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 : 3 Class: Mathematical analysis I

June, 2022

TABLE OF CONTENTS

INTRODUCTION OF PRINCIPLES – ANALYSIS 5

ANALYSIS METHODS 6

Primitives of some basic functions

Integral Calculation Methods

Sub-conclusion

FEATURES ON THE BIRTH AND DEVELOPMENT OF THE THEORY OF ANALYSIS10

1 HISTORY ANALYSIS

2 ANALYSIS APPLICATION

6

7

9

10

11

APPLICATIONS OF DEFINITION ANALYSIS………12

1 GEOLOGICAL APPLICATIONS OF DEFINITIONAL ANALYSIS 12

A General diagram using integrals

B Area of flat figure

C Bow length

12

13

18

2 Algebraic Applications of DEFINITIONAL ANALYSIS 20

A Integral inequality 20

B Integral applications prove the inequality 20

CONCLUSION……….21

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)

∫ 𝒇(𝒙)𝒅𝒙 = F(x) + C (C = const)

1 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

B Nature

2 Definite integral

A Definition

➢ Consider the function f(x) on a closed interval [a,b] and divide [a,b] into n points: a= x

0<x

1<x

2<…<x

n=b, then we take a number ɛ

i ∈ [xi–1, xi], for i = 1, 2,…, n

We have the Riemann sum:

S

n=∑𝑛𝑖=1 = (x

i–x

i-1).f(ɛ

i) = (x

1–x

0).f(ɛ

1) + (x

2–x

1).f(ɛ

2) +…+ (x

n–x

n-1).f(ɛ

n)

If max |x

i − x

i–1| → 0, then Sn → α,

1 ≤ i ≤ n

that means f(x) is an integrable function and α is called integral of f(x) on [a,b], or

∫ 𝑓(𝑥)𝑑𝑥𝑎𝑏 = α

➢ If a function f(x) is continuous on [a,b], then f(x) is integrable on [a,b]

The outline of primitives - integrals

∫𝒂𝒃𝒇(𝒙)𝒅𝒙 = F(b) – F(a) ≡ [𝑭(𝒙)]|𝒃

𝒂

➢ Instead of the above definition, we can also define an integral by Newton – Leibnitz formula:

B Properties

1 ∫ 𝑓(𝑥)𝑑𝑥𝑎𝑎 = 0

2 ∫ 𝑓(𝑥)𝑑𝑥𝑎𝑏 = -∫ 𝑓(𝑥)𝑑𝑥𝑏𝑎

3 If f(x) ≥ 0, ∀ x ∈ [a,b], then ∫ 𝑓(𝑥)𝑑𝑥𝑎𝑏 ≥ 0

4 If f(x) ≥ g(x), ∀ x ∈ [a,b], then∫ 𝑓(𝑥)𝑑𝑥𝑎𝑏 ≥ ∫ 𝑔(𝑥)𝑑𝑥𝑎𝑏

From that reduced |∫ 𝑓(𝑥)𝑑𝑥𝑎𝑏 | ≤ ∫ |𝑓(𝑥)|𝑑𝑥𝑎𝑏

5 ∫ 𝑓(𝑥)𝑑𝑥𝑎𝑏 = ∫ 𝑓(𝑥)𝑑𝑥𝑎𝑐 + ∫ 𝑓(𝑥)𝑑𝑥𝑐𝑏 (a≤ b≤ c)

3 Integral method

A The primitive of a basic function

i Elementary function

ii Inverse trigonometric function

1) ∫ 𝒙𝒂ⅆ𝒙 = 𝒙𝒂+𝟏

𝒂+𝟏 + 𝑪(𝒂 ≠ 𝟏) 5) ∫ 𝒂𝒙ⅆ𝒙 = 𝒂𝒙

𝒍𝒏 𝒂+ 𝑪

2) ∫ 𝟏

𝒙ⅆ𝒙 = 𝒍𝒏|𝒙| + 𝑪 6) ∫ 𝑐𝑜𝑠 𝑥 ⅆ𝑥 = 𝑠𝑖𝑛 𝑥 + 𝑐

3) ∫ 𝟏

√𝒙ⅆ𝒙 = 𝟐√𝒙 + 𝑪 7) ∫ 𝑠𝑖𝑛 𝑥 ⅆ𝑥 = − 𝑐𝑜𝑠 𝑥 + 𝑐

4) ∫ ⅇ𝒙ⅆ𝒙 = ⅇ𝒙+ 𝑪 8) ∫ (1 + 𝑡𝑎𝑛2𝑥) ⅆ𝑥 = ∫ 1

𝑐𝑜𝑠 2 𝑥 ⅆ𝑥 = 𝑡𝑎𝑛𝑥 + 𝐶

9) ∫ (1 + 𝑐𝑜𝑡2𝑥) ⅆ𝑥 = ∫ 1

𝑠𝑖𝑛2𝑥 ⅆ𝑥 = −𝑐𝑜𝑡𝑥 + 𝐶

1) ∫𝑥2𝑑𝑥+𝑎2 = 1

𝑎arctan (𝑥

𝑎) + 𝑐, arctan(𝑥

𝑎) ϵ (−𝜋

2 ;𝜋

2)

2) ∫𝑥2𝑑𝑥+𝑎2 = 1

𝑎arccot (𝑥

𝑎) + 𝑐, arccot(𝑥

𝑎) ϵ (0; 𝜋)

3) ∫√𝑎𝑑𝑥2−𝑥2 = 𝑎𝑟𝑐𝑠𝑖𝑛 (𝑥

𝑎) + 𝑐, arcsin(𝑥

𝑎) ϵ [−𝜋

2 ;𝜋

2]

4) ∫ 𝑑𝑥

√𝑎 2 −𝑥 2 = −𝑎𝑟𝑐𝑐𝑜𝑠 (𝑥

𝑎) + 𝑐, arccos(𝑥

𝑎) ϵ [0; 𝜋]

iii Common special functions

1) ∫ ⅆ𝒙

√𝒂+𝒙 𝟐 = 𝒍𝒏 𝒙 + √𝒂 + 𝒙𝟐+ 𝑪

2) ∫ ⅆ𝒙

𝒂𝟐−𝒙𝟐 = 𝟏

𝟐𝒂𝒍𝒏 |𝒂+𝒙

𝒂−𝒙| + 𝑪

3) ∫ √𝒂𝟐 − 𝒙𝟐ⅆ𝒙 = 𝟏

𝟐[𝒙√𝒂𝟐− 𝒙𝟐+ 𝒂𝟐𝐚𝐫𝐜𝐬𝐢𝐧 (𝒙

𝒂)] + 𝑪

4) ∫ √𝒙𝟐± 𝒂𝟐ⅆ𝒙 =𝟏

𝟐[𝒙√𝒙𝟐 ± 𝒂𝟐+ 𝒂𝟐𝐥𝐧 | 𝒙√𝒙𝟐± 𝒂𝟐 |] + 𝑪

5) ∫ 𝒙 ⅆ𝒙

(𝒙 𝟐 +𝒂𝟐)𝟑𝟐

𝒂 𝟐 √𝒙 𝟐 +𝒂 𝟐 + 𝑪

6) ∫ ⅆ𝒙

(𝒙 𝟐 +𝒂 𝟐 )𝟑𝟐

𝒂 𝟐 √𝒙 𝟐 +𝒂 𝟐 + 𝑪

B Methods of integral calculation

- As learned in high school, to determine an integral expression we usually Two methods are used: the transform method and the partial integration method Use fluently the above two methods, combined with mathematical tricks such as homogenizing coefficients, union

multiplication, trigonometric transformation We can easily solve all fundamental integrals copy Above all, for definite integral problems, based on the two bounds of the integral we can deduce guess the solution of that problem

-Indeed, we can consider examples:

Example 1: 𝐼 = ∫ 𝑥2√1 − 𝑥2

√3 2

√2 2

𝑑𝑥 Solution:

set 𝑐𝑜𝑥𝑡 = 𝑥 (𝑡 ∈ [−𝜋

2 ,𝜋

2])

→ 𝑐𝑜𝑠𝑡𝑑𝑡 = 𝑑𝑥 𝑎𝑛𝑑 𝑐ℎ𝑎𝑛𝑔𝑒 𝑡ℎ𝑒 𝑏𝑜𝑢𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒:

𝒙 √𝟐

𝟐

√𝟑 𝟐

𝒕 = 𝐚𝐫𝐜𝐬𝐢𝐧(𝒙) 𝜋

4

𝜋 3

So

I= ∫ sin2𝑡√ 1 − 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡𝑑𝑡 = ∫ sin2𝑡|𝑐𝑜𝑠𝑡| 𝑐𝑜𝑠𝑡𝑑𝑡

𝜋 3 𝜋 4

𝜋

3

𝜋

4

= 1

4∫ sin

22𝑡𝑑𝑡 (𝑠𝑖𝑛𝑐𝑒 𝑡 ∈ [0,𝜋

2] 𝑠𝑜 𝑐𝑜𝑠𝑡 > 0 => |cos 𝑡| = 𝑐𝑜𝑠𝑡 )

𝜋

3

𝜋

4

= 1

8 ∫ (1 − 𝑐𝑜𝑠4𝑡)𝑑𝑡 =

1 8

𝜋

3

𝜋

4

(𝑡 +𝑠𝑖𝑛4𝑡

4 )|𝜋

4

𝜋 3

= 𝜋

96+

√3

64 Example 2: 𝐼 = ∫ 𝑠𝑖𝑛√𝑥𝑑𝑥0𝜋2

Solution:

𝑆𝑒𝑡 𝑡 = √𝑥 → 𝑑𝑡 = 1

2√𝑥𝑑𝑥 =

1 2𝑡𝑑𝑥

→ 𝑑𝑥 = 2𝑡𝑑𝑡 𝑎𝑛𝑑 𝑐ℎ𝑎𝑛𝑔𝑒 𝑡ℎ𝑒 𝑏𝑜𝑢𝑛𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒:

=> 𝐼 = 2 ∫ 𝑡 𝑠𝑖𝑛𝑡𝑑𝑡 𝑆𝑒𝑡 { 𝑢 = 𝑡 → 𝑑𝑢 = 𝑑𝑡

𝑑𝑣 = 𝑠𝑖𝑛𝑡𝑑𝑡 𝑐ℎ𝑜𝑜𝑠𝑒 𝑣 = − cos 𝑡

𝜋 0

𝑆𝑜 𝐼 = 2 [(𝑡 𝑐𝑜𝑠 𝑡 )|0𝜋 − ∫ (− cos 𝑡 ) 𝑑𝑡

𝜋 0

] = 2𝜋 + (𝑠𝑖𝑛𝑡 )|0𝜋 = 2𝜋

Example 3: 𝐼 = ∫ sin 𝑥−𝑐𝑜𝑠𝑥

√1+sin 2𝑥

𝜋 2 𝜋 4

𝑑𝑥

we have:

𝐼 = ∫ sin 𝑥 − 𝑐𝑜𝑠𝑥

√1 + sin 2𝑥 𝑑𝑥 = ∫

𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠𝑥

√(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)2𝑑𝑥 = ∫ sin 𝑥 − cos 𝑥

|𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥|𝑑𝑥

𝜋 2 𝜋 4

𝜋 2 𝜋 4

𝜋 2 𝜋 4

𝐵𝑢𝑡 𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 = √2 sin (𝑥 +𝜋

4) 𝑤𝑖𝑡ℎ

−𝜋

4 ≤ 𝑥 ≤

𝜋

2( ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠)  𝜋

2 ≤ 𝑥 +

𝜋

4 ≤

3𝜋 4

sin (𝑥 +𝜋

4) > 0; cos (𝑥 +

𝜋

4 ) < 0 𝑎𝑛𝑑 |sin( 𝑥 +

𝜋

4 )| > |cos ( 𝑥 +

𝜋

4 )|

Therefore: |𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠𝑥 | = sin 𝑥 + cos 𝑥 𝑤𝑖𝑡ℎ 𝜋

2 ≤ 𝑥 +𝜋

4 ≤ 3𝜋

4

We get:

(𝑐𝑜𝑠𝑥 − sin 𝑥 )𝑑𝑥 − 𝑑(𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠𝑥 )

So : 𝐼 = ∫ −𝑑(sin 𝑥+𝑐𝑜𝑠𝑥) sin 𝑥+cos 𝑥

𝜋

2

𝜋

4

= −𝑙𝑛|𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥 ||

𝜋 2 𝜋 4

= −[𝑙𝑛1 − 𝑙𝑛√2] = 12ln 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:

1 (Acosx ∓ Bsinx)dx = d(Asinx ± Bcosx + C)

2 (A ∓ B)sin2xdx = d(Asin2x ± Bcos2x + C)

3 −sin4xdx = d(sin4x + cos4x + C)

1 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!

2 Integral application

Some details on the birth and development of integral theory

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

I Geometrical applications of definite integrals

1 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:

𝑆 = ∫ 𝑓(𝑥)𝑑𝑥

𝑏

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

VD

1 : Consider the area of the plane bounded by: horizontal axis, y = f(x) = cosx + 2 and two lines x = 0; x = 9.5

Application of definite integrals

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

Total area: 20 trapezoids

Integral value:18.92 Total area: 100 trapezoids

If the function y = f(x) is continuous on [a,

b] then the area S of a curved trapezoid

bounded by

{ (𝐶): 𝑦 = 𝑓(𝑥), ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑥𝑖𝑠

𝑥 = 𝑎, 𝑥 = 𝑏 𝑖𝑠

𝑆 = ∫ |𝑓(𝑥)|𝑑𝑥

𝑏 𝑎

y

(C): y = f(x)

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:

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 − 2

Equation of the intersection point:

1, √𝑥 = 𝑥 − 2  { 𝑥 ≥ 2

𝑥2− 5𝑥 + 4 = 0  { 𝑥 ≥ 2

[ 𝑥 = 1(𝑐𝑎𝑡𝑒𝑔𝑜𝑟𝑦)

𝑥 = 4(𝑟𝑒𝑐𝑒𝑖𝑣𝑒)

(𝐶) ∩ (𝐷) = 𝐵(4; 2)

2, √𝑥 = 0  𝑥 = 0 (𝐶) ∩ (𝑂𝑥) = 𝑂(0; 0)

3, 𝑥 − 2 = 0  𝑥 = 2  (𝐷) ∩ (𝑂𝑥) = 𝐴(2; 0)

We can easily see the area of the plane figure (H):

(H) = 𝑆(𝑂𝐴𝐵)= (𝐻1) + (𝐻2) = ∫ (√𝑥 − 0)𝑑𝑥02 +

∫ [√𝑥 − (𝑥 − 2)𝑑𝑥]24

= ∫ √𝑥𝑑𝑥 + ∫ (√𝑥 − 𝑥 + 2)𝑑𝑥 =02 24

(2

3𝑥√𝑥)|2

0+ (

2

3𝑥√𝑥 −𝑥22+ 2𝑥)|4

2=

10

3

2 Area of a plane bounded by a line with parametric equations

If curve (C): y = f(x) has parametric equation {𝑥 = 𝜑 (𝑡)

𝑦 = 𝜔(𝑡)

In the formula to calculate the area 𝑆 = ∫ |𝑓(𝑥)| 𝑑𝑥𝑎𝑏 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 =∫𝑆𝛼𝛽|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 {𝑥 = 𝜑 (𝑡)

𝑦 = 𝜔(𝑡)with 0 ≤ t ≤ T where T is the periodic period of it

Then: S = 𝟏

𝟐 ∫ [𝝎(𝒕)𝝋𝟎𝑻 ′(𝒕) − 𝝋(𝒕)𝝎′(𝒕)] 𝒅𝒕 Example 5 : Calculate the area of an ellipse limited by (E):𝑥2

𝑎 2+𝑦2

𝑏 2 = 1

We have the parametric equation of (E) consider the area of (E) in the first quadrant on the plane (Oxy)

(𝐸): {𝑥 = 𝑎 cos 𝑡𝑦 = 𝑏 sin 𝑡 ; 0 ≤ 𝑡 ≤ 2𝜋

Converting the integral we get:

{0 = 𝑎𝑐𝑜𝑠𝑡

𝑎 = 𝑎𝑐𝑜𝑠𝑡⇒ {

𝑡 =𝜋

2

𝑡 = 0

So

𝑆 = 4𝑆1= 4 ∫ 𝑏 𝑠𝑖𝑛 𝑡(− 𝑎𝑠𝑖𝑛 𝑡) 𝑑𝑡

0 𝜋 2

4𝑎𝑏 ∫ sin 2 𝑡 𝑑𝑡

𝜋 2 0

= 4𝑎𝑏 ∫ 1 − 𝑐𝑜𝑠2𝑡

2

𝜋 2 0

𝑑𝑡 = 𝜋𝑎𝑏

3 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 > 0 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 2 points A and B as shown

in figure 3 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 formula to calculate the curve length limited by { (𝑆): 𝑦 = 𝑓(𝑥)

𝑥 = 𝑎 𝑎𝑛𝑑 𝑥 = 𝑏 is L

Then: L= ∫ √[𝒇’(𝒙)]𝒂𝒃 𝟐+ 𝟏𝒅𝒙

Example 7: Calculate the length of the arc OA lying on

the parapol y= 𝑥

2

2𝑎 with a≠0, which O is the origin, A is a point on the parabol whose latitude is t

We have: A(t,t

2

2a) and y’(x) = 2𝑥

2𝑎 =𝑥

𝑎

Therefore L = ∫ √[𝑦0𝑡 ′(𝑥)]2+ 1𝑑𝑥 = ∫ √(0𝑡 𝑥𝑎)2+ 1𝑑𝑥

= 1

𝑎√𝑥2+ 𝑎2𝑑𝑥 = 1

2𝑎[x√𝑥2+ 𝑎2 + 𝑎2ln|𝑥 + √𝑥2+ 𝑎2|]|0𝑡

= 𝑡

2𝑎√𝑡2+ 𝑎2 +𝑎

2ln (𝑡+√𝑡2+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)

From then, instead of (5) we get: l = ∫ √[𝑥𝑏 ′(𝑡)]2+ [𝑦′(𝑡)]2

= ∫ √[𝑥𝑏 ′(𝑡)]2+ [𝑦′(𝑡)]2+ [𝑧′(𝑡)]2

Example 8: Calculate the length of the arc of Archimedes’spiral

{

𝑥 = 𝑎𝑐𝑜𝑠𝑡

𝑦 = 𝑎𝑠𝑖𝑛𝑡 𝑤𝑖𝑡ℎ 0 ≤ 𝑡 ≤ 2𝜋 𝑎𝑛𝑑 𝑎 > 0

𝑧 = 𝑎𝑡

We have: x’(t) = -asint, y’(t) = acost, z’(t) = a

Therefore L = ∫2𝜋√[𝑥′(𝑡)]2+ [𝑦′(𝑡)]2+ [𝑧′(𝑡)]2

= ∫2𝜋√(−𝑎𝑠𝑖𝑛𝑡)2+ (𝑎𝑐𝑜𝑠𝑡)2+ 𝑎2

= a∫02𝜋√𝑠𝑖𝑛𝑡2+ 𝑐𝑜𝑠𝑡2+ 1𝑑𝑡

= a√2 ∫02𝜋𝑑𝑡 = a√2 |2𝜋0 = 2𝜋a√2

II ALGEBRAIC APPLICATIONS OF DEFINITE INTEGRAL

1 Integral inequality

We evalute the inequality in terms of functions and integrals