Definition of an antiderivative function. Antiderivative and integrals

Indefinite integral

The main task of differential calculus was the calculation of the derivative or differential 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 from 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 the indefinite integral is a function of general form, 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 some function is equal to this function up to a constant term, that is (7.5)

There are three basic rules for finding antiderivative functions. They are very similar to the corresponding differentiation rules.

Rule 1

If F is an antiderivative for some function f, and G is an antiderivative for some function g, then F + G will be an antiderivative for f + g.

By definition of an antiderivative, F’ = f. G' = g. And since these conditions are met, then according to the rule for calculating the derivative for the sum of functions we will have:

(F + G)’ = F’ + G’ = f + g.

Rule 2

If F is an antiderivative for some function f, and k is some constant. Then k*F is the antiderivative of the function k*f. This rule follows from the rule for calculating the derivative of a complex function.

We have: (k*F)’ = k*F’ = k*f.

Rule 3

If F(x) is some antiderivative for the function f(x), and k and b are some constants, and k is not equal to zero, then (1/k)*F*(k*x+b) will be an antiderivative for the function f (k*x+b).

This rule follows from the rule for calculating the derivative of a complex function:

((1/k)*F*(k*x+b))’ = (1/k)*F’(k*x+b)*k = f(k*x+b).

Let's look at a few examples of how these rules apply:

Example 1. Find general view antiderivatives for the function f(x) = x^3 +1/x^2. For the function x^3 one of the antiderivatives will be the function (x^4)/4, and for the function 1/x^2 one of the antiderivatives will be the function -1/x. Using the first rule, we have:

F(x) = x^4/4 - 1/x +C.

Example 2. Let's find the general form of antiderivatives for the function f(x) = 5*cos(x). For the function cos(x), one of the antiderivatives will be the function sin(x). If we now use the second rule, we will have:

F(x) = 5*sin(x).

Example 3. Find one of the antiderivatives for the function y = sin(3*x-2). For the function sin(x) one of the antiderivatives will be the function -cos(x). If we now use the third rule, we obtain an expression for the antiderivative:

F(x) = (-1/3)*cos(3*x-2)

Example 4. Find the antiderivative for the function f(x) = 1/(7-3*x)^5

The antiderivative for the function 1/x^5 will be the function (-1/(4*x^4)). Now, using the third rule, we get.

Previously, given a given function, guided by various formulas and rules, we found its derivative. The derivative has numerous uses: it is the speed of movement (or, more generally, the speed of any process); slope tangent to the graph of a function; using the derivative, you can examine a function for monotonicity and extrema; it helps solve optimization problems.

But along with the problem of finding the speed according to a known law of motion, there is also an inverse problem - the problem of restoring the law of motion according to a known speed. Let's consider one of these problems.

Example 1. Moves in a straight line material point, the speed of its movement at time t is given by the formula v=gt. Find the law of motion.
Solution. Let s = s(t) be the desired law of motion. It is known that s"(t) = v(t). This means that to solve the problem you need to select a function s = s(t), the derivative of which is equal to gt. It is not difficult to guess that \(s(t) = \frac(gt^ 2)(2)\).
\(s"(t) = \left(\frac(gt^2)(2) \right)" = \frac(g)(2)(t^2)" = \frac(g)(2) \ cdot 2t = gt\)
Answer: \(s(t) = \frac(gt^2)(2) \)

Let us immediately note that the example is solved correctly, but incompletely. We got \(s(t) = \frac(gt^2)(2) \). In fact, the problem has infinitely many solutions: any function of the form \(s(t) = \frac(gt^2)(2) + C\), where C is an arbitrary constant, can serve as a law of motion, since \(\left (\frac(gt^2)(2) +C \right)" = gt \)

To make the problem more specific, we had 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 s(t) = (gt 2)/2 + C we get: s(0) = 0 + C, i.e. C = s 0. Now the law of motion is uniquely defined: s(t) = (gt 2)/2 + s 0.

In mathematics, reciprocal operations are assigned different names, come up with special notations, for example: squaring (x 2) and extracting square root(\(\sqrt(x) \)), sine (sin x) 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, is integration.

The term “derivative” itself can be justified “in everyday life”: the function y = f(x) “produces” 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) , primary image, or primitive.

Definition. The function y = F(x) is called antiderivative for the function y = f(x) on the interval X if the equality F"(x) = f(x) holds for \(x \in X\)

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

Let's give examples.
1) The function y = x 2 is antiderivative for the function y = 2x, since for any x the equality (x 2)" = 2x is true
2) The function y = x 3 is antiderivative for the function y = 3x 2, since for any x the equality (x 3)" = 3x 2 is true
3) The function y = sin(x) is antiderivative for the function y = cos(x), since for any x the equality (sin(x))" = cos(x) is true

When finding antiderivatives, as well as derivatives, not only formulas are used, 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 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. If F(x) is an antiderivative for f(x), then kF(x) is an antiderivative for kf(x).

Theorem 1. 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 \(y=\frac(1)(k)F(kx+m) \)

Theorem 2. 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 they all have the form y = F(x) + C.

Integration methods

Variable replacement method (substitution method)

The method of integration by substitution involves introducing a new integration variable (that is, substitution). In this case, the given integral is reduced to a new integral, which is tabular or reducible to it. There are no general methods for selecting substitutions. The ability to correctly determine substitution is acquired through practice.
Let it be necessary to calculate the integral \(\textstyle \int F(x)dx \). Let's make the substitution \(x= \varphi(t) \) where \(\varphi(t) \) is a function that has a continuous derivative.
Then \(dx = \varphi " (t) \cdot dt \) and based on the invariance property of the integration formula for the indefinite integral, we obtain the integration formula by substitution:
\(\int F(x) dx = \int F(\varphi(t)) \cdot \varphi " (t) dt \)

Integration of expressions of the form \(\textstyle \int \sin^n x \cos^m x dx \)

If m is odd, m > 0, then it is more convenient to make the substitution sin x = t.
If n is odd, n > 0, then it is more convenient to make the substitution cos x = t.
If n and m are even, then it is more convenient to make the substitution tg x = t.

Integration by parts

Integration by parts - applying the following formula for integration:
\(\textstyle \int u \cdot dv = u \cdot v - \int v \cdot du \)
or:
\(\textstyle \int u \cdot v" \cdot dx = u \cdot v - \int v \cdot u" \cdot dx \)

Table of indefinite integrals (antiderivatives) of some functions

$$ \int 0 \cdot dx = C $$ $$ \int 1 \cdot dx = x+C $$ $$ \int x^n dx = \frac(x^(n+1))(n+1 ) +C \;\; (n \neq -1) $$ $$ \int \frac(1)(x) dx = \ln |x| +C $$ $$ \int e^x dx = e^x +C $$ $$ \int a^x dx = \frac(a^x)(\ln a) +C \;\; (a>0, \;\; a \neq 1) $$ $$ \int \cos x dx = \sin x +C $$ $$ \int \sin x dx = -\cos x +C $$ $ $ \int \frac(dx)(\cos^2 x) = \text(tg) x +C $$ $$ \int \frac(dx)(\sin^2 x) = -\text(ctg) x +C $$ $$ \int \frac(dx)(\sqrt(1-x^2)) = \text(arcsin) x +C $$ $$ \int \frac(dx)(1+x^2 ) = \text(arctg) x +C $$ $$ \int \text(ch) x dx = \text(sh) x +C $$ $$ \int \text(sh) x dx = \text(ch ) x +C $$

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

Additional materials
Dear users, do not forget to leave your comments, reviews, wishes! All materials have been checked by an anti-virus program.

Teaching aids and simulators in the Integral online store for grade 11
Algebraic problems with parameters, grades 9–11
"Interactive tasks on building in space for grades 10 and 11"

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 antiderivative 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 of 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. Document

Some interval X. If For any xХ F"(x) = f(x), then function F calledantiderivativeForfunctions f on the interval X. AntiderivativeForfunctions you can try to find...

Antiderivative for function

Document

... . Function F(x) calledantiderivativeForfunctions f(x) on the interval (a;b), if For all x(a;b) the equality F(x) = f(x) holds. For example, Forfunctions x2 antiderivative will function x3...

Fundamentals of Integral Calculus Study Guide

Tutorial

... ; 5. Find the integral. ; B) ; C) ; D) ; 6. Functioncalledantiderivative To functions on a set if: For everyone; at some point; For everyone; at some... interval. Definition 1. FunctioncalledantiderivativeForfunctions on many...

Antiderivative Indefinite integral

Document

Integration. Antiderivative. Continuous function F(x) calledantiderivativeForfunctions f (x) on the interval X if For each F’ (x) = f (x). EXAMPLE Function F(x) = x 3 is antiderivativeForfunctions f(x) = 3x...

SPECIAL EDUCATION OF THE USSR Approved by the Educational and Methodological Directorate for Higher Education HIGHER MATHEMATICS METHODICAL INSTRUCTIONS AND CONTROL TASKS (WITH THE PROGRAM) for part-time students of engineering and technical specialties

Guidelines

Questions For self-test Define antiderivativefunctions. Indicate the geometric meaning of the aggregate primitivefunctions. What called uncertain...