home PC software Variable substitution method in a fraction. The method of changing a variable in an indefinite integral. Solution examples. Basic properties of a definite integral

Variable substitution method in a fraction. The method of changing a variable in an indefinite integral. Solution examples. Basic properties of a definite integral

When calculating certain integrals using the Newton-Leibniz formula, it is preferable not to strictly distinguish between the stages of solving the problem (finding the antiderivative integrand, finding the increment of the antiderivative). This approach, which uses, in particular, the formulas for the change of variable and integration by parts for a certain integral, usually makes it possible to simplify the writing of the solution.

THEOREM. Let the function φ(t) have a continuous derivative on the segment [α,β], a=φ(α), b=φ(β) and the function f(x) be continuous at each point x of the form x=φ(t), where t[α,β].

Then the following equality is true:

This formula is called the change of variable formula in a definite integral.

Just as it was in the case of the indefinite integral, the use of a change of variable allows you to simplify the integral, bringing it closer to the tabular (tabular). In this case, unlike the indefinite integral, in this case there is no need to return to the original integration variable. It is enough to find the limits of integration of α and β with respect to the new variable t as a solution to the equations φ(t)=а and φ(t)=в with respect to the variable t. In practice, when changing a variable, one often starts by specifying the expression t=ψ(x) of the new variable through the old one. In this case, finding the limits of integration with respect to the variable t is simplified: α=ψ(a), β=ψ(c).

Example 19. Calculate

Let's put t=2-x 2 . Then dt=d(2-x 2)=(2-x 2)"dx=-2xdx and xdx=-dt. If x=0, then t=2-0 2 =2, and if x=1, then t=2-1 2 = 1. Therefore:

Example 20 Calculate

Let's use the change of variable. Then and . If x=0 then t=1 and if x=5 then t=4. By substituting, we get

We turn to the consideration of the general case - the method of changing variables in the indefinite integral.

Example 5

As an example, I took the integral, which we considered at the very beginning of the lesson. As we have already said, to solve the integral, we liked the tabular formula, and I would like to reduce the whole thing to it.

The idea behind the replacement method is to replace a complex expression (or some function) with one letter.
In this case it asks:
The second most popular replacement letter is the letter .
In principle, you can use other letters, but we still stick to the traditions.

So:
But when replacing, we have left! Probably, many have guessed that if a transition is made to a new variable, then in the new integral everything must be expressed through the letter, and there is no place for the differential at all.
It follows a logical conclusion that it is necessary turn into some expression that depends only on.

The action is the following. After we have selected a replacement, this example, , we need to find the differential . With differentials, I think, friendship has already been established for everyone.

Since then

After the showdown with the differential, I recommend rewriting the final result as briefly as possible:
Now, according to the rules of proportion, we express the one we need:

Eventually:
In this way:

And this is already the most tabular integral ( table of integrals, of course, is also valid for the variable ).

In conclusion, it remains to carry out the reverse replacement. We remember that .

Ready.

The final design of this example should look something like this:

“

Let's replace:

“

The icon does not carry any mathematical meaning, it means that we have interrupted the solution for intermediate explanations.

When making an example in a notebook, it is better to superscript the reverse substitution with a simple pencil.

Attention! In the following examples, finding the differential will not be described in detail.

And now it's time to remember the first solution:

What is the difference? There is no fundamental difference. It's actually the same thing. But from the point of view of the design of the task, the method of bringing the function under the sign of the differential is much shorter.

The question arises. If the first way is shorter, then why use the replace method? The fact is that for a number of integrals it is not so easy to "fit" the function under the sign of the differential.

Example 6

Find the indefinite integral.

Let's make a replacement: (it's hard to think of another replacement here)

As you can see, as a result of the replacement, the original integral has been greatly simplified - reduced to an ordinary power function. This is the purpose of the replacement - to simplify the integral.

Lazy advanced people can easily solve this integral by bringing the function under the differential sign:

Another thing is that such a solution is not obvious to all students. In addition, already in this example, the use of the method of bringing a function under the differential sign significantly increases the risk of confusion in the decision.

Example 7

Find the indefinite integral. Run a check.

Example 8

Find the indefinite integral.

Replacement:
It remains to be seen what will become

Well, we have expressed, but what to do with the “X” remaining in the numerator?!
From time to time, in the course of solving integrals, the following trick occurs: we will express from the same replacement !

Example 9

Find the indefinite integral.

This is an example for independent decision. Answer at the end of the lesson.

Example 10

Find the indefinite integral.

Surely some have noticed that my reference table does not have a variable substitution rule. It was done deliberately. The rule would confuse the explanation and understanding, since it does not appear explicitly in the above examples.

It's time to talk about the basic premise of using the variable substitution method: the integrand must contain some function and its derivative : (functions may not be in the product)

In this regard, when finding integrals, one often has to look into the table of derivatives.

In this example, we notice that the degree of the numerator is one less than the degree of the denominator. In the table of derivatives we find the formula, which just lowers the degree by one. And, therefore, if you designate for the denominator, then there are great chances that the numerator will turn into something good.

Replacement:

By the way, here it is not so difficult to bring the function under the differential sign:

It should be noted that for fractions like , such a trick will no longer work (more precisely, it will be necessary to apply not only the substitution technique). You can learn how to integrate some fractions in the lesson Integration of some fractions.

Here are a couple more typical examples for an independent solution from the same opera:

Example 11

Find the indefinite integral.

Example 12

Find the indefinite integral.

Solutions at the end of the lesson.

Example 13

Find the indefinite integral.

We look at the table of derivatives and find our arc cosine: . We have in the integrand the arccosine and something similar to its derivative.

General rule:
Per denote the function itself(and not its derivative).

In this case: . It remains to find out what the rest of the integrand will turn into.

In this example, I will describe the finding in detail because it is a complex function.

Or shorter:
According to the rule of proportion, we express the remainder we need:

In this way:

Here it is not so easy to bring the function under the sign of the differential.

Example 14

Find the indefinite integral.

An example for an independent solution. The answer is very close.

Attentive readers will notice that I have considered few examples with trigonometric functions. And this is not accidental, because integrals of trigonometric functions given a separate lesson. Moreover, the specified lesson gives some useful guidelines for changing a variable, which is especially important for dummies, who do not always and immediately understand what kind of replacement should be carried out in a particular integral. Also, some types of replacements can be found in the article Definite integral. Solution examples.

More experienced students can familiarize themselves with a typical replacement in integrals with irrational functions. Root integration substitution is specific, and its execution technique differs from that which we have considered in this lesson.

Wish you success!

Example 3:Solution :

Example 4:Solution :

Example 7:Solution :

Example 9:Solution :

Replacement:

Example 11:Solution :

Let's replace:

Example 12:Solution :

Let's replace:

Example 14:Solution :

Let's replace:

Integration by parts. Solution examples

Hello again. Today in the lesson we will learn how to integrate by parts. The method of integration by parts is one of the cornerstones of integral calculus. At the test, exam, the student is almost always offered to solve integrals of the following types: the simplest integral (see articleIndefinite integral. Solution examples ) or an integral to change the variable (see articleVariable change method in indefinite integral ) or the integral just on method of integration by parts.

As always, on hand should be: Table of integrals and Derivative table. If you still do not have them, then please visit the storeroom of my site: Mathematical formulas and tables. I will not get tired of repeating - it is better to print everything. I will try to present all the material in a consistent, simple and accessible way; there are no particular difficulties in integrating by parts.

What problem does integration by parts solve? The method of integration by parts solves a very important problem, it allows you to integrate some functions that are not in the table, work functions, and in some cases - and private. As we remember, there is no convenient formula: . But there is this one: is the formula for integration by parts in person. I know, I know, you are the only one - with her we will work the whole lesson (it’s already easier).

4) , are inverse trigonometric functions("arches"), "arches" multiplied by some polynomial.

Also, some fractions are taken in parts, we will also consider the corresponding examples in detail.

Integrals of logarithms

Example 1

Find the indefinite integral.

Classic. From time to time, this integral can be found in tables, but it is undesirable to use a ready-made answer, since the teacher has beriberi in the spring and he will scold a lot. Because the integral under consideration is by no means tabular - it is taken in parts. We decide:

We interrupt the solution for intermediate explanations.

We use the formula for integration by parts:

Calculate a given integral by direct integration

does not always succeed. One of the most effective methods

is a method of substituting or replacing a variable of integration.

The essence of this method lies in the fact that by introducing a new integration variable, it is possible to reduce the given integral to

new integral, which is taken by direct integration.

Consider this method:

Let be a continuous function

need to find: (1)

Let's change the integration variable:

where φ (t) is a monotonic function that has a continuous derivative

and there is a complex function f(φ(t)).

Applying to F (x) = F (φ (t)) the differentiation formula of the complex

functions, we get:

﴾F (φ (t))﴿′ = F′(x) ∙ φ′ (t)

But F (x) = f (x) = f (φ (t)), so

﴾F (φ (t))﴿′ = f (φ (t)) ∙ φ′ (t) (3)

Thus, the function F(φ (t)) is an antiderivative for the function

f (φ (t)) ∙ φ′ (t), so:

∫ f (φ (t)) ∙ φ′ (t) dt = F (φ (t)) + C (4)

Considering that F (φ (t)﴿ = F (x), from (1) and (4) follows the replacement formula

variable in the indefinite integral:

∫ f (x)dx = ∫ f(φ (t)) φ′ (t)dt (5)

Formally, formula (5) is obtained by replacing x with φ (t) and dx with φ′ (t)dt

In the result obtained after integration by formula (5) follows

go back to x. This is always possible, since

position, the function x = φ (t) is monotonic.

A good choice of substitution usually represents well-known work.

ness. To overcome them, it is necessary to master the technique of differentiation

quoting and have a good knowledge of tabular integrals.

However, you can still set general rules and some tricks

integration.

Rules for integration by the substitution method:

1. Determine to which table integral this integral is reduced (having previously converted the integrand, if necessary).

2. Determine which part of the integrand should be replaced

new variable, and record this replacement.

3. Find the differentials of both parts of the record and express the differential

the old variable (or an expression containing this differential

rencial) through the differential of the new variable.

4. Make a replacement under the integral.

5. Find the resulting integral.

6. As a result, go to the old variable.

Examples of solving integrals by the substitution method:

1. Find: ∫ x²(3+2x) dx

Solution:

let's make a substitution 3+2x = t

Find the differential of both parts of the substitution:

6x dx = dt, whence

Consequently:

∫ x (3+2x ) dx = ∫ t ∙ dt = ∫ t dt = ∙ + C = t + C

Replacing t with its substitution expression, we get:

∫ x (3+2x) dx = (3+2x) + C

Solution:

= ∫ = e = e + C = e + C

Solution:

The concept of a definite integral.

The difference in values for any antiderivative function when the argument changes from to is called the integral of this function in the range from a to b and is denoted:

a and b are called the lower and upper limits of integration.

To calculate the definite integral you need:

1. Find the corresponding indefinite integral

2. Substitute in the resulting expression instead of x, first the upper limit of integration in, and then the lower limit - a.

3. Subtract the second result from the first substitution result.

Briefly, this rule is written in the form of formulas as follows:

This formula is called the Newton-Leibniz formula.

The main properties of the definite integral:

1. , where K=const

3. If , then

4. If the function is non-negative on the interval , where , then

When replacing the old integration variable with a new one in a definite integral, it is necessary to replace the old integration limits with new ones. These new limits are determined by the chosen substitution.

Application of a definite integral.

Area of a curvilinear trapezoid bounded by a curve, an x-axis and two straight lines and calculated by the formula:

The volume of a body formed by rotation around the abscissa axis of a curvilinear trapezoid, bounded by a curve that does not change its sign by , the abscissa axis and two straight lines and calculated by the formula:

A number of physical problems can also be solved with the help of a definite integral.

For example:

If the speed of a rectilinearly moving body is a known function of time t, then the path S traveled by this body from time t \u003d t 1 to time t \u003d t 2 is determined by the formula:

If the variable force is a known function of the path S (it is assumed that the direction of the force does not change), then the work A performed by this force on the path from to is determined by the formula:

Examples:

1. Calculate the area of a figure bounded by lines:

y = ; y = (x-2) 2 ; 0x.

Solution:

a) Let's build graphs of functions: y = ; y=(x-2)2

b) Determine the figure whose area you want to calculate.

c) Determine the limits of integration by solving the equation: = (x-2) 2 ; x = 1

d) Calculate the area of the given figure:

S = dx + 2 dx = 1 unit 2

2. Calculate the area of the figure bounded by lines:

Y = x 2 ; x = y 2 .

Solution:

x 2 = ; x 4 \u003d x;

x (x 3 - 1) = 0

x 1 = 0; x2 = 1

S = - x 2) dx = ( x 3 \ 2 - ) │ 0 1 = unit 2

3. Calculate the volume of the body obtained by rotating around the 0x axis of the figure bounded by lines: y = ; x = 1 .

Solution:

V = π dx = π ) 2 dx = π = π │ = π/2 units 3

Home test mathematics
Task options.

Option number 1

y = (x + 1) 2 ; y \u003d 1 - x; 0x

Option number 2