Antiderivative
Definition antiderivative function
- Function y=F(x) is called the antiderivative of the function y=f(x) at a given interval X, if for everyone X ∈X equality holds: F′(x) = f(x)
Can be read in two ways:
- f derivative of a function F
- F antiderivative of a function f
Property of antiderivatives
- If F(x)- antiderivative of a function f(x) on a given interval, then the function f(x) has infinitely many antiderivatives, and all these antiderivatives can be written in the form F(x) + C, where C is an arbitrary constant.
Geometric interpretation
- Graphs of all antiderivatives of a given function f(x) are obtained from the graph of any one antiderivative parallel transfers along the O axis at.
Rules for calculating antiderivatives
- The antiderivative of the sum is equal to the sum of the antiderivatives. If F(x)- antiderivative for f(x), and G(x) is an antiderivative for g(x), That F(x) + G(x)- antiderivative for f(x) + g(x).
- Constant multiplier can be taken out of the sign of the derivative. If F(x)- antiderivative for f(x), And k- constant, then k·F(x)- antiderivative for k f(x).
- If F(x)- antiderivative for f(x), And k, b- constant, and k ≠ 0, That 1/k F(kx + b)- antiderivative for f(kx + b).
Remember!
Any function F(x) = x 2 + C , where C is an arbitrary constant, and only such a function is an antiderivative for the function f(x) = 2x.
- For example:
F"(x) = (x 2 + 1)" = 2x = f(x);
f(x) = 2x, because F"(x) = (x 2 – 1)" = 2x = f(x);
f(x) = 2x, because F"(x) = (x 2 –3)" = 2x = f(x);
Relationship between the graphs of a function and its antiderivative:
- If the graph of a function f(x)>0 F(x) increases over this interval.
- If the graph of a function f(x)<0 on the interval, then the graph of its antiderivative F(x) decreases over this interval.
- If f(x)=0, then the graph of its antiderivative F(x) at this point changes from increasing to decreasing (or vice versa).
To denote the antiderivative, the sign of the indefinite integral is used, that is, the integral without indicating the limits of integration.
Indefinite integral
Definition:
- The indefinite integral of the function f(x) is the expression F(x) + C, that is, the set of all antiderivatives of a given function f(x). The indefinite integral is denoted as follows: \int f(x) dx = F(x) + C
- f(x)- called the integrand function;
- f(x) dx- called the integrand;
- x- called the variable of integration;
- F(x)- one of the antiderivatives of the function f(x);
- WITH- arbitrary constant.
Properties of the indefinite integral
- The derivative of the indefinite integral is equal to the integrand: (\int f(x) dx)\prime= f(x) .
- The constant factor of the integrand can be taken out of the integral sign: \int k \cdot f(x) dx = k \cdot \int f(x) dx.
- The integral of the sum (difference) of functions is equal to the sum (difference) of the integrals of these functions: \int (f(x) \pm g(x)) dx = \int f(x) dx \pm \int g(x) dx.
- If k, b are constants, and k ≠ 0, then \int f(kx + b) dx = \frac(1)(k) \cdot F(kx + b) + C.
Table of antiderivatives and indefinite integrals
| Function f(x) | Antiderivative F(x) + C | Indefinite integrals \int f(x) dx = F(x) + C |
| 0 | C | \int 0 dx = C |
| f(x) = k | F(x) = kx + C | \int kdx = kx + C |
| f(x) = x^m, m\not =-1 | F(x) = \frac(x^(m+1))(m+1) + C | \int x(^m)dx = \frac(x^(m+1))(m+1) + C |
| f(x) = \frac(1)(x) | F(x) = l n \lvert x \rvert + C | \int \frac(dx)(x) = l n \lvert x \rvert + C |
| f(x) = e^x | F(x) = e^x + C | \int e(^x )dx = e^x + C |
| f(x) = a^x | F(x) = \frac(a^x)(l na) + C | \int a(^x )dx = \frac(a^x)(l na) + C |
| f(x) = \sin x | F(x) = -\cos x + C | \int \sin x dx = -\cos x + C |
| f(x) = \cos x | F(x) =\sin x + C | \int \cos x dx = \sin x + C |
| f(x) = \frac(1)(\sin (^2) x) | F(x) = -\ctg x + C | \int \frac (dx)(\sin (^2) x) = -\ctg x + C |
| f(x) = \frac(1)(\cos (^2) x) | F(x) = \tg x + C | \int \frac(dx)(\sin (^2) x) = \tg x + C |
| f(x) = \sqrt(x) | F(x) =\frac(2x \sqrt(x))(3) + C | |
| f(x) =\frac(1)( \sqrt(x)) | F(x) =2\sqrt(x) + C | |
| f(x) =\frac(1)( \sqrt(1-x^2)) | F(x)=\arcsin x + C | \int \frac(dx)( \sqrt(1-x^2))=\arcsin x + C |
| f(x) =\frac(1)( \sqrt(1+x^2)) | F(x)=\arctg x + C | \int \frac(dx)( \sqrt(1+x^2))=\arctg x + C |
| f(x)=\frac(1)( \sqrt(a^2-x^2)) | F(x)=\arcsin \frac (x)(a)+ C | \int \frac(dx)( \sqrt(a^2-x^2)) =\arcsin \frac (x)(a)+ C |
| f(x)=\frac(1)( \sqrt(a^2+x^2)) | F(x)=\arctg \frac (x)(a)+ C | \int \frac(dx)( \sqrt(a^2+x^2)) = \frac (1)(a) \arctg \frac (x)(a)+ C |
| f(x) =\frac(1)( 1+x^2) | F(x)=\arctg + C | \int \frac(dx)( 1+x^2)=\arctg + C |
| f(x)=\frac(1)( \sqrt(x^2-a^2)) (a \not= 0) | F(x)=\frac(1)(2a)l n \lvert \frac (x-a)(x+a) \rvert + C | \int \frac(dx)( \sqrt(x^2-a^2))=\frac(1)(2a)l n \lvert \frac (x-a)(x+a) \rvert + C |
| f(x)=\tg x | F(x)= - l n \lvert \cos x \rvert + C | \int \tg x dx =- l n \lvert \cos x \rvert + C |
| f(x)=\ctg x | F(x)= l n \lvert \sin x \rvert + C | \int \ctg x dx = l n \lvert \sin x \rvert + C |
| f(x)=\frac(1)(\sin x) | F(x)= l n \lvert \tg \frac(x)(2) \rvert + C | \int \frac (dx)(\sin x) = l n \lvert \tg \frac(x)(2) \rvert + C |
| f(x)=\frac(1)(\cos x) | F(x)= l n \lvert \tg (\frac(x)(2) +\frac(\pi)(4)) \rvert + C | \int \frac (dx)(\cos x) = l n \lvert \tg (\frac(x)(2) +\frac(\pi)(4)) \rvert + C |
Newton–Leibniz formula
Let f(x) this function F its arbitrary antiderivative.
\int_(a)^(b) f(x) dx =F(x)|_(a)^(b)= F(b) - F(a)
Where F(x)- antiderivative for f(x)
That is, the integral of the function f(x) on an interval is equal to the difference of antiderivatives at points b And a.
Area of a curved trapezoid
Curvilinear trapezoid is a figure bounded by the graph of a function that is non-negative and continuous on an interval f, Ox axis and straight lines x = a And x = b.
The area of a curved trapezoid is found using the Newton-Leibniz formula:
S= \int_(a)^(b) f(x) dx
The calculation of area is fundamental to area theory. The question arises of finding it when the figure has an irregular shape or it is necessary to resort to calculating it through an integral.
This article talks about calculating the area of a curved trapezoid in geometric terms. This makes it possible to identify the relationship between the integral and the area of a curvilinear trapezoid. If a function f (x) is given, and it is continuous on the interval [ a ; b ] , the sign in front of the expression does not change.
Yandex.RTB R-A-339285-1 Definition 1
A figure designated as G, bounded by lines of the form y = f(x), y = 0, x = a and x = b, is called curved trapezoid. It takes the designation S(G).
Let's look at the figure below.
To calculate a curved trapezoid, you need to split the segment [a; b ] for the number n of parts x i - 1 ; x i, i = 1, 2, . . . , n with points defined at a = x 0< x 1 < x 2 < . . . < x n - 1 < x n = b , причем дать обозначение λ = m a x i = 1 , 2 , . . . , n x i - x i - 1 с точками x i , i = 1 , 2 , . . . , n - 1 . Необходимо выбрать так, чтобы λ → 0 при n → + ∞ , тогда фигуры, которые соответствуют нижней и верхней частям Дарбу, считаются входящей Р и объемлющей Q многоугольными фигурами для G . Рассмотрим рисунок, приведенный ниже.
From here we have that P ⊂ G ⊂ Q, and with an increase in the number of partition points n, we obtain an inequality of the form S - s< ε , где ε является малым положительным числом, s и S являются верхними и нижними суммами Дабру из отрезка [ a ; b ] . Иначе это запишется как lim λ → 0 S - s = 0 . Значит, при обращении к понятию определенного интеграла Дарбу, получим, что lim λ → 0 S = lim λ → 0 s = S G = ∫ a b f (x) d x .
From the last equality we obtain that a definite integral of the form ∫ a b f (x) d x is the area of a curvilinear trapezoid for a given continuous function of the form y = f (x) . This is the geometric meaning of a definite integral.
When calculating ∫ a b f (x) d x, we obtain the area of the desired figure, which is limited by the lines y = f (x), y = 0, x = a and x = b.
Comment: When the function y = f (x) is non-positive from the interval [ a ; b ], then we find that the area of a curvilinear trapezoid is calculated based on the formula S (G) = - ∫ a b f (x) d x.
Example 1
Calculate the area of the figure, which is limited by given lines of the form y = 2 · e x 3, y = 0, x = - 2, x = 3.
Solution
In order to solve, it is necessary to first construct a figure on a plane where there is a straight line y = 0, coinciding with O x, with lines of the form x = - 2 and x = 3, parallel to the o y axis, where the curve y = 2 e x 3 is constructed using geometric transformations of the graph of the function y = e x. Let's build a graph.
This shows that it is necessary to find the area of a curved trapezoid. Recalling the geometric meaning of the integral, we find that the desired area will be expressed by a certain integral, which must be resolved. This means that it is necessary to apply the formula S (G) = ∫ - 2 3 2 · e x 3 d x . This indefinite integral is calculated based on the Newton-Leibniz formula
S (G) = ∫ - 2 3 2 e x 3 d x = 6 e x 3 - 2 3 = 6 e 3 3 - 6 e - 2 3 = 6 e - e - 2 3
Answer: S (G) = 6 e - e - 2 3
Comment: To find the area of a curved trapezoid, it is not always possible to construct a figure. Then the solution is carried out as follows. Given a known function f (x) that is non-negative or non-positive on the interval [ a ; b ] , a formula of the form S G = ∫ a b f (x) d x or S G = - ∫ a b f (x) d x is used.
Example 2
Calculate the area bounded by lines of the form y = 1 3 (x 2 + 2 x - 8), y = 0, x = - 2, x = 4.
Solution
To construct this figure, we find that y = 0 coincides with O x, and x = - 2 and x = 4 are parallel to O y. The graph of the function y = 1 3 (x 2 + 2 x - 8) = 1 3 (x + 1) 2 - 3 is a parabola with the coordinates of the point (- 1 ; 3) being its vertex with branches pointing upward. To find the intersection points of the parabola with O x, you need to calculate:
1 3 (x 2 + 2 x - 8) = 0 ⇔ x 2 + 2 x - 8 = 0 D = 2 2 - 4 1 (- 8) = 36 x 1 = - 2 + 36 2 = 2 , x 2 = - 2 - 36 2 = - 4
This means that the parabola intersects oh at points (4; 0) and (2; 0). From this we get that the figure designated as G will take the form shown in the figure below.
This figure is not a curvilinear trapezoid, because a function of the form y = 1 3 (x 2 + 2 x - 8) changes sign on the interval [ - 2 ; 4 ] . The figure G can be represented as a union of two curvilinear trapezoids G = G 1 ∪ G 2, based on the property of area additivity, we have that S (G) = S (G 1) + S (G 2). Consider the graph below.
Segment [ - 2 ; 4 ] is considered a non-negative area of the parabola, then from this we obtain that the area will have the form S G 2 = ∫ 2 4 1 3 (x 2 + 2 x - 8) d x . Segment [ - 2 ; 2 ] is non-positive for a function of the form y = 1 3 (x 2 + 2 x - 8), which means, based on the geometric meaning of the definite integral, we obtain that S (G 1) = - ∫ - 2 2 1 3 (x 2 + 2 x - 8) d x . It is necessary to make calculations using the Newton-Leibniz formula. Then the definite integral will take the form:
S (G) = S (G 1) + S (G 2) = - ∫ - 2 2 1 3 (x 2 + 2 x - 8) d x + ∫ 2 4 1 3 (x 2 + 2 x - 8) d x = = - 1 3 x 3 3 + x 2 - 8 x - 2 2 + 1 3 x 3 3 + x 2 - 8 x 2 4 = = - 1 3 2 3 3 + 2 2 - 8 2 - - 2 3 3 + (- 2) 2 - 8 · (- 2) + + 1 3 4 3 3 + 4 3 - 8 · 4 - 2 3 3 + 2 2 - 8 · 2 = = - 1 3 8 3 - 12 + 8 3 - 20 + 1 3 64 3 - 16 - 8 3 + 12 = 124 9
It is worth noting that finding the area is not correct according to the principle S (G) = ∫ - 2 4 1 3 (x 2 + 2 x - 8) d x = 1 3 x 3 3 + x 2 - 8 x - 2 4 = = 1 3 4 3 3 + 4 3 - 8 4 - - 2 3 3 + - 2 2 - 8 - 2 = 1 3 64 3 - 16 + 8 3 - 20 = - 4
Since the resulting number is negative and represents the difference S (G 2) - S (G 1).
Answer: S (G) = S (G 1) + S (G 2) = 124 9
If the figures are limited by lines of the form y = c, y = d, x = 0 and x = g (y), and the function is equal to x = g (y), and is continuous and has a constant sign on the interval [ c; d ], then they are called curvilinear tarpeziums. Consider in the figure below.
∫ c d g (y) d y is that its value is the area of a curvilinear trapezoid for a continuous and non-negative function of the form x = g (y) located on the interval [ c ; d] .
Example 3
Calculate the figure, which is limited by the ordinate axis and the lines x = 4 ln y y + 3, y = 1, y = 4.
Solution
Plotting a graph of x = 4 ln y y + 3 is not easy. Therefore, it is necessary to solve without a drawing. Recall that the function is defined for all positive values of y. Let's consider the function values available on the interval [ 1 ; 4 ] . From the properties of elementary functions we know that the logarithmic function increases over the entire domain of definition. Then not the segment [ 1 ; 4 ] is non-negative. This means that ln y ≥ 0. The existing expression ln y y , defined on the same segment, is non-negative. We can conclude that the function x = 4 ln y y + 3 is positive on the interval equal to [ 1 ; 4 ] . We find that the figure on this interval is positive. Then its area should be calculated using the formula S (G) = ∫ 1 4 4 ln y y + 3 d y .
It is necessary to calculate the indefinite integral. To do this, you need to find the antiderivative of the function x = 4 ln y y + 3 and apply the Newton-Leibniz formula. We get that
∫ 4 ln y y + 3 d y = 4 ∫ ln y y d y + 3 ∫ d y = 4 ∫ ln y d (ln y) + 3 y = = 4 ln 2 y 2 + 3 y + C = 2 ln 2 y + 3 y + C ⇒ S (G) = ∫ 1 4 4 ln y y + 3 d y = 2 ln 2 + y + 3 y 1 4 = = 2 ln 2 4 + 3 4 - (2 ln 2 1 + 3 1) = 8 ln 2 2 + 9
Consider the drawing below.
Answer: S (G) = 8 ln 2 2 + 9
Results
In this article, we identified the geometric meaning of a definite integral and studied the relationship with the area of a curvilinear trapezoid. It follows that we have the opportunity to calculate the area of complex figures by calculating the integral for a curved trapezoid. In the section on finding areas and figures that are bounded by lines y = f (x), x = g (y), these examples are discussed in detail.
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Antiderivative function and indefinite integral
Fact 1. Integration is the inverse action of differentiation, namely, restoring a function from the known derivative of this function. The function thus restored F(x) is called antiderivative for function f(x).
Definition 1. Function F(x f(x) on some interval X, if for all values x from this interval the equality holds F "(x)=f(x), that is, this function f(x) is the derivative of the antiderivative function F(x). .
For example, the function F(x) = sin x is an antiderivative of the function f(x) = cos x on the entire number line, since for any value of x (sin x)" = (cos x) .
Definition 2. Indefinite integral of a function f(x) is the set of all its antiderivatives. In this case, the notation is used
∫
f(x)dx
,where is the sign ∫ called the integral sign, the function f(x) – integrand function, and f(x)dx – integrand expression.
Thus, if F(x) – some antiderivative for f(x) , That
∫
f(x)dx = F(x) +C
Where C - arbitrary constant (constant).
To understand the meaning of the set of antiderivatives of a function as an indefinite integral, the following analogy is appropriate. Let there be a door (traditional wooden door). Its function is to “be a door.” What is the door made of? Made of wood. This means that the set of antiderivatives of the integrand of the function “to be a door”, that is, its indefinite integral, is the function “to be a tree + C”, where C is a constant, which in this context can denote, for example, the type of tree. Just as a door is made from wood using some tools, a derivative of a function is “made” from an antiderivative function using formulas we learned while studying the derivative.
Then the table of functions of common objects and their corresponding antiderivatives (“to be a door” - “to be a tree”, “to be a spoon” - “to be metal”, etc.) is similar to the table of basic indefinite integrals, which will be given below. The table of indefinite integrals lists common functions with an indication of the antiderivatives from which these functions are “made”. In part of the problems on finding the indefinite integral, integrands are given that can be integrated directly without much effort, that is, using the table of indefinite integrals. In more complex problems, the integrand must first be transformed so that table integrals can be used.
Fact 2. When restoring a function as an antiderivative, we must take into account an arbitrary constant (constant) C, and in order not to write a list of antiderivatives with various constants from 1 to infinity, you need to write a set of antiderivatives with an arbitrary constant C, for example, like this: 5 x³+C. So, an arbitrary constant (constant) is included in the expression of the antiderivative, since the antiderivative can be a function, for example, 5 x³+4 or 5 x³+3 and when differentiated, 4 or 3, or any other constant goes to zero.
Let us pose the integration problem: for this function f(x) find such a function F(x), whose derivative equal to f(x).
Example 1. Find the set of antiderivatives of a function
Solution. For this function, the antiderivative is the function
Function F(x) is called an antiderivative for the function f(x), if the derivative F(x) is equal to f(x), or, which is the same thing, differential F(x) is equal f(x) dx, i.e.
(2)
Therefore, the function is an antiderivative of the function. However, it is not the only antiderivative for . They also serve as functions
Where WITH– arbitrary constant. This can be verified by differentiation.
Thus, if there is one antiderivative for a function, then for it there is an infinite number of antiderivatives that differ by a constant term. All antiderivatives for a function are written in the above form. This follows from the following theorem.
Theorem (formal statement of fact 2). If F(x) – antiderivative for the function f(x) on some interval X, then any other antiderivative for f(x) on the same interval can be represented in the form F(x) + C, Where WITH– arbitrary constant.
In the next example, we turn to the table of integrals, which will be given in paragraph 3, after the properties of the indefinite integral. We do this before reading the entire table so that the essence of the above is clear. And after the table and properties, we will use them in their entirety during integration.
Example 2. Find sets of antiderivative functions:
Solution. We find sets of antiderivative functions from which these functions are “made”. When mentioning formulas from the table of integrals, for now just accept that there are such formulas there, and we will study the table of indefinite integrals itself a little further.
1) Applying formula (7) from the table of integrals for n= 3, we get
2) Using formula (10) from the table of integrals for n= 1/3, we have
3) Since
then according to formula (7) with n= -1/4 we find
It is not the function itself that is written under the integral sign. f, and its product by the differential dx. This is done primarily in order to indicate by which variable the antiderivative is sought. For example,
,
;
here in both cases the integrand is equal to , but its indefinite integrals in the cases considered turn out to be different. In the first case, this function is considered as a function of the variable x, and in the second - as a function of z .
The process of finding the indefinite integral of a function is called integrating that function.
Geometric meaning of the indefinite integral
Suppose we need to find a curve y=F(x) and we already know that the tangent of the tangent angle at each of its points is a given function f(x) abscissa of this point.
According to the geometric meaning of the derivative, the tangent of the angle of inclination of the tangent at a given point of the curve y=F(x) equal to the value of the derivative F"(x). So we need to find such a function F(x), for which F"(x)=f(x). Function required in the task F(x) is an antiderivative of f(x). The conditions of the problem are satisfied not by one curve, but by a family of curves. y=F(x)- one of these curves, and any other curve can be obtained from it by parallel translation along the axis Oy.
Let's call the graph of the antiderivative function of f(x) integral curve. If F"(x)=f(x), then the graph of the function y=F(x) there is an integral curve.
Fact 3. The indefinite integral is geometrically represented by the family of all integral curves , as in the picture below. The distance of each curve from the origin of coordinates is determined by an arbitrary integration constant C.
Properties of the indefinite integral
Fact 4. Theorem 1. The derivative of an indefinite integral is equal to the integrand, and its differential is equal to the integrand.
Fact 5. Theorem 2. Indefinite integral of the differential of a function f(x) is equal to the function f(x) up to a constant term , i.e.
(3)
Theorems 1 and 2 show that differentiation and integration are mutually inverse operations.
Fact 6. Theorem 3. The constant factor in the integrand can be taken out of the sign of the indefinite integral , i.e.
This lesson is the first in a series of videos on integration. In it we will analyze what an antiderivative of a function is, and also study the elementary methods of calculating these very antiderivatives.
In fact, there is nothing complicated here: essentially it all comes down to the concept of derivative, which you should already be familiar with. :)
I will immediately note that since this is the very first lesson in our new topic, today there will be no complex calculations and formulas, but what we will learn today will form the basis for much more complex calculations and constructions when calculating complex integrals and areas.
In addition, when starting to study integration and integrals in particular, we implicitly assume that the student is already at least familiar with the concepts of derivatives and has at least basic skills in calculating them. Without a clear understanding of this, there is absolutely nothing to do in integration.
However, here lies one of the most common and insidious problems. The fact is that, when starting to calculate their first antiderivatives, many students confuse them with derivatives. As a result, stupid and offensive mistakes are made during exams and independent work.
Therefore, now I will not give a clear definition of an antiderivative. In return, I suggest you see how it is calculated using a simple specific example.
What is an antiderivative and how is it calculated?
We know this formula:
\[((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))\]
This derivative is calculated simply:
\[(f)"\left(x \right)=((\left(((x)^(3)) \right))^(\prime ))=3((x)^(2))\ ]
Let's look carefully at the resulting expression and express $((x)^(2))$:
\[((x)^(2))=\frac(((\left(((x)^(3)) \right))^(\prime )))(3)\]
But we can write it this way, according to the definition of a derivative:
\[((x)^(2))=((\left(\frac(((x)^(3)))(3) \right))^(\prime ))\]
And now attention: what we just wrote down is the definition of an antiderivative. But to write it correctly, you need to write the following:
Let us write the following expression in the same way:
If we generalize this rule, we can derive the following formula:
\[((x)^(n))\to \frac(((x)^(n+1)))(n+1)\]
Now we can formulate a clear definition.
An antiderivative of a function is a function whose derivative is equal to the original function.
Questions about the antiderivative function
It would seem a fairly simple and understandable definition. However, upon hearing it, the attentive student will immediately have several questions:
- Let's say, okay, this formula is correct. However, in this case, with $n=1$, we have problems: “zero” appears in the denominator, and we cannot divide by “zero”.
- The formula is limited to degrees only. How to calculate the antiderivative, for example, of sine, cosine and any other trigonometry, as well as constants.
- Existential question: is it always possible to find an antiderivative? If yes, then what about the antiderivative of the sum, difference, product, etc.?
I will answer the last question right away. Unfortunately, the antiderivative, unlike the derivative, is not always considered. There is no universal formula by which from any initial construction we will obtain a function that will be equal to this similar construction. As for powers and constants, we’ll talk about that now.
Solving problems with power functions
\[((x)^(-1))\to \frac(((x)^(-1+1)))(-1+1)=\frac(1)(0)\]
As you can see, this formula for $((x)^(-1))$ does not work. The question arises: what works then? Can't we count $((x)^(-1))$? Of course we can. Let's just remember this first:
\[((x)^(-1))=\frac(1)(x)\]
Now let's think: the derivative of which function is equal to $\frac(1)(x)$. Obviously, any student who has studied this topic at least a little will remember that this expression is equal to the derivative of the natural logarithm:
\[((\left(\ln x \right))^(\prime ))=\frac(1)(x)\]
Therefore, we can confidently write the following:
\[\frac(1)(x)=((x)^(-1))\to \ln x\]
You need to know this formula, just like the derivative of a power function.
So what we know so far:
- For a power function - $((x)^(n))\to \frac(((x)^(n+1)))(n+1)$
- For a constant - $=const\to \cdot x$
- A special case of a power function is $\frac(1)(x)\to \ln x$
And if we start multiplying and dividing the simplest functions, how then can we calculate the antiderivative of a product or quotient. Unfortunately, analogies with the derivative of a product or quotient do not work here. There is no standard formula. For some cases, there are tricky special formulas - we will get acquainted with them in future video lessons.
However, remember: there is no general formula similar to the formula for calculating the derivative of a quotient and a product.
Solving real problems
Task No. 1
Let's calculate each of the power functions separately:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
Returning to our expression, we write the general construction:
Problem No. 2
As I already said, prototypes of works and particulars “to the point” are not considered. However, here you can do the following:
We broke down the fraction into the sum of two fractions.
Let's do the math:
The good news is that knowing the formulas for calculating antiderivatives, you can already calculate more complex structures. However, let's go further and expand our knowledge a little more. The fact is that many constructions and expressions, which, at first glance, have nothing to do with $((x)^(n))$, can be represented as a power with a rational exponent, namely:
\[\sqrt(x)=((x)^(\frac(1)(2)))\]
\[\sqrt[n](x)=((x)^(\frac(1)(n)))\]
\[\frac(1)(((x)^(n)))=((x)^(-n))\]
All these techniques can and should be combined. Power expressions can be
- multiply (degrees add);
- divide (degrees are subtracted);
- multiply by a constant;
- etc.
Solving power expressions with rational exponent
Example #1
Let's calculate each root separately:
\[\sqrt(x)=((x)^(\frac(1)(2)))\to \frac(((x)^(\frac(1)(2)+1)))(\ frac(1)(2)+1)=\frac(((x)^(\frac(3)(2))))(\frac(3)(2))=\frac(2\cdot (( x)^(\frac(3)(2))))(3)\]
\[\sqrt(x)=((x)^(\frac(1)(4)))\to \frac(((x)^(\frac(1)(4))))(\frac( 1)(4)+1)=\frac(((x)^(\frac(5)(4))))(\frac(5)(4))=\frac(4\cdot ((x) ^(\frac(5)(4))))(5)\]
In total, our entire construction can be written as follows:
Example No. 2
\[\frac(1)(\sqrt(x))=((\left(\sqrt(x) \right))^(-1))=((\left(((x)^(\frac( 1)(2))) \right))^(-1))=((x)^(-\frac(1)(2)))\]
Therefore we get:
\[\frac(1)(((x)^(3)))=((x)^(-3))\to \frac(((x)^(-3+1)))(-3 +1)=\frac(((x)^(-2)))(-2)=-\frac(1)(2((x)^(2)))\]
In total, collecting everything into one expression, we can write:
Example No. 3
To begin with, we note that we have already calculated $\sqrt(x)$:
\[\sqrt(x)\to \frac(4((x)^(\frac(5)(4))))(5)\]
\[((x)^(\frac(3)(2)))\to \frac(((x)^(\frac(3)(2)+1)))(\frac(3)(2 )+1)=\frac(2\cdot ((x)^(\frac(5)(2))))(5)\]
Let's rewrite:
I hope I will not surprise anyone if I say that what we have just studied is only the simplest calculations of antiderivatives, the most elementary constructions. Let's now look at slightly more complex examples, in which, in addition to the tabular antiderivatives, you will also need to remember the school curriculum, namely, abbreviated multiplication formulas.
Solving more complex examples
Task No. 1
Let us recall the formula for the squared difference:
\[((\left(a-b \right))^(2))=((a)^(2))-ab+((b)^(2))\]
Let's rewrite our function:
We now have to find the prototype of such a function:
\[((x)^(\frac(2)(3)))\to \frac(3\cdot ((x)^(\frac(5)(3))))(5)\]
\[((x)^(\frac(1)(3)))\to \frac(3\cdot ((x)^(\frac(4)(3))))(4)\]
Let's put everything together into a common design:
Problem No. 2
In this case, we need to expand the difference cube. Let's remember:
\[((\left(a-b \right))^(3))=((a)^(3))-3((a)^(2))\cdot b+3a\cdot ((b)^ (2))-((b)^(3))\]
Taking this fact into account, we can write it like this:
Let's transform our function a little:
We count as always - for each term separately:
\[((x)^(-3))\to \frac(((x)^(-2)))(-2)\]
\[((x)^(-2))\to \frac(((x)^(-1)))(-1)\]
\[((x)^(-1))\to \ln x\]
Let us write the resulting construction:
Problem No. 3
At the top we have the square of the sum, let's expand it:
\[\frac(((\left(x+\sqrt(x) \right))^(2)))(x)=\frac(((x)^(2))+2x\cdot \sqrt(x )+((\left(\sqrt(x) \right))^(2)))(x)=\]
\[=\frac(((x)^(2)))(x)+\frac(2x\sqrt(x))(x)+\frac(x)(x)=x+2((x) ^(\frac(1)(2)))+1\]
\[((x)^(\frac(1)(2)))\to \frac(2\cdot ((x)^(\frac(3)(2))))(3)\]
Let's write the final solution:
Now attention! A very important thing, which is associated with the lion's share of errors and misunderstandings. The fact is that until now, counting antiderivatives using derivatives and bringing transformations, we did not think about what the derivative of a constant is equal to. But the derivative of a constant is equal to “zero”. This means that you can write the following options:
- $((x)^(2))\to \frac(((x)^(3)))(3)$
- $((x)^(2))\to \frac(((x)^(3)))(3)+1$
- $((x)^(2))\to \frac(((x)^(3)))(3)+C$
This is very important to understand: if the derivative of a function is always the same, then the same function has an infinite number of antiderivatives. We can simply add any constant numbers to our antiderivatives and get new ones.
It is no coincidence that in the explanation of the problems that we just solved, it was written “Write down the general form of antiderivatives.” Those. It is already assumed in advance that there is not one of them, but a whole multitude. But, in fact, they differ only in the constant $C$ at the end. Therefore, in our tasks we will correct what we did not complete.
Once again we rewrite our constructions:
In such cases, you should add that $C$ is a constant - $C=const$.
In our second function we get the following construction:
And the last one:
And now we really got what was required of us in the original condition of the problem.
Solving problems of finding antiderivatives with a given point
Now that we know about constants and the peculiarities of writing antiderivatives, it is quite logical that the next type of problem arises when, from the set of all antiderivatives, it is required to find the one and only one that would pass through a given point. What is this task?
The fact is that all antiderivatives of a given function differ only in that they are shifted vertically by a certain number. And this means that no matter what point on the coordinate plane we take, one antiderivative will definitely pass, and, moreover, only one.
So, the problems that we will now solve are formulated as follows: not just find the antiderivative, knowing the formula of the original function, but choose exactly the one that passes through the given point, the coordinates of which will be given in the problem statement.
Example #1
First, let’s simply count each term:
\[((x)^(4))\to \frac(((x)^(5)))(5)\]
\[((x)^(3))\to \frac(((x)^(4)))(4)\]
Now we substitute these expressions into our construction:
This function must pass through the point $M\left(-1;4 \right)$. What does it mean that it passes through a point? This means that if instead of $x$ we put $-1$ everywhere, and instead of $F\left(x \right)$ - $-4$, then we should get the correct numerical equality. Let's do this:
We see that we have an equation for $C$, so let's try to solve it:
Let's write down the very solution we were looking for:
Example No. 2
First of all, it is necessary to reveal the square of the difference using the abbreviated multiplication formula:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
The original construction will be written as follows:
Now let's find $C$: substitute the coordinates of point $M$:
\[-1=\frac(8)(3)-12+18+C\]
We express $C$:
It remains to display the final expression:
Solving trigonometric problems
As a final touch to what we have just discussed, I propose to consider two more complex problems that involve trigonometry. In them, in the same way, you will need to find antiderivatives for all functions, then select from this set the only one that passes through the point $M$ on the coordinate plane.
Looking ahead, I would like to note that the technique that we will now use to find antiderivatives of trigonometric functions is, in fact, a universal technique for self-test.
Task No. 1
Let's remember the following formula:
\[((\left(\text(tg)x \right))^(\prime ))=\frac(1)(((\cos )^(2))x)\]
Based on this, we can write:
Let's substitute the coordinates of point $M$ into our expression:
\[-1=\text(tg)\frac(\text( )\!\!\pi\!\!\text( ))(\text(4))+C\]
Let's rewrite the expression taking this fact into account:
Problem No. 2
This will be a little more difficult. Now you'll see why.
Let's remember this formula:
\[((\left(\text(ctg)x \right))^(\prime ))=-\frac(1)(((\sin )^(2))x)\]
To get rid of the “minus”, you need to do the following:
\[((\left(-\text(ctg)x \right))^(\prime ))=\frac(1)(((\sin )^(2))x)\]
Here is our design
Let's substitute the coordinates of point $M$:
In total, we write down the final construction:
That's all I wanted to tell you about today. We studied the very term antiderivatives, how to calculate them from elementary functions, and also how to find an antiderivative passing through a specific point on the coordinate plane.
I hope this lesson will help you understand this complex topic at least a little. In any case, it is on antiderivatives that indefinite and indefinite integrals are constructed, so it is absolutely necessary to calculate them. That's all for me. See you again!