Target:

• Formation of the concept of antiderivative.
• Preparation for the perception of the integral.
• Formation of computing skills.
• Cultivating a sense of beauty (the ability to see beauty in the unusual).

Mathematical analysis is a set of branches of mathematics devoted to the study of functions and their generalizations by methods of differential and integral calculus.

Until now we have studied a branch of mathematical analysis called differential calculus, the essence of which is the study of a function in the “small”.

Those. study of a function in sufficiently small neighborhoods of each definition point. One of the operations of differentiation is finding the derivative (differential) and applying it to the study of functions.

The inverse problem is no less important. If the behavior of a function in the vicinity of each point of its definition is known, then how can one reconstruct the function as a whole, i.e. throughout the entire scope of its definition. This problem is the subject of study of the so-called integral calculus.

Integration is the inverse action of differentiation. Or restoring the function f(x) from a given derivative f`(x). The Latin word “integro” means restoration.

Example No. 1.

Let (x)`=3x 2.
Let's find f(x).

Solution:

Based on the rule of differentiation, it is not difficult to guess that f(x) = x 3, because (x 3)` = 3x 2
However, you can easily notice that f(x) is not found uniquely.
As f(x) we can take
f(x)= x 3 +1
f(x)= x 3 +2
f(x)= x 3 -3, etc.

Because the derivative of each of them is equal to 3x 2. (The derivative of a constant is 0). All these functions differ from each other by a constant term. Therefore, the general solution to the problem can be written as f(x) = x 3 + C, where C is any constant real number.

Any of the found functions f(x) is called PRIMODIUM for the function F`(x)= 3x 2

Definition. A function F(x) is called antiderivative for a function f(x) on a given interval J if for all x from this interval F`(x)= f(x). So the function F(x)=x 3 is antiderivative for f(x)=3x 2 on (- ∞ ; ∞).
Since for all x ~R the equality is true: F`(x)=(x 3)`=3x 2

As we have already noticed, this function has an infinite number of antiderivatives (see example No. 1).

Example No. 2. The function F(x)=x is antiderivative for all f(x)= 1/x on the interval (0; +), because for all x from this interval, equality holds.
F`(x)= (x 1/2)`=1/2x -1/2 =1/2x

Example No. 3. The function F(x)=tg3x is an antiderivative for f(x)=3/cos3x on the interval (-n/ 2; P/ 2),
because F`(x)=(tg3x)`= 3/cos 2 3x

Example No. 4. The function F(x)=3sin4x+1/x-2 is antiderivative for f(x)=12cos4x-1/x 2 on the interval (0;∞)
because F`(x)=(3sin4x)+1/x-2)`= 4cos4x-1/x 2

Lecture 2.

Topic: Antiderivative. The main property of an antiderivative function.

When studying the antiderivative, we will rely on the following statement. Sign of constancy of a function: If on the interval J the derivative Ψ(x) of the function is equal to 0, then on this interval the function Ψ(x) is constant.

This statement can be demonstrated geometrically.

It is known that Ψ`(x)=tgα, γde α is the angle of inclination of the tangent to the graph of the function Ψ(x) at the point with abscissa x 0. If Ψ`(υ)=0 at any point in the interval J, then tanα=0 δfor any tangent to the graph of the function Ψ(x). This means that the tangent to the graph of the function at any point is parallel to the abscissa axis. Therefore, on the indicated interval, the graph of the function Ψ(x) coincides with the straight line segment y=C.

So, the function f(x)=c is constant on the interval J if f`(x)=0 on this interval.

Indeed, for an arbitrary x 1 and x 2 from the interval J, using the theorem on the mean value of a function, we can write:
f(x 2) - f(x 1) = f`(c) (x 2 - x 1), because f`(c)=0, then f(x 2)= f(x 1)

Theorem: (The main property of the antiderivative function)

If F(x) is one of the antiderivatives for the function f(x) on the interval J, then the set of all antiderivatives of this function has the form: F(x)+C, where C is any real number.

Proof:

Let F`(x) = f (x), then (F(x)+C)`= F`(x)+C`= f (x), for x Є J.
Suppose there exists Φ(x) - another antiderivative for f (x) on the interval J, i.e. Φ`(x) = f (x),
then (Φ(x) - F(x))` = f (x) – f (x) = 0, for x Є J.
This means that Φ(x) - F(x) is constant on the interval J.
Therefore, Φ(x) - F(x) = C.
From where Φ(x)= F(x)+C.
This means that if F(x) is an antiderivative for a function f (x) on the interval J, then the set of all antiderivatives of this function has the form: F(x)+C, where C is any real number.
Consequently, any two antiderivatives of a given function differ from each other by a constant term.

Example: Find the set of antiderivatives of the function f (x) = cos x. Draw graphs of the first three.

Solution: Sin x is one of the antiderivatives for the function f (x) = cos x
F(x) = Sin x+C – the set of all antiderivatives.

F 1 (x) = Sin x-1
F 2 (x) = Sin x
F 3 (x) = Sin x+1

Geometric illustration: The graph of any antiderivative F(x)+C can be obtained from the graph of the antiderivative F(x) using parallel transfer of r (0;c).

Example: For the function f (x) = 2x, find an antiderivative whose graph passes through t.M (1;4)

Solution: F(x)=x 2 +C – the set of all antiderivatives, F(1)=4 - according to the conditions of the problem.
Therefore, 4 = 1 2 +C
C = 3
F(x) = x 2 +3

Let's consider the movement of a point along a straight line. Let it take time t from the beginning of the movement the point has traveled a distance s(t). Then the instantaneous speed v(t) equal to the derivative of the function s(t), that is v(t) = s"(t).

In practice, we encounter the inverse problem: given the speed of movement of a point v(t) find the path she took s(t), that is, find such a function s(t), whose derivative is equal to v(t). Function s(t), such that s"(t) = v(t), is called the antiderivative of the function v(t).

For example, if v(t) = аt, Where A is a given number, then the function
s(t) = (аt 2) / 2v(t), because
s"(t) = ((аt 2) / 2) " = аt = v(t).

Function F(x) called the antiderivative of the function f(x) on some interval, if for all X from this gap F"(x) = f(x).

For example, the function F(x) = sin x is the antiderivative of the function f(x) = cos x, because (sin x)" = cos x; function F(x) = x 4 /4 is the antiderivative of the function f(x) = x 3, because (x 4 /4)" = x 3.

Let's consider the problem.

Task.

Prove that the functions x 3 /3, x 3 /3 + 1, x 3 /3 – 4 are antiderivatives of the same function f(x) = x 2.

Solution.

1) Let us denote F 1 (x) = x 3 /3, then F" 1 (x) = 3 ∙ (x 2 /3) = x 2 = f(x).

2) F 2 (x) = x 3 /3 + 1, F" 2 (x) = (x 3 /3 + 1)" = (x 3 /3)" + (1)" = x 2 = f( x).

3) F 3 (x) = x 3 /3 – 4, F" 3 (x) = (x 3 /3 – 4)" = x 2 = f (x).

In general, any function x 3 /3 + C, where C is a constant, is an antiderivative of the function x 2. This follows from the fact that the derivative of the constant is zero. This example shows that for a given function its antiderivative is determined ambiguously.

Let F 1 (x) and F 2 (x) be two antiderivatives of the same function f(x).

Then F 1 "(x) = f(x) and F" 2 (x) = f(x).

The derivative of their difference g(x) = F 1 (x) – F 2 (x) is equal to zero, since g"(x) = F" 1 (x) – F" 2 (x) = f(x) – f (x) = 0.

If g"(x) = 0 on a certain interval, then the tangent to the graph of the function y = g(x) at each point of this interval is parallel to the Ox axis. Therefore, the graph of the function y = g(x) is a straight line parallel to the Ox axis, i.e. e. g(x) = C, where C is some constant. From the equalities g(x) = C, g(x) = F 1 (x) – F 2 (x) it follows that F 1 (x) = F 2 (x) + S.

So, if the function F(x) is an antiderivative of the function f(x) on a certain interval, then all antiderivatives of the function f(x) are written in the form F(x) + C, where C is an arbitrary constant.

Let's consider the graphs of all antiderivatives of a given function f(x). If F(x) is one of the antiderivatives of the function f(x), then any antiderivative of this function is obtained by adding to F(x) some constant: F(x) + C. Graphs of functions y = F(x) + C are obtained from the graph y = F(x) by shift along the Oy axis. By choosing C, you can ensure that the graph of the antiderivative passes through a given point.

Let us pay attention to the rules for finding antiderivatives.

Recall that the operation of finding the derivative for a given function is called differentiation. The inverse operation of finding the antiderivative for a given function is called integration(from the Latin word "restore").

Table of antiderivatives for some functions it can be compiled using a table of derivatives. For example, knowing that (cos x)" = -sin x, we get (-cos x)" = sin x, from which it follows that all antiderivative functions sin x are written in the form -cos x + C, Where WITH– constant.

Let's look at some of the meanings of antiderivatives.

1) Function: x p, p ≠ -1. Antiderivative: (x p+1) / (p+1) + C.

2) Function: 1/x, x > 0. Antiderivative: ln x + C.

3) Function: x p, p ≠ -1. Antiderivative: (x p+1) / (p+1) + C.

4) Function: e x. Antiderivative: e x + C.

5) Function: sin x. Antiderivative: -cos x + C.

6) Function: (kx + b) p, р ≠ -1, k ≠ 0. Antiderivative: (((kx + b) p+1) / k(p+1)) + C.

7) Function: 1/(kx + b), k ≠ 0. Antiderivative: (1/k) ln (kx + b)+ C.

8) Function: e kx + b, k ≠ 0. Antiderivative: (1/k) e kx + b + C.

9) Function: sin (kx + b), k ≠ 0. Antiderivative: (-1/k) cos (kx + b).

10) Function: cos (kx + b), k ≠ 0. Antiderivative: (1/k) sin (kx + b).

Integration rules can be obtained using differentiation rules. Let's look at some rules.

Let F(x) And G(x)– antiderivatives of the corresponding functions f(x) And g(x) at some interval. Then:

1) function F(x) ± G(x) is the antiderivative of the function f(x) ± g(x);

2) function аF(x) is the antiderivative of the function аf(x).

website, when copying material in full or in part, a link to the source is required.

Definition of antiderivative.

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.

Definition of an indefinite integral.

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 integrand, and f(x) – integrand function. The integrand represents the differential of the function f(x) .

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

Based on the properties of the derivative, one can formulate and prove properties of the indefinite integral(properties of an antiderivative).

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.

Let's look at an example.

Example.

Find the antiderivative of the function whose value is equal to one at x = 1.

Solution.

We know from differential calculus that (just look at the table of derivatives of basic elementary functions). Thus, . By the second property . That is, we have many antiderivatives. For x = 1 we get the value . According to the condition, this value must be equal to one, therefore, C = 1. The desired antiderivative will take the form .

Example.

Find the indefinite integral and check the result by differentiation.

Solution.

Using the double angle sine formula from trigonometry , That's why

For every mathematical action there is an inverse action. For the action of differentiation (finding derivatives of functions), there is also an inverse action - integration. Through integration, a function is found (reconstructed) from its given derivative or differential. The found function is called antiderivative.

Definition. Differentiable function F(x) is called the antiderivative of the function f(x) on a given interval, if for all X from this interval the following equality holds: F′(x)=f (x).

Examples. Find antiderivatives for the functions: 1) f (x)=2x; 2) f (x)=3cos3x.

1) Since (x²)′=2x, then, by definition, the function F (x)=x² will be an antiderivative of the function f (x)=2x.

2) (sin3x)′=3cos3x. If we denote f (x)=3cos3x and F (x)=sin3x, then, by definition of an antiderivative, we have: F′(x)=f (x), and, therefore, F (x)=sin3x is an antiderivative for f ( x)=3cos3x.

Note that (sin3x +5 )′= 3cos3x, and (sin3x -8,2 )′= 3cos3x, ... in general form we can write: (sin3x +C)′= 3cos3x, Where WITH- some constant value. These examples indicate the ambiguity of the action of integration, in contrast to the action of differentiation, when any differentiable function has a single derivative.

Definition. If the function F(x) is an antiderivative of the function f(x) on a certain interval, then the set of all antiderivatives of this function has the form:

F(x)+C, where C is any real number.

The set of all antiderivatives F (x) + C of the function f (x) on the interval under consideration is called the indefinite integral and is denoted by the symbol ∫ (integral sign). Write down: ∫f (x) dx=F (x)+C.

Expression ∫f(x)dx read: “integral ef from x to de x.”

f(x)dx- integrand expression,

f(x)— integrand function,

X is the integration variable.

F(x)- antiderivative of a function f(x),

WITH- some constant value.

Now the considered examples can be written as follows:

1) ∫ 2xdx=x²+C. 2) ∫ 3cos3xdx=sin3x+C.

What does the sign d mean?

d— differential sign - has a dual purpose: firstly, this sign separates the integrand from the integration variable; secondly, everything that comes after this sign is differentiated by default and multiplied by the integrand.

Examples. Find the integrals: 3) ∫ 2pxdx; 4) ∫ 2pxdp.

3) After the differential icon d costs XX, A R

∫ 2хрdx=рх²+С. Compare with example 1).

Let's do a check. F′(x)=(px²+C)′=p·(x²)′+C′=p·2x=2px=f (x).

4) After the differential icon d costs R. This means that the integration variable R, and the multiplier X should be considered some constant value.

∫ 2хрдр=р²х+С. Compare with examples 1) And 3).

Let's do a check. F′(p)=(p²x+C)′=x·(p²)′+C′=x·2p=2px=f (p).