Engineering Mathematics 2_ Pocket Notes eBook

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Published by:
Ministry of Higher Education
Mathematics, Science & Computer Department

Politeknik Port Dickson
http://www.polipd.edu.my
©Politeknik Port Dickson 2021

Perpustakaan Negara Malaysia Cataloguing-in-Publications Data

Engineering Mathematics 2, Pocket Notes eBook/ JUNALIZA BINTI ISHAK,

SUHANA BINTI RAMLI, SYAFARIZAN BINTI NASRODDIN

Mode of access : Internet
eISBN 978-967-2246-89-3

1. Engineering Mathematics.
2. Government publications—Malaysia

3. Electronic books

Graphic Designer:
Mathematics, Science & Computer Department
Polytechnic Port Dickson Graphic Designer :
Mathematics, Science & Computer Department
Polytechnic Port Dickson



This Engineering Mathematics 2, Pocket Notes eBook is written by the experienced lecturers from
Mathematics, Science and Computer Department, Polytechnic Port Dickson. It is written based on
the following topics

Chapter 1 : Indices & Logarithm
Chapter 2 : Differentiation
Chapter 3 : Integration
However, this pocket notes eBook only touches on the basics for the following topics. The purpose
is to strengthen students' understanding of the basics of the topics studied before applying them
and to make it easier for students to make revision before facing quizzes, test and exam.
This pocket notes eBook is written based on the latest syllabus of the Malaysian Polytechnic for the
Engineering Mathematics 2 course.
By the way, we would like to thank all the lecturers involved for their contribution and efforts in
producing this great pocket notes eBook. Any views or suggestions for improvement are most
welcome. Thank You Very Much.

Suhana Binti Ramli
(i) Editor

▪Laws of Indices
▪Define Index
▪Laws of Logarithm
▪Define Logarithm
▪Simplify Logarithm
Expressions

RULES EXAMPLES

a) × = + 5 × 2 = 5+2 = 7

b) ÷ = − 5 ÷ 2 = 5−2 = 3

c) = 2 3 = 6

d) 2 = 2 2 2 2 = 22 2 = 4 2

e) − = 1 2−3 = 1 = 1
23 8

g) 0 = 1 20 = 1

f) = 5 23 3

= 25

There are 2 methods for finding
the value of an index

Example 1 Find the value of x for the equation 8 = 32.

Method 1 Change to the SAME BASE

By comparing base and index of the terms on the left and on the right

Simplify using law of indices 8 = 32 When the BASE is the SAME then
= 23 = 25 the INDEX can be equated

23 = 25
∴ 3 = 5

= 5

3

Example 1 Find the value of x for the equation 8 = 32.

Method 2 Place the base log 10 on the left and right
of the equation

8 = 32

108 = 1025

8 = 5 2

5 2 10 base logs can be
∴ = 8 calculated using a calculator

= 5
3

Example 2 Given 82 ∙ 4 = 16 −1. Find the value of .

Method 1 Change to the SAME
BASE

By comparing base and index of the terms on the left and on

the right 82 ∙ 4 = 16 −1

23 2 ∙ 22 = 24 −1 When the BASE is the
SAME then the INDEX
Simplify using law of indices 26 +2 = 24 −1
× = + can be equated
∴ 6 + 2 = 4 − 1

6 + 2 = 4 − 4

6 − 4 = −4 − 2

2 = −6

= −3

RULES EXAMPLES

1) = 1 3 3 = 1

2) 1 = 0 3 1 = 0

3) = 3 5 = 10 5
103

4) = + 3 2 ∙ 5 = 3 2 + 3 5
2
5) = −
3 5 = 3 2 − 3 5

6) = 325 = 5 3 2

7) = 3 325 = 25

There are 3 methods for finding
the value of logarithm

Example 1 Find the value of 28

Method 1 Change to the Method 2 Change to INDEX
SIMPLEST FORM FORM

= 2 8 changeto theform 28 = changefromlog formto
= 223 movethe power 8 = 2 index form
= 3 22 Law: = 1 23 = 2
=3 1 Changetothesame BASE

=3 ∴ 3 = WhentheBASEistheSAMEthenthe
INDEX can be equated

Example 1 Find the value of 28

Method 3 Change to the HOW TO PRESS THE CALCULATOR
BASE 10 OR

= 2 8 Rule #3

108 changeto thebase10
102
=

0.903 Calculate by using
= 0.301 calculator

=3

Example 2 Simplify each of the following

a) 2 + 2 2 + 1 c) 39 − 35 + 315

= 2 2 + 1 Rule #4 9 ∙ 15 Rule #4
= 2 2 2 + = 3 5 Rule #5

= 327

b) 525 − 55 = 333 Rule #6

25 Rule #5 = 3 33 Rule #1
= 5 5 =3 1

= 55 Rule #1

=1 =3

▪Constant Rule
▪Constant Multiple Rule
▪Power Function Rule
▪Sum & Difference Rule
▪Composite Function
▪Chain Rule
▪Product Rule
▪Quotient Rule

Basic Rules Of Differentiation

Constant Rule Constant Multiple Rule Power Function Rule

= = =

= = = −


Sum & Difference Rule =

= ± = −

= ′ ± ′

Basic Rules Of Differentiation

How to differentiate ?

Example 1 Example 2

Differentiate Differentiate
= 5 = 3 4

2. POWER is subtracted by 1 = 5 5−1 = 4 ∙ 3 4−1


= 5 4 = 12 3

Basic Rules Of Differentiation

CHAIN RULE PRODUCT RULE QUOTIENT RULE

= × = + = −


2

1. Find u & y 1. Find u & v 1. Find u& v
o u in the bracket & y the remain 2. Differentiate u & v 2. Differentiate u & v
3. Substitute into the formula 3. Substitute into the formula
2. Differentiate u & y 4. Factorize 4. Factorize
3. Substitute into the formula 5. Solve/Simplify 5. Solve/Simplify
4. Solve / simplify
5. Replace u

a) Must have POWER = 5 3 3

b) POWER must be numerator 2 = 5
= 4 = 2 −4

c) 0 = 1 = 2 − − 2
= 2 + 4 + 4
d) Expand = + 2 1 −
2 5 2 5
d) Separate = + 2 2 = 2 − 2 + 2
2 5 − 2 + 5 = 2 3 − 1 + 5 −2

= 2

Example 1

Differentiate = 7 3 − 2 + 5 + 1 = 3 + 2 2 ATTENTION!
∴ = 9 2 + 12 + 4 3 + 2 2
3
= 3 + 2 3 + 2
ATTENTION! = 9 2 + 6 + 6 + 4
= 9 2 + 12 + 4

2 = 1 2
3 3

3 7 3−1 2−1 + 1 5 1−1 + 0 2 9 2−1 + 1 12 1−1 + 0
= −2 3 =

2 2 = 18 + 12
3
= 21 − + 5

REMEMBER!
0 = 1

Example 2

Differentiate = 6 4 + 3 3 − 5 + 1 ATTENTION!

5 = 5

1

= 5 2

6 4 3 3 5 1 = 6 3 − 3 2 − 5 −21 + −1
= + − +

Simplify using law of indices 3 6 3−1 − 2 3 2−1 − 1 5 −21−1 + −1 −1−1
= −2
= − & 1 = −

5 −23
= 6 4−1 − 3 3−1 − 5 21−1 + −1 = 18 2 − 6 − 2 − −2

= 6 3 − 3 2 − 5 −12 + −1 = 18 2 − 6 − 1 51
2 3 − 2

Basic Rules Of Differentiation

Composite Function Example

= + Differentiate = 7 − 6 2 8

= + −1 ∙ ∙ −1 = 8 7 − 6 2 8−1 ∙ −12


= −96 7 − 6 2 7

Composite Function

1) Differentiate OUTSIDE(power value)
2) Differentiate INSIDE(value in bracket)

Rules Of Differentiation

Chain Rule Example

= × Differentiate = 7 − 6 2 8

1. Find u & y = 7 − 6 2 = 8
o uin the bracket & y the remain
= 8 7
2. Differentiate u & y = −12
3. Substitute into the formula
4. Solve/simplify
5. Replace u = ×

= 8 7 ∙ −12


= −96 7

= −96 7 − 6 2 7

Example CHAIN RULE

Differentiate = 3 5 4 2 changetothe = 4 5 − 2 −31 = 4 −13
− form = 5 − 2

= 4 5 − 2 −31 4 −34
= −2 3
COMPOSITE FUNCTION = −

1 4 5 − 2 −13−1 ∙ 2 − 2−1 = − 4 −43 ∙ −2
= −3 3

4 5 − 2 −34 ∙ −2 = 8 −34
= −3 3

= 8 5 − 2 −43 8 5 − 2 −43
3 = 3

8 8
= 33 5 − 2 4 = 33 5 − 2 4

Rules Of Differentiation

Product Rule Example 1 Differentiate = 5 2 + 8 6

= 5 = 2 + 8 6
= +
= 5 4 = 6 2 +8 5 2
1. Find u & v
2. Differentiate u & v
3. Substitute into the formula = 12 2 + 8 5
4. Factorize
5. Solve/Simplify
= +

2 + 8 6 ∙ 5 4 + 5 ∙ 12 2 + 8 5
=

= 5 4 2 + 8 6 + 12 5 2 + 8 5

= 4 2 + 8 5 5 2 + 8 + 12

= 4 2 + 8 5 10 + 40 + 12

= 4 2 + 8 5 22 + 40

Example 2 Differentiate = 2 + 3 2 − 3 2 5

1 = 2 − 3 2 5

= 2 + 3 2

= 1 3 −12 2 = 5 2 − 3 2 4 −6
= 2 2 +
2 + 3 −12
= −30 2 − 3 2 4


= +

2 − 3 2 5 ∙ 2 + 3 −12 + 1 4
= 2 − 3 2 5 ∙
2 + 3 2 −30 2 − 3 2
=
2 + 3 −12 − 30 2 + 3 1 2 − 3 2 4
2

= 2 − 3 2 4 2 + 3 −12 2 − 3 2 − 30 2 + 3

= 2 − 3 2 4 2 + 3 −12 2 − 3 2 − 60 2 − 90

= 2 − 3 2 4 2 + 3 −21 2 − 63 2 − 90

Rules Of Differentiation

Quotient Rule Example 1 Differentiate = 5
2 + 8 6

= 5 = 2 + 8 6

− = 5 4 =6 2 + 8 5 2

=
2 = 12 2 + 8 5

= −


2

1. Find u& v 2 + 8 6 ∙ 5 4 − 5 ∙ 12 2 + 8 5
2. Differentiate u & v =
3. Substitute into the formula ( 2 + 8 6)2
4. Factorize
5. Solve/Simplify 4 2 + 8 5 5 2 + 8 − 12
= 2 + 8 12

4 10 + 40 − 12 4 40 − 2
= 2 + 8 7 = 2 + 8 7

Example 2 Differentiate = 7 +2 2
4 3+1 3

= 7 + 2 2 = 4 3 + 1 3

= 12 3 +1 3 3 2
= 2 7 + 2 7

= 14 7 + 2 = 36 2 3 + 1 3

= −
2

4 3 + 1 3 ∙ 14 7 + 2 − 7 + 2 2 ∙ 36 2 3 + 1 3
=
(4 3 + 1 3)2

56 3 + 1 3 ∙ 7 + 2 − 36 2 3 + 1 3 7 + 2 2
= 16 3 + 1 6

4 3 + 1 2 7 + 2 14 3 + 1 − 9 2 7 + 2 7 + 2 14 − 18 2 − 49 3
= 16 3 + 1 6 = 4 3 + 1 4

4 3 + 1 2 7 + 2 14 3 + 14 − 63 3 − 18 2
= 16 3 + 1 6

Constant
Basic Algebraic Function
Addition & Subtraction
Composite Function
Substitution Method

Basic of Integration Function

▪ Integration is the inverse or reverse process of differentiation
▪ The symbol for integration is ‫… ׬‬
▪ There are no limits of integration in indefinite integral
▪ In indefinite integral, the constant c must be stated

Basic of Integration Function

CONSTANT FUNCTION ALGEBRAIC FUNCTION = න , ≠ −1


= න ′ ± න ′ = න , ≠ −1 +1
= + 1 +
ADDITION &SUBTRACTION 1+1
= 1 + 1 +
′( ) = න ±
= න , ≠ −1


+1
= + 1 +

Basic of Integration Function

HOW TO INTEGRATE ?

Formula

න
+1

= + 1 +

Basic of Integration Function

Constant Multiple Rule Power Function Rule

= න ≠ −1
= න ≠ −1
+1
1+1 = + 1 +
= 1 + 1 +

= න ≠ −1


+1
= + 1 +

Basic of Integration Function

Example 1 Example 2 Example 3

න 5 න 5 = න 3 5
5 1+1 5+1

= 1 + 1 + = 5 + 1 + 5+1
5 2 6 = 3 ∙ 5 + 1 +

= 2 + = 6 + 3 6
= 6 +

6
= 2 +

Basic of Integration Function

Addition & Subtraction Example 1
′ = න ±
න + 3
= න ′ ± න ′ 1+1

= 1 + 1 + 3 +
2

= 2 + 3 +

Example 2 Example 3

න 2 3 + 5 + 3 න 3 + 2 − 7
2 3+1 5 1+1
expand
= 3 + 1 + 1 + 1 + 3 +
2 4 5 2 = න 3 2 − 21 + 2 − 14

= 4 + 2 + 3 + = න 3 2 − 19 − 14
4 5 2
3 2+1 19 1+1
= 2 + 2 + 3 + = 2 + 1 − 1 + 1 − 14 +

3 3 19 2
= 3 − 2 − 14 +

= 3 − 19 2 − 14 +
2

Basic of Integration Function

Composite Function Substitution Method

න + , ≠ −1 ‫ ׬‬ × ′

+ −1 OR
= + +
‫׬‬ ′( ) ]
[

Basic of Integration Function

Composite Function Example 1

= න + න 7 − 6 8
7 − 6 8+1

+ +1 = −6 8 + 1 +
= + 1 + 7 − 6 9

= −54 +

1. POWERisadded by1
2. Denominator : ×

3. ‘+c’attheend

Example 2

7 changethe
න 4 1 − 4 form
= න 7 1 − 4 −41

7 1 − 4 −41+1
= +
−4 1
− 4 + 1

3

7 1 − 4 4
= −3 +

Basic of Integration Function

Substitution Method How To Integrate ?

= න × ′ 1. Substitute =
2. Differentiate = = ′
i. Inthebracket ➔‫ ׬‬5 2 2 + 5 3

ii. Denominator ➔‫׬‬ 2
1− 2 o Then make =

iii. Highest power➔ ‫ ׬‬ 3 43 ′

3. Substitute = and = into theoriginal

′

function
4. Solvetheintegral
5. Substitute = →forthefinalanswer

Example 1

න 7 − 6 8

= 7 − 6

1. Substitute =
= −6


= −6

2. Differentiate = = ′ = න 8 ∙
−6
o Then make =
′
1
3. Substitute = and = into the = −6 න 8

′ 1 8+1
= −6 ∙ 8 + 1 +
original function

4. Solve the integral

5. Substitute = → for the final answer 9
= −54 +

7 − 6 9
= −54 +

Example 2

4 change
න 2 + 1 5 the form 2
= 4 න 2 + 1 −5
= 4 න −5 ∙

= 2 + 1 = 2 න −5

−4
= 2 = 2 ∙ −4 +

1
= 2 = 2 2 ∙ +

1
= 2 2 + 1 2 ∙ +

Example 3 Substitution Method

= න3 1
න 3 4 + 5 4+5
2

changetheform
= 4 + 5
1
= න3 4+5 1
2 = 4

Composite Function = 4

1 1
4+5
2 = 3 න 2 ∙ 4

= න3 21+1

3 + 5 12+1 = 3 ∙ 1 +
4 2
= 1 1 + + 1
4 2
+ 1 1

1 = 8 2 ∙ +

=8 2 + =8 1
4+5 4+5
2 +

BIBLIOGRAPHY

Abd Wahid Raji, H. R. (1999). Matematik Asas Jilid 2. Skudai: Universiti Teknologi Malaysia.
Anton, H. a. (2002). Calculus. New York: John Wiley and Sons.
Bird, J. (2010). Engineering mathematics (6th Edition). UK: Newnes.
Larson, R. &. (2013). Calculus (10th Edition). USA: Cengage Learning.
Stroud, K. A. (2011). Advanced Engineering Mathematics (5th Edition). New York: Industrial Press Inc.
Stroud, K. A. (2013). Engineering mathematics (7th Edition). New York: Industrial Press Inc.

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