Finding the antiderivative function. Antiderivative and Indefinite Integral - Knowledge Hypermarket

There are three basic rules for finding antiderivative functions. They are very similar to the corresponding differentiation rules.

Rule 1

If F is an antiderivative for some function f, and G is an antiderivative for some function g, then F + G will be an antiderivative for f + g.

By definition of antiderivative F' = f. G' = g. And since these conditions are met, then according to the rule for calculating the derivative for the sum of functions, we will have:

(F + G)' = F' + G' = f + g.

Rule 2

If F is an antiderivative for some function f and k is some constant. Then k*F is the antiderivative for the function k*f. This rule follows from the rule for calculating the derivative of a complex function.

We have: (k*F)’ = k*F’ = k*f.

Rule 3

If F(x) is some antiderivative for the function f(x), and k and b are some constants, and k is not equal to zero, then (1/k)*F*(k*x+b) will be the antiderivative for the function f (k*x+b).

This rule follows from the rule for calculating the derivative of a complex function:

((1/k)*F*(k*x+b))’ = (1/k)*F’(k*x+b)*k = f(k*x+b).

Let's look at a few examples of how these rules apply:

Example 1. Find the general form of antiderivatives for the function f(x) = x^3 +1/x^2. For the function x^3 one of the antiderivatives will be the function (x^4)/4, and for the function 1/x^2 one of the antiderivatives will be the function -1/x. Using the first rule, we have:

F(x) = x^4/4 - 1/x +C.

Example 2. Let's find the general form of antiderivatives for the function f(x) = 5*cos(x). For the cos(x) function, one of the antiderivatives will be the sin(x) function. If we now use the second rule, we will have:

F(x) = 5*sin(x).

Example 3 Find one of the antiderivatives for the function y = sin(3*x-2). For the sin(x) function, one of the antiderivatives will be the -cos(x) function. If we now use the third rule, we get an expression for the antiderivative:

F(x) = (-1/3)*cos(3*x-2)

Example 4. Find the antiderivative for the function f(x) = 1/(7-3*x)^5

The antiderivative for the function 1/x^5 will be the function (-1/(4*x^4)). Now, using the third rule, we get.

Solving integrals is an easy task, but only for the elite. This article is for those who want to learn to understand integrals, but know little or nothing about them. Integral... Why is it needed? How to calculate it? What are definite and indefinite integrals? If the only use of the integral you know is to get something useful from hard-to-reach places with a hook in the shape of an integral icon, then welcome! Learn how to solve integrals and why you can't do without it.

We study the concept of "integral"

Integration was known in ancient Egypt. Of course, not in a modern form, but still. Since then, mathematicians have written a great many books on the subject. Particularly distinguished newton and Leibniz but the essence of things has not changed. How to understand integrals from scratch? No way! To understand this topic, you will still need a basic knowledge of the basics of mathematical analysis. Information about , which is also necessary for understanding integrals, is already in our blog.

Indefinite integral

Let's have some function f(x) .

The indefinite integral of the function f(x) such a function is called F(x) , whose derivative is equal to the function f(x) .

In other words, an integral is a reverse derivative or antiderivative. By the way, about how to read in our article.

An antiderivative exists for all continuous functions. Also, a constant sign is often added to the antiderivative, since the derivatives of functions that differ by a constant coincide. The process of finding an integral is called integration.

Simple example:

In order not to constantly calculate the antiderivatives of elementary functions, it is convenient to bring them into a table and use ready-made values.

Complete table of integrals for students

Definite integral

When dealing with the concept of an integral, we are dealing with infinitesimal quantities. The integral will help calculate the area of the figure, the mass of an inhomogeneous body, the path traveled during uneven movement, and much more. It should be remembered that the integral is the sum of an infinitely large number of infinitely small terms.

As an example, imagine a graph of some function. How to find the area of a figure bounded by a graph of a function?

With the help of an integral! Let's break the curvilinear trapezoid, bounded by the coordinate axes and the graph of the function, into infinitesimal segments. Thus, the figure will be divided into thin columns. The sum of the areas of the columns will be the area of the trapezoid. But remember that such a calculation will give an approximate result. However, the smaller and narrower the segments, the more accurate the calculation will be. If we reduce them to such an extent that the length tends to zero, then the sum of the areas of the segments will tend to the area of the figure. This is the definite integral, which is written as follows:

The points a and b are called the limits of integration.

Bari Alibasov and the group "Integral"

By the way! For our readers there is now a 10% discount on

Rules for Calculating Integrals for Dummies

Properties of the indefinite integral

How to solve indefinite integral? Here we will consider the properties of the indefinite integral, which will be useful in solving examples.

The derivative of the integral is equal to the integrand:

The constant can be taken out from under the integral sign:

The integral of the sum is equal to the sum of the integrals. Also true for the difference:

Properties of the Definite Integral

Linearity:

The sign of the integral changes if the limits of integration are reversed:

At any points a, b and with:

We have already found out that the definite integral is the limit of the sum. But how to get a specific value when solving an example? For this, there is the Newton-Leibniz formula:

Examples of solving integrals

Below we consider several examples of finding indefinite integrals. We offer you to independently understand the intricacies of the solution, and if something is not clear, ask questions in the comments.

To consolidate the material, watch a video on how integrals are solved in practice. Do not despair if the integral is not given immediately. Turn to a professional student service, and any triple or curvilinear integral over a closed surface will be within your power.

This lesson is the first in a series of videos on integration. In it, we will analyze what the antiderivative of a function is, and also study the elementary methods for 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 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 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, stupid and offensive mistakes are made in exams and independent work.

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 would seem that a fairly simple and understandable 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 you can see, this formula for $((x)^(-1))$ does not 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 begin to multiply and divide 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. There is no standard formula. For some cases, there are tricky special formulas - we will get to know them in future video tutorials.

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 #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:

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 degree 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

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 slightly more complex examples, in which, in addition to tabular antiderivatives, you still need to remember the school curriculum, namely, abbreviated 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! A very important thing, which is associated with the lion's share of errors and misunderstandings. 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 the general appearance of antiderivatives." 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 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 some 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 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 a final chord to what we have just analyzed, I propose to consider two more complex problems that contain 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 the antiderivatives of trigonometric functions is, in fact, a universal technique for self-checking.

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 studied the very term antiderivatives, how to count 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 at least a little to understand this complex 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!

Definition. The function F (x) is called antiderivative for the function f (x) on a given interval, if for any x from the given interval F "(x) \u003d f (x).

The main property of primitives.

If F (x) is the antiderivative of the function f (x), then the function F (x) + C , where C is an arbitrary constant, is also the antiderivative of the function f (x) (i.e., all antiderivatives of f(x) are written in the form F(x) + C).

Geometric interpretation.

Graphs of all antiderivatives of a given function f (x) are obtained from the graph of any one antiderivative by parallel transfers along the Oy axis.

Table of primitives.

Rules for finding antiderivatives .

Let F(x) and G(x) be the antiderivatives of the functions f(x) and g(x), respectively. Then:

1.F( x)±G( x) is antiderivative for f(x) ± g(x);

2. a F( x) is antiderivative for af(x);

3. - antiderivative for af(kx +b).

Tasks and tests on the topic "Antiprimitive"

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Lessons: 1 Assignments: 11 Tests: 1
Derivative and antiderivative - Preparation for the exam in mathematics
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Integral - Antiderivative and integral Grade 11
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Calculating areas using integrals - Antiderivative and integral Grade 11
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Having studied this topic, you should know what is called an antiderivative, its main property, geometric interpretation, rules for finding antiderivatives; be able to find all antiderivatives of functions using a table and rules for finding antiderivatives, as well as an antiderivative passing through a given point. Consider solving problems on this topic using examples. Pay attention to the design of the decisions.

Examples.

1. Find out if the function F ( x) = X 3 – 3X+ 1 antiderivative for the function f(x) = 3(X 2 – 1).

Decision: F"( x) = (X 3 – 3X+ 1)′ = 3 X 2 – 3 = 3(X 2 – 1) = f(x), i.e. F"( x) = f(x), therefore, F(x) is an antiderivative for the function f(x).

2. Find all antiderivative functions f(x) :

a) f(x) = X 4 + 3X 2 + 5

Decision: Using the table and the rules for finding antiderivatives, we get:

Answer:

b) f(x) = sin(3 x – 2)

Decision:

We have seen that the derivative has numerous applications: the derivative is the speed of movement (or, more generally, the speed of any process); the derivative is the slope of the 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, one also has to solve 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 A material point moves along a straight line, 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, since

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, mutually inverse operations are given different names, special notations are invented: for example, squaring (x 2) and extracting the 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 term "derivative" itself can be justified "in a worldly way": the function y - f (x) "produces" a new function y "= f" (x) The function y \u003d f (x) acts as if as a "parent" , but mathematicians, of course, do not call it a "parent" or "producer", they say that it 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 true.
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, respectively, y-F(x) and y-G(x), then the sum of the functions y = f(x) + g(x) has an antiderivative on the interval X, and this antiderivative is the function y = F(x) + G(x). But usually, when formulating rules (and not theorems), only keywords are left - this is more convenient for applying 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 the entire set of antiderivatives is exhausted by the indicated type of functions.

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 the 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 the hidden meaning of this notation is.
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 The integral of the sum of functions is equal to the sum of the integrals of these functions:

Rule 2 The constant factor can be taken out of the integral sign:

Rule 3 If 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, preliminary identical transformations of the expression contained under the integral sign sometimes help.

Let's use the trigonometric formula for decreasing the degree:

Then successively we find:

A.G. Mordkovich Algebra Grade 10

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