Finding antiderivative examples. Antiderivative of function and general appearance

Job type: 7
Subject: Antiderivative of function

Condition

The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight segments). Using the figure, calculate F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x).

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Solution

According to the Newton-Leibniz formula, the difference F(9)-F(5), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of ​​the curvilinear trapezoid limited by the graph of the function y=f(x), straight lines y=0 , x=9 and x=5. From the graph we determine that the indicated curved trapezoid is a trapezoid with bases equal to 4 and 3 and height 3.

Its area is equal \frac(4+3)(2)\cdot 3=10.5.

Answer

Job type: 7
Topic: Antiderivative of function

Condition

The figure shows a graph of the function y=F(x) - one of the antiderivatives of some function f(x) defined on the interval (-5; 5). Using the figure, determine the number of solutions to the equation f(x)=0 on the segment [-3; 4].

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Solution

According to the definition of an antiderivative, the equality holds: F"(x)=f(x). Therefore, the equation f(x)=0 can be written as F"(x)=0. Since the figure shows the graph of the function y=F(x), we need to find those points in the interval [-3; 4], in which the derivative of the function F(x) is equal to zero. It is clear from the figure that these will be the abscissas of the extreme points (maximum or minimum) of the F(x) graph. There are exactly 7 of them in the indicated interval (four minimum points and three maximum points).

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: Antiderivative of function

Condition

The figure shows a graph of the function y=f(x) (which is a broken line made up of three straight segments). Using the figure, calculate F(5)-F(0), where F(x) is one of the antiderivatives of the function f(x).

Show solution

Solution

According to the Newton-Leibniz formula, the difference F(5)-F(0), where F(x) is one of the antiderivatives of the function f(x), is equal to the area of ​​the curvilinear trapezoid limited by the graph of the function y=f(x), straight lines y=0 , x=5 and x=0. From the graph we determine that the indicated curved trapezoid is a trapezoid with bases equal to 5 and 3 and height 3.

Its area is equal \frac(5+3)(2)\cdot 3=12.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: Antiderivative of function

Condition

The figure shows a graph of the function y=F(x) - one of the antiderivatives of some function f(x), defined on the interval (-5; 4). Using the figure, determine the number of solutions to the equation f (x) = 0 on the segment (-3; 3].

Show solution

Solution

According to the definition of an antiderivative, the equality holds: F"(x)=f(x). Therefore, the equation f(x)=0 can be written as F"(x)=0. Since the figure shows the graph of the function y=F(x), we need to find those points in the interval [-3; 3], in which the derivative of the function F(x) is equal to zero.

It is clear from the figure that these will be the abscissas of the extreme points (maximum or minimum) of the F(x) graph. There are exactly 5 of them in the indicated interval (two minimum points and three maximum points).

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: Antiderivative of function

Condition

The figure shows a graph of some function y=f(x). The function F(x)=-x^3+4.5x^2-7 is one of the antiderivatives of the function f(x).

Find the area of ​​the shaded figure.

Show solution

Solution

The shaded figure is curved trapezoid, bounded from above by the graph of the function y=f(x), by the straight lines y=0, x=1 and x=3. According to the Newton-Leibniz formula, its area S is equal to the difference F(3)-F(1), where F(x) is the antiderivative of the function f(x) specified in the condition. That's why S= F(3)-F(1)= -3^3 +(4.5)\cdot 3^2 -7-(-1^3 +(4.5)\cdot 1^2 -7)= 6,5-(-3,5)= 10.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: Antiderivative of function

Condition

The figure shows a graph of some function y=f(x). The function F(x)=x^3+6x^2+13x-5 is one of the antiderivatives of the function f(x). Find the area of ​​the shaded figure.

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 the problem 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 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

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 with respect to 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.

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 the 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 terms”: the function y = f(x) “gives birth” to a new function y" = f"(x). The function y = f(x) acts as if it were 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 $$ Document

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