Solutions for applied tasks are calculated up to the calculation of the integral, but if not, it is possible to calculate exactly. It is necessary for others to know the meaning sing integral with a world of accuracy, for example, up to a thousandth.

Іsnuyut zavdannya, if you know the approximate value of the linear integral with the necessary accuracy, then you can perform numerical integration the same way as the Simpos method, trapezoidal, rectangular. Not all vipadki give the ability to calculate yoga with singing accuracy.

This article examines the Newton-Leibniz formulas. It is necessary for the exact calculation of the first integral. Be hovered report butts, We looked at the replacement of the change in the sing integral and we know the value of the sing integral when integrated by the parts.

Newton-Leibniz formula

Appointment 1

If the function y = y (x) is a continuous loop [a; b ] ,and F (x) Newton-Leibniz formula respect justice. We write її as follows: ∫ a b f(x) d x = F(b) - F(a) .

Qiu formula is important the main formula of the integral calculation.

In order to bring this formula, it is necessary to win the understanding of the integral with a clear changing upper boundary.

If the function y = f(x) is uninterrupted with a break [a; b], if the value of the argument x ∈ a; b and the integral may look ∫ a x f (t) d t i is taken into account by the function of the upper boundary. It is necessary to accept the value of the function in the future as ∫ a x f (t) d t = Φ (x) , it will be uninterrupted, and for it the non-uniformity of the form ∫ a x f (t) d t = Φ "(x) = f (x) is valid.

It is fixed that the incremental functions Φ (x) substantiate the incremental argument ∆ x , it is necessary to accelerate the fifth main power of the sing integral

Φ (x + ∆ x) - Φ x = ∫ a x + ∆ x f (t) d t - ∫ a x f (t) d t = = ∫ a x + ∆ x f (t) d t = f (c) x + ∆ x - x = f(c) ∆ x

de value c ∈ x; x + ∆x.

We fix the equality of the eye Φ(x + ∆x) - Φ(x) ∆x = f(c) . For the chosen function, it is necessary to go to the boundary as ∆ x → 0, then we take the formula of the form Φ "(x) = f (x).

F (x) = Φ (x) + C = ∫ a x f (t) d t + C

Let's calculate F(a) for the first power of the first integral. Todi otrimuemo that

F (a) = Φ (a) + C = ∫ a a f (t) d t + C = 0 + C = C , obviously, C = F (a) . The result can be obtained when calculating F (b) and taken:

F (b) = Φ (b) + C = ∫ a b f (t) d t + C = ∫ a b f (t) d t + F (a) a) . Equanimity to bring the Newton-Leibniz formula ∫ a b f (x) d x + F (b) - F (a)

The increment of the function is accepted as F x a b = F (b) - F (a) . For additional knowledge, the Newton-Leibniz formula looks like ∫ a b f (x) d x = F x a b = F (b) - F (a) .

In order to complete the formula, it is obligatory to know one of the primary y = F(x) of the integrand function y = f(x) of the integral function [a; b ] , calculate the cost of the primary charge. Let's take a look at the application of the calculation, vicorist formula of Newton-Leibniz.

butt 1

Calculate the simple integral ∫ 1 3 x 2 d x using the Newton-Leibnitz formula.

Solution

Let's see that the integrand function of the form y = x2 is an interruption-free loop [1; 3], but it is also integrated on the other side. Behind the table non-significant integrals Bachimo, that the function y = x 2 can be impersonal for all real values ​​x , also, x ∈ 1 ; 3 write F(x) = ∫ x 2 d x = x 3 3 + C . It is necessary to take the first s Z = 0, then it is necessary that F (x) = x 3 3 .

Accelerated by the Newton-Leibniz formula and we take into account that the calculation of the sing integral will be seen in the future ∫ 1 3 x 2 d x = x 3 3 1 3 = 3 3 3 - 1 3 3 = 26 3 .

Suggestion:∫ 1 3 x 2 d x = 26 3

butt 2

Calculate the simple integral ∫ - 1 2 x · e x 2 + 1 d x using the Newton-Leibnitz formula.

Solution

The function without interruption is set [-1; 2], later, integrated on a new one. It is necessary to know the value of the undefined integral ∫ x e x 2 + 1 d x for the additional method of adding the sign of the differential, then we can take ∫ x e x 2 + 1 d x = 1 2 ∫ e x 2 + 1 d (x 2 + 1) = 1 2 e x 2+1+C.

There are possible impersonal primary functions y = x · e x 2 + 1 , which are valid for all x , x ∈ - 1 ; 2.

It is necessary to take the primary at C = 0 and apply the Newton-Leibniz formula. Todi otrimaemo mind

∫ - 1 2 x e x 2 + 1 d x = 1 2 e x 2 + 1 - 1 2 = = 1 2 e 2 2 + 1 - 1 2 e (- 1) 2 + 1 = 1 2 e (- 1) 2 + 1 = 1 2 e 2 (e 3 - 1)

Suggestion:∫ - 1 2 x e x 2 + 1 d x = 1 2 e 2 (e 3 - 1)

butt 3

Calculate the integrals ∫ - 4 - 1 2 4 x 3 + 2 x 2 d x i ∫ - 1 1 4 x 3 + 2 x 2 d x .

Solution

Vіdrіzok - 4; - 1 2 talk about those functions that are under the sign of the integral, that are uninterrupted, that is, that they are integrated. We know the unknown primary functions y = 4 x 3 + 2 x 2 . We accept that

∫ 4 x 3 + 2 x 2 d x = 4 ∫ x d x + 2 ∫ x - 2 d x = 2 x 2 - 2 x + C

It is necessary to take the primary F (x) \u003d 2 x 2 - 2 x todi, using the Newton-Leibniz formula, we take the integral, which is calculable:

∫ - 4 - 1 2 4 x 3 + 2 x 2 d x = 2 x 2 - 2 x - 4 - 1 2 = 2 - 1 2 2 - 2 - 1 2 - 2 - 4 2 - 2 - 4 = 1 2 + 4 - 32 - 1 2 = - 28

We carry out a transition to the calculation of another integral.

Z vіdrіzka [-1; 1 ] it is possible that the integrand function is important to the non-delimited one, because lim x → 0 4 x 3 + 2 x 2 = + ∞ necessary mental integration from the wind. Then F(x) = 2 x 2 - 2 x is not the primary one for y = 4 x 3 + 2 x 2 with an additional [-1; 1 ] , the shards of the point O lie in the vіrіzku, but do not enter the region of destination. Also, є the Riemann and Newton-Leibniz sing integral for the function y = 4 x 3 + 2 x 2 z double [-1; one].

Suggestion: ∫ - 4 - 1 2 4 x 3 + 2 x 2 d x \u003d - 28,є sing integral of Riemann and Newton-Leibniz for the function y = 4 x 3 + 2 x 2 z vіdrіzka [-1; one].

Before trying the Newton-Leibnitz formulas, it is necessary to know exactly the basis of the sing integral.

Replacement of the change in the singing integral

If the function y = f(x) is continuous and uninterrupted, it is [a; b], but also obviously impersonal [a; b] the value of the function x = g (z) is assigned to the region α; β s obviously without interruption, de g (α) = a і g β = b , it is necessary, what?

The given formula can be fixed only if it is necessary to calculate the integral a b f (x) d x , the number of mismatches of the integral can look like f (x) d x , which is calculated using the additional substitution method.

butt 4

Calculate the simple integral in the form ∫ 9 18 1 x 2 x - 9 d x .

Solution

The integrand function is considered to be non-interruptible for integration, and the same integral can be used in the same place. The value is 2 x - 9 = z ⇒ x = g (z) = z 2 + 9 2 . The value x \u003d 9 means that z \u003d 2 9 - 9 \u003d 9 \u003d 3, and with x \u003d 18 it is acceptable that z \u003d 2 18 - 9 \u003d 27 \u003d 3 3 then g α \u003d g (3) \u003d 9, g β = g 3 3 = 18 . When substituting, we subtract the value of the formula ∫ a b f (x) d x = ∫ α β f (g (z)) g "(z) d z

∫ 9 18 1 x 2 x - 9 d x = ∫ 3 3 3 1 z 2 + 9 2 z z 2 + 9 2 "d z = = ∫ 3 3 3 1 z 2 + 9 2 z z d z = ∫ 3 3 3 2 z 2 + 9 d z

According to the table of non-significant integrals, it is possible that the prior functions 2 z 2 + 9 take the value 2 3 a r c t g z 3 . Then, when the formulas of Newton-Leibniz are stuck, it is necessary that

∫ 3 3 3 2 z 2 + 9 d z = 2 3 a r c t g z 3 3 3 3 = 2 3 a r c t g 3 3 3 - 2 3 a r c t g 3 3 = 2 3

The knowledge can be used without victorious formula ∫ a b f (x) d x = ∫ α β f (g (z)) g "(z) d z.

As for the method of replacing the twisted integral of the form ∫ 1 x 2 x - 9 d x , you can get the result ∫ 1 x 2 x - 9 d x = 2 3 a r c t g 2 x - 9 3 + C .

Zvіdsi zrobchimo calculus for the formula of Newton-Leibnіts and calculable sevny іntegrаl. We accept that

∫ 9 18 2 z 2 + 9 d z = 2 3 a r c t g z 3 9 18 = = 2 3 a r c t g 2 18 - 9 3 - a r c t g 2 9 - 9 3 = = 2 3 a r c t g 3 - π 4 = π 18

The results were skewed.

Suggestion: ∫ 9 18 2 x 2 x - 9 d x = π 18

Integrating by parts pid hour of calculation of the sing integral

Yakshcho on vіdrіzku [a; b ] assigned and uninterrupted functions u (x) і v (x) , similarly similar to the first order v "(x) u (x) є integrating, so the third order for the integrated function u "(x) v ( x) equality ∫ a b v "(x) u (x) d x = (u (x) v (x)) a b - ∫ a b u "(x) v (x) d x is true.

The formula can be tweaked just the same, it is necessary to calculate the integral a b f (x) d x , moreover, ∫ f (x) d x needs to be checked for additional integration by parts.

butt 5

Calculate the simple integral ∫ - π 2 3 π 2 x · sin x 3 + π 6 d x .

Solution

The function x · sin x 3 + π 6 is integrated on the vіdrіzku - π 2; 3 π 2 means there is no interruption.

Let u (x) \u003d x, then d (v (x)) \u003d v "(x) d x \u003d sin x 3 + 6 d x, moreover, d (u (x)) \u003d u "(x) d x \u003d d x, and v (x) \u003d - 3 cos π 3 + π 6. 3 formulas ∫ a b v "(x) u (x) d x = (u (x) v (x)) a b - ∫ a b u "(x) v (x) d x

∫ - π 2 3 π 2 x sin x 3 + π 6 d x = - 3 x cos x 3 + π 6 - π 2 3 π 2 - ∫ - π 2 3 π 2 - 3 cos x 3 + π 6 d x \u003d \u003d - 3 3 π 2 cos π 2 + π 6 - - 3 - π 2 cos - π 6 + π 6 + 9 sin x 3 + π 6 - π 2 3 π 2 \u003d 9 π 4 - 3 π 2 + 9 sin π 2 + π 6 - sin - π 6 + π 6 = 9 π 4 - 3 π 2 + 9 3 2 = 3 π 4 + 9 3 2

The solution to the butt can be vikonated in another way.

To know the impersonal primary functions x sin x 3 + π 6 with additional integration of parts from the Newton-Leibniz formula:

∫ x sin x x 3 + π 6 d x = u = x, d v = sin x 3 + π 6 d x ⇒ d u = d x , v = - 3 cos x 3 + π 6 = = - 3 cos x 3 + π 6 + 3 ∫ cos x 3 + π 6 d x = = - 3 x cos x 3 + π 6 + 9 sin x 3 + π 6 + C ⇒ ∫ - π 2 3 π 2 x sin x 3 + π 6 d x = - 3 cos x 3 + π 6 + 9 sincos x 3 + π 6 - - - 3 - π 2 cos - π 6 + π 6 + 9 sin - π 6 + π 6 = = 9 π 4 + 9 3 2 - 3 π 2 - 0 = 3 π 4 + 9 3 2

Suggestion: ∫ x sin x x 3 + π 6 d x = 3 π 4 + 9 3 2

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

Pokhіdnі higher orders

At this age, we learn to know the lower orders, and also write down the common formula of the lower order. In addition, Leibniz's formula will be looked at as such a similar and numerically prohannya - similar to higher orders of magnitude implicitly defined functions. I'm tempted to take a mini-test:

Axis function: i axis її persha pokhіdna:

In that mood, as you have blamed for some difficulties / incomprehensibility of a good butt, be kind, read from the two basic articles of my course: How to know if I'm going to go?і Folding function. After mastering the elementary pokhіdnyh, I recommend that you learn from the lesson The simplest task for the funeral, on which we rose, zokrema z another pokhіdny.

It doesn’t matter to guess that a friend is bad - it’s bad like the 1st bad:

In principle, I’ll leave a friend, but vzhe vvazhayut in a similar way.

Similarly: the third is worse - the same is worse than the 2nd is worse:

Fourth pokhіdna - є pokhіdna vіd 3-ї pokhіdnoї:

P'yata is good: , and it is obvious that all similarities of higher orders can equal zero:

The Crimean Roman numeration in practice often has the following designations:
, Let’s signify the “energetic” order through. With this, the superstring index needs to be laid on the shackle.- to resurrect the death of the “gravity” of the world.

Sometimes there is such a record: - Third, fourth, p'yata, ..., "Enna" is similar.

Forward without fear and sumnіvіv:

butt 1

Given a function. Know.

Solution: What can you do here ... - go ahead for the fourth good :)

Chotiri put strokes already not accepted, so let's move on to numerical indexes:

Vidpovid:

Good, but now let’s think about such food: why work, if it’s necessary for the mind to know not the 4th, but, for example, the 20th I’ll die? Yakshcho for the march 3-4-5 (maximum, 6-7th) The order of the decision is made out to finish it quickly, then we will not “get” to the next higher order, oh, yak, not soon. Do not write down the truth 20 rows! In a similar situation, it is necessary to analyze a sample of the known ones, to work out the regularity and to formulate a formula of a similar one. So, in Applied No. 1, it is easy to understand that in case of skin offensive differentiation in front of the exponent “viscosity” the additional “triika” is to be added, moreover, on the shortest step of the “troika” it is better for the number of the worse, also:

De is a fairly natural number.

Yakshcho, then go out exactly the 1st bad: yakscho - then 2-a: i etc. In such a rank, twenty pokhіdna vyznaєtsya mittevo: - And the next "kilometer stretch"!

Playing independently:

butt 2

Know functions. Write down the system

The solution is to follow the example of the lesson.

After a warm-up, what to improve, we can look at more foldable butts, in some practical ways, the solution algorithm. Tim, who has learned the lesson Between sequences, be a troch easier:

butt 3

Know functions.

Solution: to clarify the situation, we know a few of the following:

Multiplying the numbers is not quick! ;-)


Mabut, year. ... Navit trohi overdone.

On the offensive krotsі, it is best to add the formula “enї” pokhіdnoї. (if you don’t mind the mind, then you can get along with a black man). For whom we marvel at the rejection of the results, we see the regularity, with which the skin is attacked.

First of all, the stench knows the devils. The queue sign is safe "flasher", І oskіlki 1st pokhіdna is positive, then I will use the formula to see the offensive multiplier: . Pіdіyde th equivalent variant, but especially I, like an optimist, love the plus sign \u003d)

In a different way, the numerator "winds" factorial, moreover, the VIN "vіdstaє" vіd pokhіdnі numbers per one unit:

І in a third way, the steps of the “two” are growing at the numeralist, as if the number is similar. The same can be said about the steps of the bannerman. Residual:

From the method of reverification, we can substitute a couple of values ​​\u200b\u200b"en", for example, i:

Miraculously, now pardons will start - just a sin:

Vidpovid:

A simple function for an independent vision:

butt 4

Know functions.

І zavdannya tsikavіshe:

butt 5

Know functions.

Let's repeat the order once again:

1) We know a few sprats of the dead. In order to catch the regularities, ring out the triokh-chotiriokh.

2) Then I strongly recommend folding (I would like to use black)"Ennu" will be gone - it's guaranteed to be on the coast in the face of pardons. Ale can be undone and without, tobto. Think about your thoughts and write down, for example, twenty or eight I’ll die. More than that, deakі people vzagalі zdatnі vіrіshiti tsі zavdannya usno. However, the next thing to remember is that "Shvidki" can threaten, and rather be safe.

3) At the final stage, it is necessary to re-verify the “en” pokhidnoy - we take a couple of “en” values ​​(more shortly for courts) and substantiate the substitution. And what’s even better is to reconsider everything that was known earlier. If something is presented in the need for meaning, for example, and the result is accurately calculated.

Short solution 4 and 5 examples for a lesson.

In some tasks, in order to solve problems, it is necessary to fix a little bit over the function:

butt 6

Solution: I don’t want to differentiate the proponated function, I don’t want to, the shards of the “filthy” things will be removed, which will make it easier to change the coming ones.

For whom dotsilno vikonati in front of the transformation: vikoristovuemo square difference formulaі the power of the logarithm :

Zovsіm іnsha on the right:

I old friends:

I think everything is visible. Give respect, that another bad sign is drawn, and the first - no. We construct a similar system:

Control:

Well, for beauty, the factorial for the arms:

Vidpovid:

Tsіkave zavdannya for independent vyrіshennya:

butt 7

Write a formula in the same order for the function

And now about the inviolable mutual responsibility, which is to congratulate the Italian mafia:

butt 8

Given a function. Know

Vіsіmnadtsyata pokhіdna at the point. Usyogo.

Solution: back to back, obviously, it is necessary to know Let's go:

They repaired the sinus, they came to the sinus. It was clear that for a farther differentiation this cycle is trivial, and blames the same power: how to “distance” to the eighteenth century more quickly?

The “amateur” method: it is easy to write down the right-handed number of the upcoming dead ones at the counter:

In this manner:

Ale tse pratsyuє, as if the order of the pokhіdnoi is not so great. Well, I need to know, let’s say, I’ll leave the cell, I’ll speed up the podilnistyu by 4. One hundred to divide into chotiri without excess, and it is easy to bachiti, because such numbers are rolled in the bottom row, to that:.

Before the speech, 18 pokhіdnu tezh can be distinguished from similar mirkuvan:
the other row has numbers, which are subdivided by 4 out of surplus 2.

The second, more academic method of founding periodicity to sinusі guidance formulas. Koristuyemosya ready-made formula "enoi" similar to the sine , in the yak, the required number is simply displayed. For example:
(reduction formula ) ;
(reduction formula )

For our viewpoint:

(1) Since the sine is a periodic function with a period, then the argument can be painlessly “turned” to the 4th period (tobto.).

Pokhіdnu system vіd vykonannya dvoh funktsіy can be known for the formula:

Zokrema:

You don’t need to specifically remember anything, because the more you know the formulas, the less you understand. Learn more about the story Newton's binomial oskіlki Leibniz's formula is more and more similar to the new one. Well, you are lucky, how to get away from the 7th or higher order (which, however, is small), you will be embarrassed. Vtіm, if cherga didde to combinatorics- then everything will be brought up =)

We know the third similar function. Vikoristovuemo Leibnitz's formula:

At to this particular type: . Pokhіdnі easily translated orally:

Now carefully and respectfully substitution and the simple result:

Vidpovid:

A similar task for an independent vision:

butt 11

Know functions

If in the front butt the solution “on the forehead” competed with Leibniz's formula, then here it would be rightly unacceptable. And more unacceptable - in a different order, it’s worse:

butt 12

Know the exact order

Solution: first and foremost respect - turn the axis like this, singly, not necessary =) =)

Let's write down the functions and know their similarities up to the 5th order inclusive. I admit that the steps of the right side have become sleepy for you:

At the left of the “live”, the bad things “ended up” and it’s even better - at the Leibniz formula, three additions are reset to zero:

I'm reeling at the dilemma, I figured it out in the article about folding pokhіdnyh: chi ask the result? In principle, you can leave it out and so - it’s easier to reverse it. Ale vin can help bring the decision to a path. From the other side, forgiven for the power of the initiative, I threaten algebra with pardons. However, we have є vіdpovіd, otrimana "primary" way =) (Div. sent to the cob), and I agree, vin is correct:

Miraculously, everything worked out.

Vidpovid:

Happy task for independent vision:

butt 13

For function:
a) signify direct differentiation;
b) know behind the Leibniz formula;
c) calculate.

No, I'm not a sadist - point "a" here is forgiven =)

And even more seriously, then the “direct” solution to the last differentiations can also have the “right to life” - in a number of ways, the folding can be equal to the folding of the Leibniz formula. Vykoristovyte, as if you respect for the dotsіlne - it is unlikely that you will be the basis of a small task.

Briefly, the solution is to illustrate the lesson.

To raise the final paragraph, it is necessary to remember differentiate implicit functions:

Changes in higher orders of functions, jobs implicitly

Rich someone from us vitrativ dovgі years, days and tizhnі life on vvchennya kіl, parabola, hyperbole- and sometimes it was given to punishment. So let's take revenge and differentiate them like a trace!

Pochnemo zі "shkіlnoї" parabola to her canonical camp:

butt 14

Rivnyanya is given. Know.

Solution: first krok good knowledge:

Those that function that її is similar to the expression implicitly do not change the essence, the other is worse - the same is similar to the 1st worse:

However, it’s necessary to establish its own rules: only through "iks" and "iplayer". To that in the otriman 2nd pokhіdnu let's imagine:

Third pokhіdna - є pokhіdna vіd 2-ї pokhіdnoї:

Similarly, imagine:

Vidpovid:

"Shkilna" hyperbole in canonical camp– for independent work:

butt 15

Rivnyanya is given. Know.

I repeat, I’ll lose the 2nd and the result should be explained only through “iks” / “iplayer”!

Briefly, the solution is to illustrate the lesson.

After the childish coils, marveling at the German pornography @ fiyu, we look more mature butts, from which we know one more important decision:

butt 16

Elips domineering person.

Solution: we know the 1st pokhіdnu:

And now let's remember and analyze the coming moment: we can differentiate at once, so we don't have to be quiet. In this state of mind, it’s extremely simple, but in reality, the orders of such gifts are given twice and once in power. What is the best way to get rid of the bulky pokhidnoy? Ісnuє! We take equal and victorious by the very same trick, which, even when the 1st is important, is “hanging” strokes on the offending parts:

Another pokhіdna is guilty, but only expressed through this and that at the same time (at once) pokhidnoy. For whom in otriman equal is conceivable:

To get rid of the most technical difficulties, multiply the offending parts by:

І less at the final stage, we draw up drіb:

Now we marvel at the vihіdne rіvnyannya, and we note that if the result is taken away, it is supposed to be forgiven:

Vidpovid:

How to know the meaning of the 2nd pokhіdnoї in any point (yaka, zrozumіlo, lie on the elіpsu), for example, at the point ? Too easy! Tsej motive already zustrіchavsya on the lesson about equal normals: in virase 2, it is necessary to represent :

Without a doubt, in all three ways you can clearly see function assignments and differentiate them, but also morally set up the practice with two functions, as if to avenge the root. In my opinion, it would be better to make a decision by “implicit path”.

The final example for an independent vision:

butt 17

Find an implicitly defined function

The Leibniz formula for number of n-th similar work of two functions. Nadano її proof in two ways. Looked at the butt of the calculation of the n-th order.

Zmist

Div. also: Pokhіdna robot two functions

Leibniz formula

For the help of the Leibniz formula, you can calculate the n-th order loss in two functions. Vaughn may look like this:
(1) ,
de
- Binomial coefficients.

Binomial coefficients with the coefficients of the distribution of the binomial for steps i:
.
So the number is the number of the same day s n k .

Proof of the Leibniz formula

Zastosuєmo the formula of a similar addition of two functions :
(2) .
We rewrite formula (2) in such a way:
.
Therefore, we are aware that one function is deposited in the form of change x, and another - in the form of change y. For example, we respect the rozrahunka. The previous formula can be written as follows:
(3) .
Oscilki are similar to the sum of the terms, and the skin term is the addition of two functions, then for the calculation of the lower orders, the rule (3) can be successively added.

Same for similar n-th order maybe:

.
Vrahovyuchi, scho and mi otrimuemo Leibniz formula:
(1) .

Proof by induction

Let us prove the Leibniz formula by mathematical induction.

Once again we write the Leibniz formula:
(4) .
For n = 1 it is possible:
.
This formula is similar to the practice of two functions. Vaughn is right.

Assume that formula (4) is valid for similar n-th order. It can be shown that it is valid for a similar n+ 1 th order.

Differential (4):
;



.
Father, we knew:
(5) .

Let's put it in (5) and we can say that:

.
It can be seen that formula (4) can look the same for a similar n + 1 th order.

Later, formula (4) is valid for n = 1 . Worry, what is won for the decimal number n \u003d m 1 .
The Leibniz formula has been completed.

butt

Enumerate the n-th random function
.

Let's solve the Leibniz formula
(2) .
To our liking
;
.

per tables of similar maybe:
.
Zastosovuemo power of trigonometric functions :
.
Todi
.
It can be seen that the differentiation of the function sine is brought up to zsuv on . Todi
.

We know similar functions.
;
;
;
, .

Oskіlki at , then the Leibniz formula has more than three first terms in the form of zero. We know bіnomnі koefіtsієnti.
;
.

Behind the Leibniz formula can be:

.

Div. also: