Antiderivative function and indefinite integral

Fact 1. Integration is the inverse action of differentiation, namely, restoring a function from the known derivative of this function. The function thus restored F(x) is called antiderivative for function f(x).

Definition 1. Function F(x f(x) on some interval X, if for all values x from this interval the equality holds F "(x)=f(x), that is, this function f(x) is the derivative of the antiderivative function F(x). .

For example, the function F(x) = sin x is an antiderivative of the function f(x) = cos x on the entire number line, since for any value of x (sin x)" = (cos x) .

Definition 2. Indefinite integral of a function f(x) is the set of all its antiderivatives. In this case, the notation is used

∫

f(x)dx

where is the sign ∫ called the integral sign, the function f(x) – integrand function, and f(x)dx – integrand expression.

Thus, if F(x) – some antiderivative for f(x) , That

∫

f(x)dx = F(x) +C

Where C - arbitrary constant (constant).

To understand the meaning of the set of antiderivatives of a function as an indefinite integral, the following analogy is appropriate. Let there be a door (traditional wooden door). Its function is to “be a door.” What is the door made of? Made of wood. This means that the set of antiderivatives of the integrand of the function “to be a door”, that is, its indefinite integral, is the function “to be a tree + C”, where C is a constant, which in this context can denote, for example, the type of tree. Just as a door is made from wood using some tools, a derivative of a function is “made” from an antiderivative function using formulas we learned while studying the derivative .

Then the table of functions of common objects and their corresponding antiderivatives (“to be a door” - “to be a tree”, “to be a spoon” - “to be metal”, etc.) is similar to the table of basic indefinite integrals, which will be given below. The table of indefinite integrals lists common functions with an indication of the antiderivatives from which these functions are “made”. In part of the problems on finding the indefinite integral, integrands are given that can be integrated directly without much effort, that is, using the table of indefinite integrals. In more complex problems, the integrand must first be transformed so that table integrals can be used.

Fact 2. When restoring a function as an antiderivative, we must take into account an arbitrary constant (constant) C, and in order not to write a list of antiderivatives with various constants from 1 to infinity, you need to write a set of antiderivatives with an arbitrary constant C, for example, like this: 5 x³+C. So, an arbitrary constant (constant) is included in the expression of the antiderivative, since the antiderivative can be a function, for example, 5 x³+4 or 5 x³+3 and when differentiated, 4 or 3, or any other constant goes to zero.

Let us pose the integration problem: for this function f(x) find such a function F(x), whose derivative equal to f(x).

Example 1. Find the set of antiderivatives of a function

Solution. For this function, the antiderivative is the function

Function F(x) is called an antiderivative for the function f(x), if the derivative F(x) is equal to f(x), or, which is the same thing, differential F(x) is equal f(x) dx, i.e.

(2)

Therefore, the function is an antiderivative of the function. However, it is not the only antiderivative for . They also serve as functions

Where WITH– arbitrary constant. This can be verified by differentiation.

Thus, if there is one antiderivative for a function, then for it there is an infinite number of antiderivatives that differ by a constant term. All antiderivatives for a function are written in the above form. This follows from the following theorem.

Theorem (formal statement of fact 2). If F(x) – antiderivative for the function f(x) on some interval X, then any other antiderivative for f(x) on the same interval can be represented in the form F(x) + C, Where WITH– arbitrary constant.

In the next example, we turn to the table of integrals, which will be given in paragraph 3, after the properties of the indefinite integral. We do this before reading the entire table so that the essence of the above is clear. And after the table and properties, we will use them in their entirety during integration.

Example 2. Find sets of antiderivative functions:

Solution. We find sets of antiderivative functions from which these functions are “made”. When mentioning formulas from the table of integrals, for now just accept that there are such formulas there, and we will study the table of indefinite integrals itself a little further.

1) Applying formula (7) from the table of integrals for n= 3, we get

2) Using formula (10) from the table of integrals for n= 1/3, we have

3) Since

then according to formula (7) with n= -1/4 we find

It is not the function itself that is written under the integral sign. f, and its product by the differential dx. This is done primarily in order to indicate by which variable the antiderivative is sought. For example,

, ;

here in both cases the integrand is equal to , but its indefinite integrals in the cases considered turn out to be different. In the first case, this function is considered as a function of the variable x, and in the second - as a function of z .

The process of finding the indefinite integral of a function is called integrating that function.

Geometric meaning of the indefinite integral

Suppose we need to find a curve y=F(x) and we already know that the tangent of the tangent angle at each of its points is a given function f(x) abscissa of this point.

According to the geometric meaning of the derivative, the tangent of the angle of inclination of the tangent at a given point of the curve y=F(x) equal to the value of the derivative F"(x). So we need to find such a function F(x), for which F"(x)=f(x). Function required in the task F(x) is an antiderivative of f(x). The conditions of the problem are satisfied not by one curve, but by a family of curves. y=F(x)- one of these curves, and any other curve can be obtained from it by parallel translation along the axis Oy.

Let's call the graph of the antiderivative function of f(x) integral curve. If F"(x)=f(x), then the graph of the function y=F(x) there is an integral curve.

Fact 3. The indefinite integral is geometrically represented by the family of all integral curves , as in the picture below. The distance of each curve from the origin of coordinates is determined by an arbitrary integration constant C.

Properties of the indefinite integral

Fact 4. Theorem 1. The derivative of an indefinite integral is equal to the integrand, and its differential is equal to the integrand.

Fact 5. Theorem 2. Indefinite integral of the differential of a function f(x) is equal to the function f(x) up to a constant term , i.e.

(3)

Theorems 1 and 2 show that differentiation and integration are mutually inverse operations.

Fact 6. Theorem 3. The constant factor in the integrand can be taken out of the sign of the indefinite integral , i.e.

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 knowledge of the basics of 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 of elementary functions, it is convenient to put 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 »

By the way! For our readers there is now a 10% discount on

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.

>>Integration methods

Basic integration methods

Definition of integral, definite and indefinite integral, table of integrals, Newton-Leibniz formula, integration by parts, examples of calculating integrals.

Indefinite integral

A function F(x) differentiable in a given interval X is called antiderivative of the function f(x), or the integral of f(x), if for every x ∈X the following equality holds:

F " (x) = f(x). (8.1)

Finding all antiderivatives for a given function is called its integration. Indefinite integral function f(x) on a given interval X is the set of all antiderivative functions for the function f(x); designation -

If F(x) is some antiderivative of the function f(x), then ∫ f(x)dx = F(x) + C, (8.2)

where C is an arbitrary constant.

Table of integrals

Directly from the definition we obtain the main properties of the indefinite integral and a list of tabular integrals:

1) d∫f(x)dx=f(x)

2)∫df(x)=f(x)+C

3) ∫af(x)dx=a∫f(x)dx (a=const)

4) ∫(f(x)+g(x))dx = ∫f(x)dx+∫g(x)dx

List of tabular integrals

1. ∫x m dx = x m+1 /(m + 1) +C; (m ≠ -1)

3.∫a x dx = a x /ln a + C (a>0, a ≠1)

4.∫e x dx = e x + C

5.∫sin x dx = cosx + C

6.∫cos x dx = - sin x + C

7. = arctan x + C

8. = arcsin x + C

10. = - ctg x + C

Variable replacement

To integrate many functions, use the variable replacement method or substitutions, allowing you to reduce integrals to tabular form.

If the function f(z) is continuous on [α,β], the function z =g(x) has a continuous derivative and α ≤ g(x) ≤ β, then

∫ f(g(x)) g " (x) dx = ∫f(z)dz, (8.3)

Moreover, after integration on the right side, the substitution z=g(x) should be made.

To prove it, it is enough to write the original integral in the form:

∫ f(g(x)) g " (x) dx = ∫ f(g(x)) dg(x).

For example:

1)

2) .

Method of integration by parts

Let u = f(x) and v = g(x) be functions that have continuous . Then, according to the work,

d(uv))= udv + vdu or udv = d(uv) - vdu.

For the expression d(uv), the antiderivative will obviously be uv, so the formula holds:

∫ udv = uv - ∫ vdu (8.4.)

This formula expresses the rule integration by parts. It leads the integration of the expression udv=uv"dx to the integration of the expression vdu=vu"dx.

Let, for example, you want to find ∫xcosx dx. Let us put u = x, dv = cosxdx, so du=dx, v=sinx. Then

∫xcosxdx = ∫x d(sin x) = x sin x - ∫sin x dx = x sin x + cosx + C.

The rule of integration by parts has a more limited scope than substitution of variables. But there are whole classes of integrals, for example,

∫x k ln m xdx, ∫x k sinbxdx, ∫ x k cosbxdx, ∫x k e ax and others, which are calculated precisely using integration by parts.

Definite integral

The concept of a definite integral is introduced as follows. Let a function f(x) be defined on an interval. Let us divide the segment [a,b] into n parts by points a= x 0< x 1 <...< x n = b. Из каждого интервала (x i-1 , x i) возьмем произвольную точку ξ i и составим сумму f(ξ i) Δx i где
Δ x i =x i - x i-1. A sum of the form f(ξ i)Δ x i is called integral sum, and its limit at λ = maxΔx i → 0, if it exists and is finite, is called definite integral functions f(x) of a before b and is designated:

F(ξ i)Δx i (8.5).

The function f(x) in this case is called integrable on the interval, numbers a and b are called lower and upper limits of the integral.

The following properties are true for a definite integral:

4), (k = const, k∈R);

5)

6)

7) f(ξ)(b-a) (ξ∈).

The last property is called mean value theorem.

Let f(x) be continuous on . Then on this segment there is an indefinite integral

∫f(x)dx = F(x) + C

and takes place Newton-Leibniz formula, connecting the definite integral with the indefinite integral:

F(b) - F(a). (8.6)

Geometric interpretation: the definite integral is the area of ​​a curvilinear trapezoid bounded from above by the curve y=f(x), straight lines x = a and x = b and a segment of the axis Ox.

Improper integrals

Integrals with infinite limits and integrals of discontinuous (unbounded) functions are called not your own. Improper integrals of the first kind - These are integrals over an infinite interval, defined as follows:

(8.7)

If this limit exists and is finite, then it is called convergent improper integral of f(x) on the interval [a,+ ∞), and the function f(x) is called integrable over an infinite interval[a,+ ∞). Otherwise, the integral is said to be does not exist or diverges.

Improper integrals on the intervals (-∞,b] and (-∞, + ∞) are defined similarly:

Let us define the concept of an integral of an unbounded function. If f(x) is continuous for all values x segment , except for the point c, at which f(x) has an infinite discontinuity, then improper integral of the second kind of f(x) ranging from a to b the amount is called:

if these limits exist and are finite. Designation:

Examples of integral calculations

Example 3.30. Calculate ∫dx/(x+2).

Solution. Let us denote t = x+2, then dx = dt, ∫dx/(x+2) = ∫dt/t = ln|t| + C = ln|x+2| +C.

Example 3.31. Find ∫ tgxdx.

Solution.∫ tgxdx = ∫sinx/cosxdx = - ∫dcosx/cosx. Let t=cosx, then ∫ tgxdx = -∫ dt/t = - ln|t| + C = -ln|cosx|+C.

Solution.

Example3.33. Find .

Solution. =

.

Example3.34 . Find ∫arctgxdx.

Solution. Let's integrate by parts. Let us denote u=arctgx, dv=dx. Then du = dx/(x 2 +1), v=x, whence ∫arctgxdx = xarctgx - ∫ xdx/(x 2 +1) = xarctgx + 1/2 ln(x 2 +1) +C; because
∫xdx/(x 2 +1) = 1/2 ∫d(x 2 +1)/(x 2 +1) = 1/2 ln(x 2 +1) +C.

Example3.35 . Calculate ∫lnxdx.

Solution. Applying the integration by parts formula, we obtain:
u=lnx, dv=dx, du=1/x dx, v=x. Then ∫lnxdx = xlnx - ∫x 1/x dx =
= xlnx - ∫dx + C= xlnx - x + C.

Example3.36 . Calculate ∫e x sinxdx.

Solution. Let us denote u = e x, dv = sinxdx, then du = e x dx, v =∫ sinxdx= - cosx → ∫ e x sinxdx = - e x cosx + ∫ e x cosxdx. We also integrate the integral ∫e x cosxdx by parts: u = e x , dv = cosxdx, du=e x dx, v=sinx. We have:
∫ e x cosxdx = e x sinx - ∫ e x sinxdx. We obtained the relation ∫e x sinxdx = - e x cosx + e x sinx - ∫ e x sinxdx, from which 2∫e x sinx dx = - e x cosx + e x sinx + C.

Example 3.37. Calculate J = ∫cos(lnx)dx/x.

Solution. Since dx/x = dlnx, then J= ∫cos(lnx)d(lnx). Replacing lnx through t, we arrive at the table integral J = ∫ costdt = sint + C = sin(lnx) + C.

Example 3.38 . Calculate J = .

Solution. Considering that = d(lnx), we substitute lnx = t. Then J = .

Definition 1

The antiderivative $F(x)$ for the function $y=f(x)$ on the segment $$ is a function that is differentiable at each point of this segment and the following equality holds for its derivative:

Definition 2

The set of all antiderivatives of a given function $y=f(x)$, defined on a certain segment, is called the indefinite integral of a given function $y=f(x)$. The indefinite integral is denoted by the symbol $\int f(x)dx $.

From the table of derivatives and Definition 2 we obtain the table of basic integrals.

Example 1

Check the validity of formula 7 from the table of integrals:

\[\int tgxdx =-\ln |\cos x|+C,\, \, C=const.\]

Let's differentiate the right-hand side: $-\ln |\cos x|+C$.

\[\left(-\ln |\cos x|+C\right)"=-\frac(1)(\cos x) \cdot (-\sin x)=\frac(\sin x)(\cos x) =tgx\]

Example 2

Check the validity of formula 8 from the table of integrals:

\[\int ctgxdx =\ln |\sin x|+C,\, \, C=const.\]

Let's differentiate the right-hand side: $\ln |\sin x|+C$.

\[\left(\ln |\sin x|\right)"=\frac(1)(\sin x) \cdot \cos x=ctgx\]

The derivative turned out to be equal to the integrand. Therefore, the formula is correct.

Example 3

Check the validity of formula 11" from the table of integrals:

\[\int \frac(dx)(a^(2) +x^(2) ) =\frac(1)(a) arctg\frac(x)(a) +C,\, \, C=const .\]

Let's differentiate the right-hand side: $\frac(1)(a) arctg\frac(x)(a) +C$.

\[\left(\frac(1)(a) arctg\frac(x)(a) +C\right)"=\frac(1)(a) \cdot \frac(1)(1+\left( \frac(x)(a) \right)^(2) ) \cdot \frac(1)(a) =\frac(1)(a^(2) ) \cdot \frac(a^(2) ) (a^(2) +x^(2) ) \]

The derivative turned out to be equal to the integrand. Therefore, the formula is correct.

Example 4

Check the validity of formula 12 from the table of integrals:

\[\int \frac(dx)(a^(2) -x^(2) ) =\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+ C,\, \, C=const.\]

Let's differentiate the right-hand side: $\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+C$.

$\left(\frac(1)(2a) \ln \left|\frac(a+x)(a-x) \right|+C\right)"=\frac(1)(2a) \cdot \frac( 1)(\frac(a+x)(a-x) ) \cdot \left(\frac(a+x)(a-x) \right)"=\frac(1)(2a) \cdot \frac(a-x)( a+x) \cdot \frac(a-x+a+x)((a-x)^(2) ) =\frac(1)(2a) \cdot \frac(a-x)(a+x) \cdot \ frac(2a)((a-x)^(2) ) =\frac(1)(a^(2) -x^(2) ) $The derivative turned out to be equal to the integrand. Therefore, the formula is correct.

Example 5

Check the validity of formula 13" from the table of integrals:

\[\int \frac(dx)(\sqrt(a^(2) -x^(2) ) ) =\arcsin \frac(x)(a) +C,\, \, C=const.\]

Let's differentiate the right-hand side: $\arcsin \frac(x)(a) +C$.

\[\left(\arcsin \frac(x)(a) +C\right)"=\frac(1)(\sqrt(1-\left(\frac(x)(a) \right)^(2 ) ) ) \cdot \frac(1)(a) =\frac(a)(\sqrt(a^(2) -x^(2) ) ) \cdot \frac(1)(a) =\frac( 1)(\sqrt(a^(2) -x^(2) ) ) \]

The derivative turned out to be equal to the integrand. Therefore, the formula is correct.

Example 6

Check the validity of formula 14 from the table of integrals:

\[\int \frac(dx)(\sqrt(x^(2) \pm a^(2) ) =\ln |x+\sqrt(x^(2) \pm a^(2) ) |+ C,\, \, C=const.\]

Let's differentiate the right-hand side: $\ln |x+\sqrt(x^(2) \pm a^(2) ) |+C$.

\[\left(\ln |x+\sqrt(x^(2) \pm a^(2) ) |+C\right)"=\frac(1)(x+\sqrt(x^(2) \pm a^(2) ) \cdot \left(x+\sqrt(x^(2) \pm a^(2) ) \right)"=\frac(1)(x+\sqrt(x^(2) \ pm a^(2) ) \cdot \left(1+\frac(1)(2\sqrt(x^(2) \pm a^(2) ) ) \cdot 2x\right)=\] \[ =\frac(1)(x+\sqrt(x^(2) \pm a^(2) ) \cdot \frac(\sqrt(x^(2) \pm a^(2) ) +x)( \sqrt(x^(2) \pm a^(2) ) =\frac(1)(\sqrt(x^(2) \pm a^(2) ) ) \]

The derivative turned out to be equal to the integrand. Therefore, the formula is correct.

Example 7

Find the integral:

\[\int \left(\cos (3x+2)+5x\right) dx.\]

Let's use the sum integral theorem:

\[\int \left(\cos (3x+2)+5x\right) dx=\int \cos (3x+2)dx +\int 5xdx .\]

Let us use the theorem about placing a constant factor outside the integral sign:

\[\int \cos (3x+2)dx +\int 5xdx =\int \cos (3x+2)dx +5\int xdx .\]

According to the table of integrals:

\[\int \cos x dx=\sin x+C;\] \[\int xdx =\frac(x^(2) )(2) +C.\]

When calculating the first integral, we use rule 3:

\[\int \cos (3x+2) dx=\frac(1)(3) \sin (3x+2)+C_(1) .\]

Hence,

\[\int \left(\cos (3x+2)+5x\right) dx=\frac(1)(3) \sin (3x+2)+C_(1) +\frac(5x^(2) )(2) +C_(2) =\frac(1)(3) \sin (3x+2)+\frac(5x^(2) )(2) +C,\, \, C=C_(1 ) +C_(2) \]