INDETERMINATE INTEGRAL

We begin to study integrals, which are widely used in many fields of technology. Let's start our study with the indefinite integral.

Antiderivative and indefinite integral

The main task of differential calculus is the differentiation of given functions, in other words, the task of finding the rate of change of a given function. Numerous questions of science and technology lead to the formulation of the inverse problem: according to given function f (x) restore a function F(x) for which f (x) would be a derivative: F ¢ (x) = f (x).

Definition. A function F(x) is called antiderivative for f (x) if

F ¢ (x) = f (x) or dF(x) = f (x) dx.

Examples. 1) f (x) = 3x 2 , F (x) = x 3 ;

2) f (x) = cosx, F(x) = sinx.

It is easy to see that this function f (x) = 3x 2 corresponds not to one antiderivative, but to a set: x 3 ; x 3 + 1; x 3 - 1; x 3 + 5; x 3 - 100; x 3 + C.

Indeed, (x 3)¢ = 3x 2 ; (x 3 + 1)¢ = 3x 2 ; (x 3 - 1) ¢ = 3x 2 ; . . . . (x 3 + C)¢ = 3x 2.

In general, if F(x) is an antiderivative of a given function f (x), then the function F(x) + c, "СОR will also be an antiderivative function, since:

¢ = F¢(x) = f (x).

Is the set of all antiderivatives of f (x) exhausted by expressions of the form F(x) + C, or are there antiderivatives of this function that cannot be obtained from F(x) + C for any value of C? It turns out that the statement is true: there are no other antiderivatives of the function f (x). In other words, if F 1 (x) and F 2 (x) are two antiderivatives for f (x), then F 1 (x) = F 2 (x) + C,

where C is some constant.

Indeed, because F 1 (x) and F 2 (x) are antiderivatives for f (x), then

Let's consider the difference for all x.

Let x 0 be some fixed value of the argument,

x is an arbitrary other value.

According to Lagrange's formula

where is some number between x 0 and x. Because:

Does every function f (x) have an antiderivative?

Theorem. If a function f (x) is continuous on some interval, then it has an antiderivative on it (no proof).

Definition. If F (x) is some kind of antiderivative for f (x), then the expression F (x) + C, where C is an arbitrary constant, is called an indefinite integral and is denoted: , while f (x) is called an integrand function, and the expression f (x) dx - by the integrand:

The action of finding an indefinite integral, otherwise, finding all antiderivatives of a given function, is called integration this function. It is obvious that the operations of differentiation and integration are mutually inverse.

Addition and subtraction, exponentiation and root extraction, multiplication and division provide examples of inverse mathematical operations.

Definition of an antiderivative function

  • Function y=F(x) is called the antiderivative of the function y=f(x) at a given interval X, if for everyone X ∈X equality holds: F′(x) = f(x)

Can be read in two ways:

  1. f derivative of a function F
  2. F antiderivative of a function f

Property of antiderivatives

  • If F(x)- antiderivative of a function f(x) on a given interval, then the function f(x) has infinitely many antiderivatives, and all these antiderivatives can be written in the form F(x) + C, where C is an arbitrary constant.

Geometric interpretation

  • Graphs of all antiderivatives of a given function f(x) are obtained from the graph of any one antiderivative by parallel translations along the O axis at.

Rules for calculating antiderivatives

  1. The antiderivative of the sum is equal to the sum of the antiderivatives. If F(x)- antiderivative for f(x), and G(x) is an antiderivative for g(x), That F(x) + G(x)- antiderivative for f(x) + g(x).
  2. The constant factor can be taken out of the sign of the derivative. If F(x)- antiderivative for f(x), And k- constant, then k·F(x)- antiderivative for k f(x).
  3. If F(x)- antiderivative for f(x), And k, b- constant, and k ≠ 0, That 1/k F(kx + b)- antiderivative for f(kx + b).

Remember!

Any function F(x) = x 2 + C , where C is an arbitrary constant, and only such a function is an antiderivative for the function f(x) = 2x.

  • For example:

    F"(x) = (x 2 + 1)" = 2x = f(x);

    f(x) = 2x, because F"(x) = (x 2 – 1)" = 2x = f(x);

    f(x) = 2x, because F"(x) = (x 2 –3)" = 2x = f(x);

Relationship between the graphs of a function and its antiderivative:

  1. If the graph of a function f(x)>0 on the interval, then the graph of its antiderivative F(x) increases over this interval.
  2. If the graph of a function f(x) on the interval, then the graph of its antiderivative F(x) decreases over this interval.
  3. If f(x)=0, then the graph of its antiderivative F(x) at this point changes from increasing to decreasing (or vice versa).

To denote an antiderivative, the sign of an indefinite integral is used, that is, an integral without indicating the limits of integration.

Indefinite integral

Definition:

  • Not definite integral from the function f(x) the expression F(x) + C is called, that is, the set of all antiderivatives of a given function f(x). The indefinite integral is denoted as follows: \int f(x) dx = F(x) + C
  • f(x)- called the integrand function;
  • f(x) dx- called the integrand;
  • x- called the variable of integration;
  • F(x)- one of the antiderivatives of the function f(x);
  • WITH- arbitrary constant.

Properties of the indefinite integral

  1. The derivative of the indefinite integral is equal to the integrand: (\int f(x) dx)\prime= f(x) .
  2. The constant factor of the integrand can be taken out of the integral sign: \int k \cdot f(x) dx = k \cdot \int f(x) dx.
  3. The integral of the sum (difference) of functions is equal to the sum (difference) of the integrals of these functions: \int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx.
  4. If k, b are constants, and k ≠ 0, then \int f(kx + b) dx = \frac ( 1 ) ( k ) \cdot F(kx + b) + C.

Table of antiderivatives and indefinite integrals

Function

f(x)

Antiderivative

F(x) + C

Indefinite integrals

\int f(x) dx = F(x) + C
0 C \int 0 dx = C
f(x) = k F(x) = kx + C \int kdx = kx + C
f(x) = x^m, m\not =-1 F(x) = \frac ( x^ ( m+1 ) ) ( m+1 ) + C \int x ( ^m ) dx = \frac ( x^ ( m+1 ) ) ( m+1 ) + C
f(x) = \frac ( 1 ) ( x ) F(x) = l n \lvert x \rvert + C \int \frac ( dx ) ( x ) = l n \lvert x \rvert + C
f(x) = e^x F(x) = e^x + C \int e ( ^x ) dx = e^x + C
f(x) = a^x F(x) = \frac ( a^x ) ( l na ) + C \int a ( ^x ) dx = \frac ( a^x ) ( l na ) + C
f(x) = \sin x F(x) = -\cos x + C \int \sin x dx = -\cos x + C
f(x) = \cos x F(x) =\sin x + C \int \cos x dx = \sin x + C
f(x) = \frac ( 1 ) ( \sin ( ^2 ) x ) F(x) = -\ctg x + C \int \frac ( dx ) ( \sin ( ^2 ) x ) = -\ctg x + C
f(x) = \frac ( 1 ) ( \cos ( ^2 ) x ) F(x) = \tg x + C \int \frac ( dx ) ( \sin ( ^2 ) x ) = \tg x + C
f(x) = \sqrt ( x ) F(x) =\frac ( 2x \sqrt ( x ) ) ( 3 ) + C
f(x) =\frac ( 1 ) ( \sqrt ( x ) ) F(x) =2\sqrt ( x ) + C
f(x) =\frac ( 1 ) ( \sqrt ( 1-x^2 ) ) F(x)=\arcsin x + C \int \frac ( dx ) ( \sqrt ( 1-x^2 ) ) =\arcsin x + C
f(x) =\frac ( 1 ) ( \sqrt ( 1+x^2 ) ) F(x)=\arctg x + C \int \frac ( dx ) ( \sqrt ( 1+x^2 ) ) =\arctg x + C
f(x)=\frac ( 1 ) ( \sqrt ( a^2-x^2 ) ) F(x)=\arcsin \frac ( x ) ( a ) + C \int \frac ( dx ) ( \sqrt ( a^2-x^2 ) ) =\arcsin \frac ( x ) ( a ) + C
f(x)=\frac ( 1 ) ( \sqrt ( a^2+x^2 ) ) F(x)=\arctg \frac ( x ) ( a ) + C \int \frac ( dx ) ( \sqrt ( a^2+x^2 ) ) = \frac ( 1 ) ( a ) \arctg \frac ( x ) ( a ) + C
f(x) =\frac ( 1 ) ( 1+x^2 ) F(x)=\arctg + C \int \frac ( dx ) ( 1+x^2 ) =\arctg + C
f(x)=\frac ( 1 ) ( \sqrt ( x^2-a^2 ) ) (a \not= 0) F(x)=\frac ( 1 ) ( 2a ) l n \lvert \frac ( x-a ) ( x+a ) \rvert + C \int \frac ( dx ) ( \sqrt ( x^2-a^2 ) ) =\frac ( 1 ) ( 2a ) l n \lvert \frac ( x-a ) ( x+a ) \rvert + C
f(x)=\tg x F(x)= - l n \lvert \cos x \rvert + C \int \tg x dx =- l n \lvert \cos x \rvert + C
f(x)=\ctg x F(x)= l n \lvert \sin x \rvert + C \int \ctg x dx = l n \lvert \sin x \rvert + C
f(x)=\frac ( 1 ) ( \sin x ) F(x)= l n \lvert \tg \frac ( x ) ( 2 ) \rvert + C \int \frac ( dx ) ( \sin x ) = l n \lvert \tg \frac ( x ) ( 2 ) \rvert + C
f(x)=\frac ( 1 ) ( \cos x ) F(x)= l n \lvert \tg (\frac ( x ) ( 2 ) +\frac ( \pi ) ( 4 )) \rvert + C \int \frac ( dx ) ( \cos x ) = l n \lvert \tg (\frac ( x ) ( 2 ) +\frac ( \pi ) ( 4 )) \rvert + C


Newton–Leibniz formula

Let f(x) this function F its arbitrary antiderivative.

\int_ ( a ) ^ ( b ) f(x) dx =F(x)|_ ( a ) ^ ( b )= F(b) - F(a)

Where F(x)- antiderivative for f(x)

That is, the integral of the function f(x) on an interval is equal to the difference of antiderivatives at points b And a.

Area of ​​a curved trapezoid

Curvilinear trapezoid is a figure bounded by the graph of a function that is non-negative and continuous on an interval f, Ox axis and straight lines x = a And x = b.

Square curved trapezoid found using the Newton-Leibniz formula:

S= \int_ ( a ) ^ ( b ) f(x) dx

The main task of differential calculus is to find the differential of a given function or its derivative. Integral calculus solves the inverse problem: given a differential, and, consequently, the derivative of an unknown function F(x), you need to define this function. In other words, having the expression

or accordingly

,

Where f(x)– known function, need to find the function F(x). Required function F(x) it is called antiderivative function in relation to function f(x). For simplicity, we will assume that equality (1) holds on some finite or infinite interval.

Definition: Antiderivative function for a given function f(x) on a given interval such a function is called F(x), whose derivative is equal to f(x) or whose differential is equal to f(x)dx on the interval under consideration.

For example, one of the antiderivative functions for a function will be , because . The antiderivative function is not unique, since etc., and therefore the functions and so on. are also antiderivatives for the function. Consequently, this function has an infinite number of antiderivatives.

In our example, every two antiderivatives differed from each other by a certain constant term. Let us show that this will also take place in the general case.

Theorem: Two different antiderivatives of the same function defined on a certain interval differ from each other on this interval by a constant term.

Proof: In fact, let f(x)– some function defined on the interval , And F 1 (x), F 2 (x)– its primitives, i.e.

And .

From here .

y=F 1 (x)
y=F 2 (x)
F 1 (x)
F2(x)
WITH
M 2
M 1
X
α
X
α
Y
Rice. 1.

But if two functions have the same derivatives, then these functions differ from each other by a constant term. Hence,

F 1 (x) - F 2 (x) = C,

Where WITH– constant value. The theorem has been proven.

Consider a geometric illustration. If y = F 1 (x) and Y = F 2 (x)

Antiderivatives of the same function f(x), then tangents to their graphs at points with a common abscissa X parallel to each other (Fig. 1):

tgα = = f(x).

In this case, the distance between these curves along the axis OU remains constant: F 2 (x) – F 1 (x) = C, those. these curves are in a sense “parallel” to each other.

Consequence: Adding to any antiderivative function f(x), defined on the interval , all possible constants WITH, we will get all the antiderivatives for the function f(x).

In fact, if F(x) there is an antiderivative function for f(x), then the function F(x)+C, Where WITH- any constant will also be an antiderivative of the function f(x), because .


On the other hand, we have proven that every antiderivative of the function f(x) can be obtained from a function F(x) by adding to it a properly selected constant term WITH.

Therefore, the expression F(x) + C, Where , (2)

Where F(x)– any antiderivative for a function f(x), exhausts the entire set of antiderivatives for a given function f(x).

In what follows we will assume, unless otherwise stated explicitly, that the function under consideration f(x) defined and continuous on some finite or infinite interval .

Let us now introduce the basic concept of integral calculus - the concept of an indefinite integral.

Definition: General expression for all antiderivatives of a given continuous function f(x) called the indefinite integral of the function f(x) or from the differential expression f(x)dx and is indicated by the symbol .

In this case, the function f(x) is called the integrand, and the expression f(x)dx is called an integrand.

According to the definition of the indefinite integral, we can write

, (3)

C 4
C 3
C 2
C 1
X
Y
Rice. 2.
Where , constant WITH can take any value and is therefore called an arbitrary constant.

Example. As we have seen, for a function one of the antiderivatives is the function. That's why .

Geometrically indefinite integral y=F(x)+C represents a family of “parallel” curves (Fig. 2).

We have seen that the derivative has numerous uses: the derivative is the speed of movement (or, more generally, the speed of any process); derivative is the slope of the tangent to the graph of the function; using the derivative, you can examine a function for monotonicity and extrema; the derivative helps solve optimization problems.

But in real life Inverse problems also have to be solved: for example, along with the problem of finding the speed according to a known law of motion, there is also the problem of restoring the law of motion according to a known speed. Let's consider one of these problems.

Example 1. A material point moves in a straight line, its speed at time t is given by the formula u = tg. Find the law of motion.

Solution. Let s = s(t) be the desired law of motion. It is known that s"(t) = u"(t). This means that to solve the problem you need to choose function s = s(t), whose derivative is equal to tg. It's not hard to guess that

Let us immediately note that the example is solved correctly, but incompletely. We found that, in fact, the problem has infinitely many solutions: any function of the form an arbitrary constant can serve as a law of motion, since


To make the task more specific, we needed to fix the initial situation: indicate the coordinate of a moving point at some point in time, for example, at t=0. If, say, s(0) = s 0, then from the equality we obtain s(0) = 0 + C, i.e. S 0 = C. Now the law of motion is uniquely defined:
In mathematics, reciprocal operations are assigned different names, come up with special notations: for example, squaring (x 2) and extracting square root sine(sinх) and arcsine(arcsin x), etc. The process of finding the derivative of a given function is called differentiation, and the inverse operation, i.e. the process of finding a function from a given derivative - integration.
The term “derivative” itself can be justified “in everyday terms”: the function y - f(x) “produces into existence” new feature y"= f"(x) The function y = f(x) acts as a “parent”, but mathematicians, naturally, do not call it a “parent” or “producer”, they say that it is, in relation to the function y"=f"(x), the primary image, or, in short, the antiderivative.

Definition 1. The function y = F(x) is called antiderivative for the function y = f(x) on a given interval X if for all x from X the equality F"(x)=f(x) holds.

In practice, the interval X is usually not specified, but is implied (as the natural domain of definition of the function).

Here are some examples:

1) The function y = x 2 is antiderivative for the function y = 2x, since for all x the equality (x 2)" = 2x is true.
2) the function y - x 3 is antiderivative for the function y-3x 2, since for all x the equality (x 3)" = 3x 2 is true.
3) The function y-sinх is antiderivative for the function y = cosx, since for all x the equality (sinx)" = cosx is true.
4) The function is antiderivative for a function on the interval since for all x > 0 the equality is true
In general, knowing the formulas for finding derivatives, it is not difficult to compile a table of formulas for finding antiderivatives.


We hope you understand how this table is compiled: the derivative of the function, which is written in the second column, is equal to the function that is written in the corresponding row of the first column (check it, don’t be lazy, it’s very useful). For example, for the function y = x 5 the antiderivative, as you will establish, is the function (see the fourth row of the table).

Notes: 1. Below we will prove the theorem that if y = F(x) is an antiderivative for the function y = f(x), then the function y = f(x) has infinitely many antiderivatives and they all have the form y = F(x ) + C. Therefore, it would be more correct to add the term C everywhere in the second column of the table, where C is an arbitrary real number.
2. For the sake of brevity, sometimes instead of the phrase “the function y = F(x) is an antiderivative of the function y = f(x),” they say F(x) is an antiderivative of f(x).”

2. Rules for finding antiderivatives

When finding antiderivatives, as well as when finding derivatives, not only formulas are used (they are listed in the table on p. 196), but also some rules. They are directly related to the corresponding rules for calculating derivatives.

We know that the derivative of a sum is equal to the sum of its derivatives. This rule generates the corresponding rule for finding antiderivatives.

Rule 1. The antiderivative of a sum is equal to the sum of the antiderivatives.

We draw your attention to the somewhat “lightness” of this formulation. In fact, one should formulate the theorem: if the functions y = f(x) and y = g(x) have antiderivatives on the interval X, respectively y-F(x) and y-G(x), then the sum of the functions y = f(x)+g(x) has an antiderivative on the interval X, and this antiderivative is the function y = F(x)+G(x). But usually, when formulating rules (and not theorems), they leave only keywords- this makes it more convenient to apply the rule in practice

Example 2. Find the antiderivative for the function y = 2x + cos x.

Solution. The antiderivative for 2x is x"; the antiderivative for cox is sin x. This means that the antiderivative for the function y = 2x + cos x will be the function y = x 2 + sin x (and in general any function of the form Y = x 1 + sinx + C) .
We know that the constant factor can be taken out of the sign of the derivative. This rule generates the corresponding rule for finding antiderivatives.

Rule 2. The constant factor can be taken out of the sign of the antiderivative.

Example 3.

Solution. a) The antiderivative for sin x is -soz x; This means that for the function y = 5 sin x the antiderivative function will be the function y = -5 cos x.

b) The antiderivative for cos x is sin x; This means that the antiderivative of a function is the function
c) The antiderivative for x 3 is the antiderivative for x, the antiderivative for the function y = 1 is the function y = x. Using the first and second rules for finding antiderivatives, we find that the antiderivative for the function y = 12x 3 + 8x-1 is the function
Comment. As is known, the derivative of a product is not equal to the product of derivatives (the rule for differentiating a product is more complex) and the derivative of a quotient is not equal to the quotient of derivatives. Therefore, there are no rules for finding the antiderivative of the product or the antiderivative of the quotient of two functions. Be careful!
Let us obtain another rule for finding antiderivatives. We know that the derivative of the function y = f(kx+m) is calculated by the formula

This rule generates the corresponding rule for finding antiderivatives.
Rule 3. If y = F(x) is an antiderivative for the function y = f(x), then the antiderivative for the function y=f(kx+m) is the function

Indeed,


This means that it is an antiderivative for the function y = f(kx+m).
The meaning of the third rule is as follows. If you know that the antiderivative of the function y = f(x) is the function y = F(x), and you need to find the antiderivative of the function y = f(kx+m), then proceed like this: take the same function F, but instead of the argument x, substitute the expression kx+m; in addition, do not forget to write “correction factor” before the function sign
Example 4. Find antiderivatives for given functions:

Solution, a) The antiderivative for sin x is -soz x; This means that for the function y = sin2x the antiderivative will be the function
b) The antiderivative for cos x is sin x; This means that the antiderivative of a function is the function

c) The antiderivative for x 7 means that for the function y = (4-5x) 7 the antiderivative will be the function

3. Indefinite integral

We have already noted above that the problem of finding an antiderivative for a given function y = f(x) has more than one solution. Let's discuss this issue in more detail.

Proof. 1. Let y = F(x) be the antiderivative for the function y = f(x) on the interval X. This means that for all x from X the equality x"(x) = f(x) holds. Let us find the derivative of any function of the form y = F(x)+C:
(F(x) +C) = F"(x) +C = f(x) +0 = f(x).

So, (F(x)+C) = f(x). This means that y = F(x) + C is an antiderivative for the function y = f(x).
Thus, we have proven that if the function y = f(x) has an antiderivative y=F(x), then the function (f = f(x) has infinitely many antiderivatives, for example, any function of the form y = F(x) +C is an antiderivative.
2. Let us now prove that the indicated type of functions exhausts the entire set of antiderivatives.

Let y=F 1 (x) and y=F(x) be two antiderivatives for the function Y = f(x) on the interval X. This means that for all x from the interval X the following relations hold: F^ (x) = f (X); F"(x) = f(x).

Let's consider the function y = F 1 (x) -.F(x) and find its derivative: (F, (x) -F(x))" = F[(x)-F(x) = f(x) - f(x) = 0.
It is known that if the derivative of a function on an interval X is identically equal to zero, then the function is constant on the interval X (see Theorem 3 from § 35). This means that F 1 (x) - F (x) = C, i.e. Fx) = F(x)+C.

The theorem has been proven.

Example 5. The law of change of speed with time is given: v = -5sin2t. Find the law of motion s = s(t), if it is known that at time t=0 the coordinate of the point was equal to the number 1.5 (i.e. s(t) = 1.5).

Solution. Since speed is a derivative of the coordinate as a function of time, we first need to find the antiderivative of the speed, i.e. antiderivative for the function v = -5sin2t. One of such antiderivatives is the function , and the set of all antiderivatives has the form:

To find specific meaning constant C, we use the initial conditions, according to which s(0) = 1.5. Substituting the values ​​t=0, S = 1.5 into formula (1), we obtain:

Substituting the found value of C into formula (1), we obtain the law of motion that interests us:

Definition 2. If a function y = f(x) has an antiderivative y = F(x) on an interval X, then the set of all antiderivatives, i.e. the set of functions of the form y = F(x) + C is called the indefinite integral of the function y = f(x) and is denoted by:

(read: “indefinite integral ef from x de x”).
In the next paragraph we will find out what is hidden meaning the indicated designation.
Based on the table of antiderivatives available in this section, we will compile a table of the main indefinite integrals:

Based on the above three rules for finding antiderivatives, we can formulate the corresponding integration rules.

Rule 1. The integral of the sum of functions is equal to the sum of the integrals of these functions:

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

Rule 3. If

Example 6. Find indefinite integrals:

Solution, a) Using the first and second rules of integration, we obtain:


Now let's use the 3rd and 4th integration formulas:

As a result we get:

b) Using the third rule of integration and formula 8, we obtain:


c) To directly find a given integral, we have neither the corresponding formula nor the corresponding rule. In such cases, previously performed identical transformations of the expression contained under the integral sign sometimes help.

Let's take advantage trigonometric formula Degree reduction:

Then we find sequentially:

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Function F(x ) called antiderivative for function f(x) on a given interval, if for all x from this interval the equality holds

F"(x ) = f(x ) .

For example, the function F(x) = x 2 f(x ) = 2X , because

F"(x) = (x 2 )" = 2x = f(x). ◄

The main property of the antiderivative

If F(x) - antiderivative of a function f(x) on a given interval, then the function f(x) has infinitely many antiderivatives, and all these antiderivatives can be written in the form F(x) + C, Where WITH is an arbitrary constant.

For example.

Function F(x) = x 2 + 1 is an antiderivative of the function

f(x ) = 2X , because F"(x) = (x 2 + 1 )" = 2 x = f(x);

function F(x) = x 2 - 1 is an antiderivative of the function

f(x ) = 2X , because F"(x) = (x 2 - 1)" = 2x = f(x) ;

function F(x) = x 2 - 3 is an antiderivative of the function

f(x) = 2X , because F"(x) = (x 2 - 3)" = 2 x = f(x);

any function F(x) = x 2 + WITH , Where WITH - an arbitrary constant, and only such a function is an antiderivative of the function f(x) = 2X . ◄

Rules for calculating antiderivatives

  1. If F(x) - antiderivative for f(x) , A G(x) - antiderivative for g(x) , That F(x) + G(x) - antiderivative for f(x) + g(x) . In other words, the antiderivative of the sum is equal to the sum of the antiderivatives .
  2. If F(x) - antiderivative for f(x) , And k - constant, then k · F(x) - antiderivative for k · f(x) . In other words, the constant factor can be taken out of the sign of the derivative .
  3. If F(x) - antiderivative for f(x) , And k,b- constant, and k ≠ 0 , That 1 / k F( k x+ b ) - antiderivative for f(k x+ b) .

Indefinite integral

Indefinite integral from function f(x) called expression F(x) + C, that is, the set of all antiderivatives of a given function f(x) . The indefinite integral is denoted as follows:

∫ f(x) dx = F(x) + C ,

f(x)- they call integrand function ;

f(x) dx- they call integrand ;

x - they call integration variable ;

F(x) - one of the primitive functions f(x) ;

WITH is an arbitrary constant.

For example, ∫ 2 x dx =X 2 + WITH , ∫ cosx dx = sin X + WITH and so on. ◄

The word "integral" comes from the Latin word integer , which means "restored". Considering the indefinite integral of 2 x, we seem to restore the function X 2 , whose derivative is equal to 2 x. Restoring a function from its derivative, or, what is the same, finding an indefinite integral over a given integrand is called integration this function. Integration is the inverse operation of differentiation. In order to check whether the integration was performed correctly, it is enough to differentiate the result and obtain the integrand.

Basic properties of the indefinite integral

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

    2. (∫ f(x) dx )" = f(x) .

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

    3. ∫ k · f(x) dx = k · ∫ f(x) dx .

  3. The integral of the sum (difference) of functions is equal to the sum (difference) of the integrals of these functions:

    4. ∫ ( f(x) ± g(x ) ) dx = ∫ f(x) dx ± ∫ g(x ) dx .

  4. If k,b- constant, and k ≠ 0 , That

    5. ∫ f ( k x+ b) dx = 1 / k F( k x+ b ) + C .

Table of antiderivatives and indefinite integrals


f(x)
F(x) + C
∫ f(x) dx = F(x) + C
I.
$$0$$
$$C$$
$$\int 0dx=C$$
II.
$$k$$
$$kx+C$$
$$\int kdx=kx+C$$
III.
$$x^n~(n\neq-1)$$
$$\frac(x^(n+1))(n+1)+C$$
$$\int x^ndx=\frac(x^(n+1))(n+1)+C$$
IV.
$$\frac(1)(x)$$
$$\ln |x|+C$$
$$\int\frac(dx)(x)=\ln |x|+C$$
V.
$$\sin x$$
$$-\cos x+C$$
$$\int\sin x~dx=-\cos x+C$$
VI.
$$\cos x$$
$$\sin x+C$$
$$\int\cos x~dx=\sin x+C$$
VII.
$$\frac(1)(\cos^2x)$$
$$\textrm(tg) ~x+C$$
$$\int\frac(dx)(\cos^2x)=\textrm(tg) ~x+C$$
VIII.
$$\frac(1)(\sin^2x)$$
$$-\textrm(ctg) ~x+C$$
$$\int\frac(dx)(\sin^2x)=-\textrm(ctg) ~x+C$$
IX.
$$e^x$$
$$e^x+C$$
$$\int e^xdx=e^x+C$$
X.
$$a^x$$
$$\frac(a^x)(\ln a)+C$$
$$\int a^xdx=\frac(a^x)(\ln a)+C$$
XI.
$$\frac(1)(\sqrt(1-x^2))$$
$$\arcsin x +C$$
$$\int\frac(dx)(\sqrt(1-x^2))=\arcsin x +C$$
XII.
$$\frac(1)(\sqrt(a^2-x^2))$$
$$\arcsin \frac(x)(a)+C$$
$$\int\frac(dx)(\sqrt(a^2-x^2))=\arcsin \frac(x)(a)+C$$
XIII.
$$\frac(1)(1+x^2)$$
$$\textrm(arctg) ~x+C$$
$$\int \frac(dx)(1+x^2)=\textrm(arctg) ~x+C$$
XIV.
$$\frac(1)(a^2+x^2)$$
$$\frac(1)(a)\textrm(arctg) ~\frac(x)(a)+C$$
$$\int \frac(dx)(a^2+x^2)=\frac(1)(a)\textrm(arctg) ~\frac(x)(a)+C$$
XV.
$$\frac(1)(\sqrt(a^2+x^2))$$
$$\ln|x+\sqrt(a^2+x^2)|+C$$
$$\int\frac(dx)(\sqrt(a^2+x^2))=\ln|x+\sqrt(a^2+x^2)|+C$$
XVI.
$$\frac(1)(x^2-a^2)~(a\neq0)$$
$$\frac(1)(2a)\ln \begin(vmatrix)\frac(x-a)(x+a)\end(vmatrix)+C$$
$$\int\frac(dx)(x^2-a^2)=\frac(1)(2a)\ln \begin(vmatrix)\frac(x-a)(x+a)\end(vmatrix)+ C$$
XVII.
$$\textrm(tg) ~x$$
$$-\ln |\cos x|+C$$
$$\int \textrm(tg) ~x ~dx=-\ln |\cos x|+C$$
XVIII.
$$\textrm(ctg) ~x$$
$$\ln |\sin x|+C$$
$$\int \textrm(ctg) ~x ~dx=\ln |\sin x|+C$$
XIX.
$$ \frac(1)(\sin x) $$
$$\ln \begin(vmatrix)\textrm(tg) ~\frac(x)(2)\end(vmatrix)+C $$
$$\int \frac(dx)(\sin x)=\ln \begin(vmatrix)\textrm(tg) ~\frac(x)(2)\end(vmatrix)+C $$
XX.
$$ \frac(1)(\cos x) $$
$$\ln \begin(vmatrix)\textrm(tg)\left (\frac(x)(2)+\frac(\pi )(4) \right) \end(vmatrix)+C $$
$$\int \frac(dx)(\cos x)=\ln \begin(vmatrix)\textrm(tg)\left (\frac(x)(2)+\frac(\pi )(4) \right ) \end(vmatrix)+C $$
The antiderivative and indefinite integrals given in this table are usually called tabular antiderivatives And table integrals .

Definite integral

Let in between [a; b] given continuous function y = f(x) , Then definite integral from a to b functions f(x) is called the increment of the antiderivative F(x) this function, that is

$$\int_(a)^(b)f(x)dx=F(x)|(_a^b) = ~~F(a)-F(b).$$

Numbers a And b are called accordingly lower And top limits of integration.

Basic rules for calculating the definite integral

1. \(\int_(a)^(a)f(x)dx=0\);

2. \(\int_(a)^(b)f(x)dx=- \int_(b)^(a)f(x)dx\);

3. \(\int_(a)^(b)kf(x)dx=k\int_(a)^(b)f(x)dx,\) where k - constant;

4. \(\int_(a)^(b)(f(x) ± g(x))dx=\int_(a)^(b)f(x) dx±\int_(a)^(b) g(x)dx\);

5. \(\int_(a)^(b)f(x)dx=\int_(a)^(c)f(x)dx+\int_(c)^(b)f(x)dx\);

6. \(\int_(-a)^(a)f(x)dx=2\int_(0)^(a)f(x)dx\), where f(x) — even function;

7. \(\int_(-a)^(a)f(x)dx=0\), where f(x) is an odd function.

Comment . In all cases, it is assumed that the integrands are integrable on numerical intervals, the boundaries of which are the limits of integration.

Geometric and physical meaning of the definite integral

Geometric meaning
definite integral

Physical meaning
definite integral



Square S curvilinear trapezoid (a figure limited by the graph of a continuous positive on the interval [a; b] functions f(x) , axis Ox and straight x=a , x=b ) is calculated by the formula

$$S=\int_(a)^(b)f(x)dx.$$

Path s, which the material point has overcome, moving rectilinearly with a speed varying according to the law v(t) , for a period of time a ; b] , then the area of ​​the figure limited by the graphs of these functions and straight lines x = a , x = b , calculated by the formula

$$S=\int_(a)^(b)(f(x)-g(x))dx.$$

For example. Let's calculate the area of ​​the figure, limited by lines

y = x 2 And y = 2- x .


Let us schematically depict the graphs of these functions and highlight in a different color the figure whose area needs to be found. To find the limits of integration, we solve the equation:

x 2 = 2- x ; x 2 + x- 2 = 0 ; x 1 = -2, x 2 = 1 .

$$S=\int_(-2)^(1)((2-x)-x^2)dx=$$

$$=\int_(-2)^(1)(2-x-x^2)dx=\left (2x-\frac(x^2)(2)-\frac(x^3)(2) \right )\bigm|(_(-2)^(~1))=4\frac(1)(2). $$

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Volume of a body of rotation


If a body is obtained as a result of rotation about an axis Ox curvilinear trapezoid bounded by a continuous and non-negative graph on the interval [a; b] functions y = f(x) and straight x = a And x = b , then it is called body of rotation .

The volume of a body of revolution is calculated by the formula

$$V=\pi\int_(a)^(b)f^2(x)dx.$$

If a body of revolution is obtained as a result of the rotation of a figure bounded above and below by graphs of functions y = f(x) And y = g(x) , accordingly, then

$$V=\pi\int_(a)^(b)(f^2(x)-g^2(x))dx.$$

For example. Let's calculate the volume of a cone with radius r and height h .

Let us position the cone in a rectangular coordinate system so that its axis coincides with the axis Ox , and the center of the base was located at the origin. Generator rotation AB defines a cone. Since the equation AB

$$\frac(x)(h)+\frac(y)(r)=1,$$

$$y=r-\frac(rx)(h)$$
and for the volume of the cone we have

$$V=\pi\int_(0)^(h)(r-\frac(rx)(h))^2dx=\pi r^2\int_(0)^(h)(1-\frac( x)(h))^2dx=-\pi r^2h\cdot \frac((1-\frac(x)(h))^3)(3)|(_0^h)=-\pi r^ 2h\left (0-\frac(1)(3) \right)=\frac(\pi r^2h)(3).$$

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