The integration formula in parts has the form:

.

The method of integration in parts consists in applying this formula. In practical applications, it is worth noting that u and v are functions of the integration variable. Let the integration variable be denoted as x (the character after the differential sign d at the end of the integral record) . Then u and v are functions of x: u (x) and v (x).

Then

, .

And the integration formula in parts takes the form:

.

That is, the integrand must consist of the product of two functions:

,

one of which is denoted as u: g (x) = u, and the other must calculate the integral (more precisely, the antiderivative):

then dv = f (x) dx.

In some cases, f (x) = 1. That is, in the integral

,

we can put g (x) = u, x = v.

So, in this method, the integration formula in parts should be remembered and applied in two ways:

,

.

## How to decide?

The process of solving integrals in science called "mathematics" is called integration. Using integration, you can find some physical quantities: area, volume, mass of bodies and much more.

Integrals are indefinite and definite. Consider the form of a certain integral and try to understand its physical meaning. It appears in this form: $$ int ^ a _b f (x) dx $$. A distinctive feature of writing a certain integral from an indefinite one is that there are limits of integration a and b. Now we find out why they are needed, and what a definite integral does mean. In the geometric sense, such an integral is equal to the area of the figure bounded by the curve f (x), the lines a and b, and the axis Ox.

It can be seen from Fig. 1 that a certain integral is the very area that is grayed out. Let's check it out with a simple example. We find the area of the figure in the image presented below by integration, and then calculate it in the usual way of multiplying length by width.

It can be seen from Fig. 2 that $ y = f (x) = 3 $, $ a = 1, b = 2 $. Now we substitute them into the definition of the integral, we get that $$ S = int _a ^ bf (x) dx = int _1 ^ 2 3 dx = $$ $$ = (3x) Big | _1 ^ 2 = (3 cdot 2) - (3 cdot 1) = $$ $$ = 6-3 = 3 text<ед>^ 2 $$ We do the verification in the usual way. In our case, length = 3, shape width = 1. $$ S = text <длина> cdot text <ширина>= 3 cdot 1 = 3 text<ед>^ 2 $$ As you can see, everything coincided perfectly.

The question arises: how to solve the indefinite integrals and what is their meaning? The solution to such integrals is to find primitive functions. This process is the opposite of finding a derivative. In order to find the primitive, you can use our help in solving problems in mathematics, or you need to independently unmistakably memorize the properties of the integrals and the integration table of the simplest elementary functions. Finding looks like this $$ int f (x) dx = F (x) + C text <где>F (x) $ is the antiderivative of $ f (x), C = const $.

To solve the integral, we need to integrate the function $ f (x) $ with respect to the variable. If the function is tabular, then the response is written in the appropriate form. If not, then the process is reduced to obtaining a tabular function from the function $ f (x) $ by means of cunning mathematical transformations. There are various methods and properties for this, which we will consider later.

### Integrals containing the logarithm and inverse trigonometric (hyperbolic) functions

In parts, integrals are often integrated that contain the logarithm and inverse trigonometric or hyperbolic functions. Moreover, the part that contains the logarithm or inverse trigonometric (hyperbolic) functions is denoted by u, the remaining part - by dv.

Here are examples of such integrals, which are calculated by the method of integration by parts:

, , , , , , .

A detailed solution of these integrals >>>

##### Detailed solution

Here the integrand contains the logarithm. Making substitutions

u = ln x,

dv = x 2 dx.

Then

,

.

We compute the remaining integral:

.

Then

.

At the end of the calculations, it is necessary to add the constant C, since the indefinite integral is the set of all primitives. It could also be added in intermediate calculations, but this would only clutter up the calculations.

## Integral Properties

- Removing constants from under the integral sign: $$ $$ $$ int Cg (x) dx = C int g (x) dx $$
- The integral of the sum / difference of two functions is equal to the sum / difference of the integrals of these functions: $$ int (f (x) pm g (x)) dx = int f (x) dx pm int g (x) dx $$
- Change of direction of integration: $$ int _a ^ b f (x) = - int _b ^ a f (x) dx $$
- Dividing the integration interval: $$ int_a ^ b f (x) dx = int_a ^ c f (x) dx + int_c ^ b f (x) dx $$ $$ c in (a, b) $$

So, now we will make an algorithm how to solve integrals for dummies?

- We learn a certain integral or not.
- If it is indefinite, then we need to find the antiderivative function $ F (x) $ of the integrand $ f (x) $ using mathematical transformations leading to the tabular form of the function $ f (x) $.
- If certain, then you need to perform step 2, and then substitute the limits $ a $ and $ b $ in the antiderivative function $ F (x) $. You will find out what formula to do in the article "Newton's Leibniz Formula".

## Solution Examples

$$ int x dx = frac

This integral contains a table function under its sign, which means that you can immediately record the answer taken from the table.

Example 1 |

$$ int x dx $$ |

Decision |

Answer |

$$ int x dx = frac |

$$ int 3xdx = 3 int xdx = frac <3x ^ 2> <2> + C $$

We notice that under the sign of the integral there is a constant 3. By the first property, it can be taken out of the integral icon. Further, we see that the integrand is tabular and we get from it the antiderivative for f (x) = x.

Example 2 |

$$ int 3xdx $$ |

Decision |

Answer |

$$ int 3xdx = frac <3x ^ 2> <2> + C $$ |

Having analyzed the indefinite integral, we noticed that the integrands are tabular. And their sum is given. You can use property number 2. Therefore, we perform operations on the functions $ f (x) $ and $ g (x) $ according to the transformations indicated in the plate. Since the integral is indefinite, we get the antiderivative in the answer.

Example 3 |

$$ int (x ^ 3 + frac <1> <2 sqrt |

Decision |

Answer |

$$ int (x ^ 3 + frac <1> <2 sqrt |

So, you have learned how to solve integrals for dummies, examples of solving integrals have been sorted into shelves. Learned their physical and geometric meaning. The methods of solution will be described in other articles.

## The concept of a definite integral and the Newton-Leibniz formula

** Definite integral** from continuous function

*f*(

*x*) on a finite segment [

*a*,

*b*] (where) is called the increment of some of its antiderivatives on this segment. (In general, understanding will be much easier if we repeat the topic of the indefinite integral)

As can be seen in the graphs below (the increment of the antiderivative function is indicated), **a certain 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*)).

The numbers *a* and *b* are called the lower and upper limits of integration, respectively, and the segment [*a*, *b*] - a segment of integration.

So if *F*(*x*) Is some primitive function for *f*(*x*), then, by definition,

(38)

Equality (38) is called ** Newton-Leibniz formula**. Difference

*F*(

*b*) –

*F*(

*a*) are briefly written as follows:

Therefore, the Newton-Leibniz formula will be written as follows:

(39)

Let us prove that a certain integral does not depend on which antiderivative of the integrand is taken in calculating it. Let be *F*(*x*) and Ф (*x*) Are arbitrary antiderivatives of the integrand. Since these are primitives of the same function, they differ by a constant term: Φ (*x*) = *F*(*x*) + *C*. therefore

Thus, it was established that on the segment [*a*, *b*] increments of all primitive functions *f*(*x*) match up.

Thus, to calculate a certain integral, it is necessary to find any antiderivative of the integrand, i.e. you must first find the indefinite integral. Constant *WITH* excluded from subsequent calculations. Then the Newton-Leibniz formula is applied: the value of the upper limit is substituted into the antiderivative function *b* , further - the value of the lower limit *a* and the difference is calculated *F (b) - F (a)* . The resulting number will be a certain integral. .

At *a* = *b* by definition accepted

In order to practice finding certain integrals, it will take **table of basic indefinite integrals** and allowance "**Degrees and Root Actions**".

**Example 1** Calculate a definite integral

Decision. First we find the indefinite integral:

Applying the Newton-Leibniz formula to the antiderivative

(at *WITH* = 0), we obtain

However, when calculating a certain integral, it is better not to find the antiderivative separately, but immediately write the integral in the form (39).

**Example 2** Calculate a definite integral

Decision. Using the formula

## Properties of a definite integral

**Theorem 1***A definite integral with the same limits of integration is equal to zero*, i.e.

This property is contained in the definition of a definite integral. However, it can also be obtained by the Newton-Leibniz formula:

**Theorem 2***The value of a certain integral does not depend on the designation of the integration variable*, i.e.

(40)

Let be *F*(*x*) - antiderivative for *f*(*x*) For *f*(*t*) primitive is the same function *F*(*t*), in which the independent variable is only indicated differently. Consequently,

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 a certain integral*, i.e.

(41)

**Theorem 4***The definite integral of the algebraic sum of a finite number of functions is equal to the algebraic sum of the definite integrals of these functions*, i.e.

(42)

**Theorem 5***If the integration segment is divided into parts, then a certain integral over the entire segment is equal to the sum of certain integrals in its parts*, i.e. if a

(43)

**Theorem 6***When rearranging the limits of integration, the absolute value of a certain 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 by 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 a certain integral is non-negative (positive), i.e. if a*

**Theorem 9.***If the upper limit of integration is greater than the lower one and the functions are continuous, then the inequality*

*can integrate term by term*, i.e.

(46)

The properties of a certain integral make it possible to simplify the direct calculation of integrals.

**Example 5** Calculate a definite integral

Using Theorems 4 and 3, and when finding the antiderivatives - the table integrals (7) and (6), we obtain

## Defined integral with variable upper limit

Let be *f*(*x*) - continuous on the segment [*a*, *b*] function, and *F*(*x*) - its antiderivative. Consider a definite integral

(47)

,

and through *t* the integration variable is indicated 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 primitive for *f*(*x*) = *f*(*t*) Indeed, differentiating *F*(*x*), we get

Function *F*(*x*) Is one of an infinite number of primitives for *f*(*x*), namely the one that *x* = *a*vanishes. This statement is obtained if in equality (48) we put *x* = *a*and use Theorem 1 of the previous section.

## Calculation of certain integrals by the method of integration by parts and by the method of variable replacement

In deriving the integration formula in parts, we obtained the equality *u dv* = *d* (*uv*) – *v du*. Integrating it within *a* before *b* and taking into account Theorem 4 of the section of this article on the properties of a certain integral, we obtain

As follows from Theorem 2 of the section on the properties of the indefinite integral, the first term on the right-hand side is equal to the difference of the values of the product *uv* with the upper and lower limits of integration. Having written this difference briefly in the form

we get the formula **integration by parts** to calculate a certain integral:

(49)

**Example 6** Calculate a definite integral

Decision. Integrate by parts, setting *u* = ln *x* , *dv* = *dx* then *du* = (1/*x*)*dx* , *v* = *x* . By the formula (49) we find

### Find a certain integral in parts by yourself, and then see the solution

**Example 7** Find a definite integral

.

**Example 8** Find a definite integral

.

We proceed to the calculation of a certain integral **variable replacement method**. Let be

where, by definition, *F*(*x*) - antiderivative for *f*(*x*) If in the integrand to replace the variable

then in accordance with formula (16) we can write

In this expression

antiderivative function for

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

(50)

This is the formula for the transition to a new variable under the sign of a certain integral. With its help, a definite integral

after replacing a variable

converts to a certain integral with respect to the new variable *t*. Moreover, the old limits of integration *a* and *b* replaced by new limits and. To find new limits, you need to equation

put values *x* = *a*and *x* = *b*, i.e. solve equations

relative to and. After finding new limits of integration, the calculation of a certain integral reduces to applying the Newton-Leibniz formula to the integral of a new variable *t*. In the primitive function, which is obtained by finding the integral, there is no need to return to the old variable.

When calculating a certain integral by the variable replacement method, it is often convenient to express not the old variable as a function of the new one, but, on the contrary, the new one as a function of the old one.

**Example 9** Calculate a definite integral

Decision. We replace the variable by setting

Then *dt* = 2*x dx* from where *x dx* = (1/2) *dt* , and the integrand is converted like this:

We will find new limits of integration. Value substitution *x* = 4 and *x* = 5 in the equation

Using now formula (50), we obtain

After replacing the variable, we did not return to the old variable, but applied the Newton-Leibniz formula to the obtained antiderivative.