How to find an antiderivative function at a point. Integrals for dummies: how to solve, calculation rules, explanation

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 OU 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.

A-priory:

f(x) - integrand function,

f(х)dх - integrand expression

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.

From a geometric point of view 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 OU(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)

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); the angular coefficient of the tangent to the graph of the 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. A material point moves in a straight line, its speed 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, mutually inverse operations are given different names, special notations are invented, for example: squaring (x 2) and square root (\(\sqrt(x) \)), sine (sin x) and arcsine (arcsin x) and 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 terms”: the function y = f(x) “gives birth” to a new function 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"

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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 with a 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.

In our table, no initial conditions were specified. 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 an exponential function is the exponential function itself. 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 considered 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.
Given the 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.

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 we ​​also have to solve inverse problems: for example, along with the problem of finding the speed according to a known law of motion, we also encounter 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, mutually inverse operations are given different names and special notations are invented: for example, squaring (x 2) and taking the square root of 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 life”: the function y - f(x) “gives birth” to a new function 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 this, in relation to the function y"=f"(x), is 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 (not theorems), only keywords are left - this is more convenient for applying the rules 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 the specific value of the 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 get:

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 the hidden meaning of this designation is.
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 use the trigonometric formula for reducing the degree:

Then we find sequentially:

A.G. Mordkovich Algebra 10th grade

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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 (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, it can be easily seen 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 noticed, 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 task of integration is to find all its antiderivatives for a given function. 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:

1. Whatever number we put in expression (1) instead of C, we obtain the antiderivative for f on the interval I;
2. 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.

1. 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.
2. 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