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.
PRIMITIVE
Primitive functions
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 2: � kf x dc k f x dx ( ) � ( ) ( k is a constant, k � 0 )
Basic primitive table
( cos ) sin cos ax ax e a bx b bx e bxdx C a b
� 11 � cos( ax b dx ) 1 a sin( ax b ) C
� 12 � cos ( 2 1 ax b ) dx 1 a tan( ax b ) C
4 sin( ax b dx ) 1 cos( ax b ) C
� 13 � tan( ax b dx ) 1 a ln cos( ax b ) C
1 1 cot( ) sin ( ) dx ax b C ax b a
6 cot( ax b dx ) 1 ln sin( ax b ) C
9. ln( ) b ln( ) ax b dx x ax b x C a
( sin ) cos sin ax ax e a bx b bx e bxdx C a b
DEFINITE AND INDEFINITE INTERGRALS
INDEFINITE INTEGRALS
1 Some basic concepts and properties :
Definition : Given a function f x ( ) defined on the interval a b , 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
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 :
1 Every function f x ( ) that is continuous on the interval a b , has a primitive on this interval.
4 � af x ( ) bg x dx a f x dx b g x dx C ( ) � ( ) � ( ) with a b , ��
Example 2 Find the interval : � sin 2 cos 3 x xdx
1 1 1 sin 2 cos3 sin 5 sin cos cos 5
2 Methods for Calculating Uncertainty Integral :
Consider the integral � f x dx F x ( ) ( ) C Then, with the discriminant function
Case 1 : Consider the integral I � f x dx ( ) Set t u x ( ) , if we get
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 1 Find the interval : � e sin x cos xdx
Solution sin x cos sin x sin sin x e xdx e d x e C
Example 2 Find the interval : ln 2 x x dx
Example 3 Find the interval : 4 1 x dx x
Example 4 Find the interval : e x x dx
Example 5 Find the interval : � x 2 x 3 dx
Case 2 : Consider the integral I � f x dx ( ) Set x x t ( ) , hence:
In mathematical analysis, when dealing with a function \( x = t(x) \) that possesses a continuous derivative and an inverse function, it's essential to choose a substitution that simplifies the integral expression For instance, if the function \( f(x) \) includes a term like \( 2 - x^2 \), one effective substitution could be \( x = a \sin(t) \) This approach ensures that the modified integral is more straightforward to calculate compared to the original expression, facilitating easier evaluation of the integral.
1 x a , we need to set x a tan t
Example 1 Find the interval : � 1 dx x 2 3 2
1 dx x 2 3 2 cos dt 2 t tan t C t , arcsin x
Example 2 Find the interval : � x 2 1 4 2 dx
x 2 1 4 2 dx 1 8 cos 2 tdt 16 1 � � � t 1 2 sin 2 t � � � C t , arctan 2 x
With two differentiable functions u x v x ( ), ( ) , we have uv ' uv ' vu ' From that, we infer:
The method described involves integrating only one factor of the complex expression f(x), rather than the entire expression In the formula analyzed, f(x) is represented as u(x)v'(x), and the focus is on integrating the second factor This approach aims to simplify the resulting integral, ensuring that it is easier to work with than the original integral expression.
* Integral types calculated by the method of partial integration:
Integral form How to apply the integral part formula
� Let e ax into in d (.) , apply (2.1) n times
� Let sin( ax b ) into in d (.) , apply (2.1) n times
� Let cos( ax b ) into in d (.) , apply (2.1) n times
� Let P x n ( ) into in d (.) , apply (2.1) once e cos( ax ax b dx )
� Put one of the two factors in d (.) , apply (2.1) twice e cos( ax ax b dx )
� Put one of the two factors in d (.) , apply (2.1) twice
Example 1 Find the interval : � ln xdx
Solution. ln xdx x ln x 1 dx x ln x x C
Example 2 Find the interval : � arctan xdx
Example 3 Find the interval : � e x x 2 2 x 3 dx
Applying the integral by parts formula twice, we have:
Example 4 Find the interval : I � arccos 2 xdx
1 arccos 2 arccos 1 arccos 2 1 arccos 2 xar x
3 Integral of rational fraction functions
In this category, we consider integrals of the following form:
In the context of rational integrals, we consider polynomials P and Q of degrees n and m, respectively, where n is less than m If n is greater than or equal to m, we can simplify the problem by dividing the numerator by the denominator to obtain a remainder The primary approach for calculating rational integrals involves decomposing the integral's expression into a sum of simpler rational integrals, which can be categorized into two types of simplified fractions.
When a 2 4 b 0 (i.e it is not possible to decompose the quadratic trinomial x 2 ax b into a monomial ) , p q , ��
Applying transformation 1 , we easily get:
The second integral can be reduced to the form:
The final integral can be calculated by applying the lower order formula to the cos function.
Example 2 Find the interval : 2 2 2 dx x x
Example 3 Find the interval: 2 3 2 dx x x
Identifying the coefficients of x on both sides, we get the system of equations:
Solving the above system of equations, we get: A 2, B 1, C 2, D 1
* The function under the integral is even/odd for sin/cos :
Even for both sin x and cos x
Set t cos x Set t sin x Set t tan x or applying the downgrade formula
Table 3 : How to solve for even/odd function case for sin/cos
Example 1 Find the interval : I � cos 3 x sin 8 xdx
2sin 3cos sin cos 9cos x x dx
Dividing both numerator and denominator by cos 3 x and set t tan x , we have:
* The function under the integral sign is a rational function for sin and cos:
to return the integral of the rational analytic function
Example 1 Find the interval : 3sin 4 cos 5
Analysis: 2sin x cos x 3 A 3sin x 4cos x 5 B 3sin x 4 cos x 5 ' C
That implies , we have a system of equations:
Solution the system of equations , we get that : A 2 5, B 1 5, C 1
� � � where I 1 is the integral in the previous example
5 Integral of some irrational functions
In this section, we consider integrals containing roots In the program, we encounter mainly the following two types of integrals:
In this case, we set m ax b t cx d
, represent x in terms of t and then substitute the original integral to return to the rational function integral form.
Set 2 x 1 t t 4 , 0 This implies that 2 dx 4 t dt 3 We have:
Form 2 : I � R x ax , 2 bx c dx , when R , is a rational function
In this case, we find a way to return to one of the basic formulas shown in Table 1 or convert to integral trigonometric functions.
Example 1 Find the interval : 2 2 dx x x
Example 2 Find the interval : 2 2 2 dx x x
DEFINITE INTEGRALS
1 Definition, geometric meaning of definite integral
To find the area of a curved trapezoid defined by the function f(x) over the interval [a, b], we must consider the region bounded by the graph of f(x), the vertical lines x = a and x = b, and the x-axis Since there is no explicit formula for calculating the area of a curved trapezoid, we can approximate it by dividing the interval [a, b] into n segments We denote these segments with points x₀, x₁, x₂, , xₙ, where x₀ = a and xₙ = b For each segment, we select a representative point ξᵢ and construct a rectangle with height f(ξᵢ) The total area can then be approximated by summing the areas of these rectangles.
To calculate the area under a curve defined by a function \( f \) over the interval \([a, b]\), we set \( \Delta = \max\{\Delta_i : 1 \leq i \leq n\} \) The sum \( \sum_{i=1}^{n} f(\xi_i) \Delta_i \) represents the total of the integrals of the function across the segment This sum is influenced not only by the function \( f \) itself but also by how the interval \([a, b]\) is partitioned and the choice of representative points \( \xi_i \) As the lengths of these divisions decrease towards zero, the total area of the rectangles approaches the actual area of the curved trapezoid beneath the curve.
A function \( f(x) \) is considered integrable over the interval \([a, b]\) if the limit \( \lim_{\Delta x \to 0} \sum_{i=1}^{n} f(\xi_i) \Delta x_i \) exists, regardless of how the interval is divided or how the representative points \( \xi_i \) are selected The value of this limit is referred to as the definite integral of the function \( f(x) \) over the interval \([a, b]\), and is denoted as \( \int_{a}^{b} f(x) \, dx \).
Example Consider the integrability of the Dirichlet function, given by the formula
Divide segment 0,1 into n segments of equal length, on the i-th division, choose two representative points i �� and i �� � \ The respective sums of integrals are
The limits of two integral sums are different, depending on how the representative points are chosen So function f x ( ) is not integrable
In the problem of calculating the area of a curved trapezoid, we have divided segment a b , into segments of very small length and considered the function
To approximate the area of a curved trapezoid, we can sum the areas of rectangles, which is a method we will also apply to the path length problem When an object moves in a straight line with a velocity represented by v(t) at time t, we can calculate the length of its path between two points in time, t1 and t2 By dividing the interval [t1, t2] into very small segments of length Δi, we assume the object moves uniformly within each segment The distance traveled by the object during each small time interval Δi is given by f(ξi)Δi, where ξi is a point within that interval Consequently, the total distance covered by the object over the interval [t1, t2] is precisely represented by the integral of the velocity function.
2 Integralability criteria Properties of definite integrals.
Theorem 2.1 A function f x ( ) defined on a segment a b , will be integrable on that segment if one of the following conditions is satisfied:
The function f x ( ) is continuous on the interval a b ,
The function f x ( ) is bounded and has a finite number of discontinuities on
The function f x ( ) is monotonous and bounded on the interval a b ,
The function f x ( ) x 2 is continuous so it is integrable on the interval 0,1
The limit of the sum of the integrals remains constant regardless of how the segment [0, 1] is divided or which representative points are selected within each division This implies that we can divide the segment [0, 1] into n equal segments, each of equal length.
1 n On each division, we choose a representative point that coincides with the right end of that segment, i.e
Then, the total integration is obtained to be
Theorem 2.2 Let f x g x ( ), ( ) be two integrable functions on the interval a b ,
4 The function f x ( ) is also integrable on the interval a b , and
5 If f x ( ) is an even function and integrable over the interval a a , , then
6 If f x ( ) is an odd function and integrable over the interval a a , , then
7 If f x ( ) is a periodic function with period T and integrable over every segment in � , then
8 (First integral mean theorem) If m � f x ( ) � M , x � a b , , then
In particular, if f x ( ) is continuous on the segment a b , , then there exists
9 (Second integral mean theorem) Suppose
( ) ( ) ii g x does not change sign on a b ,
In particular, if f x ( ) is continuous on a b , , then there exists c � a b , such that
Since the function under the integral sign is a periodic function with period
On the other hand, the function under the integral sign is an odd function, the integral is signed on the symmetry segment ; , so we have I 0
Example 3 Determine the sign of the integral
I dx dx x x x y dx dy set y x x y x x dx dx x x x dx x x
3 Derivative formula with bounds Formula Newton-Leibniz
Currently, there appears to be no direct connection between indefinite and definite integrals Calculating a definite integral involves determining the limit of the sum of integrals, a process that can be complex and challenging In this section, we will present a formula for evaluating the definite integral of a function using its primitive.
Theorem 3.1 Let the function f x ( ) be integrable on the segment a b , For all x � a b , , set F x ( ) : � a x f t dt ( )
1 If f x ( ) is integrable on the interval a b , then F x ( ) is continuous on that interval.
2 If f x ( ) is continuous on the segment a b , then F x ( ) is a primitive of f x ( ) , that is
At the same time we have:
The Newton-Leibniz formula, as presented in Equation (2.4), holds true for any primitive F(x) of the function f(x) Additionally, the extension of formula (2.3) applies when both bounds of the integral are dependent on the variable x.
Example 1 Calculating sin cos 1 4 x x d dt dx � t
Example 2 Calculate the limit: arcsin
0 arctan 0 0 arctan lim arcsin x x x tdt tdt
It is easy to see that the limit has the form
0 Applying L'Hospital's rule, we have arcsin /
2 0 arctan arctan(arcsin ) 1 lim lim 1 1. arcsin(arctan ) 1 arcsin 1 x x x x tdt x x tdt x x
To compute the limit, we will express it as the sum of the integrals of the function f(x) over the interval [0, 1] This limit is precisely equal to the definite integral of the function within that range.
� We divide the segment 0,1 into n segments, each of length
1 n On each division, choose a representative point i that coincides with the right end of the division, i.e i i
Then, the sum of the integrals of the function f x ( ) over the segment 0,1 is equal to
Compared with the expression to calculate the limit, we need to choose the function f x ( ) 1 x Then we have
The Newton-Leibniz formula connects primitive functions and definite integrals, allowing us to calculate the definite integral of a function by finding its primitive and substituting the appropriate bounds In the following section, we will explore methods for evaluating definite integrals, which directly stem from primitive function techniques and the Newton-Leibniz formula.
Suppose we need to integrate ( ) b a f x dx
Example 1 Calculating the integral � 0 1 1 x 2 3 dx
In certain situations, formula (2.5) can be utilized in reverse, allowing us to transform the integral expression on the right side into the format of the left side for calculation.
Example 2 Calculating the integral � 0 1 2 x 3 ( 5 x 2 x 1) dx
Similar to the part in the calculation of primitives, suppose that we need to calculate the integral of the form ( ) b a f x dx
� We decompose f x ( ) into two factors
( ) ( ) ( ) f x u x w x , then integrate the second factor, put the integral to be calculated into the form ( ) ( ) b a u x dv x
I x dx x x x x dx x x x x x x x dx x t tdt set x t t t dt t t
When calculating a definite integral, it is essential to utilize the special properties of the integral, particularly the concepts of even and odd functions, to simplify the process and achieve accurate results.
Example 3 Calculating integrals a � a a ln x x 2 1 cos xdx b. sin 2 a a x dx x
The integrals presented involve functions with complex primitives; however, it can be confidently stated that these integrals equal zero due to the odd nature of the functions within the integral sign, especially since the integration is performed over a symmetric interval.
a a ; Notice that, in the second integral, although the function is undefined at
0, it is bounded in all neighborhoods of this point and thus the function is still integrable on the interval a a ;
Example 4. a Let f x ( ) be a continuous function on the interval 0,1 Prove that:
, it is easy to prove the equality in sentence a Apply this result to sentence b We have
0 0 sin cos sin cos sin cos 4 x x dx dx x x x x
Example 5 a Let f x ( ) be continuous, even on a a ; Prove
( ) a a x x x a a a x a y a a y x y a f x dx f x dx f x dx b b b f x dx f y dy set y x b b f x dx b f y dy because f is the even function b b f x x
APPLICATIONS OF DEFINITE INTEGRALS
As shown in Section 2.1, the area of a curved trapezoid D is limited by the graph y f x ( ) , the line x a x b , , and the axis Ox and ( ) ( ) b
( ) 0 f x for all x � a b , The construction of the formula for calculating the area of this curved trapezoid can be done through a two-step integral diagram:
Step 1 (Differential): Divide the trapezoid into "very small parts" by a slice perpendicular to the Ox axis For each point x � a b , , this is a rectangle with width dx and length f x ( )
Step 2 (Integral): The area of a curved trapezoid D is the sum of the areas of the "very small parts" above as x runs from a to b , i.e ( ) b
This section explores the practical applications of definite integrals in measurement, including the determination of a plane figure's area, the length of a curve, the volume of an object, and its surface area By utilizing integral diagrams, one can derive calculation formulas, akin to those used for finding the area of a curved trapezoid.
1 Compute the area of a plane figure.
Let f x g x ( ), ( ) are continuous functions on the closed interval a b , The following are the formulas for computing the area of a plane figure D in different cases.
Case 1: If D is bounded by the lines y f x y ( ), 0, x a x b , then
Case 2: If D is bounded by the lines y f x y 1 , f x x a x b 2 , , , then
Case 3: If D is limited by the lines x 1 ( ), y x 2 ( ), y y c y d , , then
Case 4: If D is limited by the lines x x t y , y t x a x b and y , , 0 , then
In which, it is further assumed that the equation x t a x t , b has a unique solution of t t 1 , 2 respectively.
Case 5: If D is limited by the lines r r , , , then
Case 6: If D is limited by the lines r r 1 , r r 2 , , , then
Example 1 Calculate the area of a plane bounded by the line y x 2 and the line y 3 x 2
The horizontal intersection of the line y x 2 and y 3 x 2 is the solution of the equation
Therefore, the area of the plane figure to be found is determined by the formula
Example 2 Calculate the area of the plane bounded by the lines x t sin t ,
1 cos y t with 0 � � t 2 and the line y 0
The area to be found is determined by the formula
Example 3 Calculate the area of a plane bounded by: a Line x y 3 and lines , 2 y x y x with y � 0 b Line x 2 y 2 y 4
S � y y dy � y y dy 4 b Note that if M x y , � C , then M ' � � � x y , C Therefore, the area to be calculated is S 4 ( ) S D , where D is the region bounded by the part of the curve
C corresponding to x y , � 0 Then, the domain D lies entirely in square 0 � � x 1 ,
0 � � y 1 , bounded by x y 2 y 4 with x 0 x 1 0 and has a maximum point in 0,1 Therefore, we can apply the formula
Example 4 Calculate the area of the plane bounded by the line given by the equation in the polar coordinate system as follows: a r 2sin b r 2sin 2 c r 2 2sin 2
In each case, survey and plot the curve in the polar coordinate system and comment on the symmetry of the figure a S 2 ( ) 2 S D 1 2 � 0 2 r 2 d � 0 2 4sin 2 d b S 8 ( ) 8 S D 1 2 � 0 4 r 2 d 4 � 0 4 4sin 2 2 d 2 c S 4 ( ) 4 S D 1 2 � 0 4 r 2 d 2 � 0 4 2sin 2 d 2
2 Calculate the length of the curve
The notation C a b , is a set of functions that are continuous on the interval
a b , , C a b 1 , is the set of functions with continuous derivatives on the interval
Given the functions f x ( ) � C a b 1 , , y � C c d x t 1 , , � C t t 1 1 , 2 , y t � C t t 1 1 , 2 and r � C 1 , Following are the formulas for calculating the length of curve C in different cases
Case 1: If C is given by the equation y f x a x b ( ), � � then
To set up the formula for calculating the curve length, we use the integral diagram as follows:
Step 1 (Differential): Divide the curve into "very small parts" The lengths of these "very small parts" are called the length differentials, denoted ds Then:
In the case y f x ( ) , we have dy f x dx '( ) Deduce ds 1 f x '( ) 2 dx
Step 2 (Integral): The curve length C is the sum of the areas of the above
"very small parts" as x runs from a to b , i.e s C � C ds Hence,
Case 2: If C is given by the equation x y c y d , � � , then
The calculus diagram is as follows:
Step 1 (Differential): Differential of length
2 2 1 '( ) 2 ds dx dy y dy , because dy '( ) y dy
Step 2 (Integral): Curve length C is s C � � C ds c d 1 '( ) y 2 dy
Case 3: If C for all parametric equations x x t y , y t t , 1 � � t t 2 then
The calculus diagram is as follows:
Step 1 (Differential): Differential of length
2 2 ' 2 ' 2 ds dx dy � � x t � � � � y t � � dt , because dx x t dt dy ' , y t dt '
Step 2 (Integral): Curve length C is
Case 4: If C is given by the equation in polar coordinates r r , � � then
To set up the formula for calculating the curve length, we use the integral diagram as follows:
Step 1 (Differential): We have x r cos and y r sin
Therefore, dx r sin r 'cos and dy r cos r ' sin From this, it can be deduced that the length differential is ds dx 2 dy 2 r 2 r ' 2 d
Step 2 (Integral): Curve length C is s C � � C ds r 2 r ' 2 d
Example 1 Calculate the curve length y cosh x with 0 � � x 1
Therefore, applying formula 2.12, we get
Example 2 Calculate the length of the curve sin
Solution Applying the formula (2.14), we have :
' cos , ' 1 sin , sin cos cos ln 3
Example 3 Calculate the length of the curve : r 1 cos
We have: r 2 r ' 2 1 cos 2 sin 2 2 2 cos 4cos 2 2
Therefore , applying the formula (2.15) , we have:
3 Calculate the volume of an object
To calculate the volume of an object D bounded by a curved surface and two planes at x = a and x = b, which are perpendicular to the coordinate axis Ox, we utilize a specific integral diagram to derive the formula for volume calculation.
To calculate the volume of an object using differential calculus, begin by dividing the object into very small cylindrical plates, each with a thickness of dx and a cross-sectional area S(x) at position x within the interval [a, b] The volume of these infinitesimal parts, referred to as volumetric differentials, is represented as dV According to the volume formula for a cylinder, the relationship is expressed as dV = S(x) dx.
Step 2: (integration): the volume of the object V is the sum of the volumes of the "very small parts" above, i.e V D � D dV Therefore, V D � a b S x dx
Case 1 : If D is defined by slices of perpendiculars to Ox and the cross- sectional area is determined at any point x where x � a b , is S x , then
Case 2 : If D is cut by slices perpendicular to Oy and the cross-sectional area is determined at any point y where y � c d , is S y , then
Example 1 Find the volume of a cone with base area B and height h
To construct a square cross-section along the Ox axis through the point (x, 0, 0), where 0 ≤ x ≤ h, we position O at the top of the pyramid In this configuration, the Ox axis represents both the height and the descent from the apex to the base Consequently, the area of the cross-section is directly proportional to the square of x multiplied by h.
Applying the formula (2.16), we have :
V S x dx dx Bh x dx Bh Bh h
Example 2 Calculate the volume of the object that is the common part of two cylinders defined by the equation x 2 y 2 a 2 and y 2 z 2 a 2 with a 0
To construct a segment perpendicular to the Oy axis through the point (0, 0, y), where -a ≤ y ≤ a, we define the cross-section S(y) using the inequality -a² - y² ≤ x ≤ a² - y² and -a² - y² ≤ z ≤ a² - y² This results in a square cross-section with a side length of 2√(a² - y²), leading to an area of S(y) = 4(a² - y²) Using formula (2.17), we can derive further results.
Case 3 : If D is a circular object created when rotating a curved trapezoid
( ) y f x , y 0, a x b � � about the axis Ox , where f � C a b , then :
In this case, the cross section perpendicular to Ox is circular with radius
( ) f x , so the cross-sectional area S x f 2 x Substituting S x into formula
(1) , we get formula (3) Similar from formula (2), we deduce formula (4) below.
Case 4 : If D is a rotating object formed when rotating a curved trapezoid
, 0, x y x c � � y d around the Oy axis, where � C c d , then
Case 5 : If D is a rotating object formed when rotating a curved trapezoid
( ) y f x , y 0,0 � � � a x b around the axis Oy , where f � C a b , then
In case 5, the curved trapezoid D is defined similarly to case 3, but the focus shifts to calculating the volume of the body around the Oy axis instead of the previous axis.
To determine the volume of the object, we divide it into concentric circular slices around the Oy axis, where each slice has a circumference of 2πx, a height of f(x), and a thickness of dx This leads to the differential volume formula dV = 2πx f(x) dx Consequently, this relationship gives rise to the derived formula (5), which is similarly applicable in case 6.
Case 6 : If D is a revolving body formed when rotating a curved trapezoid
, 0,0 x y x � � � c y d around the Ox axis, where � C c d , , then
Example 1 Calculate the volume of a sphere with radius R 0
A sphere is a special type of circular object that rotates when it rotates a half- circle defined by y R 2 x y 2 , 0, R x R � � around the Ox axis.
Applying the formula (3), we have:
To calculate the volume of the solid formed by rotating the area bounded by the lines \( y = 2 - 2x \) and \( y = 0 \), we consider two cases: a) the volume generated by one revolution around the x-axis and b) the volume generated by one revolution around the y-axis.
Solution. a) Applying the formula (3), we get:
V � x x dx b) Applying the formula (5), we get:
4 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 around the axis Ox , where f � C a b 1 , then the surface area S is calculated by the formula:
This formula originates from calculus, specifically the integral of the surface area, represented as dS = 2πf(x)ds Here, ds denotes the differential of the curve length, calculated using the formula ds = √(dx² + dy²) = √(1 + (f')²(x))dx.
2 ( ) 1 ' 2 dS f x f 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
, x y c � � y d around the axis Oy , where � C c d 1 , , then the surface area S is calculated by the formula:
Example 1 Calculate the area of a sphere with radius R 0
Solution A sphere is a special type of circle that rotates when it rotates half a circle defined by y R 2 x 2 , R x R � � around the Ox axis Applying the formula (7), we have:
Example 2 Calculate the area of the rotating circle created when turning the line y x 3 ,0 � � x 1 around the Ox axis.
Example 3 Calculate the area of the rotating circle created when rotating arctan ,0 1 y x � � x around the axis Oy
S y y dy t t dt set t y t t d t t s ds set s t s u du
IMPROPER INTEGRALS
IMPROPER INTEGRALS TYPE I
Suppose f x ( ) is a function on the interval [a, � ) and integrable over every finite interval [a, b], (a � b < �).
Definition 2.4.1 The limit of the integral ( ) b a f x dx
� when b � � is called the improper integrals type 1 of the function f x ( ) over the interval a , � and is denoted
If this limit exist we say the improper integrals a f x dx ( )
Otherwise, if this limit does not exist or the limit is infinity, we say that the integral diverges.
Similarly, we define the integral of a function f x ( ) over the interval � ,b and � � , by the following fomular:
These integral cases with close to infinity are called improper integrals type 1.
� � � 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:
Similarly, we have the extended Leibniz formula for the remaining cases of improper integrals type 1.
Using the transform method, set e x t � e dx dt x , we have
Example 2: Compute the integral 2 1 dx x
Example 3: Compute the integral 1 2 1 2 dx x x
Using the transform method, set tan cos 2 x t dx dt
4 cos 1 sin sin 2 1 1 dx tdt t t x x
Example 4: Consider the converges of integral
We have 0 x sin xdx x cos x sin x 0 � x lim � � x cos x sin x
� Because this limit does not exist, therefore the integral diverges.
IMPROPER INTEGRALS TYPE II
Assume f x ( ) is a definite function on the interval a b , and integrable over every segment in the wihth a t b and x b lim ( ) � f x � The point x b is called the in regular point of the function f x ( )
Definition The limit of the integral
� when t � b is called the improper integral type 2 of the function f x ( ) over the interval a b , and is denoted as follows:
An improper integral is said to converge if the limit exists finitely; conversely, if the limit does not exist or is infinite, the integral is considered to diverge.
Similarly, we define the improper integral of the function f x ( ) with no upper leg inervals a b , and a b , take x a and x b as outliers respectively.
These cases of integration with unbounded are called the improper integral type 2.
Note: Note that in the case of a function that is undefined at a but x lim ( ) � a f x has a finite existence then
For integrals with two outliers x a and x b , we can write:
.when two of the above three integrals converges.
Example 1: Consider of the integral converges I
� Because this limit exists when
1 a 0 it mean a 1 , if the integral I converges when a 1 Otherwise, when
So, the improper integral I converges when and only if a 1 , diverges when and only if a � 1
Using the transform method, set
Using the method of integration by parts, set arcsin , 1 2 u x dv xdx
CONVERGENCE STANDARDS
1 Assume f x and g x integrable over any finite interval a b , with b a and we have inequality :
Then, we have i) If a g x dx
2 Assume f x and g x are two positive functions integrable over all finite segments a b , with b a and x lim f x k 0 k g x
Definition 3.2 Consider improper integrals a f x dx
Let f and g are two positive function over a , � If f x and g x are VCB and VCL when x � � , and f x : g x when x � � then a f x dx
� � and either converges or diverges.
Definition 3.3 Let f and g are two positive function over a , � Then, we have
Similarly, we also have the convergence standards for the case of the improper integral of an unbounded function.
Consider the improper integrals b a f x dx
� have the same point anomaly are x a
1 Let the function f x and g x determined on a b , and we have inequality
2 Assume f x and g x are two positive functions integrable over all finite segments a b , with b a and x lim � a g x f x k 0 � k
Definition 3.5 Consider improper integrals a f x dx
Let f and g are two positive function over a b , 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 when x � a then a f x dx
� and either converges or diverges.
Definition 3.6 Let f and g are two positive function over a b , and have the same point anomaly are x a Then, we have
Note: When using the standard of comparison we usually compare the given improper integrals with the basic improper integrals: a) With 1
, converges if a 1 and diverges if a � 1 b) With a b ,
Example: Consider converges and compute the integrals: a)
Thus, the limit is not exist x lim sin � � x then intergral converges. c) x 2 dx 1 2
ABSOLUTELY CONVERGENT AND SEMI-CONVERGENT IMPROPER INTEGRALS
Definition 4.1 Let improper the integral a f x dx
� ( have the same point anomaly are a or b ) converges then
Let the improper integral a f x dx
� then we say that a f x dx
� absolutely converges and if a f x dx
� diverges then we say that a f x dx
Let the improper intergral b a f x dx
� ( have the same point anomaly are a or b ) Then:
� then we say that b a f x dx
� absolutely converges and if b a f x dx
� diverges then we say that b a f x dx
Example 1 : Cosider converges of integral 1 sin x 2cos x x x dx
� � converges then 1 sin x 2 cos x x x dx
� converges Therefore, the integral 1 sin x 2cos x x x dx
Example 2: Prove that the improper integral is semi-converges.
2 sin xdx sin xdx sin xdx x x x
� should the integral I 1 is definite integral, therefore just consider I 2
2 2 2 2 sin cos cos 1 cos 1 cos
� � converges So, we have I 2 converges.
We will prove that 0 sin x x dx
� diverges We have sin sin 2 1 cos 2
Thus, we get 0 sin x x dx
� diverges Adding, the integral 0 sin x x dx
� is definite integral 0 lim sin 0 x x x
The calculus module is a crucial component of the advanced math program at universities, introduced from the very first semester Integral calculus, in particular, holds significant importance due to its wide range of applications, including solving limit problems and calculating areas.
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