Integral and its application

THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS  PROBLEM SEMINAR PROBLEM Integral and its application Supervisors NGUYEN VAN THIN Group 3 Class Mathematical analysis I June, 2022 TABLE OF CONTENTS INTRODUCTION OF PRINCIPLES – ANALYSIS 5 PRINCIPLES AND DIFFERENTIAL ANALYSIS 5 Define 5 Nature 5 DEFINITION ANALYSIS 5 Define 5 Nature 6 ANALYSIS METHODS 6 Primitives of some basic functions Integral Calculation Methods Sub conclusion FEATURES ON THE BIRTH AND DEVELOPMENT OF THE THEORY OF ANALYSIS.

THAI NGUYEN UNIVERSITY OF EDUCATION MATHEMATICS - - PROBLEM SEMINAR PROBLEM: Integral and its application Supervisors: NGUYEN VAN THIN Group : Class: Mathematical analysis I June, 2022 TABLE OF CONTENTS INTRODUCTION OF PRINCIPLES – ANALYSIS PRINCIPLES AND DIFFERENTIAL ANALYSIS Define Nature DEFINITION ANALYSIS Define Nature ANALYSIS METHODS Primitives of some basic functions Integral Calculation Methods Sub-conclusion FEATURES ON THE BIRTH AND DEVELOPMENT OF THE THEORY OF ANALYSIS10 10 HISTORY ANALYSIS 11 ANALYSIS APPLICATION APPLICATIONS OF DEFINITION ANALYSIS………………………………………………12 GEOLOGICAL APPLICATIONS OF DEFINITIONAL ANALYSIS A General diagram using integrals 12 12 B Area of flat figure C Bow length 13 18 Algebraic Applications of DEFINITIONAL ANALYSIS 20 A Integral inequality 20 B Integral applications prove the inequality 20 CONCLUSION………………………………………………………………………………….21 The outline of primitives - integrals Primitive - uncertain integral A Define i Primitive - The function F(x) is a primitive of the function f(x) on (a, b) if F'(x) = f(x) ∀x ∈ (a, b) - Let F(x) and G(x) be two primitives of the function f(x) on (a, b), then there exists a constant C such that: F(x) = G(x) + C - So all primitives differ by only one constant ii Uncertainly integral - The set of primitives of f (x), recorded as ∫f(x)dx is called the uncertain integral of the function f(x) - If F (x) is a primitive of f (x) then we have ∫ 𝒇(𝒙)𝒅𝒙 = F(x) + C (C = const) B Nature 1) [∫ f(x)dx]’ = f(x) 2) ∫[f(x) ± g(x)]dx = ∫ f(x)dx ± ∫ g(x)dx 3) ∫ kf(x)dx = k∫ f(x)dx (với k ∈ R) Definite integral A Definition ➢ Consider the function f(x) on a closed interval [a,b] and divide [a,b] into n points: a= x0 0; cos (𝑥 + ) < 𝑎𝑛𝑑 |sin( 𝑥 + )| > |cos ( 𝑥 + )| 4 4 𝜋 𝜋 3𝜋 Therefore: |𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠𝑥 | = sin 𝑥 + cos 𝑥 𝑤𝑖𝑡ℎ ≤ 𝑥 + ≤ √1 + sin 2𝑥 𝜋 𝜋 2 4 We get: (𝑐𝑜𝑠𝑥 − sin 𝑥 )𝑑𝑥 − 𝑑(𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠𝑥 ) 𝜋 𝜋 So : 𝐼 = ∫ − 𝜋 𝑑(sin 𝑥+𝑐𝑜𝑠𝑥) sin 𝑥+cos 𝑥 = −𝑙𝑛|𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 ||𝜋2 = −[𝑙𝑛1 − 𝑙𝑛√2] = ln 2 C Sub-conclusion ➢ From this we see that in addition to applying basic trigonometric transformations, the recognition of “hidden” trigonometric differential formulas will also be an important skill If I realise them, the problem solving becomes much simpler ➢ Like some of the following common types: (Acosx ∓ Bsinx)dx = d(Asinx ± Bcosx + C) (A ∓ B)sin2xdx = d(Asin2x ± Bcos2x + C) −sin4xdx = d(sin4x + cos4x + C) Some details on the birth and development of integral theory History of intergral Calculating the area of a plane or calculating the volume of an object in space, whose shape could not apply the formulas of elementary geometry had long been devised Since ancient times, the eminent mathematician and physicist Archimède has used elementary mathematical tools to calculate the area of a number of planes limited by curves such as spheres and cones He is considered to be the person who laid the first brick to enable integral math Until the 17th century, the systematic development of the above method of area and volume calculation was done by mathematicians such as Cavalieri, Torricelli, Fermat, Pascal and many other mathematicians In 1659, Barrow established the relationship between the area calculation and the tangent finding of the relevant curve Not long after, Issac Newton and Gottfried Leibniz abstracted the above connection into the connection between differential and integral, two important shapes of calculus.Newton and Gottfried Leibniz abstracted the above connection into the connection between differential and integral, two important shapes of calculus Newton and Leibnitz, together with their students, used the relationship between differential and integral to broaden methods of integration, but methods of integral calculation are known in the equation The present degree is largely presented in Euler's work The contributions of mathematicians Tchébicheff and Ostro-gradski ended the process of developing this calculation Or we can relate to the law of transformation between "quantity - quality" of materialistic dialectic in Marx-Leninist Philosophy From the following two problems: (1) When a cow is divided in two, its substance does not change, further dividing it will remain the same, but if you continue to divide it, the cow's substance will remain the same head? (2) A grain of sand is not called a desert, adding another grain of sand still does not become a desert, but as it continues to add, it creates a desert So we can see a special correlation in micro-integral, it is like the dialectical relationship between quantity and inseparable matter! Integral application Before the advent of integrals, mathematicians were able to calculate the speed of a ship But they still could not calculate the relationship between the acceleration and the ratio of the force acting on the ship; or it is still not possible to calculate the appropriate angle of fire in the environment with resistance for the bullet to go the farthest Today, the differential - integral calculus is one of the important tools in physics, economics and subjects probability science In the 1960s, it was the vi-integral functions that first became instrumental in enabling spacecraft engineers Apollo calculate the data in the mission to conquer the moon of man But it's interesting that the first person to fly into space is a hero Yuri Gagarin (1934 1968) Soviets His space trip lasted for 108 minutes on April 12, 1964 and became an important historical milestone for all of humanity Application of definite integrals I Geometrical applications of definite integrals General diagram using integral A The first method - integral sum Suppose: We want to calculate any quantity Q such as volume, area, length, etc We divide Q into n parts in an appropriate way: Q = Q1 + Q2 +…+Qn (1) Then calculate Qkrelatively accurately, instead (1) use the limit to find the sum Q - We used the above method to calculate the area of a curved trapezoid and got: 𝑏 𝑆 = ∫ 𝑓(𝑥) 𝑑𝑥 𝑎 - Comment: The larger n is (the greater the number of Q divisions), the higher the accuracy, so the calculation is very "hard" The integral helps us to limit the error and calculate the exact quantity we are looking for VD1: Consider the area of the plane bounded by: horizontal axis, y = f(x) = cosx + and two lines x = 0; x = 9.5 Integral value:18.92 Total area: 20 trapezoids Integral value:18.92 Total area: 20 trapezoids Integral value:18.92 Total area: 20 trapezoids Integral value:18.92 Integral value:18.92 Total area: 20 trapezoids Total area: 100 trapezoids Second method – differential diagram - Suppose: Quantity to be calculated Q depends additively on the segments: 𝑎 ≤ 𝑎′ < 𝑐 < 𝑏′ ≤ 𝑏 We have: Q[a', b'] = Q[a', c] + Q[c, b'] and Q = Q[a, b] To calculate Q we set Q(x) = Q[a, x] Considering increments: ∆Q = Q(x + ∆x) − Q(x) = Q[a, x + ∆x] − Q[a, x] = Q[x, x + ∆x] And try to represent it as linear in terms of ∆x : Q = q(x)∆x + 0(∆x) => (2) dQ(x) = q(x)dx, so: Q = Q[a, b] = ∫130_𝑎^𝑏▒ⅆ𝑄=∫130_𝑎^𝑏▒𝑞(𝑥)ⅆ𝑥 (3) - So the content of the differential diagram method is to establish the relation (2), whose final result is (3) 2.Area of flat figure Area of a plane bounded by curves in Cartesian coordinates Apply the sum of integrals method to calculate the area of the plane figure S formed by two any function f1(x) and f2(x) into aninfinite number of shapes very small curved scale MPFE (Figure 1) I'm easy It is easy to prove specific cases can be as follows: If the function y = f(x) is continuous on [a, b] then the area S of a curved trapezoid bounded by { y (C): y = f(x) (𝐶): 𝑦 = 𝑓(𝑥), ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑥𝑖𝑠 𝑖𝑠 𝑥 = 𝑎, 𝑥 = 𝑏 𝑏 𝑆 = ∫ |𝑓(𝑥)|𝑑𝑥 𝑎 a O b x Example 4: Calculate the area S of the plane figure (H) bounded by the graph of the function (𝐶): 𝑦 = √𝑥 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑥𝑖𝑠 and straight line (𝐷): 𝑦 = 𝑥 − Equation of the intersection point: 1, √𝑥 = 𝑥 −  { 𝑥≥2 𝑥≥2 𝑥 = 1(𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑦)  { [ 𝑥 − 5𝑥 + = 𝑥 = 4(𝑟𝑒𝑐𝑒𝑖𝑣𝑒) (𝐶) ∩ (𝐷) = 𝐵(4; 2) 2, √𝑥 =  𝑥 = 0 (𝐶) ∩ (𝑂𝑥) = 𝑂(0; 0) 3, 𝑥 − =  𝑥 =  (𝐷) ∩ (𝑂𝑥) = 𝐴(2; 0) We can easily see the area of the plane figure (H): (H) = 𝑆(𝑂𝐴𝐵) = (𝐻1 ) + (𝐻2 ) = ∫0 (√𝑥 − 0)𝑑𝑥 + ∫2 [√𝑥 − (𝑥 − 2)𝑑𝑥] = ∫0 √𝑥𝑑𝑥 + ∫2 (√𝑥 − 𝑥 + 2)𝑑𝑥 = 2 𝑥2 10 (3 𝑥 √𝑥)| + (3 𝑥 √𝑥 − + 2𝑥)| = 2 Area of a plane bounded by a line with parametric equations 𝑥 = 𝜑 (𝑡) If curve (C): y = f(x) has parametric equation { 𝑦 = 𝜔(𝑡) 𝑏 In the formula to calculate the area 𝑆 = ∫𝑎 |𝑓(𝑥)| 𝑑𝑥 I replace + y = f(x) by y = ω(t) ′ + dx by φ (t)dt + Two bounds a and b are replaced by α and β which are solutions of a = φ(α) and b = φ(β) respectively ′ Then : S 𝛽 =∫𝑆𝛼 |𝝎(𝒕)𝝋 (𝒕)|𝒅𝒕 If the curve (C) is partially smooth, counter-clockwise and the area S is limited to the left and (C) has a parametric equation { Then: 𝑥 = 𝜑 (𝑡) with ≤ t ≤ T where T is the periodic period of it 𝑦 = 𝜔(𝑡) S= 𝟏 𝟐 ′ 𝑻 ∫𝟎 [𝝎(𝒕)𝝋 (𝒕) − 𝝋(𝒕)𝝎′(𝒕)] 𝒅𝒕 Example : Calculate the area of an ellipse limited by (E): 𝑥2 𝑦2 𝑎 𝑏2 + =1 We have the parametric equation of (E) consider the area of (E) in the first quadrant on the plane (Oxy) 𝑥 = 𝑎 cos 𝑡 (𝐸): { 𝑦 = 𝑏 sin 𝑡 ; ≤ 𝑡 ≤ 2𝜋 Converting the integral we get: 𝜋 { 𝑡= = 𝑎𝑐𝑜𝑠𝑡 ⇒{ 𝑎 = 𝑎𝑐𝑜𝑠𝑡 𝑡=0 So 𝜋 2 𝑆 = 4𝑆1 = ∫ 𝑏 𝑠𝑖𝑛 𝑡(− 𝑎𝑠𝑖𝑛 𝑡) 𝑑𝑡 4𝑎𝑏 ∫ sin 𝑡 𝑑𝑡 = 4𝑎𝑏 ∫ 𝜋 𝜋 21 − 𝑐𝑜𝑠2𝑡 𝑑𝑡 = 𝜋𝑎𝑏 Acr length A Theoretical basis ➢ We can subdivide this curve into an infinite number of “near-straight” segments Δl and sum them together Consider 𝑥0 ∈ [𝑎; 𝑏] and Δx > such that Δx∈ [𝑎; 𝑏] ➢ With Δx small enough, we consider the length of the graph curve f(x) limiting between the two lines x=𝑥0 and x=𝑥0 + Δx is the length of the line connecting points A and B as shown in figure And also because Δx is small, so AB = (d) should be considered as the tangent at x of f(x) ➢ So the arc length Δl ≈ AB = Δx 𝑐𝑜𝑠𝑎 , with a is the angle formed by the tangent AB at x of f(x) and the axis Ox, so tanα=f’(𝑥0 ) Therefore: Δl=Δx√𝒇’(𝒙𝟎 ) + 𝟏 (5) B Curve length in Cartesian coordinates From (5) we take the sum of the lengths of the small “straight” segments together, we get the (𝑆): 𝑦 = 𝑓(𝑥) formula to calculate the curve length limited by { is L 𝑥 = 𝑎 𝑎𝑛𝑑 𝑥 = 𝑏 Then: 𝒃 L= ∫𝒂 √[𝒇’(𝒙)]𝟐 + 𝟏 𝒅𝒙 Example 7: Calculate the length of the arc OA lying on 𝑥2 the parapol y= with a≠0, which O is the origin, A is a 2𝑎 point on the parabol whose latitude is t t2 2𝑥 2a 2𝑎 We have: A(t, ) and y’(x) = = 𝑥 𝑎 𝑡 𝑡 𝑥 Therefore L = ∫0 √[𝑦 ′ (𝑥)]2 + 𝑑𝑥 = ∫0 √( )2 + 𝑑𝑥 𝑎 = √𝑥 + 𝑎2 𝑑𝑥 𝑎 = = 2𝑎 [x√𝑥 + 𝑎2 + 𝑎2 ln|𝑥 + √𝑥 + 𝑎2 |]| 0𝑡 𝑡 𝑎 √𝑡 + 𝑎2 +2 ln ( 2𝑎 𝑡+√𝑡 +𝑎2 |𝑎| ) C Curve length with parametric equation We demonstrate the complete analogy from how to convert the area of a plane figure in Cartesian coordinates to the area of a plane bounded by a line with parametric equations (in section 2) 𝑏 From then, instead of (5) we get: l = ∫𝑎 √[𝑥 ′ (𝑡)]2 + [𝑦 ′ (𝑡)]2 dx in 𝑅2 𝑏 = ∫𝑎 √[𝑥 ′ (𝑡)]2 + [𝑦 ′ (𝑡)]2 + [𝑧 ′ (𝑡)]2dx in 𝑅3 Example 8: Calculate the length of the arc of Archimedes’spiral 𝑥 = 𝑎𝑐𝑜𝑠𝑡 {𝑦 = 𝑎𝑠𝑖𝑛𝑡 𝑤𝑖𝑡ℎ ≤ 𝑡 ≤ 2𝜋 𝑎𝑛𝑑 𝑎 > 𝑧 = 𝑎𝑡 We have: x’(t) = -asint, y’(t) = acost, z’(t) = a 2𝜋 Therefore L = ∫0 √[𝑥 ′ (𝑡)]2 + [𝑦 ′ (𝑡)]2 + [𝑧 ′ (𝑡)]2 𝑑𝑡 2𝜋 = ∫0 √(−𝑎𝑠𝑖𝑛𝑡)2 + (𝑎𝑐𝑜𝑠𝑡)2 + 𝑎2 𝑑𝑡 2𝜋 = a∫0 √𝑠𝑖𝑛𝑡 + 𝑐𝑜𝑠𝑡 + 1𝑑𝑡 2𝜋 = a√2 ∫0 𝑑𝑡 = a√2 | 2𝜋 = 2𝜋a√2 II ALGEBRAIC APPLICATIONS OF DEFINITE INTEGRAL Integral inequality We evalute the inequality in terms of functions and integrals Example11: Prove that: Consider f(x) = 𝜋 𝜋 < ∫0 16 𝑑𝑥 < +3(𝑐𝑜𝑠𝑥)3 𝜋 10 𝜋 +3(𝑐𝑜𝑠𝑥)3 is continuous on [0; ] 𝜋 We have ≤ cosx ≤ 1, for all x ϵ [0; ] 𝜋 Then: ≤ (𝑐𝑜𝑠𝑥)3 ≤ 1, for all x ϵ [0; ] 1 => ≤  𝜋 ≤ , for all x ϵ [0; ] +3(𝑐𝑜𝑠𝑥)3 𝜋 𝜋 1 = ∫02 𝑑𝑥 < ∫02 𝑑𝑥 16 +3(𝑐𝑜𝑠𝑥)3 𝜋 𝜋 < ∫0 𝑑𝑥 = 𝜋 10 Integral applications prove the inequality Example 12: Prove that: ln(1+a) > 2𝑎 1+𝑎 , for all a > 1 +) In the coordinate system Oxy, consider the graph of the function (C): y = and set 𝑥0 = 𝑥 1+𝑎 ϵ [1;a] Let the points A, B, C, D be A(1; 0), B(1+a; 0), C(𝑥0 ; 0), D(𝑥0 ; ) 𝑥0 We have: y’ = −1 𝑥2 => y’’ = −(−2𝑥) (𝑥 )2 = > 0, for all x > 𝑥3 Then: y = is a concave function, for all x > 𝑥 Two lines x = 1, x = 1+a intersect the tangent (d) at D(𝑥0 ; ) at two points E, F and 𝑥0 intersect the graph (C): y = 1+𝑎 Then:𝑆𝐴𝐵𝐺𝐻 = ∫1 𝑥 𝑥 at two points H, G 𝑑𝑥 = ln (1 + 𝑎) +) We have: CD = √(𝑥0 − 𝑥0 )2 + ( But 𝑥0 = 1+𝑎 => CD = Therefore, 𝑆𝐴𝐵𝐺𝐻 = 𝐴𝐵 1+𝑎 = 𝑥0 − 0)2 = 𝑥0 1+𝑎 𝐴𝐹+𝐵𝐸 (6) = 𝐴𝐵 𝐶𝐷 = 𝑎 1+𝑎 = 2𝑎 1+𝑎 From (6) and (7) with 𝑆𝐴𝐵𝐺𝐻 > 𝑆𝐴𝐵𝐸𝐹 , we have ln(1+a) > (7) 2𝑎 1+𝑎 , for all a > Conclusion Essay content"Integral Applications”Including methods of calculating primitives, definite integrals and some applications of definite integrals, the essay has achieved some important results: + The essay has demarcated and presented the method of calculating the area of a plane figure, the length arc in geometrical applications of integrals From there, apply integration to practical problems and solve some common problems such as proving inequality + Is aimed at raising the reader's awareness of the integral and its application in mathematics Through the above analysis, we will have an overview and a more positive view of integrals, about applications that are very close to reality But in any field, two methods total integral and differential diagram are always two directions for easy access and use in practical problems Today, calculus - integration requires learners and researchers to have a deep understanding of the nature so that they can come up with new and more practical applications, not just stopping at things that the group has already learned tell In the process of implementation, the team has tried to limit errors but still cannot avoid mistakes, so the essay is still incomplete and profound The student group hopes to receive more comments from readers Finally, the group would like to express their sincere thanks to Mr Nguyen Van Thin's support during the process of writing the essay  ... April 12, 1964 and became an important historical milestone for all of humanity Application of definite integrals I Geometrical applications of definite integrals General diagram using integral A... reader's awareness of the integral and its application in mathematics Through the above analysis, we will have an overview and a more positive view of integrals, about applications that are very close... a√2 ∫0