Solving integrals by direct integration method. Methods for calculating indefinite integrals

1. Integral calculus of functions of one variable

2. Antiderivative and indefinite integral.

3. Properties of the indefinite integral.

4. Table of integrals

When studying the differentiation of functions, the task was set - for a given function, find its derivative or differential. Many questions of science and technology lead to the formulation of an inverse problem - for a given function f(x) find such a function F(x), the derivative or differential of which are equal respectively f(x) or f(x)dx.

Definition 1. Function F(x) called antiderivative in relation to function f(x) at some interval (a, b), if on this interval the function F(x) is differentiable and satisfies the equation

F′ (x) = f(x)

or, what is the same, the relation

dF(x) = f(x)dx.

So, for example, the function sin 5 x- antiderivative on any interval with respect to the function f(x) = 5cos5 x, since (sin5 x)′ = 5cos5 x.

It is easy to check that the presence of one antiderivative ensures the presence of such functions in an infinite set. In fact, if F(x)- antiderivative of the function f(x), That

Ф(x) = F(x) + C,

Where WITH- any constant is also antiderivative, since

F′( X) = (F(x) + C)′ = F′( x) + 0 = f(x).

The following theorem gives the answer to the question of how to find all antiderivatives of a given function if one of them is known.

Theorem 1(about primitives). If F(x) − some antiderivative of the function f(x) on the interval ( a, b), then all its antiderivatives have the form F(x) + C, Where WITH- arbitrary constant.

Geometrically y = F(x) + C means that the graph of any antiderivative function is obtained from the graph of the function y = F(x) by simply shifting it parallel to the Oy axis by an amount WITH(see picture). Due to the fact that the same function f(x) has infinitely many antiderivatives, the problem arises of choosing an antiderivative that solves one or another practical problem.

It is known that the derivative of the path with respect to time is equal to the speed of the point: S′( t) = V(t), therefore, if the law of speed change is known V(t), the path of motion of a point is an antiderivative of the speed of the point, i.e. S(t) =F(t) +C.

To find the law of path change S(t) you need to use initial conditions, i.e. know what the distance traveled is S0 at t = t0. Let at t = t0 we have S = S0. Then

S(t 0 ) = S 0 = F(t 0 ) + C. C = S 0 - F(t 0 ) And S(t) = F(t) + S 0 - F(t 0 ).

Definition 2. If F(x)- some antiderivative of the function f(x), then the expression F(x) + C, Where WITH- an arbitrary constant, called indefinite integral and is designated

∫f(x)dx= F(x) + C,

i.e. the indefinite integral of the function f(x) there is a set of all its primitives.

In this case, the function f(x) called integrand, and the work f(x)dx- integrand; F(x)- one of the prototypes; X- integration variable. The process of finding an antiderivative is called integration.

Example 1. Find indefinite integrals:

Theorem 2(existence of an indefinite integral). If the function f(x) continuous on (a, b) , then there is an antiderivative, and hence an integral ∫ f(x)dx.

Properties of indefinite integrals:

1. (∫f(x)dx)′ = f(x) , i.e. the derivative of the indefinite integral is equal to the integrand.

2. d(∫f(x)dx) = f(x)dx, i.e. the differential of the indefinite integral is equal to the integrand.

3. ∫dF(x) = F(x) + C.

4. ∫(C 1 f 1(x) + C 2 f 2 (x))dx= C 1∫f 1(x)dx+ C 2∫f 2(x)dx− property of linearity; C1, C2- permanent.

5. If ∫ f(x)dx= F(x) + C, That

The first three properties follow from the definition of an indefinite integral. We obtain properties 4 and 5 by differentiating the left and right sides of the equations with respect to X, using property 1 of integrals and properties of derivatives.

Example 2. Find the indefinite integral: a) ∫( e x+cos5 x)dx.

Solution. Using properties 4 and 5, we find:

Let us present a table of basic integrals, which plays the same role in higher mathematics as the multiplication table in arithmetic.

Basic integration methods

There are three main integration method.

1. Direct integration− calculation of integrals using a table of integrals and the basic properties of indefinite integrals.

Example 3. Calculate the integral: ∫ tg 2 xdx.

2. Substitution method . In many cases, the introduction of a new integration variable allows one to reduce the calculation of a given integral to finding a tabular one. This method is also called variable replacement method.

Theorem 3. Let the function x = φ(t) defined, continuous and differentiable on a certain interval T let it go X- the set of values ​​of this function, on it, i.e. on T complex function defined f(φ(t)). Then if ∫ f(x)dx= F(x)+ C, That

∫f(x)dx=∫f(φ(t)) φ ′ (t)dt. (1)

Formula (1) is called formula changing a variable in an indefinite integral.

Comment. After calculating the integral ∫ f(φ(t)) φ ′ (t)dt you need to go back to the variable X.

Example 4. Find the integral: ∫cos 3 x sin xdx.

a) Replace sin xdx on (− d cos x), i.e. we introduce the function cos x under the differential sign. We get

3. Method of integration by parts

Theorem 4. Let the functions u(x) And v(x) defined and differentiable on a certain interval X let it go u′ (x)v(x) has an antiderivative on this interval, i.e. there is an integral ∫ u′( x)v(x)dx.

Then on this interval the function has an antiderivative and u(x)v′ (x) and the formula is correct:

∫u(x)v′( x)dx= u(x)v(x) −∫v(x)u′( x)dx(2)

∫udv= uv−∫vdu.(2′)

Formulas (2) and (2′) are called formulas for integration by parts in the indefinite integral.

Using the method of integration by parts, integrals of the following functions are calculated: P(x)arcsin( ax),P(x)arccos( ax), P(x)arctg( ax), P(x)arcctg( ax),P(x)ln x, P(x)e kx, P(x)sin kx, P(x)cos kx, Here P(x)- polynomial; e ax cos bx, e ax sin bx.

Of course, these functions do not exhaust all the integrals that are calculated using the method of integration by parts.

Example 6. Find the integral: ∫ arctg 3xdx.

Solution. Let's put u= arctg 3x; dv= dx. Then

According to formula (2) we have

Direct integration

Calculation of indefinite integrals using a table of integrals and their basic properties is called direct integration.

Example 1. Let's find the integral

.

Applying the second and fifth properties of the indefinite integral, we obtain

.(*)

Next, using the formulasII, Sh,IV, VIIItables and the third property of integrals, we find each of the terms of the integrals separately:

= ,

,

Let us substitute these results into (*) and, denoting the sum of all constants(3 WITH 1 +7WITH 2 +4WITH 3 +2WITH 4 +WITH 5) letter WITH, we finally get:

Let's check the result by differentiation. Let's find the derivative of the resulting expression:

We have obtained the integrand, this proves that the integration was carried out correctly.

Example 2 . We'll find

.

The table of integrals shows the corollaryIIIA from the formula III:

To use this corollary, we find the differential of a function in the exponent:

To create this differential, it is enough to multiply the denominator of the fraction under the integral by the number 2 (obviously, in order for the fraction not to change, it is necessary to multiply by 2 and numerator). After placing the constant factor outside the integral sign, it becomes ready to apply the tabular formulaIIIA:

.

Examination:

therefore, the integration is done correctly.

Example 3 . We'll find

Since the differential of a quadratic function can be constructed from the expression in the numerator, the following function should be distinguished in the denominator:

.

To create its differential it is enough to multiply the numerator by 4 (we also multiply the denominator by 4 and take this factor of the denominator out of the integral). As a result, we will be able to use the tabular formulaX:

Examination:

,

those. the integration was performed correctly.

Example 4 . We'll find

Note that now the quadratic function whose differential can be created in the numerator, is a radical expression. Therefore, it would be reasonable to write the integrand as a power function in order to use the formulaItables of integrals:

Examination:

Conclusion: the integral was found correctly.

Example 5. We'll find

Let us note that the integrand contains

function ; and its differential. But the fraction is also the differential of the entire radical expression (up to sign):

Therefore, it is reasonable to represent the fraction in the form degrees:

Then after multiplying the numerator and denominator by (-1) we get a power integral (tabular formulaI):

By differentiating the result, we make sure that the integration is performed correctly.

Example 6. We'll find

It is easy to see that in this integral from the expression the differential of the radical function cannot be obtained using numerical coefficients. Really,

,

Where k -constant. But, from experience example 3 , it is possible to construct an integral that is identical in form to the formulaXfrom the table of integrals:

Example 7. We'll find

Let us pay attention to the fact that the differential of a cubic function can easily be created in the numeratord(x 3 ) = 3 x 2 dx. After which we get the opportunity to use the tabular formulaVI:

Example 8. We'll find

It is known that the derivative of the function arcsin x is a fraction

Then

.

This leads us to the conclusion that the required integral has the form of a power integral: , in whichand = arcsin x, which means

Example 9 . To find

let's use the same table formula I and the fact that

We get

Example 10 . We'll find

Since the expression is the differential of the function, then, using the formula I tables of integrals, we get

Example 11. To find the integral

let's use sequentially: the trigonometric formula

,

by the fact that

and formula IItables of integrals:

Example 12 . We'll find

.

Since the expression

is the differential of the function , then using the same formulaII, we get

Example 13 . Let's find the integral

Note that the degree of the variable in the numerator is one less than in the denominator. This allows us to create a differential in the numeratordenominator. We'll find

.

After taking the constant factor out of the integral sign, we multiply the numerator and denominator of the integrand by (-7), we get:

(The same formula was used hereIIfrom the table of integrals).

Example 14. Let's find the integral

.

Let's imagine the numerator in a different form: 1 + 2 X 2 = (1 + X 2 )+ x 2 and perform term-by-term division, after which we use the fifth property of integrals and formulasI And VIII tables:

Example 15. We'll find

Let’s take the constant factor beyond the sign of the integral, subtract and add 5 to the numerator, then divide the numerator term by term by the denominator and use the fifth property of the integral:

To calculate the first integral, we use the third property of integrals, and present the second integral in a form convenient for applying the formulaIX:

Example 16. We'll find

Note that the exponent of the variable in the numerator is one less than in the denominator (which is typical for a derivative), which means that the differential of the denominator can be constructed in the numerator. Let's find the differential of the expression in the denominator:

d(x 2- 5)=(X 2 - 5)" dx = 2 xdx.

To obtain a constant factor of 2 in the numerator of the denominator differential, we need to multiply and divide the integrand by 2 and take out the constant factor -

for the integral sign

here we usedIItable integral.

Let's consider a similar situation in the following example.

Example 17. We'll find

.

Let's calculate the differential of the denominator:

.

Let's create it in the numerator using the fourth property of integrals:

=

A more complex similar situation will be considered in example 19.

Example 18 We'll find

.

Let us select a complete square in the denominator:

We get

.

After isolating the perfect square in the denominator, we obtained an integral close in form to the formulasVIII And IXtables of integrals, but in the denominator of the formulaVIIIthe terms of the complete squares have the same signs, and in the denominator of our integral the signs of the terms are different, although they do not coincide with the signs of the ninth formula. Achieve complete coincidence of the signs of the terms in the denominator with the signs in the formulaIXis possible by adding a coefficient (-1) to the integral. So, to apply the formulaIXtables of integrals, we will carry out the following activities:

1) put (-1) outside the brackets in the denominator and then outside the integral;

2) find the differential of the expression

3) create the found differential in the numerator;

4) imagine the number 2 in a form convenient for applying the formulaIX tables:

Then

Using IXformula of the table of integrals, we get

Example 19. We'll find

.

Using the experience gained in finding integrals in the previous two examples and the results obtained in them, we will have

.

Let us summarize some of the experience gained as a result of the solution examples 17,18,19.

So, if we have an integral of the form

(example 18 ), That, by isolating the complete square in the denominator, you can arrive at one of the tabular formulasVIII or IX.

The integral is of the form

(example 19 ) after creating the derivative of the denominator in the numerator, it splits into two integrals: the first one is of the form

( example 17 ), taken from the formulaP, and the second type

(example 18 ), taken from one of the formulasVIII or IX.

Example 20 . We'll find

.

Integral of the form

can be reduced to the form of tabular formulasX or XI, highlighting a complete square in the radical expression. IN in our case

= .

The radical expression has the form

The same is always done when calculating integrals of the form

,

if one of the exponents is a positive odd number and the second is an arbitrary real number (example 23 ).

Example 23 . We'll find

Using the experience of the previous example and the identity

2 sin 2 φ = l - cos 2 φ ,2 cos 2 φ = l + cos 2 φ

Substituting the resulting sum into the integral, we get

Since now we will only talk about the indefinite integral, for the sake of brevity we will omit the term “indefinite”.

In order to learn how to calculate integrals (or, as they say, integrate functions), you first need to learn the table of integrals:

Table 1. Table of integrals

2. ( ), u>0. 2a. (α=0); 2b. (α=1); 2c. (α= ). 3. 3a. 4. 5. 5a) 6a. 7. 7a.

8. 9. 10. 10a. 11. 11a. 12. 13. 13a.

In addition, you will need the ability to calculate the derivative of a given function, which means you need to remember the rules of differentiation and the table of derivatives of basic elementary functions:

Table 2. Table of derivatives and differentiation rules:

6.a .

(sin And) = cos And  And (cos u) = – sin And And

We also need the ability to find the differential of a function. Recall that the differential of the function
find by formula
, i.e. the differential of a function is equal to the product of the derivative of this function and the differential of its argument. It is useful to keep in mind the following known relationships:

Table 3. Differential table

1. (b= Const) 2. ( ) 3. 4. 5. (b= Const) 6. 7. 8. 9.

10. 11. 12. 14. 15. 16. 17.

Moreover, these formulas can be used either by reading them from left to right or from right to left.

Let us consider sequentially the three main methods of calculating the integral. The first of them is called by direct integration method. It is based on the use of the properties of the indefinite integral and includes two main techniques: expansion of an integral into an algebraic sum simpler and subscribing to the differential sign, and these techniques can be used both independently and in combination.

A) Let's consider algebraic sum expansion– this technique involves the use of identical transformations of the integrand and the linearity properties of the indefinite integral:
And .

Example 1. Find the integrals:

A)
; b)
;

V)
G)

d)
.

Solution.

A)Let's transform the integrand by dividing the numerator term by term:

The property of powers is used here:
.

b) First, we transform the numerator of the fraction, then we divide the numerator term by term by the denominator:

The property of degrees is also used here:
.

The property used here is:
,
.

.

Formulas 2 and 5 of Table 1 are used here.

Example 2. Find the integrals:

A)
; b)
;

V)
G)

d)
.

Solution.

A)Let's transform the integrand using the trigonometric identity:

.

Here we again use term-by-term division of the numerator by the denominator and formulas 8 and 9 of Table 1.

b) We transform similarly, using the identity
:

.

c) First, divide the numerator term by term by the denominator and take the constants out of the integral sign, then use the trigonometric identity
:

d) Apply the formula for reducing the degree:

,

e) Using trigonometric identities, we transform:

B) Let's consider the integration technique, which is called p by placing it under the differential sign. This technique is based on the invariance property of the indefinite integral:

If
, then for any differentiable function And = And(X) occurs:
.

This property allows us to significantly expand the table of simple integrals, since due to this property the formulas in Table 1 are valid not only for the independent variable And, but also in the case when And is a differentiable function of some other variable.

For example,
, but also
, And
, And
.

Or
And
, And
.

The essence of the method is to isolate the differential of a certain function in a given integrand so that this isolated differential, together with the rest of the expression, forms a tabular formula for this function. If necessary, during such a conversion, constants can be added accordingly. For example:

(in the last example written ln(3 + x 2) instead of ln|3 + x 2 | , since the expression is 3 + x 2 is always positive).

Example 3. Find the integrals:

A)
; b)
; V)
;

G)
; d)
; e)
;

and)
; h)
.

Solution.

A) .

Formulas 2a, 5a and 7a of Table 1 are used here, the last two of which are obtained precisely by subsuming the differential sign:

Integrate view functions
occurs very often within the framework of calculating integrals of more complex functions. In order not to repeat the steps described above each time, we recommend that you remember the corresponding formulas given in Table 1.

.

Formula 3 of Table 1 is used here.

c) Similarly, taking into account that , we transform:

.

Formula 2c in Table 1 is used here.

G)

.

d) ;

e)

.

and) ;

h)

.

Example 4. Find the integrals:

A)
b)

V)
.

Solution.

a) Let's transform:

Formula 3 of Table 1 is also used here.

b) We use the formula for reducing the degree
:

Formulas 2a and 7a of Table 1 are used here.

Here, along with formulas 2 and 8 of Table 1, the formulas of Table 3 are also used:
,
.

Example 5. Find the integrals:

A)
; b)

V)
; G)
.

Solution.

a) Work
can be supplemented (see formulas 4 and 5 of Table 3) to the differential of the function
, Where A And b– any constants,
. Indeed, from where
.

Then we have:

.

b) Using formula 6 of table 3, we have
, and
, which means the presence in the integrand of the product
means a hint: under the differential sign you need to enter the expression
. Therefore we get

c) Same as in point b), the product
can be extended to differential functions
. Then we get:

.

d) First we use the linearity properties of the integral:

Example 6. Find the integrals:

A)
; b)
;

V)
; G)
.

Solution.

A)Considering that
(formula 9 of table 3), we transform:

b) Using formula 12 of table 3, we get

c) Taking into account formula 11 of table 3, we transform

d) Using formula 16 of Table 3, we obtain:

.

Example 7. Find the integrals:

A)
; b)
;

V)
; G)
.

Solution.

A)All integrals presented in this example have a common feature: The integrand contains a quadratic trinomial. Therefore, the method of calculating these integrals will be based on the same transformation - isolating the complete square in this quadratic trinomial.

.

b)

.

V)

G)

The method of substituting a differential sign is an oral implementation of a more general method of calculating an integral, called the substitution method or change of variable. Indeed, each time, selecting a suitable formula in Table 1 for the one obtained as a result of subsuming the function differential sign, we mentally replaced the letter And function introduced under the differential sign. Therefore, if integration by subsuming the differential sign does not work out very well, you can directly change the variable. More details about this in the next paragraph.

The problem of finding an antiderivative function does not always have a solution, while we can differentiate any function. This explains the lack of a universal integration method.

In this article we will look at the basic methods for finding the indefinite integral using examples with detailed solutions. We will also group the types of integrand functions characteristic of each integration method.

Page navigation.

Direct integration.

Undoubtedly, the main method of finding an antiderivative function is direct integration using a table of antiderivatives and the properties of the indefinite integral. All other methods are used only to reduce the original integral to tabular form.

Example.

Find the set of antiderivatives of the function.

Solution.

Let's write the function in the form .

Since the integral of a sum of functions is equal to the sum of integrals, then

The numerical coefficient can be taken out of the integral sign:

The first of the integrals is reduced to tabular form, therefore, from the table of antiderivatives for the exponential function we have .

To find the second integral, we use the table of antiderivatives for the power function and the rule . That is, .

Hence,

Where

Integration by substitution method.

The essence of the method is that we introduce a new variable, express the integrand through this variable, and as a result we arrive at a tabular (or simpler) form of the integral.

Very often the substitution method comes to the rescue when integrating trigonometric functions and functions with radicals.

Example.

Find the indefinite integral .

Solution.

Let's introduce a new variable. Let's express x through z:

We substitute the resulting expressions into the original integral:

From the table of antiderivatives we have .

It remains to return to the original variable x:

Answer:

Very often the substitution method is used when integrating trigonometric functions. For example, using the universal trigonometric substitution allows you to transform the integrand into a fractionally rational form.

The substitution method allows you to explain the integration rule .

We introduce a new variable, then

We substitute the resulting expressions into the original integral:

If we accept and return to the original variable x, we get

Submitting the differential sign.

The method of subsuming the differential sign is based on reducing the integrand to the form . Next, the substitution method is used: a new variable is introduced and after finding the antiderivative for the new variable, we return to the original variable, that is

For convenience, place it in front of your eyes in the form of differentials to make it easier to convert the integrand, as well as a table of antiderivatives to see what form to convert the integrand to.

For example, let's find the set of antiderivatives of the cotangent function.

Example.

Find the indefinite integral.

Solution.

The integrand can be transformed using trigonometry formulas:

Looking at the table of derivatives, we conclude that the expression in the numerator can be subsumed under the differential sign , That's why

That is .

Let it be then . From the table of antiderivatives we see that . Returning to the original variable .

Without explanation, the solution is written as follows:

Integration by parts.

Integration by parts is based on representing the integrand as a product and then applying the formula. This method is a very powerful integration tool. Depending on the integrand, the method of integration by parts sometimes has to be applied several times in a row before obtaining the result. For example, let's find the set of antiderivatives of the arctangent function.

Example.

Calculate the indefinite integral.

Solution.

Let it be then

It should be noted that when finding the function v(x) do not add an arbitrary constant C.

Now we apply the integration by parts formula:

We calculate the last integral using the method of subsuming it under the differential sign.

Since then . That's why

Hence,

Where .

Answer:

The main difficulties in integrating by parts arise from the choice: which part of the integrand to take as the function u(x) and which part as the differential d(v(x)). However, there are a number of standard recommendations, which we recommend that you familiarize yourself with in the section integration by parts.

When integrating power expressions, for example or , use recurrent formulas that allow you to reduce the degree from step to step. These formulas are obtained by successive repeated integration by parts. We recommend that you familiarize yourself with the section integration using recurrence formulas.

In conclusion, I would like to summarize all the material in this article. The basis of the fundamentals is the method of direct integration. The methods of substitution, substitution under the differential sign and the method of integration by parts make it possible to reduce the original integral to a tabular one.

This method comes down to integrating the differential equation of the curved axis of the beam (9.1) with the known law of change in bending moments M(X). Assuming the bending rigidity of the beam to be constant (EJ z= const) and sequentially integrating equation (9.1), we obtain

In expressions (9.5) and below, to simplify the notation, the indices of the moments of inertia and bending moments are omitted.

Expressions (9.5) allow us to obtain analytical laws for changes in deflections and rotation angles in a beam. Integration constants included in (9.5) C 1 and C 2 are subject to determination from the kinematic boundary conditions and the conditions for connecting sections of the beam.

Kinematic boundary conditions reflect the nature of the fastening (support) of the beam and are set relative to deflections and rotation angles. For example, for a simply supported beam (Fig. 9.4), the boundary conditions characterize the absence of deflections on the supports: x = 0, x = /, v = 0. For a cantilever beam (Fig. 9.5), the boundary conditions characterize the equality of the deflection and the angle of rotation in the rigid embedment to zero: x = 0, v= 0; av = 0.

Matching conditions are set at the boundaries of sections with different laws of change in bending moments. In the absence of intermediate hinges and so-called parallelogram mechanisms (sliders), the mating conditions consist in the equality of deflections and rotation angles in the sections to the left and right of the boundary of the sections, that is, they characterize the continuity and smoothness of the curved axis of the beam. For example, for the beam in Fig. 9.4 can be written: X = A, and = and

In the presence of P sections with different laws of change in bending moments, the expression for deflection will contain 2 P integration constants. Using boundary conditions and conditions for connecting sections, we can obtain system 2 P linear algebraic equations with respect to these constants. After determining all integration constants, the laws of change u(x) and ср(х) within each section of the beam will be established. Let's look at examples of determining deflections and rotation angles in beams using the direct integration method.

Example 9.1. Let us determine analytical expressions for u(lc) and cp(x) in a cantilever beam loaded with a uniformly distributed load (Fig. 9.6), and calculate the values ​​of these quantities at the free end.

The bending moment in the beam along its entire length varies according to the law of a square parabola:

Let's substitute this expression into solution (9.5) and integrate it:

Using the boundary conditions, we determine the integration constants:

Let us write down the final expressions for deflections and rotation angles in the beam and determine the values ​​of these quantities at the free end:

Example 9.2. For a simply supported beam loaded at the end with a concentrated force (Fig. 9.7), we define expressions for y(x) and (p(x) and calculate the values ​​of these quantities in characteristic sections.

Diagram M shown in Fig. 9.7. Bending moments have different laws of change in the first and second sections of the beam. We integrate the differential equation of the curved axis within each section.

First section (0 2a):

Second section (2 A

To determine the four integration constants C, C 2, D x And D 2 we set boundary conditions and conditions for connecting sections:

From the condition for conjugating the sections, we obtain the equality of the integration constants in the first and second sections: C ( = D v C 2 = D T Using boundary conditions, we find the values ​​of the constants:

Let us write down the final expressions for u(x) and cp(x) within each section:

In these expressions, a vertical bar with a number at the bottom corresponds to the boundary of each area. Within the first section v and cp are determined by the functions up to the vertical line with the number 1, and within the second section - up to the vertical line with the number 2, that is, by all functions.

Let's calculate v and (p in characteristic sections of the beam:

Within the first section, the sign of the rotation angle changes to the opposite. Let’s set the position of the section where the rotation angle becomes zero:

In section x =x Q the deflection of the beam has an extremum. We calculate its value:

For comparison, we determine the amount of deflection of the beam in the middle of the span:

It can be noted that the extreme deflection differs very slightly (by 2.6%) from the deflection in the middle of the span.

Let's perform a numerical calculation at P= 20 kN and A= 1.6 m. Let us select the section of the beam in the form of a rolled steel I-beam, taking the load reliability factor y^= 1.2, operating conditions coefficient y c = 1, design resistance of the material R= 210 MPa = = 21 kN/cm 2 and modulus of elasticity of steel E- 2.1 10 4 kN/cm 2 .

We accept 120, W z = 184 cm 3, J= 1840 cm 4.

Let's calculate the largest values ​​of the angle of rotation and deflection in the beam. According to SNiP, we carry out calculations based on the effect of standard loads.

From the considered example it is clear that if there are several sections in the beam with different laws of change in bending moments, the direct integration method becomes cumbersome and inconvenient.