Mathematical Modelling (Math1003) Supplementary Reference Notes

MATH1013

Calculus I

Supplementary Reference Notes

Department of Mathematics

The Hong Kong University of Science and Technology

August 16, 2013

Contents

Table of Contents iii

1 Functions 1

1.1 Fundamentals of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Linear Functions and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.3 Quadratic Functions and Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.4 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.5 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.6 Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2 Limits and Continuity 29

2.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.2 Techniques for Evaluating Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.3 Existence of Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3 Differentiation 59

3.1 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.2 Tangent Lines and Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.3 Basic Rules for Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

3.4 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

3.5 Transcendental Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

3.6 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

3.7 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

4 Applications of Differentiation 103

4.1 Increasing and Decreasing Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

4.2 Relative Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

4.3 Absolute Extrema on a Closed Interval . . . . . . . . . . . . . . . . . . . . . . . . . . 109

4.4 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

4.5 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

4.6 LHopital Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

4.7 High Order Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

4.8 Business and Economics Applications . . . . . . . . . . . . . . . . . . . . . . . . . . 122

iii

CONTENTS

5 Integration 127

5.1 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1285.2 Some Elementary Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1305.3 Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1325.4 Properties of Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1375.5 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . 1385.6 Techniques of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.7 Applications of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1545.8 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

iv

Chapter 1

Functions

This chapter reviews the basic ideas you need to start calculus. The topics include the real number sys-tem, Cartesian coordinates in the plane, straight lines, parabolas, and graphs of elementary transcendentalfunctions.

1.1 Fundamentals of Functions

In mathematics, we often encounter the word functions. When we discuss functions, we cannot avoid someabstract mathematical concepts of which many students find hesitant.

We start with the concept of set notation which is very useful for specifying a particular set of numbers.

Set

1.1.1 Definition A set is a well-defined class or collection of subjects. It is usually denoted by A, B, X,Y (capital letters). The subjects which contribute the set are called the elements or the members of the set.They are usually denoted by a, b, x, y.

Let A be a set and a be an element of A, then we write a A [reads a belongs to A]. If a does notbelong to A, we write a 6 A.

Example 1.1.1 (Forms of a set)

1. Tabular form For example: Let A consists of 1, 3, 7, 10. We write A = {1, 3, 7, 10}.2. Set-builder form For example: Let B consists of all even integers. We write B = {n : n is even},

or B = {n | n is even}, where both : and | mean such that.2

Example 1.1.2 (Sets of numbers)

N = natural numbers = { 1, 2, 3, 4, 5, 6, }.Z = integers = { 0,1,2,3,4,5,6, } (eg. Z+ = N).Q = rational numbers = { p

q: p, q are integers and q 6= 0 } (eg. pi 6 Q).

R = real numbers = rational numbers + irrational numbers.

C = complex numbers = { a+ ib : a and b are real numbers, and i2 = 1 }.

In the following context, we shall use these symbols to represent the different sets of numbers. 2

1

1. Functions

Real numbers

The set of real numbers, denoted by R, consists of all numbers that can be expressed as decimals, such as

32= 1.5, 1

3= 0.33333 ,

2 = 1.41421 , pi = 3.14159 .

The dots in the above indicate that the sequence of decimal digits goes on forever.The real numbers can be represented geometrically as points along a number line with the integers

marked off at equal intervals. This line is called the real line.

6 5 4 3 2 1 0 1 2 3 4 5 6

pi 1.5 0.5 2 pi

negativelyinfinite

positivelyinfinite

R

Each point on the real line corresponds to a real number which can be rational (positive and negativeintegers, zero, and common fractions) or irrational (e.g.

2, pi, ). The symbol R denotes either the set

of real numbers, or equivalently, the real line.

In Example ??, we mentioned three special subsets of real numbers, they include the natural numbers,the integers, the rational numbers. That is, N R, Z R, Q R. In particular, the rational numbers areprecisely the real numbers with decimal expansions that are either

(a) terminating (ending in an infinite string of zeros), for example,

32= 1.5000 = 1.5 or

(b) eventually repeating (ending with a block of digits that repeats over and over), for example,

23

11= 2.090909 = 2.09.

A terminating decimal expansion is a special type of repeating decimal since the ending zeros repeat.

Real number that are not rational are called irrational numbers. They are characterized by havingnonterminating and nonrepeating decimal expansions. Examples are pi,

2, 35, and log10 2. Since every

decimal expansion represents a real number, it should be clear that there are infinitely many irrationalnumbers. Both rational and irrational numbers are found arbitrarily close to any point on the real line.

Remark. Before going on, we pause to recall two special facts that are important in algebra and calculus.

1. Division by zero is never allowed. Expressions such as

3

0,

0

0,

x

0,

4

1 + 2 3are always considered to be undefined. They are meaningless in mathematics.

2. A positive real number c always has two square roots,c and c, and c always stands for the

positive square root. Negative real numbers do not have real square roots. For each positive realnumber c,

c is positive and

c is undefined.

A subset of the real line is called an interval if it contains at least two numbers and contains all thereal numbers lying between any two of its elements. Geometrically, intervals of numbers correspond to rays(infinite intervals) and line segments (finite intervals) on the real line. We discuss the details in the following.

2


Intervals

1.1.2 Definition An interval is a point set described by inequalities of one of the following types:

a b

a 6 x 6 b

a b

a < x < b

a b

a < x 6 b

a b

a 6 x < b

R

1. Closed interval: [a, b] = {x : a 6 x 6 b}.2. Open interval: (a, b) = {x : a < x < b}.3. Halfly-open, halfly-closed interval: (a, b] = {x : a < x 6 b}; [a, b) = {x : a 6 x < b}.

If a and b are finite numbers, the above intervals are called finite intervals. The following are infinite intervals.

(a,) = {x : x > a}, [a,) = {x : x > a},(, b) = {x : x < b}, (, b] = {x : x 6 b}, (,) = R.

a

x > a

a

x > a

b

x < b

b

x 6 b

R

Example 1.1.3 (Intervals)

1. The endpoints of the interval (a, b) are a and b, whereas a is called the left hand endpoint and b iscalled the right hand endpoint. A finite interval is said to be closed if it contains both of its endpoints,halfly-open if it contains one endpoint but not the other, and open if it contains neither endpoint.

2. The symbols and , read infinity and minus infinity, do not stand for numbers; they areonly used to indicate an interval with no left hand endpoint, or no right hand endpoint (just a remark:whenever we write in the context, we always mean + for convenience). Infinite intervals areclosed if they contain a finite endpoint, and open otherwise. The entire real line R is an infiniteinterval that is both open and closed.

3. The length of the interval (a, b) is b a > 0.

4. The neighborhood of a point x0 is an open interval centered at x0:

{x R : x0 h < x < x0 + h, h > 0} = (x0 h, x0 + h).

5. The deleleted neighborhood of x0 is a neighborhood from which the point x0 itself has beenremoved: (x0 h, x0) (x0, x0 + h), where is the union of two sets.

6. The endpoints are also called boundary points, they make up the intervals boundary. The remainingpoints of the interval are interior points and together comprise the intervals interior. More specifically,the interior point of an interval S is a point x0 such that the neighborhood of x0 is entirely containedin S. For example, 3

4is an interior point of S = ( 1

2, 1] but 1 is not an interior point of S.

2

3

1. Functions

Solving inequalities

The process of finding the interval or intervals of numbers that satisfy an inequality in x is called solvingthe inequality.

Example 1.1.4 (Solving inequalities)

Solve the following inequalities and find their solution sets on the real line.

(a) 2x 1 < x+ 5, (b) x5< 2x 1, (c) 5

x+ 1> 3.

Solution

(a)

2x 1 < x+ 5,2x x < 5 + 1,

x < 6.

The solution set is the open interval (, 6).(b)

x5

< 2x 1,x < 10x 5,5 < 11x.

The solution set is the open interval (5/11, ).(c) The inequality 5/(x+1) > 3 holds only if x > 1, because otherwise 5/(x+1) is undefined or negative.

Therefore, (x+1) is positive and the inequality will be preserved if we multiply both sides by (x+1),and we have

5

x+ 1> 3,

5 > 3x+ 3,

3x 6 2.

The solution set is the open interval (1, 2/3].2

Absolute value

The absolute value of a number x, denoted by |x|, is defined by the formula

|x| =(

x, if x > 0,

x, if x < 0.

Example 1.1.5 (Absolute value)

Finding the absolute values:

|3| = 3, |0| = 0, | 4| = (4) = 4, | |a|| = |a|.2

Geometrically, the absolute value of x is the distance from x to 0 on the real line. Since distances arealways positive or zero, we see that |x| > 0 for every real number x, and |x| = 0 if and only if x = 0. Also,|x y| equals the distance between x and y on the real line.

Remark. Since the symbola always denotes the nonnegative square root of a, an alternate definition

of |x| is |x| =x2. It is also important to remember that

a2 = |a|. Do not write

a2 = a unless you

already know that a > 0.

4


Example 1.1.6 (Solving equation with absolute values)

Solve the equation |2x 4| = 6.

Solution There are two possibilities:8 1

12< x < 1.

Remark that multiplying by a negative number reverses the inequality. So does taking reciprocals in aninequality in which both sides are positive. The original inequality holds if and only if 1/2 < x < 1. Thesolution set is the open interval (1/2, 1). 2

Example 1.1.8 (Solving inequality with absolute values)

Solve the inequalities (a) |2x 3| 6 1, (b) |2x 3| > 1.

Solution

(a)

|2x 3| 6 1 1 6 2x 3 6 1 2 6 2x 6 4 1 6 x 6 2.

The solution set is the closed interval [1, 2].

(b)

|2x 3| > 1 2x 3 > 1 or 2x 3 6 1 2x > 4 or 2x 6 2 x > 2 or x 6 1.

The solution set is the union of infinite sets (, 1] [2, ).2

5

1. Functions

Constants and variables

Mathematical problems always involve two kinds of quantities: constants and variables. In any givenproblem, a constant has a fixed value, whereas a variable is a letter which stands for an arbitrary realnumber; that is, it varies over the real line.

Example 1.1.9 (Constants and variables)

1. In the definition of the interval a < x < b, the given numbers a and b are called constants. The symbolx which can represent any number greater than a and less than b is called a variable.

2. In the equation x2+y2 = r2 of a circle with centre at the origin, the radius r is constant for a particularcircle and both x and y are variables each having a range from r to r, inclusively.

2

Remark. In Example ??, a, b and r are although treated as constants, however, they may differ frominterval to interval, or from circle to circle. In other words, there are no limitations on the length of theinterval or the size of the circle (a, b can be any real number while r may have any positive value), in thiscase, they are sometimes called arbitrary constants, or parameters. Besides, real numbers such as 2,3, pi whose values are the same in all problems are called absolute constants.

Function

In the 17th century, G. W. Leibnitz, one of the inventors of calculus, introduced the term function into themathematical vocabulary. The concept of a function is one of the most basic in all of mathematics, and itis essential to the study of calculus.

Briefly, a function is a special type of relation that expresses how one quantity (the output) dependson another quantity (the input). For example, when money is invested at some interest rate, the interest I(output) depends on the length of time t (input) that the money is invested. To express this dependence,we say that I is a function of t. Other examples include the area of a circle is a function of its radius; thelength of a day is a function of the latitude and the date; the price of an object is a function of the supplyand the demand. Functional relations like these are usually specified by a formula (or a rule) that showswhat must be done to the input to find the output.

To further illustrate, suppose $100 earns simple interest at an annual rate of 6%. Then it can be shownthat interest and time are related by the formula

I = 100(0.06) t, (1.1)

where I is in dollars and t is in years. In particular, if t = 12, then I = 100(0.06)( 1

2) = 3. Thus, the

formula (??) defines a rule which assigns the output 3 to the input 12.

1.1.3 Definition A function from a set X to a set R is a rule that assigns to each input number x Xexactly one output number f(x) R. The set X of all input numbers to which the rule applies is called thedomain of the function. The set R of all output numbers is called the range.

Remark. We have been using the term function in a restricted sense because, in general, the inputs oroutputs do not have to be numbers. We may consider some examples from everyday life. 1. The functionAge is the process of subtracting the birth year from the current year. 2. The function FA defined byFA(red) = r, FA(green) = g, FA(blue) = b is the process of assigning the first alphabet of the name of thecolour. 3. The function A defined by A(red) = {d, e, r}, A(green) = {e, g, n, r}, A(blue) = {b, e, l, u} isthe process of assigning all the alphabets in the name of the colour. Therefore, the domain and range of afunction can be any sets of objects, but often in calculus they are sets of real numbers.

6


Loosely speaking, we may think of a function f as a kind of machine that produces an output value f(x)in its range whenever we feed it an input value x from its domain.

x f f(x)Input

(domain)Output(range)

A function can also be pictured as an arrow diagram. Given the sets of real numbers X and Y , each arrowassociates a number x in the domain X to a unique number y = f(x) in the codomain Y . We express it as

f : X Y [reads f maps X into Y ],where X is the domain and Y is the codomain.

xf

f(x)

a

f(a)

X = domain set Y = codomain set= set containing the range

Important terminology on functions.

1. Functions are defined by functional notation. For example, the equation f(x) = x2 defines thefunction f that assigns the output number x2 to an input number x:

f : x 7 x2 [reads f maps x to x2].2. The variable y in Y whose value depends upon the chosen value of x is called the dependent variable

while x in X is called the independent variable. The sets X and Y are called the domain andcodomain, respectively.

3. If f(x) = y, the value y associated to x X by the function f is called an image of x under thefunction f or the function value of f at x.

4. The set of images of X under f , denoted by

f [X] = {f(x) : x X},is called the range (or image set) of the function. In fact, the range is always a subset of thecodomain. That is, f [X] Y .

Example 1.1.10 (Function)

1. By the definition of a function, it is impermissible that f(x) = y and f(x) = z with y 6= z. But,however, it is permissible that f(x1) = f(x2) when x1 6= x2.

2. The formula y = f(x) = sin1 x for 1 6 x 6 1 does not yet well-define a function, as there areinfinitely many angles y whose sine is x. For examples, 0 = sin1 0, pi = sin1 0, 2pi = sin1 0, and infact, kpi = sin1 0 for any integer k.

3. If the domain and the range are clearly defined, the formula y = sin1 x can be a well-defined function.Let F (x) = sin1 x such that the domain and the range are {x : 1 6 x 6 1} and {y : pi

26 y 6 pi

2},

respectively. Then F is a function.

2

7

1. Functions

A function f is said to be defined on an interval if it is defined for every point of the interval. When afunction f is given by an expression and the domain is not prescribed explicitly, the domain of f is usuallytaken as the largest set D such that for each x in D, the expression is meaningful.

Example 1.1.11 (Finding domains)

1. For the interest function defined by (??), the input number t cannot be negative, because negativetime makes no sense. Thus, the domain consists of all non-negative numbers; that is, all t > 0.

2. Find the domain of the functionf(x) =

x

x2 x 2 .In mathematics, it is meaningless to divide by zero, so we must find any values of x that make thedenominator 0 and try to exclude them. Thus, we set the denominator equal to 0 and solve for x:

x2 x 2 = (x+ 1)(x 2) = 0which implies x = 1, 2. Therefore, the domain of f is all real numbers except 1 and 2.

3. Find the domain of the functiong(x) =

2x 1.

In mathematics, given any negative number, say 1, it is impossible to find a real number x such thatx2 = 1. Therefore, for a square-root function like g, its argument must be greater than or equal to0 (i.e., non-negative). For this example, we must have 2x 1 > 0 which means x > 1

2. Thus, the

domain is the interval [ 12,).

2

Graphs

The graph of a real function of one variable defined by y = f(x) is a pictorial representation of the function.The graph consists of the set of points defined by the ordered pairs of (x, y) or (x, f(x)) in the plane. Forthis reason, we can also describe the function with set-builder notation:

{(x, y) : a 6 x 6 b, y = f(x)}.Remark. An ordered pair of real numbers, (a, b), is given by the first number a and the second number b.For example, (1, 2), (2, 1), and (1, 1) are three different ordered pairs. Following tradition, we use the samesymbol for the open interval (a, b) and the ordered pair (a, b). However the open interval and ordered pairare completely different things. It will always be quite obvious from the context whether (a, b) stands forthe open interval or the ordered pair.

x

y

f(x)

0

Domain X

Range

a x

(x, y) y = f(x)

b

Question. How can we tell whether a set of points in the plane is the graph of some function? By readingthe definition of a function again, we have an answer.

A set of points in the plane is the graph of some function f if and only if for each vertical line one of thefollowing happens:

8


1. Exactly one point on the line belongs to the set.

2. No point on the line belongs to the set.

A vertical line crossing the x-axis at a point a will meet the set in exactly one point (a, b) if f(a) is definedand f(a) = b, and the line will not meet the set at all if f(a) is undefined.

y

x

y

x

y

x

Graphs of functions

y

x

y

x

y

x

y

x

Not graphs of functions (why?)

Elementary functions

In the following we will introduce some functions which are fundamental yet important in mathematics. Inparticular, we will present the definitions of the functions and describe their properties by showing someillustrative examples and corresponding graphs.

1. Identity functionf : R R defined by f = {(x, y) R2 : y = x}.

The conventional expression of the above function is f(x) = x. The graph is a straight line makingequal angles with the coordinate axes.

x

y

y = f(x) = x

(or simply y = x)

0

Domain = RRange = R

2. Constant function

f : R R defined by f(x) = c, for some constant c R.

The graph is a horizontal line cutting the y-axis at the point (0, c). Thus, f(5) = c, f(2) = c,f(1000) = c.

9

1. Functions

x

y

y = c

0

c Domain = RRange = {c}

3. Step function

A function, f , whose domain is an interval [a, b) R, is called a step function if there is a partitionP = {x0, x1, , xn} of [a, b) such that f is constant on each subinterval of P . That is to say, for eachk = 1, 2, , n, there is a real number yk such that

f(x) = yk, for all x [xk1, xk).For example, f(x) = k, for all x [k 1, k) where k is any integer. The graph is a step-ladderwhich jumps up one unit at each integer. At each step, both the left hand endpoint is included andright hand endpoint is not included.

x

y

y = f(x)

0 1

1

Domain = RRange = Z

4. Absolute value function

f(x) = |x| =(

x, if x > 0,

x, if x < 0.Thus, |2| = 2, |2| = 2, |0| = 0. The function assigns to each real number x the non-negative numberx2.

x

y

y = |x|

0

Domain = RRange = [0,)

5. Reciprocal function

f(x) =1

x, x R {0}.

The function is defined for all nonzero x, but is undefined at x = 0. Therefore, in the above, R {0}stands for the set of all real numbers, except, with x = 0 excluded. The graph of the reciprocalfunction has the equation y = 1/x, which can be written as xy = 1.

10


x

y

y =1

x

0

Domain = R {0}Range = R {0}

6. Even functionf(x) = f(x), for all x R.

For example, f(x) = x2 is an even function, since f(x) = (x)2 = x2 = f(x). Geometrically, evenfunctions are symmetrical about the y-axis.

x

y y = x2

0

Domain = RRange = [0,)

7. Odd functionf(x) = f(x), for all x R.

For example, f(x) = x3 is an odd function, since f(x) = x3 = f(x). Geometrically, odd functionsare not symmetrical about the y-axis, but have the origin as their center of symmetry.

x

y

y = x3

0

Domain = RRange = R

8. Bounded function

A function f(x) is said to be a bounded function on an interval if its function value is never largerthan some value M and never smaller than some value m for all values of x in the interval. That is,

there exist m,M R such that m 6 f(x) 6 M, for all x in the interval.

11

1. Functions

In this case, we say f(x) is bounded above (by M) and below (by m).

For example, f(x) = sin x is a bounded function in R = (,) since1 6 sin x 6 1, for all x R.

Its graph is shown as follows.

x

y

y = sin x

0

1

1

Domain = RRange = [1, 1]

9. Convex function

A convex function is one which has the property that a chord joining any two points A and B on itsgraph always lie above the graph of the function contained between these points.

For example, f(x) = x2, for x R is a convex function.

x

y

0

A

B Domain = RRange = [0,)

If f(x) is convex in (a, b), then we have the following inequality

f(a) + f(b)

2> f

a+ b

2

which can be proved by the following figure.

f(b)

f( a+b2)f(a)

f(a)+f(b)2

a a+b2

b

10. Concave function

A concave function is one which has the property that a chord joining any two points A and B on itsgraph always lie below the graph of the function contained between these points.

For example, f(x) = sin x, for x [0, pi] is a concave function.

12


x

y

0

A

B Domain = [0, pi]Range = [0, 1]

If f(x) is concave in (a, b), then we have the following inequality

f(a) + f(b)

26 f

a+ b

2

which can be proved by the following figure.

f(b)f( a+b

2)

f(a)

f(a)+f(b)2

a a+b2

b

11. Power function and polynomial

A function of the form f(x) = xn, where n is a constant, is called a power function. A polynomial (inone variable) is a function

f(x) = anxn + an1x

n1 + + arxr + + a1x+ a0,

where the real constant ar is called the coefficient of the r-th term, the non-negative integer n is thedegree of the polynomial and the variable x can be any real number (i.e., domain = R). Also recallthe fact that all polynomials are continuous over R.

For example, f(x) = x3 + x 13is a polynomial in x of degree 3.

12. Rational function

A rational function is one which can be expressed as a quotient of two polynomials and has the form

f(x) =anx

n + an1xn1 + + a1x+ a0bmxm + bm1xm1 + + b1x+ b0 .

A rational function is defined for all values of x for which the denominator does not vanish.

For example, f(x) =x3 + x 3x2 3x+ 2, for all x R {1, 2} (i.e., x can be any real number except 1, 2).

13

1. Functions

13. Algebraic function

An algebraic function is a function whose connection with the variable is expressed by an equationthat involves only the algebraic operations of addition, subtraction, multiplication, division, raising to agiven power, and extracting a given root. An algebraic function may be transformed into a polynomialinvolving the two variables x and y, the highest power of x, y both being greater than 1.

For example, y =xx

2 x , for all x in {x : 0 6 x < 2}. By squaring both sides of the algebraicfunction and doing some simplifications, it can be converted to a polynomial (in 2 variables) such that

2y2 xy2 x3 = 0.14. Transcendental function

A transcendental function is defined as a function which is not algebraic. This is a very wide class offunctions. Some common examples of transcendental functions include

Trigonometrical functions: f(x) = sin x, f(x) = cosx, .Exponential functions: f(x) = 10x, f(x) = ex, .Logarithmic functions: f(x) = log10 x, f(x) = lnx, .

We shall discuss the exponential and logarithmic functions in more detail in Section ?? (page ??) andSection ?? (page ??), respectively.

15. Periodic function

A function f is said to be periodic in x with T 6= 0 if its domain contains x+ T whenever it containsx and if f(x + T ) = f(x), for all x in the domain of f . Here we call T the period of the periodicfunction f . By induction, we know that f is periodic in x with a period T if and only if there existsT 6= 0 such that f(x+ nT ) = f(x) is satisfied for all x in the domain of f and for all n = 1, 2, .For example, f(x) = sin x. Then we have f(x+ 2npi) = f(x), for all x R, where n = 1, 2, .

x

y

x T0

x x+ T x+ 2T x+ 3T

16. Inverse function

Let f be a function defined on a domain D and have a range R, and be a function defined on adomain R and have a range D. If f and are related such that

f((y)) = y, for all y R and (f(x)) = x, for all x D,then is called the inverse function of f , and denoted by = f1.

For example, f : (0,) R is defined by f(x) = log10 x, for all x (0,), then : R (0,) isdefined by (y) = 10y , for all y R. Since

f((y)) = log10 10y = y, for all y R and (f(x)) = 10log10 x = x, for all x (0,),

is the inverse function of f . In fact, with the same reasoning, you can regard f as the inversefunction of . See also the mirror reflection for exponential and logarithmic functions (page ??).

14


17. Composite function

If f : X Y and g : Y Z be two functions with domains X and Y respectively. Then for each x X, there corresponds a value z Z through the intermediate variable y Y , i.e., z = g(y) = g(f(x)).Thus we have a new function of X into Z. We define the composite function g f : X Z by

(g f)(x) = g(f(x)).You may regard g f as function of function. By the definition, we see that the composition g fexists if and only if the domain of g contains the range of f . The domain of g f coincides with thedomain of f .

X Y Z

g f

f g

x

f(x)

g(f(x))

For example, let f : R R and g : R R be defined by f(x) = x2 1x2 + 1

, and g(x) = x3. Then

(g f)(x) = g(f(x)) = gx2 1x2 + 1

=

x2 1x2 + 1

3, and

(f g)(x) = f(g(x)) = f(x3) = (x3)2 1

(x3)2 + 1=

x6 1x6 + 1

.

Remark that the composition of functions is in general not commutative, i.e., g f 6= f g.

Bounds of function

1.1.4 Definition A function f is said to be bounded if there exists a positive real number M such that

|f(x)| 6 M, for all x in the domain.

1. If there is a constant K such that f(x) 6 K for all x in the domain of the function, then the functionis said to be bounded above, and K is called an upper bound of f .

2. Similarly, if there is a constant L such that f(x) > L for all x in the domain of the function, then thefunction is said to be bounded below, and L is called an lower bound of f .

3. A function which is bounded above and below is said to be bounded. In fact,

|f(x)| 6 M M 6 f(x) 6 M, for some M R+ M 6 f(x) and f(x) 6 M, for some M R+.

Example 1.1.12 (Bounds of a function)

1. f(x) =1 x2, |x| 6 1 is bounded since |f(x)| 6 1 for |x| 6 1.

2. f(x) = sin x, x R is bounded since | sin x| 6 1, for all x R.3. f(x) = x2+ x+ 3

4= 1 (x 1

2)2, x R is bounded above since f(x) 6 1, for all x R but however

f is not bounded below for x R. Hence, f is not a bounded function for all real numbers.2

15

1. Functions

Monotone function

1. A function f is said to be increasing in an interval D if

f(x1) 6 f(x2), where x1 < x2 and x1, x2 D.

2. A function f is said to be strictly increasing in an interval D if

f(x1) < f(x2), where x1 < x2 and x1, x2 D.

3. A function f is said to be decreasing in an interval D if

f(x1) > f(x2), where x1 < x2 and x1, x2 D.

4. A function f is said to be strictly decreasing in an interval D if

f(x1) > f(x2), where x1 < x2 and x1, x2 D.

5. A function is monotonic if it is either increasing or decreasing (not both) throughout D.

For example, the increasing and decreasing functions have the following graphs (left and right figures,respectively). Assume that the interval D has endpoints a and b.

x

y

0 a bx1 x2

y = f(x)

f(x1)

f(x2)

x

y

0 a bx1 x2

y = f(x)f(x2)

f(x1)

Example 1.1.13 (Monotone function)

Show that f(x) =x

1 + xis an increasing function on (1,).

Solution For any x1, x2 (1,), we have

f(x1) f(x2) = x11 + x1

x21 + x2

=x1(1 + x2) x2(1 + x1)

(1 + x1)(1 + x2)

=x1 x2

(1 + x1)(1 + x2).

From which we observe thatf(x1) < f(x2) whenever x1 < x2

for any x1, x2 (1,). Hence, f(x) is an increasing function on (1,). 2

16

1.2 Linear Functions and Lines

1.2 Linear Functions and Lines

The simplest mathematical model for relating two variables is the linear equation

y = mx+ b.

The equation is called linear because its graph is a line (in this text, we use the term line to mean straightline). When we write f(x) = mx+ b we mean f is a linear function and thus y = f(x) represents the linearequation.

As we mentioned, a line has the equation given by y = mx + b, where m and b are two constants. Ifwe know the values of m and b, then we can use the equation to determine a relationship between y and x.Together, (x, y) forms an ordered pair or geometrically a point in the xy-plane. Collectively, these pointsform a line in this plane. By letting x = 0, one can see that the line crosses the y-axis at y = b. In otherwords, The y-intercept is (0, b). The steepness or slope of the line is m.

x

yy = mx+ b

0

y-intercept(0, b)

Positive slope (m > 0), line rises

1 unit

m units

x

y

y = mx+ b

0

(0, b)y-intercept

Negative slope (m < 0), line falls

1 unit

m units

x

The positive (resp. negative) slope of a line is the number of units the line rises (resp. falls) verticallyfor each unit of horizontal change from left to right, as shown in the above figure.

In fact, if (x1, y1) and (x2, y2) are lying on a line, then

y1 = mx1 + b, y2 = mx2 + b.

Solving for b from the first equation and then substituting it into the second equation, we can obtain theexpression for m:

m =y2 y1x2 x1 =

vertical change

horizontal change.

This ratio represents the steepness of the line and is called the slope of the line (so m is the slope). Inparticular, for the line, if we put m = 0, we have the equation for the horizontal line which is y = b. Besides,a vertical line has an equation of the form x = a. Because such an equation cannot be written in the formy = mx+ b, it follows that the slope of a vertical line is undefined in such case. However, every line has anequation that can be written in the general form Ax+By + C = 0, where A and B are not both zero. Forinstance, the vertical line given by x = a can be represented by the general form (1)x+ (0)y + (a) = 0.

Main points on lines.

1. Slope: m =y2 y1x2 x1 =

y1 y2x1 x2 .

2. Slope-intercept form: y = mx+ b.

3. Point-slope form: y y1 = m(x x1).

4. Two-point form: y y1 = y2 y1x2 x1 (x x1).

5. General form: Ax+By + C = 0.

6. If two lines are parallel, then m1 = m2.

7. If two lines are perpendicular, thenm1 m2 = 1.

8. Vertical line: x = a (no slope).

9. Horizontal line: y = b (slope = 0).

17

1. Functions

Example 1.2.1 (Slope)

Find the slope of the line containing the points (1, 2) and (5,3)?

Solution Let P and Q be the two distinct points with coordinates (x1, y1) and (x2, y2), respectively. Recall

that the slope m of the line through the points P and Q is given by the formula m =y2 y1x2 x1 . Hence, the

slope of the given line is

Slope =2 (3)1 5 =

5

4.

2

Example 1.2.2 (Slope as a rate of change)

A manufacturing company determines that the total cost in dollars of producing x units of a product isC = 25x + 3500. Describe the practical significance of the y-intercept and slope of the line given by thisequation.

Solution The y-intercept (0, 3500) tells you that the cost of producing zero units is $3500. This is thefixed cost of production it includes costs that must be paid regardless of the number of units produced.The slope of m = 25 tells you that the cost of producing each unit is $25. Economists call the cost per unitthe marginal cost. If the production increases by one unit, then the margin or extra amount of cost is $25.

2

Example 1.2.3 (Point-slope form)

Find the equation of the line that has a slope of 3 and passes through the point (1,2).

Solution Use the point-slope form with m = 3 and (x1, y1) = (1,2). The equation of the line is given by

y y1 = m(x x1),y (2) = 3(x 1), y = 3x 5.

2

Example 1.2.4 (Two-point form)

Find the equation of the line passing through the points (2, 3) and (4, 5).

Solution Given the two points (x1, y1) = (2, 3), (x2, y2) = (4, 5). The equation of the line is given by

y y1 = y2 y1x2 x1 (x x1),

y 3 = 5 34 2(x 2),

y = 13x+

11

3,

which represents a straight line passing through the points (2, 3) and (4, 5), and with y-intercept b = 11/3and slope m = 1/3. 2

Example 1.2.5 (Predicting sales per share)

The sales per share for a company was $25.73 in 1996 and $30.59 in 2000. Using only this information,write a linear equation that gives the sales per share in terms of the year.

18

1.3 Quadratic Functions and Parabolas

Solution Let t = 6 represent 1996. Then the two given values are represented by the ordered pairs(6, 25.73) and (10, 30.59). The slope of the line passing through these points is

m =30.59 25.73

10 6 1.215.

Using the point-slope form, one can find the equation that relates the sales per share S and the year t tobe S = 1.215t + 18.44, where 6 6 t 6 10. In fact, one can use this model to estimate the sales per sharein other years. For instance, the model estimates the sales per share in 1997, 1998, and 1999 to be $26.95,$28.16, and $29.38, respectively. The estimation method illustrated here is called a linear interpolation(since the interpolated points lie between the two given points). 2


If y > 0, the equation x2 = y has two solutions. We denote the positive solution byy. The negative

solution is therefore y. We note again that there is no ambiguity about these symbols and that ysimply means

y or y. The general form of a quadratic equation has the form

y = ax2 + bx+ c,

where a, b and c are constants (we assume a 6= 0, otherwise it is just linear). For any quadratic functions ofthe form f(x) = ax2 + bx+ c, we have the so-called completing square method (shown as follow):

ax2 + bx+ c = a

x2 +

b

ax+

c

a

= a

"x+

b

2a

2 b

2

4a2+

c

a

#

= a

x+

b

2a

2 b

2 4ac4a

. (1.2)

One of the importance of this method is the application to the maximization/minimization problem. If

a > 0 (resp. a < 0), it follows from (??) that ax2 + bx+ c > b24ac4a

(resp. ax2 + bx+ c 6 b24ac4a

) since`x+ b

2a

2is a perfect square ( > 0) and the first term of (??) is non-negative (resp. non-positive). Thus

the minimized (resp. maximized) value of f(x) = ax2 + bx+ c is b24ac4a

which occurs when x = b2a.

Example 1.3.1 (Quadratic function)

The demand function for a manufacturers product is p = f(q) = 1200 3q, where p is the price (indollars) per unit when q units are demanded (per week). Find the level of production that maximizes themanufacturers total revenue and determine this revenue.

Solution Using the completing square method for quadratic functions, we have

pq = (1200 3q) q = 3q2 + 1200q = 3(q 200)2 + 120000.Thus, the maximized revenue is r = $120, 000 when q = 200. 2

The general quadratic equation ax2 + bx + c = 0 can be regarded as finding the intersection points (ifany) between the graphs of y = ax2 + bx + c and y = 0 (i.e., the x-axis). Based on the formula (??), we

have 4a2`x+ b

2a

2= b2 4ac. Since the left side cannot be negative, it follows that the quadratic equation

has no real solution if b2 4ac < 0, one real solution if b2 4ac = 0 and two real solutions if b2 4ac > 0.If b2 4ac > 0, the two roots of the equation ax2 + bx+ c = 0 are therefore

=bb2 4ac

2aand =

b+b2 4ac2a

.

19

1. Functions

It is a simple matter to check that, for all values of x, f(x) = ax2+ bx+ c = a(x)(x ). With the helpof this formula, we can sketch the graph of the quadratic function which is a parabola in the xy-plane.

x

y

y = ax2 + bx+ c

a > 0

x

y

y = ax2 + bx+ c

a < 0

The above figure shows the graphs of the quadratic functions when a > 0 (parabola open upwards) andwhen a < 0 (open downwards). This can be easily verified by knowing that x2 is the dominant term in thequadratic function.

20


Main points on quadratic functions.

1. The quadratic equation ax2 + bx+ c = 0 has two roots:

x =bb2 4ac

2a. (1.3)

Eq (??) is called the quadratic formula.

2. If a > 0, then ax2 + bx+ c > b2

4a+ c. If a < 0, then ax2 + bx+ c 6 b

2

4a+ c.

3. The graph of y = ax2 + bx+ c is a parabola which is symmetric to the line x = b2a

.

4. The vertex is

b2a, c b

2

4a

.

5. If a > 0, then the curve opens upward. If a < 0, then it opens downward.

6. The y-intercept is c.

The following figure shows the graphs of y = ax2 + bx + c (a > 0) for the cases when b2 4ac > 0,b2 4ac = 0, b2 4ac < 0 (from left to right).

2 intersection points 1 intersectionpoint

no intersectionpoint

(a) a > 0 and b2 4ac > 0 (b) a > 0 and b2 4ac = 0 (c) a > 0 and b2 4ac < 0

The following figure shows the graphs of y = ax2 + bx + c (a < 0) for the cases when b2 4ac > 0,b2 4ac = 0, b2 4ac < 0 (from left to right).

2 intersectionpoints 1 intersection

point

no intersectionpoint

(a) a < 0 and b2 4ac > 0 (b) a < 0 and b2 4ac = 0 (c) a < 0 and b2 4ac < 0

As we mentioned earlier, when we look for any intersection point(s) between the graphs of the quadraticfunction and the x-axis, then based on the quadratic formula, we know that, when y = 0, the quadraticequation ax2 + bx+ c = 0 (where a, b, c are real) may only have

(a) two distinct real roots the parabola cuts the x-axis; or(b) one double root (i.e., repeated roots) the parabola touches the x-axis; or(c) no real root the parabola does not intersect the x-axis.

21

1. Functions

Example 1.3.2 (Quadratic Function)

A nice application of the work on quadratic equations described above is the proof of the Cauchy-Schwarzinequality. This asserts that, if a1, a2, , an and b1, b2, , bn are any real numbers, then

(a1b1 + a2b2 + + anbn)2 6 (a21 + a22 + + a2n)(b21 + b22 + + b2n).

Solution For any x,

0 6 (a1x+ b1)2 + (a2x+ b2)

2 + + (anx+ bn)2= (a21 + a

22 + + a2n)x2 + 2(a1b1 + a2b2 + + anbn) x+ (b21 + b22 + + b2n)

= Ax2 + 2Bx+ C.

Since y = Ax2 +Bx+C > 0 for all values of x, it follows that the equation Ax2+Bx+C = 0 cannot havetwo (distinct) real roots. Hence, (2B)2 4AC 6 0, i.e. B2 6 AC which is what we need to prove. 2

1.4 Exponential Functions

Given a real constant a > 0 (but a 6= 1), the function f with domain R defined by

f(x) = ax

is called an exponential function. Here a is called the base and x is called the exponent. Thus,

f(x) = 2x, f(x) = 3x, f(x) = 10x+1, f(x) = 53x+2, f(x) = pix

are all exponential functions of x. In general, exponential functions have the following properties:

aman = am+n,am

an= amn, (am)n = amn, (ab)n = anbn,

ab

n=

an

bn, a1 = a, a0 = 1, an =

1

an.

The graphs of the exponential function f(x) = ax (for a > 1 and 0 < a < 1) are shown as follows.

x

y

1

exponentiallygrows

y = ax

(a > 1)

0x

y

1

exponentiallydecays

y = ax

(0 < a < 1)

0

The function value will be positive because a positive base raised to any power is positive. This means thatthe graph of any exponential function will be located in the first and second quadrants.

22

1.4 Exponential Functions

Main points on exponential functions.

1. The domain of f(x) = ax is the whole real line R, i.e., x can be any real number.

2. The range of f(x) = ax is the positive real line R+ since f(x) > 0 for all real values of x.

3. The y-intercept is always 1, i.e., f(0) = 1.

4. f(x) = ax is a strictly monotone function (increasing when a > 1; decreasing when 0 < a < 1).

5. If a > 1, then as x , y = ax 0; as x, y = ax .In this case, f(x) increases (very rapidly) as x increases, it models exponential growth.

6. If a < 1, then as x , y = ax ; as x, y = ax 0.In this case, f(x) decreases (very rapidly) as x increases, it models exponential decay.

In calculus, the most convenient choice for a base of exponential functions is a certain irrational number,denoted by e,

e = 2.71828 .The function f(x) = ex is called the natural exponential function, or simply call it exponential function.

Example 1.4.1 (Exponents)

1. 21 22 23 24 25 2100 = 21+2+3+4+5++100 = 250101 = 25050 .2. 1 4 9 16 25 36 = 12 22 32 42 52 62 = (1 2 3 4 5 6)2 = 7202 = 518400.

2

Example 1.4.2 (Modeling a population)

A bacterial culture is growing according to the logistic growth model

y =1.25

1 + 0.25e0.4t, t > 0

where y is the culture weight (in grams) and t is the time (in hours). Find the weight of the culture after 0hour, 1 hour, and 10 hours. What is the limit of the model as t increases without bound?

Solution

y =1.25

1 + 0.25e0.4(0)= 1gram, y =

1.25

1 + 0.25e0.4(1) 1.071 gram, y = 1.25

1 + 0.25e0.4(10) 1.244 gram.

As t approaches infinity, the limit of y is

limt

1.25

1 + 0.25e0.4t=

1.25

1 + 0= 1.25.

So, as t increases without bound, the weight of the culture approaches 1.25 grams. 2

23

1. Functions

1.5 Logarithmic Functions

Given a real constant a > 0 (but a 6= 1) and for any x > 0, the word logarithm means power of exponent

ay = x if and only if y = loga x. (1.4)

Here loga x means the exponent of a that gives x which we call it the logarithmic function with base a.For example, one can immediately tell the answers (i.e., the values of y) for the exponent equations suchas 2y = 16 or 3y = 27. However, the question will become more difficult if, for example, you are given anexponent equation likes this: 4y = 56. In this situation, logarithm will tell you the answer: y = log4 56 2.9.

So it should not be difficult to see that exponential functions and logarithmic functions are closely related.In fact, logarithmic functions are just inverse functions of exponential functions, or vice versa. To view this,we have the following identities:

aloga x = x, loga(ay) = y.

The above identities can be deduced by eliminating y and x, respectively, in (??). In calculus, the logarithmplays an important role to convert products to sums which very often can simplify the computationwork of differentiation. In particular, one of the most well-known techniques is called the logarithmicdifferentiation. We shall discuss the topic of differentiation in Chapter ??.

In general, logarithmic functions have the following properties:

loga(mn) = logam+ loga n, logam

n= logam loga n, logamr = r logam,

loga1

m= logam, loga 1 = 0, loga a = 1,

loga ar = r, alogam = m, logam =

logcm

logc a.

The graphs of the logarithmic function f(x) = loga x (for a > 1 and 0 < a < 1) are shown as follows.

x

y

10

graduallyincreasing

y = loga x

(a > 1)

x

y

10

graduallydecreasing

y = loga x

(0 < a < 1)

Depending on the value of the given base a (a > 1 or 0 < a < 1), the graph of the logarithmic functionwill be either strictly increasing or strictly decreasing. However, we should remark that, very often, we referlogarithmic function to the case when a > 1 for a practical reason. In calculus and its applications it isfound advantageous to use e (= 2.71828 > 1) as the base of logarithms. The choice of e matches withwhat we discussed the exponential functions in Section ??.

24

1.5 Logarithmic Functions

Main points on logarithmic functions.

1. The domain of f(x) = loga x is R+ = (0,) and the range is the whole real line R.

2. They all pass through the point (1, 0), since a0 = 1, i.e., the x-intercept is 1.

3. The function increases very gradually for higher value of x.

4. The logarithm graph is the mirror image about y = x of the exponential graph, i.e., if you fold thepaper along the line y = x, both graphs will match exactly.

5. If a > 1, then as x 0+, f(x) = loga x ; as x, f(x) = loga x.In this case, f(x) increases as x increases, hence f is an increasing function.

6. If 0 < a < 1, then as x 0+, f(x) = loga x; as x, f(x) = loga x .In this case, f(x) decreases as x increases, hence f is a decreasing function.

Very often, for convenience, we use log x for log10 x (called the common logarithmic function) and lnxfor loge x (called the natural logarithmic function, for which we choose a = e = 2.71828 ). The graphsof the logarithmic function f(x) = ln x as well as the exponential function f(x) = ex are shown as follows.

x

y y = ex

y = ln x

y = x

0

1

1

Since the logarithmic function y = loga x has been regarded as the inverse function of the exponentialfunction x = ay, therefore, the logarithmic curve is a mirror reflection of the graph of the exponentialfunction about the line y = x. This logarithmic curve lies wholly to the right of the y-axis which isconsistent with the fact that only positive numbers (i.e., x > 0) have real logarithms.

Example 1.5.1 (Logarithms)

Without using a calculator, determine the values of the expressions.

1. log6 54 log6 9 = log654

9= log6 6 = 1.

2. log10 0.01 log2 4 = log10 102 log2 22 = 2 log10 10 2 log2 2 = 2(1) 2(1) = 4.

3. e4 ln 33 ln 4 = eln 34ln 43 =

eln 34

eln 43=

eloge 34

eloge 43=

34

43=

81

64.

2

25

1. Functions

Example 1.5.2 (Logarithms)

Simplify the following expressions by the properties of exponential and logarithmic functions.

1. e lnx = eln1x =

1

x.

2. ln ex = x, by the identity ln eu = u.

3. (e2)lnx =èlnx

2= x2. Here, we have used the laws (eu)v = euv = (ev)u and elnu = u.

4. (3e)lnx =èln 3e

lnx=èln 3+1

lnx=èlnx

ln 3+1= xln 3+1.

5. e1lnx = e1 e lnx = eelnx

=e

x.

6. ln(ex

x) = ln ex ln x = x lnx. Here, we have used the laws ln(u

v) = ln u ln v and ln eu = u.

2

Example 1.5.3 (Logarithmic equation)

Solve3 + log1/2 2 = 2 logx(x

2 x 3).Solution Since

log1/2 2 = log1/2(1

2)1 = (1) log1/2

1

2= 1,

we have2 logx(x

2 x 3) = 3 1 = 2,or

logx(x2 x 3) = 1.

The last equation is equivalent to x2 x 3 = x, orx2 2x 3 = 0.

Solve the last quadratic equation, we have two roots

x = 1, x = 3.Since base of logarithmic function must be positive, we have x > 0. Thus, we have the only solution x = 3.It is easy to verify that x = 3 is indeed a solution of the original equation. 2

Example 1.5.4 (Exponential equation)

Solve92x3 = 7x+1.

Solution We take the natural logarithm of both sides,

(2x 3) ln 9 = (x+ 1) ln 7.Solving the equation gives

x =3 ln 9 + ln 7

2 ln 9 ln 7 3.48681.2

26

1.6 Trigonometric Functions

1.6 Trigonometric Functions

The Greek letters (theta) and (phi) are often used for angles. In calculus it is convenient to measureangles in radians instead of degrees. An angle in radians is defined as the length of the arc of the angleon a circle of radius one. Since a circle of radius one has circumference 2pi,

360 degrees = 2pi radians.

Thus a right angle is90 degrees = pi/2 radians.

To define the sine and cosine functions, we consider a point P (x, y) on the unit circle x2 + y2 = 1. Let be the angle measured counterclockwise in radians from the point (1, 0) to the point P (x, y) as shown inthe following figure. Both coordinates x and y depend on . The value of x is called the cosine of , andthe value of y is called the sine of . In symbols,

x = cos , y = sin .

1

x = cos

y = sin

(1, 0)

P (x, y)

The tangent of is defined by

tan =sin

cos .

In addition we have the secant, the cosecant, and the cotangent of . They are respectively defined as

sec def.=

1

cos , csc

def.=

1

sin , cot

def.=

1

tan .

Negative angles and angles greater than 2pi radians are also allowed.

The trigonometric functions can also be defined using the sides of a right (angle) triangle, but thismethod only works for between 0 and pi/2. Let be one of the acute angles (i.e., 0 < < pi/2) of a righttriangle as shown in the following.

ca

b

Then

sin =a

c, cos =

b

c, tan =

a

b.

The two definitions, with circles and right triangles, can be seen to be equivalent using similar triangles.

27

1. Functions

The following table gives the values of sin and cos for some common values of .

in degrees 0 30 45 60 90 180 270 360

in radians 0 pi/6 pi/4 pi/3 pi/2 pi 3pi/2 2pi

sin 0 1/22/2

3/2 1 0 1 0

cos 13/2

2/2 1/2 0 1 0 1

A useful identity which follows from the unit circle equation x2 + y2 = 1 is

sin2 + cos2 = 1.

Here we note that sin2 means (sin )2.

The following figure shows the graphs of sin and cos , which look like waves that oscillate between 1and 1 and repeat every 2pi radians.

y = sin

4pi 2pi 0 2pi 4pi

1

1

y = cos

4pi 2pi 0 2pi 4pi

1

1

28

Chapter 2

Limits and Continuity

2.1 Limits of Functions

To say that the concept of limits is important to your understanding of calculus is an understatement becauseliterally everything we do for calculus is based on this concept. This is fundamental to building up moreadvanced topics such as continuity (Section ??), derivatives (Section ??), definite integrals (Section ??).

Basically, limit is to functions as square root is to numbers. We take square roots of numbers; wetake limits of functions. Sometimes we say the square root of a number does not exist. We will often saythat a limit does not exist. The reasons for why we will say limits do not exist will vary greatly, dependingupon what we see and calculate, but mostly depending upon how we understand the process of limit-taking.First things first. Lets experience limits intuitively.

The idea of finding (or taking) limits of functions always involves a specific point of interest. We alwaysknow the x-coordinate of this point. All we need to do is to find the y-coordinate (if it exists). Practically,when we find limits, we are endeavoring to describe the behavior of the function for all points on its graphnear the point of interest. How near? Nearer than you can imagine. To describe this abstract concept,we shall introduce several phrases

1. x gets closer and closer to a;

2. x approaches a;

3. x tends to a;

4. x is sufficiently close to a,

which, basically, have the same meaning. For convenience, we often use the arrow notation to representthe meaning of gets closer and closer to. That is to say, x a means that x gets closer and closer to a.

x

y

from the left

(a

)

from the right

f(a)

y = f(x)

point of interest

29

2. Limits and Continuity

As we stated before, using the concept of limits, we may describe the value of f(x) when the value of x getscloser and closer to a particular value a. In this case, since x is the only independent variable, the domainof f can only include points lying on the (one-dimensional) real number line. In view of this, when we saya variable is getting closer and closer to a fixed constant, we actually mean the variable is approaching theconstant from two possible directions: from the left and from the right.

For illustrative purpose, let us consider in more detail the variation of f(x) =1

2(3x1) when x is getting

closer and closer to 4 (i.e., when x 4).

x

yy =

1

2(3x 1)

is small is small

0 4 4 4+

5.5+5.5

5.5

x

from the left4

from the right

3.99

4 4 4 + 4.01

infinitesimal

By direct calculations, we have the following numerical results.

If 3.9 < x < 4.1 then 5.35 < f(x) < 5.65.

If 3.99 < x < 4.01 then 5.485 < f(x) < 5.515.

If 3.999 < x < 4.001 then 5.4985 < f(x) < 5.5015.

If 3.9999 < x < 4.0001 then 5.49985 < f(x) < 5.50015.

If 3.99999 < x < 4.00001 then 5.499985 < f(x) < 5.500015.

Basically we can continue to make f(x) as close to 5.5 as we wish by taking x closer and closer to 4. In fact,each of the above has the following form

If 4 < x < 4 + then 5.5 < f(x) < 5.5 + ,or equivalently (since x 6= 4), if x is in the open interval (4 , 4 + ), then f(x) is in (5.5 , 5.5 + ):

If 0 < |x 4| < then |f(x) 5.5| < . (2.1)Hence, when x is getting closer and closer to 4 (or equivalently, when is chosen to be smaller and smaller),f(x) is getting closer and closer to 5.5 (or equivalently, is getting smaller and smaller and to be negligible).Mathematically, we also use the phrases

1. f(x) gets closer and closer to 5.5;

2. f(x) approaches 5.5;

3. f(x) tends to 5.5;

4. f(x) is sufficiently close to 5.5;

5. f(x) converges to 5.5;

6. f(x) has 5.5 as a limit (in other words, the limit of f(x) is 5.5)

to describe the (limiting) behavior of the function values f(x) as x approaches 4.

Before giving the informal definition of limits, we shall present two more illustrative examples in thefollowing.

30


Example 2.1.1 (Limits)

Consider the function f(x) = x22x+2. What happens to the value of f(x) when the value of x gets closerand closer to 3?

x

y

y = x2 2x+ 2

0 3

5

Solution The graph of the function f(x) is a parabola and it is evident from the graph that the values off(x) get closer and closer to 5 as x gets closer and closer to 3 from both left and right sides. We say thatthe limit of f(x) is 5 as x approaches 3 and denote lim

x3f(x) = 5. 2

In Example ??, one may observe that the function value 5 can easily be obtained by typically pluggingx = 3 in the function f(x), i.e., f(3) = 5. This is true for a nice function like this example (here a nicefunction we mean that its graph is continuous (i.e., without break) at the point, we will discuss continuityin more detail in Section ?? (page ??)). However, sometimes we may encounter such a case that a functiondoes not have a value at a given point (plugging in that point would cause division by zero) we are stillinterested in how the function values behave near that point. Read the following example.


For another example, consider the function f(x) =x3 1x 1 . Although this function is not defined at x = 1,

can you tell what happens to the value of f(x) when the value of x gets closer and closer to 1?

x

yy =

x3 1x 1

0 1

3

Solution Notice that as x takes on values closer and closer to 1, regardless of whether x approaches itfrom the left (x < 1) or from the right (x > 1), the corresponding values of f(x) get closer and closer to oneand only one number 3. Even though the function is not defined at x = 1 (as indicated by the hollow dotin the figure), the function values get closer and closer to 3 as x gets closer and closer to 1. To express this,we say that the limit of f(x) as x approaches 1 is 3 and write

limx1

x3 1x 1 = 3.

We can make f(x) as close to 3 as we wish by taking x sufficiently close to, but not equal to, 1. We emphasizeagain that the limit exists at 1, even though 1 is not in the domain of f . 2

31


2.1.1 Definition The limit of the function f(x) as x approaches a is the number L, written as

limxa

f(x) = L, (2.2)

provided that f(x) is arbitrarily close to L for all x sufficiently close to, but not equal to, a.

For a given function f(x), if there is a number L satisfying the above informal definition, we say that

1. the limit of f(x) exists and equals L; or

2. f(x) converges to L; or

3. f(x) has the value L as its limit,

when x is sufficiently close to, but not equal to, a. Otherwise, if there exists no such L, we say that

1. the limit of f(x) does not exist; or

2. f(x) diverges; or

3. f(x) has no limit.

Remark. We emphasize that, when finding a limit, we are concerned not with what happens to f(x) whenx equals a, but only with what happens to f(x) when x is close to a. Be very careful of this difference.Moreover, a limit must be independent of the way in which x approaches a. In other words, the limit mustbe the same whether x approaches a from the left or from the right (i.e., for x < a or for x > a, respectively).


x

y y = f(x)

0 a

L

x

y y = g(x)

0 a

L

x

y y = h(x)

0 a

L

h(a)

The functions f , g and h all approach the limit L when x approaches a. Note that g(x) is not defined atx = a and both g(x) and h(x) have a point of discontinuity at x = a (indicated by the hollow dot in thefigures). However, both functions g(x) and h(x) still get closer and closer to the value L as x gets closer andcloser to a from the left and from the right. Thus,

limxa

f(x) = L, limxa

g(x) = L, limxa

h(x) = L.

The crucial point here is that each limit depends on the values of the function sufficiently close but not equalto a. The function value at x = a itself has no influence at all on the limit. 2

In summary there are three important points about limits of functions.

1. Saying that the limit of f(x) approaches L as x approaches a means that the value of f(x) may bemade arbitrary close to the number L by choosing x closer and closer to a.

2. For a limit to exist, you must allow x to approach a from either side of a. If f(x) approaches adifferent number as x approaches a from the left than it does as x approaches a from the right,then the limit does not exist.

3. The value of f(x) when x = a has no influence on the existence or non-existence of the limit of f(x)as x approaches a. In fact, a may not even be in the domain of f . However, the function must bedefined on both sides of a.

32


Let us explain why we called Definition ?? as an informal definition of limit of function. In fact, we havea more rigorous definition which often appears in more advanced calculus books. This is sometimes calledthe - definition of limit. Please note that this definition of limit is supposed to be optional stuff which willnot be tested in examinations. Recall that the limit process is the tendency for a changing number towardcertain value. For a function f(x) defined near (but not necessarily at) a, we can consider its tendency asx a. We first give the more rigorous definition of the limit lim

xaf(x) of functions in the following.

2.1.2 Definition The limit of the function f(x) as x approaches a is the number L, denoted limxa

f(x) = L,

if for any > 0, there is > 0, such that

0 < |x a| < = |f(x) L| < . (2.3)

In the sense of limit taking process, the predetermined smallness of |f(x)L| is arbitrarily given, whilethe size for |x a| has to be found after is given. Thus the choice of usually depends on and is oftenexpressed as a function of . Note also that because the limit only concerns what happens when numbersare close, only small and need to be considered.

At first glance, it is not that trivial what the definition is telling us. However, recall the previousdiscussion in (??), you will find this definition much easier to understand.

x

y

y = f(x)

0

a

L

Example 2.1.4 (Limits) The graph for the function f(x) = x2 suggests that limx2

x2 = 4. Rigorously

following the definition, for any > 0, choose = min1,

5

. Then

0 < |x 2| < = |x 2| < 1, |x 2| < 5

= |x+ 2| < 5, |x 2| < 5

= |x2 4| = |x+ 2| |x 2| < 5 5= .

The choice of above comes from analyzing the problem. We try to achieve |x2 4| = |x + 2| |x 2| < by requiring 0 < |x 2| < . Note that when x is close to 2, |x+ 2| is close to 4 and |x2 4| is close to 4.If we give some room for x by choosing 6 1 so that |x+ 2| is not more than 5. Then we end up with therequirement 5 6 . Combining 6 1 and 5 6 together yields our choice for . 2

33


Example 2.1.5 (Limits) The graph for the function f(x) = 1/x suggests that limx1

1/x = 1. Rigorously

following the definition, for any > 0, choose = min12, 2

. Then

0 < |x 1| < = |x 1| < 12, |x 1| <

2

= |x| > 12, |x 1| <

2

= 1x 1

= |x 1||x| < /21/2 = .

The choice of is motivated by the formula 1x 1

= |x 1||x| .

When x is close to 1, the numerator is small and the denominator |x| is also close to 1, and the key for thefraction to be small is that the denominator is not too small. Specifically, by choosing 6 1

2, we can make

sure |x| > 1/2 so that |1/x 1| < 2. Thus to get the desired estimation, it is sufficient to further require2 < . 2

Example 2.1.6 (Limits) The graph for the function f(x) =x suggests that lim

x4x = 2. Rigorously

following the definition, for any > 0, choose = . Then

0 < |x 4| < = |x 4| <

= |x 2| = |x 2| |x+ 2||x+ 2| =

|x 4||x+ 2| 0, if f(x) > 0 for all x in [a, b].

7.

Z ba

f(x) dx 6

Z ba

g(x)dx, if f(x) 6 g(x) for all x in [a, b].

8.

Z aa

f(x) dx = 2

Z a0

f(x) dx, if f(x) is an even function: f(x) = f(x).

9.

Z aa

f(x) dx = 0, if f(x) is an odd function: f(x) = f(x).

10.

Z ba

f(x) dx

6

Z ba

|f(x)| dx.

11. m(b a) 6Z ba

f(x) dx 6 M(b a), if m 6 f(x) 6 M for all x in [a, b].

Example 5.4.1 (Properties of definite integral)

Without attempting to evaluate them, determine whether the following integrals are positive, negative or zero.

1.

Z 10

x3(1 x)3 dx. 2.Z pi0

ex sin x dx. 3.Z 112

ex ln xdx.

Solution

1. Since f(x) = x3(1 x)3 > 0 for all x in (0, 1), we have R 10x3(1 x)3 dx > 0.

2. Since f(x) = ex sin x > 0 for all x in (0, pi), we haveR pi0ex sin x dx > 0.

3. Since f(x) = ex lnx < 0 for all x in ( 12, 1), we have

R 112ex ln xdx < 0.

2

5.5 The Fundamental Theorem of Calculus

As we mentioned in Section ??, the approach provided in Section ?? to find definite integrals is only anillustration. In practice we never use the limit definition to evaluate a definite integral. In this section, weshall introduce the real method, the so-called fundamental theorem of calculus. We give the detailed proofin the following.

Suppose that A(x) is the function for the net area of the region bounded by the curve y = f(x) and thex-axis from a to x, i.e.,

A(x)def.=

Z xa

f(x) dx

138

5.5 The Fundamental Theorem of Calculus

which is in general an indefinite integral. In particular, A(a) = 0, A(b) =R baf(x) dx. Consider the area of

the region bounded by the curve y = f(x) and the x-axis from x to x+ h (h is a small number).

a

x x+ h b

y = f(x)

In particular, the shaded region can be approximated by a rectangle, so the area of the region is approx-imately hy, where y is a number between f(x) and f(x + h). On the other hand, the area of the shadedregion is A(x+ h) A(x), by the definition of A(x). Thus,

A(x+ h)A(x) = area of shaded region = hy, or A(x+ h) A(x)h

= y.

As h 0, y approaches f(x) since y is a number between f(x) and f(x+ h). Hence,

A(x) def.= limh0

A(x+ h) A(x)h

= f(x),

or A(x) = f(x). This implies that A(x) is a primitive function of f(x). Now let F (x) be any primitivefunction of f(x). Then

A(x) = F (x) +C.

Since A(a) = 0, we have 0 = F (a) + C. This gives C = F (a). Hence,A(x) = F (x) F (a) = A(b) = F (b) F (a).

By definition, A(b) =R baf(x) dx and hence we have

Z ba

f(x) dx = F (b) F (a), F (x) = f(x). (5.3)

In other words, the definite integral can be evaluated as the difference of any primitive function evaluatedat the endpoints. This is known as the fundamental theorem of calculus. This formula (??) tells usthat as long as we can find a primitive function of f , then we can use it to evaluate the definite integralR baf(x) dx. However, we should emphasize that the fundamental theorem of calculus only describes a way

of evaluating a definite integral, not a procedure for finding primitive functions. That is to say, in order toevaluate a given definite integral, one should have to know more techniques for finding primitive functions(see Section ??).

Due to the importance of the formula (??) for evaluating definite integrals, and because it uncovers theconnection between differentiation and integration, the formula (??) is also known as the Newton-Leibnizformula, which was discovered independently by Sir Isaac Newton (1643-1727) in England and by GottfriedWilhelm von Leibniz (1646-1716) in Germany. As you might know these two famous mathematicians arecredited with the invention of calculus.

The task of evaluating a definite integral by means of definition is quite difficult even in the simplestcases (see Example ??). The fundamental theorem of calculus is then applicable in finding definite integralswithout using the limit definition.

Example 5.5.1 (Evaluate a definite integral)

Evaluate the definite integral Z 31

x2 dx.

139

5. Integration

Solution By the fundamental theorem of calculus, it follows that

Z 31

x2 dx = F (3) F (1), where F is a primitive function of x2.

One primitive function of x2 is x3/3. Therefore,

Z 31

x2 dx =x3

3

x=3

x3

3

x=1

=(3)3

3 (1)

3

3=

26

3.

Very often we also write Z 31

x2 dx =

x3

3

31

=(3)3

3 (1)

3

3=

26

3.

Integration by this method is much faster and easier than finding the integral by summing areas of rectangles.2

There are other forms of the fundamental theorem of calculus.

1. If f is differentiable (continuously) on (a, b), then

Z ba

f (x) dx = f(b) f(a).

2. If x [a, b], F is a primitive function of f , then from the fundamental theorem of calculus,Z xa

f(t) dt = F (x) F (a),

whereR xaf(t) dt is a differentiable function of x. Thus,

d

dx

Z xa

f(t) dt = F (x) 0 = f(x).

Example 5.5.2 (Other forms)

Verify the other forms of the fundamental theorem of calculus for the function f(x) = x+ 1.

1.

Z 31

f (x) dx =Z 31

1 dx = [x]31 = 3 (1) = 4 and f(3) f(1) = 4 0 = 4.

2.d

dx

Z x1

f(t) dt =d

dx

Z x1

(t+1) dt =d

dx

(t+ 1)2

2

t=xt=1

!=

d

dx

(x+ 1)2

2 2

2

2

= (x+1)0 = f(x).

2

140

5.6 Techniques of Integration


Formal integration depends ultimately upon the use of a table of integrals. If, in a given case, no formula isfound in the table resembling the given integral, it is often possible to transform the later so as to make itdepend upon formulae in the table.

If we cannot formally evaluate a given indefinite integral by table, there are three important techniquesavailable for practical usage.

1. Integration by substitution, a method based on a change of variables and the chain rule.

2. Integration by parts, a method based on the product rule for differentiation.

3. Integration by partial fractions, whose integrand is a rational function.

Integration by substitution (of a new variable)

Suppose that, in the indefinite integral,

I =

Zf(x) dx,

we wish to change the variable from x to u by means of the substitution x = (u) which transforms f(x)into f((u)). By using the chain rule,

dI

dx= f(x) = dI

du=

dI

dx dxdu

= f((u)) (u) = I =Z

f((u)) (u) du.

Hence, when changing the variable from x to u we replace f(x) by f((u)) and dx by (u) du.

Example 5.6.1 (Integration by substitution)

Find the integral Z(x+ 1)20 dx.

Solution Let u = x+ 1. Thendu

dx= 1, or dx = du.

Hence, Z(x+ 1)20 dx =

Zu20 du =

u21

21+C =

(x+ 1)21

21+ C.

2


Find the integral Zx2(x3 + 7)3 dx.

Solution Let u = x3 + 7. Thendu

dx= 3x2, or du = 3x2 dx.

Hence, Zx2(x3 + 7)3 dx =

1

3

Z(x3 + 7)3 3x2 dx = 1

3

Zu3 du =

1

3 u

4

4+C =

(x3 + 7)4

12+C.

2

Once you get familiar with the technique of substitution, you might skip the new variable u and directlyevaluate the integral by putting the integral in a suitable form to which the technique can be applied.

141

5. Integration


1.

Z(x 1)2000 dx =

Z(x 1)2000 d(x 1) = (x 1)

2001

2001+ C.

2.

Z(x3 + 2)1/2 x2 dx =

1

3

Z(x3 + 2)1/2 d(x3 + 2) =

1

3 (x

3 + 2)3/2

3/2+ C =

2

9(x3 + 2)3/2 + C.

3.

Zeax dx =

1

a

Zeax d(ax) =

1

aeax + C, a 6= 0.

4.

Zx2ex

3

dx =1

3

Zex

3

d(x3) =1

3ex

3

+ C.

2

The following are some difficult examples for which we shall also need the table of integrals (page ??).


Find the integral Zx1/2

x 1 dx.

Solution Let u = x1/2. Then x = u2 and dx = 2u du.

Zx1/2

x 1 dx =Z

u

u2 1 2u du = 2Z

(u2 1) + 1u2 1 du

= 2

Z 1 +

1

u2 1du = 2

u+

1

2ln

u 1u+ 1

+ C

= 2x+ ln

|x 1|x+ 1

+C.

2


Find the integral Z1

x8(1 + x2)dx.

Solution Let u =1

x(reciprocal substitution). Then x =

1

uand dx = 1

u2du.

Z1

x8(1 + x2)dx =

Zu8

u2 + 1du =

Z u6 u4 + u2 1 + 1

u2 + 1

du

= 17u7 +

1

5u5 1

3u3 + u tan1 u+ C

= 17x7

+1

5x5 1

3x3+

1

x tan1 1

x+ C.

2



x+x2 + 1

dx.

142


Solution Let u = x+x2 + 1. Then x =

u2 12u

and dx =(2u)(2u) (u2 1)(2)

4u2du =

u2 + 1

2u2du.

Z1

x+x2 + 1

dx =

Z1

u u

2 + 1

2u2du =

1

2

Z 1

u+

1

u3

du

=1

2

ln u 1

2u2

+ C

=1

2ln(x+

px2 + 1) 1

4(x+x2 + 1)2

+C.

2



1 + exdx.

Solution Let u =1 + ex. Then ex = u2 1 and ex dx

du= 2u = dx = 2u

u2 1 du.

Z1

1 + exdx =

Z1

u 2uu2 1 du = 2

Z1

u2 1 du = 2 1

2ln

u 1u+ 1

+ C = ln

1 + ex 11 + ex + 1

+ C.

2

Integration by substitution (of trigonometric functions)

Integrands involving forms such aspa2 x2,

px2 + a2,

px2 a2.

Then the following trigonometric substitutions will help to eliminate the square roots.

1. Whena2 x2 appears, let x = a sin u = a2 x2 = a cosu and dx = a cos u du.

2. Whenx2 + a2 appears, let x = a tanu = x2 + a2 = a sec u and dx = a sec2 u du.

3. Whenx2 a2 appears, let x = a secu = x2 a2 = a tan u and dx = a secu tan u du.

Example 5.6.8 (Trigonometric substitutions)

1.

Z1

a2 x2 dx =Z

1

a cosu a cosu du =

Z1 du = u+ C = sin1

x

a+ C.

2.

Z1

x2 + a2dx =

Z1

a secu a sec2 u du =

Zsec u du = ln | secu+ tan u|+ c

= ln |ptan2 u+ 1 + tan u|+ c = ln

rx2

a2+ 1 +

x

a

+ C + ln a = ln |px2 + a2 + x|+ C.

3.

Z1

x2 a2 dx =Z

1

a tan u a secu tan u du =

Zsecu du = ln | sec u+ tan u|+ c

= ln | secu+psec2 u 1|+ c = ln

xa+

rx2

a2 1

+ C + ln a = ln |x+px2 a2|+ C.

143

5. Integration

2

II. Integrands as a reciprocal of a linear function of sin x and/or cos x, the substitution tanx

2= t is

recommended. By this substitution, we also have

1. sin x = 2 sinx

2cos

x

2= 2 tan

x

2cos2

x

2=

2 tan x2

1 + tan2 x2

=2t

1 + t2.

2. cos x = cos2x

2 sin2 x

2=

cos2 x2 sin2 x

2

cos2 x2+ sin2 x

2

=1 tan2 x

2

1 + tan2 x2

=1 t21 + t2

.

3. dt =1

2sec2

x

2dx =

1

2

1 + tan2

x

2

dx =

1

2

`1 + t2

dx, or dx =

2

1 + t2dt.

Hence, any rational function of sin x and cos x can be expressed as a rational function of t. In particular,integrals

Rsec xdx,

Rcscx dx and integrals of the form

Rdx

a+b sinx,R

dxa+b cos x

,R

dxa+b cos x+c sinx

may allbe evaluated in this way.



5 + 4 sin xdx.

Solution Let t = tanx

2. Then sin x =

2t

1 + t2and dx =

2

1 + t2dt.

Z1

5 + 4 sin xdx =

Z1

5 + 8t1+t2

21 + t2

dt = 2

Z1

5t2 + 8t+ 5dt =

2

5

Z1

(t+ 45)2 + 9

25

dt

=2

5 53tan1

t+ 4

535

+ C =

2

3tan1

5 tan x

2+ 4

3

+C.

2

III. Integration of expressions containing sinm x cosn x, we can make use of the following trigonometricidentities.

1. sin x cosx =1

2sin 2x. 2. sin2 x =

1

2(1 cos 2x). 3. cos2 x = 1

2(1 + cos 2x).


Find the integral Zsin4 x cos2 x dx.

Solution Zsin4 x cos2 xdx =

Z(sin x cosx)2 sin2 xdx =

Z(1

4sin2 2x) 1

2(1 cos 2x) dx

=1

8

Zsin2 2x dx 1

8

Zsin2 2x cos 2xdx

=1

8

Z1

2(1 cos 4x) dx 1

8

Z1

2sin2 2xd(sin 2x)

=x

16 sin 4x

64 sin

3 2x

48+ C.

2

144


IV. Integration of expressions containing sinmx cosnx or sinmx sinnx or cosmx cosnx, we canmake use of the following trigonometric identities.

1. sinmx cosnx =1

2sin(m+ n)x+

1

2sin(m n)x.

2. sinmx sinnx = 12cos(m+ n)x+

1

2cos(m n)x.

3. cosmx cosnx =1

2cos(m+ n)x+

1

2cos(m n)x.


Find the integral Zcos 4x cos 3x dx.

Solution Zcos 4x cos 3x dx =

Z1

2(cos 7x+ cos x) dx

=1

2

sin 7x

7+ sin x

+C

=sin 7x

14+

sin x

2+ C.

2

Integration by parts

We shall introduce another integration technique called integration by parts. This formula frequentlyallows us to compute a difficult integral by computing a much simpler integral. Now we show how thistechnique works.

If u and v are functions of x, the product rule for differentiation gives

d

dx(uv) = u

dv

dx+ v

du

dx.

Integrating with respect to x gives

uv =

Zudv

dxdx+

Zvdu

dxdx =

Zu dv +

Zv du,

or Zu dv = uv

Zv du.

Note that the formula for integration by parts expresses the original integral in terms of another integral.Depending on the choices of functions u and v, it is supposed to be easier to evaluate the second integralthan the original one. In fact, this formula is particularly useful when the integrand contains (i) a product oftwo factors, (ii) an inverse trigonometric function, (iii) a logarithmic function, (iv) an exponential function.

145

5. Integration

Example 5.6.12 (Integration by parts)

Find the integral Zln xdx.

Solution By identifying u = ln x and v = x and integrating by parts, we haveZln xdx =

Z(ln x)| {z }

u

d(x)|{z}v

= (ln x)| {z }u

(x)|{z}v

Z

(x)|{z}v

d(ln x)| {z }u

= x ln xZ

x 1xdx = x ln x x+ C.

2


Find the integral Zxex dx.

Solution By identifying u = x and v = ex and integrating by parts, we haveZxex dx =

Zx d(ex) =

Z(x)|{z}u

d(ex)|{z}v

= (x)|{z}u

(ex)|{z}v

Z

(ex)|{z}v

d(x)|{z}u

= xex Z

ex dx = xex ex + C.

2

Remark. We are going to give some additional examples. Most of the following examples are average. Afew are however challenging. When you read the examples, make careful and precise use of the differentialnotation and be careful when arithmetically and algebraically simplifying expressions. Also in the examples,whenever you find a symbol over the equal sign (i.e., =) we mean the technique of integration by partshas been used at that particular step of calculations.


Find the integral Zx3 ln xdx.

Solution Zx3 ln xdx =

1

4

Zln xd(x4)

=

1

4

x4 ln x

Zx4 d(lnx)

=1

4

x4 ln x x

4

4

+C =

x4

16(4 ln x 1) +C.

2


Find the integral Zln x

x2dx.

146


Solution Zln x

x2dx =

Zlnx d(

1

x)

= ln x

x+

Z1

xd(lnx)

= ln xx

+

Zx2 dx = lnx

x 1x+ C = 1

x(ln x+ 1) + C.

2


Find the integral Zxe2x dx.

Solution Zxe2x dx =

1

2

Zxd(e2x)

=

1

2

xe2x

Ze2x dx

=1

2xe2x 1

2 e

2x

2+ C =

e2x

4(2x 1) + C.

2


Find the integral Zx2ex dx.

Solution Zx2ex dx =

Zx2 d(ex)

= x2ex

Zex d(x2)

= x2ex 2Z

xex dx = x2ex 2Z

x d(ex)

= x2ex 2

xex

Zex dx

= x2ex 2xex + 2ex +C

= ex`x2 2x+ 2+C.

2


Find the integral Zxf (x) dx.

Solution Zxf (x) dx =

Zxd(f (x)) = xf (x)

Zf (x) dx

= xf (x) f(x) +C.2


Find the integral Zsin(ln x) dx.

147

5. Integration

Solution Using integration by parts twice,Zsin(ln x) dx

= x sin(ln x)

Zxd(sin(lnx))

= x sin(ln x)Z

x cos(lnx) 1xdx

= x sin(ln x)Z

cos(ln x) dx

= x sin(ln x) x cos(ln x)

Zsin(ln x) dx.

Therefore,

2

Zsin(lnx) dx = x sin(ln x) x cos(ln x) + 2C,

Zsin(lnx) dx =

x

2sin(ln x) x

2cos(ln x) + C.

2


Find the integral Zxex

(x 1)2 dx.

Solution Zxex

(x 1)2 dx =Z

(x 1 + 1) ex(x 1)2 dx =

Z ex

x 1 +ex

(x 1)2dx

= Z

1

x 1 d(ex) +

Zex

(x 1)2 dx

=

ex

x 1 Z

ex d(1

x 1)+

Zex

(x 1)2 dx

= ex

x 1 +Z

ex1

(x 1)2 dx+Z

ex

(x 1)2 dx

= ex

x 1 + C.2

Example 5.6.21 (Inverse trigonometric functions)

Recall the facts (the following derivatives can be found by logarithmic differentiation)

1.d

dx

`sin1 x

=

11 x2 . 2.

d

dx

`cos1 x

=

11 x2 . 3.

d

dx

`tan1 x

=

1

x2 + 1.

Hence, find the integrals

1.

Zsin1 xdx. 2.

Zcos1 xdx. 3.

Ztan1 x dx.

Solution

1.

Zsin1 xdx

= x sin1 x

Zx 1

1 x2 dx = x sin1 x+

1

2

Z1

1 x2 d(1 x2)

= x sin1 x+p1 x2 + C.

148


2.

Zcos1 x dx = x cos1 x

Zx 1

1 x2 dx = x cos1 x 1

2

Z1

1 x2 d(1 x2)

= x cos1 xp1 x2 + C.

3.

Ztan1 x dx

= x tan1 x

Zx 1

x2 + 1dx = x tan1 x 1

2

Z1

x2 + 1d(x2 + 1)

= x tan1 x 12ln(x2 + 1) + C.

2

Integration by partial fractions

Recall that a rational function is a quotient of two polynomials. In this subsection, we shall provide a generalmethod for integrating rational functions that is based on the idea of decomposing a rational function intoa sum of simple rational functions that can be integrated by the methods studied earlier, in principle, theonly primitive functions involved are rational functions, the logarithm and the arc tangent functions.

In algebra we learn to combine two or more fractions into a single fraction by finding common denomi-nator. For example,

3

x 4

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Mathematical Modelling (Math1003) Supplementary Reference Notes