Integration is not difficult to learn. To do this, you just need to learn a certain, fairly small set of rules and develop a kind of instinct. It is, of course, easy to learn the rules and formulas, but it is quite difficult to understand where and when to apply this or that rule of integration or differentiation. This, in fact, is the ability to integrate.
1. Antiderivative. Indefinite integral.
It is assumed that by the time of reading this article the reader already has some differentiation skills (i.e., finding derivatives).
Definition 1.1: A function is called an antiderivative of a function if the equality holds:
Comments:> The emphasis in the word “primordial” can be placed in two ways: first O figurative or prototype A knowing.
Property 1: If a function is an antiderivative of a function, then the function is also an antiderivative of a function.
Proof: Let us prove this from the definition of an antiderivative. Let's find the derivative of the function:
The first term in definition 1.1 is equal to , and the second term is the derivative of the constant, which is equal to 0.
.
Summarize. Let's write down the beginning and end of the chain of equalities: 
Thus, the derivative of a function is equal to , and therefore, by definition, is its antiderivative. The property has been proven.
Definition 1.2: The indefinite integral of a function is the entire set of antiderivatives of this function. This is indicated as follows: 
.
Let's look at the names of each part of the record in detail:
— general designation of the integral,
— integrand (integral) expression, integrable function.
is a differential, and the expression after the letter , in this case it is , will be called the variable of integration.
Comments: The key words in this definition are “the whole set.” Those. If in the future this same “plus C” is not written down in the answer, then the examiner has every right not to count this assignment, because it is necessary to find the entire set of antiderivatives, and if C is missing, then only one is found.
Conclusion: In order to check whether the integral is calculated correctly, it is necessary to find the derivative of the result. It must coincide with the integrand.
Example:
Exercise: Calculate the indefinite integral and check.
Solution:
The way this integral is calculated does not matter in this case. Let's assume that this is a revelation from above. Our task is to show that the revelation did not deceive us, and this can be done through verification.
Examination:
When differentiating the result, we obtained an integrand, which means that the integral was calculated correctly.
2. Beginning. Table of integrals.
To integrate, you do not need to remember each time the function whose derivative is equal to the given integrand (i.e., use the definition of the integral directly). Each collection of problems or textbook on mathematical analysis contains a list of properties of integrals and a table of the simplest integrals.
Let's list the properties.
Properties:
1.
The integral of the differential is equal to the variable of integration.
2. , where is a constant.
The constant multiplier can be taken out of the integral sign.
3.
The integral of a sum is equal to the sum of integrals (if the number of terms is finite).
Table of integrals:
1.  |
10. |
| 2. | 11.  |
| 3. | 12.  |
4.  |
13.  |
5.  |
14.  |
| 6. | 15. |
7.  |
16. |
| 8. | 17.  |
| 9. | 18.  |
Most often, the task is to reduce the integral under study to a tabular one using properties and formulas.
Example:
[Let's use the third property of integrals and write it as a sum of three integrals.]
[Let's use the second property and move the constants beyond the integration sign.]
[ In the first integral we will use table integral No. 1 (n=2), in the second we will use the same formula, but n=1, and for the third integral we can either use the same table integral, but with n=0, or the first property. ]
.
Let's check by differentiation:
The original integrand was obtained, therefore, the integration was performed without errors (and the addition of an arbitrary constant C was not even forgotten).
Table integrals must be learned by heart for one simple reason - in order to know what to strive for, i.e. know the purpose of transforming a given expression.
Here are a few more examples:
1) 
2) 
3) 
Tasks for independent solution:
Exercise 1. Calculate the indefinite integral:
+ Show/hide hint #1.
1) Use the third property and represent this integral as the sum of three integrals.
+ Show/hide hint #2.
+ Show/hide hint #3.
3) For the first two terms, use the first tabular integral, and for the third, use the second tabular integral.
+ Show/hide Solution and Answer.
4) Solution: 
Answer: 
In earlier material, the issue of finding the derivative was considered and its various applications were shown: calculating the slope of a tangent to a graph, solving optimization problems, studying functions for monotonicity and extrema. $\newcommand(\tg)(\mathop(\mathrm(tg))\nolimits)$ $\newcommand(\ctg)(\mathop(\mathrm(tg))\nolimits)$ $\newcommand(\arctg)( \mathop(\mathrm(arctg))\nolimits)$ $\newcommand(\arcctg)(\mathop(\mathrm(arcctg))\nolimits)$
Picture 1.
The problem of finding the instantaneous velocity $v(t)$ using the derivative along a previously known path traveled, expressed by the function $s(t)$, was also considered.
Figure 2.
The inverse problem is also very common, when you need to find the path $s(t)$ traversed by a point in time $t$, knowing the speed of the point $v(t)$. If we recall, the instantaneous speed $v(t)$ is found as the derivative of the path function $s(t)$: $v(t)=s’(t)$. This means that in order to solve the inverse problem, that is, calculate the path, you need to find a function whose derivative will be equal to the speed function. But we know that the derivative of the path is the speed, that is: $s’(t) = v(t)$. Velocity is equal to acceleration times time: $v=at$. It is easy to determine that the desired path function will have the form: $s(t) = \frac(at^2)(2)$. But this is not quite a complete solution. The complete solution will have the form: $s(t)= \frac(at^2)(2)+C$, where $C$ is some constant. Why this is so will be discussed further. For now, let's check the correctness of the solution found: $s"(t)=\left(\frac(at^2)(2)+C\right)"=2\frac(at)(2)+0=at=v( t)$.
It is worth noting that finding a path based on speed is the physical meaning of an antiderivative.
The resulting function $s(t)$ is called the antiderivative of the function $v(t)$. Quite an interesting and unusual name, isn’t it. It contains a great meaning that explains the essence of this concept and leads to its understanding. You will notice that it contains two words “first” and “image”. They speak for themselves. That is, this is the function that is the initial one for the derivative we have. And using this derivative we are looking for the function that was at the beginning, was “first”, “first image”, that is, antiderivative. It is sometimes also called a primitive function or antiderivative.
As we already know, the process of finding the derivative is called differentiation. And the process of finding the antiderivative is called integration. The operation of integration is the inverse of the operation of differentiation. The converse is also true.
Definition. An antiderivative for a function $f(x)$ on a certain interval is a function $F(x)$ whose derivative is equal to this function $f(x)$ for all $x$ from the specified interval: $F'(x)=f (x)$.
Someone may have a question: where did $F(x)$ and $f(x)$ come from in the definition, if initially we were talking about $s(t)$ and $v(t)$. The fact is that $s(t)$ and $v(t)$ are special cases of function designation that have a specific meaning in this case, that is, they are a function of time and a function of speed, respectively. It's the same with the variable $t$ - it denotes time. And $f$ and $x$ are the traditional variant of the general designation of a function and a variable, respectively. It is worth paying special attention to the notation of the antiderivative $F(x)$. First of all, $F$ is capital. Antiderivatives are indicated in capital letters. Secondly, the letters are the same: $F$ and $f$. That is, for the function $g(x)$ the antiderivative will be denoted by $G(x)$, for $z(x)$ – by $Z(x)$. Regardless of the notation, the rules for finding an antiderivative function are always the same.
Let's look at a few examples.
Example 1. Prove that the function $F(x)=\frac(1)(5)\sin5x$ is an antiderivative of the function $f(x)=\cos5x$.
To prove this, we will use the definition, or rather the fact that $F'(x)=f(x)$, and find the derivative of the function $F(x)$: $F'(x)=(\frac(1)(5 ) \sin5x)'=\frac(1)(5)\cdot 5\cos5x= \cos5x$. This means $F(x)=\frac(1)(5) \sin5x$ is the antiderivative of $f(x)=\cos5x$. Q.E.D.
Example 2. Find which functions correspond to the following antiderivatives: a) $F(z)=\tg z$; b) $G(l) = \sin l$.
To find the required functions, let's calculate their derivatives:
a) $F’(z)=(\tg z)’=\frac(1)(\cos^2 z)$;
b) $G(l) = (\sin l)’ = \cos l$.
Example 3. What will be the antiderivative for $f(x)=0$?
Let's use the definition. Let's think about which function can have a derivative equal to $0$. Recalling the table of derivatives, we find that any constant will have such a derivative. We find that the antiderivative we are looking for is: $F(x)= C$.
The resulting solution can be explained geometrically and physically. Geometrically, it means that the tangent to the graph $y=F(x)$ is horizontal at each point of this graph and, therefore, coincides with the $Ox$ axis. Physically it is explained by the fact that a point with a speed equal to zero remains in place, that is, the path it has traveled is unchanged. Based on this, we can formulate the following theorem.
Theorem. (Sign of constancy of functions). If on some interval $F’(x) = 0$, then the function $F(x)$ on this interval is constant.
Example 4. Determine which functions are antiderivatives of a) $F_1 = \frac(x^7)(7)$; b) $F_2 = \frac(x^7)(7) – 3$; c) $F_3 = \frac(x^7)(7) + 9$; d) $F_4 = \frac(x^7)(7) + a$, where $a$ is some number.
Using the definition of an antiderivative, we conclude that to solve this problem we need to calculate the derivatives of the antiderivative functions given to us. When calculating, remember that the derivative of a constant, that is, of any number, is equal to zero.
a) $F_1 =(\frac(x^7)(7))"= 7 \cdot \frac(x^6)(7) = x^6$;
b) $F_2 =\left(\frac(x^7)(7) – 3\right)"=7 \cdot \frac(x^6)(7)= x^6$;
c) $F_3 =(\frac(x^7)(7) + 9)’= x^6$;
d) $F_4 =(\frac(x^7)(7) + a)’ = x^6$.
What do we see? Several different functions are primitives of the same function. This suggests that any function has infinitely many antiderivatives, and they have the form $F(x) + C$, where $C$ is an arbitrary constant. That is, the operation of integration is multivalued, unlike the operation of differentiation. Based on this, let us formulate a theorem that describes the main property of antiderivatives.
Theorem. (The main property of antiderivatives). Let the functions $F_1$ and $F_2$ be antiderivatives of the function $f(x)$ on some interval. Then for all values from this interval the following equality is true: $F_2=F_1+C$, where $C$ is some constant.
The fact of the presence of an infinite number of antiderivatives can be interpreted geometrically. Using parallel translation along the $Oy$ axis, one can obtain from each other the graphs of any two antiderivatives for $f(x)$. This is the geometric meaning of the antiderivative.
It is very important to pay attention to the fact that by choosing the constant $C$ you can ensure that the graph of the antiderivative passes through a certain point.
Figure 3.
Example 5. Find the antiderivative for the function $f(x)=\frac(x^2)(3)+1$, the graph of which passes through the point $(3; 1)$.
Let's first find all antiderivatives for $f(x)$: $F(x)=\frac(x^3)(9)+x + C$.
Next, we will find a number C for which the graph $y=\frac(x^3)(9)+x + C$ will pass through the point $(3; 1)$. To do this, we substitute the coordinates of the point into the graph equation and solve it for $C$:
$1= \frac(3^3)(9)+3 + C$, $C=-5$.
We obtained a graph $y=\frac(x^3)(9)+x-5$, which corresponds to the antiderivative $F(x)=\frac(x^3)(9)+x-5$.
Table of antiderivatives
A table of formulas for finding antiderivatives can be compiled using formulas for finding derivatives.
| Functions | Antiderivatives |
| $0$ | $C$ |
| $1$ | $x+C$ |
| $a\in R$ | $ax+C$ |
| $x^n, n\ne1$ | $\displaystyle \frac(x^(n+1))(n+1)+C$ |
| $\displaystyle \frac(1)(x)$ | $\ln|x|+C$ |
| $\sin x$ | $-\cos x+C$ |
| $\cos x$ | $\sin x+C$ |
| $\displaystyle \frac(1)(\sin^2 x)$ | $-\ctg x+C$ |
| $\displaystyle \frac(1)(\cos^2 x)$ | $\tg x+C$ |
| $e^x$ | $e^x+C$ |
| $a^x, a>0, a\ne1$ | $\displaystyle \frac(a^x)(\ln a) +C$ |
| $\displaystyle \frac(1)(\sqrt(1-x^2))$ | $\arcsin x+C$ |
| $\displaystyle -\frac(1)(\sqrt(1-x^2))$ | $\arccos x+C$ |
| $\displaystyle \frac(1)(1+x^2)$ | $\arctg x+C$ |
| $\displaystyle -\frac(1)(1+x^2)$ | $\arcctg x+C$ |
You can check the correctness of the table in the following way: for each set of antiderivatives located in the right column, find the derivative, which will result in the corresponding functions in the left column.
Some rules for finding antiderivatives
As is known, many functions have a more complex form than those indicated in the table of antiderivatives, and can be any arbitrary combination of sums and products of functions from this table. And here the question arises: how to calculate antiderivatives of such functions. For example, from the table we know how to calculate the antiderivatives of $x^3$, $\sin x$ and $10$. How, for example, can one calculate the antiderivative $x^3-10\sin x$? Looking ahead, it is worth noting that it will be equal to $\frac(x^4)(4)+10\cos x$.
1. If $F(x)$ is antiderivative for $f(x)$, $G(x)$ for $g(x)$, then for $f(x)+g(x)$ the antiderivative will be equal to $ F(x)+G(x)$.
2. If $F(x)$ is an antiderivative for $f(x)$ and $a$ is a constant, then for $af(x)$ the antiderivative is $aF(x)$.
3. If for $f(x)$ the antiderivative is $F(x)$, $a$ and $b$ are constants, then $\frac(1)(a) F(ax+b)$ is the antiderivative for $f (ax+b)$.
Using the obtained rules we can expand the table of antiderivatives.
| Functions | Antiderivatives |
| $(ax+b)^n, n\ne1, a\ne0$ | $\displaystyle \frac((ax+b)^n)(a(n+1)) +C$ |
| $\displaystyle \frac(1)(ax+b), a\ne0$ | $\displaystyle \frac(1)(a)\ln|ax+b|+C$ |
| $e^(ax+b), a\ne0$ | $\displaystyle \frac(1)(a) e^(ax+b)+C$ |
| $\sin(ax+b), a\ne0$ | $\displaystyle -\frac(1)(a)\cos(ax+b)+C$ |
| $\cos(ax+b), a\ne0$ | $\displaystyle \frac(1)(a)\sin(ax+b)+C$ |
Example 5. Find antiderivatives for:
a) $\displaystyle 4x^3+10x^7$;
b) $\displaystyle \frac(6)(x^5) -\frac(2)(x)$;
c) $\displaystyle 5\cos x+\sin(3x+15)$;
d) $\displaystyle \sqrt(x)-2\sqrt(x)$.
a) $4\frac (x^(3+1))(3+1)+10\frac(x^(7+1))(7+1)+C=x^4+\frac(5)( 4) x^8+C$;
b) $-\frac(3)(2x^4) -2\ln|x|+C$;
c) $5 \sin x - \frac(1)(3)\cos(3x + 15) + C$;
d) $\frac(2)(3)x\sqrt(x) - \frac(3)(2) x\sqrt(x) + C$.
Antiderivative
Definition of an 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 by parallel translations 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).
- The constant factor 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
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 concrete 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 mistakes 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!
Integration is one of the main operations in mathematical analysis. Tables of known antiderivatives can be useful, but now, after the advent of computer algebra systems, they are losing their significance. Below is a list of the most common primitives.
Table of basic integrals
Another, compact option
Table of integrals of trigonometric functions
From rational functions
From irrational functions
Integrals of transcendental functions
"C" is an arbitrary integration constant, which is determined if the value of the integral at any point is known. Each function has an infinite number of antiderivatives.
Most schoolchildren and students have problems calculating integrals. This page contains integral tables from trigonometric, rational, irrational and transcendental functions that will help in the solution. A table of derivatives will also help you.
