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Contents

1 The Real Number System 11.1 Types of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Properties of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Mathematical problems involving inequality . . . . . . . . . . . . . . 4

1.3.1 Solving a Linear Inequality . . . . . . . . . . . . . . . . . . . . 4

1.3.2 Solving a Twosided Inequality . . . . . . . . . . . . . . . . . 5

1.3.3 Solving an Inequality Involving a Fraction . . . . . . . . . . . 5

1.3.4 Solving a Quadratic Inequality . . . . . . . . . . . . . . . . . . 6

1.3.5 Solving an Inequality Containing an Absolute Value . . . . . . 8

1.4 Number in Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Matrix and Determinant 15

2.1 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 System of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . 16

2.3 Row Echelon Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.4 Augmented Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.5 Matrix Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5.1 Properties of Determinants . . . . . . . . . . . . . . . . . . . . 20

2.5.2 Finding Determinants: The Method of Cofactors . . . . . . . 22

2.5.3 Cramers Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Vector Algebra 25

3.0.4 Introduction to vector . . . . . . . . . . . . . . . . . . . . . . 25

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3.0.5 Vectors in the Plane (2D vector) . . . . . . . . . . . . . . . . . 26

3.0.6 Vectors in the Space (3D vector) . . . . . . . . . . . . . . . . 30

3.0.7 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.0.8 The Cross Product (or Vector Product) . . . . . . . . . . . . . 35

4 One Variable Function 39

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.1.1 Vertical Line Test . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.1.2 Polynomial and Rational Functions . . . . . . . . . . . . . . . 44

4.1.3 Function Evaluation . . . . . . . . . . . . . . . . . . . . . . . 44

4.1.4 Finding Roots of a Function . . . . . . . . . . . . . . . . . . . 45

4.1.5 Finding Domain and Range of a Function . . . . . . . . . . . 46

4.1.6 Inverse Function . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.2 Trigonometric Function . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.2.1 Trigonometric Function as a Periodic Function . . . . . . . . . 49

4.2.2 Definitions of Trigonometric Function . . . . . . . . . . . . . . 51

4.2.3 Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . 53

4.2.4 Solving Trigonometric Equations . . . . . . . . . . . . . . . . 54

4.2.5 The Inverse Trigonometric Functions . . . . . . . . . . . . . . 55

4.3 Hyperbolic Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5 Function Derivative 57

5.1 Definition of Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.1.1 Derivative as Slope of Tangent Line . . . . . . . . . . . . . . . 60

5.2 Derivative Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.2.1 The Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.2.2 General Derivative Rules . . . . . . . . . . . . . . . . . . . . . 62

5.2.3 The Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.2.4 The Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . 64

5.2.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 65

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5.3 Derivative of Special Functions . . . . . . . . . . . . . . . . . . . . . 65

5.3.1 Derivative of the Trigonometric Function . . . . . . . . . . . . 65

5.3.2 Derivative of the Exponential and Logarithmic Functions . . . 66

5.3.3 Derivative of the Hyperbolic Function . . . . . . . . . . . . . . 66

5.4 Higher Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.5 Implicit Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.6 Applications of Derivatives . . . . . . . . . . . . . . . . . . . . . . . 78

6 Integration 816.1 Introduction to Integration . . . . . . . . . . . . . . . . . . . . . . . . 81

6.2 Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6.2.1 The Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 83

6.2.2 The Trigonometric Functions . . . . . . . . . . . . . . . . . . 86

6.2.3 The Exponential and Logarithmic Functions . . . . . . . . . . 86

6.3 Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

6.3.1 Properties of the Definite Integral . . . . . . . . . . . . . . . . 926.3.2 The Fundamental Theorem of Calculus . . . . . . . . . . . . . 94

6.4 Integration Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . 98

6.4.1 Integration by Substitution . . . . . . . . . . . . . . . . . . . 98

6.4.2 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . 101

6.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

7 Differential Equations 107

7.1 Introduction to Differential Equations . . . . . . . . . . . . . . . . . . 107

7.2 Analytical Solution of First Order ODE . . . . . . . . . . . . . . . . . 108

7.2.1 Linear Differential Equations . . . . . . . . . . . . . . . . . . 108

7.2.2 Separable Differential Equations . . . . . . . . . . . . . . . . . 116

7.2.3 Exact Differential Equations . . . . . . . . . . . . . . . . . . . 120

7.2.4 Bernoulli Differential Equations . . . . . . . . . . . . . . . . . 124

7.2.5 Homogeneous Differential Equation . . . . . . . . . . . . . . . 125

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List of Tables

1 Values of sin and cos functions at a few angles in both degrees and

radians . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

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List of Figures

1 Subsets representing real number system . . . . . . . . . . . . . . . . 3

2 An open interval (, 1) . . . . . . . . . . . . . . . . . . . . . . . . 53 An open interval (2, 3

2) . . . . . . . . . . . . . . . . . . . . . . . . . 6

4 Number lines to represent each term in the fraction . . . . . . . . . . 6

5 An open interval (, 2) [1, ) . . . . . . . . . . . . . . . . . . . 76 Number lines to represent each term of the quadratic expression . . . 7

7 An open interval (, 3) (2, ) . . . . . . . . . . . . . . . . . . . 7

8 The distance between a and b . . . . . . . . . . . . . . . . . . . . . . 89 |x 3| < 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 910 Points A, B, C and D plotted on a cartesian coordinate . . . . . . . 10

11 Sketching lines on a cartesian coordinate . . . . . . . . . . . . . . . . 12

12 Shading the region for inequalities . . . . . . . . . . . . . . . . . . . . 13

13 A vector represented by a directed line segment . . . . . . . . . . . . 26

14 Equivalent vectors, a = b = c . . . . . . . . . . . . . . . . . . . . . . 26

15 Vector in the plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

16 Vector in the plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

17 Types of relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

18 Vertical line tests on various curves . . . . . . . . . . . . . . . . . . . 42

19 Inverse function, g = f1 . . . . . . . . . . . . . . . . . . . . . . . . . 48

20 Graphs for sin and cos functions . . . . . . . . . . . . . . . . . . . . . 51

21 Right triangle to define trigonometric functions . . . . . . . . . . . . 51

22 Definition of cos(x) and sin(x) on a unit circle . . . . . . . . . . . . . 53

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23 Derivative Rules (Source: math.arizona.edu/ calc/Rules.pdf) . . . . . 67

24 f(y) = y2 + y2 is not a continuous function . . . . . . . . . . . . . . 96

25 Graph for f(x) = x2 + 16 and f(x) = 2x 8 . . . . . . . . . . . . . 104

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Chapter 1

The Real Number System

1.1 Types of Real Numbers

Natural numbers

Natural numbers are also known as counting numbers.

The numbers are 1, 2, 3, 4, 5, 6, .

The set of all natural numbers is denoted as N.

Natural numbers together with zero are called whole numbers (i.e 0, 1, 2, 3, 4, 5, 6, )

Integers

Integers are the positive and negative whole numbers.

The numbers are , 3, 2, 1, 0, 1, 2, 3, .

The set of all integers is denoted as Z

Rational numbers

Rational numbers are quotients of integers (i.e All numbers of the form a

bwith

a and b are integers and b = 0).

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1.1. TYPES OF REAL NUMBERS

The set of all rational numbers is denoted as Q.

All integers are rational numbers because for any integer i Z, i can bewritten as i

1.

Real numbers with terminating decimal or non-terminating decimals that re-peat are also rational number.

Real numbers with non-terminating decimals that do not repeat are irra-tional numbers. = 3.14159265

and

2 = 1.41421356

are irrational

numbers.

Real numbers

Real numbers are the set of all decimals, both terminating and non-terminating.

The set of all real numbers is denoted as R.

The above mentioned set of numbers (i.e Natural numbers, Integers, Rationaland Irrational numbers) are subsets of real numbers.

The relationships of the sets consist in real numbers can be illustrated in Figure1

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THE REAL NUMBER SYSTEM

Figure 1: Subsets representing real number system

Exercises 1.1

1 Tick the correct type for the following numbers. Each number may fall into

more than one type.

Number Natural Whole Integer Rational Irrational Real

2

0

0.25

-5

2

8

1.342234223422...

1.234567

45

2 Tick the correct type of the following decimal number. Then, determined

whether the number is a rational or irrational number. If it is a rational

number, rewrite the decimal number in the form of integer fraction, ab

.

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1.2. PROPERTIES OF REAL NUMBERS

Number Terminating Nonterminating decimal Rational\ Fractiondecimal Repeat Not Repeat Irrational form

3.16792

10.121212...

4.275191919191...

3.14159265...

3.41287548754875...

1.2 Properties of Real Numbers

If a and b are real numbers and a < b, then

i For any real number c, a + c < b + c.

ii For any real numbers c and d, if c < d, then a + c < b + d.

iii For any real number c > 0, a

c < b

c.

iv For any real number c < 0, a c > b c

1.3 Mathematical problems involving inequality

1.3.1 Solving a Linear Inequality

Problem: Solve the linear inequality 5x + 1 < 6.

Solution: Substract 1 from both sides. Thus we have

(5x + 1) 1 < 6 15x < 5

Then divide the resulting inequality by 5

5x5

< 55

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Finally we have x < 1 or also can be written as an interval (, 1)

Figure 2: An open interval (, 1)

1.3.2 Solving a Twosided Inequality

Problem: Solve the twosided inequality 2 < 8 4x < 16.Solution: We work with both inequalities simultaneously. First, substract 8 from

each term. Thus we have

2 8 < (8 4x) 8 < 16 8

6 < 4x < 8

Then divide the resulting inequalities by -4. Note that, since 4 < 0, our inequali-ties are reversed.

64 >

4x4 >

8

43

2 > x > 2

Rearrange the inequalities, finally we have 2 < x < 32

or also can be written as an

interval (2, 32

)

1.3.3 Solving an Inequality Involving a Fraction

Problem: Solve the inequality x1x+2

0

Solution: In order to visualize the function of the fraction, we draw separate number

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1.3. MATHEMATICAL PROBLEMS INVOLVING INEQUALITY

Figure 3: An open interval (2, 32

)

lines for the numerator and the denominator

Figure 4: Number lines to represent each term in the fraction

From the above number lines, we may conclude that the fraction is satisfying theinequality (i.e nonnegative in this case) whenever x < 2 or x 0. The solutionalso can be written in interval notation as (, 2) [1, )

1.3.4 Solving a Quadratic Inequality

Problem: Solve the quadratic inequality x2 + x 6 > 0Solution: By factorising the quadratic term, we have the following equivalent in-

equality to the original problem:

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Figure 5: An open interval (,2) [1,)

(x + 3)(x 2) > 0 (1)

From here, we may draw the number lines to represent the two terms: (x + 3) and

(x 2), and finally the combination of both.

Figure 6: Number lines to represent each term of the quadratic expression

The number lines show that the product of the two terms is positive whenever

x < 3 or x > 2. In interval notation, this can be written as (, 3) (2, ).

Figure 7: An open interval (,3) (2,)

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1.3. MATHEMATICAL PROBLEMS INVOLVING INEQUALITY

Alternatively, you may sketch the quadratic graph and find the interval

whenever the quadratic graph is positive. Try this method for the same

problem above!

1.3.5 Solving an Inequality Containing an Absolute Value

Definition of Absolute Value

The absolute value of a real number x is |x| =

x, if x 0

x if x < 0

.

For any real numbers a and b,

1. |a b| = |a| |b|.

2. |a + b| = |a| + |b| in general.

3. |a + b| |a| + |b| (the triangle inequality).

4. |a b| is referred as the distance between a and b (Figure 8)

Figure 8: The distance between a and b

Problem: Solve the inequality |x 3| < 5Solution: The LHS of the inequality refers to the distance between 3 and point x.

Considering the value 5 on the RHS, the inequality shows that the distance between

x and 3 must be less than 5. We may visualise this inequality using the following

figure: Obviously, the solution for x is 2 < x < 8 or in interval notation:(2, 8).

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Figure 9: |x 3| < 5

Problem: Solve the inequality |x + 4| 7Solution:

(Hint: |x + 4| = |x (4)|)

There is an alternative method for solving inequalities involv-ing absolute values.

|x a| < d

also can be written as the two-sided inequality

d < x a < d

. Try this method to solve the previous problems!

1.4 Number in Plane

Our previous discussions only considered real numbers as laid out on a single

line (1 dimension coordinate). Now, we extend our discussion to coordinate in

2-dimensions. We locate points in the plane by using two coordinate lines: the hor-

izontal real line is usually called the x

axis and the vertical real line is usually

called the y axis. Instead of just one real number, the point now is represent

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1.4. NUMBER IN PLANE

by an ordered pair (x, y) of real numbers, called coordinates of the point. See the

following example to understand how to locate a point based on the given coordinate.

Example: Plotting points

Plot the points A = (3, 2), B = (3, 6), C = (2, 4) and D = (1, 2).Ans:Figure 10

Figure 10: Points A, B, C and D plotted on a cartesian coordinate

Example: Drawing lines

Sketch the following lines:

i x = 1

ii y = 4

iii 4x + 3y = 24

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Ans:Figure 11

Line x = 1 and y = 4 are easy to skecth.The third line, 4x + 3y = 24 can be skecthed by finding 2 points on the line and

connecting those 2 points to have a straight line. The easiest 2 points that can be

considered are points when x = 0 and when y = 0.

Point 1: When x = 0,

4(0) + 3y = 24

3y = 24

y = 8

Hence, the first point is (0, 8).

Point 2: When y = 0,

4x + 3(0) = 24

4x = 24

x = 6

Hence, the second point is (6, 0)

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Figure 11: Sketching lines on a cartesian coordinate

Example: Graphing Inequalities

Shade the region which contains the points that satisfy the following inequalities:

i y 4

ii 4x + 3y > 24

Ans:Figure 12. In order to graph the region for an inequality, we have to determine

the edge of the region first. This can be done by sketching a line representing the

given equation (replace the inequality sign (, or ) to an equal sign (=)). Inthis example, both lines y = 4 and 4x + 3y > 24 are already sketched in Figure11. We use solid line to represent edge for the first inequality because of the sign

. However, dashed line is used for the second inequality because of the sign > (i.edoes not include equal sign). The reqion for the first inequality is obvious. However,

for the second inequality, we have to choose which side to be shaded by testing one

point at each side.

Side 1: Point to test is (0,0):

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4(0) + 3(0) = 0 < 24does not satisfies the inequality.

Side 2: Point to test is (5,5):4(5) + 3(5) = 35 > 24satisfies the inequality. So we choose this side to be

shaded.

Figure 12: Shading the region for inequalities

Your Tasks

Find the formulae and examples for the following topics:

1. The distance and midpoint between 2 points.

2. Equation of a line.

3. Equation of a circle

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Chapter 2

Matrix and Determinant

2.1 Matrices

(Adams, 1999, pg 636)

An m n matrix A is a rectangular array of mn numbers arranged in m rows

and n columns. If aij is the element in the ith row and the jthe column, then wemay write matrix A as follows:

A =

a11 a12 a1na21 a22 a2n

......

......

am1 am2 amn

.

In short, we may also write A = (aij) where i = 1, 2, , m and j = 1, 2, , n.What is the transpose of matrix A i.e AT ?

What is square matrix?

What is symmetric matrix?

How to multiply matrices?

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2.2. SYSTEM OF LINEAR EQUATIONS

2.2 System of Linear Equations

A linear equation can be defined as

a1x1 + a2x2 + + anxn = b (2)

where

x1, x2, , xn is the variables (unknowns) (3)(4)

A system of linear equations may consists a number of linear equations. As an

example:

a11x1 + a12x2 + . . . + a1nxn = b1

a21x1 + a22x2 + . . . + a2nxn = b2...

......

......

am1x1 + am2x2 + . . . + amnxn = bm

is a system of linear equations consists of m equations and n unknown variables.

Our aim is to determine the value of the unknown variables x1, x2,

, xn when

a11, , amn and b1, b2, bm are given.

Consider a set of n linear equations:

a11x1 + a12x2 + . . . + a1nxn = b1

a21x1 + a22x2 + . . . + a2nxn = b2...

......

......

an1x1 + an2x2 + . . . + annxn = bn

(5)

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MATRIX AND DETERMINANT

We then could represent (5) with the following matrix equation:

Ax = b (6)

where

b =

b1

b2...

bn

A =

a11 a12 . . . a1n

a21 a22 . . . a2n...

......

...

an1 an2 . . . ann

x =

x1

x2...

xn

(7)

2.3 Row Echelon Form

A matrix is in row echelon form if it satisfies:

i All nonzero rows are above any rows of all zeroes.

ii Each leading entry (i.e. left most nonzero entry ) of a row is in a column to

the right of the leading entry of the row above it.

iii All entries in a column below a leading entry are zero.

Example

Reduce the following matrix to its row echelon form:

0 3 6 4 91 2 1 3 12 3 0 3 11 4 5 9 7

Solution:

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2.4. AUGMENTED MATRICES

ans:

1 4 5 9 70 2 4 6 60 0 0 5 00 0 0 0 0

2.4 Augmented Matrices

If given matrices A and B, the augmented matrix of the two matrices can be written

as (A...B).

Example

Given

A =

2 2 1

4 1 2

1 2 3

B =

1

2

4

show the augmented matrix (A...B)

Solution:

(A...B) =

2 2 1

..

. 14 1 2

... 2

1 2 3... 4

Example: Solving system of linear equations using augmented

matrix and Gauss Jordan Elimination

Solve the following system:

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3x 2y = 14x + 3y = 1

Solution:

The augmented matrix of the system is:

3 2... 14

1 3... 1

In Gauss-Jordan Elimination, our aim is to reduce the LHS matrix

3 2

1 3

to be

1 0

0 1

So, first we want to make the leading entry in the first row as 1

3 2

... 14

1 3... 1

In order to achieve that, we may simply interchange the the first and the second

row

1 3..

. 13 2 ... 14

Then, our aim is to make the element below the first leading entry as 0. The second

row minus 3 times the first row

R2 3R1 >

1 3

... 1

0 11 ... 11

Now, we can make the leading entry of the second row as 1 by multiplying the

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2.5. MATRIX DETERMINANTS

second row by 111

. Now we have

111

R2 >

1 3

... 1

0 1... 1

Our final task is to make 3 in the first row to become 0:

R1 3R2 >1 0

... 4

0 1... 1

We have reached to the augmented matrix that we required. The solution is

x = 4 and y = 1

Exercise

Using the method explained above, solve the following system:

2x + y = 3x 4y = 2

2.5 Matrix Determinants

2.5.1 Properties of Determinants

Theorem 2.5.1. LetA be an n n matrix and c be a scalar, then,

det(cA) = cndet(A)

Theorem 2.5.2. Suppose that A, B and C are all n n matrices and that theydiffer by only a row, say the kth row. Lets further suppose that the kth row of C can

be found by adding the corresponding entries from the kth rows of A and B. Then

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in this case we will have that

det(C) = det(A) + det(B)

The same result will hold if we replace the word row with column above.

Theorem 2.5.3. If A and B are matrices of the same size, then

det(AB) = det(A)det(B)

Theorem 2.5.4. Suppose that A is an invertible matrix, then

det(A1) =1

det(A)

Theorem 2.5.5. A suare matrixA is invertible if and only ifdet(A) = 0. A matrixthat is invertible is oftern called non-singular and a matrix that is not invertible is

oftern called singular

Theorem 2.5.6. If A is a square matrix, then

det(A) = det(AT)

Theorem 2.5.7. If A is a square matrix with a row or column of all zeroes, then

det(A) = 0

and so A will be singular.

Theorem 2.5.8. Suppose that A is an n n triangular matrix, then,

det(A) = a11a22 ann

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2.5.2 Finding Determinants: The Method of Cofactors

For a matrix with only one entry, the determinant of the matrix is the value of the

entry. (i.e. If A = (a), then det(A) = a).

For a 2 2 matrix,

A =

a11 a12

a21 a22

the determinant can be calculated as follows:

det(A) = a11a22 a21a12

Try this:

Find the determinant of the following matrix:

A =

2 11 4

Solution:

ans=7

This method in finding determinant can be generalise (known as the method of

cofactors) to solve larger size matrices. Before we proceed in utilising this method,

let first get to know 2 important terms in this method: minor and cofactor.

Definition 2.5.9. (Minor) If A is a square matrix, then the minor of aij, denoted

by Mij , is the determinant of the submatrix that results from removing the ith row

and the jth column of A

Definition 2.5.10. (Cofactor) If A is a square matrix, then the cofactor of aij ,

denoted by Sij, is the number (1)i+jMij

Note: Basically cofactor is just Mij

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Exercise

For the following matrix, compute the cofactors C11 and C12

A =

4 2 1

2 6 37 5 0

Solution:

(ans=C11 = 15 and C12 = 21)Now, we can use cofactors to find determinant as in the following theorem:

Theorem 2.5.11. If A is an n n matrix,

(a) Choose any row, say rowi, then

det(A) = ai1Ci1 + ai2Ci2 + + ainCin

(b) Choose any column, say columnj,then

det(A) = a1jC1j + a2jC2j + + anjCnj

Exercises

1 Recall the matrix:

A =

4 2 1

2 6 37 5 0

Reconsider the previous matrix, compute the determinant using:

(i) Row 1

(ii) Column 3

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Solution:

(ans det(A) = 154)

2 Find the determinant for the following matrix:

B =

4 0 10 4

1 2 3 95

5

1 6

3 7 1 2

2.5.3 Cramers Rule

In this section we will looking back to the system of linear equations. Now, having

understand the calculation of the determinant, we may use Cramers rule in order

to find the solution of the system.

Theorem 2.5.12. Suppose that A is ann n invertible matrix. Then the solutionto the system Ax = b is given by,

x1 =det(A1)

det(A), x2 =

det(A2)

det(A), , xn = det(An)

det(A)

where Aj is the matrix found by replacing the jth column A with b

Try this: Use Cramers rule to determine the solution to the following system

of equations:

2x1 + 3x2 x3 = 1x1 + 2x2 x3 = 4

2x1 + x2 + x3 = 3

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Chapter 3

Vector Algebra

3.0.4 Introduction to vector

Definition 3.0.13. (Scalar) Scalar is a quantity that only has the magnitude.

Examples: Volume, distance, mass.

Definition 3.0.14. (Vector) Vector is a quantity that has the magnitude as well

as the direction.

Examples: Velocity, force, acceleration.

What about weight? Is it a scalar or a vector?Weight is a force F = ma W = mg. The direction is downward .

Therefore, weight is a vector.

A vector V in the plane or in the space is represented by a directed line segment

where the direction is shown by the arrow and the magnitude is determined by the

length of the line, |V|.In Figure 13, there are 2 ways to denote the vector: OA or a. O is called the

initial point and A is called the terminal point. The magnitude of the vector is

the length of the line from O to A, denoted as | OA| or |a|.Two directed line segments represent the same vector if they have the same

length and are parallel. In Figure 14, three vectors that are all considered to be

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Figure 13: A vector represented by a directed line segment

equivalent are shown, even though their initial points are different. We may write

for this case, a = b = c.

Figure 14: Equivalent vectors, a = b = c

Vectors can appear in the plane or in the space. A vector in the plane consists

two components, whereas a vector in the space consists three components. We will

begin our discussion for vectors in the plane first, follow by vectors in the space.

Most of the discussion for vectors in the plane are applicable to vectors in the space.

3.0.5 Vectors in the Plane (2D vector)

Vectors are movable and not tied to any particular system, but for vectors in plane

it might be easier for us to imagine the coordinate system of x

and y

axes. The

vector in the plane consists 2 components a =< a1, a2 > as shown in Figure 15.

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VECTOR ALGEBRA

Figure 15: Vector in the plane

Position Vector

Since the location of the initial point is irrelevent, we usually draw vectors with

the initial point located at the origin, O. This vector is called a position vector.

Vector a in Figure 15 is an example of a position vector.

Example: Finding a Position Vector

(Refer Smith and Minton, 2012, pg 700-Example 1.3).

Exercices

In Exercises 15-18 (Refer Smith and Minton, 2012, pg 703-Exercises 10.1), find the

position vector with initial point A and terminal point B. Then, sketch the vector

in a Cartesian coordinate

(Hint: AB =< b1 a1, b2 a2 >)

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Unit Vector

Any vector with the magnitude 1 is called a unit vector. In 2D, the magnitude of

a vector, a =< a1, a2 > can be calculated as follows:

|a| =

a21 + a22

We may find a unit vector that having the same direction as any vector a using the

following theorem.

Theorem 3.0.15. (Unit Vector ) For any nonzero position vector a =< a1, a2 >,

a unit vector having the same direction as a is given by

u =1

|a|a

Standard basis vectors i and j are defined by

i =< 1, 0 > and j =< 0, 1 >

Notice that |i| = |j| = 1, so both vectors are unit vectors. These vectors form abasis for vector in 2D. Therefore, we may write any 2D vector a in the following

form:

a =< a1, a2 >= a1i + a2j

Example: Finding a Unit Vector


Exercises

(Refer Smith and Minton, 2012, pg 703-Exercises 10.1).

In Exercises 19-24, Find a unit vector in the same direction as the given vector.

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VECTOR ALGEBRA

Vector Arithmetic

The following theorem conclude the arithmetic operations in vector

Theorem 3.0.16. For any 2D vectors a, b and c and any scalars d and e inR,

the following hold:

i (Commutativity)

a + b = b + a

ii (Associativity)

a + (b + c) = (a + b) + c

iii (Zero Vector)

a + 0 = a

iv (Additive Inverse)

a + (a) = 0v (Distributive Law)

d(a + b) = da + db

vi (Distributive Law)

(d + e)a = da + ea

vii (Multiplication by 1)

(1)a = a

viii (Multiplication by 0)

(0)a = 0

Example: Vector Arithmetic


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Exercises


In Exercises 3-6, compute a + b, a 2b, 3a and |5b 2a|Additional Task: For the given vectors in the above exercises, sketch the

vectors a, b and the resulting vectors a + b and a b

Parallel Vector

Definition 3.0.17. Two vectors having the same or opposite direction are called

parallel. The zero vector is considered parallel to every vector

Mathematically we may write

a and b are parallel b = ca for some scalar c

Example: Determining When 2 Vectors are Parallel


Exercises


In exercises 9-14, determine whether the vectors a and b are parallel.

3.0.6 Vectors in the Space (3D vector)

We now extend the ideas in the 2D vector to the 3D vector. Coordinate axes in

R3 is now used (3 coordinate axes: x , y, z axes). Similar to the vector in 2D, we

may draw the position vector a =< a1, a2, a3 > by connecting the origin to a point

(a1, a2, a3) in the space. Figure 16 shows how to sketch a vector in the space.

Basically, the idea extensions from vector in 2D to vector in 3D are straight

forward. Theorem 3.0.5 that was introduced while discussing vector in the plane is

also satisfied for vector in the space. In what follows, we will consider a few examples

to see the manipulations of vector in the space.

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VECTOR ALGEBRA

Figure 16: Vector in the plane

Example: Finding a Unit Vector

(Refer to Smith and Minton, 2012, pg 709-Example 2.3)

Find a unit vector in the same direction as < 1, 2, 3 > and write < 1, 2, 3 > asthe product of its magnitude and a unit vector.

Solution:

Let a =< 1, 2, 3 >. Recall the formula for unit vector u that has the samedirection as vector a

u = 1|a|a

In order to calculate this unit vector, we need the magnitude of the vector a i.e |a|:

|a| =

a21 + a22 + a

23 =

12 + (2)2 + 32 =

14

Therefore, the unit vector for vector a =< 1, 2, 3 > is:

u = 114

< 1, 2, 3 >=< 114

, 214

, 314

>

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The vector a is now can be written in the following form:

a = |a|u< 1, 2, 3 > = 14 1

14,

214

,314

Example: Arithmetic Operations

(Smith and Minton, 2012, pg 711-Exercises 10.2 (7))

Compute a + b, a 3b and |4a + 2b| for a =< 2, 1, 2 > and b =< 1, 3, 0 >

Solution:

a + b =< 2, 1, 2 > + < 1, 3, 0 >=< 2 + 1, 1 + 3, 2 + 0 >

=< 3, 4, 2 >

a 3b =< 2, 1, 2 > 3 < 1, 3, 0 >=< 2, 1, 2 > < 3 1, 3 3, 3 0 >

=< 2, 1, 2 > < 3, 9, 0 >=< 2

3, 1

9,

2

0 >

=< 1, 8, 2 >

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VECTOR ALGEBRA

4a + 2b = 4 < 2, 1, 2 > +2 < 1, 3, 0 >=< 4 2, 4 1, 4 (2) > + < 2 1, 2 3, 2 0 >

=< 8, 4, 8 > + < 2, 6, 0 >=< 8 + 2, 4 + 6, 8 + 0 >

=< 10, 10,

8 >

Then, the magnitude:

|4a + 2b| = | < 10, 10, 8 > |=

102 + 102 + (8)2

=

264

=

4(66)=

4

66

= 2

66

Exercises

(Smith and Minton, 2012, pg 711-Exercises 10.2 (8-20))

3.0.7 The Dot Product

The dot product of two vectors is a scalar. Hence this product is also known as

scalar product.

Definition 3.0.18. The dot product of two vectors of dimension n, a =< a1, a2, , an >

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and b =< b1, b2, , bn > is defined by

a b =< a1, a2, , an > < b1, b2, , bn >= a1b1 + a2b2 + + anbn

Examples: Computing a Dot Product in R3 and R2

(Smith and Minton, 2012, pg 713-Example 3.1 and 3.2)

Theorem 3.0.19. For vectors a, b and c and any scalar d, the following hold:

i a b = b a

ii a (b + c) = a b + a c

iii (da) b = d(a b) = a (db)

iv 0 a = 0 and

v a.a = a2

Theorem 3.0.20. Let be the angle between nonzero vectors a and b. Then,

a b = |a||b| cos .

Example: Finding the Angle between Two Vectors

(Smith and Minton, 2012, pg 714-Example 3.3)

Note that cos(

2) = 0.

Therefore, if angle between vector a and b is

2(i.e a and b are orthogonal),

the dot product of the two vectors will become 0, a b = 0because a b = |a||b| cos(

2) = |a||b|(0) = 0

Exercises

(Smith and Minton, 2012, pg 719-Exercises 10.3, 1-18)

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VECTOR ALGEBRA

15 Find a 3D vector perpendicular to vector < 2, 1, 0 >

Solution:

We are looking for vector a =< a1, a2, a3 > that perpendicular/ orthogonal to

vector < 2, 1, 0 >. We have the dot product of the two vectors as follows

< a1, a2, a3 > < 2, 1, 0 > = a1 2 + a2 (1) + a3 0= 2a1 a2

Then, using the fact that a b = 0 if a and b are orthogonal, we now have

2a1 a2 = 0

Since we only have 1 equation despite three unknowns, we may choose any

value for 2 of the unknown (a3 and a1 or a2). Let a3 = 3, a2 = 2, then find

a1 using the above equation:

2a1 a2 = 02a1 2 = 0

2a1 = 2

a1 = 1

Our suggested vector a that perpendicular to the given vector < 2, 1, 0 > is

< 1, 2, 3 >.

Verify this answer!

3.0.8 The Cross Product (or Vector Product)

The cross product is also known as vector product since it produces a vector. The

cross product is not defined for vectors in 2D. So our remainder discussion will only

consider vectors in the plane (3D vector).

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Definition 3.0.21. (The Cross Product) For two vectors a =< a1, a2, a3 > and

b =< b1, b2, b3 > in space, we define the cross product of a and b to be

a b =

i j k

a1 a2 a3

b1 b2 b3

=

a2 a3

b2 b3

i a1 a3

b1 b3

j +a1 a2

b1 b2

k

Example: Computing a Cross Product

(Smith and Minton, 2012, pg 724-Example 4.3)

Exercises: Computing a Cross Product

(Smith and Minton, 2012, pg 731-732-Exercises 10.4 (5-10))

Theorem 3.0.22. For any 3D vector a, a a = 0 and a 0 = 0

Theorem 3.0.23. For any 3D vectors a and b, a

b is orthogonal to both a and b

Theorem 3.0.24. For any 3D vectors a, b and c, and any scalar d, the following

hold:

i a b = (b a) (anticommutativity)

ii (da) b = d(a b) = a (db)

iii a

(b + c) = a

b + a

c (distributive law)

iv (a + b) c = a c + b c (distributive law)

v a (b c) = (a b) c (scalar triple product)

vi a (b c) = (a c)b (a b)c (vector triple product)

Theorem 3.0.25. For any nonzero 3D vectors a and b, if is the angle between a

and b (0

), then

|a b| = |a||b| sin().

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VECTOR ALGEBRA

Note that from Theorem 3.0.25, when = 0 we have

|a b| = |a||b| sin(0)= |a||b|(0) = 0.

Thus, we can conclude two nonzero 3D vectors a and b are parallel if and only if

a b = 0.

You may refer to Example 4.4 until 4.8 (Smith & Minton,2012) page 728731 to

see the uses of cross product.

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Chapter 4

One Variable Function

4.1 Introduction

Definition 4.1.1. (Function) A function f is a rule that assigns exactly one

element y in a set called Range to each element x in a set called Domain. In this

case we write y = f(x).

We refer to x as the independent variable and y as the dependent variable.

Figure 17(a) and 17(b) show 2 types of relations that considered as functions.

The relation shown in Figure 17(c) however is not a functions because one element

in the domain has more than one value in the range.

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4.1. INTRODUCTION

(a)Relation one to one: f is a function

(b)Relation many to one: f is a function

(c)Relation one to many: f is not a function

Figure 17: Types of relations

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ONE VARIABLE FUNCTION

4.1.1 Vertical Line Test

We may determine whether or not a curve is the graph of a function by using the

Vertical Line Test. In this test, if any vertical line intersects the graph in more

than one point, the curve is not the graph of a function. Figure 18 shows a few

examples of vertical line tests on curves. In Figure 18 (a) and (c), the vertical line

intersects on two points on the curve. This means that both curves in Figure 18 (a)

and (c) are not the graph of a function. However, in 18 (b) and (d), the vertical line

only intersescts on one point on each curve. So, the curves in Figure 18 (b) and (d)

are the graph of a function.

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4.1. INTRODUCTION

(a) (b)

(c) (d)

Figure 18: Vertical line tests on various curves

Exercises

1 Answer question (34) until (38) in (Smith & Minton, 2012) page 17.

2 Determine if each of the following are functions:

a) y = x2 + 1

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4.1. INTRODUCTION

4.1.2 Polynomial and Rational Functions

There are many types of functions. In this early section (Section 4.1), we will discuss

about polynomial and rational functions. Later, in Section 4.2 we will discuss

about trigonometric function and finally we will discuss about hyperbolic function

in Section 4.3

Polynomial Function

Definition 4.1.2. (Polynomial) A polynomial is any function that can be written

in the form

f(x) = anxn + an1x

n1 + + a1x + a0,

where a0, a1, a2, , an are real numbers (the coefficients of the polynomial) withan = 0 and n 0 is an integer (the degree of the polynomial).

(For examples, refer to Smith & Minton, 2012 page 12)

(Rational Function)

Definition 4.1.3. Rational Function Any function that can be written in the form

f(x) =p(x)

q(x)

where p and q are polynomials, is called a rational function.

(For example, refer to Smith & Minton, 2012 page 13)

4.1.3 Function Evaluation

Example(Pauls note)

Given f(x) = x2 + 6x 11, find each of the following:

a) f(2)

b) f(10)

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c) f(t)

d) f(t 3)

e) f(x 3)

f) f(4x 1)

Solution:

(ans:(a)

3, (b)

171, (c)

t2 +6t

11, (d)

t2 +12t

38, (e)

x2 +12x

38 ,(f)

16x2 +

32x 18 )

4.1.4 Finding Roots of a Function

A solution of the equation f(x) = 0 is called a zero of the function f or a root of the

equation f(x) = 0. The zero/root of the function f corresponds to an xinterceptof the graph of y = f(x).

Example(Smith & Minton, 2012)

Find all x and y intercept of f(x) = x2 4x + 3Solution:

Example(Smith & Minton, 2012)

Find the zeroes of f(x) = x2 5x 12Solution:

Example(Pauls note)

Determine all the roots of f(t) = 9t3 18t2 + 6t = 0Solution:

(ans: t = 0, t =3 +

(3)

3, t =

3 33

)

You have learned about the method of Long Division in order to find

roots of polynomial during your secondary school. You may revise this

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4.1. INTRODUCTION

method so it much easier for you to find roots of polynomial with order

higher than 2

4.1.5 Finding Domain and Range of a Function

It is important to find the domain of a function where the function is defined. In

the early section we have been introduced to the definition of the domain and range

of a function. In this section we will go through a few examples how to find them

when a function is given. Finding the range can be difficult, thus we only discuss a

few simple examples on this and more examples that involve finding the domain.

Examples (Finding the domain and range)

Find the domain and range of the following functions:

a) f(x) = 5x 3

b) g(t) =

4

7t

c) h(x) = 2x2 + 12x + 5

d) f(z) = |z 6| 3

e) g(x) = 8

Solution:

(ans: (a)Domain:(

,

), Range:(

,

),(b)Domain:(

,

4

7

], Range:[0,

), (c)Domain:(

,

),

Range:(, 23], (d)Domain:(, ), Range:[3, ), (e)Domain:(, ), Range:8,)

Examples (Finding the domain)

Find the domain of the following functions:

a) f(x) =x 4

x2

2x

15

b) g(t) = 6 + t t2

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c) h(x) =x

x2

9

Solution:

a) The given function f(x) =x 4

x2 2x 15 is not defined when the denominatoris zero. So now, we have to find the value of x when x2 2x 15 = 0.

x2 2x 15 = 0(x

5)(x + 3) = 0

x = 5 or x = 3

Therefore, the function is undefined at point x = 5 and x = 3. The domainfor this function is all the real numbers except 5 and -3.

(ans: (a)Domain:All real numbers except x = 3 & x = 5 , (b)Domain:2 t 3 , (c)Domain:x < 3orx > 3)

4.1.6 Inverse Function

Figure 19 illustrates the basic idea of inverse function. Given a value in the range of

f, we may find the respected value in the domain of f by using function g. Function

f and g reverse the action of each other.

The formal definition of inverse function is stated as follows:

Definition 4.1.4. (Inverse Function) Assume that f has domain A and g has do-

main B. Assume further that f(g(x)) is defined for all x B and g (f(x)) is definedfor all x A. If

f(g(x)) = x, for all x B, andg (f(x)) = x, for all x A,

we say that g is the inverse of f, written g = f1. Equivalently, f is the inverse of

g, f = g1.

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4.1. INTRODUCTION

(

Figure 19: Inverse function, g = f1

Example (Smith & Minton, 2012-pg 26, Example 3.1)

Exercises

1 Given f(x) = 3x 2 find f1(x)Solution:

Let y = f(x) and g = f1(x)

We want to find g such that g(f(x)) = x

y = 3x 2y + 2 = 3x

x = y + 23

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Therefore, g(f(x)) = x =f(x) + 2

3, or g(x) = f1(x) =

x + 2

3

Check the answer:

We may check whether g(x) =x + 2

3is the inverse function for f(x) = 3x 2

using the definition of the inverse function (f(g(x)) = x).

f(g(x)) = f(x + 2

3

)

= 3

x + 2

3

2

= (x + 2) 2= x

We have proven that f(g(x)) = x, therefore our g(x) is the inverse function

for f(x).

2 Given g(x) =

x 3, find g1(x)Solution:

4.2 Trigonometric Function

4.2.1 Trigonometric Function as a Periodic Function

Definition 4.2.1. (Periodic Function) A function f is periodic of period T if

f(x + T) = f(x)

for all x such that x and x + T are in the domain of f. The smallest such number

T > 0 is called the fundamental period.

Table 1 shows the values of sin(x) and cos(x) functions at a few points x. x

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4.2. TRIGONOMETRIC FUNCTION

x (in degree) -360 -180 0 30 45 60 90 180 270 360

x (in radian) 2 0

6

4

3

2

3

2 2

f(x) = sin(x) 0 0 01

2

2

2

3

21 0 -1 0

f(x) = cos(x) 1 -1 1

3

2

2

2

1

20 -1 0 1

Table 1: Values of sin and cos functions at a few angles in both degrees and radians

is an angle that can be measured in degree or radians. The given table also show

the measurement in both units. Most of the time later in this lecture note, we will

consider the angle in radians. From the table, we have:

sin(2) = 0 = sin(0) = sin(2 + 2)

sin(0) = 0 = sin(2) = sin(0 + 2)

sin() = 0 = sin() = sin( + 2)

and so on. We can see that sin(x) = sin(x + 2). Therefore we may conclude that

sin(x) is a periodic function with period T = 2.

for the cos function, we have

cos(2) = 1 = cos(0) = cos(2 + 2)cos(0) = 1 = cos(2) = cos(0 + 2)

cos() = 1 = cos() = cos( + 2)

and so on. This shows that cos(x) = cos(x + 2). Thus, cos(x) is also a periodic

function with period T = 2. We can see the repetition of function sin(x) and cos(x)

for every 2 more clearly in Figure 20.

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(a) (b)

Figure 20: Graphs for sin and cos functions

4.2.2 Definitions of Trigonometric Function

We have been introduced to sin and cos functions. The following definitions are for

the other four trigonometric functions:

Definition 4.2.2. (Tangent, Cotangent, Secant and Cosecant)

The tangent function is defined by tan(x) =sin(x)

cos(x)

The cotangent function is defined by cot(x) =cos(x)

sin(x)=

1

tan(x)

The secant function is defined by sec(x) =1

cos(x)

The cosecant function is defined by csc(x) =1

sin(x)

Trigonometric Function in terms of a Right Triangle

Figure 21: Right triangle to define trigonometric functions

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From the right triangle as illustrated in Figure 21, we have the following definition

for the six trigonometric functions:

cos(x) =adjacent

hypotenusesin(x) =

opposite

hypotenuse

tan(x) =opposite

adjacentcot(x) =

adjacent

opposite

sec(x) =hypotenuse

adjacentcot(x) =

hypotenuse

opposite

Trigonometric Function in terms of a Unit Circle

Imagine a circle with radius 1. For every given angle, we may draw a line from

the center towards the point on the circle. The point where the line and the circle

intersect has coordinate (cos(x), sin(x)) where x is the counterclockwise angle be-

tween the positive xaxis to the drawn line. Based on Figure 22, we can see thatcos(4 ) =

22 and cos(

4 ) =

22 .

Your Task: Draw lines for another angles (

6,

3etc) in the unit circle

with the appropriate coordinates for the point of intersection

Exercises

1) Without using calculator, evaluate each of the following:

a) sin( 23

) and sin(23

) (Given cos(3

) = 12

)

b) cos( 76

) and cos(76

) (Given sin(6

) = 12

)

c) tan(4

) and tan( 74

) (Given sin(4

) = 22

)

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Figure 22: Definition of cos(x) and sin(x) on a unit circle

4.2.3 Trigonometric Identities

Theorem 4.2.3. For any real numbers and , the following identities hold:

sin( + ) = sin()cos() + sin() cos()

cos( + ) = cos()cos() sin() sin()sin2() =

1

2(1 cos(2))

cos2() = 12(1 + cos(2))

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Exercises(Smith & Minton) page 40

Prove that the following trigonometric identities are true:

a) sin(2) = 2 sin()cos()

b) sin( ) = sin() cos() sin() cos()

c) cos( ) = cos() cos() + sin() sin()

Solution:

4.2.4 Solving Trigonometric Equations

Example(Smith & Minton, 2012) page 33

Find all solutions of the equations:

a) 2sin(x) 1 = 0

b) cos2(x) 3 cos(x) + 2 = 0

Solution:

a) 2sin(x) 1 = 0

2 sin(x) 1 = 0 (8)2 sin(x) = 1 (9)

sin(x) =1

2(10)

From the unit circle, we know that sin(x) =1

2when x =

6or x =

6. Since

sin(x) has period 2, other possible solutions are x =

6

+ 2n or x =

6

+ 2n

for any interger n.

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b) cos2(x) 3 cos(x) + 2 = 0

4.2.5 The Inverse Trigonometric Functions

Originally the trigonometric functions is not one to one functions. For example, if

we refer to f = sin(t) graph, keep repeating the same value after a certain period.

Thus,this function does not has an inverse function. However we can retrict our

domain to certain interval so the function will become one to one function. For

example, for the sine function, we can choose interval (2

, 2

). The definitions of

the inverse trigonometric functions are then as follow:

Definition 4.2.4. (Inverse sine)

y = sin1(x) if and only if sin(y) = x and 2

y 2

Definition 4.2.5. (Inverse cosine)

y = cos1(x) if and only if cos(y) = x and 0

y

Definition 4.2.6. (Inverse tangent)

y = tan1(x) if and only if tan(y) = x and 2

< y 0,d

dxxn = nxn1.

(General) For any real number r = 0,

d

dxxr = rxr1.

Example (See Example 3.2 (Smith & Minton,2012) pg 148)

5.2.2 General Derivative Rules

If f(x) and g(x) are differentiable at x and c is any constant, then

ddx

[f(x) g(x)] = f(x) g(x),

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FUNCTION DERIVATIVE

ddx

[cf(x)] = cf(x)

Example (See Examples (Smith & Minton,2012) pg 149)

Exercises (Pauls Note)

1 Differentiate each of the following functions:

a) f(x) = 15x100 3x12 + 5x 46

b) g(t) = 2t6 + 7t6

c) y = 8z3 13z5

+ z 23

2 Rewrite each of the following functions before computing the derivatives:

a) y = 3

x2(2x

x2)

b) h(t) =2t5 + t2 5

t2

5.2.3 The Product Rule

If the two functions f(x) and g(x) are differentiable (i.e the derivative exist) then

the product is also differentiable and,

d

dx(f(x)g(x)) = f(x)g(x) + f(x)g(x)



Differentiate each of the following functions using product rule:

a) y =3x2 (2x x2)

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5.2. DERIVATIVE RULES

b) f(x) = (6x3 x)(10 20x)

Solution:

a) y = 3

x2 (2x x2)Let f =

3

x2 and g = (2x x2).The problem now can be written as y = f g

Thus, using the product rule, we have:

y = fg + f g

=

d

dx3

x2

(2x x2) + 3

x2

d

dx(2x x2)

=

d

dxx

2

3

(2x x2) + 3

x2

d

dx(2x x2)

=

2

3x

2

31

(2x x2) + 3

x2 (2 2x21)

= 2

3

x1

3 (2x x2) +

3

x2 (2

2x)

=

2

3x

1

3

(2x x2) + x 23 (2 2x)

=4

3x

2

3 23

x5

3 + 2x2

3 2x 53

=

4

3+ 2

x

2

3

2

3+ 2

x

5

3

=10

3x

2

3 83

x5

3

b) f(x) = (6x3 x)(10 20x)

5.2.4 The Quotient Rule

If the two functions f(x) and g(x) are differentiable (i.e the derivative exist) then

the quotient is also differentiable and,

d

dxf(x)

g(x)

=f(x)g(x)

f(x)g(x)

g2(x)

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FUNCTION DERIVATIVE



a) W(z) =3z+ 9

2 z

b) h(x) =4

x

x2 2

c) f(x) =4

x6

d) y =w6

5

5.2.5 The Chain Rule

If g is differentiable at x and f is differentiable at g(x), then

d

dx[f(g(x))] = f (g(x)) g(x)

Example (See Example 5.1, 5.2 and 5.3 (Smith & Minton,2012) pg 162)


Differentiate f(x) =

5z 8 using the chain rule.

5.3 Derivative of Special Functions

5.3.1 Derivative of the Trigonometric Function

We can find the derivatives of trigonometric functions using the definition of deriva-

tive discussed in the early section and the following lemmas:

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5.3. DERIVATIVE OF SPECIAL FUNCTIONS

lim0 sin = 0

lim0 cos = 1

lim0sin

= 1

lim01 cos

= 0

From the lemmas stated above, we may prove that the following Theorems hold

for derivative of trigonometric functions:

ddx sin x = cos x ddx cos x = sin xd

dxtan x = sec2 x

d

dxcot x = csc2 x

d

dxsec x = sec x tan x

d

dxcsc x = csc x cot x

You may refer to (Smith & minton, 2012) page 170 for proofs of a few trigonometric

functions stated above.

5.3.2 Derivative of the Exponential and Logarithmic Func-tions

d

dxax = ax ln a, a > 0

d

dxex = ex

d

dxlogb x =

1

x ln bd

dx(ln x) =

1

x

5.3.3 Derivative of the Hyperbolic Function

d

dxsinh x = cosh x

d

dxcosh x = sinh x

d

dxtanh x = sech2x

d

dxcoth x = csch2x

d

dxsechx = sechxtanhx d

dxcschx = cschx coth x

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FUNCTION DERIVATIVE

Figure 23: Derivative Rules (Source: math.arizona.edu/ calc/Rules.pdf)

Exercises

1) Use chain rule to differentiate the following special functions problems

(www.math.ucdavis.edu/kouba):

a) y = sin(5x)

b) y = e5x2

+7x13

c) y = 2cotx

d) y = 3 tan

x

e) y = ln(17 x)

f) y = log(4 + cos x)

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Solutions:

a) y = sin(5x)

Let f(g(x)) = sin(g(x)) and g(x) = 5x

y = f(g(x))g(x) =df

dg

dg

gx

= cos(g(x))g(x)

= cos(5x)(5)

= 5 cos(5x)

b) y = e5x2+7x13

Let f(g(x)) = eg(x) and g(x) = 5x2 + 7x 13

y = f(g(x))g(x) =

df

dg

dg

gx

= eg(x)g(x)

= e5x2+7x13(10x + 7)

= (10x + 7)e5x2+7x13

c) y = 2cotx

Let f(g(x)) = 2

g(x)

and g(x) = cot x

y = f(g(x))g(x) =df

dg

dg

gx

= ln 22g(x)g(x)

= ln 22cotx(csc2x)= 2cot x(ln2)(csc2x)

d) y = 3 tan x

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FUNCTION DERIVATIVE

Let f(g(x)) = 3 tan(g(x)) and g(x) =

x

y = f(g(x))g(x) =df

dg

dg

gx

= 3 sec2(g(x))g(x)

= 3 sec2(

x)(x1

2 )

= 3 sec2(

x)(1

2x

1

2 1)

= 3 sec2(

x)(1

2x

1

2 )

= 3 sec2(x)(1

2

1

x1

2)

= 3 sec2(

x)(1

2

x

= (3

2

xsec2(

x)

e) y = ln(17 x)Let f(g(x)) = ln(g(x)) and g(x) = (17 x)

y = f(g(x))g(x) =df

dg

dg

gx

=1

g(x)g(x)

=1

(17 x)(

1)

= 1(17 x)

=1

x 17

f) y = log(4 + cos x)

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Let f(g(x)) = log(g(x)) and g(x) = 4 + cos x

y = f(g(x))g(x) =df

dg

dg

gx

=1

g(x)ln10g(x)

=1

(4 + cos x)ln10( sin(x))

= sin(x)(4 + cos x)ln10

2) Differentiate each of the following functions. You will need to apply

the chain rule more than once. (www.math.ucdavis.edu/kouba):

a) y = cos2(x3)

b) y =1

5 sec4(4 + x3)

c) y =

sin(7x + ln(5x))

d) y = 10(1 + (2 (6 + 7x4)9)3)5

e) y = 4 ln(ln(ln(sec x)))

f) y = tan3

cot(7x)

Solutions:

a) y = cos2(x3) = (cos(x3))2

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FUNCTION DERIVATIVE

Let f(g) = g2, g(h) = cos(h) and h(x) = x3

y =d

dxf(g((h(x))))

=df

dg

dg

dh

dh

dx

= 2g(x)( sin(h(x)))3x2

= 2(cos(x3))( sin(x3))3x2

= 6x2 cos(x3) sin(x3)

b) y =1

5sec4(4 + x3)

Let f(g) =

1

5g, g(h) = h4

, h(i) = sec(i), and i(x) = 4 + x3

y =d

dxf(g((h(i(x)))))

=df

dg

dg

dh

dh

di

di

dx

=1

5(4h)5 (sec(i) tan(i))3x2

=12x2

5(sec(4 + x3))5 sec(4 + x3) tan(4 + x3)

= 125

x2 sec4(4 + x3) tan(4 + x3)

c) y =

sin(7x + ln(5x))

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Let f(g) =

g = g1

2 , g(h) = sin(h) and h(x) = 7x + ln(5x)

y =d

dxf(g ((h(x))))

=df

dg

dg

dh

dh

dx

=1

2g

1

2 (cos(h))h(x)

Evaluate h(x) :dh

dx=

d

dx(7x) +

d

dx(ln(5x)) = 7 +

1

5x(5) =

7x + 1

x

Therefore

y =1

2g

1

2 (cos(h))7x + 1

x

=1

2

g(cos(h))

7x + 1

x

=1

2

sin(7x + ln(5x))(cos(7x + ln(5x)))

7x + 1

x

=(7x + 1) cos(7x + ln(5x))

2xsin(7x + ln(5x))

d) y = 10(1 + (2 (6 + 7x4)9)3)5

y = 37800x3(1 + (2 (6 + 7x4)9)3)4(2 (6 + 7x4)9)2(6 + 7x4)8

e) y = 4 ln(ln(ln(sec x)))

y =4tan x

ln(ln(sec x)) ln(sec x)

f) y = tan3

cot(7x)

y = 21csc2

(7x)(tan

2cot(7x))(sec

2cot(7x))

2

cot(7x)

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FUNCTION DERIVATIVE

5.4 Higher Order Derivatives

In the previous sections, we have learned how to differentiate various functions to

obtain the functions derivatives of order one only. We are able to find the sec-

ond derivatives of the functions by simply differentiate the first derivatives of the

functions. Then, the third derivatives can be obtained by differentiating the second

derivatives. Consider the following simple example to have a clear idea about higher

order derivatives:

Example

Find the first, second, third and fourth derivative of y = 5x6

Solution:

The function is y = 5x6

Differentiate once to get the first derivative: y = y(1) =dy

dx= 30x5

Differentiate the first derivative to get the second derivative: y = y(2) =d2y

dx2= 150x4

Differentiate the second derivative to get the third derivative: y = y(3) =d3y

dx3

= 600x3

Differentiate the third derivative to get the fourth derivative: y = y(4) =d4ydx4

= 1800x2

Exercises

1) Find the first four derivatives for each of the following (Pauls Note)

a) R(t) = 3t2 + 8t1

2 + et

b) y = cos x

c) f(y) = sin(3y) + e2y + ln(7y)

Solution:

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5.4. HIGHER ORDER DERIVATIVES

a) R(t) = 3t2 + 8t1

2 + et

R(t) = 6t + 4t1

2 + et

R(t) = 6 2t 32 + et

R(t) = 3t5

2 + et

R(t) =15

2t

7

2 + et

b) y = cos x

y = sin(x)y = cos(x)y = sin(x)

y = cos(x)

c) f(y) = sin(3y) + e2y

+ ln(7y)

f = 3 cos(3y) 2e2y + y1

f =

f =

f = 81 sin(3y) + 16e2y 6y4

2) Find the second derivative for each of the following functions (Pauls Note):

a) Q(t) = sec(5t)

b) e12w3

c) f(t) = ln(1 + t2)

Solution:

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FUNCTION DERIVATIVE

a) Q(t) = sec(5t)

Q(t) = 25sec(5t)tan2(5t) + 25 sec3(5t)

b) e12w3

g(w) = 12we12w3 + 36w4e12w3

c) f(t) = ln(1 + t2)

f(t) =2 2t2

(1 + t2)2

5.5 Implicit Derivative

Usually we have been asking to find the derivative, i.e y =df

dx when given an equa-

tion in the form ofy = f(x). However, sometimes we cannot write a function in the

form of y = f(x). In this case, implicit differentiation might be needed. Consider

the following example, discussed in (Smith & Minton):

Example (Smith & Minton) page 186

Find y(x) implicitly for y2 + 2exy = 6

Solution:

Differentiate both sides of the equations with respect to x:

d

dx

y(x)2 + 2exy(x)

=

d

dx(6)

2y(x)y(x) + 2d

dx(exyx) = 0

Solving ddx

(exy(x)) using chain rule:

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5.5. IMPLICIT DERIVATIVE

Let f(g) = eg and g(x) = xy(x)

d

dx(exy(x)) =

df

dg

dg

dx

= eg (xy(x) y(x)) Using product rule= exy(x) (xy(x) + y(x))

Substitute back this derivatives to the original problem,

2y(x)y(x) 2exy(x) (xy(x) + y(x)) = 0

Simplify the equation by dividing both sides by 2, we have

y(x)y(x) exy(x)

(xy(x) + y(x)) = 0

Differentiate once more

d

dx

y(x)y(x) exy(x) (xy(x) + y(x)) = 0

y(x)y(x) + y(x)y(x) ddxexy(x) (xy(x) + y(x)) = 0

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FUNCTION DERIVATIVE

Solved

dxexy(x) (xy(x) + y(x)):

d

dxexy(x) (xy(x) + y(x))

= exy(x)d

dx(xy(x) + y(x)) +

d

dxexy(x)

(xy(x) + y(x))

= exy(x) ((xy(x) + y(x)) + y(x)) exy(x) (xy(x) + y(x)) (xy(x) + y(x))

Then, by substituting this expression to the solution, now we have

y(x)y(x) + y(x)y(x) exy(x) ((xy(x) + y(x)) + y(x)) exy(x) (xy(x) + y(x)) (xy(x) + y(x))

= 0

Grouping all the terms involving y(x) on one side of the equation gives us:

yy (x) xexyy(x) = [y(x)]2 exy[y + xy(x)]2 + 2exyy(x)(y

xexy)y(x) =

[y(x)]2

exy[y + xy

(x)]2 + 2e

xyy

(x)

y(x) =[y(x)]2 exy[y + xy(x)]2 + 2exyy(x)

(y xexy)

Exercises

1) Do Exercise 2.8 (Smith & Minton) page 191, no 5-19.

2) Find y(x) for y = xx.

Solution:

We cant differentiate y = xx using the power rule for y = xa because our a is not

a constant in this case. Similarly we cannot use the formula for y = ax. Therefore,

we apply ln to both sides of the equation:

ln(y) = ln(xx)

= x ln(x)

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5.6. APPLICATIONS OF DERIVATIVES

Now, in order to find y, we have to use implicit differentiation. (Please continue

until you get y = xx(1 + ln x))

5.6 Applications of Derivatives

In this section we will go through a few examples on applications of derivatives. You

have been given a task to find another applications so you will have a broader view

about this topic.

Rate of Changes

Example: Pumping a Sphere Ballooon (Pauls Note)

Air is being pumped into a spherical balloon at a rate of 5cm3/min. Determine

the rate at which the radius of the balloon is increasing when the diameter of the

balloon is 20cm.

Solution:

Let V(t) represent the volume of the balloon and r(t) is the radius of the balloon.

Since the balloon is spherical, we have

V(t) = 43

[r(t)]3 (11)

The question ask us to finddr

dtwhen r =

20cm

2= 10cm. The given information

is, the rate of the volume isdV

dt= 5cm3/min. Therefore, we need to implicitly

differentiate Equation 11:

V(t) =d

dt

4

3

[r(t)]3

= 4[r(t)]2r(t)

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FUNCTION DERIVATIVE

Subtitute V(t) = 5cm3 min and r(t) = 10cm to the above equation, we have

5cm3min1 = 4(10cm)2r(t)

r(t) =5cm3min1

4(10cm)2

=5cm3min1

400cm2

=1

80cm/min

Maximum and minimum

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5.6. APPLICATIONS OF DERIVATIVES

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Chapter 6

Integration

6.1 Introduction to Integration

In the previous chapter, we have been asking to find the derivative of a given func-

tion. In this section we will work the other way around. When a function is given,

we need to find what is the function that we need to differentiate in order to obtain

the given function. In other words, we will try to find the anti-derivative of the

given function. You may refer to the following simple example to understand this

idea.

Example

Given that f(x) = 3x2. Find the anti-derivative of this function.

Solution:

We want to find F(x) such that F(x) = f(x). So, what is the function that we

need to differentiate in order to get 3x2.

Let F(x) = xn

ddx

xn = (n)xn1 = 3x2 (12)

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6.1. INTRODUCTION TO INTEGRATION

Therefore we have n = 3. Thus, the answer is F(x) = x3.

We may check this answer by differentiating this function.

F(x) =d

dxx3 = 3x2

Note that, the differentiation of any constant will lead to zero ( ddx

c = 0). There-

fore, we can write a more general answer as:

The antiderivative of 3x2 is

F(x) = x3 + c

where c is a constant.

Definition 6.1.1. Given a function, f(x), an anti-derivative off(x) is any function

F(x) such that

F(x) = f(x)

If F(x) is any anti-derivative of f(x) then the most general anti-derivative of f(x)

is called an indefinite integral and denoted as,

f(x)dx = F(x) + c, c is any constant (13)

In this definition, the

is called the integral symbol, f(x) is called the inte-

grand, x is called the integration variable and the c is called the constant of

integration.

Basically there are 2 types of integral, called the Indefinite Integral and the

Definite Integral. We will first discuss the indefinite integral because most of the

theory is applicable to the definite integral. However, we will also specifically discuss

the definite integral afterwards.

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INTEGRATION

6.2 Indefinite Integral

Properties of the Indefinite Integral

(Pauls Note)

1)

kf(x)dx = k

f(x)dx where k is any number. So we can factor multiplica-

tive constants out of indefinite integrals.

2) f(x) g(x)dx = f(x)dx

g(x)dx. In other words, the integral of a sum

of difference of functions is the sum or difference of the individual integrals.

This rule can be extended to as many functions as we need.

In the next section we will show how to evaluate the indefinite integral.

6.2.1 The Power Rule

The integral of a power of x can be evaluated as follows:

xndx =

xn+1

n + 1+ c n = 1

Example

Evaluate

4x7 + 2x + 1dx.

Solution:

4x7 + 2x + 1dx = 4

x7dx + 2

xdx +

x0dx

= 4x7+1

7 + 1+ 2

x1+1

1 + 1+

x0+1

0 + 1+ c = 4

x8

8+ 2

x2

2+

x1

1+ c

=x8

2+ x2 + x + c

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6.2. INDEFINITE INTEGRAL

Exercises

Evaluate each of the following indefinite integrals.

a)

3t5 4t2 + 5dt

b)

x7 + x7dx

c)

2x3

3

x7

+

2

8dx

d)

dx

e)

x(x2 + 2x + 1)dx

f)

t2dx

Solution:

a)

3t5 4t2 + 5dt

3t5 4t2 + 5dt = 3t6

6 4t

1

1

+ 5t + c

=t6

2+

4

t+ 5t + c

b)

x7 + x7dx

x7 + x7dx =

x8

8+

x6

6 + c

=x8

8

1

6x6

+ c

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6.2.2 The Trigonometric Functions

The indefinite integrals of the trigonometric functions are listed as follow:

sin xdx = cos x + c cos xdx = sin x + c

sec2 xdx = tan x + c sec x tan xdx = sec x + c

csc2 xdx = cot x + c csc x cot xdx = csc x + c

6.2.3 The Exponential and Logarithmic Functions

The following list shows the integrals that involve exponential and logarithm func-

tions:

1)

exdx = ex + c

2)

axdx =ax

ln a+ c

3) 1

x

dx = x1dx = ln |x

|+ c

4)f(x)

f(x)dx = ln |f(x)| + c in any interval in which f(x) = 0

Try This

Prove the no. (4) rule in the above formula using the chain rule of derivative.

Solution:

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INTEGRATION

Using chain rule, we may find the derivative of ln |f(x)|

d

dxln |f(x)| = d

dxln

(f(x))2

=1

(f(x))2d

dx((f(x))2)

1

2

=1

(f(x))21

2((f(x))2)

12 2f(x)f(x)

=1

(f(x))21

((f(x))2)1

2

f(x)f(x)

=1

(f(x))2

(f(x))2 f(x)f(x)

=f(x)

(f(x))2f(x)

=f(x)

(f(x))

It is now shown that derivative of ln |f(x)| is equal to f(x)

(f(x)). Therefore, the an-

tiderivative of

f(x)

(f(x)) is as stated in rule no. (4).

Exercises

1) Find the general antiderivative (Exercises 4.1 page 307 (Smith & Minton, 2012))

a)

(2 sin x + cos x)dx

b) (3cos x sin x)dx

c)

2sec x tan xdx

d)

5sec2 xdx

e)

(3ex 2)dx

f)

(4x 2ex)dx

g)

(2x1 + sin x)dx

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h)

2cos x

e2x

dx

i) ex

ex + 3

Solution:

a)

(2 sin x + cos x)dx

b)

(3cos x sin x)dx

(3cos x sin x)dx = 3 cos xdx sin xdx= 3(sin x) ( cos x) + c

= 3 sin x + cos x + c

c)

2sec x tan xdx

d)

5sec2 xdx

e)

(3ex 2)dx

f)

(4x 2ex)dx

(4x 2ex)dx = 4x

2

2 2ex + c

= 2x2 2ex + c

g)

(2x1 + sin x)dx

(2x1 + sin x)dx = 2

1

xdx + sin xdx

= 2 ln x cos(x) + c

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INTEGRATION

h)

2cos x

e2x

dx

2cos x

e2x

dx = 2

cos xdx (e2x) 12 dx= 2

cos xdx (ex)dx

= 2 sin x ex + c

i) ex

ex + 3dx

Let f(x) = ex

+ 3. Hence, f(x) = ex

. Therefore, we may rewrite the aboveintegrand in the form of

f(x)

f(x). It is known that

f(x)

f(x)dx = ln |f(x)| + c

In what follows,

exex + 3 dx = ln |e

x

+ 3| + c

2) Given the following information, determine the function f(x) (Pauls Note)

a) f(x) = 4x3 9 + 2 sin x + 7ex, f(0) = 15

b) f(x) = 15

x + 5x3 + 6, f(1) = 54

, f(4) = 404

Solution:

Recall that

f(x) =

f(x)dx

Since the values of the function is given at specific points, we are able to find the

constant of the integration.

a) f(x) = 4x3 9 + 2 sin x + 7ex, f(0) = 15

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6.3. DEFINITE INTEGRAL

Integrate the derivative:

4x3 9 + 2 sin x + 7exdx = 4x

4

4 9x + 2( cos x) + 7ex + c

= x4 2cos x + 7ex + c

Substitute the value at the given point f(0) = 15:

f(x) = x4 9x 2cos x + 7ex + c

f(0) = 04

9(0) 2 cos(0) + 7e(

0) + c = 15

0 0 2(1) + 7(1) + c = 155 + c = 15

c = 15 5c = 10

Therefore,

f(x) = x4 9x 2cos x + 7ex + 10

b) f(x) = 15

x + 5x3 + 6, f(1) = 54

, f(4) = 404

ans:

f(x) = 4x5

2 + 14x5 + 3x2 132

x 2

6.3 Definite Integral

Definition 6.3.1. Given a function f(x) that is continuous on the interval [a, b] we

divide the interval into n subintervals of equal width, x, and from each interval

choose a point, xi . Then, the definite integral of f(x) from a to b is

ba

f(x)dx = limn

ni=1

f(xi )x

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INTEGRATION

Example (Pauls Note)

Using the definition of the definite integral, compute the following

20

x2 + 1dx

Solution:

2

0

x2 + 1dx = limnn

i=1 f(xi )x

The interval [0, 2] is divided into n subinterval. Therefore, we may define the width

x as follows:

x =2 0

n=

2

n

Our subintervals are then

[0,2

n], [

2

n,

4

n], [

4

n,

6

n], , [ 2(i 1)

n,

2i

n], [2(n 1)

n, 2]

We may choose xi as the right endpoint, the left endpoint or the midpoint of the

ith subinterval. For simplicity, in this solution we choose xi as the right endpoint of

the ith subinterval:

xi =2in

The summation in the definition of the definite integral is then

ni=1

f(xi )x =n

i=1 f(2i

n)

2

n

=n

i=1((2i

n)2 + 1)

2

n

=

ni=1

8i

2

n3+ 2

n

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We can simplify the above summation as follows:

ni=1

8i2

n3+

2

n

=n

i=1

8i2

n3

+n

i=1

2

n

=8

n3n

i=1 i2 +

2

n

ni=1 1

=8

n3(12 + 22 + 32 + + n2) + 2

n(1 + 1 + 1 + + 1)

=8

n3

n(n + 1)(2n + 1)

6

+

2

nn(1)

=

4

n2(n + 1)(2n + 1)

3

+ 2

=

4(n + 1)(2n + 1)

3n2

+ 2

=14n2 + 12n + 4

3n2

Finally, we may compute the definite integral:

2

0

x2 + 1dx = limnni=1 f(x

i

)x = limn

14n2 + 12n + 4

3n2

= limn

14n2

3n2+

12n

3n2+

4

3n2

= limn

14

3+

4

n+

4

3n2

=14

3+ 0 + 0

=14

3

6.3.1 Properties of the Definite Integral

(Pauls Note)

1)ba

f(x)dx = ab

f(x)dx

2)aa

f(x)dx = 0

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INTEGRATION

3)

b

acf(x)dx = c

b

af(x)dx

4)ba

f(x) g(x)dx ba

f(x)dx ba

g(x)dx

5)ba

f(x)dx =ca

f(x)dx +bc

f(x)dx

6)ba

f(x)dx =ba

f(t)dt

7)ba

cdx = c(b a), c is any number.

8) Iff(x)

0 for a

x

b then ba

f(x)dx

0.

9) Iff(x) g(x) for a x b then ba

f(x)dx ba

g(x)dx

10) Ifm f(x) M for a x b then m(b a) ba

f(x)dx M(b a)

11) | ba

f(x)dx| ba

|f(x)|dx

Exercises

Using the given properties, answer the following questions (Pauls Note):

1) Given that the definite integral2

0x2 + 1dx =

14

3. Find

a)0

2x2 + 1dx

b)2

010x2 + 10dx

c)2

0t2 + 1dt

2)178

1784x5 + cos2 x x2 sin2x

x2 + 1dx

3) Given that10

6f(x)dx = 23 and

610 g(x)dx = 9, determine the value of

610

2f(x) 10g(x)dx

4) Given that 10

12f(x)dx = 6,

10

100f(x)dx = 2, and

5100

f(x)dx = 4 determine

the value of

125 f(x)dx

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INTEGRATION

a)

2

0(2x 3)dx

b)1

0(6e3x + 4)dx

c)

2

(2 sin x cos x)dx

2)

a)

y2 + y2dy

y2 + y2dy = y

3

3 + y

1

1 + c

=y3

3 1

y+ c

b)2

1y2 + y2dy

21 y

2

+ y2

dy =

y3

3 1

y |2

1

=

23

3 1

2

13

3 1

1

= (8

3 1

2) ( 1

3 1)

=16 3 2 + 6

6=

17

6

c) 2

1

y2 + y2dy

21

y2 + y2dy =y3

3 1

y|21

=

23

3 1

2

13

3 11

= (8

3 1

2) (1

3+ 1)

=16 3 + 2 6

6=

9

6=

3

2

This is a wrong answer!!! We cannot apply the fundamental theorem

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of calculus to this integral because f(y) is not continuous in interval

[1, 2] (f(y) at y = 0 is undefined) Refer to Figure 24

4 3 2 1 0 1 2 3 40

20

40

60

80

100

120

y

f(y)

f(y)=y2

+y(2)

Figure 24: f(y) = y2 + y2 is not a continuous function

Part II

Theorem 6.3.3. The Fundamental Theorem of Calculus, Part II

If f is continuous on [a, b] and F(x) =xa

f(t)dt, then F(x) = f(x), on [a, b].

Example (Smith & Minton, 2012 page 338)

For F(x) = x

1(t2

2t + 3)dt, compute F(x). Solution:

The integrand is f(t) = t2 2t +3 which is continuous on [1,b] where b (, ]).

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INTEGRATION

Using Theorem 6.3.2, the derivative is

F(x) = f(x) = x2 2x + 3.


If F(x) =x2

2cos tdt, compute F(x).

Solution:

In order to use Theorem 6.3.2, we have to rewrite the integral so the limit in the

integral match the theorem.

So, our first step is to changex2

2dt u

2dt.

Let u(x) = x2, then

F(x) =

u(x)2

cos tdt

Using the theorem, we have F(u(x)) = cos u(x)

Then, using the chain rule

F(x) = F(u)u(x) = cos u(x)d

dxx2 = cos u(x) (2x) = 2x cos x2

Exercises (Pauls note)

1)Differentiate each of the following

a) g(x) = x

4

e2t cos2(1

5t)dt

b)1x2

t4 + 1

t2 + 1dt

2) Prove that

d

dx

u(x)v(x)

f(t)dt = v(x)f(v(x)) + u(x)f(u(x))

(Hint: Rewrite the integral so the limit will involve a constant u(x)

v(x)f(t)dt =a

v(x)f(t)dt +

u(x)a

f(t)dt)

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6.4. INTEGRATION TECHNIQUES

6.4 Integration Techniques

Up to this stage we were only dealing with simple integral problems. Sometimes

we have to evaluate integral of a very complicated integrands. We need special

techniques in order to simplify such integral problems, thus solve them. In this

section, 2 techniques are presented: Integration by Substitution and Integration by

Parts.

6.4.1 Integration by Substitution

Substitution Rule:

For indefinite integral:

f(u(x))u(x)dx =

f(u)du

For definite integral:

ba

f(u(x))u(x)dx =

u(b)u(a)

f(u)du

The basic idea is to change the variable x to a new variable u so the integration

is much simpler.

Example (Pauls Note)

Evaluate x

1 4x2 dxSolution:

Let u = 1 4x2, so we have

du

dx= 8x

dx = 18x

du

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INTEGRATION

Now, we can write

x

1 4x2 dx = x

u

1

8xdu

= 1

u

1

8du

=

1

8

u

1

2 du

=

1

8

u

1

2

12

+ c

=1

8

2u 12 + c

=

1

4

u

1

2 + c

=

1

4

(1 4x2) 12 + c


Compute15

0te

t2

2 dt

Solution:

Let u(t) = t2

2, so we have

du

dt=

d

dt

t

2

2

= t

dt = 1

t du

Substitute u(t) and dt into the original integral,

150

tet2

2 dt = 152

2

022

teu

1t

du

=

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