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Presentation on the topic: Newton-Leibniz formula

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Newton and Leibniz From surviving documents, historians of science have found out that Newton discovered differential and integral calculus back in 1665-1666, but did not publish it until 1704. Leibniz developed his version of the calculus independently (from 1675), although the initial impetus for his thought probably came from rumors that Newton already had such a calculus, as well as through scientific conversations in England and correspondence with Newton. Unlike Newton, Leibniz immediately published his version, and later, together with Jacob and Johann Bernoulli, widely propagated this epoch-making discovery throughout Europe. Most scientists on the continent had no doubt that Leibniz had discovered analysis.

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Heeding the persuasion of friends who appealed to his patriotism, Newton, in the 2nd book of his Elements (1687), said: In letters that about ten years ago I exchanged with the very skilled mathematician Mr. Leibniz, I informed him that I had a method for determining maxima and minima, drawing tangents and solving similar questions, equally applicable to both rational and irrational terms, and I hid the method by rearranging the letters of the following sentence: “when given an equation containing any number of current quantities, find the fluxions and back". The most famous man answered me that he also attacked such a method and told me his method, which turned out to be barely different from mine, and then only in terms and outline of formulas.

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In 1693, when Newton finally published the first summary his version of the analysis, he exchanged friendly letters with Leibniz. Newton said: Our Wallis added to his “Algebra”, which had just appeared, some of the letters that I wrote to you at one time. At the same time, he demanded that I openly state the method that I at that time hid from you by rearranging the letters; I made it as short as I could. I hope that I didn’t write anything that was unpleasant for you, but if this happened, then please let me know, because friends are dearer to me than mathematical discoveries.

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After the first detailed publication of Newton's analysis (mathematical appendix to Optics, 1704) appeared in Leibniz's journal Acta eruditorum, an anonymous review appeared with insulting allusions to Newton. The review clearly indicated that the author of the new calculus was Leibniz. Leibniz himself strongly denied that he had written the review, but historians were able to find a draft written in his handwriting. Newton ignored Leibniz's paper, but his students responded indignantly, after which a pan-European priority war broke out, "the most shameful squabble in the entire history of mathematics."

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On January 31, 1713, the Royal Society received a letter from Leibniz containing a conciliatory formulation: he agreed that Newton came to the analysis on his own, “on general principles similar to ours." An angry Newton demanded the creation of an international commission to clarify priority. The commission did not need much time: after a month and a half, having studied Newton’s correspondence with Oldenburg and other documents, it unanimously recognized Newton’s priority, and in wording, this time offensive to Leibniz. The decision of the commission was published in the proceedings of the Society with all supporting documents attached.

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In response, from the summer of 1713, Europe was flooded with anonymous pamphlets that defended Leibniz's priority and argued that "Newton arrogates to himself the honor that belongs to another." The pamphlets also accused Newton of stealing the results of Hooke and Flamsteed. Newton's friends, for their part, accused Leibniz himself of plagiarism; according to their version, during his stay in London (1676) Leibniz Royal Society got acquainted with Newton’s unpublished works and letters, after which Leibniz published the ideas expressed there and passed them off as his own. The war did not subside until December 1716, when Abbot Conti informed Newton: “Leibniz is dead - the dispute is over

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Let's set an arbitrary value x € (a.b) and define a new function. It is defined for all values ​​of x € (a.b) because we know that if there is an integral of ʄ on (a,b) then there is also an integral from ʄ on (a ,b) , where Let us recall that we consider by definition

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Thus, F is continuous on (a,b) regardless of whether ʄ has or does not have discontinuities; it is important that ʄ is integrable on (a,b) The figure shows the graph of ʄ . The area of ​​the variable figure aABx is equal to F (X). Its increment F (X+h)-F(x) is equal to the area of ​​the figure xBC(x+h), which, due to the Boundedness of ʄ, obviously tends to zero as h→ 0, regardless of will x be a point of continuity or discontinuity ʄ e.g. point x-d

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Passing to the limit in as h→0 shows the existence of the derivative of F at the point and the validity of the equality. For x=a,b we are talking here about the right and left derivatives, respectively. If the function ʄ is continuous on (a,b), then, based on what was proven above, the corresponding function has a derivative equal to Therefore, the function F(x) is an antiderivative for ʄ (a,b)

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We have proved that an arbitrary function ʄ, continuous on the interval (a,b), has an antiderivative on this interval defined by equality. This proves the existence of an antiderivative for any function continuous on an interval. Let now there be an arbitrary antiderivative of the function ʄ(x) on (a,b) . We know that Where C is some constant. Assuming x=a in this equality and taking into account that F(a)=0 we obtain Ф(a)=C Thus, But

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Integral The integral of a function is a natural analogue of the sum of a sequence. According to the main theorem of analysis, integration is the inverse operation of differentiation. The process of finding the integral is called integration. There are several different definitions integration operations that differ in technical details. However, they are all compatible, that is, any two methods of integration, if they can be applied to a given function, will give the same result.

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History The signs of the integral ʃ differentiation dx were first used by Leibniz at the end of the 17th century. The integral symbol is formed from the letter S - an abbreviation of the Latin word. summa (sum). Integral in antiquityIntegration can be traced back to ancient Egypt, around 1800 BC. e., the Moscow mathematical papyrus demonstrates knowledge of the formula for the volume of a truncated pyramid. First known method for calculating integrals is the exhaustion method of Eudoxus (approximately 370 BC), who tried to find areas and volumes by breaking them into infinite set parts for which the area or volume is already known. This method was taken up and developed by Archimedes, and was used to calculate the areas of parabolas and approximate the area of ​​a circle. Similar methods were developed independently in China in the 3rd century AD by Liu Hui, who used them to find the area of ​​a circle. This method was later used by Ju Chongshi to find the volume of a sphere.

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Historical significance and philosophical meaning of the Newton-Leibniz formula One of the most important research tools of this series is the Newton-Leibniz formula, and the method behind it for finding the antiderivative function by integrating its derivative. The historical significance of the formula is in the use of infinitesimal quantities and an absolutely accurate answer to the question posed. The advantages of using this method for solving mathematical, physical and other natural science problems are well known, for example, the classical problem of squaring a circle - constructing a square equal in size to a given circle. The philosophical meaning - in the possibility of obtaining information about the whole from its infinitesimal part, noted earlier - is clearly realized in medicine and biology, as exemplified by the successes genetic engineering in cloning - the creation of mutually similar living beings. History remains a rare exception in the list of sciences that have used the Newton-Leibniz formula. Inability to provide information historical sources in the form of numbers - formula arguments - is traditional. Thus, until now the philosophical meaning of the formula is not entirely philosophical, since it is realized only in natural science knowledge, leaving social and humanitarian knowledge without such a powerful tool. Although, if you stick to traditional features social and humanitarian knowledge, its weaknesses, so to speak, and what suits it.

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But further scientific analysis gives in our time a new, different picture of the ongoing process. The atomic views currently dominant in science decompose matter into a heap of tiny particles or regularly located centers of force, which are in eternal various movements. In exactly the same way, the ether that penetrates matter is constantly excited and oscillates in waves. All these movements of matter and ether are in the closest and continuous connection with the world space, which is infinite for us. This idea, inaccessible to our concrete imagination, follows from the data of physics.

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Even mystical and magical movements must take this situation into account, although they can, by giving a different meaning to the concept of time, completely destroy the meaning of this fact in the general worldview. Thus, as long as the question concerns phenomena perceived by the senses, even these areas of philosophy and religion that are most far from exact knowledge must take into account the scientifically proven fact, just as they must take into account the fact that two and two are four in the area that is subject to the knowledge of the senses and reason.

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At the same time, the volume of knowledge accumulated by humanity is already quite enough to break this tradition. In fact, there is no need, in a Pythagorean manner, to look for a digital correspondence to the statements “Peter I visited Venice during the Great Embassy” and “Peter I was not in Venice during the Great Embassy,” when these expressions themselves can easily serve as arguments in the algebra of George Boole’s logic. The result of each historical research is essentially a set of such arguments. Thus, it is justified, in my opinion, to use as an integrand function a set of historical studies presented in the form of arguments of the algebra of logic, with the aim of correspondingly obtaining as an antiderivative - the most probable reconstruction of the studied historical event. There are many problems along this path. In particular: the presentation of a specific historical research - a derivative of the reconstructed event - in the form of a set of logical expressions - an operation that is obviously more complex than, for example, electronic cataloging of a simple library archive. However, the information breakthrough of the late 20th century beginning of the XXI century (extremely high degree integration of the element base and an increase in information power) make the implementation of such a task quite realistic.

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In light of the above, on modern stage historical analysis is a mathematical analysis with the theory of probability and the algebra of logic, and the sought-for antiderivative function is the probability of a historical event, which in general is quite consistent with and even complements the idea of ​​science at the present stage, because replacing the concept of essence with the concept of function is the main thing in understanding science in the New time - supplemented by an assessment of this function. Therefore, modern historical meaning formulas in the possibility of realizing Leibniz’s dream “of the time when two philosophers, instead of endless disputes, will, like two mathematicians, take up pens in their hands and, sitting down at the table, replace the argument with calculation.” Each historical research - conclusion has the right to exist, reflects the actual event and complements the informational historical picture. Danger of degeneration historical science into a set of colorless phrases-statements - the result of applying the proposed method, there is no greater danger of music degenerating into a set of sounds, and painting into a set of colors at the present stage of human development. This is how I see the new philosophical meaning of the Newton-Leibniz formula, first given at the end of the 17th - beginning of the 18th centuries.

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Actually, the formula, in view of the peculiarity of the perception of mathematical symbols by carriers of social and humanitarian knowledge, which is expressed in the panic fear of these carriers of any representation of such signs, we present in verbal form: the definite integral of the derivative of a function is the antiderivative of this function. Some formal difference between the given example of the problem of squaring a circle and the usual educational and mathematical example of calculating the area located under an arbitrary curve in the Cartesian coordinate system does not, of course, change the essence.

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REFERENCES USED: 1. Brodsky I.A. Works in four volumes. T.3. St. Petersburg, 1994. 2. Vernadsky V.I. Biosphere and noosphere. M., 2003. 3. Wundt, Wilhelm. Introduction to Philosophy. M., 2001. 4. Gaidenko P.P. Evolution of the concept of science. M., 1980. 5. Descartes, Rene. Reflections on original philosophy. St. Petersburg, 1995. 6. Karpov G.M. Great Embassy of Peter I. Kaliningrad, 1998. 7. Kunzman P., Burkard F.-P., Widman F. Philosophy: dtv-Atlas. M., 2002. 8. Malakhovsky V.S. Selected chapters in the history of mathematics. Kaliningrad, 2002. 9. Nathanson I.P. Short course higher mathematics. St. Petersburg, 2001. 10. Engels F. Anti-Dühring. M., 1988. 11. Sheremetevsky V.P. Essays on the history of mathematics. M., 2004 Internet resources http://ru.wikipedia.org

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Newton-Leibniz formula

Main theorem of analysis or Newton - Leibniz formula gives a relationship between two operations: taking a definite integral and calculating the antiderivative

Formulation

Consider the integral of the function y = f(x) within a constant number a up to the number x, which we will consider variable. Let's write the integral in the following form:

This type of integral is called an integral with a variable upper limit. Using the mean value theorem in a definite integral, it is easy to show that this function continuous and differentiable. And also the derivative of a given function at point x is equal to the integrable function itself. From this it follows that any continuous function has an antiderivative in the form of a quadrature: . And since the class of antiderivative functions of the function f differs by a constant, it is easy to show that: the definite integral of the function f is equal to the difference in the values ​​of the antiderivatives at points b and a

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See what the “Newton-Leibniz formula” is in other dictionaries: Newton-Leibniz formula

- The main theorem of analysis or Newton’s Leibniz formula gives the relationship between two operations: taking a definite integral and calculating the antiderivative Formulation Let’s consider the integral of the function y = f(x) in the range from a constant number a to... ... Wikipedia Finite Increment Formula

- This term has other meanings, see Lagrange’s Theorem. The finite increment formula or Lagrange's mean value theorem states that if a function is continuous on an interval and... Wikipedia Stokes formula - Stokes theorem is one of the main theorems of differential geometry and mathematical analysis

on the integration of differential forms, which generalizes several theorems of analysis. Named after J. G. Stokes. Contents 1 General formulation 2… … Wikipedia NEWTON - LEIBNITZ FORMULA - a formula expressing the value of a definite integral of given function f along a segment in the form of the difference of values ​​at the ends of a segment of any antiderivative F of this function. Named after I. Newton and G. Leibniz, because the rule is... ...

Mathematical Encyclopedia NEWTON-LEIBNITZ FORMULA - the basic formula of integral calculus. Expresses the connection between a definite integral of a function f(x) and any of its antiderivatives F(x) ...

Big Encyclopedic Dictionary- This term has other meanings, see List of objects named after Leibniz. This term has other meanings, see Leibniz Formula (meanings). The Leibniz formula in integral calculus is the rule... ... Wikipedia

Newton-Leibniz formula- Newton Leibniz formula, the basic formula of integral calculus. Expresses the connection between the definite integral of the function f(x) and any of its antiderivatives F(x). . * * * NEWTON LEIBNITZ FORMULA NEWTON LEIBNITZ FORMULA, basic formula... ... encyclopedic Dictionary

Rectangle formula

Trapezoid formula - Definite integral as the area of ​​the figure Numerical integration ( historical name: quadrature) calculation of the value of a certain integral (usually approximate), based on the fact that the value of the integral is numerically equal to the area ... ... Wikipedia

Newton's theorem- Newton's Leibniz formula or the fundamental theorem of analysis gives the relationship between two operations: taking a definite integral and calculating the antiderivative. If it is continuous on a segment and any antiderivative of it on this segment has ... Wikipedia

Let some continuous function f be given on a certain segment of the Ox axis. Let us assume that this function does not change its sign throughout the entire segment.

If f is a continuous and non-negative function on a certain segment, and F is some antiderivative of it on this segment, then the area of ​​the curvilinear trapezoid S is equal to the increment of the antiderivative on this segment.

This theorem can be written as follows:

S = F(b) - F(a)

The integral of the function f(x) from a to b will be equal to S. Here and further, to denote the definite integral of some function f(x), with the limits of integration from a to b, we will use the following notation (a;b)∫f( x). Below is an example of how it will look.

Newton-Leibniz formula

This means that we can equate these two results. We obtain: (a;b)∫f(x)dx = F(b) - F(a), provided that F is an antiderivative for the function f on . This formula is called Newton - Leibniz formulas. It will be true for any continuous function f on an interval.

The Newton-Leibniz formula is used to calculate integrals. Let's look at a few examples:

Example 1: calculate the integral. Find the antiderivative for the integrand function x 2 . One of the antiderivatives will be the function (x 3)/3.

Now we use the Newton-Leibniz formula:

(-1;2)∫x 2 dx = (2 3)/3 - ((-1) 3)/3 = 3

Answer: (-1;2)∫x 2 dx = 3.

Example 2: calculate the integral (0;pi)∫sin(x)dx.

Find the antiderivative for the integrand function sin(x). One of the antiderivatives will be the -cos(x) function. Let's use the Newton-Leibniz formula:

(0;pi)∫cos(x)dx = -cos(pi) + cos(0) = 2.

Answer: (0;pi)∫sin(x)dx=2

Sometimes, for simplicity and convenience of recording, the increment of the function F on the segment (F(b)-F(a)) is written as follows:

Using this notation for the increment, the Newton-Leibniz formula can be rewritten as follows:

As noted above, this is just an abbreviation for ease of recording; this recording does not affect anything else. This notation and the formula (a;b)∫f(x)dx = F(b) - F(a) will be equivalent.

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Integral. Newton–Leibniz formula. Compiled by: Mathematics teacher of State Educational Institution of Educational Institution PU No. 27 Shchelyayur Semyashkina Irina Vasilievna

Objective of the lesson: Introduce the concept of an integral and its calculation using the Newton-Leibniz formula, using knowledge about the antiderivative and the rules for its calculation; Illustrate the practical application of the integral using examples of finding the area of ​​a curved trapezoid; Reinforce what you have learned during the exercises.

Definition: Let it be given positive function f(x) defined on the finite segment [ a;b ] . The integral of a function f(x) on [ a;b ] is the area of ​​its curvilinear trapezoid. y=f(x) b a 0 x y

Designation:  “integral from a to b ef from x de x”

Historical reference: Leibniz derived the notation for the integral from the first letter of the word “Summa”. Newton did not propose an alternative symbolism for the integral in his works, although he tried various options. The term integral itself was coined by Jacob Bernoulli. S umma Isaac Newton Gottfried Wilhelm von Leibniz Jacob Bernoulli

Euler introduced the notation for the indefinite integral. Jean Baptiste Joseph Fourier Leonard Euler The design of the definite integral in the form we are familiar with was invented by Fourier.

Newton-Leibniz formula

Example 1. Calculate the definite integral: = Solution:

Example 2. Calculate definite integrals: 5 9 1

Example 3. S y x Calculate the area of ​​the figure bounded by the lines and the x-axis. To start let's find the points intersection of the x-axis with the graph of the function. To do this, let's solve the equation. = Solution: S =

y x S A B D C Example 4. Calculate the area of ​​the figure bounded by the lines and Find the intersection points (abscissa) of these lines by solving the equation S=S BADC - S BAC S BADC = = S BAC = S = 9 – 4.5 = 4.5 see example 1 Solution:

SINCWAIN RULES 1st line – theme of syncwine 1 word 2nd line – 2 adjectives describing the signs and properties of the theme 3rd line – 3 verbs describing the nature of the action 4th line – short sentence of 4 words, showing your personal attitude to the topic 5 line - 1 word, synonym or your association the topic of the subject.

Integral 2. Definite, positive Count, add, multiply 4. Calculate using the Newton-Leibniz formula 5. Area

List of used literature: textbook by A.N. Kolmagorov. and others. Algebra and beginnings of analysis 10 - 11 grades.

Thank you for your attention! “TALENT is 99% of labor and 1% of ability” folk wisdom

Example 1. Calculate the definite integral: = Solution: example 4

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Subject: mathematics (algebra and beginnings of analysis), grade: 11th grade.

Lesson topic: "Integral. Newton-Leibniz formula."

Lesson type: Learning new material.

Lesson duration: 45 minutes.

Lesson objectives: introduce the concept of an integral and its calculation using the Newton-Leibniz formula, using knowledge about the antiderivative and the rules for its calculation; illustrate the practical application of the integral using examples of finding the area of ​​a curved trapezoid; consolidate what you have learned during the exercises.

Lesson objectives:

Educational:

1. form the concept of integral;
2. developing skills in calculating a definite integral;
3. formation of skills practical application integral to find the area of ​​a curved trapezoid.

Educational:

1. development cognitive interest students, develop mathematical speech, the ability to observe, compare, and draw conclusions;
2. develop interest in the subject using ICT.

Educational:

1. to intensify interest in acquiring new knowledge, developing accuracy and accuracy when calculating the integral and making drawings.

Equipment: PC, operating system Microsoft Windows 2000/XP, MS Office 2007: Power Point, Microsoft Word; multimedia projector, screen.

Literature: textbook by Kolmagorov A.N. and others. Algebra and beginnings of analysis 10-11 grades.

Technologies: ICT, individual training.

DURING THE CLASSES

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Basic summary on the topic “Integral. Newton-Leibniz formula."

Solving applied problems comes down to calculating the integral, but it is not always possible to do this accurately. Sometimes it is necessary to know the value of a certain integral with a certain degree of accuracy, for example, to the thousandth.

There are problems when it would be necessary to find the approximate value of a certain integral with the required accuracy, then numerical integration such as the Simposny method, trapezoids, and rectangles is used. Not all cases allow us to calculate it with a certain accuracy.

This article examines the application of the Newton-Leibniz formula. This is necessary for accurate calculation of the definite integral. Will be given detailed examples, changes of variable in the definite integral are considered and we find the values ​​of the definite integral when integrating by parts.

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Newton-Leibniz formula

When the function y = y (x) is continuous from the interval [ a ; b ] , and F (x) is one of antiderivative functions this segment, then Newton-Leibniz formula considered fair. Let's write it like this: ∫ a b f (x) d x = F (b) - F (a) .

This formula consider the basic formula of integral calculus.

To produce a proof of this formula, it is necessary to use the concept of an integral with an available variable upper limit.

When the function y = f (x) is continuous from the interval [ a ; b ], then the value of the argument x ∈ a; b , and the integral has the form ∫ a x f (t) d t and is considered a function upper limit. It is necessary to take the notation of the function will take the form ∫ a x f (t) d t = Φ (x) , it is continuous, and an inequality of the form ∫ a x f (t) d t " = Φ " (x) = f (x) is valid for it.

Let us fix that the increment of the function Φ (x) corresponds to the increment of the argument ∆ x , it is necessary to use the fifth main property of the definite integral and we obtain

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

where value c ∈ x; x + ∆ x .

Let us fix the equality in the form Φ (x + ∆ x) - Φ (x) ∆ x = f (c) . By definition of the derivative of a function, it is necessary to go to the limit as ∆ x → 0, then we obtain a formula of the form Φ " (x) = f (x). We find that Φ (x) is one of the antiderivatives for a function of the form y = f (x), located on [a; b]. Otherwise, the expression can be written

F (x) = Φ (x) + C = ∫ a x f (t) d t + C, where the value of C is constant.

Let's calculate F (a) using the first property of the definite integral. Then we get that

F (a) = Φ (a) + C = ∫ a a f (t) d t + C = 0 + C = C, hence we get that C = F (a). The result is applicable when calculating F (b) and we get:

F (b) = Φ (b) + C = ∫ a b f (t) d t + C = ∫ a b f (t) d t + F (a), in other words, F (b) = ∫ a b f (t) d t + F ( a) . The equality is proved by the Newton-Leibniz formula ∫ a b f (x) d x + F (b) - F (a) .

We take the increment of the function as F x a b = F (b) - F (a) . Using the notation, the Newton-Leibniz formula takes the form ∫ a b f (x) d x = F x a b = F (b) - F (a) .

To apply the formula, it is necessary to know one of the antiderivatives y = F (x) of the integrand function y = f (x) from the segment [ a ; b ], calculate the increment of the antiderivative from this segment. Let's look at a few examples of calculations using the Newton-Leibniz formula.

Example 1

Calculate the definite integral ∫ 1 3 x 2 d x using the Newton-Leibniz formula.

Solution

Consider that the integrand of the form y = x 2 is continuous from the interval [ 1 ; 3 ], then it is integrable on this interval. According to the table indefinite integrals we see that the function y = x 2 has a set of antiderivatives for all real values ​​of x, which means x ∈ 1; 3 will be written as F (x) = ∫ x 2 d x = x 3 3 + C . It is necessary to take the antiderivative with C = 0, then we obtain that F (x) = x 3 3.

We use the Newton-Leibniz formula and find that the calculation of the definite integral takes the form ∫ 1 3 x 2 d x = x 3 3 1 3 = 3 3 3 - 1 3 3 = 26 3.

Answer:∫ 1 3 x 2 d x = 26 3

Example 2

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

Solution

The given function is continuous from the segment [ - 1 ; 2 ], which means it is integrable on it. It is necessary to find the value of the indefinite integral ∫ x · e x 2 + 1 d x using the method of subsuming under the differential sign, then we obtain ∫ x · e x 2 + 1 d x = 1 2 ∫ e x 2 + 1 d (x 2 + 1) = 1 2 e x 2 + 1 + C .

Hence we have a set of antiderivatives of the function y = x · e x 2 + 1, which are valid for all x, x ∈ - 1; 2.

It is necessary to take the antiderivative at C = 0 and apply the Newton-Leibniz formula. Then we get an expression of the form

∫ - 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)

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

Example 3

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

Solution

Segment - 4; - 1 2 says that the function under the integral sign is continuous, which means it is integrable. From here we find the set of antiderivatives of the function y = 4 x 3 + 2 x 2. We get 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 antiderivative F (x) = 2 x 2 - 2 x, then, applying the Newton-Leibniz formula, we obtain the integral, which we calculate:

∫ - 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 proceed to the calculation of the second integral.

From the segment [ - 1 ; 1 ] we have that the integrand function is considered unbounded, because lim x → 0 4 x 3 + 2 x 2 = + ∞ , then it follows that a necessary condition integrability from a segment. Then F (x) = 2 x 2 - 2 x is not antiderivative for y = 4 x 3 + 2 x 2 from the interval [ - 1 ; 1 ], since point O belongs to the segment, but is not included in the domain of definition. This means that there is a definite Riemann and Newton-Leibniz integral for the function y = 4 x 3 + 2 x 2 from the interval [ - 1 ; 1 ] .

Answer: ∫ - 4 - 1 2 4 x 3 + 2 x 2 d x = - 28 , there is a definite Riemann and Newton-Leibniz integral for the function y = 4 x 3 + 2 x 2 from the interval [ - 1 ; 1 ] .

Before using the Newton-Leibniz formula, you need to know exactly about the existence of a definite integral.

Changing a variable in a definite integral

When the function y = f (x) is defined and continuous from the interval [ a ; b], then the available set [a; b] is considered to be the range of values ​​of the function x = g (z), defined on the segment α; β with the existing continuous derivative, where g (α) = a and g β = b, we obtain from this that ∫ a b f (x) d x = ∫ α β f (g (z)) g " (z) d z.

This formula is used when you need to calculate the integral ∫ a b f (x) d x, where the indefinite integral has the form ∫ f (x) d x, we calculate using the substitution method.

Example 4

Calculate a definite integral of the form ∫ 9 18 1 x 2 x - 9 d x .

Solution

The integrand function is considered continuous on the interval of integration, which means that a definite integral exists. Let's give the notation that 2 x - 9 = z ⇒ x = g (z) = z 2 + 9 2. The value x = 9 means that z = 2 9 - 9 = 9 = 3, and for x = 18 we get that z = 2 18 - 9 = 27 = 3 3, then g α = g (3) = 9, g β = g 3 3 = 18. When substituting the obtained values ​​into the formula ∫ a b f (x) d x = ∫ α β f (g (z)) g " (z) d z we obtain that

∫ 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 indefinite integrals, we have that one of the antiderivatives of the function 2 z 2 + 9 takes the value 2 3 a r c t g z 3 . Then, when applying the Newton-Leibniz formula, we obtain 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 a r c t g 3 - a r c t g 1 = 2 3 π 3 - π 4 = π 18

The finding could be done without using the formula ∫ a b f (x) d x = ∫ α β f (g (z)) · g " (z) d z .

If using the replacement method we use an integral of the form ∫ 1 x 2 x - 9 d x, then we can come to the result ∫ 1 x 2 x - 9 d x = 2 3 a r c t g 2 x - 9 3 + C.

From here we will carry out calculations using the Newton-Leibniz formula and calculate the definite integral. We get 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 - a r c t g 1 = 2 3 π 3 - π 4 = π 18

The results were the same.

Answer: ∫ 9 18 2 x 2 x - 9 d x = π 18

Integration by parts when calculating a definite integral

If on the segment [ a ; b ] the functions u (x) and v (x) are defined and continuous, then their first-order derivatives v " (x) · u (x) are integrable, thus from this segment for the integrable function u " (x) · v ( x) the 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 used then, it is necessary to calculate the integral ∫ a b f (x) d x, and ∫ f (x) d x it was necessary to look for it using integration by parts.

Example 5

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

Solution

The function x · sin x 3 + π 6 is integrable on the interval - π 2 ; 3 π 2, which means it is continuous.

Let u (x) = x, then d (v (x)) = v " (x) d x = sin x 3 + π 6 d x, and d (u (x)) = u " (x) d x = d x, and v (x) = - 3 cos π 3 + π 6 . From the formula ∫ a b v " (x) · u (x) d x = (u (x) · v (x)) a b - ∫ a b u " (x) · v (x) d x we ​​obtain that

∫ - 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 = = - 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 = 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 example can be solved in another way.

Find the set of antiderivatives of the function x · sin x 3 + π 6 using integration by parts using 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

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

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