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THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS PROBLEM SEMINAR PROBLEM INTEGRAL AND APPLICATION OF INTEGRAL Subject Mathematical analysis I Supervisors NGUYEN VAN THIN Class English of mathematics (NO1) Members NGUYEN NHU QUYNH June, 2022 CONTENTS INTRODUCTION 3 CHAPTER 1 PRIMITIVE 4 1 Primitive functions 4 2 Basic primitive table 4 CHAPTER 2 DEFINITE AND INDEFINITE INTERGRALS 6 I INDEFINITE INTEGRALS 6 1 Some basic concepts and properties 6 2 Methods for Calculating Uncertainty Integral 8 3 In.
THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS - - PROBLEM SEMINAR PROBLEM: INTEGRAL AND INTEGRAL APPLICATION OF Subject: Mathematical analysis I Supervisors: NGUYEN VAN THIN Class: English of mathematics (NO1) Members: NGUYEN NHU QUYNH June, 2022 CONTENTS INTRODUCTION CHAPTER 1: PRIMITIVE .4 Primitive functions Basic primitive table .4 CHAPTER 2: DEFINITE AND INDEFINITE INTERGRALS I INDEFINITE INTEGRALS: Some basic concepts and properties : Methods for Calculating Uncertainty Integral : Integral of rational fraction functions 11 Integral of trigonometric functions: 14 Integral of some irrational functions 16 II DEFINITE INTEGRALS: 18 Definition, geometric meaning of definite integral 18 Integralability criteria Properties of definite integrals 19 Derivative formula with bounds Formula Newton-Leibniz 22 Calculation methods .24 III APPLICATIONS OF DEFINITE INTEGRALS: .28 Compute the area of a plane figure .29 Calculate the length of the curve 31 Calculate the volume of an object 34 Calculate the area of the rotating circle 37 CHAPTER 3: IMPROPER INTEGRALS 39 I IMPROPER INTEGRALS TYPE I: .39 II IMPROPER INTEGRALS TYPE II: 41 III CONVERGENCE STANDARDS: .43 IV ABSOLUTELY CONVERGENT AND SEMI-CONVERGENT IMPROPER INTEGRALS: .46 CONCLUSION 49 REFERENCES .50 INTRODUCTION Mathematics is the science that studies the spatial structure of numbers and transformations In other words, it is said that it is the subject of "figures and numbers." Mathematics is the foundation for all other natural sciences It can be said that without mathematics, there would be no science at all Mathematics is divided into many sub-disciplines, including analysis Calculus is a branch of Mathematics that studies concepts: limits, derivatives, primitives, integrals The basic math of calculus is taking limits Most learners are very confused and have difficulty learning calculus in general and primes, integrals, and practical problems that need to use integration in particular Integration of applications in some problems about finding limits, proving inequality, or calculating area Besides, in the graduation exam of high school, university and college, there are always problems related to integration With the desire to systematize knowledge about integrals and its applications, we have chosen the topic "Integrals and applications" as the research topic CHAPTER 1: PRIMITIVE Primitive functions a Definition: Let f ( x) be defined on K The function F ( x) is called a primitive of f ( x) on K if F '( x) f ( x) , x �K - Theorem 1.1: If F ( x) is a primitive of the function f ( x) on K , then with each of constant C , the function G ( x) F ( x) C is also a primitive of f ( x) on K - Theorem 1.2: Let f ( x ) be an integrable map on I and F a primitive Then any primitive of f is of the form F ( x) c , with the constant c varying in � - Theorem 1.3: Every function f ( x) that is continuous on K has a primitive on K b Proposition: - Proposition 1: f '( x )dx f ( x ) C � - Proposition 2: kf ( x)dc k � f ( x)dx � ( k is a constant, k �0 ) f ( x)dx �� g ( x) dx f ( x) �g ( x) dx � - proposition 3: � Basic primitive table 1 ( ax b) C ( �1) 1 1 dx ln ax b C � a ax b e ax b dx e ax b C � a ( ax b) � dx 10 � sin e ax (a cos bx ) b sin bx C a b2 cos(ax b)dx sin(ax b) C � a 11 12 sin(ax b)dx cos(ax b) C � a e ax cos bxdx � dx tan( ax b) C � cos (ax b) a tan( ax b)dx ln cos(ax b) C � a 13 1 dx cot(ax b) C (ax b) a dx 14 � a x 2 x arctan C a a cot(ax b)dx ln sin(ax b) C � a � a x dx �a dx x2 ax ln C 2a a x arcsin dx 15 �x 16 � x x a 2 dx x C a 17 � b� ln( ax b) dx �x � ln( ax b) x C � � a� 18 ln x x a C a2 x arccos C a a 2 �a x dx x a2 x2 a2 x arcsin C 2 a e ax sin bxdx � e ax (a sin bx) b cos bx C a2 b2 CHAPTER 2: DEFINITE AND INDEFINITE INTERGRALS I INDEFINITE INTEGRALS: Some basic concepts and properties : a, b Definition : Given a function f ( x ) defined on the interval A function F ( x) that is differentiable on a, b is said to be a primitive of f ( x ) if F '( x ) f ( x) It is easy to see that if F ( x) is a primitive of f ( x ) , then : Every function of the form F ( x) C (with C �� ) is also a primitive of f ( x) All other primitives of f ( x) have the form F ( x ) C The family of all primitives of f ( x) , is called the uncertainty integral of f ( x ) , denoted by : f ( x )dx F ( x) C � Theorem a, b Every function f ( x ) that is continuous on the interval has a primitive on this interval F '( x)dx F ( x ) � �f ( x)dx / f ( x) f ( x )dx b � g ( x )dx C af ( x) bg ( x) dx a � � with a, b �� * Basic primitive table f ( x) F ( x) C x a xa ( x a ) 1 ( �1) 1 ln x a sin ax b cos ax b cos x sin x sin x cos x x2 b x2 b a x2 a2 x2 x a2 eax b cos ax b a sin ax b a tan x cot x ln tan �x � ln tan � � �2 � ln x x b b x x b ln x x b 2 x arcsin , a a a2 �x � x a x arcsin � � 2 �a � x arctan a a ax b e a 2� � � x �dx � x� Example Find the interval : � Solution 2� � x 12 x 3 x 3 dx � x �dx � � x� � x x 24 x 1 x 2 C Example Find the interval : x sin x cos xdx � Solution 1 sin x cos xdx � sin x sin x dx cos x cos x C � 2 10 Methods for Calculating Uncertainty Integral : * Variation method: Consider the integral f ( x )dx F ( x) C � Then, with the discriminant function u ( x) , we have: f (u ) du F (u ) C � Case : Consider the integral I � f ( x)dx Set t u ( x) , if we get f ( x)dx g (u )du , then: f ( x)dx � g u ( x) du ( x) G (u ) C , � (when G is a primitive of g ) The main idea of the above method is that we put a factor of f ( x) into d (.) , so that the integral expression can be easily computed Example Find the interval : e � sin x cos xdx Solution e � sin x cos xdx � esin x d sin x esin x C ln x �x dx Example Find the interval : Solution ln x ln xd ln x ln x C �x dx � x Example Find the interval : dx � x 1 Solution x dx dx dx arctan x C � � x 1 x2 e Example Find the interval : x �x dx Solution e x e d �x dx 2� x x 2e x C Example Find the interval : x x 3dx � Solution t2 x , dx tdt Set t = t x Hence: 1 x x 3dx � t 3 t dt 10 t � 10 2 x 3 Case : Consider the integral t3 C x 3 C I � f ( x)dx Set x x(t ) , hence: f ( x)dx � f x (t ) x '(t )dt , � Where x x(t ) is a function with continuous derivative and inverse function Setting x x(t ) needs to satisfy the requirement that the integral expression after setting must be simpler and easier to calculate than the original expression For 2 example, if in the expression of f ( x) there is a x , we put x a sin t ; and in 2 f ( x ) case the expression contains x a , we need to set x a tan t dx Example Find the interval : �1 x 32 Solution � � x sin t , t �� ; � 2 � � Set dx �1 x 32 dt � tan t C, t arcsin x cos t Example Find the interval : �x 4 dx Solution Set x tan t We have: �x 4 dx 1� x � cos tdt � t sin 2t � C , t arctan � 16 � 2 � * Integral by parts method: uv ' uv ' vu ' With two differentiable functions u ( x), v( x) , we have From that, we infer: f ( x)dx � u ( x)v '( x)dx � u ( x)dv( x) u ( x)v ( x) � v( x )u '( x )dx � (2.1) The meaning of the above method is: instead of integrating the whole expression f ( x) (complex), we only need to integrate one of its factors In the above formula, we analyzed f ( x) u( x)v '( x) and integrated the second factor This analysis must ensure that the requirements of the resulting integral are simpler than the original integral expression * Integral types calculated by the method of partial integration: Integral form P ( x)e � n ax dx How to apply the integral part formula ax Let e into in d (.) , apply (2.1) n times P ( x )sin(ax b)dx � Let sin( ax b) into in d (.) , apply (2.1) n times P ( x ) cos(ax b)dx � P ( x) lnQ ( x)dx � Let cos(ax b) into in d (.) , apply (2.1) n times n n n m e cos( ax b)dx � ax Let Pn ( x) into in d (.) , apply (2.1) once Put one of the two factors in d (.) , apply (2.1) twice 10 Example Calculate the volume of a sphere with radius R Solution A sphere is a special type of circular object that rotates when it rotates a half2 circle defined by y R x , y 0, R �x �R around the Ox axis Applying the formula (3), we have: R V R2 x2 � R dx R R � R x dx R Example Calculate the volume of the rotating object created when rotating the figure bounded by the lines y x x and y a) one revolution around the Ox axis b) one revolution around the Oy axis Solution a) Applying the formula (3), we get: VOx � x x dx 16 15 b) Applying the formula (5), we get: VOy 2 � x x x dx 8 Calculate the area of the rotating circle Case 1: If S is a rotating circle formed when the curve y f ( x), a �x �b is f �C1 a, b around the axis Ox , where then the surface area S is calculated by the formula: b S 2 �f ( x ) f '2 x dx a (7) This formula is established from the calculus diagram where the integral of the surface area dS 2 f ( x) ds where ds is the differential of the curve length is 38 calculated by the formula dx ds dy f '2 x dx 2 Therefore, dS 2 f ( x ) f '2 x dx From that, the formula (7) is derived Similarly, changing the roles of x and y , we derive the formula (8) Case 2: If S is a rotating circle created when rotating the curve �C1 c, d around the axis Oy , where , then the surface area S is calculated by the formula: x y , c �y �d d S 2 � y '2 y dy c (8) Example Calculate the area of a sphere with radius R Solution A sphere is a special type of circle that rotates when it rotates half a 2 circle defined by y R x , R �x �R around the Ox axis Applying the formula (7), we have: S 2 R �R R x R x2 1 dx 2 � Rdx 4 R R x R Example Calculate the area of the rotating circle created when turning the line y x , �x �1 around the Ox axis Solution Applying formula (7), we have: S 2 � x3 3x dx 9x4 d x4 10 10 � 27 Example Calculate the area of the rotating circle created when rotating y arctan x, �x �1 around the axis Oy Solution We have: y �� arctan �� x x tan y,, x y 39 Applying formula (8), we have: S 2 � tan y tan y dy 2 � t 1 1 t2 1 1 t2 � 1 t2 dt 1 t2 ( set t tan y ) d t 1 s2 ds ( set s t ) s 5� � �� 1 � du � u 1 � � � � � 1 1� � � �5 � ln ln � � � 1 1� � � CHAPTER 3: IMPROPER INTEGRALS In section of this chapter, we consider definite integrals with the region of integration being a finite segment [a, b] and the integral function bounded on [a,b] In this section, we extend the integration domain to the case where the sagment [a, b] has bounds going to infinity, or the integral function is not bounded on [a, b] I IMPROPER INTEGRALS TYPE I: Suppose f ( x) is a function on the interval [a, �) and integrable over every finite interval [a, b], (a �b < �) b Definition 2.4.1 The limit of the integral f ( x)dx � a when b � �is called the a, � improper integrals type of the function f ( x ) over the interval and is denoted � b f ( x)dx �f ( x)dx lim � a b �� a 40 � �f ( x)dx If this limit exist we say the improper integrals a converges Otherwise, if this limit does not exist or the limit is infinity, we say that the integral diverges �,b Similarly, we define the integral of a function f ( x) over the interval �, � and by the following fomular: b f ( x)dx �f ( x)dx lim � � � b a � � a and �f ( x)dx � b lim f ( x)dx � a ��,b �� a These integral cases with close to infinity are called improper integrals type We can write � �f ( x)dx � a � � a �f ( x)dx �f ( x)dx when two of the above three integrals converges Through the above definition, we see that the improper integral iss the limit of the definite integral (understood in the usual sense) when the integral approaches infinity Therefore, the formula Leibniz can be extended to calculate the integral for the case approaching infinity as: � �f ( x)dx F ( x) � a F (�) F (a) a It where, symbol F (�) lim F ( x ) x �� Similarly, we have the extended Leibniz formula for the remaining cases of improper integrals type e2 x �e x 1dx � Example 1: Compute the integral 41 Solution: x x Using the transform method, set e t � e dx dt , we have 1 t � � I � dt � 1 dt t ln t � � t 1 t 1� 0� � ln dx �x Example 2: Compute the integral � Solution: � dx �x arctan t � � � � � � � � �2� dx �x Example 3: Compute the integral x2 Solution: Using the transform method, set � �x dx cos tdt � sin t 1 x sin t x tan t � dx dt cos t , we have 1 Example 4: Consider the converges of integral x sin xdx � � Solution: x sin xdx x cos x sin x � � lim x cos x sin x x � � We have limit does not exist, therefore the integral diverges � 42 Because this II IMPROPER INTEGRALS TYPE II: a, b Assume f ( x) is a definite function on the interval and integrable over lim f ( x) � every segment in the wihth a t b and x �b The point x b is called the in regular point of the function f ( x) t Definition The limit of the integral f ( x)dx � a when t � b is called the a, b improper integral type of the function f ( x) over the interval and is denoted as follows: b t f ( x ) dx lim � f ( x)dx � t �b a a If the limit on must exists finitely, we say the improper integral converges Otherwise, if this limit does not exist, or the limit is infinity, we say the integral diverges Similarly, we define the improper integral of the function f ( x ) with no upper a, b a, b leg inervals and take x a and x b as outliers respectively b b f ( x) dx lim � f ( x)dx � x �a a t b and f ( x)dx � a t' lim t �a ,t ' �b f ( x)dx � t These cases of integration with unbounded are called the improper integral type Note: Note that in the case of a function that is undefined at a but b has a finite existence then f ( x) dx � a is a definite product For integrals with two outliers x a and x b , we can write: b c b a a c f ( x )dx � f ( x )dx � f ( x )dx � when two of the above three integrals converges 43 lim f ( x) x �a Example 1: Consider of the integral converges I dx � x a Solution: dx x1 a � xa a 1 1 lim x1 a x �0 1 a If a �1 then Because this limit exists when a it mean a , if the integral I converges when a Otherwise, when a , the integral I diverges If a then dx �x ln x � Therefore, we get I diverges So, the improper integral I converges when and only if a , diverges when and only if a �1 Example 2: Compute the integral x �2 x dx Solution: �� x 2sin t , t �� 0; � �, we get � Using the transform method, set x 2sin t tan t 2 x sin t dx 4sin t cos tdt , Therefore, we have x I � dx � sin tdt 2t sin 2t 2 x 0 x arcsin x I � dx x Example 3: Compute the integral Solution Using the method of integration by parts, set 44 u arcsin x, dv xdx x2 Then, we count dx du 1 x ; v x2 We have I x arcsin x 10 � dx x arcsin x x 1 III CONVERGENCE STANDARDS: Definition 3.1 (Comparative standards) � Consider improper integrals �f x dx a � and g x dx � a f x g x a, b Assume and integrable over any finite interval with b a and we have inequality : �f x �g x , x � a, � Then, we have � i) If g x dx � a � converges then � ii) If �f x dx a Assume �f x dx a a converges � diverges then f x g x dx � a diverges g x and are two positive functions integrable over all finite a, b segments with b a and � �f x dx lim f x x �� g x k k � Then, the integrals � and g x dx � a either converges or diverges � Definition 3.2 Consider improper integrals �f x dx a � and g x dx � a Let f a, � f x g x and g are two positive function over If and are VCB and VCL when x � �, and f x : g x when x � � 45 � �f x dx then a and either converges or diverges a, � Definition 3.3 Let f and g are two positive function over Then, we have lim If x � � lim x �� f x g x f x g x 0 � and g x dx � a � converges then �f x dx a � g x dx � � converges � �f x dx If and diverges then a diverges Similarly, we also have the convergence standards for the case of the improper integral of an unbounded function Definition 3.4 (Comparative standards) a b Consider the improper integrals anomaly are x a Let the function inequality f x f x dx � a b and g x dx � a have the same point g x a, b and determined on and we have �f x �g x , x � a, b Then � i) If g x dx � a � converges then � ii) If �f x dx a Assume �f x dx a a converges � diverges then f x a, b segments with � �f x dx g x dx � a diverges g x and are two positive functions integrable over all finite b a and lim x �a f x g x k k � � and g x dx � a either converges or diverges 46 Then, the integrals � Definition 3.5 Consider improper integrals �f x dx a � and g x dx � a Let f a, b and g are two positive function over and have the same point anomaly are x a If f x and g x are VCB and VCL when x � a , and f x : g x x � a when � then �f x dx a and either converges or diverges a, b Definition 3.6 Let f and g are two positive function over and have the same point anomaly are x a Then, we have If lim x �a lim x �a f x g x f x g x � 0 and g x dx � a � converges then �f x dx � converges � g x dx � � a �f x dx If and diverges then a diverges Note: When using the standard of comparison we usually compare the given improper integrals with the basic improper integrals: a 0, I1 a) With b) With a b , a � dx �x a a , converges if a and diverges if a �1 b dx I2 � a a x a , converges if a , diverges if a �1 , b dx I � a a b x ' , converges if a , diverges if a �1 Example: Consider converges and compute the integrals: xe x dx � a) Solution � � � b) cos xdx � c) 47 �x � dx 1 xe dx � e x 1 x x a) � � 1 � cos xdx � sin x b) converges � �x � c) � dx 1 = � Thus, the limit is not exist dx 2� 2 x 1 lim sin x x �� Set x tan t then then intergral I 2� cos tdt IV ABSOLUTELY CONVERGENT AND SEMI-CONVERGENT IMPROPER INTEGRALS: � Definition 4.1 Let improper the integral � If �f x dx a �f x dx a Then � �f x dx converges then a converges b If �f x dx a ( have the same point anomaly are a or b ) converges then � �f x dx a converges Definition 4.2 � Let the improper integral � �f x dx If a �f x dx a then we say that but a � �f x dx absolutely converges and if a � �f x dx Then: � �f x dx a � diverges then we say that �f x dx a semi-converges b Let the improper intergral b ) Then: f x dx � a 48 ( have the same point anomaly are a or b �f x dx If a b then we say that f x dx � a b but �f x dx a b absolutely converges and if f x dx � a b f x dx � diverges then we say that a � semi-converges sin x cos x dx x x � Example : Cosider converges of integral Solution: x � 1, � , we have sin x cos x 0� �3 x x x � Because � integral �x � dx converges then sin x cos x dx x x converges Therefore, the � sin x cos x dx x x absolutely converges � Example 2: Prove that the improper integral is semi-converges � sin x � x dx Solution: We have � sin xdx sin xdx � � x x 0 � sin xdx x � Set sin xdx I1 � , I2 x � sin xdx x � 49 Because lim x �0 sin x 0 x should the integral I1 is definite integral, therefore just consider I I2 � d cos x sin x cos x dx � � x x x � 2 � cos x Because x � �3 x should � We will prove that sin x �x cos x � x � cos x �x dx � cos x �x dx dx converges So, we have I converges dx diverges We have sin x cos x � �0 x x x sin x � dx � x Because diverges and � cos 2x dx � x converges, should � cos 2x dx x � diverges � sin x dx � x Thus, we get diverges Adding, the integral � sin x � lim 0� � x �0 x �, we get integral � � sin x �x dx diverges 50 � sin x �x dx is definite CONCLUSION In the current university's advanced math program, the calculus module plays an important role and is distributed right from the first semester In this, integral and its applications play an important role Integral has many applications such as in problems of limits, calculating areas, In the process of studying the document due to the limited time, after completing it, we will continue to study more deeply about integrals and its applications We look forward to receiving your comments to improve our essay We sincerely thank you 51 REFERENCES [1] Bùi Xuân Diệu, Trịnh Ngọc Hải, Nguyễn Hải Sơn, Nguyễn Thị Tồn, Bài tập Giải tích 1, NXB Bách khoa Hà Nội (2020) [2] Nguyễn Đình Trí (chủ biên), Tạ Văn Đĩnh, Nguyễn Hồ Quỳnh, Toán học cao cấp, tập hai, NXB Giáo dục, Hà Nội (2006) [3] Nguyễn Đình Trí (chủ biên), Tạ Văn Đĩnh, Nguyễn Hồ Quỳnh, Tốn học cao cấp, tập ba, NXB Giáo dục, Hà Nội (2006) [4] Nguyễn Đình Trí (chủ biên), Tạ Văn Đĩnh, Nguyễn Hồ Quỳnh, Bài tập toán cao cấp, tập hai, NXB Giáo dục, Hà Nội (2006) [5] Nguyễn Đình Trí (chủ biên), Tạ Văn Đĩnh, Nguyễn Hồ Quỳnh, Bài tập toán cao cấp, tập ba, NXB Giáo dục, Hà Nội (2006) [6] Nguyễn Ngọc Cừ, Lê Huy Đạm, Trịnh Danh Đằng, Trần Thanh Sơn, Giải tích I, NXB Đại học Quốc gia Hà Nội (2005) [7] Trần Bình, Hướng dẫn giải tập giải tích tốn học, tập một, NXB Đại học Quốc gia Hà Nội (2001) [8] James Stewart, Calculus, Early Transcendentals, 7th Ed Brooks Cole Cengage Learning, (2012) 52 ... between the indefinite integral and the definite integral To compute a definite integral, we need to calculate the limit of the sum of the integrals This is cumbersome and difficult to In this... 0,1 From this, we infer that the value of the limit of the sum of the integrals does not 0,1 depend on the division of segment and the choice of representative point in 0,1 each division... 11 Integral of trigonometric functions: 14 Integral of some irrational functions 16 II DEFINITE INTEGRALS: 18 Definition, geometric meaning of definite integral