Integrals and their properties. Basic properties of the indefinite integral. Basic properties of the definite integral

Let the function y = f(x) is defined on the interval [ a, b ], a < b. Let's perform the following operations:

1) let's split [ a, b] dots a = x 0 < x 1 < ... < x i- 1 < x i < ... < x n = b on n partial segments [ x 0 , x 1 ], [x 1 , x 2 ], ..., [x i- 1 , x i ], ..., [x n- 1 , x n ];

2) in each of the partial segments [ x i- 1 , x i ], i = 1, 2, ... n, choose an arbitrary point and calculate the value of the function at this point: f(z i ) ;

3) find the works f(z i ) · Δ x i , where is the length of the partial segment [ x i- 1 , x i ], i = 1, 2, ... n;

4) let's make up integral sum functions y = f(x) on the segment [ a, b ]:

WITH geometric point From a visual perspective, this sum σ is the sum of the areas of rectangles whose bases are partial segments [ x 0 , x 1 ], [x 1 , x 2 ], ..., [x i- 1 , x i ], ..., [x n- 1 , x n ], and the heights are equal f(z 1 ) , f(z 2 ), ..., f(z n) accordingly (Fig. 1). Let us denote by λ length of the longest partial segment:

5) find the limit of the integral sum when λ → 0.

Definition. If there is a finite limit of the integral sum (1) and it does not depend on the method of partitioning the segment [ a, b] to partial segments, nor from the selection of points z i in them, then this limit is called definite integral from function y = f(x) on the segment [ a, b] and is denoted

Thus,

In this case the function f(x) is called integrable on [ a, b]. Numbers a And b are called the lower and upper limits of integration, respectively, f(x) – integrand function, f(x ) dx– integrand expression, x– integration variable; line segment [ a, b] is called the integration interval.

Theorem 1. If the function y = f(x) is continuous on the interval [ a, b], then it is integrable on this interval.

The definite integral with the same limits of integration is equal to zero:

If a > b, then, by definition, we assume

2. Geometric meaning of the definite integral

Let on the segment [ a, b] a continuous non-negative function is specified y = f(x ) . Curvilinear trapezoid is a figure bounded above by the graph of a function y = f(x), from below - along the Ox axis, to the left and right - straight lines x = a And x = b(Fig. 2).

Definite integral of a non-negative function y = f(x) from a geometric point of view equal to area curvilinear trapezoid bounded above by the graph of the function y = f(x) , left and right – line segments x = a And x = b, from below - a segment of the Ox axis.

3. Basic properties of the definite integral

1. Meaning definite integral does not depend on the designation of the integration variable:

2. The constant factor can be taken out of the sign of the definite integral:

3. The definite integral of the algebraic sum of two functions is equal to the algebraic sum of the definite integrals of these functions:

4.If function y = f(x) is integrable on [ a, b] And a < b < c, That

5. (mean value theorem). If the function y = f(x) is continuous on the interval [ a, b], then on this segment there is a point such that

4. Newton–Leibniz formula

Theorem 2. If the function y = f(x) is continuous on the interval [ a, b] And F(x) is any of its antiderivatives on this segment, then the following formula is valid:

which is called Newton–Leibniz formula. Difference F(b) - F(a) is usually written as follows:

where the symbol is called a double wildcard.

Thus, formula (2) can be written as:

Example 1. Calculate integral

Solution. For the integrand f(x ) = x 2 an arbitrary antiderivative has the form

Since any antiderivative can be used in the Newton-Leibniz formula, to calculate the integral we take the antiderivative that has the simplest form:

5. Change of variable in a definite integral

Theorem 3. Let the function y = f(x) is continuous on the interval [ a, b]. If:

1) function x = φ ( t) and its derivative φ "( t) are continuous for ;

2) a set of function values x = φ ( t) for is the segment [ a, b ];

3) φ ( a) = a, φ ( b) = b, then the formula is valid

which is called formula for changing a variable in a definite integral .

Unlike indefinite integral, V in this case not necessary to return to the original integration variable - it is enough just to find new limits of integration α and β (for this you need to solve for the variable t equations φ ( t) = a and φ ( t) = b).

Instead of substitution x = φ ( t) you can use substitution t = g(x) . In this case, finding new limits of integration over a variable t simplifies: α = g(a) , β = g(b) .

Example 2. Calculate integral

Solution. Let's introduce a new variable using the formula. By squaring both sides of the equality, we get 1 + x = t 2 , where x = t 2 - 1, dx = (t 2 - 1)"dt= 2tdt. We find new limits of integration. To do this, let’s substitute the old limits into the formula x = 3 and x = 8. We get: , from where t= 2 and α = 2; , where t= 3 and β = 3. So,

Example 3. Calculate

Solution. Let u= log x, Then , v = x. According to formula (4)

Antiderivative and indefinite integral.

An antiderivative of a function f(x) on the interval (a; b) is a function F(x) such that the equality holds for any x from the given interval.

If we take into account the fact that the derivative of the constant C is equal to zero, then the equality is true . Thus, the function f(x) has a set of antiderivatives F(x)+C, for an arbitrary constant C, and these antiderivatives differ from each other by an arbitrary constant value.

The entire set of antiderivatives of the function f(x) is called the indefinite integral of this function and is denoted .

The expression is called the integrand, and f(x) is called the integrand. The integrand represents the differential of the function f(x).

The action of finding an unknown function given its differential is called indefinite integration, because the result of integration is not one function F(x), but a set of its antiderivatives F(x)+C.

Table integrals

The simplest properties of integrals

1. The derivative of the integration result is equal to the integrand.

2. Indefinite integral of the differential function equal to the sum the function itself and an arbitrary constant.

3. The coefficient can be taken out of the sign of the indefinite integral.

4. The indefinite integral of the sum/difference of functions is equal to the sum/difference of the indefinite integrals of functions.

Intermediate equalities of the first and second properties of the indefinite integral are given for clarification.

To prove the third and fourth properties, it is enough to find the derivatives of the right-hand sides of the equalities:

These derivatives are equal to the integrands, which is a proof due to the first property. It is also used in the last transitions.

Thus, the problem of integration is the inverse of the problem of differentiation, and there is a very close connection between these problems:

the first property allows one to check integration. To check the correctness of the integration performed, it is enough to calculate the derivative of the result obtained. If the function obtained as a result of differentiation turns out to be equal to the integrand, this will mean that the integration was carried out correctly;

the second property of the indefinite integral allows one to find its antiderivative from a known differential of a function. The direct calculation of indefinite integrals is based on this property.

1.4.Invariance of integration forms.

Invariant integration is a type of integration for functions whose arguments are elements of a group or points of a homogeneous space (any point in such a space can be transferred to another by a given action of the group).

function f(x) reduces to calculating the integral of the differential form f.w, where

An explicit formula for r(x) is given below. The agreement condition has the form .

here Tg means the shift operator on X using gОG: Tgf(x)=f(g-1x). Let X=G be a topology, a group acting on itself by left shifts. I. and. exists if and only if G is locally compact (in particular, on infinite-dimensional groups I.I. does not exist). For a subset of I. and. characteristic function cA (equal to 1 on A and 0 outside A) specifies the left Xaar measure m(A). The defining property of this measure is its invariance under left shifts: m(g-1A)=m(A) for all gОG. The left Haar measure on a group is uniquely defined up to a positive scalar factor. If the Haar measure m is known, then I. and. function f is given by the formula . The right Haar measure has similar properties. There is a continuous homomorphism (map preserving the group property) DG of the group G into the group (with respect to multiplication) posit. numbers for which

where dmr and dmi are the right and left Haar measures. The function DG(g) is called module of the group G. If , then the group G is called. unimodular; in this case the right and left Haar measures coincide. Compact, semisimple and nilpotent (in particular, commutative) groups are unimodular. If G is an n-dimensional Lie group and q1,...,qn is a basis in the space of left-invariant 1-forms on G, then the left Haar measure on G is given by the n-form. In local coordinates for calculation

forms qi, you can use any matrix realization of the group G: the matrix 1-form g-1dg is left invariant, and its coefficient. are left-invariant scalar 1-forms from which the required basis is selected. For example, the complete matrix group GL(n, R) is unimodular and the Haar measure on it is given by the form. Let X=G/H is a homogeneous space for which the locally compact group G is a transformation group, and the closed subgroup H is the stabilizer of a certain point. In order for an i.i. to exist on X, it is necessary and sufficient that for all hОH the equality DG(h)=DH(h) holds. In particular, this is true in the case when H is compact or semisimple. Complete theory of I. and. does not exist on infinite-dimensional manifolds.

Replacing variables.

Solving integrals is an easy task, but only for a select few. This article is for those who want to learn to understand integrals, but know nothing or almost nothing about them. Integral... Why is it needed? How to calculate it? What are definite and indefinite integrals?

If the only use you know of for an integral is to use a crochet hook shaped like an integral icon to get something useful out of hard-to-reach places, then welcome! Find out how to solve the simplest and other integrals and why you can’t do without it in mathematics.

We study the concept « integral »

Integration was known back in Ancient Egypt. Of course, not in its modern form, but still. Since then, mathematicians have written many books on this topic. Especially distinguished themselves Newton And Leibniz , but the essence of things has not changed.

How to understand integrals from scratch? No way! To understand this topic you will still need a basic understanding of the basics. mathematical analysis. We already have information about , necessary for understanding integrals, on our blog.

Indefinite integral

Let us have some function f(x) .

Indefinite integral function f(x) this function is called F(x) , whose derivative is equal to the function f(x) .

In other words, an integral is a derivative in reverse or an antiderivative. By the way, read about how in our article.

An antiderivative exists for all continuous functions. Also, a constant sign is often added to the antiderivative, since the derivatives of functions that differ by a constant coincide. The process of finding the integral is called integration.

Simple example:

In order not to constantly calculate antiderivatives elementary functions, it is convenient to summarize them in a table and use ready-made values.

Complete table of integrals for students

Definite integral

When dealing with the concept of an integral, we are dealing with infinitesimal quantities. The integral will help to calculate the area of ​​a figure, the mass of a non-uniform body, the distance traveled during uneven movement, and much more. It should be remembered that an integral is the sum of an infinitely large number of infinitesimal terms.

As an example, imagine a graph of some function.

How to find the area of ​​a figure bounded by the graph of a function? Using an integral! Let us divide the curvilinear trapezoid, limited by the coordinate axes and the graph of the function, into infinitesimal segments. This way the figure will be divided into thin columns. The sum of the areas of the columns will be the area of ​​the trapezoid. But remember that such a calculation will give an approximate result. However, the smaller and narrower the segments, the more accurate the calculation will be. If we reduce them to such an extent that the length tends to zero, then the sum of the areas of the segments will tend to the area of ​​the figure. This is a definite integral, which is written like this:

Points a and b are called limits of integration.

« Integral »

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Rules for calculating integrals for dummies

Properties of the indefinite integral

How to solve an indefinite integral? Here we will look at the properties of the indefinite integral, which will be useful when solving examples.

• The derivative of the integral is equal to the integrand:

• The constant can be taken out from under the integral sign:

• The integral of the sum is equal to the sum of the integrals. This is also true for the difference:

Properties of a definite integral

• Linearity:

• The sign of the integral changes if the limits of integration are swapped:

• At any points a, b And With:

We have already found out that a definite integral is the limit of a sum. But how to get a specific value when solving an example? For this there is the Newton-Leibniz formula:

Examples of solving integrals

Below we will consider the indefinite integral and examples with solutions. We suggest you figure out the intricacies of the solution yourself, and if something is unclear, ask questions in the comments.

To reinforce the material, watch a video about how integrals are solved in practice. Don't despair if the integral is not given right away. Contact a professional service for students, and any triple or curved integral over a closed surface will be within your power.

This article talks in detail about the main properties of the definite integral. They are proved using the concept of the Riemann and Darboux integral. The calculation of a definite integral takes place thanks to 5 properties. The remaining ones are used to evaluate various expressions.

Before moving on to the main properties of the definite integral, it is necessary to make sure that a does not exceed b.

Basic properties of the definite integral

The function y = f (x) defined at x = a is similar to the fair equality ∫ a a f (x) d x = 0.

Evidence 1

From this we see that the value of the integral with coinciding limits is equal to zero. This is a consequence of the Riemann integral, because every integral sum σ for any partition on the interval [ a ; a ] and any choice of points ζ i equals zero, because x i - x i - 1 = 0 , i = 1 , 2 , . . . , n , which means we find that the limit of integral functions is zero.

Definition 2

For a function that is integrable on the interval [a; b ] , the condition ∫ a b f (x) d x = - ∫ b a f (x) d x is satisfied.

Evidence 2

In other words, if you swap the upper and lower limits of integration, the value of the integral will change to the opposite value. This property is taken from the Riemann integral. However, the numbering of the partition of the segment starts from the point x = b.

Definition 3

∫ a b f x ± g (x) d x = ∫ a b f (x) d x ± ∫ a b g (x) d x applies to integrable functions of type y = f (x) and y = g (x) defined on the interval [ a ; b ] .

Evidence 3

Write down the integral sum of the function y = f (x) ± g (x) for partitioning into segments with a given choice of points ζ i: σ = ∑ i = 1 n f ζ i ± g ζ i · x i - x i - 1 = = ∑ i = 1 n f (ζ i) · x i - x i - 1 ± ∑ i = 1 n g ζ i · x i - x i - 1 = σ f ± σ g

where σ f and σ g are the integral sums of the functions y = f (x) and y = g (x) for partitioning the segment. After passing to the limit at λ = m a x i = 1, 2, . . . , n (x i - x i - 1) → 0 we obtain that lim λ → 0 σ = lim λ → 0 σ f ± σ g = lim λ → 0 σ g ± lim λ → 0 σ g .

From Riemann's definition, this expression is equivalent.

Definition 4

Extending the constant factor beyond the sign of the definite integral. Integrated function from the interval [a; b ] with an arbitrary value k has a fair inequality of the form ∫ a b k · f (x) d x = k · ∫ a b f (x) d x .

Proof 4

The proof of the definite integral property is similar to the previous one:

σ = ∑ i = 1 n k · f ζ i · (x i - x i - 1) = = k · ∑ i = 1 n f ζ i · (x i - x i - 1) = k · σ f ⇒ lim λ → 0 σ = lim λ → 0 (k · σ f) = k · lim λ → 0 σ f ⇒ ∫ a b k · f (x) d x = k · ∫ a b f (x) d x

Definition 5

If a function of the form y = f (x) is integrable on an interval x with a ∈ x, b ∈ x, we obtain that ∫ a b f (x) d x = ∫ a c f (x) d x + ∫ c b f (x) d x.

Evidence 5

The property is considered valid for c ∈ a; b, for c ≤ a and c ≥ b. The proof is similar to the previous properties.

Definition 6

When a function can be integrable from the segment [a; b ], then this is feasible for any internal segment c; d ∈ a ; b.

Proof 6

The proof is based on the Darboux property: if points are added to an existing partition of a segment, then the lower Darboux sum will not decrease, and the upper one will not increase.

Definition 7

When a function is integrable on [a; b ] from f (x) ≥ 0 f (x) ≤ 0 for any value x ∈ a ; b , then we get that ∫ a b f (x) d x ≥ 0 ∫ a b f (x) ≤ 0 .

The property can be proven using the definition of the Riemann integral: any integral sum for any choice of points of partition of the segment and points ζ i with the condition that f (x) ≥ 0 f (x) ≤ 0 is non-negative.

Evidence 7

If the functions y = f (x) and y = g (x) are integrable on the interval [ a ; b ], then the following inequalities are considered valid:

∫ a b f (x) d x ≤ ∫ a b g (x) d x , f (x) ≤ g (x) ∀ x ∈ a ; b ∫ a b f (x) d x ≥ ∫ a b g (x) d x , f (x) ≥ g (x) ∀ x ∈ a ; b

Thanks to the statement, we know that integration is permissible. This corollary will be used in the proof of other properties.

Definition 8

For an integrable function y = f (x) from the interval [ a ; b ] we have a fair inequality of the form ∫ a b f (x) d x ≤ ∫ a b f (x) d x .

Proof 8

We have that - f (x) ≤ f (x) ≤ f (x) . From the previous property we found that the inequality can be integrated term by term and it corresponds to an inequality of the form - ∫ a b f (x) d x ≤ ∫ a b f (x) d x ≤ ∫ a b f (x) d x . This double inequality can be written in another form: ∫ a b f (x) d x ≤ ∫ a b f (x) d x .

Definition 9

When the functions y = f (x) and y = g (x) are integrated from the interval [ a ; b ] for g (x) ≥ 0 for any x ∈ a ; b , we obtain an inequality of the form m · ∫ a b g (x) d x ≤ ∫ a b f (x) · g (x) d x ≤ M · ∫ a b g (x) d x , where m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) .

Evidence 9

The proof is carried out in a similar way. M and m are considered to be the largest and smallest values ​​of the function y = f (x) defined from the segment [a; b ] , then m ≤ f (x) ≤ M . It is necessary to multiply the double inequality by the function y = g (x), which gives the value double inequality of the form m · g (x) ≤ f (x) · g (x) ≤ M · g (x) . It is necessary to integrate it on the interval [a; b ] , then we get the statement to be proved.

Consequence: For g (x) = 1, the inequality takes the form m · b - a ≤ ∫ a b f (x) d x ≤ M · (b - a) .

First average formula

For y = f (x) integrable on the interval [ a ; b ] with m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) there is a number μ ∈ m; M , which fits ∫ a b f (x) d x = μ · b - a .

Consequence: When the function y = f (x) is continuous from the interval [ a ; b ], then there is a number c ∈ a; b, which satisfies the equality ∫ a b f (x) d x = f (c) b - a.

The first average formula in generalized form

When the functions y = f (x) and y = g (x) are integrable from the interval [ a ; b ] with m = m i n x ∈ a ; b f (x) and M = m a x x ∈ a ; b f (x) , and g (x) > 0 for any value x ∈ a ; b. From here we have that there is a number μ ∈ m; M , which satisfies the equality ∫ a b f (x) · g (x) d x = μ · ∫ a b g (x) d x .

Second average formula

When the function y = f (x) is integrable from the interval [ a ; b ], and y = g (x) is monotonic, then there is a number that c ∈ a; b , where we obtain a fair equality of the form ∫ a b f (x) · g (x) d x = g (a) · ∫ a c f (x) d x + g (b) · ∫ c b f (x) d x

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In differential calculus the problem is solved: under this function ƒ(x) find its derivative(or differential). Integral calculus solves the inverse problem: find the function F(x), knowing its derivative F "(x)=ƒ(x) (or differential). The sought function F(x) is called the antiderivative of the function ƒ(x).

The function F(x) is called antiderivative function ƒ(x) on the interval (a; b), if for any x є (a; b) the equality

F " (x)=ƒ(x) (or dF(x)=ƒ(x)dx).

For example, the antiderivative of the function y = x 2, x є R, is the function, since

Obviously, any functions will also be antiderivatives

where C is a constant, since

Theorem 29. 1. If the function F(x) is an antiderivative of the function ƒ(x) on (a;b), then the set of all antiderivatives for ƒ(x) is given by the formula F(x)+C, where C is a constant number.

▲ The function F(x)+C is an antiderivative of ƒ(x).

Indeed, (F(x)+C) " =F " (x)=ƒ(x).

Let Ф(х) be some other antiderivative of the function ƒ(x), different from F(x), i.e. Ф "(x)=ƒ(х). Then for any x є (а; b) we have

And this means (see Corollary 25.1) that

where C is a constant number. Therefore, Ф(x)=F(x)+С.▼

The set of all antiderivative functions F(x)+С for ƒ(x) is called indefinite integral of the function ƒ(x) and is denoted by the symbol ∫ ƒ(x) dx.

Thus, by definition

∫ ƒ(x)dx= F(x)+C.

Here ƒ(x) is called integrand function, ƒ(x)dx — integrand expression, X - integration variable, ∫ -sign of the indefinite integral.

The operation of finding the indefinite integral of a function is called integrating this function.

Geometrically, the indefinite integral is a family of “parallel” curves y=F(x)+C (each numerical value of C corresponds to a specific curve of the family) (see Fig. 166). The graph of each antiderivative (curve) is called integral curve.

Does every function have an indefinite integral?

There is a theorem stating that “every function continuous on (a;b) has an antiderivative on this interval,” and, consequently, an indefinite integral.

Let us note a number of properties of the indefinite integral that follow from its definition.

1. The differential of the indefinite integral is equal to the integrand, and the derivative of the indefinite integral is equal to the integrand:

d(∫ ƒ(x)dx)=ƒ(x)dх, (∫ ƒ(x)dx) " =ƒ(x).

Indeed, d(∫ ƒ(x) dx)=d(F(x)+C)=dF(x)+d(C)=F " (x) dx =ƒ(x) dx

(∫ ƒ (x) dx) " =(F(x)+C)"=F"(x)+0 =ƒ (x).

Thanks to this property, the correctness of integration is checked by differentiation. For example, equality

∫(3x 2 + 4) dx=х з +4х+С

true, since (x 3 +4x+C)"=3x 2 +4.

2. The indefinite integral of the differential of a certain function is equal to the sum of this function and an arbitrary constant:

∫dF(x)= F(x)+C.

Really,

3. The constant factor can be taken out of the integral sign:

α ≠ 0 is a constant.

Really,

(put C 1 / a = C.)

4. The indefinite integral of the algebraic sum of a finite number of continuous functions is equal to the algebraic sum of the integrals of the summands of the functions:

Let F"(x)=ƒ(x) and G"(x)=g(x). Then

where C 1 ±C 2 =C.

5. (Invariance of the integration formula).

If , where u=φ(x) is an arbitrary function with a continuous derivative.

▲ Let x be an independent variable, ƒ(x) - continuous function and F(x) is its antigen. Then

Let us now set u=φ(x), where φ(x) is a continuously differentiable function. Consider the complex function F(u)=F(φ(x)). Due to the invariance of the form of the first differential of the function (see p. 160), we have

From here▼

Thus, the formula for the indefinite integral remains valid regardless of whether the variable of integration is the independent variable or any function of it that has a continuous derivative.

So, from the formula by replacing x with u (u=φ(x)) we get

In particular,

Example 29.1. Find the integral

where C=C1+C 2 +C 3 +C 4.

Example 29.2. Find the integral Solution:

• 29.3. Table of basic indefinite integrals

Taking advantage of the fact that integration is the inverse action of differentiation, one can obtain a table of basic integrals by inverting the corresponding formulas of differential calculus (table of differentials) and using the properties of the indefinite integral.

For example, because

d(sin u)=cos u . du

The derivation of a number of formulas in the table will be given when considering the basic methods of integration.

The integrals in the table below are called tabular. They should be known by heart. In integral calculus there are no simple and universal rules for finding antiderivatives of elementary functions, as in differential calculus. Methods for finding antiderivatives (i.e., integrating a function) are reduced to indicating techniques that bring a given (sought) integral to a tabular one. Therefore, it is necessary to know table integrals and be able to recognize them.

Note that in the table of basic integrals, the integration variable can denote both an independent variable and a function of the independent variable (according to the invariance property of the integration formula).

The validity of the formulas below can be verified by taking the differential on the right side, which will be equal to the integrand on the left side of the formula.

Let us prove, for example, the validity of formula 2. The function 1/u is defined and continuous for all values ​​of and other than zero.

If u > 0, then ln|u|=lnu, then That's why

If u<0, то ln|u|=ln(-u). НоMeans

So, formula 2 is correct. Similarly, let's check formula 15:

Table of main integrals

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