￼

Contents

Contents

1. Introduction

About This Book

Revision History

Book Summary : You Must Read This!

Chapter 2. Anti-differentiation

§2.1 Antiderivatives

● Definition : Antiderivatives

● Additive Constant (or Constant of Integration)

● Integration Variable

● Algebra of Antiderivatives

[Example 1] Find the indefinite integral

[Example 2] Find the indefinite integral

[Example 3] Find the antiderivative

§2.2 Definite Integral

● Riemann Sum

● Integrability

● Definite Integral

● Fundamental Theorem of Calculus

● Properties of Definite Integral

[Example 1] Find the derivative of integral function

[Example 2] Find the function

[Example 3] Infinite Series and Definite Integral

[Example 4] Properties of definite integral

[Example 5] Solve the following problems

§2.3 Geometric Interpretation of Integration

Chapter 3. Integration Rules

§3.1 Integration Formula

● Integration of Power Functions

● Integration of Exponential Functions

● Integration of Trigonometric Functions

● Integration related to Inverse Trigonometric Functions

● Integration of Hyperbolic Functions

● Integration related to Inverse Hyperbolic Functions

[Example 1] Find indefinite integral

[Example 2] Show the following is true

§3.2 Integration By Parts

[Example 1] Find the indefinite integral

[Example 2] Evaluate the definite integral

§3.3 Integration By Substitutions

[Example 1] Evaluate the indefinite integral

[Example 2] Evaluate the definite integral

[Example 3] Evaluate the definite integral

[Example 4] Solve the following problems

§3.4 Trigonometric Substitution

● Trigonometric Substitution

● Hyperbolic Substitution

[Example 1] Find the indefinite integral

[Example 2] Evaluate the definite integral

§3.5 Integration By Partial fractions

● Partial fractions Decomposition

● Integration By Partial fractions

[Example 1] Find partial fractions decomposition

[Example 2] Evaluate the integrals.

[Example 3] Evaluate the definite integrals.

§3.6 Improper Integral

● Basic Concepts

● Comparison Test for Convergence

[Example 1] Evaluate the improper integrals.

[Example 2] Test for convergence of the improper integrals.

4. Applications of Integration

§4.1 Area between Curves

[Example 1] Find the area

[Example 2] Find the area

[Example 3] Find the area

[Example 4] Find the area

[Example 5] Find the area

§4.2 Volume of Solids

● Volume of a Solid with known cross-section

● Volume of a Cuboid

● Volume of a Right Circular Cone.

● Volume of a Sphere

● Volume of a Solid of Revolution

[Example 1] Find the volume

[Example 2] Find the volume

[Example 3] Find the volume

§4.3 Length of Curve

[Example 1] Find the length

§4.4 Accumulated Change and Average of Function

● Position, Velocity and Acceleration

● Distance Traveled

● Accumulated Change of Physical Quantities.

● Average Value of a Function.

[Example 1] Solve the following problems

[Example 2] Solve the following problems

[Example 3] Find the average value

Chapter 5. Approximation of Definite Integral

§5.1 Rectangular Approximation Methods

● Left Riemann Sum

● Right Riemann Sum

● Midpoint Riemann Sum

§5.2 Trapezoidal Rule

[Example 1] Approximation of Definite Integral

Chapter 6. Introduction to Differential Equation

§6.1 Basic Concepts

● Definition and General Forms

● Properties

§6.2 Separation of Variables for Solving ODEs

[Example 1] Find the position

[Example 2] Find the solution

[Example 3] Find the solution

§6.3 Slope Fields

§6.4 Euler Method

Chapter 7. Infinite Series

§7.1 Convergence Test of Nonnegative Series

● Comparison Test

● Limit Comparison Test

● Integral Test

● p-Series Test

● Ratio Test

● Root Test

[Example 1] Comparison Test

[Example 2] Limit Comparison Test

[Example 3] Integral Test

[Example 4] Ratio Test and Root Test

§7.2 Convergence Test of Alternating Series

[Example 1] Convergence test of alternating series

[Example 2] Error bound of alternating series

§7.3 Absolute Convergence and Conditional Convergence

● Absolute Convergence

● Conditional Convergence

● Rearrangement of Terms

[Example 1] Absolute Convergence

Chapter 8. Power Series

§8.1 Definition and Key Properties

● Definition of Power Series

● Interval of Convergence

[Example 1] Find the interval of convergence.

§8.2 Taylor Series (Maclaurin Series)

● Definition of Taylor Series

● Taylor Series of Elementary Functions

● Obtaining Taylor Series Expansion with Known Series

Appendix: Term by Term Differentiation and Integration

● Approximation By Taylor Polynomials and Error Bounds

● Error Bounds of Alternating Taylor Series

[Example 1] Find the Taylor series

[Example 2] Taylor Polynomial and Error Bound

Chapter 9. Exercises

Exercise 1. Indefinite Integral

Solution 1.1

Solution 1.2

Solution 1.3

Solution 1.4

Solution 1.5

Solution 1.6

Solution 1.7

Solution 1.8

Solution 1.9

Solution 1.10

Solution 1.11

Solution 1.12

Solution 1.13

Solution 1.14

Solution 1.15

Solution 1.16

Solution 1.17

Solution 1.18

Solution 1.19

Solution 1.20

Solution 1.21

Solution 1.22

Solution 1.23

Solution 1.24

Solution 1.25

Solution 1.26

Solution 1.27

Solution 1.28

Solution 1.29

Solution 1.30

Exercise 2. Definite Integral

Solution 2.1

Solution 2.2

Solution 2.3

Solution 2.4

Solution 2.5

Solution 2.6

Solution 2.7

Solution 2.8

Solution 2.9

Solution 2.10

Solution 2.11

Solution 2.12

Solution 2.13

Solution 2.14

Solution 2.15

Solution 2.16

Solution 2.17

Solution 2.18

Solution 2.19

Solution 2.20

Solution 2.21

Solution 2.22

Solution 2.23

Solution 2.24

Solution 2.25

Exercise 3. Improper Integral

Solution 3.1

Solution 3.2

Solution 3.3

Solution 3.4

Solution 3.5

Solution 3.6

Solution 3.7

Exercise 4. Properties of Integration

Solution 4.1

Solution 4.2

Solution 4.3

Solution 4.4

Solution 4.5

Solution 4.6

Solution 4.7

Exercise 5. Recursive Formula

Solution 5.1

Solution 5.2

Solution 5.3

Exercise 6. Fundamental Theorem of Calculus

Solution 6.1

Solution 6.2

Solution 6.3

Solution 6.4

Solution 6.5

Solution 6.6

Solution 6.7

Solution 6.8

Exercise 7. Find the area enclosed

Solution 7.1

Solution 7.2

Solution 7.3

Solution 7.4

Solution 7.5

Exercise 8. Find the volume of revolution

Solution 8.1

Solution 8.2

Solution 8.3

Solution 8.4

Solution 8.5

Solution 8.6

Exercise 9. Find the volume

Solution 9.1

Solution 9.2

Exercise 10. Find the length of curve

Solution 10.1

Solution 10.2

Solution 10.3

Solution 10.4

Exercise 11. Find the average value

Solution 11.1

Solution 11.2

Solution 11.3

Exercise 12. Position, Velocity, and Acceleration

Solution 12.1

Solution 12.2

Solution 12.3

Solution 12.4

Solution 12.5

Exercise 13. Approximation of Integral

Solution 13.1

Solution 13.2

Exercise 14. Differential Equations

Solution 14.1

Solution 14.2

Solution 14.3

Solution 14.4

Exercise 15. Convergence Test of Infinite Series

Solution 15.1

Solution 15.2

Solution 15.3

Solution 15.4

Solution 15.5

Exercise 16 Absolute or Conditional Convergence

Solution 16.1

Solution 16.2

Solution 16.3

Solution 16.4

Solution 16.5

Exercise 17 Infinite Series

Solution 17.1

Solution 17.2

Solution 17.3

Exercise 18 Power Series

Solution 18.1

Solution 18.2

Solution 18.3

Solution 18.4

Solution 18.5

Exercise 19 Taylor Series

Solution 19.1

Solution 19.2

Solution 19.3

Solution 19.4

Solution 19.5

Exercise 20 Taylor Series

Solution 20.1

Solution 20.2

Solution 20.3

Solution 20.4

Solution 20.5

Solution 20.6

10. Epilogue

1. Introduction

This three-volume edition, Calculus: Fundamentals of Limits, Differentiation, and Integration (feat. AP Calculus), extensively delves into the essential principles and applications of calculus for beginners. This book is the third volume of the series. It's specifically designed for individuals who are either grasping the fundamentals of calculus or preparing for the AP calculus test. Notably, this edition follows the previously published "Calculus Basics," now in its second edition.

In the first volume, you delve into the following topics concerning limits: ▪︎ Definitions of Functions ▪︎ Algebraic and Transcendental Functions ▪︎ Definitions of Limits ▪︎ Theorems on Limits

▪︎ Evaluations of Limits ▪︎ Continuity of Functions ▪︎ Infinite Sequence

▪︎ Infinite Series

In the second volume, you learn the following topics about differentiation: ▪︎ Definitions of Differentiation ▪︎ Derivatives ▪︎ Rules of Differentiation ▪︎ Analysis of Function Graphs ▪︎ Applications of Differential Calculus In the third volume, you learn the following topics about integration: ▪︎ Definitions of Integral ▪︎ Antidifferentiation

▪︎ Definite Integrals

▪︎ Fundamental Theorem of Calculus ▪︎ Rules of Antidifferentiation ▪︎ Applications of Integral Calculus ▪︎ Introduction to Differential Equations ▪︎ Infinite Series and Power Series ● Flowchart

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About This Book

✔︎ While solving problems or validating theorems in this book, I've included detailed intermediate steps whenever feasible, ensuring none are omitted. This encompasses all essential intermediate steps, encompassing numerical computations, thereby facilitating comprehension of the answer-finding process as you progress through the text.

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Revision History

Version 1.0 (September, 2018 )

- Published

Version 2.0 (July, 2024)

- Published

To those who seek a little truth beyond

Book Summary : You Must Read This!

￼

In this book, you will study the definitions, theorems, and principles about the integral calculus. After reading this book, I wish you get back here to find what you have learned so far, which will be good chance to reorganize the basic concepts about the differential calculus.

Chapter 2. Anti-differentiation This chapter focuses on anti-differentiation, an essential concept in calculus that deals with finding integrals of functions. Anti-differentiation, also known as integration, is the reverse process of differentiation. It plays a critical role in solving problems related to accumulation and areas under curves.

Section 2.1 begins with a discussion on antiderivatives. An antiderivative of a function ￼ is a function ￼ such that the derivative of ￼ is ￼. This concept introduces the idea of a function whose rate of change is known, allowing us to reconstruct the original function up to an arbitrary constant. The notion of the additive constant highlights that antiderivatives are not unique; they differ by a constant value. The integration variable is used to express the antiderivative. Additionally, the algebra of antiderivatives involves methods and rules for finding antiderivatives of combined functions, such as sums and differences, providing a foundation for more complex integration tasks.

Next, section 2.2 covers the definite integral, which extends the concept of antiderivatives to calculating the total accumulation of a function over an interval. The definite integral is introduced through the Riemann sum, a method for approximating the accumulation of a function by dividing the interval into small segments and summing up. This approach leads to the idea of partitions, which are divisions of the interval into subintervals, crucial for refining the approximation of the integral. Integrability is then discussed, focusing on the conditions under which the Riemann sum converges to a well-defined value as the partitions become finer. This leads to the formal definition of the definite integral, explaining how it represents the accumulated quantity of the function over a specified interval. Next subsections addresses the Fundamental Theorem of Calculus, which establishes a critical connection between differentiation and integration and the properties of definite integral.

Section 2.3 then explores the geometric interpretation of integrals. This section explains how the definite integral can be understood as the area under a curve. By providing a geometric perspective, it illustrates the practical applications of integrals in measuring areas, volumes, and other geometric quantities.

In this chapter, you will learn: §2.1 Antiderivatives

§2.2 Definite Integral

§2.3 Geometric Interpretation of Integration

§2.1 Antiderivatives

In this section, you will study:

▪︎ Definition : Antiderivatives

▪︎ Additive Constant

▪︎ Integration Variable

▪︎ Algebra of Antiderivatives

● Definition : Antiderivatives

An antiderivative of a function ￼ is a function ￼ such that the derivative of ￼ is equal to ￼. In other words, if ￼, then ￼ is an antiderivative of ￼.

The process of finding the antiderivative of a function is called anti-differentiation or integration. Unlike differentiation, which yields a unique result, integration can yield a family of functions, all differing by a constant. This constant is known as the additive constant or constant of integration, usually denoted as ￼. Thus, if ￼ is an antiderivative of ￼, then any function of the form ￼, where ￼ is a constant, is also an antiderivative of ￼.

Integration is denoted using the integral symbol ￼ and is a fundamental operation in calculus. There are two main types of integrals: the indefinite integral and the definite integral. Each has a distinct denotation.

(i) Indefinite Integral

The indefinite integral of a function ￼ is denoted by ￼. It represents a family of functions whose derivative is ￼. The result of an indefinite integral is called the antiderivative of ￼, and it includes an arbitrary constant ￼, known as the constant of integration.

- Integral Symbol ￼: This symbol indicates the operation of integration.

- Integrand ￼: The function to be integrated.

- Differential ￼: Indicates the variable with respect to which the function is integrated.

- Constant of Integration ￼: Represents the arbitrary constant added to the antiderivative.

The general form is:

￼

where ￼ is the antiderivative of ￼.

Example: For ￼,

Since ￼, ￼

(ii) Definite Integral

The definite integral of a function ￼ over an interval ￼ is denoted by ￼. It represents the net accumulation of ￼ from ￼ to ￼.

- Integral Symbol ￼: Indicates the operation of integration.

- Limits of Integration ￼ and ￼: The endpoints of the interval over which the function is integrated.

- Integrand ￼: The function to be integrated.

- Differential ￼: Indicates the variable of integration.

The general form is:

￼

Here, ￼ is a constant. The result of a definite integral with constant limits is a constant, whereas the result of an indefinite integral is a family of functions.

We will examine the definite integral in detail in section 2.2.

In this section, you will study:

▪︎ Definition : Antiderivatives

▪︎ Additive Constant

▪︎ Integration Variable

▪︎ Algebra of Antiderivatives

● Additive Constant (or Constant of Integration)

The additive constant ￼ in integration arises from the fundamental nature of differentiation and integration in calculus.

When finding the antiderivative or indefinite integral of a function ￼, we are essentially looking for a function ￼ such that ￼. However, differentiation eliminates any constant term, because the derivative of a constant is zero. This means:

￼￼

where ￼ is any constant. Therefore, if ￼ is an antiderivative of ￼, any function of the form ￼ (where ￼ is an arbitrary constant) will also be an antiderivative of ￼.

The presence of the constant ￼ reflects the fact that there is not just one but an entire family of functions that serve as antiderivatives for ￼. Each function in this family differs from the others only by a constant. This is because the process of differentiation removes the constant, and thus we lose the information about it when we reverse the process (integration). That is, even when the two derivatives are the same, we cannot determine the primitive as one function.

Consider integrating the function ￼. We seek a function ￼ such that ￼. The antiderivative of ￼ is:

￼

Here, ￼ is the general antiderivative, and ￼ is the additive constant. Different values of ￼ yield different functions (e.g., ￼, ￼), but all have the same derivative ￼.

Consider the specific case where we need the antiderivative to pass through a particular point, say ￼. We can solve for ￼:

￼

Thus, the specific antiderivative is ￼.

The additive constant ￼ exists in integration because differentiation eliminates constant terms, leading to a loss of information about them. When we integrate, we recover a family of functions differing by a constant ￼, reflecting the fact that any constant term would have been lost during differentiation.

In this section, you will study:

▪︎ Definition : Antiderivatives

▪︎ Additive Constant

▪︎ Integration Variable

▪︎ Algebra of Antiderivatives

● Integration Variable

In calculus, the integration variable is essential in integrals. It denotes the variable with respect to which we integrate a function. The role and usage of this variable differ slightly between indefinite and definite integrals.

For example, the following are integrals with respect to the variable ￼, and the result are functions with respect to ￼.

￼, ￼

In indefinite integrals, changing the integration variable can alter the result of the integral. In contrast, for definite integrals, the result remains the same regardless of the integration variable, as long as the integration limits are the same.

In definite integrals, the integration variable is referred to as a dummy variable. This term means that the specific letter used for the variable of integration is arbitrary and has no effect on the result of the integral, as long as the integrand and the integration limits are properly defined. The role of the dummy variable is solely to indicate the variable with respect to which integration is performed over a given interval.

Let's consider the definite integral of the function ￼ from ￼ to ￼.

The following integrals are evaluated in the same way:

￼￼

￼

Both integrals yield the same result. This demonstrates that the choice of the integration variable (whether ￼, ￼, or any other symbol) does not affect the outcome of a definite integral, as long as the limits and the function are correctly applied.

In this section, you will study:

▪︎ Definition : Antiderivatives

▪︎ Additive Constant

▪︎ Integration Variable

▪︎ Algebra of Antiderivatives

● Algebra of Antiderivatives

The algebra of antiderivatives refers to the rules and properties used to find antiderivatives (indefinite integrals) of more complex functions by using known antiderivatives of simpler functions. These rules facilitate the integration of sums, differences, and certain types of products and compositions of functions.

(i) Sum and Difference Rules

The sum and difference rules allow us to integrate the sum or difference of functions separately:

￼￼

Find the integral of ￼:

￼

(ii) Multiplying Constant with Antiderivatives

When integrating a function, a constant factor can be factored out of the integrand. Specifically, if a function ￼ is multiplied by a constant ￼, the antiderivative of ￼ can be obtained by first integrating ￼ and then multiplying the result by ￼:

￼

where ￼ is a constant, and ￼ is the integrand.

Consider the function ￼:

￼

(iii) Linearity of Integration

The linearity property allows us to integrate sums and scalar multiples of functions. Specifically, for any functions ￼ and ￼, and any constants ￼ and ￼:

￼￼

Find the integral of ￼:

￼￼￼

These rules allow us to decompose complex integrals into simpler parts, integrate them, and then combine the results to find the antiderivative of the original function.

* The evaluation of integrals are described later.

[Example 1] Find the indefinite integral

￼

(Solution)

By using the property of the integral, we can get: ￼

Here, we are going to use the derivative rules to find the antiderivatives.

Since:

￼

Then: ￼

￼

￼

Since:

￼

Then: ￼

￼

￼

Thus:

￼

￼

In this context, since the integration constant ￼ is arbitrary, the sum of the integration constants obtained from two different indefinite integrals or the product of an integration constant by another constant can still be represented as ￼.

￼

[Example 2] Find the indefinite integral

￼

(Solution)

￼

Here, we are going to use the derivative rules to find the antiderivatives.

Since:

￼

Then ￼

￼

Therefore,

￼

[Example 3] Find the antiderivative

For the given derivative, find the antiderivative function passing through the given point.

￼

(Solution)

We find the antiderivative and determine the integral constant by the given point.

￼

￼￼￼

Then:

￼

The point (5, 4) is on this function.

￼

￼

Therefore:

￼

In this chapter, you will learn:

§2.1 Antiderivatives

§2.2 Definite Integral

§2.3 Geometric Interpretation of Integration

§2.2 Definite Integral

In this section, you will study:

▪︎ Riemann Sum

▪︎ Integrability

▪︎ Definite Integral

▪︎ Fundamental Theorem of Calculus

▪︎ Properties of Definite Integral

● Riemann Sum

The Riemann sum is a mathematical tool used to approximate the value of a definite integral. It is defined by partitioning an interval and summing the values of the function at certain points within these subintervals, weighted by the widths of the subintervals. Here is the definition:

Consider a closed interval ￼. A partition ￼ of ￼ is a finite sequence of points:

￼

such that:

￼

These points divide the interval ￼ into ￼ subintervals: ￼ for ￼.

The width of each subinterval ￼ is:

￼

The norm of the partition ￼ is the length of the largest subinterval:

￼

For each subinterval ￼, choose a sample point ￼ such that:

￼

Given a function ￼ defined on ￼, the Riemann sum of ￼ with respect to the partition ￼ and sample points ￼ is defined as:

￼

Here:

- ￼ is the value of the function at the sample point ￼.

- ￼ is the width of the ￼-th subinterval.

The Riemann sum is a way to approximate the definite integral of a function by summing the values of the function at chosen sample points within subintervals of a partition of the integration interval. The more refined the partition (i.e., the smaller the norm ￼, the closer the Riemann sum approximates the exact value of the integral.

In this section, you will study:

▪︎ Riemann Sum

▪︎ Integrability

▪︎ Definite Integral

▪︎ Fundamental Theorem of Calculus

▪︎ Properties of Definite Integral

● Integrability

Integrability refers to whether a function can be integrated over a given interval, meaning whether its definite integral exists. The formal concept of integrability is often framed in terms of Riemann sums, partitions, and related concepts like upper and lower Riemann sums. A function is Riemann integrable if the Riemann sums converge to a single value as the partitions of the interval become finer.

(i) Partition and Refinement

A partition ￼ of an interval ￼ is a finite sequence of points ￼ such that:

￼

A refinement ￼ of ￼ is a partition that includes all points of ￼ and possibly more, creating smaller subintervals. Formally, ￼ is a refinement of ￼ if every subinterval defined by ￼ is subdivided by ￼.

(ii) Upper and Lower Riemann Sums

- Upper Riemann Sum: Given a partition ￼ of ￼, the upper Riemann sum ￼ is the sum of the areas of rectangles that overestimate the function. It uses the maximum (least upper bound) ￼ of the function on each subinterval ￼:

￼

where ￼.

(iii) Lower Riemann Sum: The lower Riemann sum ￼ is the sum of the areas of rectangles that underestimate the function. It uses the minimum (greatest lower bound) ￼ of the function on each subinterval:

￼

where ￼.

￼

￼

(iv) Refinement of Partitions

Suppose that function ￼ is bounded on the interval [a, b]. Let a certain partition P be the following.

￼

Furthermore, Let another partition Q be the following. The partition ￼ has one more point ￼ than partition P.

￼

By definition:

￼￼￼￼

And on the interval ￼, the minimum is ￼. This minimum occurs either on ￼ or ￼.

Suppose that it does on ￼, and let ￼ be the minimum on ￼. Then, ￼

￼￼￼￼

￼￼￼￼

￼￼￼

Here,

￼￼￼

￼￼￼

Thus

￼

This also holds true in the upper Riemann sum.

￼

In the end, by the refinement of the partition, the following inequality holds.

￼￼

(v) Integrability

Suppose that function ￼ is bounded on ￼. For any positive number ￼, the function ￼ is integrable on the interval if and only if there exists a partition such that

￼

That is, we can say that function ￼ is integrable on ￼ when, by choosing specific partitions, we can make the difference between the upper Riemann sum and the lower Riemann sum less than any positive number ￼.

In other words, the definite integral of ￼ from ￼ to ￼ is defined as the limit of the Riemann sums as the norm of the partition ￼ approaches zero:

￼

This limit exists if ￼ is integrable on ￼, meaning that for any ￼, there exists a ￼ such that for any partition ￼ with ￼, the difference between the Riemann sum and the integral is less than ￼.

Note that a function ￼ is integrable if it is continuous on a closed interval. Additionally, ￼ is integrable if it is bounded and has only a finite number of discontinuities within a given interval.

Consider a function ￼ defined on ￼ with a simple partition ￼. Partition ￼ divides ￼ into subintervals ￼ and ￼.

Let ￼:

- For the subinterval ￼:

- ￼

- ￼

- ￼

- For the subinterval ￼:

- ￼

- ￼

- ￼

Upper Riemann sum:

￼￼

Lower Riemann sum:

￼￼

Suppose we refine ￼ to ￼. This introduces additional points, making the subintervals ￼, ￼, ￼, and ￼.

Let ￼:

- For the subinterval ￼:

- ￼

- ￼

- ￼

- For the subinterval ￼:

- ￼

- ￼

- ￼

- For the subinterval ￼:

- ￼

- ￼

- ￼

- For the subinterval ￼:

- ￼

- ￼

- ￼

- For the subinterval ￼:

- ￼

- ￼

- ￼

Upper Riemann sum:

￼￼￼￼

Lower Riemann sum:

￼￼￼￼

We can verify ￼￼

If we evaluate the definite integral, ￼

Integrability of a function ￼ on ￼ involves checking if the upper and lower Riemann sums converge to the same value as the partitions are refined (i.e., as the widths of the subintervals become arbitrarily small). If they do, the function is Riemann integrable, and the common value of the upper and lower integrals is the definite integral of ￼ over ￼.

As explained previously, a function ￼ is integrable if it is continuous on a closed interval. Additionally, ￼ is integrable if it is bounded and has only a finite number of discontinuities within a given interval.

You should note that this conditions for integrability are necessary, but not sufficient. There are instances where a function can be integrable over an interval, even though it remains unbounded within that interval. Such cases include improper integrals, which we will discuss further.

In this section, you will study: ▪︎ Riemann Sum

▪︎ Integrability

▪︎ Definite Integral

▪︎ Fundamental Theorem of Calculus ▪︎ Properties of Definite Integral

● Definite Integral

When function ￼ is integrable on ￼, the definite integral is defined with the limit of the Riemann sum.

￼￼ ￼

In the left side, ￼ and ￼ are the upper limit and the lower limit of the integral respectively.

If the every subinterval ￼ has the same size, then ￼

By using ￼ definition, the function is integrable if the following proposition holds true.

￼￼

In this section, you will study:

▪︎ Riemann Sum

▪︎ Integrability

▪︎ Definite Integral

▪︎ Fundamental Theorem of Calculus

▪︎ Properties of Definite Integral

● Fundamental Theorem of Calculus

The Fundamental Theorems of Calculus link the concepts of differentiation and integration, providing a bridge between the process of finding the derivative of a function and the process of finding the accumulation of a function. There are two main parts:

(i) First Fundamental Theorem of Calculus

If ￼ is a continuous function on the closed interval ￼, then the function ￼ defined by:

￼

is continuous on ￼, differentiable on ￼, and its derivative is ￼:

￼

This theorem asserts that if you define a function ￼ as the integral of ￼ from a fixed point ￼ to a variable point ￼, then ￼ is differentiable, and its derivative is ￼. In other words, integrating ￼ and then differentiating the result returns the original function ￼.

(ii) Second Fundamental Theorem of Calculus

If ￼ is a continuous function on the closed interval ￼ and ￼ is an antiderivative of ￼ on ￼, then:

￼

This theorem states that the definite integral of a function ￼ from ￼ to ￼ is equal to the difference in the values of its antiderivative ￼ evaluated at the endpoints of the interval. In other words, if you know an antiderivative ￼ of ￼, we can compute the definite integral by simply evaluating ￼ at the boundaries of the interval and subtracting.

If we consider that a definite integral is the limit of a Riemann sum, then originally we would need to find the sum of an infinite series. However, according to the Fundamental Theorem of Calculus, we can calculate the definite integral without summing an infinite series by using the antiderivative and the endpoints of the integration interval.

Consider ￼. To find ￼:

￼ (since ￼)

￼￼￼

The Fundamental Theorems of Calculus link the processes of differentiation and integration. the First Fundamental Theorem shows that differentiation and integration are inverse processes, while The Second Fundamental Theorem allows the evaluation of definite integrals using antiderivatives. Together, they provide a powerful framework for understanding and applying integral calculus.

The notation ￼ is commonly used in calculus to represent the evaluation of a definite integral using the Fundamental Theorem of Calculus:

￼￼

- ￼: This is an antiderivative of ￼, meaning ￼.

- ￼: This notation means the evaluation of ￼ at the upper limit ￼ minus the evaluation at the lower limit ￼:

￼

Suppose we want to evaluate the integral ￼:

For ￼, an antiderivative is ￼.

￼￼￼￼￼

In this section, you will study:

▪︎ Riemann Sum

▪︎ Partitions and Integrability

▪︎ Definite Integral

▪︎ Fundamental Theorem of Calculus

▪︎ Properties of Definite Integral

● Properties of Definite Integral

Definite integrals have several important properties that make them useful in mathematical analysis and applications. These properties facilitate the computation of integrals and the understanding of the behavior of functions over intervals. Here are the key properties:

(i) Linearity

The definite integral is linear, which means it adheres to the properties of addition and scalar multiplication. For functions ￼ and ￼, and constants ￼, ￼, ￼, and ￼:

￼￼

For ￼ and ￼:

￼￼

(ii) Splitting Integral

The integral of a function over an interval that can be split into two adjacent subintervals is equal to the sum of the integrals over each subinterval:

￼￼

For ￼ from ￼ to ￼:

￼

(iii) Reversal of Limits

Reversing the limits of integration changes the sign of the integral:

￼

For ￼ from ￼ to ￼:

￼

(iv) Zero-Width Interval

When the upper and lower limits of definite integral are the same:

￼

(v) Non-Negativity

If ￼ for all ￼ in ￼, then the integral of ￼ is non-negative:

￼

(vi) Comparison Property

If ￼ for all ￼ in ￼, then:

￼

(vii) Minimum and Maximum Property

If ￼ for all ￼ in ￼, then:

￼

￼

￼

(viii) Absolute Value Inequality

The absolute value of the integral of ￼ is less than or equal to the integral of the absolute value of ￼:

￼

(ix) Mean Value Theorem for Integrals

If ￼ is continuous on ￼, then there exists some ￼ such that:

￼

￼

The properties of definite integrals provide tools for simplifying and understanding the behavior of integrals. They include linearity, splitting integration limits, reversal of limits, non-negativity, comparison, and absolute value inequality. These properties help in evaluating integrals, analyzing functions, and solving problems in calculus and applied mathematics.

[Example 1] Find the derivative of integral function

Find the derivative of the following functions defined by definite integral.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

By Fundamental Theorem of Calculus: ￼

(Solution 2)

Since the definite integral is a function with respect to ￼, not ￼: ￼

￼

Let ￼

￼

￼

Since ￼￼

￼

By chain rule:

￼￼￼

[Example 2] Find the function

Find the function ￼ and positive number ￼ when the following formula holds for ￼.

￼

(Solution)

By plugging ￼ in the formula, we get:

￼

￼

By differentiating both sides:

￼

Let’s verify the answer by substituting this function into given formula.

￼￼￼￼

[Example 3] Infinite Series and Definite Integral

Prove the formula by using the definition of the definite integral and infinite sum of ￼ partitions with equal length.

(1) ￼

(2) ￼

(3)￼

(Solution)

(Solution 1)

Let ￼ be the equal length of partitions on the interval ￼.

￼

And let ￼ be the right endpoint of each partition. Then:

￼

By the definition of the definite integral:

￼￼

￼￼

(Solution 2)

Let ￼ be the equal length of partitions on the interval ￼.

￼

And let ￼ be the right endpoint of each partition. Then:

￼

By the definition of the definite integral:

￼￼

￼￼

Here, we used the following summation formula.

￼

(Solution 3)

Let ￼ be the equal length of partitions on the interval ￼.

￼

And let ￼ be the right endpoint of each partition. Then:

￼

By the definition of the definite integral:

￼￼

￼￼

Here, we used the following summation formula.

￼

[Example 4] Properties of definite integral

Evaluate the following definite integral by using the properties of the integral and the result of Example 3.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

￼￼

(Solution 2)

￼￼￼

[Example 5] Solve the following problems

(1) When￼ and ￼, evaluate the following integral.

￼

(2) Prove the following inequality by using the properties of the integral and the result of Example 3.

￼

(Solution)

(Solution 1)

￼

￼

￼

(Solution 2) On the given interval, the following inequality holds: ￼

Thus:

￼

Here:

￼￼￼￼

Therefore:

￼

In this chapter, you will learn:

§2.1 Antiderivatives

§2.2 Definite Integral

§2.3 Geometric Interpretation of Integration

§2.3 Geometric Interpretation of Integration

Differential calculus was studied to determine the angle between two curves at their intersection, while integral calculus was developed to calculate the area between two curves. The fundamental principle of integral calculus was to compute the total area by dividing it into smaller, more manageable sections, such as rectangles, whose areas can be easily determined and then summed.

The definite integral of a function provides a way to compute the total area under the curve of that function between two points on the x-axis. This concept is deeply related to the idea of summing an infinite series of infinitesimally small rectangles. Here’s a detailed explanation:

(i) Definite Integral as Area Under the Curve

When we talk about the definite integral ￼, we are calculating the net area under the curve ￼ from ￼ to ￼:

- Above the x-axis: If ￼ is positive, this area is above the x-axis.

- Below the x-axis: If ￼ is negative, this area is below the x-axis and counted as negative.

(ii) Approximating with Rectangles

To understand this area in a practical sense, we approximate it by summing up the areas of many thin rectangles:

- Divide the Interval: Split the interval ￼ into ￼ subintervals of equal width ￼.

- Calculate Heights: At each subinterval, the height of the rectangle is the function value ￼ at some chosen point in the subinterval (left endpoint, right endpoint, or midpoint).

- Sum Areas: The total area under the curve is approximated by summing the areas of these rectangles.

(iii) Riemann Sum and Infinite Series

As the number of rectangles ￼ increases and ￼ gets smaller, the sum of the areas of these rectangles becomes a more accurate approximation of the true area. In the limit as ￼ approaches infinity and ￼ approaches zero, this sum becomes the exact area under the curve. This limiting process is formally expressed as:

￼

where ￼ is a sample point in the ￼-th subinterval.

￼

￼

￼

- Red Curve: Represents the function ￼.

- Blue Rectangles: Represent the Riemann sum approximation. Each rectangle's height is determined by the function value at the left endpoint of each subinterval.

- Blue Shaded Area: Represents the actual area under the curve between ￼ and ￼.

- Interval Markers: Vertical dashed lines mark the limits ￼ and ￼.

This geometric interpretation helps visualize how integration works and how it relates to the summation of areas under a curve.

Chapter 3. Integration Rules

Integration is a fundamental concept in calculus, representing the accumulation of quantities and the reverse process of differentiation. This chapter introduces several key methods and techniques for finding antiderivatives and solving integrals of various functions.

In the upcoming sections, we will explore the following integration methods: Section 3.1 covers basic integration formulas, focusing on power functions and elementary transcendental functions such as exponential, logarithmic, and trigonometric functions. These foundational formulas are essential for handling a wide range of integrals.

Integration by parts is a technique derived from the product rule of differentiation. It is useful for integrating the product of two functions, especially when one function is easily integrable, and the other is easily differentiable.

Integration by substitution simplifies integrals by transforming them into a different variable. This technique is especially effective for integrals involving composite functions.

Trigonometric substitution is a method used to evaluate integrals involving square roots and other expressions that can be simplified using trigonometric identities. This approach often transforms complex integrals into simpler trigonometric forms.

Integration by partial fractions decomposes rational functions into simpler fractions that can be integrated more easily. This method is particularly useful for integrating rational functions where the degree of the numerator is less than the degree of the denominator.

Improper integrals extend the concept of definite integrals to unbounded intervals or integrands with infinite discontinuities. This section covers the techniques for evaluating these integrals, including limits and convergence criteria.

In this chapter, you will learn: §3.1 Integration formula

§3.2 Integration By Parts

§3.3 Integration By Substitutions §3.4 Trigonometric Substitutions §3.5 Integration By Partial fractions

§3.6 Improper Integral

§3.1 Integration Formula

In this section, you will study:

▪︎ Integration of Power Functions

▪︎ Integration of Exponential Functions

▪︎ Integration of Trigonometric Functions

▪︎ Integration related to Inverse Trigonometric Functions

▪︎ Integration of Hyperbolic Functions

▪︎ Integration related to Inverse Hyperbolic Functions

● Integration of Power Functions

Power functions are of the form ￼, where ￼ is a real number. Integrating power functions involves finding the antiderivative, which is the reverse process of differentiation. The general formula for integrating a power function is:

￼

where ￼ is the constant of integration, and ￼.

To understand the formula, consider the differentiation rule for power functions. If ￼, then differentiating ￼ gives:

￼

This shows that ￼ is the antiderivative of ￼.

The formula does not apply when ￼, as it would involve division by zero. Instead, the integral of ￼ or ￼ is given by:

Since:

￼

￼

When integrating the function ￼, the result is ￼, where ￼ is the constant of integration.

The function ￼ is defined for all ￼, which includes both positive and negative values of ￼. The natural logarithm function ￼, however, is only defined for positive values of ￼. By using ￼, we extend the domain of the integral to include both positive and negative values of ￼.

Using ￼ rather than ￼ as the antiderivative of ￼ ensures that the integral is defined for all non-zero values of ￼, maintains continuity, and accurately reflects the behavior of the function across its entire domain.

Let’s consider some examples:

￼

￼￼

￼￼

￼

In this section, you will study:

▪︎ Integration of Power Functions

▪︎ Integration of Exponential Functions ▪︎ Integration of Trigonometric Functions ▪︎ Integration related to Inverse Trigonometric Functions ▪︎ Integration of Hyperbolic Functions ▪︎ Integration related to Inverse Hyperbolic Functions

● Integration of Exponential Functions

By using the antiderivative, we can get the integration formula for ￼, ￼.

Since ￼: ￼

￼

When ￼: ￼

In this section, you will study:

▪︎ Integration of Power Functions

▪︎ Integration of Exponential Functions ▪︎ Integration of Trigonometric Functions ▪︎ Integration related to Inverse Trigonometric Functions ▪︎ Integration of Hyperbolic Functions

▪︎ Integration related to Inverse Hyperbolic Functions

● Integration of Trigonometric Functions

We can find the indefinite integrals of the following trigonometric functions by using their antiderivatives.

￼

￼

￼

￼

￼

￼

￼

￼

￼

￼

￼

￼

To find the indefinite integrals of the functions ￼, ￼, ￼, and ￼, it is necessary to learn various integration techniques.

In this section, you will study:

▪︎ Integration of Power Functions

▪︎ Integration of Exponential Functions

▪︎ Integration of Trigonometric Functions ▪︎ Integration related to Inverse Trigonometric Functions ▪︎ Integration of Hyperbolic Functions

▪︎ Integration related to Inverse Hyperbolic Functions

● Integration related to Inverse Trigonometric Functions

We can find the indefinite integrals of the following inverse trigonometric functions by using their antiderivatives.

￼

￼

￼

￼

Note that the following formula holds true: ￼

￼

￼

You do not need to memorize the integration formulas for these functions here. As will be explained later, the formulas can be derived by using substitution methods with trigonometric or hyperbolic functions.

In this section, you will study:

▪︎ Integration of Power Functions

▪︎ Integration of Exponential Functions

▪︎ Integration of Trigonometric Functions

▪︎ Integration related to Inverse Trigonometric Functions

▪︎ Integration of Hyperbolic Functions

▪︎ Integration related to Inverse Hyperbolic Functions

● Integration of Hyperbolic Functions

We can find the indefinite integrals of the following hyperbolic functions by using their antiderivatives.

￼

￼

￼

￼

￼

￼

To find the indefinite integrals of the functions ￼, ￼, ￼, and ￼, it is necessary to learn various integration techniques.

In this section, you will study:

▪︎ Integration of Power Functions

▪︎ Integration of Exponential Functions

▪︎ Integration of Trigonometric Functions ▪︎ Integration related to Inverse Trigonometric Functions ▪︎ Integration of Hyperbolic Functions

▪︎ Integration related to Inverse Hyperbolic Functions

● Integration related to Inverse Hyperbolic Functions

We can find the indefinite integrals of the following inverse hyperbolic functions by using their antiderivatives.

￼

￼

￼

￼

￼

￼

You do not need to memorize the integration formulas for these functions here. As will be explained later, the formulas can be derived by using substitution methods with trigonometric or hyperbolic functions.

[Example 1] Find indefinite integral

(1) ￼

(2) ￼

(Solution)

(Solution 1)

￼￼

￼￼

(Solution 2)

￼￼

￼

[Example 2] Show the following is true

Show that the following is true. (￼, ￼)

￼

(Solution)

￼

￼

￼￼

￼

Since (1) = (2), the given formula is true.

In this chapter, you will learn:

§3.1 Integration formula

§3.2 Integration By Parts

§3.3 Integration By Substitutions

§3.4 Trigonometric Substitutions

§3.5 Integration By Partial fractions

§3.6 Improper Integral

§3.2 Integration By Parts

Integration by parts is a technique for finding the integral of the product of two functions. It is derived from the product rule of differentiation and is particularly useful when one function is easily integrable, and the other is easily differentiable.

The formula for integration by parts is:

￼

Here:

- ￼ and ￼ are functions of ￼.

- ￼ and ￼ are the differentials of ￼ and ￼, respectively.

The formula is derived from the product rule of differentiation:

￼

Integrating both sides with respect to ￼, we get:

￼￼

This simplifies to:

￼￼

Rearranging, we obtain the integration by parts formula:

￼￼

To use integration by parts, follow these steps:

(i) Identify parts of the integrand to set as ￼ and ￼. Ideally, ￼ should be a function that becomes simpler when differentiated, and ￼ should be easily integrable.

(ii) Differentiate ￼ to find ￼

(iii) Integrate ￼ to find ￼

(iv) Substitute ￼, ￼, ￼, and ￼ into the integration by parts formula ￼.

(v) Simplify the resulting expression and integrate ￼.

When explaining the method of integration by parts, the integration constant is typically omitted because the focus is usually on deriving the integral formula itself, rather than solving a specific indefinite integral. Integration by parts is given by the formula:

￼

Here, both sides of the equation are indefinite integrals, which inherently include an arbitrary constant of integration. However, when we apply the integration by parts formula, we often focus on manipulating the functions ￼ and ￼ to simplify the integral, and the constant is not written out explicitly during these steps.

When solving a specific indefinite integral using integration by parts, the integration constant should indeed be included in the final answer to account for all possible antiderivatives. For clarity and completeness in specific solutions, the constant ￼ is added to the result:

￼

This ensures the solution encompasses all antiderivatives of the integrand.

Let’s integrate ￼:

- Choose ￼ and ￼:

￼

￼

- Apply the formula:

￼￼￼￼

Let’s integrate ￼:

- Choose ￼ and ￼:

￼

￼

- Apply the formula:

￼￼￼

Generally but not always, functions earlier in this list make good choices for ￼, and those later in the list are better for ￼:

① Logarithmic: ￼

② Inverse Trigonometric: ￼, ￼, etc.

③ Algebraic: ￼

④ Trigonometric: ￼, ￼, etc.

⑤ Exponential: ￼

Integration by parts is a versatile technique in integral calculus that simplifies the integration of products of functions by breaking them into simpler parts. It relies on the product rule for differentiation and is particularly useful when one function is easier to integrate, and the other is easier to differentiate. Properly choosing ￼ and ￼ can simplify the integration process and make otherwise complex integrals more manageable.

[Example 1] Find the indefinite integral

Evaluate the following integral by using integration by parts.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

Let ￼. Then: ￼

By the formula of the integration by parts:

￼

Thus:

￼￼￼

(Solution 2)

Let ￼. Then: ￼

Therefore:

￼￼

￼￼

￼

[Example 2] Evaluate the definite integral

Evaluate the following definite integral by using integration by parts.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

￼￼

￼￼

Let's integrate the formula denoted in red by using the integration by parts.

Let ￼￼. Then:

￼￼

Let's integrate the formula denoted in blue by using the integration by parts.

Let ￼￼. Then:

￼￼

￼￼￼

￼￼￼

(Solution 2)

￼

￼￼

Let's integrate the formula denoted in red by using the integration by parts.

Let ￼￼. Then:

￼￼￼

￼￼￼￼

In this chapter, you will learn:

§3.1 Integration formula

§3.2 Integration By Parts

§3.3 Integration By Substitutions

§3.4 Trigonometric Substitutions

§3.5 Integration By Partial fractions

§3.6 Improper Integral

§3.3 Integration By Substitutions

Integration by substitution, also known as ￼-substitution, is a technique used to simplify the process of finding the integral of more complex functions. It involves changing the variable of integration to make the integral easier to solve. This method is particularly effective for integrals involving composite functions or where a part of the integrand can be easily substituted.

The basic idea behind substitution is to transform an integral of the form:

￼

into a simpler integral by setting ￼. Then, ￼, and the integral becomes:

￼

Here are the steps to perform integration by substitution:

(i) Choose a Substitution: Identify a part of the integrand to set as ￼. Typically, choose ￼ where ￼ is a function inside the integrand whose derivative ￼ is also present in the integrand.

(ii) Compute ￼: Differentiate ￼ to find ￼. Express ￼ in terms of ￼.

(iii) Rewrite the Integral: Substitute ￼ and ￼ into the original integral, transforming it into an integral in terms of ￼.

(iv) Integrate: Solve the integral in terms of ￼.

(v) Substitute Back: Replace ￼ with the original variable ￼ to express the integral in its original variable.

Let’s evaluate ￼.

(i) Let ￼.

(ii) Then ￼, so ￼.

(iii) Rewrite the Integral:

￼

(iv) Integrate:

￼

(v) ￼

So, the integral is:

￼

Let’s evaluate ￼.

(i) Let ￼.

(ii) Then ￼, so ￼.

(iii) Rewrite the Integral:

￼￼

(iv) Integrate:

￼￼

(v) Substitute Back:

￼

So, the integral is:

￼

Integration by substitution is a versatile technique that simplifies the process of integrating complex functions by transforming them into a simpler form. It is especially effective for integrals involving composite functions, and when the integrand can be matched with a derivative within the integral. Properly applying substitution can make solving otherwise difficult integrals much more manageable.

When using integration by substitution to evaluate definite integrals, additional care is required compared to indefinite integrals.

After substituting ￼ in the integral, you need to change the limits of integration from ￼ to ￼. The new limits correspond to the values of ￼ when ￼ is at the original limits.

￼￼

Let’s evaluate ￼.

Let ￼, then ￼ or ￼.

For ￼, ￼.

For ￼, ￼.

￼

￼

To successfully evaluate definite integrals using substitution:

(i) Identify the substitution ￼.

(ii) Adjust the limits of integration from ￼ to ￼.

(iii) Rewrite the integral in terms of ￼ and ￼.

(iv) Integrate and apply the new limits directly, or back-substitute and use the original limits.

(v) Carefully handle the differential and make sure the integral is transformed correctly.

By following these steps, you can simplify and accurately evaluate definite integrals using the substitution method.

Moreover, when performing integration by substitution, the choice of the substitution function ￼ (i.e., ￼) is crucial for simplifying the integral. One important requirement for ￼ is that it should ideally be a one-to-one function on the interval of integration. Here’s why this is important and what other requirements need to be considered:

(i) One-to-One Correspondence

A function ￼ is one-to-one and monotonic if it maps each value in its domain to a unique value in its range without overlap. This property is important for substitution because:

- Inverse Function: A one-to-one function has an inverse, meaning each ￼ corresponds to exactly one ￼. This makes it possible to switch back and forth between ￼ and ￼ without ambiguity.

- Well-Defined Integral: Ensures that the substitution does not create overlapping intervals or multiple values for ￼ corresponding to a single ￼, which can complicate the integration process and lead to incorrect results.

(ii) Differentiability and Continuity

The function ￼ should be differentiable and continuous over the interval of integration. This ensures that the differential ￼ is well-defined and that we can properly substitute and manipulate the integral.

[Example 1] Evaluate the indefinite integral

￼

￼

(Solution)

(Solution 1)

Let ￼. Then ￼ and ￼:

￼￼

￼

(Solution 2)

Let ￼. Then ￼ and ￼:

￼￼

￼

[Example 2] Evaluate the definite integral

(1) ￼

(2) ￼

(Solution)

(Solution 1)

Let ￼. Then ￼￼ and ￼:

￼:

￼￼￼￼

(Solution 2)

Let ￼. Then, ￼￼ and ￼:

￼

￼￼

￼￼￼

[Example 3] Evaluate the definite integral

Let ￼ be the inverse function of ￼. Evaluate the following definite integral by using substitution.

￼

Let ￼. Then, ￼￼ and ￼: ￼

Therefore:

￼￼

￼

Let's obtain the definite integral by directly getting the inverse function.

￼

￼

Thus:

￼

￼￼

[Example 4] Solve the following problems

(1) Verify that the following formula is valid for ￼ when ￼ is an even function.

￼

(2) Show that the following formula is true when ￼ is the odd function.

￼

(3) Show that the following formula is true.

￼

(4) When ￼ is the inverse function of ￼, evaluate the following definite integral.

￼

(Solution)

(Solution 1)

If ￼ is the even function, then ￼.

By using the property of the integral:

￼

Let ￼. Then ￼ and ￼:

￼

￼

For definite integrals, ￼ and ￼ are dummy variables so the following holds true.

￼

Therefore:

￼

(Solution 2)

If ￼ is an odd function, then ￼.

By using the property of the integral:

￼

Let ￼. Then ￼ and ￼.

￼

￼￼

For definite integrals, ￼ and ￼ are dummy variables so the following holds true.

￼

Thus:

￼

(Solution 3)

￼

In the left side, let ￼. Then:

￼

￼

￼

Therefore:

￼

￼￼

￼

(Solution 4)

￼￼

￼￼

Let ￼. Then, ￼ and ￼:

￼

￼￼￼￼

Therefore:

￼

In this chapter, you will learn:

§3.1 Integration formula

§3.2 Integration By Parts

§3.3 Integration By Substitutions

§3.4 Trigonometric Substitutions

§3.5 Integration By Partial fractions

§3.6 Improper Integral

§3.4 Trigonometric Substitution

In this section, you will study:

▪︎ Trigonometric Substitution

▪︎ Hyperbolic Substitution

● Trigonometric Substitution

Trigonometric substitution is a technique for evaluating integrals involving algebraic expressions, especially when dealing with expressions involving squares and square roots. This method transforms the integrand into a trigonometric form, simplifying the integration process.

Trigonometric substitution relies on substituting algebraic expressions with trigonometric functions. The goal is to simplify integrals of the form: - ￼

- ￼

- ￼

These expressions suggest the use of the Pythagorean identity. Each substitution corresponds to a specific trigonometric function: - ￼ for ￼

- ￼ for ￼

- ￼ for ￼

(i) Case 1: ￼

Use ￼, ￼.

￼￼

Example:

￼

Substitute ￼: ￼

Back-substitute: ￼: ￼

(ii) Case 2: ￼

Use ￼, ￼.

￼￼

Example:

￼

Substitute ￼:

￼

￼

￼

Let ￼. Then ￼: ￼￼￼￼

Back-substitute: ￼: ￼

￼￼

￼

Case 3: ￼

Use ￼, ￼.

￼￼

Example:

￼

Substitute ￼:

￼

￼

Back-substitute: ￼: ￼￼

￼

￼

In this section, you will study:

▪︎ Trigonometric Substitution

▪︎ Hyperbolic Substitution

● Hyperbolic Substitution

Integration by hyperbolic substitution is a technique used to simplify integrals involving certain types of algebraic expressions, especially those containing squares and square roots. This method uses hyperbolic functions, which are analogs of trigonometric functions, to transform the integrand into a simpler form.

(i) Case 1: ￼

Use ￼, ￼.

￼￼

Example:

￼

Substitute ￼: ￼

Back-substitute: ￼: ￼

(ii) Case 2: ￼

Use ￼, ￼.

￼￼

Example:

￼

Substitute ￼: ￼

Back-substitute: ￼: ￼

[Example 1] Find the indefinite integral

(1) ￼

(2) ￼

(3) ￼

(Solution)

(Solution 1)

Use ￼, ￼:

￼￼￼￼

￼￼

Back-substitute: ￼:

￼

(Solution 2)

Use ￼, ￼:

￼￼￼￼

￼

Back-substitute: ￼:

￼

(Solution 3)

￼

Use ￼, ￼:

￼

￼￼￼

[Example 2] Evaluate the definite integral

(1) ￼

(2) ￼

(3) ￼

(Solution)

(Solution 1)

Use ￼, ￼: ￼

￼

￼￼

￼￼

￼￼

Here, we used the trigonometric identity.

￼

(Solution 2)

Use ￼, ￼: ￼

￼

￼

￼￼

￼￼

￼

Here, we used the trigonometric identity.

￼

(Solution 3)

Use ￼, ￼: ￼

￼￼

￼￼

￼￼

￼

￼

In this chapter, you will learn:

§3.1 Integration formula

§3.2 Integration By Parts

§3.3 Integration By Substitutions

§3.4 Trigonometric Substitutions

§3.5 Integration By Partial fractions

§3.6 Improper Integral

§3.5 Integration By Partial fractions

In this section, you will study:

▪︎ Partial fractions Decomposition

▪︎ Integration By Partial fractions

● Partial fractions Decomposition

Partial fractions decomposition is a method used to express a rational function as a sum of simpler fractions. Given a rational function ￼, where ￼ and ￼ are polynomials, we decompose it into a sum of simpler rational functions whose denominators are factors of ￼.

Here's a detailed explanation of partial fractions decomposition according to different cases based on the nature of the roots of the denominator ￼. Suppose the degree of ￼ must be less than the degree of ￼. If not, perform polynomial long division first to reduce the degree of ￼.

(i) Case 1: Distinct Linear Factors

If ￼ can be factored into distinct linear factors: ￼

Decomposition:

￼￼

Where ￼ are constants to be determined.

Example:

￼

Decomposition:

￼

Determine Constants:

Multiply both sides by ￼: ￼

Solve for ￼ and ￼ by plugging in convenient values for ￼.

(ii) Case 2: Repeated Linear Factors

If ￼ has repeated linear factors: ￼

Decomposition:

￼￼

Where ￼ are constants.

Example:

￼

Decomposition:

￼

Determine Constants:

Multiply both sides by ￼: ￼

Solve for ￼ and ￼.

(iii) Case 3: Distinct Irreducible Quadratic Factors If ￼ has distinct irreducible quadratic factors: ￼

Decomposition:

￼￼

Where ￼ and ￼ are constants.

Example:

￼

Decomposition:

￼

Determine Constants:

Multiply both sides by ￼: ￼

Solve for ￼ and ￼.

Mastering partial fractions decomposition requires practice, especially in setting up the form of the decomposition and solving for constants. This method simplifies integration, making it a crucial tool in calculus.

In this section, you will study:

▪︎ Partial fractions Decomposition

▪︎ Integration By Partial fractions

● Integration By Partial fractions

Integration by partial fractions is a method used to evaluate integrals of rational functions, which are ratios of polynomials. The idea is to decompose the integrand into simpler fractions that are easier to integrate. Here's a step-by-step explanation of the process:

(i) Ensure Proper Fraction: If the degree of the numerator is greater than or equal to the degree of the denominator, perform polynomial long division first to rewrite the fraction as a sum of a polynomial and a proper fraction.

(ii) Factor the Denominator: Completely factor the denominator of the rational function into linear factors and/or irreducible quadratic factors.

(iii) Set Up Partial Fractions: Write the integrand as a sum of partial fractions. The form of the partial fractions depends on the factors of the denominator:

- For each linear factor ￼, include a term of the form ￼.

- For each repeated linear factor ￼, include terms of the form

￼.

- For each irreducible quadratic factor ￼, include a term of the form

￼.

(iv) Determine Coefficients: Find the coefficients (￼, ￼, etc.) by equating the numerator of the original fraction to the numerator of the partial fractions and solving the resulting system of equations.

(v) Integrate Each Term: Integrate each of the simpler fractions separately.

Let’s consider the integral:

￼

Write the integrand as a sum of partial fractions:

￼

Multiply both sides by the denominator ￼ to clear the fractions:

￼

Expand and combine like terms:

￼

￼

Thus:

￼

￼￼￼

￼

This is the result of integrating by partial fractions.

When we have the partial fractions of the rational functions, we can use the following integration formula.

￼￼

￼￼, ￼

Let’s consider another integral:

￼￼￼

Next, in case the denominator has multiple roots:

￼

By using the partial fractions, we can get:

￼￼￼

￼

￼￼

By multiplying ￼ to both sides and substituting ￼, then:

￼

By multiplying ￼ to both sides and substituting ￼, then:

￼

By substituting ￼, then:

￼

￼

Therefore:

￼￼

￼

￼

Next, in the case the denominator has factors of second order:

￼

By using the partial fractions, we can get:

￼

By multiplying ￼ to both sides and substituting ￼, then:

￼

By substituting ￼, then:

￼

￼

By substituting ￼, then:

￼

￼

Thus:

￼￼

￼

￼

￼

Let ￼. Then:

￼

￼￼￼

￼￼

Let ￼. Then:

￼

￼￼

Therefore:

￼

￼￼

In the end:

￼￼

￼

￼

￼

[Example 1] Find partial fractions decomposition

(1) ￼

(2) ￼

(3) ￼

(4) ￼

(Solution)

(Solution 1)

￼

Multiply both sides by the common denominator ￼ to get rid of the fractions:

￼

￼

￼, ￼

￼￼￼

(Solution 2)

￼

Multiply both sides by the common denominator ￼ to get rid of the fractions:

￼

￼

￼

(Solution 3)

￼

Multiply both sides by the common denominator ￼ to get rid of the fractions:

￼￼

￼, ￼, ￼

￼￼￼

(Solution 4)

￼

￼

Multiply both sides by the common denominator ￼ to get rid of the fractions:

￼￼

￼, ￼, ￼

￼￼￼

[Example 2] Evaluate the integrals.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

￼

Therefore:

￼￼

￼

Here,

￼

￼

￼

￼

￼

￼

In the end,

￼￼￼

(Solution 2)

￼

￼

￼

By substituting ￼, then ￼

￼

By substituting ￼, then ￼

￼

Thus:

￼

In the end: ￼

￼

[Example 3] Evaluate the definite integrals.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

￼

￼

￼

￼￼

￼￼￼

￼

￼

￼

￼￼

￼￼

￼

￼

￼

￼

Thus, ￼￼￼

From (3) and (5),

￼

(Solution 2)

￼

￼

￼

￼

￼

In this chapter, you will learn:

§3.1 Integration formula

§3.2 Integration By Parts

§3.3 Integration By Substitutions

§3.4 Trigonometric Substitutions

§3.5 Integration By Partial fractions

§3.6 Improper Integral

§3.6 Improper Integral

In this section, you will study:

▪︎ Basic Concepts

▪︎ Comparison Test for Convergence

● Basic Concepts

Improper integrals extend the concept of definite integrals to functions that have infinite limits or unbounded integrands. Here’s a detailed explanation of the basic concepts related to improper integrals: Improper integrals are categorized based on the nature of their limits or integrands: (i) Improper Integrals with Infinite Limits

These integrals involve limits that approach infinity. They come in two forms: - Single Infinite Limit: Integrals where one limit of integration is infinite.

￼

- Both Limits Infinite: Integrals where both limits of integration are infinite.

￼

(ii) Improper Integrals with Unbounded Integrands These integrals involve functions that have unbounded behavior (i.e., go to infinity) at some point within the interval. They can occur in two forms: - Unbounded at an Endpoint: The integrand is unbounded as it approaches one or both endpoints of the interval.

￼

- Unbounded at an Interior Point: The integrand is unbounded at some point within the interval.

￼￼

To evaluate an improper integral, we convert it into a limit problem. Here’s how: (i) Infinite Limits of Integration ￼

￼

￼￼

(ii) Unbounded Integrands

Evaluate as ￼ when ￼ where ￼ is unbounded near ￼.

Evaluate as￼ when ￼ where ￼ is unbounded at ￼.

An improper integral converges if the limit used to evaluate it exists and is finite. It diverges if the limit does not exist or is infinite.

Examples:

(i) Improper Integral with Infinite Limits

￼￼￼￼

This integral converges to 1.

(ii) Improper Integral with Unbounded Integrand ￼￼￼￼

This integral converges to 2.

In this section, you will study:

▪︎ Basic Concepts

▪︎ Comparison Test for Convergence

● Comparison Test for Convergence

The comparison test is a method to determine the convergence or divergence of an improper integral by comparing it to another integral whose behavior is known. It is especially useful when dealing with integrals that are difficult to evaluate directly.

The comparison test involves comparing the integrand of the given improper integral to another function with known integral behavior. By establishing inequalities, we can infer the convergence or divergence of the original integral based on the comparison function, similar to how we analyze infinite sequences and infinite series.

There are two primary types of comparison tests: the Direct Comparison Test and the Limit Comparison Test.

(i) Direct Comparison Test

This test compares the integrand of the improper integral with a simpler function that bounds it from above or below.

Positive Function Comparison: If ￼ and ￼ are continuous on ￼ (or ￼ if we are dealing with an unbounded integrand), and ￼ for all ￼ in ￼ (or ￼):

- If ￼ converges, then ￼ also converges.

- If ￼ diverges, then ￼ also diverges.

Similarly, for unbounded integrands:

- If ￼ converges and ￼ near the point where ￼ or ￼ is unbounded, then ￼ also converges.

- If ￼ diverges, then ￼ also diverges.

Example:

Test the convergence of￼ by comparing it to ￼.

- Compare ￼ to ￼:

￼

- Since ￼ converges (it equals 1), by the direct comparison test, ￼ also converges.

Test the convergence of ￼ by comparing it to ￼.

- Compare ￼ to ￼:

￼

Since ￼, by comparison test, ￼ converges.

(ii) Limit Comparison Test

This test is used when the direct comparison test is difficult to apply. It compares the asymptotic behavior of the integrand to a simpler function.

If ￼ and ￼ are continuous on ￼ and ￼ for ￼ large enough:

- Compute ￼.

- If ￼, then ￼ and ￼ both converge or both diverge.

Example:

Determine if ￼ converges by comparing it to ￼.

- Let ￼ and ￼.

- Compute ￼￼￼￼.

- Since ￼, ￼ converges, so ￼ also converges.

The comparison tests for convergence of improper integrals are tools to establish the behavior of an integral without evaluating it directly. By comparing the integrand to a function with known behavior, these tests simplify the analysis of complex integrals, making them essential techniques in calculus.

[Example 1] Evaluate the improper integrals.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

￼￼￼

The improper integral doesn't converge.

(Solution 2)

￼

￼

￼

￼

￼

[Example 2] Test for convergence of the improper integrals.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

￼

Therefore:

￼

(Solution 2)

￼

￼

Therefore:

￼

4. Applications of Integration

In this chapter, we explore practical applications of integration in various mathematical and physical contexts. Integration is a fundamental concept in calculus, providing tools for quantifying and analyzing different properties of functions and geometrical shapes. Each section focuses on a specific application area where integration plays a crucial role.

In section 4.1, we explore how integration can be used to calculate the area enclosed between two curves. The focus will be on setting up and evaluating definite integrals to find these areas. We will cover the basic principles for determining the methodology for computing the area between two curves by integrating the difference of their functions. Examples will demonstrate the application of these techniques to various types of functions and intervals, providing a practical approach to area calculation using integration.

In section 4.2, we examine methods for calculating the volume of solids using integration. Topics include finding volumes by integrating cross-sectional areas. We will also cover volume determination for solids of revolution.

Section 4.3 addresses the calculation of the length of curves using integrals. We will derive the formula for arc length and apply it to various types of curves, including parametric curves. The focus will be on setting up integrals for curve length and evaluating them accurately.

Section 4.4 discusses the application of integration to measure accumulated change and calculate the average value of functions. We will explore how integrals represent accumulated quantities such as total distance traveled or total growth over an interval. Additionally, the concept of the average value of a function over an interval and its computation using integrals will be covered.

In this chapter, you will study:

§4.1 Area between Curves

§4.2 Volume of Solid

§4.3 Length of Curve

§4.4 Accumulated Change and Average of Function

§4.1 Area between Curves

To find the area between two curves ￼ and ￼ over an interval ￼, we use the integral of the absolute value of the difference between the functions. This ensures that the area calculation remains positive regardless of which function is above or below the other over the interval of integration.

￼

￼

This approach accounts for any changes in which function is on top and ensures that the area is correctly measured as a positive quantity.

The Steps for Calculating the Area:

(i) Identify the Interval: Determine the interval ￼ where you want to calculate the area between the curves.

(ii) Find Points of Intersection: Solve ￼ to find the points where the curves intersect. These points will help divide the integral into segments where one function is consistently above the other.

(iii) Determine Which Function is on Top: Over each subinterval, identify which function is greater to properly set up the integral.

(iv) Set Up the Integral: Integrate the absolute value of the difference between ￼ and ￼ over each segment:

￼

where ￼ are subintervals where one function is consistently above the other.

(v) Evaluate the Integral: Compute the integral over each subinterval and sum the results to get the total area.

Example 1: Area Between Two Curves:

Find the area between ￼ and ￼ from ￼ to ￼.

- Find Points of Intersection: Solve ￼:

￼ ￼￼

The intersection points are at ￼ and ￼.

- Determine Which Function is on Top: Over ￼, ￼ is above ￼.

- Set Up the Integral:

￼￼

Since ￼ on ￼, we can directly integrate ￼.

- Evaluate the Integral:

￼￼

￼￼

Example 2: Area Enclosed by Two Curves

Find the area enclosed by the curves ￼ and ￼.

- Find Points of Intersection: Solve ￼:

￼￼

The intersection points are at ￼ and ￼.

- Determine Which Function is on Top: Over ￼, ￼ is above ￼.

- Set Up the Integral:

￼

Since ￼ on ￼, we can integrate ￼.

- Evaluate the Integral:

￼￼

￼￼￼

Horizontal slicing is an alternative approach to calculating the area between curves, particularly useful when dealing with functions that are easier to express as functions of ￼ rather than ￼. This method involves integrating with respect to ￼, which can simplify the problem if the curves intersect horizontally.

￼

For two curves given by ￼ and ￼, the area ￼ between the curves is:

￼

This approach is particularly useful when the curves intersect in such a way that integrating vertically would require splitting the integral into multiple sections.

Steps for Calculating Area Using Horizontal Slicing:

(i) Rewrite the Functions: Express the curves as functions of ￼, i.e., ￼ and ￼.

(ii) Identify the Interval: Determine the interval ￼ over which you want to calculate the area between the curves.

(iii) Find Points of Intersection: Solve ￼ to find the intersection points in terms of ￼.

(iv) Determine Which Function is on the Right: Identify ￼ and ￼ to determine which function is to the right of the other over each subinterval.

(v) Set Up the Integral: Integrate the difference ￼ with respect to ￼ over the interval ￼.

(vi) Evaluate the Integral: Compute the integral to find the area.

Example : Area Enclosed by Two Curves (Horizontal Slicing)

Find the area between ￼ and ￼.

- Rewrite the Functions: Express each curve as a function of ￼:

￼ remains ￼.

- Find Points of Intersection: Solve ￼ and ￼:

￼

- Substitute ￼ into ￼:

￼￼￼

Corresponding ￼-values are ￼ and ￼. Intersection points are ￼ and ￼.

- Determine Which Function is on the Right:

For ￼, ￼ is to the left and ￼ is to the right.

- Set Up the Integral:

￼

- Evaluate the Integral:

￼￼

￼￼

￼￼

￼

Horizontal slicing is preferable when vertical integration becomes complicated due to intersecting curves, or when the functions are more easily expressed and handled as functions of ￼. Horizontal slicing provides a tool for calculating areas in cases where the vertical slicing method is less convenient.

[Example 1] Find the area

Find the area enclosed between the two curves and lying between ￼ and ￼.

￼, ￼

(Solution)

The graph of each function is shown below.

￼

￼￼

[Example 2] Find the area

Obtain the area enclosed by x-axis and the following curve.

￼

(Solution)

The ￼-intercepts are ￼ and ￼ from ￼.

￼￼

Here,

￼

￼￼

Thus,

￼

[Example 3] Find the area

Find the area enclosed by ￼, ￼-axis and ￼.

￼

(Solution)

Using the definite integral in this case results in a negative value because ￼ is negative over the specified interval, as illustrated in the figure below.

￼

￼

We need to use the absolute value to obtain the correct answer.

￼

￼

[Example 4] Find the area

Find the area enclosed by ￼ and ￼

￼

￼

(Solution)

To calculate the area, we must plot the graphs of each function. As depicted in the figure below, ￼ and ￼ intersect at three points.

￼

Let's determine the ￼-coordinates of the intersections.

￼

￼

￼

￼

First, we calculate the area (￼) over the interval. ￼.

￼

Since:

￼

￼

￼

￼

Note that ￼ is greater than ￼ on the interval ￼. Then: ￼

￼

￼

Therefore:

￼

[Example 5] Find the area

Find the area enclosed by the given graphs.

￼

(Solution)

Let's find the x-coordinates and y-coordinates of the intersections.

￼

￼

From these values, we get:

￼

￼

Typically, the x-coordinates serve as the upper and lower limits; however, this does not apply in this example. Calculating the definite integral to find the area as such would be incorrect.

￼

As depicted in the figure below, we calculate the area only for the interval between ￼ and ￼.

￼

To get the correct answer, we need to integrate on the interval ￼.

￼￼

￼

￼

￼

By the way, calculating the area using horizontal slicing simplifies the process and yields the same result as above.

￼

￼

￼

In this chapter, you will study:

§4.1 Area between Curves

§4.2 Volume of Solid

§4.3 Length of Curve

§4.4 Accumulated Change and Average of Function

§4.2 Volume of Solids

In this section, you will study:

▪︎ Volume of a Solid with Known Cross-Sections ▪︎ Volume of a Cuboid

▪︎ Volume of a Right Circular Cone

▪︎ Volume of a Sphere

▪︎ Volume of a Solid of Revolution

We can calculate the area of a region enclosed by curves by dividing it into smaller rectangles. Similarly, this method can be used to determine the volume of a solid. Here's a step-by-step explanation: (i) Understand the Solid and Cross-Section: - Imagine the solid as being "sliced" perpendicular to one of its axes (usually the ￼-axis) into many thin slices.

- Each slice is a cross-section of the solid at a particular value of ￼.

(ii) Define the Cross-Sectional Area:

- Let ￼ represent the area of the cross-section at position ￼.

- This area function ￼ varies along the length of the solid, depending on the shape of the solid.

(iii) Thickness of Each Slice:

- Consider a small slice at position ￼ with a tiny thickness ￼. The thickness ￼ is an infinitesimally small change in ￼.

(iv) Volume of a Thin Slice:

- The infinitesimal volume ￼ of this thin slice can be approximated as the product of the cross-sectional area and the thickness: ￼

- This represents a very small volume element of the solid.

(v) Summing Up All Slices:

- To find the total volume of the solid, sum up the volumes of all the thin slices from one end of the solid to the other.

- This summation is achieved through integration. If the solid extends from ￼ to ￼, the total volume ￼ is: ￼

Let's learn more through a detailed example.

● Volume of a Solid with known cross-section

To calculate the volume of a solid with known cross-sections using integrals, follow these steps: (i) Identify the Axis and Interval

Determine the axis along which the cross-sections are taken (usually ￼, ￼, or ￼) and identify the interval ￼ over which the solid extends along that axis.

(ii) Find the Cross-Sectional Area Function Determine the shape of the cross-sections and express the area of a cross-section as a function of the variable along the axis. Let's denote this cross-sectional area function as ￼ if the cross-sections are perpendicular to the ￼-axis. Similarly, use ￼ or ￼ if perpendicular to the ￼-axis or ￼-axis, respectively.

(iii) Set Up the Integral

Set up an integral of the cross-sectional area function over the interval to find the volume ￼ of the solid. The general formula is: ￼

where:

- ￼ is the area of the cross-section at a position ￼.

- ￼ are the limits of integration along the ￼-axis.

(iv) Compute the Integral

Evaluate the integral to find the volume. Depending on the complexity of the area function ￼, this can be done analytically (by hand) or numerically (using numerical integration methods).

Example: Volume of a Solid with Semicircular Cross-Sections Imagine a solid whose cross-sections perpendicular to the ￼-axis are semicircles with a radius that varies along ￼ from ￼ to ￼.

￼

(i) Determine the Cross-Sectional Area Function: If the radius of each semicircle at a position ￼ is ￼, the area of a semicircular cross-section is: ￼

(ii) Set Up the Integral:

Suppose the radius ￼ changes linearly from 0 to ￼ over the interval ￼. The radius function is ￼. Therefore, the area function becomes: ￼

Thus:

￼

(iii) Evaluate the Integral

￼￼￼￼

This same approach can be applied to solids with different types of cross-sections: (i) Rectangles: If cross-sections are rectangles, then ￼, where ￼ and ￼ are the width and height of the rectangle at position ￼.

(ii) Triangles: For triangular cross-sections, ￼, where ￼ and ￼ are the base and height of the triangle at position ￼.

(iii) Ellipses: For elliptical cross-sections, ￼, where ￼ and ￼ are the semi-major and semi-minor axes at position ￼.

In this section, you will study: ▪︎ Volume of a Solid with Known Cross-Sections ▪︎ Volume of a Cuboid

▪︎ Volume of a Right Circular Cone ▪︎ Volume of a Sphere

▪︎ Volume of a Solid of Revolution

● Volume of a Cuboid

Let ￼ and ￼ be the width and length of the base of a cuboid, respectively, and let ￼ be the height of the cuboid. The height is perpendicular to the cross-section (which is parallel with the base). The cross-sectional area ￼ at position ￼ can be expressed as follows.

￼

From this, we can get the volume using integral.

￼

In this section, you will study: ▪︎ Volume of a Solid with Known Cross-Sections ▪︎ Volume of a Cuboid

▪︎ Volume of a Right Circular Cone ▪︎ Volume of a Sphere

▪︎ Volume of a Solid of Revolution

● Volume of a Right Circular Cone.

A right circular cone is depicted in the figure below.

￼

The cross-section vertical to the direction of height is a circle. Let ￼ be the area of the cross-section whose vertical distance is x from the bottom surface. Then: ￼

From this, we can get the infinitesimal volume ￼.

￼

Where:

- ￼

Thus:

￼

￼￼￼

￼

In this section, you will study:

▪︎ Volume of a Solid with Known Cross-Sections ▪︎ Volume of a Cuboid

▪︎ Volume of a Right Circular Cone ▪︎ Volume of a Sphere

▪︎ Volume of a Solid of Revolution

● Volume of a Sphere

A sphere, whose radius is R, is given in the following figure.

￼

The cross-section vertical to the radius is a circle. Let ￼ be the cross-section area whose vertical distance is ￼ from the center. Then: ￼

From this, we can get the infinitesimal volume ￼.

￼

Where:

- ￼

Thus:

￼

￼

￼

In this section, you will study:

▪︎ Volume of a Solid with Known Cross-Sections ▪︎ Volume of a Cuboid

▪︎ Volume of a Right Circular Cone ▪︎ Volume of a Sphere

▪︎ Volume of a Solid of Revolution

● Volume of a Solid of Revolution

When a graph revolves around a specific axis, the volume of the resulting solid of revolution can be determined by integrating the area of circular cross-sections along that axis.

(i) Volume of a Solid Revolving About the ￼-axis As illustrated in the figure below, when the graph of ￼ between ￼ and ￼ revolves around the ￼-axis, the area of the cross-section can be determined as follows.

￼

￼

With this bottom surface, we can find the infinitesimal volume.

￼

Thus:

￼

￼

(ii) Volume of a Solid Revolving About the ￼-axis As depicted in the figure below, when the graph of ￼ lying between ￼ and ￼ revolves around the y-axis, the area of the cross-section can be calculated as follows.

￼

￼

With this bottom surface, we can find the infinitesimal volume.

￼

Thus:

￼

[Example 1] Find the volume

Let ￼ represent the region enclosed by ￼, ￼, and the ￼-axis, which serves as the base of a solid. When the solid has square cross-sections perpendicular to the x-axis, calculate the volume.

￼

(Solution)

The region R is shown in the following figure.

￼

Let's find the area of cross-section vertical to the ￼-axis.

￼

From this, we can get the infinitesimal volume ￼.

￼

Thus:

￼￼

￼

￼

[Example 2] Find the volume

Consider the solid of revolution obtained by revolving around the ￼-axis the region enclosed by the given graphs. Find the volume.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

From the cross-section vertical to the ￼-axis, we can get the infinitesimal volume ￼.

￼

Therefore:

￼￼￼

￼

(Solution 2)

From the cross-section vertical to the ￼-axis, we can get the infinitesimal volume ￼.

￼

Therefore:

￼￼￼

￼

[Example 3] Find the volume

Consider the solid of revolution obtained by revolving around the ￼-axis the region enclosed by the given graphs. Find the volume.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

From the cross-section vertical to the ￼-axis, we can get the infinitesimal volume ￼.

￼

Therefore:

￼￼￼

￼

(Solution 2)

The cross-sections vary according to the range of ￼.

When ￼, then the infinitesimal volume is:

￼

Therefore:

￼

When ￼, then then the infinitesimal volume is:

￼

Therefore:

￼

￼

￼

In this chapter, you will study:

§4.1 Area between Curves

§4.2 Volume of Solid

§4.3 Length of Curve

§4.4 Accumulated Change and Average of Function

§4.3 Length of Curve

By summing an infinite number of line segments, we can determine the length of a curve. As illustrated in the figure below, let's calculate the length of the graph between ￼ and ￼. When the interval ￼ is divided into ￼ subintervals, the length of the ￼-th line segment can be calculated using the Pythagorean theorem.

￼

￼

From this, we can get the infinitesimal length ￼.

￼

Therefore:

￼

We denote this integration with the variable ￼ when the integration interval is ￼.

￼

We denote this integration with the variable ￼ when the integration interval is ￼.

￼

Or we can use another parameter ￼ to denote the integration. Let the integration interval be ￼. Then: ￼

[Example 1] Find the length

For the given function and the interval, obtain the length of curve.

(1) ￼

(2) ￼

(3) ￼

(Solution)

(Solution 1)

We use the following formula.

￼

￼

Therefore:

￼

(Solution 2)

We use the following formula.

￼

￼

Therefore:

￼

Let ￼. Then: ￼

￼

￼

￼

￼

￼

￼

￼

(Solution 3)

We use the following formula.

￼

￼

￼

￼

￼

￼

In this chapter, you will study:

§4.1 Area between Curves

§4.2 Volume of Solid

§4.3 Length of Curve

§4.4 Accumulated Change and Average of Function

§4.4 Accumulated Change and Average of Function

In this section, you will study:

▪︎ Position, Velocity and Acceleration

▪︎ Distance Traveled

▪︎ Accumulated Change of Physical Quantities

▪︎ Average Value of a Function

● Position, Velocity and Acceleration

Position, velocity, and acceleration are fundamental concepts in physics and calculus that describe motion.

(i) Position

Position describes the location of an object in space relative to a reference point or coordinate system. Typically represented as a vector in one, two, or three dimensions. For instance, in one-dimensional motion, position might be represented as ￼, where ￼ is time. In two dimensions, position could be represented as ￼. The unit of position is the meter. If a particle moves along a straight line, its position at time ￼ might be given by ￼, where ￼ is time in seconds.

(ii) Velocity

Velocity is the rate of change of position with respect to time. It describes how fast and in what direction an object is moving. Velocity is a vector quantity. In one-dimensional motion, it’s often represented as ￼, the derivative of the position function with respect to time. In two dimensions, it's ￼. The unit of velocity is meters per second (m/s). There are two types of velocity:

- Instantaneous Velocity: The velocity of an object at a specific moment in time ￼.

- Average Velocity: The total displacement divided by the total time taken, ￼.

For the position function ￼, the velocity would be ￼.

(iii) Acceleration

Acceleration is the rate of change of velocity with respect to time. It indicates how quickly an object’s velocity is changing. Acceleration is also a vector. In one-dimensional motion, it's represented as ￼, the second derivative of the position function. In two dimensions, it’s ￼. The unit of acceleration is meters per second squared (m/s²). There are two types of acceleration.

- Instantaneous Acceleration: The acceleration of an object at a specific moment in time ￼.

- Average Acceleration: The change in velocity divided by the time taken, ￼.

For the velocity function ￼, the acceleration would be ￼, indicating a constant acceleration.

(iv) Relationships

- Velocity from Position: Velocity is the derivative of position.

￼

- Acceleration from Velocity: Acceleration is the derivative of velocity.

￼

- Position from Velocity: Position is the integral of velocity.

￼

- Velocity from Acceleration: Velocity is the integral of acceleration.

￼

(v) Examples and Applications

- Free Fall: In free fall under gravity, the position function might be ￼ (assuming initial velocity and position at the staring point are zero), where ￼ is the acceleration due to gravity. The velocity function is ￼, and the acceleration is ￼, constant.

- Projectile Motion: The position of a projectile can be described by separate equations for horizontal and vertical motion. The velocity and acceleration are derived accordingly, considering both components.

In this section, you will study:

▪︎ Position, Instantaneous Velocity and Acceleration

▪︎ Distance Traveled

▪︎ Accumulated Change of Physical Quantities

▪︎ Average Value of a Function

● Distance Traveled

Distance traveled is the total length of the path that an object covers as it moves, regardless of direction. It measures the total length of the path taken by the object, not necessarily its change in position (which could be described by displacement). Unlike displacement, distance traveled is always a positive quantity and does not consider direction; it’s a scalar quantity.

When dealing with continuous motion described by a velocity function, the process involves integrating the absolute value of the velocity function over the time interval:

If the velocity of an object is given as a function of time ￼:

￼

Here, ￼ and ￼ are the initial and final times.

If an object’s velocity is ￼, find the distance traveled from ￼ to ￼.

Let ￼ to find ￼. The velocity changes sign at ￼.

￼￼￼

If the position ￼ as a function of time is known:

If ￼, find the distance traveled from ￼ to ￼.

￼

￼

If we just find the displacement of position from ￼ to ￼, Then:

￼￼￼

If the object’s motion is described by parametric or vector functions ￼:

￼

￼

￼

If ￼, find the distance from ￼ to ￼.

￼

￼￼

￼￼￼

If we just find the displacement of position, Then:

￼￼￼

Note that the displacement is a vector.

In this section, you will study:

▪︎ Position, Instantaneous Velocity and Acceleration

▪︎ Distance Traveled

▪︎ Accumulated Change of Physical Quantities

▪︎ Average Value of a Function

● Accumulated Change of Physical Quantities.

The concept of accumulated change refers to the total change in a physical quantity over a given interval, often represented mathematically by definite integrals. This accumulated change can apply to various contexts such as distance traveled, area under a curve, total mass, and more.

Accumulated change represents the net result of continuously adding up small, instantaneous changes in a physical quantity over a certain interval. In mathematical terms, this is often computed using integration, which sums up infinitesimal changes over an interval to find the total change.

Let the function ￼ that describes the rate of change of the quantity you're interested in. We can compute the definite integral of the rate function over the interval ￼:

￼

This integral sums up all the infinitesimal changes ￼ from ￼ to ￼.

The rate of change function describes how the physical quantity changes instantaneously. The interval specifies the range over which the total change is calculated. The definite integral sums these instantaneous changes to find the total accumulated change over the interval.

In this section, you will study:

▪︎ Position, Instantaneous Velocity and Acceleration

▪︎ Distance Traveled

▪︎ Accumulated Change of Physical Quantities

▪︎ Average Value of a Function

● Average Value of a Function.

The average value of a function over a given interval ￼ is a single value that represents the "mean" height of the function over that interval. It provides an average measurement of the function’s output values, giving an idea of the function's overall behavior across the interval.

Mathematically, if ￼ is a continuous function on the interval ￼, the average value of ￼ over ￼ is given by:

￼

Consider the function ￼ over the interval ￼. Let's find the average value of this function on the given interval.

￼￼

The length of the interval ￼ is ￼. So, the average value is:

￼￼

[Example 1] Solve the following problems

In the number line, a particle lies in ￼ at ￼. And the velocity of the particle is given with the following when ￼.

￼

(1) Evaluate the change of the position on the interval ￼.

(2) Find the position at ￼.

(3) Evaluate the distance traveled during the interval ￼.

(Solution)

(Solution 1)

We can get the change of the position with the definite integral. Let ￼ be the position of time ￼. Then: ￼

Let ￼. Then: ￼

￼

￼

￼

(Solution 2)

Let ￼ be the function of the position. Then: ￼

Therefore:

￼

(Solution 3)

The distance traveled on the interval ￼ is: ￼

When ￼, ￼: ￼￼￼

When ￼, ￼: ￼￼￼

When ￼, ￼: ￼￼￼

When ￼, ￼.

￼

When ￼, ￼.

￼

When ￼, ￼.

￼

When ￼, ￼.

￼

When ￼, ￼.

￼

When ￼, ￼.

￼

Thus, the distance traveled is:

￼

[Example 2] Solve the following problems

(1) When a particle is moving along a line, find the distance traveled by a particle during the time interval ￼. The velocity of the particle is given as follows.

￼

(2) Water in the pipe whose radius is 2cm is flowing with the velocity of ￼[cm/sec]. Find the volume of water flowing during the time interval ￼.

(Solution)

(Solution 1)

When the velocity is given, we can get the distance.

￼￼

Therefore:

￼￼

￼

￼￼￼

(Solution 2)

We can obtain the volume by multiplying the velocity, elapsed time and the area of the cross-section. Suppose that the time interval is ￼, then: ￼

From this, we can get the infinitesimal volume ￼.

￼

Therefore:

￼￼￼

[Example 3] Find the average value

Find the average value of the given function for the interval.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

￼￼

Let ￼. Then: ￼

￼

￼

￼

Therefore:

￼￼￼￼

(Solution 2)

￼

￼

Chapter 5. Approximation of Definite Integral When the exact value of a definite integral is difficult to calculate analytically, numerical approximation methods are used. These methods approximate the area under a curve by dividing it into simple geometric shapes, such as rectangles or trapezoids. Here, we discuss the use of rectangles (left sum, right sum, midpoint sum) and the trapezoidal rule.

In this chapter, you will learn: §5.1 Rectangular Approximation Methods

§5.2 Trapezoidal Rule

§5.1 Rectangular Approximation Methods

Riemann sums are techniques for approximating the area under a curve or the value of a definite integral. They use different methods for sampling points within subintervals to construct sums that estimate this area. The main types are the Left Riemann Sum, Right Riemann Sum, and Midpoint Riemann Sum.

Each Riemann sum gives a different approach to approximating the area under a curve. The accuracy improves as the number of subintervals increases. These methods form the foundation for more advanced numerical approximation techniques and are essential tools for understanding definite integrals in calculus.

In this section, you will learn:

▪︎ Left Riemann Sum

▪︎ Right Riemann Sum

▪︎ Midpoint Riemann Sum

● Left Riemann Sum

The Left Riemann Sum approximates the area under a curve using the left endpoints of subintervals.

(i) Divide the Interval: Split the interval ￼ into ￼ subintervals of equal width ￼:

￼

Define points ￼ where ￼.

(ii) Construct Rectangles: For each subinterval ￼, the height of the rectangle is the function value at the left endpoint ￼, so the area is ￼.

(iii) Sum the Areas The Left Riemann Sum is:

￼

(iv) Example

To approximate ￼ using ￼:

- ￼

- Heights: ￼, ￼, ￼, ￼

- Left Riemann Sum:

￼

In this section, you will learn: ▪︎ Left Riemann Sum

▪︎ Right Riemann Sum

▪︎ Midpoint Riemann Sum

● Right Riemann Sum

The Right Riemann Sum uses the right endpoints of subintervals to construct the rectangles for the approximation.

(i) Divide the Interval: Partition ￼ into ￼ subintervals, each of width ￼, with points ￼.

(ii) Construct Rectangles: For each subinterval ￼, the height of the rectangle is the function value at the right endpoint ￼, giving an area of ￼.

(iii) Sum the Areas: The Right Riemann Sum is: ￼

(iv) Example

For ￼ with ￼: - ￼

- Heights: ￼, ￼, ￼, ￼

- Right Riemann Sum: ￼

In this section, you will learn:

▪︎ Left Riemann Sum

▪︎ Right Riemann Sum

▪︎ Midpoint Riemann Sum

● Midpoint Riemann Sum

The Midpoint Riemann Sum uses the midpoint of each subinterval to determine the height of the rectangles.

(i) Divide the Interval: Partition ￼ into ￼ subintervals of equal width ￼, and let the sample point ￼ be the midpoint of each subinterval ￼:

￼￼

(ii) Construct Rectangles: The height of the rectangle is the function value at the midpoint ￼, leading to an area of ￼.

(iii) Sum the Areas: The Midpoint Riemann Sum is:

￼

(iv) Example

For ￼ with ￼:

- ￼

- Midpoints: ￼, ￼, ￼, ￼

- Heights: ￼, ￼, ￼, ￼

- Midpoint Riemann Sum:

￼￼￼

§5.2 Trapezoidal Rule

The Trapezoidal Rule approximates the area under a curve by dividing it into trapezoids instead of rectangles. This method typically provides a more accurate approximation than the Riemann sums for the same number of subdivisions.

(i) Divide the interval ￼ into ￼ equal subintervals, each of width ￼.

(ii) For each subinterval ￼, form a trapezoid with the heights ￼ and ￼.

(iii) Sum the areas of all trapezoids.

￼￼￼

where ￼.

(iv) Example

For ￼ with ￼:

- ￼

- Midpoints: ￼, ￼, ￼, ￼, ￼

- Heights: ￼, ￼, ￼, ￼, ￼

- Trapezoid Rule:

￼￼

[Example 1] Approximation of Definite Integral

For a given function and the partition, answer the following.

￼

￼

* You may need a calculator for this example.

(1) Evaluate the following definite integral.

￼

(2) Approximate the given integral with left Riemann sum.

(3) Approximate the given integral with right Riemann sum.

(4) Approximate the given integral with midpoint Riemann sum.

(5) Approximate the given integral with trapezoid rule.

(Solution)

(Solution 1)

￼

￼

For given partition, let's find the ￼.

Since ￼, then: ￼, ￼, ￼

￼, ￼, ￼

￼

￼

(Solution 2)

￼

￼￼￼￼

(Solution 3)

￼

￼￼￼￼

(Solution 4)

￼

￼

￼

￼

￼

￼￼￼

￼

(Solution 4)

￼

￼

￼

￼

￼

￼

￼

￼

￼

Chapter 6. Introduction to Differential Equation In this chapter, you will learn: §6.1 Basic Concepts

§6.2 Separation of Variables for Solving ODEs §6.3 Slope Fields

§6.4 Euler Method

§6.1 Basic Concepts

In this section, you will study: ▪︎ Definition and General Forms ▪︎ Properties

● Definition and General Forms

A differential equation is a mathematical equation involving a function and its derivatives, expressing how the function changes in relation to one or more variables. These equations are essential in modeling dynamic systems in science, engineering, economics, and many other fields.

(i) General Form:

- ￼, ￼

- ￼, ￼

(ii) Example:

- ￼, ￼

- ￼, ￼

● Properties

(i) Order of a Differential Equation: The highest derivative present in the equation. For example, ￼ is a second-order differential equation.

(ii) Degree of a Differential Equation: The exponent of the highest derivative, provided the equation is a polynomial in derivatives. For instance, ￼ is a degree 2 differential equation.

(iii) Linear vs. Non-linear:

- Linear: The dependent variable and its derivatives appear to the first power and are not multiplied by each other.

Example: ￼.

- Non-linear: Involves products or non-linear functions of the dependent variable or its derivatives.

Example: ￼.

(iv) Ordinary Differential Equations (ODEs): Involve functions of a single variable and their derivatives.

Example: ￼.

(v) Initial Conditions:

- Initial Conditions: Specify the value of the function and its derivatives at a specific point. For example, for ￼, an initial condition could be ￼.

(vi) Differential Equations and Their Solutions - Exponential Growth/Decay:

￼

Solution: ￼

- Simple Harmonic Oscillator: ￼

Solution: ￼

§6.2 Separation of Variables for Solving ODEs

Separation of Variables is a technique used to solve certain ordinary differential equations by separating the variables ￼ and ￼ into different sides of the equation.

Steps for Separation of Variables:

(i) Rewrite the Equation: Start with an equation that can be separated. The typical form is:

￼

(ii) Separate Variables: Rearrange to isolate ￼ terms on one side and ￼ terms on the other:

￼

(iii) Integrate Both Sides: Integrate each side with respect to its variable:

￼

(iv) Solve for ￼: Solve the resulting equation for ￼ if possible, including the constant of integration.

Let’s consider the ODE:

￼￼

￼￼

￼

where ￼ is a new constant ￼.

Let’s solve ￼:

￼

￼

￼

￼

￼

Differential equations are crucial in understanding the behavior of dynamic systems. The method of separation of variables is a tool for solving ODEs that can be decomposed into separate functions of their variables. This technique simplifies complex differential relationships, making them more tractable for analysis and solution.

[Example 1] Find the position

Let ￼ be the velocity of a particle moving on a line. Obtain the position ￼. Here ￼ is the variable denoting the time. ( ￼ ) ￼

(Solution)

The velocity is the instantaneous change of the position. Thus: ￼

￼

By integration, we can obtain ￼.

￼

Since ￼: ￼

[Example 2] Find the solution

Solve the following differential equation. (￼) ￼

(Solution)

We can separate the variable ￼ and ￼.

￼

￼

￼

￼

Since ￼: ￼

￼

We can verify that ￼ is the solution to the given equation.

￼

[Example 3] Find the solution

Solve the following differential equation.

￼

(Solution)

By using the separation of the variables:

￼

￼

Then:

￼

Since ￼:

￼

￼

§6.3 Slope Fields

A slope field (or direction field) is a graphical representation of a first-order differential equation. It consists of short line segments or arrows drawn at a grid of points in the plane, each representing the slope given by the differential equation at that point. This provides a visual way to understand the behavior of solutions to the differential equation without solving it analytically.

To construct a slope field for a differential equation of the form ￼, follow these steps:

(i) Choose a Grid of Points: Select a set of points ￼ in the plane over the region of interest.

(ii) Calculate Slopes: For each point ￼ in the grid, compute the slope ￼ given by the differential equation.

(iii) Draw Line Segments: At each point ￼, draw a short line segment with the slope ￼.

For instance, in the slope field for ￼:

- At ￼, the slope is ￼. A segment with slope ￼ is drawn at ￼.

- At ￼, the slope is ￼. A horizontal segment is drawn at ￼.

- At ￼, the slope is ￼. A horizontal segment is drawn at ￼.

we can draw the line segments denoting the slope at each coordinate (x, y) geometrically. The collection of the line segments is the slope field.

￼

The solution to the given equation is:

￼

Here, we can compare the solution and slope field. Refer to the following figure.

￼

By using the slope field, we can solve the differential equation geometrically instead of analytically.

Let's draw the slope fields in the following differential equations.

￼

￼

￼

￼

￼

￼

￼

￼

￼

§6.4 Euler Method

For the differential equation ￼, we can rewrite this with the definition of the derivative.

￼

Here, we can get the approximation of ￼ numerically with the following formula.

￼

That is, we can obtain ￼. This is the principle of the Euler method.

￼

For example ￼, the approximation formula is as follows.

￼

Here, let ￼ and the starting point be ￼. Then: ￼

￼

Therefore, ￼ reaches from the point (0, 1) to the next point (0.1, 1.1). Let's find the next point.

Since ￼ and ￼. Then: ￼

The approximation gets close to the exact solution as ￼ approaches to zero. In the following figure, we can see the approximation of the solution when ￼, ￼ and ￼ respectively.

￼

Chapter 7. Infinite Series

§7.1 Convergence Test of Nonnegative Series

§7.2 Convergence Test of Alternating Series

§7.3 Absolute Convergence and Conditional Convergence

§7.1 Convergence Test of Nonnegative Series

In this section, you will study: ▪︎ Comparison Test ▪︎ Limit Comparison Test ▪︎ Integral Test

▪︎ p-Series Test

▪︎ Ratio Test

▪︎ Root Test

● Comparison Test

The Comparison Test is a straightforward technique used to determine the convergence or divergence of an infinite series by directly comparing its terms with those of a known series.

The basic idea behind the Comparison Test is to compare the terms of a given series ￼ with the terms of another series ￼ whose convergence or divergence is already known. If the terms ￼ and ￼ have a consistent relationship (one is always less than or equal to the other for sufficiently large ￼), the behavior of ￼ can be inferred from ￼.

Let ￼ and ￼ be series with positive terms ￼ and ￼.

If ￼ for all ￼ sufficiently large (i.e., ￼ for some ￼), and ￼ converges, then ￼ also converges.

If ￼ for all ￼ sufficiently large, and ￼ diverges, then ￼ also diverges.

The Comparison Test provides a method to infer the behavior of an unknown series by directly comparing it to a series with known properties. This test is particularly effective when you can establish a clear inequality between the terms of the two series for sufficiently large ￼.

In this section, you will study:

▪︎ Comparison Test

▪︎ Limit Comparison Test

▪︎ Integral Test

▪︎ p-Series Test

▪︎ Ratio Test

▪︎ Root Test

● Limit Comparison Test

The Limit Comparison Test is a method used in calculus and mathematical analysis to determine the convergence or divergence of an infinite series. This test is particularly useful when you have a series that is difficult to analyze directly but can be compared to another series whose convergence behavior is known.

Given two series ￼ and ￼ with positive terms ￼ and ￼:

￼

If ￼, then either both series ￼ and ￼ converge or both diverge.

If ￼ and ￼ converges, then ￼ also converges.

If ￼ and ￼ diverges, then ￼ also diverges.

Consider the series ￼. Let's determine its convergence by comparing it to the series ￼.

(i) Identify ￼ and choose ￼.

(ii) Calculate the limit:

￼￼￼￼

(iii) Since ￼ (a positive finite number), the Limit Comparison Test tells us that ￼ and ￼ either both converge or both diverge.

(iv) The series ￼ (the harmonic series) diverges, so ￼ also diverges.

The Limit Comparison Test is a tool for series comparison, providing a clear pathway to determine the behavior of complex series by relating them to simpler, well-understood ones.

In this section, you will study:

▪︎ Comparison Test

▪︎ Limit Comparison Test

▪︎ Integral Test

▪︎ p-Series Test

▪︎ Ratio Test

▪︎ Root Test

● Integral Test

The Integral Test is a method used to determine the convergence or divergence of an infinite series by comparing it to an improper integral. This test is particularly useful for series whose terms decrease monotonically and are continuous, positive functions.

Consider a series ￼, where ￼ for a continuous, positive, and monotonically decreasing function ￼ defined for ￼. The Integral Test states:

The series ￼ converges if and only if the improper integral ￼ converges.

Conversely, the series ￼ diverges if and only if the improper integral ￼ diverges.

Consider the series ￼. Let's use the Integral Test to determine its convergence.

￼

If ￼:

￼￼￼

￼

If ￼:

￼

Thus:

If ￼, then ￼ converges.

If ￼, then ￼ diverges.

To visualize the Integral Test, consider plotting the function ￼ along with the rectangles representing the terms ￼. The area under ￼ from 1 to infinity represents the integral ￼, while the sum of the areas of the rectangles corresponds to the series ￼.

If the total area under the curve is finite (integral converges), then the sum of the areas of the rectangles (series) is also finite.

If the area under the curve is infinite (integral diverges), then the sum of the rectangles’ areas (series) is also infinite.

￼

In this section, you will study: ▪︎ Comparison Test ▪︎ Limit Comparison Test ▪︎ Integral Test

▪︎ p-Series Test

▪︎ Ratio Test

▪︎ Root Test

● p-Series Test

The p-series test is a method used to determine the convergence or divergence of an infinite series. An infinite series is an expression of the form: ￼

where ￼ is a real number. This specific form of series is called a p-series.

The p-series test states that the series ￼ converges if and only if ￼. Conversely, the series diverges if ￼.

For ￼, the terms ￼ decrease rapidly enough such that their sum approaches a finite limit. Intuitively, as ￼ becomes large, ￼ grows very fast when ￼, making ￼ very small very quickly.

A more formal proof involves comparing the p-series to an integral test, which will be discussed later. The function ￼ is continuous, positive, and decreasing for ￼.

By evaluating the improper integral ￼, we find that this integral converges if and only if ￼.

For ￼, the terms decrease even more slowly, making divergence even more apparent. Similarly, using the integral test for ￼, the integral ￼ diverges, thus the series does too.

In this section, you will study: ▪︎ Comparison Test ▪︎ Limit Comparison Test ▪︎ Integral Test

▪︎ p-Series Test

▪︎ Ratio Test

▪︎ Root Test

● Ratio Test

The ratio test is a widely used method to determine the convergence or divergence of an infinite series: Given a series ￼, the ratio test involves examining the limit: ￼

If ￼, the series ￼ converges absolutely.

If ￼, the series ￼ diverges.

If ￼, the ratio test is inconclusive, and the series may converge or diverge; other tests must be used to determine the behavior of the series.

The ratio test is particularly useful for series where the terms ￼ involve factorials, exponential functions, or other expressions where taking the ratio of successive terms simplifies the analysis.

In this section, you will study: ▪︎ Comparison Test ▪︎ Limit Comparison Test ▪︎ Integral Test

▪︎ p-Series Test

▪︎ Ratio Test

▪︎ Root Test

● Root Test

The root test, also known as the n-th root test, is another useful method to determine the convergence or divergence of an infinite series. It is particularly effective for series where each term is raised to a power that grows with ￼.

Given a series ￼, the root test involves examining the limit: ￼

If ￼, the series ￼ converges absolutely.

If ￼, the series ￼ diverges.

If L = 1, the root test is inconclusive, and the series may converge or diverge; other tests must be used to determine the behavior of the series.

[Example 1] Comparison Test

Verify the convergence of the series by the comparison test.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

The following inequality holds true.

￼

Therefore:

￼

The series of the right side converges by the p-series test, so the given series converges.

(Solution 2)

Since ￼ for ￼: ￼

Thus:

￼

Therefore, the given series is convergent.

[Example 2] Limit Comparison Test

Verify the convergence of the following series by the limit comparison test.

(1) ￼

(2) ￼

(3) ￼

(Solution)

(Solution 1)

Let's do the limit comparison test with the harmonic series.

￼

Since the harmonic series is divergent, the given series is not convergent.

(Solution 2)

Let's do the limit comparison test with the following series.

￼

￼

Therefore, the given series is convergent.

(Solution 3)

Let's do the limit comparison test with the harmonic series.

￼

Therefore, the given series diverges.

[Example 3] Integral Test

Verify the convergence of the following series by the integral test.

(1) ￼

(2) ￼

(3) ￼

(Solution)

(Solution 1)

Let ￼. Then: ￼

Therefore, the given series is divergent.

(Solution 2)

Let ￼. Then: ￼

￼

￼

￼

Thus:

￼

￼

Therefore, the given series is convergent.

(Solution 3)

Let ￼. Then: ￼￼

Therefore, the given series is convergent.

[Example 4] Ratio Test and Root Test

Verify the convergence of the following series by the ratio test or the root test.

(1) ￼

(2) ￼

(3) ￼

(Solution)

(Solution 1)

By the ratio test, the series is convergent.

￼

(Solution 2)

By the ratio test, the series is not convergent.

￼

(Solution 3)

By the root test, the series is convergent.

￼

￼

Here, ￼

§7.2 Convergence Test of Alternating Series

The Alternating Series Test, also known as the Leibniz Test, is a criterion used to determine the convergence of alternating series. An alternating series is a series whose terms alternate in sign. This test provides a straightforward way to assess the convergence of such series under specific conditions.

Consider an alternating series of the form:

￼￼

where ￼ is a sequence of positive terms.

The Alternating Series Test states that the series ￼ converges if the following two conditions are met:

(i) Monotonic Decrease: The sequence ￼ is monotonically decreasing for all ￼ (i.e., ￼ for all ￼).

(ii) Limit to Zero: The sequence ￼ approaches zero as ￼ approaches infinity (i.e., ￼).

Consider the series ￼.

Since:

￼ and ￼,

The series converges.

Consider another series ￼.

Since:

￼ and ￼

The series converges.

* Refer to Volume 1 for the error bound of the partial sum of an alternating series.

[Example 1] Convergence test of alternating series

Verify the convergence of the following series.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

Let ￼. Then: ￼

￼

Therefore, the series is convergent.

(Solution 2)

Let ￼. Then: ￼

Since the sequence is increasing, the series does not converge. Additionally, the individual terms do not converge to zero.

[Example 2] Error bound of alternating series

How many terms should be added to approximate the sum of the following series to three decimal places?

￼

(Solution)

The error bound between the sum of the alternating series and the partial sum is given: ￼

If:

￼

then the error bound is less than 0.001, which means that the partial sum approximates the exact value to three decimal places.

￼

From this, we get ￼.

We can verify the result.

￼

￼

￼

§7.3 Absolute Convergence and Conditional Convergence

When studying infinite series, it's essential to understand different types of convergence and how rearranging terms can affect the series. Here’s a detailed look into absolute convergence, conditional convergence, and the rearrangement of terms.

In this section, you will study:

▪︎ Absolute Convergence

▪︎ Conditional Convergence

▪︎ Rearrangement of Terms

● Absolute Convergence

Absolute convergence means that a series converges when you take the absolute value of each term. For a series ￼:

The series ￼ converges absolutely if ￼ converges.

Absolute convergence implies regular convergence of the original series ￼, and it also has the property that rearranging the terms does not affect the sum of the series.

Consider the series ￼.

Since ￼ converges, the original series ￼ converges absolutely.

In this section, you will study: ▪︎ Absolute Convergence ▪︎ Conditional Convergence ▪︎ Rearrangement of Terms

● Conditional Convergence

Conditional convergence occurs when a series converges, but it does not converge absolutely. That is, the series ￼ converges, but ￼ diverges.

Conditional convergence indicates that the series converges due to a delicate balance of positive and negative terms. Rearranging the terms of a conditionally convergent series can change its sum or even make it diverge.

Consider the alternating harmonic series ￼.

Since the original series converges but the absolute series diverges, the alternating harmonic series converges conditionally.

In this section, you will study:

▪︎ Absolute Convergence

▪︎ Conditional Convergence

▪︎ Rearrangement of Terms

● Rearrangement of Terms

Rearrangement of terms refers to changing the order in which terms are added in a series. For absolutely convergent series, rearranging the terms does not change the sum. However, for conditionally convergent series, rearrangement can significantly affect the result.

(i) Absolutely Convergent Series: Rearranging the terms of an absolutely convergent series does not change its sum.

(ii) Conditionally Convergent Series: Rearranging the terms of a conditionally convergent series can change the sum or even make the series diverge. This phenomenon is captured by the Riemann Rearrangement Theorem.

The Riemann Rearrangement Theorem states that if a series ￼ is conditionally convergent, then:

- For any real number ￼, there is a rearrangement of the terms of ￼ that converges to ￼.

- There is a rearrangement of the terms of ￼ that diverges to ￼.

- There is a rearrangement of the terms of ￼ that diverges to ￼.

This shows that the sum of a conditionally convergent series is not fixed and depends on the order of the terms.

Consider the series ￼ (alternating harmonic series).

- Original sum: Converges to ￼.

- Rearrange by pairing: Consider grouping the terms in pairs

￼￼

￼

This rearrangement will converge to a different sum.

By appropriately rearranging the terms, you can make the series sum to different values or diverge.

[Example 1] Absolute Convergence

Verify if the following series is convergent absolutely or conditionally.

(1) ￼

(2) ￼

(3) ￼

(Solution)

(Solution 1)

Let's test for the absolute convergence first.

￼

Since:

￼

￼

by the comparison test, the given series is not absolutely convergent.

The following sequence is decreasing and its limit is zero.

￼

￼

￼

By the alternating series test, the given series is conditionally convergent.

(Solution 2)

Let's test for the absolute convergence first.

￼

Let ￼.

￼

￼

So the given series is absolutely convergent.

(Solution 3)

Let's test for the absolute convergence first.

￼

Let ￼.

For a series to be convergent, the limit of its terms should be zero.

￼

Thus, the given series is not convergent. That is, the series is neither absolutely convergent nor conditionally convergent.

Chapter 8. Power Series

This chapter introduces power series, an essential concept in calculus used for expressing functions as infinite sums of terms involving powers of a variable. Power series provide a tool for analyzing and approximating functions, and they form the foundation for Taylor and Maclaurin series, which are widely used in mathematical analysis and applications.

In section 8.1, we begin by defining a power series and exploring its fundamental properties. Understanding the convergence of power series is crucial, and we discuss the interval of convergence, or radius of convergence, which indicates the range of ￼ values for which the series converges absolutely.

Section 8.2 delves into Taylor and Maclaurin series, specific types of power series that approximate functions near a point. The Maclaurin series is a special case of the Taylor series centered at zero. We present the Taylor series expansions for elementary functions, providing examples and methods for obtaining these expansions. The section also covers approximation using Taylor polynomials and discusses the error bound associated with such approximations. This understanding is critical for practical applications where exact functions are approximated by polynomial expressions to simplify calculations and predictions.

This chapter provides a comprehensive overview of power series, from their definition and properties to their use in representing and approximating functions. Through the exploration of Taylor and Maclaurin series, readers will gain a solid foundation in using power series to solve mathematical problems and analyze functions in various contexts.

In this chapter, you will learn:

§8.1 Definition and Key Properties

§8.2 Taylor Series (Maclaurin Series)

§8.1 Definition and Key Properties

In this section, you will study:

▪︎ Definition of Power Series

▪︎ Interval of Convergence (Radius of Convergence)

● Definition of Power Series

A power series is an infinite series of the form:

￼￼￼

where:

- ￼ are the coefficients of the series,

- ￼ is the variable,

- ￼ is the center or point of expansion of the series.

- ￼ is the constant term.

- ￼ is the linear term.

- ￼ is the quadratic term, and so on.

(i) Coefficients ￼: These are the constants that determine the contribution of each term to the overall sum of the series. The coefficients can be any real numbers.

(ii) Variable ￼: This is the variable around which the series is expressed. As ￼ varies, the value of the series sum changes.

(iii) Center ￼: This is the point about which the series is expanded.

(iv) Powers ￼: These are non-negative integers representing the exponents in the series. The power ￼ increases sequentially, starting from 0.

Power series are crucial in calculus and mathematical analysis because they allow complex functions to be expressed as sums of simpler polynomial terms. This makes it easier to:

- Approximate functions with polynomials.

- Analyze functions using their series expansions.

- Solve differential equations by converting them into series form.

In this section, you will study:

▪︎ Definition of Power Series

▪︎ Interval of Convergence (Radius of Convergence)

● Interval of Convergence

The interval of convergence of a power series is the set of ￼-values for which the series converges to a finite sum. Determining this interval is essential to understanding where a power series representation of a function is valid and useful.

Consider a power series centered at ￼:

￼

where ￼ are the coefficients, ￼ is the center, and ￼ is the variable.

The radius of convergence ￼ is the distance from the center ￼ within which the series converges absolutely. It is a non-negative real number or infinity, and it determines the extent of the interval of convergence. The radius of convergence can be found using tests such as the Ratio Test or Root Test.

- If ￼, the series converges only at ￼.

- If ￼, the series converges for all ￼.

- For a finite ￼, the series converges for ￼ and diverges for ￼.

To find the interval of convergence using the Ratio Test, compute:

￼

Here, let ￼

The radius of convergence ￼ is:

￼ (If ￼ and ￼).

Then, the interval of convergence given by:

￼

￼

Alternatively, using the Root Test:

￼￼

￼

The radius of convergence ￼ is:

￼ (If ￼ and ￼).

The interval of convergence is the set of ￼ values for which the series converges. It is determined by:

￼

- Open Interval: The series converges absolutely for all ￼ within ￼.

- Endpoints: The behavior of the series at the endpoints ￼ and ￼ must be checked separately. The series may converge, diverge, or conditionally converge at these points.

Consider the series ￼:

￼

￼

- At ￼: ￼, which diverges.

- At ￼: ￼, which diverges.

- The series converges if ￼.

- The interval of convergence is ￼.

Consider the series ￼:

￼￼￼

￼

- At ￼: ￼, which converges (alternating series test).

- At ￼: ￼, which converges (p-series with ￼).

- The series converges for ￼.

- The interval of convergence is ￼.

[Example 1] Find the interval of convergence.

(1) ￼

(2) ￼

(Solution)

(Solution 1)

By using the ratio test:

￼

At ￼, the series is divergent.

￼

(Solution 2)

￼

At ￼, conditionally convergent.

￼

At ￼, conditionally convergent.

￼

Thus:

￼

In this chapter, you will learn:

§8.1 Definition and Key Properties

§8.2 Taylor Series (Maclaurin Series)

§8.2 Taylor Series (Maclaurin Series)

In this section, you will study:

▪︎ Definition of Taylor Series

▪︎ Taylor Series of Elementary Functions

▪︎ Obtaining Taylor Series Expansion with Known Series ▪︎ Approximation By Taylor Polynomials and Error Bounds ▪︎ Error Bounds of Alternating Taylor Series

● Definition of Taylor Series

The Taylor series and Maclaurin series are tools used in calculus to approximate functions with an infinite sum of polynomial terms. They express a function as an infinite series of terms derived from the function's derivatives at a single point.

(i) Taylor Series

For a function ￼ that is infinitely differentiable around a point ￼, the Taylor series is: ￼￼￼

More formally, it can be written as:

￼

where:

- ￼ is the ￼-th derivative of ￼ evaluated at ￼, - ￼ is the factorial of ￼, - ￼ is the term expressing the power of the difference between ￼ and ￼.

(ii) Maclaurin Series

The Maclaurin series is a special case of the Taylor series where the expansion is done around ￼. This simplifies the series to: ￼￼￼

Or in summation form:

￼

(iii) Taylor Series of Elementary Functions - Exponential Function ￼: - Taylor series at ￼ (Maclaurin series): ￼

- Sine Function ￼: - Maclaurin series:

￼

- Cosine Function ￼: - Maclaurin series:

￼

In this section, you will study:

▪︎ Definition of Taylor Series

▪︎ Taylor Series of Elementary Functions

▪︎ Obtaining Taylor Series Expansion with Known Series

▪︎ Approximation By Taylor Polynomials and Error Bounds

▪︎ Error Bounds of Alternating Taylor Series

● Taylor Series of Elementary Functions

The Taylor series of elementary functions provide a way to represent these functions as infinite sums of their derivatives at a particular point. Here we discuss the Taylor series for various elementary functions, their general forms, and their intervals of convergence.

(i) Exponential Function

￼￼

Interval of Convergence:

- The series converges for all ￼.

- Radius of Convergence: ￼.

(ii) Natural Logarithm

￼￼

Interval of Convergence:

- The series converges for ￼.

- Radius of Convergence: ￼.

Endpoints:

- At ￼: Converges to ￼.

- At ￼: Diverges.

(iii) Sine Function

￼￼

Interval of Convergence:

- The series converges for all ￼.

- Radius of Convergence: ￼.

(iv) Cosine Function

￼￼

Interval of Convergence:

- The series converges for all ￼.

- Radius of Convergence: ￼.

(v) Hyperbolic Sine Function

￼￼

Interval of Convergence:

- The series converges for all ￼.

- Radius of Convergence: ￼.

(vi) Hyperbolic Cosine Function

￼

Interval of Convergence:

- The series converges for all ￼.

- Radius of Convergence: ￼.

(vii) Geometric Series

￼￼

Interval of Convergence:

- The series converges for ￼.

- Radius of Convergence: ￼.

Endpoints:

- At ￼: Diverges.

- At ￼: Diverges.

Understanding the Taylor series and their intervals of convergence is essential for analyzing and approximating the behavior of elementary functions in calculus.

In this section, you will study:

▪︎ Definition of Taylor Series

▪︎ Taylor Series of Elementary Functions

▪︎ Obtaining Taylor Series Expansion with Known Series

▪︎ Approximation By Taylor Polynomials and Error Bounds

▪︎ Error Bounds of Alternating Taylor Series

● Obtaining Taylor Series Expansion with Known Series

Using known Taylor series expansions for fundamental functions can simplify the process of finding the Taylor series for related functions. This involves manipulating existing series by substitution, differentiation, or integration.

(i) Finding the Taylor Series for ￼ Using ￼

Known Series for ￼:

￼

To find the Taylor series for ￼, substitute ￼ into the known series for ￼:

￼￼

Interval of Convergence:

- The series converges for all ￼.

- Radius of Convergence: ￼.

(ii) Finding the Taylor Series for ￼ Using ￼

Known Series for ￼:

￼

To find the Taylor series for ￼, substitute ￼ into the known series for ￼:

￼￼

Interval of Convergence:

- The original series ￼ converges for ￼. When substituting ￼, the convergence is for ￼.

- Radius of Convergence: The series for ￼ converges for ￼.

(iii) Finding ￼ Using ￼

Known Series for ￼:

￼￼

To find the Taylor series for ￼, differentiate the series for ￼:

￼￼

Differentiate term-by-term:

￼￼

So, the Taylor series for ￼ is:

￼￼

Interval of Convergence:

- The series converges for all ￼.

- Radius of Convergence: ￼.

(iv) Finding ￼ from ￼

Known Series for ￼:

￼ for ￼

To find the Taylor series for ￼, substitute ￼ for ￼ in ￼:

￼

To find the Taylor series for ￼, integrate ￼:

￼￼

Integrate term-by-term:

￼￼￼

So, the Taylor series for ￼ is:

￼

Interval of Convergence:

- The series converges for ￼.

- Radius of Convergence: ￼.

These techniques leverage known series to efficiently find new Taylor series expansions, making them tools for function approximation and analysis.

Appendix: Term by Term Differentiation and Integration

Uniform Convergence of Power Series

Note: The content in this appendix is supplementary and not required for mastering the essential topics of calculus for beginners. It is provided for enrichment and may be useful for those pursuing further studies in mathematics or related fields.

1. What is Uniform Convergence?

Uniform convergence of a power series occurs when the series converges in such a way that the partial sums approach the full sum uniformly over the entire interval of convergence. This means that the convergence is not just pointwise, but the series converges at a consistent rate across the interval.

￼

A power series ￼ converges uniformly on a set ￼ if, for every ￼, there exists an integer ￼ such that for all ￼ and for all ￼:

￼

2. Why is Uniform Convergence Important?

Uniform convergence allows differentiation and integration to be performed term-by-term on the series:

- Differentiation: If ￼ converges uniformly, ￼ converges uniformly to the derivative.

￼￼￼

- Integration: Similarly, ￼ can be integrated term-by-term within the interval of convergence.

￼￼￼

3. Uniform Convergence of Power Series

In general, a power series is not necessarily uniformly convergent throughout the entire interval of convergence. However, it is uniformly convergent on any closed subinterval within the open interval of convergence.

- A power series ￼ has a radius of convergence ￼. Within the interval ￼, the series converges absolutely.

- The interval ￼ is called the open interval of convergence. The series converges within this interval, but the behavior at the endpoints ￼ needs separate analysis.

- Within the open interval ￼, the series is uniformly convergent on any subinterval ￼, where ￼.

- This means that for any closed and bounded subinterval strictly inside ￼, the series converges uniformly.

- At the endpoints ￼, the series may converge or diverge, and uniform convergence is not guaranteed. Each endpoint must be checked separately to determine whether the series converges and, if so, whether it converges uniformly.

In this section, you will study:

▪︎ Definition of Taylor Series

▪︎ Taylor Series of Elementary Functions

▪︎ Obtaining Taylor Series Expansion with Known Series

▪︎ Approximation By Taylor Polynomials and Error Bounds

▪︎ Error Bounds of Alternating Taylor Series

● Approximation By Taylor Polynomials and Error Bounds

Taylor polynomials provide a method for approximating functions near a specific point using a finite number of terms from their Taylor series. Understanding how to approximate functions and estimate the error of these approximations is crucial in calculus and applied mathematics.

A Taylor polynomial of degree ￼ for a function ￼ centered at ￼ is the partial sum of the Taylor series up to the ￼-th term:

￼￼￼

- ￼ is the ￼-th derivative of ￼ evaluated at ￼.

This polynomial provides an approximation to the function ￼ near ￼.

Taylor polynomials are used to approximate functions for values of ￼ near the center ￼. The polynomial ￼ provides a finite approximation of ￼, making it easier to evaluate and analyze the function.

Approximation of ￼ at ￼:

- Taylor polynomial of degree 2:

￼

This polynomial approximates ￼ near ￼.

When approximating a function ￼ using a Taylor polynomial ￼, the error ￼ represents the difference between the actual function and the approximation:

￼

The error term helps quantify the accuracy of the approximation.

The Lagrange form of the remainder provides an estimate of the error:

￼

- ￼ is the ￼-th derivative of ￼ evaluated at some point ￼ between ￼ and ￼.

- The remainder term ￼ indicates the error in approximating ￼ by ￼.

The Error Bound (the maximum possible error) is:

￼

where ￼ is the maximum value of ￼ on the interval.

For example, consider approximating a function ￼ using the first-degree Taylor polynomial ￼. We can find the remainder term as follows:

￼

Assume ￼. Then,

￼￼

We will integrate both sides again over the interval ￼:

￼

￼

￼

Therefore:

￼

Here, ￼. Using this, we rewrite the inequality:

￼

Considering the case when ￼ as well, we obtain the inequality:

￼

Using this derivation process, we can similarly find the error bounds when approximating a function with higher-degree Taylor polynomials ￼, ￼, and so on.

Once again, let’s consider approximating a function ￼ using the second-degree Taylor polynomial ￼. The remainder term can be derived as follows:

￼

Assume ￼. We begin by considering the Taylor polynomial ￼ and its remainder:

￼￼

By integrating ￼ over the interval ￼, we get:

￼

Since ￼, we can bound the integral of ￼:

￼￼

Now, we integrate this inequality again over the interval ￼:

￼

￼

￼

￼

We integrate this inequality again over the interval ￼:

￼

￼

￼

￼

For ￼, we have:

￼

Rewriting the inequality using ￼:

￼

Taking the magnitude of both sides, and considering ￼:

￼

Using this process, the Lagrange remainder formula can be generalized for higher-order Taylor polynomials to estimate the error bound in function approximation.

Consider approximation of ￼ at ￼ using ￼:

￼

- Find the fourth derivative: ￼.

- Maximum value of ￼ on ￼ is ￼.

- Error bound for ￼ close to 0:

￼

For the interval ￼:

￼

This error estimate tells us that the approximation ￼ is accurate to within ￼ when ￼.

This understanding of Taylor polynomials and error bounds allows for effective approximation of functions, which is essential for many applications in calculus and beyond.

Additionally, it is important to note that when using Taylor polynomials, we are considering the bound of the error, that is, the maximum possible error, rather than calculating the exact error itself.

In this section, you will study:

▪︎ Definition of Taylor Series

▪︎ Taylor Series of Elementary Functions

▪︎ Obtaining Taylor Series Expansion with Known Series

▪︎ Approximation By Taylor Polynomials and Error Bounds

▪︎ Error Bounds of Alternating Taylor Series

● Error Bounds of Alternating Taylor Series

The error bound of an alternating Taylor series, also known as an alternating series remainder, provides an estimate of the error when approximating a function with a finite number of terms of its Taylor series. An alternating series is one in which the signs of the terms alternate, typically in the form ￼ where ￼ are positive and decreasing to 0.

For an alternating Taylor series, the error bound is particularly simple and useful. The error ￼ in approximating a function ￼ by the sum of the first ￼ terms of its Taylor series can be bounded by the absolute value of the first omitted term:

￼

￼

The intuition behind this error bound is that the alternating nature of the series means the error at each step is approximately given by the next term in the series. This is because each term alternates and partially cancels out the error from the previous terms.

Consider approximation of ￼ at ￼ using ￼:

￼

- Find the first omitted term: ￼.

- Error bound for ￼ close to 0:

￼

For ￼:

￼

This error estimate tells us that the approximation ￼ is accurate to within ￼ when ￼.

For an alternating Taylor series that meets the conditions mentioned, the error bound when approximating the function by the sum of the first ￼ terms is given by the magnitude of the ￼-th term. This provides a simple and practical way to estimate the accuracy of such approximations.

[Example 1] Find the Taylor series

By using the first given function, obtain the power series of the second function.

￼, ￼

(Solution)

￼

By term by term differentiation, we can get the power series of ￼ since the interval of convergence of the sine function is all real numbers.

￼

Now, we can utilize substitution.

￼￼

[Example 2] Taylor Polynomial and Error Bound

Approximate ￼ using the fourth-degree Taylor polynomial ￼ centered at ￼. Calculate the Taylor polynomial and estimate the error bound for ￼ in the interval ￼.

(Solution)

The Taylor series for ￼ centered at ￼ (Maclaurin series) is given by:

￼￼

The fourth-degree Taylor polynomial ￼ is:

￼

The Lagrange remainder formula for the error ￼ when using ￼ is:

￼

where ￼ is some value between 0 and ￼.

Fifth derivative:

￼

We need to find ￼.

For ￼:

The maximum value of ￼ on the interval occurs at ￼, where:

￼.

Compute the error bound:

￼

Let’s calculate the error bound for ￼:

￼

This means that the approximation ￼ is accurate to within ￼.

Chapter 9. Exercises

Exercises begin from the next page.

Exercise 1. Indefinite Integral

Evaluate the following integral.

(1) ￼

(2) ￼

(3) ￼

(4) ￼

(5) ￼

(6) ￼

(7) ￼

(8) ￼

(9) ￼

(10) ￼

(11) ￼

(12) ￼

(13) ￼

(14) ￼

(15) ￼

(16) ￼

(17) ￼

(18) ￼

(19) ￼

(20) ￼

(21) ￼

(22) ￼

(23) ￼

(24) ￼

(25) ￼

(26) ￼

(27) ￼

(28) ￼

(29) ￼

(30) ￼

The solutions are on the next pages.

Solution 1.1

￼￼

Solution 1.2

￼￼

Solution 1.3

￼￼

￼

Solution 1.4

￼

(Solution)

By using substitution:

￼

￼￼

￼

Solution 1.5

￼

(Solution)

By using substitution:

￼

￼￼

￼

Solution 1.6

￼

(Solution)

By using substitution:

￼

￼￼

￼

Solution 1.7

￼

(Solution)

By using substitution:

￼

￼

￼￼

Solution 1.8

￼￼

Solution 1.9

￼

(Solution)

￼

￼

￼￼

￼

Solution 1.10

￼

(Solution)

￼

￼￼

￼

Solution 1.11

￼

(Solution)

￼

￼￼￼

￼

Solution 1.12

￼

(Solution)

￼

￼

￼￼

Solution 1.13

￼

(Solution)

￼

￼

￼￼

￼

Solution 1.14

￼

(Solution)

￼

￼

￼

￼

Solution 1.15

￼

(Solution)

￼

Thus:

￼￼

Solution 1.16

￼

(Solution)

Let's use the integration by parts:

￼

￼￼

Thus:

￼

￼

Solution 1.17

￼

(Solution)

￼

Thus:

￼

￼

Solution 1.18

￼

(Solution)

Let's use the integration by parts:

￼

￼￼

Then:

￼￼￼￼

From (1):

￼￼

Then:

￼

￼

Therefore:

￼￼￼

Solution 1.19

￼

(Solution)

￼

Then:

￼

￼￼

Here, ￼ is always less than 1.

Solution 1.20

￼

(Solution)

Let's use the partial fraction:

￼

Here, ￼ and ￼. Thus:

￼￼

￼

Solution 1.21

￼

(Solution)

￼

￼￼￼

Solution 1.22

￼

(Solution)

Since:

￼

Then:

￼

￼

￼￼

￼￼

Solution 1.23

￼

(Solution)

Let's use the integration by parts:

￼

￼￼

Then:

￼

￼

Here:

￼￼

Therefore:

￼

￼

Solution 1.24

￼

(Solution)

Since ￼:

￼

￼

Thus:

￼

￼

Solution 1.25

￼

(Solution)

￼

￼

￼

￼

￼

￼

Solution 1.26

￼

(Solution)

Let's use the integration by parts:

￼

￼￼

Then:

￼￼￼￼

From (1):

￼￼

Then:

￼￼￼￼

Thus:

￼￼

￼

Solution 1.27

￼

(Solution)

Let's use the partial fractions:

￼

Here:

￼

Therefore:

￼￼

￼

￼

Solution 1.28

￼

(Solution)

￼

￼

Then, we can use the integration by parts:

￼￼

Thus:

￼

￼

￼

￼

Solution 1.29

￼

(Solution)

Using Integration by substitution:

￼

￼

￼

￼

￼

Solution 1.30

￼

(Solution)

￼

Thus:

￼

￼

Exercise 2. Definite Integral

Evaluate the following definite integral.

(1) ￼

(2) ￼

(3) ￼

(4) ￼

(5) ￼

(6) ￼

(7) ￼

(8) ￼

(9) ￼

(10) ￼

(11) ￼

(12) ￼

(13) ￼

(14) ￼

(15) ￼

(16) ￼

(17) ￼

(18) ￼

(19) ￼

(20) ￼

(21) ￼

(22) ￼

(23) ￼

(24) ￼

(25) ￼

The solutions are on the next pages.

Solution 2.1

￼

(Solution)

￼

Solution 2.2

￼

(Solution)

￼

￼

Solution 2.3

￼

(Solution)

￼

￼

Solution 2.4

￼

(Solution)

￼

￼

￼

￼

Solution 2.5

￼

(Solution)

￼

￼

Solution 2.6

￼

(Solution)

￼

Then:

￼￼￼￼

Solution 2.7

￼

(Solution)

￼

We need to divide the integration interval.

￼￼

Thus:

￼

￼

￼￼￼

Solution 2.8

￼

(Solution)

￼

￼￼￼

Solution 2.9

￼

(Solution)

Let's use the partial fractions.

￼

￼

￼

￼

￼

Solution 2.10

￼

(Solution)

￼

￼

￼

￼￼￼

Solution 2.11

￼

(Solution)

According to the interval of the integration, we need to change the integrand.

￼￼

￼￼

Solution 2.12

￼

(Solution)

￼￼￼

Solution 2.13

￼

(Solution)

￼￼

Solution 2.14

￼

(Solution)

￼￼￼

Solution 2.15

￼

(Solution)

Let's use the integration by parts:

￼

￼

￼￼

￼￼

￼

Solution 2.16

￼

(Solution)

￼

￼

Solution 2.17

￼

(Solution)

￼

￼

￼

￼

Solution 2.18

￼

(Solution)

￼￼￼

￼

Solution 2.19

￼

(Solution)

￼

￼

￼

￼

Solution 2.20

￼

(Solution)

￼

￼￼￼

Solution 2.21

￼

(Solution)

￼

￼

￼

Solution 2.22

￼

(Solution)

￼

￼

￼￼￼

Solution 2.23

￼

(Solution)

￼

￼

￼

￼

￼

Solution 2.24

￼

(Solution)

￼

Thus:

￼

(Here, ￼)

￼

￼

￼

￼

￼

￼

Solution 2.25

￼

(Solution)

￼

￼

￼

￼

￼￼

￼￼

The first term of (1) is integrated in the following way.

￼

￼

The second term of (1) is integrated as follows.

￼￼

￼￼￼

Therefore:

￼￼

￼

￼

In another way, we can substitute x with hyperbolic function.

￼

￼

￼

(Here, ￼) ￼

￼￼

Let's obtain ￼: ￼

￼

￼

￼

Therefore:

￼

Exercise 3. Improper Integral

Evaluate the following improper integral if it converges.

(1) ￼

(2) ￼

(3) ￼

(4) ￼

(5) ￼

(6) Verify the convergence of the following integral.

￼

(7) Verify the convergence of the following integral.

￼

The solutions are on the next pages.

Solution 3.1

￼

(Solution)

￼

￼

Solution 3.2

￼

(Solution)

￼

￼

The integral is not convergent.

Solution 3.3

￼

(Solution)

￼

The integral is not convergent.

Solution 3.4

￼

(Solution)

￼￼

Solution 3.5

￼

(Solution)

￼

￼

The integral is not convergent.

Solution 3.6

Verify the convergence of the following integral.

￼

(Solution)

For the given integration interval:

￼

Thus:

￼

Here, the left side is divergent, so the right side approaches to the infinity.

￼

￼

Solution 3.7

Verify the convergence of the following integral.

￼

(Solution)

For the interval [1, ∞)

￼

And we can divide the given improper interval as follows.

￼

Here, the red formula is finite. (￼ is a real number) ￼

And

￼

Therefore, the given integral is convergent.

Exercise 4. Properties of Integration

(1) Show that the following formula is true. (￼ is the positive real number.) When ￼, ￼

(2) Evaluate the following integral.

￼

Use the following formula.

￼

Here, ￼ is continuous for its domain.

(3) Show that following is true on the interval ￼.

￼

(4) Show that following is true.

￼

(5) Show that the following is true when ￼ is continuous on the interval ￼. (￼ : constant) ￼

(6) For given ￼, evaluate ￼.

￼

(7) For the function ￼, evaluate the following integral.

￼

The solutions are on the next pages.

Solution 4.1

Show that the following formula is true. (￼ is the positive real number.)

When ￼

￼

(Solution)

￼￼

Let ￼. Then the second term of (1) is as follows.

￼￼

Therefore:

￼

Solution 4.2

Evaluate the following integral.

￼

Use the following formula.

￼

Here, ￼ is continuous for its domain.

(Solution)

Since

￼￼

Then:

￼

￼￼

￼￼

By the way, the left side of (2) is as follows.

￼.

Since ￼:

￼

￼￼

By using (3), we can get:

￼

Solution 4.3

Show that following is true on the interval ￼.

￼

(Solution)

For ￼:

￼

￼

Thus, for a real number c between 0 and 1,

￼

￼

So for any value on the interval ￼:

￼

Solution 4.4

Show that following is true.

￼

(Solution)

For the integration interval:

￼

Therefore:

￼

Solution 4.5

Show that the following is true when ￼ is continuous on the interval ￼. (￼ : constant)

￼

(Solution)

Let ￼. Then:

￼￼

Since ￼ is dummy variable:

￼

Solution 4.6

For given ￼, evaluate ￼.

￼

(Solution)

The result of the definite integral is a number.

￼

Then:

￼

￼￼￼

￼

￼

￼

Solution 4.7

For the given function ￼, evaluate the following integral.

￼

(Solution)

￼

￼

￼

￼

￼

Exercise 5. Recursive Formula

Show that the recursive formula is true.

(1) ￼

￼

(2) ￼

￼

(3) ￼

￼

The solutions are on the next pages.

Solution 5.1

Show that the recursive formula is true.

￼

￼

(Solution)

Let's use the integration by parts:

￼

￼

￼

￼￼￼

￼￼

￼

Thus:

 ￼

￼

￼

Solution 5.2

￼

￼

(Solution)

Let's use the integration by parts:

￼

￼

￼

￼

￼

Thus:

￼

￼

￼

Solution 5.3

￼

￼

(Solution)

Let's use the integration by parts:

￼

￼

￼

￼

￼

Thus:

￼

￼

￼

￼

Exercise 6. Fundamental Theorem of Calculus

Solve the following problems.

(1) Obtain the derivative of ￼.

￼

(2) Obtain the derivative of ￼.

￼

(3) Obtain the derivative of ￼.

￼

(4) For ￼, the following is true. Find ￼ and the constant ￼.

￼

(5) When ￼ is given as follows, Solve (a) ~ (b).

￼

￼

(a) Find the critical values of ￼.

(b) Find the interval of concave down of ￼.

(6) Obtain the following limit.

￼

(7) Obtain the sum of the local maximum and the local minimum.

￼

(8) For the given function,

￼

Evaluate the following limit.

￼

The solutions are on the next pages.

Solution 6.1

Obtain the derivative of ￼.

￼

(Solution)

By the fundamental theorem of calculus:

￼

Solution 6.2

Obtain the derivative of ￼.

￼

(Solution)

We need to use the chain rule of the derivative. Let ￼:

￼￼

￼

Solution 6.3

Obtain the derivative of ￼.

￼

(Solution)

￼

￼

￼

Solution 6.4

For ￼, the following is true. Find ￼ and the constant ￼.

￼

(Solution)

By substituting ￼ to ￼:

￼

￼

By differentiating both sides with respect to ￼:

￼

￼

Let's substitute ￼ into the original equation:

￼￼￼￼

Solution 6.5

When ￼ is given as follows, Solve (a) ~ (b).

￼

￼

(a) Find the critical points of ￼ for ￼.

(b) Find the interval of concave down of ￼.

(Solution)

(Solution 6.5a)

Let's find the derivative of ￼.

￼

￼

￼￼￼

￼￼

Critical points are:

￼, ￼

(Exercise 6.5b)

Let's find the second derivative of ￼.

￼

￼

￼

Therefore:

￼￼￼

￼

Solution 6.6

Obtain the following limit.

￼

(Solution)

By using the L'Hospital rule:

￼

￼

Solution 6.7

Obtain the sum of the local maximum and the local minimum.

￼

(Solution)

￼

Thus, the critical values are:

￼

Therefore:

￼￼￼￼

Solution 6.8

For the given function,

￼

Evaluate the following limit.

￼

(Solution)

By using the L'Hospital rule:

￼￼

￼

￼

￼

￼

Exercise 7. Find the area enclosed

Evaluate the area enclosed by the given curves.

(1) ￼

(2) ￼

(3) ￼, ￼

(4) ￼, ￼

(5) When dividing the square, whose vertices are the points (0, 0), (0, 1), (1, 0), (1, 1), with ￼, find the ratio of the two divided area. (Here, ￼ is the positive integer.)

The solutions are on the next pages.

Solution 7.1

Evaluate the area enclosed by the given curves.

￼

(Solution)

￼

￼

￼

￼￼

￼

Solution 7.2

￼, ￼

(Solution)

￼

￼

￼

￼

￼￼

￼￼

￼

Solution 7.3

￼

(Solution)

￼

￼

￼

￼

￼

Solution 7.4

￼, ￼

(Solution)

￼￼

￼

￼

The area enclosed are:

￼￼￼￼

Let's calculate each term of the right side:

￼

￼

￼

￼

￼

￼

￼

￼

Therefore:

￼

￼

Solution 7.5

When dividing the square, whose vertices are the points (0, 0), (0, 1), (1, 0), (1, 1), with ￼, find the ratio of the two divided area. (Here, n is the positive integer.)

(Solution)

Let ￼ be the area of the square. Then:

￼

When we divide the square with ￼, let U be the area enclosed by ￼, ￼ and ￼. Then:

￼

Therefore,

￼

￼

￼

Exercise 8. Find the volume of revolution

Determine the volume of the solid formed by revolving the region enclosed by the given curves around the specified axis.

(1) axis: ￼

￼

(2) axis: ￼

￼

(3) axis: ￼

￼

(4) axis: ￼

￼

(5) axis: ￼

￼

(6) axis: ￼

￼

The solutions are on the next pages.

Solution 8.1

axis: ￼

￼

(Solution)

We can get the infinitesimal volume ￼ as follows:

￼

Therefore:

￼

￼

￼

Solution 8.2

axis: ￼

￼

(Solution)

We can get the infinitesimal volume ￼ as follows.

￼

Therefore:

￼

￼

￼

￼

Solution 8.3

axis: ￼

￼

(Solution)

￼￼

￼

Solution 8.4

axis: ￼

￼

(Solution)

We can get the infinitesimal volume ￼ as follows:

￼

Therefore:

￼￼￼￼￼

￼

Solution 8.5

axis: ￼

￼

(Solution)

￼

￼

￼

￼

Solution 8.6

axis: ￼

￼

(Solution)

Let ￼ be the volume of revolution with ￼ about ￼-axis.and ￼ be the volume of revolution with ￼ about the y-axis. Then we can the volume of doughnut by subtracting ￼.

￼

￼

By using the substitution: ￼

then:

￼

￼

￼

Exercise 9. Find the volume

Evaluate the volume of a solid whose area of the cross-section is known.

(1) The bottom surface of a solid is the circle whose radius is 4. Let its diameter be the axis perpendicular to the cross-section which is the isosceles triangle. The bottom angle of the triangle is 45 degree. Evaluate the volume.

(2) The bottom surface of a solid is the ellipse whose semi-major axis is 10 and semi-minor is 8. When the cross-section perpendicular to the semi-major axis is the equilateral triangle, Evaluate the volume.

The solutions are on the next pages.

Solution 9.1

Evaluate the volume of a solid whose area of the cross-section is known.

The bottom surface of a solid is the circle whose radius is 4. Let its diameter be the axis perpendicular to the cross-section which is the isosceles triangle. The bottom angle of the triangle is 45 degree. Evaluate the volume.

(Solution)

Let the center of the bottom circle be the origin of the coordinates. Let's find the bottom length (￼) of the triangle: ￼

Since the bottom angle is 45 degree, we get the height (￼) as follows: ￼

Therefore, the cross-section area, whose perpendicular distance is x from the origin, is as follows: ￼

Therefore:

￼

￼

Solution 9.2

Evaluate the volume of a solid whose area of the cross-section is known.

The bottom surface of a solid is the ellipse whose semi-major axis is 10 and the semi-minor axis is 8. When the cross-section perpendicular to the semi-major axis is the equilateral triangle, Evaluate the volume.

(Solution)

Let the center of the ellipse be the origin of the coordinates. And let the semi-major be the x-axis. The bottom length (L) of the regular triangle is as follows: ￼

Then, we can get the area of the triangle: ￼

Thus:

￼

By using the substitution:

￼

￼

￼

￼

Exercise 10. Find the length of curve

Evaluate the curve length.

(1) ￼, ￼

(2) ￼, ￼

(3) ￼, ￼

(4) ￼, ￼, ￼

The solutions are on the next pages.

Solution 10.1

￼, ￼

(Solution)

￼￼

￼￼

￼

￼

Solution 10.2

￼, ￼

(Solution)

￼￼￼

By using the substitution:

￼

￼

Then:

￼

￼

￼

￼

Solution 10.3

￼, ￼

(Solution)

￼￼

￼￼￼

￼

Solution 10.4

￼, ￼, ￼

(Solution)

￼￼

￼￼

￼

￼

Exercise 11. Find the average value

(1) Find the average value of ￼ on the interval ￼.

(2) Find the average value of ￼ on the interval ￼.

(3) Find the average value of ￼ on the interval ￼.

The solutions are on the next pages.

Solution 11.1

Find the average value of ￼ on the interval ￼.

(Solution)

￼

Solution 11.2

Find the average value of ￼ on the interval ￼.

(Solution)

￼

By using the substitution:

￼

Then:

￼￼

Here, let's use the integration by parts.

￼￼

Then:

￼

￼

Solution 11.3

Find the average value of ￼ on the interval ￼.

(Solution)

￼￼

Exercise 12. Position, Velocity, and Acceleration

Solve the following problems.

(1) The velocity of a particle moving along the x-axis is given as follows. (Here, ￼ is the time in second.)

￼

Obtain the distance the particle moved from ￼ to ￼.

(2) The acceleration of a particle moving along the x-axis is given as follows. (Here, ￼ is the time in second.)

￼

Obtain the distance the particle moved from ￼ to ￼. (the velocity at ￼ is zero.)

(3) Find the position function of time ￼ when an object is thrown vertically upwards with the velocity ￼. (the gravitational constant is ￼. The direction upwards perpendicular to the surface is the positive direction.)

(4) When an object is thrown with the angle ￼ (measured from the ground) and the velocity ￼, find the position function of the time. (Let the throwing point be the origin of the coordinates, horizontal direction be the x-axis, and the vertical direction be the y-axis. And the gravitational constant is ￼. The direction upwards perpendicular to the surface is the positive direction.)

(5) A particle is moving in the plane and its acceleration is given as follows.

￼, ￼

When ￼, the particle is stationary. and the position is ￼ in the Cartesian coordinates.

(a) Find the position function of the time.

(b) Obtain the distance traveled during the given time interval.

The solutions are on the next pages.

Solution 12.1

The velocity of a particle moving along the x-axis is given as follows. (Here, ￼ is the time in second.)

￼

Obtain the distance the particle traveled from ￼ to ￼.

(Solution)

￼

￼

Solution 12.2

The acceleration of a particle moving along the x-axis is given as follows. (Here, ￼ is the time in second.) ￼

Obtain the distance the particle traveled from ￼ to ￼. (the velocity at ￼ is zero.) (Solution)

Let ￼ be the position. Then: ￼

Therefore, the velocity is as follows:

￼

Here, the velocity is zero at ￼, so ￼.

￼

Thus:

￼￼

Solution 12.3

Find the position function of time ￼ when an object is thrown vertically upwards with the velocity ￼. (the gravitational constant is ￼.) (Solution)

Let the upward direction be positive and ￼ be the position function. (￼) Since the direction of gravitational force is negative: ￼

By integrating both sides with respect to the time, ￼

￼

Thus:

￼

Let's integrate once more to get the change of the position: ￼

￼

￼

Solution 12.4

When an object is thrown with the angle ￼ (measured from the ground) and the velocity ￼, find the position function of the time. (Let the throwing point be the origin of the coordinates, horizontal direction be the x-axis, and the vertical direction be the y-axis. And the gravitational constant is ￼.)

(Solution)

Let the coordinates be ￼ and ￼. Then:

￼

￼

Let's integrate both sides with respect to the time:

￼

￼

The velocity of the object with respect to the x-axis is constant with ￼. Thus, ￼

￼

￼

The initial velocity of the object with respect to the y-axis is ￼. Thus:

￼

After integrating (1) and (2) once more, we get the position function. The starting point is the origin. Thus, ￼, ￼.

￼

￼

Solution 12.5

A particle is moving in the plane and its acceleration is given as follows.

￼, ￼

When ￼, the particle is stationary. and the position is ￼ in the Cartesian coordinates.

(a) Find the position function of the time.

(b) Obtain the distance traveled during the given time interval.

(Solution)

(Solution 12.5a)

Let ￼ be the velocity with respect to the x-axis.

￼￼

￼

Let ￼ be the velocity with respect to the y-axis.

￼￼

￼

Therefore:

￼

￼

￼

￼

￼

￼

(Solution 12.5b)

The distance the particle traveled is the length of the curve.

￼

￼

￼

￼

￼

Exercise 13. Approximation of Integral

Solve the following problems.

(1) Function values of ￼ for specific ￼ are shown in the following. And ￼ is continuous on the interval ￼.

￼

Find the approximation of the following definite integral by using trapezoid rule with the eight subintervals of equal length.

￼

(2) By using the five subintervals of equal length, answer (a) ~ (c). (Use the calculator.) ￼

(a) Find the approximation of the definite integral with the midpoint rule.

(b) Find the approximation of the definite integral with the trapezoid rule.

(c) Evaluate the integral.

The solutions are on the next pages.

Solution 13.1

Function values of ￼ for specific ￼ are shown in the following. And ￼ is continuous on the interval ￼.

￼

Find the approximation of the following definite integral by using trapezoid rule with the eight subintervals of equal length.

￼

(Solution)

Recall the formula of the Riemann sum.

￼

Therefore:

￼

Here ￼: ￼

￼

￼

￼

￼

Solution 13.2

By using the five subintervals of equal length, answer (a) ~ (c). (Use the calculator.) ￼

(a) Find the approximation of the definite integral with the midpoint rule.

(b) Find the approximation of the definite integral with the trapezoid rule.

(c) Evaluate the integral.

(Solution)

When we divide the interval into five subintervals of equal length, then: ￼

￼

￼

(Solution 13.2a)

￼

￼

￼

￼

￼

The result of the midpoint rule is as follows.

￼

(Solution 13.2b)

￼

￼

￼

￼

￼

The result of the trapezoid rule is as follows.

￼

(Solution 13.2c)

￼

Exercise 14. Differential Equations

Solve the following differential equations.

(1) Solve the following equation.

￼

(2) Find ￼.

￼

(3) Draw the slope field.

(a) ￼

(b) ￼

(4) Find the approximation of ￼ by using Euler method.

￼

The solutions are on the next pages.

Solution 14.1

Solve the following equation.

￼

(Solution)

￼

￼

￼

￼

Solution 14.2

Find ￼.

￼

(Solution)

￼

￼

￼

￼

￼

￼

Solution 14.3

Draw the slope field.

(Solution)

(Solution 14.3a)

￼

￼

(Solution 14.3b)

￼

￼

Solution 14.4

Find the approximation of ￼ by using Euler method.

￼

(Solution)

￼

￼

Since ￼:

￼

￼

￼

￼

￼

Let's find out the exact value:

￼

￼

￼

Exercise 15. Convergence Test of Infinite Series

Test the convergence of the following series.

(1) ￼

(2) ￼

(3) ￼

(4) ￼

(5) ￼ for ￼

The solutions are on the next pages.

Solution 15.1

￼

(Solution)

￼￼

￼￼￼

￼

Since the limit is less than 1, the series ￼ converges by the Ratio Test.

Solution 15.2

￼

(Solution)

For ￼:

￼

The series ￼ is a p-series with ￼, which converges.

By the Comparison Test, since ￼ for large ￼, and ￼ converges, the series ￼ also converges.

Solution 15.3

￼

(Solution)

￼

For ￼, ￼

For all ￼, ￼

A monotonically decreasing sequence that is bounded below by zero converges to zero. Thus: ￼

Since the limit of ￼ is 0. Therefore, the series ￼ converges by the Root Test.

Solution 15.4

￼

(Solution)

Check the two conditions of the Alternating Series Test:

- Monotonic Decrease:

   ￼

   The sequence ￼ is monotonically decreasing because:

   ￼

- Limit to Zero:

￼

Since both conditions are satisfied, the series ￼ converges by the Alternating Series Test.

Solution 15.5

￼ for ￼

(Solution)

To apply the Integral Test, evaluate the integral: ￼

Let ￼, so ￼.

Transform the integral:

￼

Evaluate the integral:

￼

For ￼: ￼￼￼

Since the improper integral converges for ￼, the series ￼ converges by the Integral Test.

Exercise 16 Absolute or Conditional Convergence

Determine whether the series is absolutely convergent or conditionally convergent.

(1) ￼

(2) ￼

(3) ￼

(4) ￼

(5) ￼

The solutions are on the next pages.

Solution 16.1

￼

(Solution)

To test for absolute convergence, consider the series of absolute values: ￼

Use the Ratio Test to check for absolute convergence: ￼

where ￼.

￼￼￼

Take the limit: ￼

Since the limit is less than 1, the series ￼ converges absolutely. Therefore, the original series ￼ is absolutely convergent.

Solution 16.2

￼

(Solution)

To test for absolute convergence, consider the series of absolute values:

￼

Use the Integral Test to determine convergence of ￼:

￼

Let ￼, so ￼.

￼￼

Since the improper integral diverges, the series ￼ diverges, indicating that ￼ is not absolutely convergent.

Next, apply the Alternating Series Test:

￼

The sequence ￼ is monotonically decreasing for large ￼ because:

Let ￼. Then ￼

Since ￼ for ￼, ￼ decreases for large ￼:

￼

Limit to Zero:

￼

Since both conditions are satisfied, the series ￼ is conditionally convergent.

Solution 16.3

￼

(Solution)

To test for absolute convergence, consider the series of absolute values: ￼

Use the Ratio Test to check for absolute convergence:

￼

where ￼.

￼￼￼￼

￼￼￼

Since the limit is 0, the series ￼ converges absolutely. Therefore, the original series ￼ is absolutely convergent.

Solution 16.4

￼

(Solution)

￼

Use the Integral Test to determine convergence:

￼

Let ￼, so ￼.

￼

￼

Since the improper integral diverges, the series ￼ diverges, indicating that ￼ is not absolutely convergent.

Next, apply the Alternating Series Test:

￼

The sequence ￼ is monotonically decreasing for large ￼ because:

Let ￼. Then ￼

Since ￼ for ￼. ￼ decreases for large ￼:

￼￼

Using L’Hopital rule:

￼￼￼

Since both conditions are satisfied, the series ￼ is conditionally convergent.

Solution 16.5

￼

(Solution)

To test for absolute convergence, consider the series of absolute values:

￼

For large ￼, the dominant term in the denominator is ￼. We compare with the simpler series ￼.

To use the Limit Comparison Test, set ￼ and ￼. Compute the limit:

￼￼

Because ￼ converges (p-series with ￼), ￼ also converges.

Therefore, the original series is absolutely convergent.

Exercise 17 Infinite Series

Solve the following problems.

(1) How many terms should be added to approximate the sum of the following series with the error less than 0.01?

￼

(2) When ￼ is convergent and ￼ is positive for every positive integer, test the convergence of the following series.

(a) ￼

(b) ￼

(3) When ￼ is convergent, test the convergence of ￼. Here, ￼ and ￼ for every positive integer.

The solutions are on the next pages.

Solution 17.1

How many terms should be added to approximate the sum of the following series with the error less than ￼?

￼

(Solution)

This is a p-series with ￼. The general p-series ￼ converges for ￼.

For a convergent p-series ￼ with ￼, the remainder ￼ after summing the first ￼ terms can be estimated by the integral test, which provides an upper bound for the error. Specifically, ￼￼

By using the definite integral and infinite series, we get: ￼

To find the error bound:

￼￼￼

We want the error to be less than 0.01:

￼

￼

Thus, to ensure the error is less than 0.01, we need ￼. So, at least 600 terms are required.

Solution 17.2

When ￼ is convergent and ￼ is positive for every positive integer, test the convergence of the following series.

(Solution)

(Solution 17.2a)

￼

Since ￼: ￼

By comparison test: ￼ is convergent so the given series is convergent.

(Solution 17.2b) ￼

By using the limit comparison test, we get: ￼

￼ is convergent so the given series is convergent.

Solution 17.3

When ￼ is convergent, test the convergence of ￼. Here, ￼ and ￼ for every positive integer..

(Solution)

By using the limit comparison test, we get:

￼

￼ is convergent so￼ is also convergent.

Exercise 18 Power Series

Find the interval of the convergence.

(1) ￼

(2) ￼

(3) ￼

(4) ￼

(5) ￼

The solutions are on the next pages.

Solution 18.1

￼

(Solution)

(Answer) ￼, All real numbers.

￼

￼

￼

Solution 18.2

￼

(Solution)

Ratio test:

￼

￼

￼

￼

Substitution of ￼:

When ￼,  the series is convergent.

When ￼, the series is convergent.

Therefore, the interval of convergence is ￼.

Solution 18.3

￼

(Solution)

Ratio Test:

￼

￼

￼

Substitution of ￼:

When ￼,  the series is not convergent.

When ￼, the series is conditionally convergent.

Therefore, the interval of convergence is ￼.

Solution 18.4

￼

(Solution)

Ratio test:

￼

￼

￼

Substitution of ￼:

When ￼, the series is convergent.

When ￼, the series is conditionally convergent.

Therefore, the interval of convergence is ￼.

Solution 18.5

￼

(Solution)

Ratio test:

￼

￼

￼

￼

When ￼, the series is conditionally convergent.

Therefore, the interval of convergence is ￼

Exercise 19 Taylor Series

Find the first three nonzero terms of the Taylor series about ￼ of the following function.

(1) ￼

(2) ￼

(3) ￼

(4) ￼

(5) ￼

The solutions are on the next pages.

Solution 19.1

￼

(Solution)

(i) By using the Taylor series formula:

￼

Let ￼. Then: ￼

￼

￼

￼

￼

￼

￼

￼

￼

Therefore:

￼

(ii) By using the Taylor series of ￼: ￼

￼

￼

￼￼￼

￼

(iii) By using the identity of the trigonometric function:

￼

￼

￼￼

Therefore:

￼￼

Solution 19.2

￼

(Solution)

(i) By using the Taylor series formula:

￼

Let ￼. Then: ￼

￼

￼

￼

￼

￼

￼￼

￼

￼

￼

￼

￼

￼

￼

Therefore:

￼

(ii) By using the Taylor series of ￼: ￼

Here, we can substitute ￼ to ￼. Then: ￼

(iii) By using the term-by-term integration:

￼

If we multiply 2x to both sides, then: ￼

Let's integrate both sides:

￼

￼

Solution 19.3

￼

(Solution)

By using the term-by-term integration:

￼

Let's integrate both sides:

￼

￼

Solution 19.4

￼

(Solution)

Let ￼. Then:

￼

￼

￼

￼

￼

￼

￼

Solution 19.5

￼

(Solution)

We use the function with known Taylor series.

￼

Therefore:

￼

￼

Exercise 20 Taylor Series

Solve the following problems.

(1) Approximate ￼ using Taylor polynomial ￼ of ￼ centered at ￼. Calculate the Taylor polynomial and estimate the error bound.

(2) To accurately determine the value of ￼ to the third decimal place, how many terms of the Maclaurin series for ￼ must be used?

(3) For the following function,

￼

(a) Find the Maclaurin series.

(b) By using the result of (a), obtain the sum of the following series.

￼

(4) Obtain the sum of the following series using the given function as a hint.

(a) ￼ (Hint. ￼) (b) ￼ (Hint. ￼) (5) Evaluate the limit by using the power series.

￼

(6) Show that the following is true.

￼

The solutions are on the next pages.

Solution 20.1

Approximate ￼ using Taylor polynomial ￼ of ￼ centered at ￼. Calculate the Taylor polynomial and estimate the error bound.

(Solution)

The Taylor series for ￼ centered at ￼ is:

￼￼

The third degree of Taylor polynomial ￼ is:

￼

￼￼￼

So, the approximation is:

￼

The error bound for the Taylor polynomial ￼ is given by the next term in the series, evaluated at some point ￼ in the interval ￼:

￼

For ￼, the fourth derivative is:

￼

Evaluating the maximum value in ￼:

￼￼￼

So the error bound is:

￼￼

Solution 20.2

To accurately determine the value of ￼ to the third decimal place, how many terms of the Maclaurin series for ￼ must be used?

(Solution)

The Maclaurin series for ￼ is given by: ￼

For ￼, this becomes: ￼

To find the number of terms required to approximate ￼ to third decimal places, we use the error bound for the Taylor polynomial. The error bound ￼ for the ￼-th degree Taylor polynomial is: ￼

where ￼ is some value in the interval between ￼ and ￼. For the Maclaurin series, ￼ and ￼. The ￼-th derivative of ￼ is still ￼, and since ￼ is an increasing function, the maximum value of ￼ in the interval ￼ is ￼. Therefore, the error bound is: ￼

To proceed, we'll test successive values of ￼ until the inequality is satisfied. Since ￼, the error bound needs to be less than 0.0007 because the approximation belong to the range of 1.6480 to 1.6489.

￼

1. For ￼: ￼

This is greater than 0.0007, so it is not sufficient.

2. For ￼: ￼

This is still greater than 0.0007, so it is not sufficient.

3. For ￼: ￼

This is less than 0.0007, so it is sufficient.

Thus, at ￼, the error bound is less than 0.0007, meaning we need at least 4 terms of the Maclaurin series to ensure the value of ￼ is accurate to the third decimal place.

Therefore, using the error bound of the Taylor polynomial, we find that the number of terms needed is 4.

￼

￼

Solution 20.3

For the following function,

￼

(Solution)

(Solution 20.3a) Find the Maclaurin series.

For ￼: ￼

By differentiating both sides, we get: ￼

Therefore:

￼

(Solution 20.3b) By using the result of (a), obtain the sum of the following series.

￼

From the result of (a), let's substitute ￼.

￼￼

Therefore:

￼

Solution 20.4

Evaluate the sum of the following series.

(Solution)

(Solution 20.4a)

￼

￼

Let's differentiate once more:

￼

Let's multiply ￼ to both sides:

￼

By substituting ￼, we get:

￼￼

Therefore:

￼

(Solution 20.4b)

￼

￼￼￼

Here, we can utilize ￼:

￼

￼

Therefore:

￼￼￼￼

Solution 20.5

Evaluate the limit by using the power series.

￼

(Solution)

￼

Thus:

￼

￼

Solution 20.6

Show that the following is true.

￼

(Solution)

￼

Thus:

￼

Here:

￼

￼

10. Epilogue

Thank you very much for your hard work in studying up to this point. I sincerely appreciate your support in learning the basics of calculus together. I believe that if you continue to strive towards your goals, you will definitely achieve the results you desire. I always wish for that to be the case.

Finally, I would be grateful if you could leave a review of the book wherever you purchased it! Your valuable feedback is a great source of encouragement for me.

June, 2024