An antiderivative of a function and a general view.
a)Direct integration.
Finding integrals of functions based on the direct application of the properties of indefinite integrals and a table of basic integration formulas. Consider an example of finding the integral of a function by direct integration.
Example:
∫(X–3) 2d X= ∫(X 2 –6X+9)d X= ∫X 2d X- 6∫X d X+9∫d X=X 3 ∕3 -3X 2 +9X+C.
In the vast majority of cases, we are dealing with integrals of functions that cannot be found by direct integration. In this case, it is necessary to make a substitution (replace the variable).
b)Integration by substitution (change of variable).
Substitution integration, or as it is often called, the change-of-variable method, is one of the more efficient and common integration methods. The substitution method is to pass from a given integration variable to another variable in order to simplify the subintegral expression and bring it to one of the tabular integrals. In this case, the choice of substitution is decided by the performer individually, because there are no general rules specifying which substitution in this case take.
Example: Find the integral ∫ e 2x+3d X.
Let us introduce a new variable t associated with X next dependency 2 X+ 3 =t.
Take the differentials of the left and right sides of this equality: 2d X=dt;d X=dt/2.
Now instead of 2 X+ 3 and d X we substitute their values into the integrand. Then we get: ∫ e 2x+3d X=∫e tdt= e t + C. Returning to the previous variable, we finally obtain the expression:
∫e 2x+3d X=e 2x+3 + C.
To make sure that the integral is taken correctly, it is necessary to use the antiderivative function e 2x+ 3 differentiate and check whether whether its derivative is equal to the integrand:
(e 2x+ 3)" =e 2x+ 3 (2 X+3)" =e 2x+ 3 .
3. Definite integral and its properties.
The concept of a definite integral is widely used in many fields of science and technology. With its help, areas bounded by curves, volumes of arbitrary shape, power and work of a variable force, the path of a moving body, moments of inertia, and many other quantities are calculated.
AT 
In the overwhelming majority of cases, the concept of a definite integral is introduced when solving problems of determining the area of a curvilinear trapezoid. Let there be a continuous function y =f( X) on the segment [ a, c]. The figure bounded by the curve y \u003d f ( X) ordinates a Ah oh in BUT P and segment [ a, c] the abscissa axis is called a curvilinear trapezoid (Fig. 1).
Let's set ourselves the task: to determine the area S of a curvilinear trapezoid a A o A P in. To do this, we split the segment [ a, c] on the P not necessary equal parts and denote the division points as follows: a=X about < X one < X 2 ‹ … ‹ X P = in.
From the division points, we restore the perpendiculars to the intersection with the curve y \u003d f ( X). Thus, we have divided the entire area bounded by the curve into P elementary curvilinear trapezoids. Let's restore from arbitrary points each segment ∆ X i ordinatesf(C i) until it intersects with the curve y =f( X). Next, we construct a stepped figure consisting of rectangles with base ∆ X i and height f(C i). elemental area ith rectangle will be S i =f(C i)(X i -X i -1 ), and the whole area S P the resulting stepped figure will be equal to the sum of the areas of the rectangles:
S P=f(C o)( X 1 -X o) + f (C 1) ( X 2 -X 1 ) + … +f(С P- 1)(X P -X P- 1).
To shorten the record of this amount, enter the symbol 
(sigma) - a sign meaning the summation of quantities. Then
S P
=
.
This sum S P, which is called the integral sum, can be either greater or less than the true value of the given area. The closest value to the true value of the area will be the sum limit, provided that the elementary segments will be split ( p→
), and the length of the big segment ∆X max will tend to zero, i.e.:
S=
(4)
This limit of the integral sum (if it exists) is called definite integral from function f( X) on the segment [ a,in] and denote: 
=
(5)
(read - “definite integral of a before in ef ot x de x”).
Numbers a and in are called the lower and upper limits of integration, respectively, f( X) is an integrand; X is the integration variable. Applying formulas (4) and (5) can be written. That the area of a curvilinear trapezoid is numerically equal to the integral of the function that bounds the trapezoid, taken over the integration interval [a,in]:
.
This fact expresses the geometric meaning of the definite integral.
Consider the properties of the definite integral.
1. The definite integral does not depend on the designation of the variable, i.e.: 
=
.
2. The definite integral of the algebraic sum is equal to the algebraic sum of the definite integrals of each term:
=
f 1 ( X)d x +
f 2 ( X)d X+ ….
We have seen that the derivative has numerous applications: the derivative is the speed of movement (or, more generally, the speed of any process); derivative is slope tangent to the graph of the function; using the derivative, you can investigate the function for monotonicity and extrema; The derivative helps to solve optimization problems.
But in real life have to decide and inverse problems: for example, along with the problem of finding the speed from a known law of motion, there is also the problem of restoring the law of motion from 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 u = tg. Find the law of motion.
Decision. Let s = s(t) be the desired law of motion. It is known that s"(t) = u"(t). So, in order to solve the problem, we need to choose function s = s(t), whose derivative is equal to tg. It's easy to guess that
We note right away that the example is solved correctly, but incompletely. We have obtained that In fact, the problem has infinitely many solutions: any function of the form 
arbitrary constant, can serve as a law of motion, because
To make the task more specific, we had to fix the initial situation: indicate the coordinate of the moving point at some point in time, for example, at t=0. If, say, s (0) \u003d s 0, then from the equality we obtain s (0) \u003d 0 + C, i.e. S 0 \u003d C. Now the law of motion is uniquely defined:
In mathematics, reciprocal operations are assigned different names, come up with special notation: for example, squaring (x 2) and extracting square root sine (sinx) and arcsine(arcsin x), etc. The process of finding the derivative with respect to a given function is called differentiation, and the inverse operation, i.e. the process of finding a function by a given derivative - by integration.
The very term "derivative" can be justified "in a worldly way": the function y - f (x) "brings into the world" new feature y "= f" (x) The function y \u003d f (x) acts as if as a "parent", but mathematicians, of course, do not call it "parent" or "producer", they say that this is, in relation to the function y"=f"(x), the primary image, or, in short, the antiderivative.
Definition 1. The function y \u003d F (x) is called the antiderivative for the function y \u003d f (x) on a given interval X, if for all x from X the equality F "(x) \u003d f (x) is true.
In practice, the interval X is usually not specified, but implied (as the natural domain of the function).
Here are some examples:
1) The function y \u003d x 2 is an antiderivative for the function y \u003d 2x, since for all x the equality (x 2) "\u003d 2x is true.
2) the function y - x 3 is the antiderivative for the function y-3x 2, since for all x the equality (x 3)" \u003d 3x 2 is true.
3) The function y-sinx is an antiderivative for the function y=cosx, since for all x the equality (sinx) "=cosx is valid.
4) The function is antiderivative for the 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 that is written in the second column is equal to the function that is written in the corresponding line of the first column (check it out, don't be lazy, it's very useful). For example, for the function y \u003d x 5, the antiderivative, as you establish, is the function (see the fourth row of the table).
Notes: 1. Below we prove the theorem that if y = F(x) is an antiderivative for a 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 the antiderivative for the function y = f(x)", they say F(x) is the antiderivative for f(x)".
2. Rules for finding antiderivatives
When searching for antiderivatives, as well as when searching for 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 computing derivatives.
We know that the derivative of a sum is equal to the sum of the derivatives. This rule generates a corresponding rule for finding antiderivatives.
Rule 1 The antiderivative of a sum is equal to the sum of antiderivatives.
We draw your attention to some "lightness" of this wording. In fact, it would be necessary to formulate a theorem: if the functions y = f(x) and y=g(x) have antiderivatives on the interval X, y-F(x) and y-G(x), respectively, 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 (and not theorems), one leaves only keywords- so it is more convenient to apply the rule in practice
Example 2 Find the antiderivative for the function y = 2x + cos x.
Decision. The antiderivative for 2x is x "; the antiderivative for cosx is sin x. Hence, the antiderivative for the function y \u003d 2x + cos x will be the function y \u003d x 2 + sin x (and in general any function of the form Y \u003d x 1 + sinx + C) .
We know that the constant factor can be taken out of the sign of the derivative. This rule generates a corresponding rule for finding antiderivatives.
Rule 2 The constant factor can be taken out of the antiderivative sign.
Example 3
Decision. a) The antiderivative for sin x is -cos x; hence, for the function y \u003d 5 sin x, the antiderivative will be the function y \u003d -5 cos x.
b) The antiderivative for cos x is sin x; hence, for the antiderivative function there will be a function
c) The antiderivative for x 3 is the antiderivative for x is the antiderivative for the function y \u003d 1 is the function y \u003d x. Using the first and second rules for finding antiderivatives, we get that the antiderivative for the function y \u003d 12x 3 + 8x-1 is the function
Comment. As you know, the derivative of a product is not equal to the product of derivatives (the rule for differentiating a product is more complicated) 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!
We obtain one more rule for finding antiderivatives. We know that the derivative of the function y \u003d f (kx + m) is calculated by the formula
This rule generates a corresponding rule for finding antiderivatives.
Rule 3 If y \u003d F (x) is the antiderivative for the function y \u003d f (x), then the antiderivative for the function y \u003d f (kx + m) is the function
Indeed,
This means that it is an antiderivative for the function y \u003d f (kx + m).
The meaning of the third rule is as follows. If you know that the antiderivative for the function y \u003d f (x) is the function y \u003d F (x), and you need to find the antiderivative of the function y \u003d f (kx + m), then proceed as follows: take the same function F, but instead of the argument x, substitute the expression xx+m; in addition, do not forget to write the “correction factor” before the sign of the function
Example 4 Find antiderivatives for given functions:
Decision, a) The antiderivative for sin x is -cos x; this means that for the function y \u003d sin2x, the antiderivative will be the function
b) The antiderivative for cos x is sin x; hence, for the antiderivative function there will be a function
c) The antiderivative for x 7 is therefore, for the function y \u003d (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 \u003d F (x) be the antiderivative for the function y \u003d f (x) on the interval X. This means that for all x from X the equality x "(x) \u003d f (x) is true. Find the derivative of any function of the form y \u003d F (x) + C:
(F (x) + C) \u003d F "(x) + C \u003d f (x) + 0 \u003d f (x).
So, (F(x)+C) = f(x). This means that y \u003d F (x) + C is an antiderivative for the function y \u003d f (x).
Thus, we have proved that if the function y \u003d f (x) has an antiderivative y \u003d F (x), then the function (f \u003d f (x) has infinitely many antiderivatives, for example, any function of the form y \u003d F (x) +C is antiderivative.
2. Let us now prove that specified type functions, the entire set of antiderivatives is exhausted.
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) \u003d f (x).
Consider the function y \u003d F 1 (x) -.F (x) and find its derivative: (F, (x) -F (x)) "\u003d F [(x) - F (x) \u003d 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 in § 35). Hence, F 1 (x) -F (x) \u003d C, i.e. Fx) \u003d F (x) + C.
The theorem has been proven.
Example 5 The law of change of speed from time v = -5sin2t is set. Find the law of motion s = s(t) if it is known that at the time t=0 the coordinate of the point was equal to the number 1.5 (i.e. s(t) = 1.5).
Decision. Since the speed is the 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 a specific value of the constant C, we use initial conditions, according to which, s(0) = 1.5. Substituting in formula (1) the values t=0, S = 1.5, we get:
Substituting the found value C into formula (1), we obtain the law of motion of interest to us:
Definition 2. If a function y = f(x) has an antiderivative y = F(x) on the interval X, then the set of all antiderivatives, i.e. the set of functions of the form y \u003d F (x) + C, is called the indefinite integral of the function y \u003d f (x) and denoted:
(they read: “the indefinite integral ef of x de x”).
In the next section, we will find out what is hidden meaning the indicated designation.
Based on the table of antiderivatives available in this paragraph, we will compile a table of basic indefinite integrals:
Based on the above three rules for finding antiderivatives, we can formulate the corresponding integration rules.
Rule 1 Integral of the sum of functions is equal to the sum integrals of these functions:
Rule 2 The constant factor can be taken out of the integral sign:
Rule 3 If a
Example 6 Find indefinite integrals:
Decision, a) Using the first and second integration rules, we obtain:
Now we use the 3rd and 4th integration formulas:
As a result, we get:
b) Using the third integration rule and formula 8, we get:
c) For the direct determination of the given integral, we have neither the corresponding formula nor the corresponding rule. In such cases, sometimes pre-executed identical transformations expression contained under the integral sign.
Let's use trigonometric formula downgrading:
Then successively we find:
A.G. Mordkovich Algebra Grade 10
Calendar-thematic planning in mathematics, video in mathematics online , Math at school
This lesson is the first in a series of videos on integration. In it, we will understand what is antiderivative of a function, and also we will study the elementary methods of calculating these very antiderivatives.
In fact, there is nothing complicated here: in essence, everything comes down to the concept of a derivative, which you should already be familiar with. :)
I note right away 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 study today will form the basis of much more complex calculations and structures when calculating complex integrals and squares.
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 the derivative and has at least elementary 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 frequent and insidious problems. The fact is that, starting to calculate their first antiderivatives, many students confuse them with derivatives. As a result, in exams and independent work stupid and offensive mistakes are made.
Therefore, now I will not give a clear definition of antiderivative. And in return, I suggest you look at how it is considered on a simple concrete example.
What is primitive and how is it considered
We know this formula:
\[((\left(((x)^(n)) \right))^(\prime ))=n\cdot ((x)^(n-1))\]
This derivative is considered elementary:
\[(f)"\left(x \right)=((\left(((x)^(3)) \right))^(\prime ))=3((x)^(2))\ ]
Let's look closely at the resulting expression and express $((x)^(2))$:
\[((x)^(2))=\frac(((\left(((x)^(3)) \right))^(\prime )))(3)\]
But we can also write it this way, according to the definition of the derivative:
\[((x)^(2))=((\left(\frac(((x)^(3)))(3) \right))^(\prime ))\]
And now attention: what we just wrote down is the definition of the antiderivative. But to write it correctly, you need to write the following:
Let's 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 seems to be quite simple and clear definition. However, upon hearing it, the attentive student will immediately have several questions:
- Let's say, well, this formula is correct. However, in this case, when $n=1$, we have problems: “zero” appears in the denominator, and it is impossible to divide by “zero”.
- The formula is only limited to powers. How to calculate the antiderivative, for example, sine, cosine and any other trigonometry, as well as constants.
- An existential question: is it always possible to find an antiderivative at all? If so, what about the antiderivative 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 such universal formula, according to 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 we see, given formula for $((x)^(-1))$ doesn't work. The question arises: what then works? Can't we count $((x)^(-1))$? Of course we can. Let's just start with this:
\[((x)^(-1))=\frac(1)(x)\]
Now let's think: the derivative of which function is equal to $\frac(1)(x)$. Obviously, any student who has been at least a little engaged in this topic 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:
\[\frac(1)(x)=((x)^(-1))\to \ln x\]
This formula needs to be known, 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 - $\frac(1)(x)\to \ln x$
And if we start multiplying and dividing the simplest functions, how then to calculate the antiderivative of a product or a quotient. Unfortunately, analogies with the derivative of a product or a quotient do not work here. Any standard formula does not exist. For some cases, there are tricky special formulas - we will get to know them in future video tutorials.
However, remember: general formula, there is no similar formula for calculating the derivative of a quotient and a product.
Solving real problems
Task #1
Let's each power functions count separately:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
Returning to our expression, we write the general construction:
Task #2
As I have already said, primitive works and private "blank through" are not considered. However, here you can do the following:
We have broken the fraction into the sum of two fractions.
Let's calculate:
The good news is that once you know the formulas for calculating antiderivatives, you are already able to calculate more complex structures. However, let's go ahead and expand our knowledge a little more. The fact is that many constructions and expressions that, at first glance, have nothing to do with $((x)^(n))$ can be represented as a power with rational indicator, 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
- multiply (the powers are added);
- divide (degrees are subtracted);
- multiply by a constant;
- etc.
Solving expressions with a degree with a rational exponent
Example #1
Let's count 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 #2
\[\frac(1)(\sqrt(x))=((\left(\sqrt(x) \right))^(-1))=((\left(((x)^(\frac( 1)(2))) \right))^(-1))=((x)^(-\frac(1)(2)))\]
Therefore, we will 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 in one expression, we can write:
Example #3
First, 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 a little more complex examples, in which, in addition to tabular antiderivatives, it will also be necessary to recall school curriculum, namely, the reduced multiplication formulas.
Solving More Complex Examples
Task #1
Recall the formula for the square of the difference:
\[((\left(a-b \right))^(2))=((a)^(2))-ab+((b)^(2))\]
Let's rewrite our function:
We now have to find the antiderivative of such a function:
\[((x)^(\frac(2)(3)))\to \frac(3\cdot ((x)^(\frac(5)(3))))(5)\]
\[((x)^(\frac(1)(3)))\to \frac(3\cdot ((x)^(\frac(4)(3))))(4)\]
We collect everything in a common design:
Task #2
In this case, we need to open the difference cube. Let's remember:
\[((\left(a-b \right))^(3))=((a)^(3))-3((a)^(2))\cdot b+3a\cdot ((b)^ (2))-((b)^(3))\]
Given this fact, it can be written as follows:
Let's modify our function a bit:
We consider, 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's write the resulting construction:
Task #3
On top we have the square of the sum, let's open 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:
And now attention! Highly important thing, with which the lion's share of errors and misunderstandings is associated. The fact is that until now, counting antiderivatives with the help of derivatives, giving 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". And 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 primitives and get new ones.
It is no coincidence that in the explanation of the tasks that we have just solved, it was written “Write down general form primitives." Those. it is already assumed in advance that there is not one, but a whole multitude of them. But, in fact, they differ only in the constant $C$ at the end. Therefore, in our tasks, we will correct what we have not completed.
Once again, we rewrite our constructions:
In such cases, one 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 initial condition of the problem.
Solving problems on finding antiderivatives with a given point
Now that we know about constants and about the peculiarities of writing antiderivatives, the following type of problems quite logically arises, when from the set of all antiderivatives it is required to find one and only such that would pass through 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 some number. And this means that no matter what point on coordinate plane we didn’t take it, one primitive will definitely pass, and, moreover, only one.
So, the tasks that we will now solve are formulated as follows: it is not easy to find the antiderivative, knowing the formula of the original function, but to choose exactly one of them that passes through a given point, the coordinates of which will be given in the condition of the problem.
Example #1
First, let's just calculate 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 #2
First of all, it is necessary to open the square of the difference using the abbreviated multiplication formula:
\[((x)^(2))\to \frac(((x)^(3)))(3)\]
The original structure will be written as follows:
Now let's find $C$: substitute the coordinates of the point $M$:
\[-1=\frac(8)(3)-12+18+C\]
We express $C$:
It remains to display the final expression:
Solving trigonometric problems
As final chord In addition to what we have just analyzed, I propose to consider two more challenging tasks containing trigonometry. In them, in the same way, it will be necessary to find the antiderivatives for all functions, then choose 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 from trigonometric functions, in fact, is universal reception for self-test.
Task #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 with this fact in mind:
Task #2
Here it will be a little more difficult. Now you will 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 must do the following:
\[((\left(-\text(ctg)x \right))^(\prime ))=\frac(1)(((\sin )^(2))x)\]
Here is our design
Substitute the coordinates of the point $M$:
Let's write down the final construction:
That's all I wanted to tell you today. We have studied the very term antiderivatives, how to count them from elementary functions, as well as how to find the antiderivative passing through a specific point on the coordinate plane.
I hope this lesson will help you a little to understand this difficult topic. In any case, it is on the antiderivatives that indefinite and indefinite integrals are built, so it is absolutely necessary to consider them. That's all for me. See you soon!
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