Basic characteristics of an undefined integral. Integrals for dummies: how to virishuvati, calculation rules, explanation

Primary and non-significance integral.

The primitive function f(x) for an interval (a; b) is such a function F(x) that wins equality for any x from a given interval.

If we take into account the fact that the constant C is equal to zero, then equality is fair . In this way, the function f(x) can be impersonal primary F(x) + C, for a fairly constant value, moreover, q primary values ​​are one and the same for a fairly constant value.

All impersonal primary functions f(x) are called the non-significant integral of this function and are assigned .

Viraz is called integrand virase, and f(x) is called piintegral function. The virase integrand is the differential function f(x).

Diya znahodzhennya nevodomoї ї ї ї on the given differential is called non-trivial іntegrіvannym, so the result of іntegrіvannya є not one function F (x), but її primaries F (x) + C.

Tabular integrals

The simplest power of integration

1. Similar to the result of integrating a better pidintegral function.

2. Non-values ​​of the integral of the differential of a function is equal to the sum of the function itself and an adequate constant.

3. The coefficient can be blamed for the sign unassigned integral.

4. Non-significances of the integral of the sum / difference of functions of the cost of the sum / difference of non-significance of the integrals of functions.

Intermediate equivalence of the first and other powers of the undefined integral is introduced for explanation.

To prove the third and fourth powers, it is sufficient to know the reverse of the right parts of equalities:

The numbers are similar to pidintegral functions, which is proof of the first quality. It won't win in the rest of the transitions.

In this order, the task of integration is the turning task of differentiation, moreover, there is a close connection between them:

First, the authorities allow the re-verification of integration. In order to reverse the correctness of the past integration, it is enough to calculate the exact result. If the results of the differentiation function are taken away, they appear equal to the integrand function, it is significant that the integration is carried out correctly;

The other power of the undefined integral allows to know the primary behind the given differential of the function. To whom the authorities are founded without any middle calculation of insignificant integrals.

1.4. Invariance of forms of integration.

Invariant integration - a kind of integration for functions, the argument of which are elements of a group or a point of a homogeneous space (whether a point of such a space can be translated into another given group).

functions f(x) are built up to the calculation of the integral in the form of a differential form f.w, de

It is clear that the f-la for r(x) should be guided lower. Umova the weather may be looking .

here Tg means the operator zsuvu X for the help gОG: Tgf(x)=f(g-1x). Let X = G - topology, a group that has its own damage. I. in. Use only if G is locally compact (so-creme, on infinitesimal groups I. i. not available). For submultiple I. in. The characteristic functions cA (equal 1 on A and 0 pose A) define the left world of Haar m(A). The initial power in the first approach is invariant in case of left destructions: m(g-1A)=m(A) for gОG. The nature of Haar's world on the group is assigned unambiguously, up to a scalar multiplier. If you see the world of Haar m, then I. in. functions f are given by the formula . Analogous to the power of May, the rightful Zahid Haar. Use non-permanent homomorphism (increment, which takes group power) DG group G group (alternatively plural) way. numbers, for which

de dmr and dmi - right and left come in Haar. Function DG(g) sv. group G module. unimodular; at this point of right and left, come in Haara run away. Compact, simple and nilpotent (zocrema, commutative) groups are unimodular. If G is an n-world group Li and q1, ..., qn is a basis for the space of left-invariant 1-forms on G, then the left Haar world on G is given by the n-form . In local coordinates for calculation

forms qi can be accelerated by any matrix implementation of the group G: the matrix 1-form g-1dg is left-invariant, and її coef. є left-invariant scalar 1-forms, including the choice of a joking basis. For example, the real matrix group GL(n, R) is unimodular and the Haar world is given by a form on it. Come on X=G/H is a uniform space, which is a locally compact group G, a transformation group, and a closed subgroup H is a stabilizer of a single point. In order for X to fall into I. i., it is necessary and sufficient, so that for all hОH the equality DG(h)=DH(h) was calculated. Zocrema, tse is correct in times, if H is compact chi napіvprosta. New theory I. in. on the inexhaustible rіznomanіttyah there is nothing.

Replacement of change.

This article reports on the main power of the singing integral. The stench is brought to the aid of the understanding of the integral of Riemann and Darboux. Calculation of the first integral to pass through 5 powers. Ti, scho lost, zastosovuyutsya for the evaluation of different viruses.

Before passing to the main powers of the sing integral, it is necessary to switch over to the fact that a does not outweigh b.

The main power of the singing integral

The function y = f (x) is assigned at x = a, similarly to fair equality ∫ a a f (x) d x = 0.

Proof 1

It turns out that the values ​​of the integral zbіgayutsya interami dovnyuє zero. The price of Riemann's integral, that's why the skin integral sum for any kind of break-up on the floor [ a ; a ] i be-any choice point ζ i is closer to zero, so x i - x i - 1 = 0, i = 1, 2,. . . , n , also, we assume that between the integral functions is zero.

Appointment 2

For a function that can be integrated on the widget [a; b ] , Umov's win ∫ a b f (x) d x = - ∫ b a f (x) d x .

Proof 2

Otherwise, it seems that if you change the upper and lower between the integration points, then the value of the integral will change the value on the length. Qia power is taken from the Riemann integral. However, the numbering of the split in the section goes from the point x = b.

Appointment 3

∫ a b f x ± g (x) d x = ∫ a b f (x) d x ± ∫ a b g (x) d x b].

Proof 3

Write down the integral sum of the function y = f (x) ± g (x) for splitting the points ζ i into sections with the given choice: σ = ∑ i = 1 n f ζ i ± g ζ i x i - x i - 1 = = ∑ i = 1 n f (ζ i) x i - x i - 1 ± ∑ i = 1 n g ζ i x i - x i - 1 = σ f ± σ g

de f і g є integrand sums of functions y = f (x) і y = g (x) for splitting the wedge. After the transition to the boundary at λ = m a x i = 1, 2,. . . , n (x i - x i - 1) → 0 is taken as lim λ → 0 σ = lim λ → 0 σ f ± σ g = lim λ → 0 σ g ± lim λ → 0 σ g .

Z vznachennya Rіman tsey vislіv є equally strong.

Appointment 4

The fault of the constant multiplier for the sign of the sing integral. The function is integrated over the interval [a; b] from sufficient values ​​of k, there can be a fair unevenness of the form ∫ a b k · f (x) d x = k · ∫ a b f (x) d x .

Proof 4

The proof of the strength of the linear integral is similar to the previous one:

σ = ∑ i = 1 n k f ζ i (xi - x i - 1) = = k ∑ i = 1 n f ζ i (xi - x i - 1) = k σ f ⇒ lim λ → 0 σ = lim λ → 0 (k σ f) = k lim λ → 0 σ f ⇒ ∫ a b k f (x) d x = k ∫ a b f (x) d x

Appointment 5

Since a function of the form y = f (x) is integrated over the interval x , a ∈ x , b ∈ x , it is assumed that ∫ a b f (x) d x = ∫ a c f (x) d x + ∫ c b f (x) d x .

Proof 5

Power is considered fair for c ∈ a; b for c ≤ a i c ≥ b . The proof is carried out similarly to the previous authorities.

Appointment 6

If the function can be integrated with the vіdrіzka [a; b], the same can be done for any internal lining c; d ∈ a; b.

Proof 6

The proof is rooted in Darboux's power: just as a blatant strife has to add points, then the lower sum of Darboux will not change, and the upper one will not change.

Appointment 7

If the function is integrated on [a; b ] s f (x) ≥ 0 f (x) ≤ 0 for any value x ∈ a ; b , then it is possessed that ∫ a b f (x) d x ≥ 0 ∫ a b f (x) ≤ 0 .

Power can be brought to the aid of Riemann's integral: be it an integral sum for a choice of a point of conflict between that point ζ i of the mind, that f (x) ≥ 0 f (x) ≤ 0 is taken by the indeterminate.

Proof 7

Likewise, the functions y = f(x) and y = g(x) are integrated on the subsection [a; b], the same unevenness is considered fair:

∫ a b f (x) d x ≤ ∫ a b g (x) d x , so f (x) ≤ g (x) ∀ x ∈ a ; b ∫ a b f (x) d x ≥ ∫ a b g (x) d x so f (x) ≥ g (x) ∀ x ∈ a ; b

We know for sure that integration is admissible. This investigation will be vindicated in proving other authorities.

Appointment 8

When integrating the function y = f (x) z vіdrіzka [a; b ] may be valid unevenness of the form ∫ a b f (x) d x ≤ ∫ a b f (x) d x .

Proof 8

Maybe - f(x) ≤ f(x) ≤ f(x) . From the front power was taken away, that the unevenness can be integrated term by term and in the form of unevenness in the form - ∫ a b f (x) d x ≤ ∫ a b f (x) d x ≤ ∫ a b f (x) d x . This underlying irregularity can be written in another form: ∫ a b f (x) d x ≤ ∫ a b f (x) d x .

Appointment 9

If the functions y = f(x) and y = g(x) are integrated with the contraction [a; b] for g (x) ≥ 0 for any x ∈ a; b , an obsessed irregularity of the form m ∫ a b g (x) d x ≤ ∫ a b f (x) g (x) d x ≤ M ∫ a b g (x) d x , de m = min x ∈ a ; b f (x) i M = m a x x ∈ a; bf(x).

Proof 9

A similar rank to pass the proof. M и m b], then m ≤ f (x) ≤ M . It is necessary to multiply the subvariance by the function y = g (x), which will give the value of the subvariant unevenness of the form m · g (x) ≤ f (x) · g (x) ≤ M · g (x) . It is necessary to integrate yoga into a vіdrіzku [a; b], then we take away the hardness that is brought.

Last: For g (x) = 1, the unevenness looks like m · b - a ≤ ∫ a b f (x) d x ≤ M · (b - a) .

Persian formula for the average

For y = f (x) it is integrated on the cross section [a; b] s m = m i n x ∈ a; b f (x) i M = m a x x ∈ a; b f (x) is the number μ ∈ m; M , to fit ∫ a b f (x) d x = μ · b - a.

Last: If the function y = f(x) is uninterrupted with a break [a; b ] , є is a number c ∈ a ; b , to satisfy the equanimity ∫ a b f (x) d x = f (c) b - a.

The first formula of the mean value in a condensed form

If the functions y = f (x) and y = g (x) are integrated with the contraction [ a ; b] s m = m i n x ∈ a; b f (x) i M = m a x x ∈ a; b f (x) , and g (x) > 0 for any value x ∈ a ; b. It is possible that the number μ ∈ m; M , to satisfy the evenness ∫ a b f (x) g (x) d x = μ ∫ a b g (x) d x .

Another mean value formula

If the function y = f (x) is integrated with the cross [a; b], while y = g (x) is monotonic, i.e., is a number, such that c ∈ a; b , where it is possessed by fair equality of the form ∫ a b f (x) g (x) d x = g (a) ∫ a c f (x) d x + g (b) ∫ c b f (x) d x

How did you remember the pardon in the text, be kind, see it and press Ctrl + Enter

The data of authority are victorious for the creation of the transformation of the integral with the method of yogo reduction to one of the elementary integrals and the farthest calculation.

1. Pokhіdna neznachennoy іntegrа dоrіvnyuє pіdіntegrаlіnіy funkії:

2. The differential of the undefined integral is better than the pintegral virase:

3. Non-values ​​of the integral in the form of the differential of the current function of the healthy sum of the functions and the sufficient post:

4. A constant multiplier can be blamed for the sign of the integral:

Moreover, a ≠ 0

5. Integral of the sum (retail) of the additional sum (retail) of the integral:

6. Powerfulness - combination of powers 4 and 5:

Moreover, a ≠ 0 ˄ b ≠ 0

7. The power of the invariance of the undefined integral:

Yakscho something

8. Power:

Yakscho something

In fact, power was given to us, we will call it a step of integration for the additional method of replacing the change, which is more reportable in the attacking division.

Let's look at the example:

On the back of the head, we scored power 5, then power 4, scored the first table and took the result.

The algorithm of our online calculator іntegrіlіv pіdtrimuє sі vіlіchenі vyshe vіsshego і vіstіvі and і easily znayde rіshenі prіshennia іn ііntegrаіv.

At the statistics, we override the main power of the singing integral. The greater part of these powers is brought on the basis of understanding the singing integral of Rimmann and Darboux.

Calculation of the first integral is often carried out with the help of the first five authorities, so we will ask for the demand. The rest of the power of the singing integral is mainly used to evaluate different viruses.

First down go to the main powers of the sing integral, we think that a does not change b.

For the function y = f(x) , assigned for x = a, equality is valid.

That is why the values ​​of the sing integral z zbіgayutsya interami іntegruvannya doіvnyuє zero. This power is consequently assigned to the Riemann integral, to that in which case the skin integral sum for any kind of splitting the gap and whether or not the choice of the point is equal to zero, to that, then, the boundary of the integral sums is zero.

For a function that can be integrated on a widget, win .

In other words, when changing the upper and lower between the integration points, the value of the sing integral changes on the length. The power of the sing integral is also clear from the understanding of the Riemann integral, it is less the numbering of the split in the next line starting from the point x = b.

for integrating the functions y = f(x) and y = g(x) on a cross-sectional basis.

Bringing.

We write down the integral sum of the function for a given connection, a window and a choice of points:

de-integral sum of functions y = f(x) і y = g(x) for a given distribution in the opposite case.

Passing to the border with we take into account that the designation of the Riemann integral is equal to the assertion of the brought value.

The constant multiplier can be blamed for the sign of the sing integral. Therefore, for a function y = f(x) integrated on a cross-section and an appropriate number k, equality is true .

The proof of the sovereignty of the sing integral is absolutely similar to the previous one:


Let the function y = f(x) be integrated over the interval X, moreover and then .

Tsya vlastіvіst fair as for, so for or.

The proof can be carried out by spiraling on the forward power of the sing integral.

Just as the function is integrated on the windshield, it is integrated and whether it is the internal windshield.

The proof rests on the power of Darboux's sum: if you add new points to the obvious breakdown, then the lower sum of Darboux will not change, and the upper one will not increase.

If the function y = f(x) is integrated on the subdivision y of any value of the argument , then .

Tsya power is brought through the designation of Riemann's integral: whether the sum is integral for whether there is a choice between a point of division and a point when it is invisible (not positive).

Consequence.

For integrating the functions y = f(x) and y = g(x) on the subsection, the following inconsistencies are true:


Tse firm means that the integration of irregularities is permissible. In the meantime, we are worthy of the hour to prove the impending dominance.

Let the function y = f(x) be integrated on the reverse, then the non-uniformity is valid .

Bringing.

Obviously what . It was stated in front of the authorities that inconsistency can be integrated term by term, it is fair . You can write down how .

Let the functions y = f(x) and y = g(x) be integrated on the subsection i for any value of the argument , then , de і .

The proof is carried out in a similar way. So m and M are the least the most significant function y = f(x) on the opposite side, then . Multiplying the subvariant unevenness by the unknown function y = g(x) is brought to the offensive subvariant unevenness. Integrating yoga into the air, we will come to a firmness, what to bring.

Consequence.

If we take g(x) = 1, then I will see the unevenness in the future .

First formula for the average.

Let the function y = f(x) be integrated on the reverse, і the same is the same number, scho.

Consequence.

If the function y = f(x) is interruptless, then there is a number such that .

The first formula of the mean value of the narrowed form.

Let the functions y = f(x) and y = g(x) be integrated on the reverse side, і , and g(x) > 0 no matter what the value of the argument . Then you have such a number that .

Another formula for the average.

In addition, the function y = f(x) is integrated, and y = g(x) is monotonic, so there is a number such that equality is valid .

The solution of integrations is easier, but more difficult. This article is for those who want to learn to understand the integration, but do not know anything about them or anything. Integral... Need more wine? How to calculate yoga? What is the difference between the integrals?

If only you can see the integration of the integral - put it in the shape of the icon of the integral to the right of the most important places, then we kindly ask you! Find out how to solve the simplest and the other integrals and why you can’t do without them in mathematics.

We understand « integral »

Integration was in the home Ancient Egypt. Obviously, not in a modern way, but all the same. In the meantime, mathematicians have written a lot of books on this topic. Particularly noted newton і Leibnitz but the essence of the speeches did not change.

How to understand integrated from zero? Niyak! For the sake of understanding, all the same, one needs basic knowledge of the basics mathematical analysis. Vіdomostі pro, nebhіdnі і for razumіnnya іntegraіv, є vzhe є vzhe vzhe blozі.

Non-significance integral

Let us have it as a function f(x) .

The non-significant integral of the function f(x) such a function is called F(x) , similar to the old functions f(x) .

Tobto іntegral - tse pokhіdna navpak or primary. Before speech, about those, like read in our article.

Primary source for everyone uninterrupted functions. Also, the symbol of a constant is often added to the primary one, the shards of similar functions, which differ by a constant, are avoided. The process of learning the integral is called integration.

Simple example:

To avoid violating the primary elementary functions, manually add them to the table and use the already prepared values.

Povna table of integrals for students

Integral value

To the right of the understanding of the integral, we may be to the right of inexorably small quantities. Integral additionally calculate the area of ​​the figure, the mass of a heterogeneous body, passages in uneven Russia, and a rich other. A trace of memory, that the integral is the sum of an infinitely large number of infinitely small additions.

As a butt, you can see the schedule as a function.

How to know the area of ​​a figure surrounded by a graph of a function? For the help of the integral! Rozіb'єmo curvilinear trapezium, surrounded by axes of coordinates and a graph of the function, on infinitely small ribs. In this way, the figure appears to be divided into thin pillars. The sum of the squares of the stovpchikiv and we become the square of the trapezium. But remember that such an calculation will give a clear result. However, if there are less and already there will be charges, then there will be more exact charges. If we change their flooring, if the dovzhina will sag to zero, then the sum of the area of ​​the vіdrіzkіv will sag to the area of ​​the figure. Tse i є pevny integrand, which is written like this:

The points a and b are called integration boundaries.

« Integral »

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Rules for calculating integrals for teapots

Power of the undefined integral

How to solve the insignificances of the integral? Here we can look at the power of the unmeasured integral, like to become in good fortune with perfect applications.

• Pokhіdna vіd іntegrа dоrіvnyuє pіdіntegrіalіy funktsії:

• The constant can be blamed on the integral sign:

• Іntegrаl vіd sumi dorivnyuє sumі іntegrіlіv. The same is true for retail:

The power of the singing integral

• Linearity:

• The sign of the integral is changed to remember the points between integration:

• At whatever points a, bі h:

We have already explained that the song integral is the sum between the sums. Ale, how to take into account the specific meaning of the pіd hour of the virіshennya butt? For which one use the Newton-Leibniz formula:

Apply the perfection of integration

Below we look at the inconsistencies of the integral and apply it to the solutions. It is proponable to independently expand in the subtleties of the decision, but if it is unreasonable, put the question in the comments.

To consolidate the material, take a look at the video about those, how integrals are violated in practice. Do not fall into the void, as the integral is not given to the eye. Return to a professional service for students curvilinear integral along the closed surface you will be given strength.