The set of all primitives. Antiderivative function and indefinite integral

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 using the 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). 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 unique.
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. That's why general solution problems can be written in the form 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 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

One of the operations of differentiation is finding the derivative (differential) and applying it to the study of functions.

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 (f(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), you 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 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 antiderivative 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 noted, this function has an infinite number of antiderivatives.

Example No. 2.

The function is antiderivative for all on the interval (0; +∞), because for all h from this interval, equality holds.

The problem of integration is to given function find all its antiderivatives. When solving this problem, the following statement plays an important role:

A sign of constancy of function. If F"(x) = 0 on some interval I, then the function F is constant on this interval.

Proof.

Let us fix some x 0 from the interval I. Then for any number x from such an interval, by virtue of the Lagrange formula, we can indicate a number c contained between x and x 0 such that

F(x) - F(x 0) = F"(c)(x-x 0).

By condition, F’ (c) = 0, since c ∈1, therefore,

F(x) - F(x 0) = 0.

So, for all x from the interval I

that is, the function F maintains a constant value.

All antiderivative functions f can be written using one formula, which is called general form of antiderivatives for the function f. The following theorem is true ( main property of antiderivatives):

Theorem. Any antiderivative for a function f on the interval I can be written in the form

F(x) + C, (1) where F (x) is one of the antiderivatives for the function f (x) on the interval I, and C is an arbitrary constant.

Let us explain this statement, in which two properties of the antiderivative are briefly formulated:

Whatever number we put in expression (1) instead of C, we obtain the antiderivative for f on the interval I;
no matter what antiderivative Ф for f on the interval I is taken, it is possible to select a number C such that for all x from the interval I the equality

Proof.

By condition, the function F is antiderivative for f on the interval I. Therefore, F"(x)= f (x) for any x∈1, therefore (F(x) + C)" = F"(x) + C"= f(x)+0=f(x), i.e. F(x) + C is the antiderivative for the function f.
Let Ф (x) be one of the antiderivatives for the function f on the same interval I, i.e. Ф "(x) = f (х) for all x∈I.

Then (Ф(x) - F (x))" = Ф"(x)-F' (x) = f(x)-f(x)=0.

From here it follows c. the power of the sign of constancy of the function, that the difference Ф(х) - F(х) is a function that takes some constant value C on the interval I.

Thus, for all x from the interval I the equality Ф(x) - F(x)=С is true, which is what needed to be proved. The main property of the antiderivative can be given a geometric meaning: graphs of any two antiderivatives for the function f are obtained from each other by parallel translation along the Oy axis

Questions for notes

The function F(x) is an antiderivative of the function f(x). Find F(1) if f(x)=9x2 - 6x + 1 and F(-1) = 2.

Find all antiderivatives for the function

For the function (x) = cos2 * sin2x, find the antiderivative of F(x) if F(0) = 0.

For a function, find an antiderivative whose graph passes through the point

Lesson and presentation on the topic: "An antiderivative function. Graph of a function"

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Antiderivative function. Introduction

Guys, you know how to find derivatives of functions using various formulas and rules. Today we will study the inverse operation of calculating the derivative. The concept of derivative is often used in real life. Let me remind you: the derivative is the rate of change of a function at a specific point. Processes involving motion and speed are well described in these terms.

Let's look at this problem: “The speed of an object moving in a straight line is described by the formula $V=gt$. It is required to restore the law of motion.
Solution.
We know the formula well: $S"=v(t)$, where S is the law of motion.
Our task comes down to finding a function $S=S(t)$ whose derivative is equal to $gt$. Looking carefully, you can guess that $S(t)=\frac(g*t^2)(2)$.
Let's check the correctness of the solution to this problem: $S"(t)=(\frac(g*t^2)(2))"=\frac(g)(2)*2t=g*t$.
Knowing the derivative of the function, we found the function itself, that is, we performed the inverse operation.
But it’s worth paying attention to this moment. The solution to our problem requires clarification; if we add any number (constant) to the found function, then the value of the derivative will not change: $S(t)=\frac(g*t^2)(2)+c,c=const$.
$S"(t)=(\frac(g*t^2)(2))"+c"=g*t+0=g*t$.

Guys, pay attention: our problem has an infinite number of solutions!
If the problem does not specify an initial or some other condition, do not forget to add a constant to the solution. For example, our task may specify the position of our body at the very beginning of the movement. Then it is not difficult to calculate the constant; by substituting zero into the resulting equation, we obtain the value of the constant.

What is this operation called?
The inverse operation of differentiation is called integration.
Finding a function from a given derivative – integration.
The function itself will be called an antiderivative, that is, the image from which the derivative of the function was obtained.
It is customary to write the antiderivative capital letter$y=F"(x)=f(x)$.

Definition. The function $y=F(x)$ is called the antiderivative of the function $у=f(x)$ on the interval X if for any $хϵХ$ the equality $F’(x)=f(x)$ holds.

Let's make a table of antiderivatives for various functions. It should be printed out as a reminder and memorized.

There are none in our table initial conditions was not asked. This means that a constant should be added to each expression on the right side of the table. We will clarify this rule later.

Rules for finding antiderivatives

Let's write down a few rules that will help us in finding antiderivatives. They are all similar to the rules of differentiation.

Rule 1. The antiderivative of a sum is equal to the sum of the antiderivatives. $F(x+y)=F(x)+F(y)$.

Example.
Find the antiderivative for the function $y=4x^3+cos(x)$.
Solution.
The antiderivative of the sum is equal to the sum of the antiderivatives, then we need to find the antiderivative for each of the presented functions.
$f(x)=4x^3$ => $F(x)=x^4$.
$f(x)=cos(x)$ => $F(x)=sin(x)$.
Then the antiderivative of the original function will be: $y=x^4+sin(x)$ or any function of the form $y=x^4+sin(x)+C$.

Rule 2. If $F(x)$ is an antiderivative for $f(x)$, then $k*F(x)$ is an antiderivative for the function $k*f(x)$.(We can easily take the coefficient as a function).

Example.
Find antiderivatives of functions:
a) $y=8sin(x)$.
b) $y=-\frac(2)(3)cos(x)$.
c) $y=(3x)^2+4x+5$.
Solution.
a) The antiderivative of $sin(x)$ is minus $cos(x)$. Then the antiderivative of the original function will take the form: $y=-8cos(x)$.

B) The antiderivative of $cos(x)$ is $sin(x)$. Then the antiderivative of the original function will take the form: $y=-\frac(2)(3)sin(x)$.

C) The antiderivative for $x^2$ is $\frac(x^3)(3)$. The antiderivative for x is $\frac(x^2)(2)$. The antiderivative of 1 is x. Then the antiderivative of the original function will take the form: $y=3*\frac(x^3)(3)+4*\frac(x^2)(2)+5*x=x^3+2x^2+5x$ .

Rule 3. If $у=F(x)$ is an antiderivative for the function $y=f(x)$, then the antiderivative for the function $y=f(kx+m)$ is the function $y=\frac(1)(k)* F(kx+m)$.

Example.
Find antiderivatives of the following functions:
a) $y=cos(7x)$.
b) $y=sin(\frac(x)(2))$.
c) $y=(-2x+3)^3$.
d) $y=e^(\frac(2x+1)(5))$.
Solution.
a) The antiderivative of $cos(x)$ is $sin(x)$. Then the antiderivative for the function $y=cos(7x)$ will be the function $y=\frac(1)(7)*sin(7x)=\frac(sin(7x))(7)$.

B) The antiderivative of $sin(x)$ is minus $cos(x)$. Then the antiderivative for the function $y=sin(\frac(x)(2))$ will be the function $y=-\frac(1)(\frac(1)(2))cos(\frac(x)(2) )=-2cos(\frac(x)(2))$.

C) The antiderivative for $x^3$ is $\frac(x^4)(4)$, then the antiderivative of the original function $y=-\frac(1)(2)*\frac(((-2x+3) )^4)(4)=-\frac(((-2x+3))^4)(8)$.

D) Slightly simplify the expression to the power $\frac(2x+1)(5)=\frac(2)(5)x+\frac(1)(5)$.
The antiderivative of the exponential function is itself exponential function. The antiderivative of the original function will be $y=\frac(1)(\frac(2)(5))e^(\frac(2)(5)x+\frac(1)(5))=\frac(5)( 2)*e^(\frac(2x+1)(5))$.

Theorem. If $y=F(x)$ is an antiderivative for the function $y=f(x)$ on the interval X, then the function $y=f(x)$ has infinitely many antiderivatives, and all of them have the form $y=F( x)+С$.

If in all the examples discussed above it was necessary to find the set of all antiderivatives, then the constant C should be added everywhere.
For the function $y=cos(7x)$ all antiderivatives have the form: $y=\frac(sin(7x))(7)+C$.
For the function $y=(-2x+3)^3$ all antiderivatives have the form: $y=-\frac(((-2x+3))^4)(8)+C$.

Example.
According to the given law of change in the speed of a body over time $v=-3sin(4t)$, find the law of motion $S=S(t)$ if at the initial moment of time the body had a coordinate equal to 1.75.
Solution.
Since $v=S’(t)$, we need to find the antiderivative for a given speed.
$S=-3*\frac(1)(4)(-cos(4t))+C=\frac(3)(4)cos(4t)+C$.
In this problem, an additional condition is given - the initial moment of time. This means that $t=0$.
$S(0)=\frac(3)(4)cos(4*0)+C=\frac(7)(4)$.
$\frac(3)(4)cos(0)+C=\frac(7)(4)$.
$\frac(3)(4)*1+C=\frac(7)(4)$.
$C=1$.
Then the law of motion is described by the formula: $S=\frac(3)(4)cos(4t)+1$.

Problems to solve independently

1. Find antiderivatives of functions:
a) $y=-10sin(x)$.
b) $y=\frac(5)(6)cos(x)$.
c) $y=(4x)^5+(3x)^2+5x$.
2. Find antiderivatives of the following functions:
a) $y=cos(\frac(3)(4)x)$.
b) $y=sin(8x)$.
c) $y=((7x+4))^4$.
d) $y=e^(\frac(3x+1)(6))$.
3. According to the given law of change in the speed of a body over time $v=4cos(6t)$, find the law of motion $S=S(t)$ if at the initial moment of time the body had a coordinate equal to 2.

Indefinite integral

The main task of differential calculus was to calculate the derivative or differential of a given function. Integral calculus, to the study of which we are moving on, solves the inverse problem, namely, finding the function itself from its derivative or differential. That is, having dF(x)= f(x)d (7.1) or F ′(x)= f(x),

Where f(x)- known function, need to find the function F(x).

Definition:The function F(x) is called antiderivative function f(x) on the segment if the equality holds at all points of this segment: F′(x) = f(x) or dF(x)= f(x)d.

For example, one of the antiderivative functions for the function f(x)=3x 2 will F(x)= x 3, because ( x 3)′=3x 2. But a prototype for the function f(x)=3x 2 there will also be functions and , since .

So this function f(x)=3x 2 has an infinite number of primitives, each of which differs only by a constant term. Let us show that this result also holds in the general case.

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

Proof

Let the function f(x) defined on the interval (a¸b) And F 1 (x) And F 2 (x) - antiderivatives, i.e. F 1 ′(x)= f(x) and F 2 ′(x)= f(x).

Then F 1 ′(x)=F 2 ′(x)Þ F 1 ′(x) - F 2 ′(x) = (F 1 ′(x) - F 2 (x))′= 0. Þ F 1 (x) - F 2 (x) = C

From here, F 2 (x) = F 1 (x) + C

Where WITH - constant (a corollary of Lagrange’s theorem is used here).

The theorem is thus proven.

Geometric illustration. If at = F 1 (x) And at = F 2 (x) – antiderivatives of the same function f(x), then the tangent to their graphs at points with a common abscissa X parallel to each other (Fig. 7.1).

In this case, the distance between these curves along the axis Oh remains constant F 2 (x) - F 1 (x) = C , that is, these curves in some understanding"parallel" to one another.

Consequence .

Adding to some antiderivative F(x) for this function f(x), defined on the interval X, all possible constants WITH, we get all possible antiderivatives for the function f(x).

So the expression F(x)+C , where , and F(x) – some antiderivative of a function f(x) includes all possible antiderivatives for f(x).

Example 1. Check if functions are antiderivatives of the function

Solution:

Answer: antiderivatives for a function there will be functions And

Definition: If the function F(x) is some antiderivative of the function f(x), then the set of all antiderivatives F(x)+ C is called indefinite integral of f(x) and denote:

∫f(х)dх.

By definition:

f(x) - integrand function,

f(х)dх - integrand

It follows from this that indefinite integral is a function general view, the differential of which is equal to the integrand, and the derivative of which with respect to the variable X is equal to the integrand at all points.

WITH geometric point vision an indefinite integral is a family of curves, each of which is obtained by shifting one of the curves parallel to itself up or down, that is, along the axis Oh(Fig. 7.2).

The operation of calculating the indefinite integral of a certain function is called integration this function.

Note that if the derivative of an elementary function is always an elementary function, then the antiderivative of an elementary function may not be represented by a finite number of elementary functions.

Let's now consider properties of the indefinite integral.

From Definition 2 it follows:

1. The derivative of the indefinite integral is equal to the integrand, that is, if F′(x) = f(x) , That

2. The differential of the indefinite integral is equal to the integrand

. (7.4)

From the definition of differential and property (7.3)

3. The indefinite integral of the differential of a certain function is equal to this function up to a constant term, that is (7.5)

Antiderivative.

The antiderivative is easy to understand with an example.

Let's take the function y = x 3. As we know from the previous sections, the derivative of X 3 is 3 X 2:

(X 3)" = 3X 2 .

Therefore, from the function y = x 3 we get new feature: at = 3X 2 .
Figuratively speaking, the function at = X 3 produced function at = 3X 2 and is its “parent”. In mathematics there is no word “parent”, but there is a related concept: antiderivative.

That is: function y = x 3 is an antiderivative of the function at = 3X 2 .

Definition of antiderivative:

In our example ( X 3)" = 3X 2 therefore y = x 3 – antiderivative for at = 3X 2 .

Integration.

As you know, the process of finding the derivative of a given function is called differentiation. And the inverse operation is called integration.

Example-explanation:

at = 3X 2 + sin x.

Solution :

We know that the antiderivative for 3 X 2 is X 3 .

Antiderivative for sin x is –cos x.

We add two antiderivatives and get the antiderivative for the given function:

y = x 3 + (–cos x),

y = x 3 – cos x.

Answer :
for function at = 3X 2 + sin x y = x 3 – cos x.

Example-explanation:

Let's find an antiderivative for the function at= 2 sin x.

Solution :

We note that k = 2. The antiderivative for sin x is –cos x.

Therefore, for the function at= 2 sin x the antiderivative is the function at= –2cos x.
Coefficient 2 in the function y = 2 sin x corresponds to the coefficient of the antiderivative from which this function was formed.

Example-explanation:

Let's find an antiderivative for the function y= sin 2 x.

Solution :

We notice that k= 2. Antiderivative for sin x is –cos x.

We apply our formula to find the antiderivative of the function y= cos 2 x:

1
y= - · (–cos 2 x),
2

cos 2 x
y = – ----
2

cos 2 x
Answer: for a function y= sin 2 x the antiderivative is the function y = – ----
2

(4)

Example-explanation.

Let's take the function from the previous example: y= sin 2 x.

For this function, all antiderivatives have the form:

cos 2 x
y = – ---- + C.
2

Explanation.

Let's take the first line. It reads like this: if the function y = f( x) is 0, then its antiderivative is 1. Why? Because the derivative of unity is zero: 1" = 0.

The remaining lines are read in the same order.

How to write data from a table? Let's take line eight:

(-cos x)" = sin x

We write the second part with the derivative sign, then the equal sign and the derivative.

We read: antiderivative for functions sin x is the -cos function x.

Or: function -cos x is antiderivative for the function sin x.