral means finding a number.
How to solve a definite integral? Using the Newton-Leibniz formula familiar from school:
It is better to rewrite the formula on a separate piece of paper; it should be in front of your eyes throughout the entire lesson.
The steps for solving a definite integral are as follows:
1) First we find the antiderivative function (indefinite integral). Note that the constant in the definite integral not added. The designation is purely technical, and the vertical stick does not carry any mathematical meaning; in fact, it is just a marking. Why is the recording itself needed? Preparation for applying the Newton-Leibniz formula.
2) Substitute the value of the upper limit into the antiderivative function: .
3) Substitute the value of the lower limit into the antiderivative function: .
4) We calculate (without errors!) the difference, that is, we find the number.
Does a definite integral always exist? No not always.
For example, the integral does not exist because the segment of integration is not included in the domain of definition of the integrand (values under the square root cannot be negative). Here's a less obvious example: . Such an integral also does not exist, since there is no tangent at the points of the segment. By the way, who hasn't read it yet? methodological material Graphs and basic properties of elementary functions– the time to do it is now. It will be great to help throughout the course of higher mathematics.
For that for a definite integral to exist at all, it is sufficient that the integrand is continuous on the interval of integration.
From the above, the first important recommendation follows: before you begin solving ANY definite integral, you need to make sure that the integrand function is continuous on the interval of integration. When I was a student, I repeatedly had an incident when I struggled for a long time with finding a difficult antiderivative, and when I finally found it, I racked my brains over another question: “What kind of nonsense did it turn out to be?” In a simplified version, the situation looks something like this:
???! You cannot substitute negative numbers under the root! What the hell is this?! Initial inattention.
If to solve (in test work, on a test, exam) You are offered a non-existent integral like , then you need to give an answer that the integral does not exist and justify why.
Can the definite integral be equal to negative number? Maybe. And a negative number. And zero. It may even turn out to be infinity, but it will already be improper integral, which are given a separate lecture.
Can the lower limit of integration be greater than the upper limit of integration? Perhaps this situation actually occurs in practice.
– the integral can be easily calculated using the Newton-Leibniz formula.
What you can't do without higher mathematics? Of course, without all sorts of properties. Therefore, let's consider some properties of the definite integral.
In a definite integral, you can rearrange the upper and lower limits, changing the sign:
For example, in a definite integral, before integration, it is advisable to change the limits of integration to the “usual” order:
– in this form it is much more convenient to integrate.
– this is true not only for two, but also for any number of functions.
In a definite integral one can carry out replacement of integration variable, however, compared to the indefinite integral, this has its own specifics, which we will talk about later.
For a definite integral the following holds true: integration by parts formula:
Example 1
Solution:
(1) We take the constant out of the integral sign.
(2) We integrate over the table using the most popular formula. It is advisable to separate the emerging constant from and put it outside the bracket. It is not necessary to do this, but it is advisable - why the extra calculations?
. First we substitute the upper limit, then the lower limit. We carry out further calculations and get the final answer.
Example 2
Calculate definite integral
This is an example for you to solve on your own, the solution and answer are at the end of the lesson.
Let's complicate the task a little:
Example 3
Calculate definite integral 
Solution: 
(1) We use the linearity properties of the definite integral.
(2) We integrate according to the table, while taking out all the constants - they will not participate in the substitution of the upper and lower limits.
(3) For each of the three terms we apply the Newton-Leibniz formula: 
THE WEAK LINK in the definite integral is calculation errors and the common CONFUSION IN SIGNS. Be careful! Special attention I focus on the third term: 
– first place in the hit parade of errors due to inattention, very often they write automatically 
(especially when the substitution of the upper and lower limits is carried out verbally and is not written out in such detail). Once again, carefully study the above example.
It should be noted that the considered method of solving a definite integral is not the only one. With some experience, the solution can be significantly reduced. For example, I myself am used to solving such integrals like this:
Here I verbally used the rules of linearity and verbally integrated using the table. I ended up with just one bracket with the limits marked out: 
(unlike three brackets in the first method). And into the “whole” antiderivative function, I first substituted 4, then –2, again performing all the actions in my mind.
What are the disadvantages of the short solution? Everything here is not very good from the point of view of the rationality of calculations, but personally I don’t care - common fractions I count on a calculator.
In addition, there is an increased risk of making an error in the calculations, so it is better for a tea student to use the first method; with “my” method of solving, the sign will definitely be lost somewhere.
However, the undoubted advantages of the second method are the speed of solution, compactness of notation and the fact that the antiderivative is in one bracket.
Advice: before using the Newton-Leibniz formula, it is useful to check: was the antiderivative itself found correctly?
So, in relation to the example under consideration: before substituting the upper and lower limits into the antiderivative function, it is advisable to check on the draft whether the indefinite integral was found correctly? Let's differentiate:
The original integrand function has been obtained, which means that the indefinite integral has been found correctly. Now we can apply the Newton-Leibniz formula.
Such a check will not be superfluous when calculating any definite integral.
Example 4
Calculate definite integral 
This is an example for you to solve yourself. Try to solve it in a short and detailed way.
Changing a variable in a definite integral
For a definite integral, all types of substitutions are valid as for the indefinite integral. Thus, if you are not very good with substitutions, you should carefully read the lesson Substitution method in indefinite integral.
There is nothing scary or difficult in this paragraph. The novelty lies in the question how to change the limits of integration when replacing.
In examples, I will try to give types of replacements that have not yet been found anywhere on the site.
Example 5
Calculate definite integral
Main question here it is not at all about the definite integral, but about how to correctly carry out the replacement. Let's look at table of integrals and figure out what our integrand function looks like most? Obviously, for the long logarithm: 
. But there is one discrepancy, in the table integral under the root, and in ours - “x” to the fourth power. The idea of replacement also follows from the reasoning - it would be nice to somehow turn our fourth power into a square. It's real.
First, we prepare our integral for replacement:
From the above considerations, a replacement quite naturally arises:
Thus, everything will be fine in the denominator: .
We find out what the remaining part of the integrand will turn into, for this we find the differential:
Compared to replacement in the indefinite integral, we add an additional step.
Finding new limits of integration.
It's quite simple. Let's look at our replacement and the old limits of integration, .
First, we substitute the lower limit of integration, that is, zero, into the replacement expression:
Then we substitute the upper limit of integration into the replacement expression, that is, the root of three:
Ready. And just...
Let's continue with the solution.
(1) According to replacement write a new integral with new limits of integration.
(2) This is the simplest table integral, we integrate over the table. It is better to leave the constant outside the brackets (you don’t have to do this) so that it does not interfere with further calculations. On the right we draw a line indicating the new limits of integration - this is preparation for applying the Newton-Leibniz formula.
(3) We use the Newton-Leibniz formula 
.
We strive to write the answer in the most compact form; here I used the properties of logarithms.
Another difference from the indefinite integral is that, after we have made the substitution, there is no need to carry out any reverse replacements.
And now a couple of examples for you to decide for yourself. What replacements to make - try to guess on your own.
Example 6
Calculate definite integral
Example 7
Calculate definite integral
These are examples for you to decide on your own. Solutions and answers at the end of the lesson.
And at the end of the paragraph important points, the analysis of which appeared thanks to site visitors. The first one concerns legality of replacement. In some cases it cannot be done! Thus, Example 6, it would seem, can be solved using universal trigonometric substitution, however, the upper limit of integration ("pi") not included in domain this tangent and therefore this substitution is illegal! Thus, the “replacement” function must be continuous in all points of the integration segment.
In another email, the following question was received: “Do we need to change the limits of integration when we subsume a function under the differential sign?” At first I wanted to “dismiss the nonsense” and automatically answer “of course not,” but then I thought about the reason for such a question and suddenly discovered that there was no information lacks. But it, although obvious, is very important:
If we subsume the function under the differential sign, then there is no need to change the limits of integration! Why? Because in this case no actual transition to new variable. For example: 
And here the summation is much more convenient than the academic replacement with the subsequent “painting” of new limits of integration. Thus, if the definite integral is not very complicated, then always try to put the function under the differential sign! It's faster, it's more compact, and it's commonplace - as you'll see dozens of times!
Thank you very much for your letters!
Method of integration by parts in a definite integral
There is even less novelty here. All calculations of the article Integration by parts in the indefinite integral are fully valid for the definite integral.
There is only one detail that is a plus; in the formula for integration by parts, the limits of integration are added:
The Newton-Leibniz formula must be applied twice here: for the product and after we take the integral.
For the example, I again chose the type of integral that has not yet been found anywhere on the site. The example is not the simplest, but very, very informative.
Example 8
Calculate definite integral
Let's decide. 
Let's integrate by parts: 
Anyone having difficulty with the integral, take a look at the lesson Integrals of trigonometric functions, it is discussed in detail there.
(1) We write the solution in accordance with the formula of integration by parts.
(2) For the product we apply the Newton-Leibniz formula. For the remaining integral we use the properties of linearity, dividing it into two integrals. Don't get confused by the signs!
(4) We apply the Newton-Leibniz formula for the two found antiderivatives.
To be honest, I don't like the formula. 
and, if possible, ... I do without it at all! Let's consider the second solution; from my point of view, it is more rational.
Calculate definite integral
At the first stage I find the indefinite integral:
Let's integrate by parts:
The antiderivative function has been found. There is no point in adding a constant in this case.
What is the advantage of such a hike? There is no need to “carry around” the limits of integration; indeed, it can be exhausting to write down the small symbols of the limits of integration a dozen times
At the second stage I check(usually in draft).
Also logical. If I found the antiderivative function incorrectly, then I will solve the definite integral incorrectly. It’s better to find out immediately, let’s differentiate the answer:
The original integrand function has been obtained, which means that the antiderivative function has been found correctly.
The third stage is the application of the Newton-Leibniz formula:
And there is a significant benefit here! In “my” solution method there is a much lower risk of getting confused in substitutions and calculations - the Newton-Leibniz formula is applied only once. If the teapot solves a similar integral using the formula 
(in the first way), then he will definitely make a mistake somewhere.
The considered solution algorithm can be applied for any definite integral.
Dear student, print and save:
What to do if you are given a definite integral that seems complicated or it is not immediately clear how to solve it?
1) First we find the indefinite integral (antiderivative function). If at the first stage there was a bummer, there is no point in further rocking the boat with Newton and Leibniz. There is only one way - to increase your level of knowledge and skills in solving indefinite integrals.
2) We check the found antiderivative function by differentiation. If it is found incorrectly, the third step will be a waste of time.
3) We use the Newton-Leibniz formula. We carry out all calculations EXTREMELY CAREFULLY - this is the weakest link of the task.
And, for a snack, an integral for independent solution.
Example 9
Calculate definite integral
The solution and the answer are somewhere nearby.
The next recommended lesson on the topic is How to calculate the area of a figure using a definite integral?
Let's integrate by parts:
Are you sure you solved them and got these answers? ;-) And there is porn for an old woman.
