Mathematics N5 Study Guide S1 hlayiso.com

STUDY GUIDE NATIONAL N DIPLOMA IN CHEMICAL, CIVIL, ELECTRICAL AND MECHANICAL ENGINEERING MATHEMATICS N5 SUBJECT CODE: 16030175 CENTRAL TECHNICAL COLLEGE 2019 Downloaded from hlayiso.com

TABLE OF CONTENTS Welcome Note 3 Mission and Vision Statement of Central Technical College 4 Student Support 5 Academic Adivisng Centre 6 Assessments 7 Course Objectives 8 Prescribed and recommended Books and Websites 9 Module 1: Limits an Continuity 10 Module 2: Differentiation 14 Module 3: Applications of Differentiation 23 Module 4: Integration Techniques 28 Module 5: Partial Fractions 38 Module 6: Applications of Definite Integrals 42 Module 7: Areas and Volumes 45 Module 8: The Secound Moment of Area 51 Module 9: The Moment 53 Module10: Differential Equations 56 Formula Sheet Maths N5 60 Copyright: In terms of the Copyright Act, no 98 of 1978, no part of this manual may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording or by any other information storage and retrieval system without permission in writing from Central Technical College. Downloaded from hlayiso.com Page 2 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

WELCOME NOTE Dear Student, Central Technical College Management and Staff take pride in welcoming you as a student of this institution We hope that you find both your time of studies to be an informative and exciting experience. This course is developed to prepare you for a future career, equipping you with the necessary competencies required in your chosen career field. We would like to encourage you to interact with other students and staff as you can build lasting friendships and future contacts. This study guide aims at assisting you in and giving you a better understanding all the content and information in order to grasp the subject. This study guide is not to be used in isolation of a recommended textbook and recommended reading and research. The study guide was developed with view to assisting you and giving you a better understanding all the content of your course. Wishing you all the best for your studies CTC MANAGEMENT AND STAFF Downloaded from hlayiso.com Page 3 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

CENTRAL TECHNICAL COLLEGE MISSION STATEMENT To provide quality education in Engineering and Business studies to enable each student to attain his/her potential career skill. In doing so the Institution seeks to prepare students for employment in the competitive Labour Force, as well as for future self-employment. Objective: To extend our recruitment drive to enlist learners who have the potential attitude to succeed. To engage in Staff Development programs to refine and improve didactics’ of its lecturers. To foster the inter-disciplinary and inter-institutional networks. To liaise with industry to keep abreast with relevant curriculum requirements. To institute and develop quality assurance and assessment standards to entrench Central Technical College as a reputable provider of quality learning. VISION STATEMENT Central Technical College seeks to provide relevant education and training to students who will be well equipped to succeed in their Engineering and Business careers. In order to facilitate this, the following objective has been identified by the institution: 1. Provision of a culture and atmosphere of learning and teaching 2. Provide academic teaching and training to equip learners’ employability in a global economy. 3. Forster a community of lifelong learners and productive citizens through a rigorous preparatory programme that meets industry needs. 4. Inculcate in learners habits and techniques for personal development and positive societal values. Downloaded from hlayiso.com Page 4 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

STUDENT SUPPORT The support systems which we at Central Technical College have put in place to support to our students, aims to have a significant influence on how well our students will achieve. Students Support will assist students to achieve academically, to make the correct programme choices, and be oriented into Further Education and Training opportunities at Central Technical College. Central Technical College envisages to provide holistic services to students that embraces the full range of any single student’s interaction with the College.  Selection and Placement into Pre-entry appropriate programmes Admission  Contract of enrolment  Orientation and Code of Conduct  Academic Support Performance monitoring and STUDENT On course feedback, Workshops, Study SUPPORT Academic Guides, Academic Advising programme Centre  Personal Support Life skills, counselling, Health and Wellness Exit  Higher Education  Self-employment Work  Employment readiness  In-service Training Students are encouraged to make use of the Student Support Services available on each of our campuses. Each campus has Lecturer consultation times where you will be able to have a consultation with your subject lecturer should you require additional support. Downloaded from hlayiso.com Page 5 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Educor has embarked on an exciting new initiative – Academic Advising Centre. This centre provides the vital additional student support in order to ensure that our students are given the optimum potential to manage their studies in order to achieve their dreams. The Academic Advising Centre (AAC) is your college’s way of providing quality student support. Academic Advising is a deliberate and collaborative partnership between the AAC, the Academic Adviser and the student. It is also a systematic process of student-adviser relationship intended to support students in developing to their maximum potential. Academic Advisers provide academic assistance and individualised attention to promote each student's success. However, successful Academic Advising requires equal commitment, dedication and engagement of all partners. Each partner has specific roles and responsibilities. The Academic Advising Centre’s Role The AAC’s key role is to nurture a campus community that promotes student success. The Adviser's role is to mentor the student to reach his/her full potential.  Schedule advising sessions with students.  Identifying challenges/gaps in a student’s learning experience.  Providing ways to overcome challenges/gaps.  Deliver a passion and desire for learning. Student’s Role The Student is an equal partner in the advising process. As a Student you are ultimately responsible for your educational choices and decisions. You are expected to:  Clarify personal values, abilities, interests, and goals for your academics and life.  Schedule regular appointments with your Adviser as required or when in need of assistance.  Be well prepared for advising sessions and have the appropriate resources or materials. It is useful to maintain your own Advising Portfolio including your educational plan and other details.  Build a list of questions and have them ready for the Adviser to ensure meaningful discussions.  Become knowledgeable and adhere to institutional policies, procedures, and requirements.  Be open and receptive to talking with your Adviser as all information is treated with strict confidence.  Create and maintain an interactive environment encouraging mutua l trust.  Refer all administrative queries to your Academic Manager. VERY IMPORTANT: AAC calls will be reflected as 0873590976 on your mobile phone. It is an outbound centre so you will not be able to dial in using that number. Contact details will be exchanged upon contact with your Adviser. Email info@academicadvisingcentre.com or visit our website www.academicadvisingcentre.com for more information. Downloaded from hlayiso.com Page 6 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

ASSESSMENTS FORMATIVE ASSESSMENTS: These are compulsory internal assessments which are compulsory and contribute towards your Term Mark. 1. NATIONAL TEST  DURATION OF TEST: 1½ HOURS  TOTAL MARKS: 50  WEIGHTING TOWARD TERM MARK: 30% 2. NATIONAL DP EXAMINATION  DURATION OF DP EXAM: 3 HOURS  TOTAL MARKS: 100  WEIGHTING TOWARD TERM MARK: 70% SUMMATIVE ASSESSMENTS: These are external assessments from the DHET. Students must be registered with the DHET in order to write these assessments. ENTRANCE TO SUMMATIVE EXAMINTION REQUIREMENT: 80% ATTENDANCE RATE AND MINIMUM TERM MARK OF 40%  DURATION OF NATIONAL EXAMINATION: 3 HOURS  TOTAL MARKS: 100  WEIGHTING OF FINAL MARKS:  TERM MARK: 40%  EXAMINATION MARK: 60% Downloaded from hlayiso.com Page 7 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

MATHEMATICS N5 LEARNING OUTCOMES On completion of this course the students should be able to: 1. Solve limits 2. Solve derivatives 3. Apply differentiation 4. Solve integrals 5. Solve by partial fractions 6. Solve definite integrals 7. Use integration to solve areas and volumes 8. Solve the second moment of area 9. Solve the moment of inertia 10. Solve differential equations. Downloaded from hlayiso.com Page 8 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

PRESCRIBED BOOK TITLE AUTHOR/S EDITION ISBN NO PUBLISHER Mathematics N5 MJJ van Rensburg 978 1 919780 84 9 Macmillan WEBSITES 1 https://www.khanacademy.org/math/differential-calculus 2 https://www.mathsisfun.com/calculus/integration-introduction.html 3 https://www.khanacademy.org/math/integral-calculus/volume-using-calculus-ic 4 https://www.whitman.edu/mathematics/calculus_online/chapter09.html 5 https://www.khanacademy.org/math/differential-equations MANDATORY COURSE REQUIREMENTS Students must meet all internal and external assessment requirements in order to pass this subject. A term mark of 40% and mandatory class attendance of 80% of all lectures per programme is required to gain entrance into the Summative Assessment. Downloaded from hlayiso.com Page 9 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

MODULE 1: LIMITS AND CONTINUITY LEARNING OUTCOMES On completion of this module the students should be able to: 0 ∞ 1. Apply L ’Hospital’s rule if the quotient is in one of the following forms: ; ; ∞ − ∞ 0 ∞ 2. Express the conditions for continuity 3. Determine the continuity or discontinuity of a given function at a given point. 1.1 Theorem of limits The following are the theorems used on limits 1. lim 𝑘(𝑓)𝑥 = 𝑘 lim 𝑓(𝑥) 𝑥→𝑎 𝑥→𝑎 2. lim [𝑓(𝑥) ± 𝑔(𝑥)] = lim 𝑓(𝑥) ± lim 𝑔(𝑥) 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 3. lim 𝑘 = 𝑘 𝑥→𝑎 4. lim[𝑓(𝑥). 𝑔(𝑥)] = lim 𝑓(𝑥) . lim 𝑔(𝑥) 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 lim 𝑓(𝑥) 𝑓(𝑥) 𝑥→𝑎 5. lim = 𝑤ℎ𝑒𝑟𝑒 lim 𝑔(𝑥) ≠ 0 𝑥→𝑎 𝑔(𝑥) lim 𝑔(𝑥) 𝑥→𝑎 𝑥→𝑎 𝑛 6. lim √𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥) 𝑥→𝑎 𝑥→𝑎 𝑐 7. lim =∞ 𝑥→0 𝑥 𝑛 𝑐 8. lim =0 𝑥→∞ 𝑥 𝑛 From theorems 2 to 6 we can deduce that it does not make any difference whether we:  First simplify and then take the limit (left-hand side of the theorem); or  First take the limit and then simplify. Example 4 lim (3𝑥 + ) 𝑥→2 𝑥 1 = 3 lim 𝑥 + 4 lim 𝑥→2 𝑥→2 𝑥 1 = 3(2) + 4 ( ) 2 =8 Downloaded from hlayiso.com Page 10 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

2.1 Limits of the indeterminate forms 0 ∞ ; ; ∞ − ∞; 0 × ∞ 0 ∞ 𝟎 1.2.1 Factorising or dividing limits of the form 𝟎 𝑥2 − 4 0 lim [𝑓𝑜𝑟𝑚 ] 𝑥→2 𝑥 − 2 0 (𝑥 + 2)(𝑥 − 2) = lim 𝑥→2 𝑥−2 = lim 𝑥 + 2 𝑥→2 =2+2 =4 ∞ 1.2.2 Dividing limits in the form ∞ To eliminate division by ∞ we must divide the numerator and denominator by the highest power 𝑥 in the denominator. From this we can deduce that this method can only be applied to algebraic expression. 𝑥 4 + 2𝑥 2 − 1 lim 𝑥→∞ 3𝑥 4 − 6 𝑥 4 2𝑥 2 1 4 + 𝑥4 − 𝑥4 = lim 𝑥 𝑥→∞ 3𝑥 4 6 − 4 𝑥4 𝑥 2 1 1+ 2− 4 = lim 𝑥 𝑥 𝑥→∞ 6 3− 4 𝑥 1+0−0 = lim 𝑥→∞ 3 − 0 1 = 3 1.3 L Hospital’s rule 𝒇(𝒙) 𝟎 𝑰𝒇 𝐥𝐢𝐦 = , 𝒕𝒉𝒆𝒏 𝒚𝒐𝒖 𝒄𝒂𝒏 𝒖𝒔𝒆 𝒙→𝒂 𝒈(𝒙) 𝟎 𝒇(𝒙) 𝒇/ (𝒙) 𝒇// (𝒙) 𝐥𝐢𝐦 = 𝐥𝐢𝐦 / = 𝐥𝐢𝐦 // = 𝒙→𝒂 𝒈(𝒙) 𝒙→𝒂 𝒈 (𝒙) 𝒙→𝒂 𝒈 (𝒙) Downloaded from hlayiso.com Page 11 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Example 3𝑥 2 + 2𝑥 − 1 0 lim = 𝑥→−1 4𝑥 2 − 4 0 6𝑥 + 2 −4 1 = lim = = 𝑥→−1 8𝑥 −8 2 1.4 Continuity Continuous means without interruption. If there is an interruption at a certain point in a graph, the function becomes discontinuous at that point. A function 𝑓 is defined as continuous at 𝑎 if the lim 𝑓(𝑥) = 𝑓(𝑎) 𝑥→𝑎 Therefore 𝑓 is continuous at 𝑎 if  𝑓 is defined at 𝑎  lim 𝑓(𝑥) exists 𝑥→𝑎  lim 𝑓(𝑥) = 𝑓(𝑎) 𝑥→𝑎 𝒈(𝒙) A function of the form is continuous for all cases except, h(x)=0. 𝒉(𝒙) 1. Example Determine whether 𝑓 with 𝑓(𝑥) = 3𝑥 2 − 4 at 𝑥 = 2 1. Solution 𝑓(𝑥) = 3𝑥 2 − 4 ∴ 𝑓(2) = 3(2)2 − 4 =8 lim(3𝑥 2 − 4) 𝑥→2 = 3(2)2 − 4 = 8 ∴ 𝑓 is continuous at 𝑥 = 2 2. Example 3𝑥 2 −4 Determine whether 𝑓(𝑥) = is a continuous function? 𝑥 2 +5𝑥+6 2. Solution Downloaded from hlayiso.com Page 12 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

𝐼𝑡 𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑤ℎ𝑒𝑛 ℎ(𝑥) = 0 𝑖. 𝑒 𝑥 2 + 5𝑥 + 6 = 0 (𝑥 + 3)(𝑥 + 2) = 0 ∴ 𝐼𝑡 𝑖𝑠 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 2 𝑎𝑛𝑑 𝑎𝑡 𝑥 = 3 TEST YOUR KNOWLEDGE 1. Determine the following limits 𝑡𝑎𝑛𝜃 a. 𝑙𝑖𝑚𝜃→0 𝜃 𝑥 b. 𝑙𝑖𝑚𝑥→0 2𝑥 𝑒 −1 c. 𝑙𝑖𝑚𝑥→𝜋 (sec 𝑥 − tan 𝑥) 2 cos 𝑥−1 d. lim 𝑥→0 𝑥 sin 𝑥 2. Determine whether the following functions are continuous or discontinuous at the specified point 𝑙𝑛 𝑥 a. 𝑦 = 𝑎𝑡 𝑥 = 1 𝑥 2 −1 𝑥 𝑛 −𝑎𝑛 b. 𝑓(𝑥) = 𝑎𝑡 𝑥 = 𝑎 𝑥−𝑎 3. Determine the values where the following function is discontinuous. 2𝑥−1 a. 𝑓(𝑥) = 𝑥 2 +2𝑥−15 Downloaded from hlayiso.com Page 13 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

MODULE 2: DIFFERENTIATION LEARNING OUTCOMES On completion of this module the students should be able to: Determine the differential coefficients (or derivatives) of: 1. Algebraic functions of the form 𝑓(𝑥) = 𝑎𝑥 𝑛 where n ∈ R ; 𝑎+𝑏𝑥 2. Algebraic functions of the form 𝑓(𝑥) = where a, b, c and d are constants 𝑐+𝑑𝑥 3. Sin x and Cos x 2.1 Differentiation from first principles The formula to calculate the gradient of a tangent to a curve or the differential coefficient of a function is given by: 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓 ′ (𝑥) = lim ℎ→0 ℎ The binomial theorem This is used when we are calculating the differentiation from first principles of an algebraic function with an exponent with a negative number or a fraction. 𝑛(𝑛 − 1) 𝑛−2 2 𝑛(𝑛 − 1)(𝑛 − 2) 𝑛−3 3 (𝑎 + 𝑏)𝑛 = 𝑎𝑛 + 𝑛𝑎𝑛−1 𝑏 + 𝑎 𝑏 + 𝑎 𝑏 +⋯ 2! 3! 2.1.1 Algebraic functions (𝑓(𝑥) = 𝑥 𝑛 , 𝑛 ∈ ℝ) Example 1 Differentiate the following from first principles 𝑓(𝑥) = 2√𝑥 3 Solution 1 3 𝑓(𝑥) = 2√𝑥 3 = 2𝑥 2 3 3 3 3 3 3 3 3 1 ( − 1) 3 ( − 1) ( − 2) 3 𝑓(𝑥 + ℎ) = 2(𝑥 + ℎ)2 = 2 (𝑥 2 + 𝑥 2 . ℎ + 2 2 𝑥 2−2 ℎ2 + 2 2 2 𝑥 2−3 ℎ3 + ⋯ ) 2 2! 3! 3 3 1 3 1 1 −3 3 = 2 (𝑥 2 + 𝑥 2 . ℎ + 𝑥 −2 ℎ2 − 𝑥 2ℎ + ⋯ ) 2 8 16 3 1 3 1 1 3 = 2𝑥 2 + 3𝑥 2 . ℎ + 𝑥 −2 ℎ2 − 𝑥 −2 ℎ3 + ⋯ 4 8 Downloaded from hlayiso.com Page 14 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

3 1 1 3 3 3 1 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = (2𝑥 2 + 3𝑥 2 . ℎ + 𝑥 −2 ℎ2 − 𝑥 −2 ℎ3 + ⋯ ) − 2𝑥 2 4 8 1 1 3 3 1 = 3𝑥 2 . ℎ + 𝑥 −2 ℎ2 − 𝑥 −2 ℎ3 + ⋯ 4 8 1 3 −1 2 1 −3 3 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 3𝑥 2 . ℎ + 4 𝑥 2 ℎ − 8 𝑥 2 ℎ + ⋯ = ℎ ℎ 1 3 1 1 3 ℎ (3𝑥 2 + 𝑥 −2 ℎ − 𝑥 −2 ℎ2 + ⋯ ) 4 8 = ℎ 1 3 1 1 3 = 3𝑥 2 + 𝑥 −2 ℎ − 𝑥 −2 ℎ2 4 8 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 1 3 1 1 3 1 ∴ 𝑓 ′ (𝑥) = lim = lim (3𝑥 2 + 𝑥 −2 ℎ − 𝑥 −2 ℎ2 ) = 3𝑥 2 ℎ→0 ℎ ℎ→0 4 8 Example 2 Differentiate the following from first principles 2+𝑥 𝑓(𝑥) = 3 − 2𝑥 Solution 2 2+𝑥 𝑓(𝑥) = 3 − 2𝑥 2 + (𝑥 + ℎ) ∴ 𝑓(𝑥 + ℎ) = 3 − 2(𝑥 + ℎ) 2+𝑥+ℎ = 3 − 2𝑥 − 2ℎ ∴ 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 2+𝑥+ℎ 2+𝑥 = − 3 − 2𝑥 − 2ℎ 3 − 2𝑥 2+𝑥+ℎ 3 − 2𝑥 2 + 𝑥 3 − 2𝑥 − 2ℎ = × − × 3 − 2𝑥 − 2ℎ 3 − 2𝑥 3 − 2𝑥 3 − 2𝑥 − 2ℎ 6 − 4𝑥 + 3𝑥 − 2𝑥 2 + 3ℎ − 2ℎ𝑥 − (6 − 4𝑥 − 4ℎ + 3𝑥 − 2𝑥 2 − 2ℎ𝑥) = (3 − 2𝑥)(3 − 2𝑥 − 2ℎ) 7ℎ = (3 − 2𝑥)(3 − 2𝑥 − 2ℎ) 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓′(𝑥) = lim ℎ→0 ℎ Downloaded from hlayiso.com Page 15 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

7ℎ = lim ℎ→0 ℎ(3 − 2𝑥)(3 − 2𝑥 − 2ℎ) 7 7 = lim = ℎ→0 (3 − 2𝑥)(3 − 2𝑥 − 2ℎ) (3 − 2𝑥)2 2.2 Differentiation techniques 2.2.1 Standard differential coefficients (and rules) Example 2 𝑓(𝑥) = 2𝑒 𝑥 + + 3 sin 𝑥 − ln 𝑥 2 𝑥 = 2𝑒 𝑥 + 2𝑥 −1 + 3 sin 𝑥 − 2 ln 𝑥 1 𝑓 ′ (𝑥) = 2𝑒 𝑥 + 2(−1𝑥 −2 ) + 3 cos 𝑥 − 2 ( ) 𝑥 2 2 = 2𝑒 𝑥 + 2 + 3 cos 𝑥 − 𝑥 𝑥 2.2.2 The chain rule If 𝑦 is a function of 𝑢 and 𝑢 is a function of 𝑥, then: 𝑑𝑦 𝑑𝑦 𝑑𝑢 = ∗ 𝑑𝑥 𝑑𝑢 𝑑𝑥 Example 𝑑𝑦 Find if 𝑦 = ln(sin √𝑒 2𝑥 ) 𝑑𝑥 Solution 𝑦 = ln (sin √𝑒 2𝑥 ) 𝑑𝑦 1 1 1 =( ) . (cos √𝑒 2𝑥 ) . ( (𝑒 2𝑥 )−2 ) . (2𝑒 2𝑥 ) 𝑑𝑥 sin √𝑒 2𝑥 2 2.2.3 Standard differential coefficients (continued) 𝑑 𝑑 (𝑓𝑥)𝑛 = 𝑛[𝑓(𝑥)]𝑛−1 . 𝑓(𝑥) 𝑑𝑥 𝑑𝑥 𝑑 𝑓(𝑥) 𝑑 𝑒 = 𝑒 𝑓(𝑥) 𝑓(𝑥) 𝑑𝑥 𝑑𝑥 Downloaded from hlayiso.com Page 16 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

𝑑 𝑓(𝑥) 𝑑 𝑎 = 𝑎 𝑓(𝑥) ln 𝑎. 𝑓(𝑥) 𝑑𝑥 𝑑𝑥 2.2.4 The product rule and the quotient rule Definition of the product rule If 𝑢 and 𝑣 are functions of 𝑥 and𝑦 = 𝑢. 𝑣, then: 𝑑𝑦 𝑑𝑣 𝑑𝑢 =𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥 Another notation is: If 𝑦 = 𝑓(𝑥). 𝑔(𝑥) 𝑑𝑦 Then = 𝑓(𝑥). 𝑔′ (𝑥) + 𝑔(𝑥). 𝑓 ′(𝑥) 𝑑𝑥 Example 𝑑𝑦 𝐹𝑖𝑛𝑑 𝑖𝑓 𝑦 = 𝑥 2 . sin 2𝑥 𝑑𝑥 Solution 𝑑𝑦 𝑑𝑣 𝑑𝑢 ∴ =𝑢 +𝑣 𝑙𝑒𝑡 𝑢 = sin 2𝑥 𝑎𝑛𝑑 𝑣 = 𝑥 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 = sin 2𝑥(2𝑥) + 𝑥 2 (2cos 2𝑥) = 2𝑥 sin 2𝑥 + 2𝑥 2 cos 2𝑥 Definition of the quotient rule 𝑢 If 𝑢 and 𝑣 are functions of 𝑥 and 𝑦 = , then: 𝑣 𝑑𝑢 𝑑𝑣 𝑑𝑦 𝑣 𝑑𝑥 − 𝑢 𝑑𝑥 = 𝑑𝑥 𝑣2 Example sin 2𝑥 𝑦= 𝑥2 𝑑𝑢 𝑑𝑣 𝑑𝑦 𝑣 𝑑𝑥 − 𝑢 𝑑𝑥 ∴ = 𝑑𝑥 𝑣2 𝑥 2 . (2cos 2𝑥) − sin 2𝑥. (2𝑥) = (𝑥 2 )2 2𝑥 (𝑥 cos 2𝑥 − sin 2𝑥) = 𝑥4 2 = 3 [𝑥 cos 2𝑥 − sin 2𝑥] 𝑥 Downloaded from hlayiso.com Page 17 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

NB: When the questions are about differentiation using the product, quotient and/or chain rule, they generally state that simplification is NOT required. Hence you are NOT supposed to simplify. 2.2.5 Implicit differentiation A function such as y  x  3x3  x  5 is called an explicit function as y is explicitly given in terms of x . 5 A function such as x  3x y  xy  5 y  3x  15 is called an implicit function as y is not given explicitly in 5 3 2 terms of x nor x in terms of y . An implicit function can be differentiated with respect to x as it stands. Example 1 𝑑𝑦 𝐹𝑖𝑛𝑑 𝑖𝑓 𝑥 2 + 𝑥𝑦 = 𝑦 2 𝑑𝑥 Solution 1 𝑑 2 𝑑 𝑑 2 ∴ 𝑥 + (𝑥. 𝑦) = 𝑦 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑 𝑑 𝑑𝑦 𝑑 ∴ 2𝑥 + 𝑥 (𝑦) + 𝑦 (𝑥) = 2𝑦 [𝑛𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 (𝑥. 𝑦) 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑠 𝑢𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑟𝑢𝑙𝑒] 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑦 𝑑𝑦 ∴ 2𝑥 + 𝑥 ( ) + 𝑦(1) = 2𝑦 𝑑𝑥 𝑑𝑥 𝑑𝑦 𝑑𝑦 ∴𝑥 − 2𝑦 = −2𝑥 − 𝑦 𝑑𝑥 𝑑𝑥 𝑑𝑦 ∴ (𝑥 − 2𝑦) = −2𝑥 − 𝑦 𝑑𝑥 −2𝑥 − 𝑦 = 𝑥 − 2𝑦 Example 2 Find the equation of the gradient of the curve 𝑥 2 + 2𝑥𝑦 − 𝑥 + 3𝑦 = 10 at the point where x  1 Solution 2 First we need to find the value of y when x  1 Putting x  1 we get 12 + 2(1)𝑦 − (1) + 3𝑦 = 10 which gives y = 2 Differentiating the function gives 𝑑𝑦 𝑑𝑦 2𝑥 + 2𝑥 𝑑𝑥 + 2𝑦 − 1 + 3 𝑑𝑥 = 0 𝑑𝑦 1 − 2𝑥 − 2𝑦 = 𝑑𝑥 −2𝑥 − 3 Since the point is (1; 2) Downloaded from hlayiso.com Page 18 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

𝑑𝑦 1 − 2(1) − 2(2) 3 = = 𝑑𝑥 −2(1) − 3 5 𝑑𝑦 3 𝑦 = 𝑚𝑥 + 𝑐 𝑠𝑖𝑛𝑐𝑒 𝑚 = 𝑑𝑥 = 5 𝑎𝑛𝑑 𝑥 = 1 𝑤ℎ𝑒𝑛 𝑦 = 2, 𝑤𝑒 𝑐𝑎𝑛 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑐 5 1 2 = 3 (1) + 𝑐 𝑐 = 3 5 1 ∴ 𝑦 = 3𝑥 + 3 2.2.6 Logarithmic differentiation Consider the following four expressions: 𝑥 𝑛 , 𝑒 𝑥 , 𝑎 𝑥 and𝑥 𝑥 . We have already done the first three types. 𝑥 𝑥 differs because the variable 𝑥 exits in the base and the exponent. We can easily change the expressions of the form 𝑥 𝑥 to a form that we can differentiate by using logarithms. Remember:  log(𝑎. 𝑏) = log 𝑎 + log 𝑏 𝑎  log = log 𝑎 − log 𝑏 𝑏  log 𝑎𝑚 = 𝑚 log 𝑎 Example 1 Differentiate 𝑦 = 𝑥 sin 𝑥 with respect to 𝑥 Solution 1 𝑦 = 𝑥 sin 𝑥 ∴ ln 𝑦 = ln 𝑥 sin 𝑥 [𝑡𝑎𝑘𝑒 𝑙𝑜𝑔𝑠 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠] = sin 𝑥 ln 𝑥 [log 𝑎𝑚 = 𝑚 log 𝑎 ] 𝑑 𝑑 ∴ ln 𝑦 = sin 𝑥 ln 𝑥 𝑑𝑥 𝑑𝑥 1 𝑑𝑦 1 ∴ = sin 𝑥 . + ln 𝑥 . cos 𝑥 [𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑟𝑢𝑙𝑒] 𝑦 𝑑𝑥 𝑥 𝑑𝑦 1 ∴ = 𝑦 [sin 𝑥 . + ln 𝑥 . cos 𝑥] 𝑑𝑥 𝑥 sin 𝑥 1 =𝑥 [sin 𝑥 . + ln 𝑥 . cos 𝑥] 𝑥 In logarithmic differentiation we simply change expressions that contain variables in the exponents to a form that we can differentiate. Downloaded from hlayiso.com Page 19 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Example 2 dy sin x Find given the function y  (i.e. tan x) dx cos x Solution 2 Taking logs gives ln y  ln sin x  ln cos x 1 dy cos x  sin x cos 2 x  sin2 x 1 Differentiate     y dx sin x cos x sin x cos x sin x cos x dy 1 1 sin x 1  y    sec2 x dx sin x cos x sin x cos x cos x cos x 2 d (tan x )  sec2 x The result should be known dx Example 3 dy x sin x Find given the function y  dx x  1 cos x Solution  x sin x  ln y   ln   lnx sin x   lnx  1 cos x   x  1 cos x   lnx   lnsin x   lnx  1  lncos x  Differentiating gives 1 dy 1 cos x 1 sin x     y dx x sin x x  1 cos x dy  1 cos x 1 sin x   y     dx  x sin x x  1 cos x  x sin x  1 cos x 1 sin x       x  1 cos x  x sin x x  1 cos x  2.2.7 Differentiating inverse trigonometry functions sin−1 𝑥 Represents the inverse sine function of 𝑥. It can also be written as 𝑎𝑟𝑐 sin 𝑥. Generally, we say that the natural trigonometric functions and inverse trigonometric functions “cancel” each other. Downloaded from hlayiso.com Page 20 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

For example  sin(sin−1 𝑥) = 𝑥  sin−1 (sin 𝑥) = 𝑥 Example 𝑦 = sin−1 𝑥 ∴ sin 𝑦 = sin sin−1 𝑥 sin 𝑦 = 𝑥 . . . 1 𝑑 𝑑 ∴ sin 𝑦 = 𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑦 ∴ cos 𝑦 =1 𝑑𝑥 𝑑𝑦 1 ∴ = . . .2 𝑑𝑥 cos 𝑦 𝑑𝑦 1 = 𝑑𝑥 cos 𝑦 1 = [cos 𝑦 = √1 − sin2 𝑦] √1 − sin2 𝑦 1 = [𝑓𝑟𝑜𝑚 1] √1 − 𝑥 2 Downloaded from hlayiso.com Page 21 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

TEST YOUR KNOWLEDGE dy 1. In the following find in terms of x and y dx (i) x 2  y 2  10 (ii) 2 x 2  2 y 2  3x  10  7 y (iii) x 2  y 2  3xy  6 (iv) 2 x 3  3xy 2  y3  0 2. Use logarithmic differentiation to differentiate the following: a. r  2 b. y  x x c. s  sin t t  sin t  t xe x  1   d . v  sin u u e. y  x e x  1 f. y  sin 2 x 1  cos x g. y  x  1 2 x  32 x  4 𝑑𝑦 3. Determine from first principles of the following equation 𝑑𝑥 𝑦 = tan 𝑒 2𝑥 −𝟐 𝑑𝑦 4. Given that 𝑦 = determine from first principles 𝒙−𝟐 𝑑𝑥 𝑑𝑦 5. Determine of the following equations 𝑑𝑥 a. 𝑦 = ln 𝑐𝑜𝑠𝑒𝑐2𝑥 b. 𝑦 = √1 + √ 𝑥 c. 𝑦 = 𝑎𝑟𝑐𝑐𝑜𝑠𝑙𝑛 sin 𝑥 d. 𝑦 = [𝑠𝑖𝑛(𝑥 2 )]𝑐𝑜𝑠2𝑥 3−2𝑥 e. 𝑦= ln √𝑥 f. 𝑦 = √ln(sin 𝑥) g. 𝑦 = 𝑒 −𝑥 . 102𝑥 𝑑𝑦 −1 6. Prove that if y =arc cosec x, then = 𝑑𝑥 𝑥√𝑥 2 −1 𝑑𝑦 7. Determine with the aid of logarithmic differentiation if: 𝑑𝑥 𝑦 = (cos 𝑥)ln 𝑥 8. Given the implicit function: 𝑥 3 − 𝑦 3 = 3𝑥𝑦 2 Downloaded from hlayiso.com Page 22 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

MODULE 3: APPLICATIONS OF DIFFERENTIATION LEARNING OUTCOMES On completion of this module the students should be able to: 1. Determine an estimated value of any root of a given function from a table and/or a sketch 𝑓(𝑥1 ) 2. Use the formulae 𝑒 = − [ ] and 𝑥2 = 𝑥1 + 𝑒 to determine a better estimation of the roots. 𝑓𝑥2 3. Solve problems involving optimisation. 4. Apply differentiation to problems involving rate of change. 5. Apply differentiation to problems involving the rate of change of two related variables with respect to time where it is not necessary to express any of the variables directly as a function of the time. 3.1 Estimating irrational roots 3.1.1 Use of tables We can use a table of values to determine where an irrational root lies. When there is a change of sign on the y- value (from + to – or vice versa), then there is a root between the corresponding x-values. Example 1 Use the table method (with x ranging from -3 to 3) to determine where the roots of the following equation lie: 𝑦 = 𝑥 3 − 𝑥 2 − 4𝑥 + 2 Solution 1 x -3 -2 -1 0 1 2 3 y -22 -2 4 2 -2 -2 8 Since there is a change of sign in y (-2 to 4), it means there is a root between x = -2 and x = -1 Also, there is a root between x = 0 and x = 1(as there is a sign change in y from 2 to -2) There is a 3rd root between x = 2 and x = 3. 3.1.2 Use of sketch When you sketch a graph, the point when the graph cuts the x- axis is where a root lies. Example 2 Sketch the graph of 𝑦 = 𝑥 3 − 𝑥 2 − 4𝑥 + 2 and estimate the roots of the equation: 𝑥 3 − 𝑥 2 = 4𝑥 − 2 Downloaded from hlayiso.com Page 23 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Solution 2 Since the graph cuts the x-axis very close to -2, first estimate can be ≈ -1,9. 2nd and 3rd estimates can be 0,5 and 2,1 respectively. 3.1.3 Taylor/Newton method A better approximation can be obtained by using the Taylor/Newton method 𝑓(𝑥𝑛 ) 𝑥𝑛+1 = 𝑥𝑛 − 𝑤ℎ𝑒𝑟𝑒 𝑛 = 1,2,3 … 𝑓 ′(𝑥𝑛 ) 3.2 Maxima and minima The following steps represent a possible procedure for solving this type of problem. 1. Allocate symbols to the variable involved 2. Determine a formula for the required variable 3. Use conditions in the problem to reduce the problem to two variables 4. Determine the interval of possible values for the variable from physical restrictions 5. Use the techniques of maximum and minimum values to solve the equation. Example Divide 80 into two parts such that the product of the one and the square of the other is a maximum. Solution Let one part be 𝑥. Downloaded from hlayiso.com Page 24 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Then the other part is 80 − 𝑥 Product = (80 − 𝑥)𝑥 2 ∴ 𝑃 = 80𝑥 2 − 𝑥 3 For a maximum or minimum product: 𝑑𝑃 =0 𝑑𝑥 𝑑 ∴ (80𝑥 2 − 𝑥 3 ) = 0 𝑑𝑥 ∴ 160𝑥 − 3𝑥 2 = 0 ∴𝑥=0 𝑜𝑟 160 − 3𝑥 = 0 ∴ 3𝑥 = 160 ∴ 𝑥 = 53.333 Test for maximum or minimum: 𝑑2𝑃 𝑑 2 = 160𝑥 − 3𝑥 2 𝑑𝑥 𝑑𝑥 ∴ 𝑓′′(𝑥) = 160 − 6𝑥 ∴ 𝑓′′(0) = 160 − 0 > 0 ∴ a minimum product And 𝑓 ′′ (53.333) = 160 − 6(53.333) < 0 ∴ a maximum product ∴ 𝑥 = 53.333 And 80 − 𝑥 = 80 − 53.333 = 26.667 3.3 Rate of change and related ratios If 𝑠 = displacement, 𝑣 = velocity, 𝑎 = acceleration and 𝑡 = time, then: 𝑑𝑠 𝑑𝑣 𝑑2 𝑠 = 𝑣 And 𝑑𝑡 = 𝑑𝑡 2 = 𝑎 𝑑𝑡 Example An object is projected vertically upwards (against gravity). Its displacement (𝑠) in meters during a time (𝑡) in seconds is given by: 𝑠 = 40𝑡 − 5𝑡 2 Calculate 1. The velocity after 2 s. 2. The acceleration after 2 s. 3. The maximum height Downloaded from hlayiso.com Page 25 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

4. The time taken to reach a maximum height. 5. The displacement after 6 s. 6. The distance travelled during 6 s. Solution 1. 𝑠 = 40𝑡 − 5𝑡 2 ∴ 𝑓(𝑡) = 40𝑡 − 5𝑡 2 ∴ 𝑓′(𝑡) = 40 − 10𝑡 ∴ 𝑓′(2) = 40 − 10(2) = 20 𝑚/𝑠 2. 𝑓 ′(𝑡) = 40 − 10𝑡 ∴ 𝑓 ′′(𝑡) = −10 ∴ 𝑓 ′′(2) = −10 𝑚/𝑠 3. For a maximum height 𝑓 ′(𝑡) = 0 ∴ 40 − 10𝑡 = 0 ∴ 𝑡 = 4𝑠 4. 𝑓(𝑡) = 40𝑡 − 5𝑡 2 ∴ 𝑓(4) = 40(4) − 5(4)2 = 160 − 80 = 80 𝑚 5. 𝑓(𝑡) = 40𝑡 − 5𝑡 2 ∴ 𝑓(6) = 40(6) − 5(6)2 = 240 − 180 = 60 𝑚 6. Distance = 80 + 20 = 100 𝑚 Downloaded from hlayiso.com Page 26 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

TEST YOUR KNOWLEDGE 1. The top a 2.6m pipe rest against a vertical wall. The lower end is pulled away at a rate of 0,2 m/s a. How fast is the top of the pipe descending when the lower end is 0.6 m from the wall? b. When will the top of the pipe descend at a rate 0.3m/s? c. When will the top and lower end move at same rate? d. At what rate is the area formed by the pipe, the wall and the ground changing if the bottom end is 1m from the wall? 2. A weight on the ground is attached to a 15 m length of rope. One end of the rope passes over a pulley 5m above weight. The other end of the rope is attached to a tractor’s tow bar 1m above the ground. How fast will weight rise if it is 2m above the ground and the tractor moves at 2 m/s? 3. The length of the two equal sides of a rectangle increases by 0.2 m/s while the length of the other two sides decrease at a constant rate to keep the figure a rectangle with a constant area 6 m² . Calculate the rate of change of the perimeter when the length of the increasing sides is 2.4m. 4. Let the volume of a cylinder be V, its radius be r and height be h. Assume that r and h vary with time. CALCULATE the rate at which the volume of the cylinder will be changing at an instant when the radius is 20 cm and increasing at a rate of 1 cm/s while the height is 15 cm and decreasing at a 0, 5 cm/s. Hint: 𝑉 = 𝜋𝑟 2 ℎ 5. The sum of the diameter and height of a cylinder is 6 cm. CALCULATE the dimensions of the cylinder to ensure maximum volume. Let the height be h and radius r. Downloaded from hlayiso.com Page 27 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

MODULE 4: INTEGRATION TECHNIQUES LEARNING OUTCOMES On completion of this module the students should be able to: 𝑓 ′(𝑥) 1. Use the rules ∫[𝑓(𝑥)]𝑛 𝑓 ′(𝑥)𝑑𝑥 and ∫ 𝑑𝑥 𝑓(𝑥) 2. Integrate fractions where the degree of the numerator is higher than the degree of the denominator 3. Integrate by using algebraic substitution 4. Integrate basic trigonometric functions 4.1 Basic integration and algebraic substitutions 4.1.1 Application  dx  x  C  sin xdx   cos x  C  a f ( x) dx  a f ( x) dx  C  cos xdx  sin x  C  u( x)  v( x)dx   u( x)dx   v( x)dx  C  tan xdx   ln cos x  C  ln sec x  C x n 1  x dx  n  1  C n 1  sec(ax)dx  a ln sec(ax)  tan(ax)  C  u dv  u v   v du  C  cot xdx   ln csc x  C  ln sin x  C dx 1  ax  b  a ln ax  b  C 1  sec axdx  a tan(ax)  C 2 ax  a dx  ln a  C x  sec( x) tan(x)dx  sec( x)  C e ax  e dx  C ax a  csc( x) cot(x)dx   csc( x)  C Downloaded from hlayiso.com Page 28 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Example ∫ √2𝑥 𝑑𝑥 1 = √2 ∫ 𝑥 2 𝑑𝑥 [𝑟𝑢𝑙𝑒 2] 1 𝑥 2+1 = √2 +𝐶 [𝑟𝑢𝑙𝑒 3] 1 +1 2 1 3 2 = 22 . 𝑥 2 . + 𝐶 3 3 (2𝑥)2 = +𝐶 3 Application of: [𝑓(𝑥)]𝑛+1 ∫[𝑓(𝑥)]𝑛 𝑓 ′(𝑥)𝑑𝑥 = +𝐶 𝑛+1 Example ∫ √2𝑥 + 4 𝑑𝑥 1 = ∫(2𝑥 + 4)4 𝑑𝑥 ...1 1 𝑛= , 𝑓(𝑥) = 2𝑥 + 4 𝑎𝑛𝑑 𝑓 ′(𝑥) = 2 2 If we sub the above into ∫[𝑓(𝑥)]𝑛 𝑓 ′(𝑥)𝑑𝑥 , we get: 1 ∫(2𝑥 + 4)4 . 2 𝑑𝑥 ...2 If …1 differs from the only by a constant, we can use the rule. In this case the constant is 2 If we compare …2 and …1, we see that …2 are multiplied by 2. So, we must allow for it by dividing by 2. 1 ∫(2𝑥 + 4)4 𝑑𝑥 1 1 = ∫(2𝑥 + 4)4 . 2 2 1 1 [2𝑥 + 4]2+1 = +𝐶 2 1 +1 2 3 1 [2𝑥 + 4]2 = +𝐶 2 3 2 Downloaded from hlayiso.com Page 29 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

1 2 3 = × [2𝑥 + 4]2 + 𝐶 2 3 1 3 = [2𝑥 + 4]2 + 𝐶 3 If 𝑛, 𝑓(𝑥), 𝑓 ′(𝑥), differ from the rule by more than a constant, we cannot use the rule. Application of: 𝑓 ′(𝑥) ∫ 𝑑𝑥 = ln 𝑓(𝑥) + 𝐶 𝑓(𝑥) 1 ∫ 𝑑𝑥 𝑓(𝑥) = 2𝑥 2𝑥 1 2 = ∫ 𝑑𝑥 ∴ 𝑓 ′(𝑥) = 2 2 2𝑥 1 = ln 2𝑥 + 𝐶 2 4.1.2 Algebraic substitution We can also integrate by using substitution, in this method we let 𝑢 = 𝑓(𝑥)and then change everything to 𝑢 , which will normally be an easier integral. Example ∫(2𝑥 2 − 4)5 𝑥. 𝑑𝑥 Let 𝑢 = 2𝑥 2 − 4 𝑑𝑢 𝑑𝑢 = ∫ 𝑢5 . 𝑥. Then = 4𝑥 4𝑥 𝑑𝑥 1 𝑑𝑢 = ∫ 𝑢5 𝑑𝑢 ∴ = 𝑑𝑥 4 4𝑥 1 𝑢6 = +𝐶 4 6 1 = (2𝑥 2 − 4)6 + 𝐶 24 4.1.3 Fractions where the degree of the numerator is higher than or equal to the degree of the denominator If the highest power in the numerator is greater than or equal to the highest power in the denominator, we must first divide. Example 𝑥+3 ∫ 𝑑𝑥 𝑥+1 Downloaded from hlayiso.com Page 30 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

1 𝑥+1 𝑥+3 𝑥+1 . 2 2 = ∫ (1 + ) 𝑑𝑥 𝑥+1 1 = ∫ 𝑑𝑥 + 2 ∫ 𝑑𝑥 𝑥+1 = 𝑥 + 2 ln(𝑥 + 1) + 𝐶 4.1.4 Integration as the inverse of differentiation It is easier to do some problems by using this method, rather than other methods or rules. Example ∫ 2𝑥 sin 2𝑥 2 𝑑𝑥 From ∫ 𝑓 ′(𝑥) sin 𝑓(𝑥)𝑑𝑥 = − cos 𝑓(𝑥) + 𝐶 It follows that: 𝑓(𝑥) = 2𝑥 2 ∴ 𝑓 ′(𝑥) = 4𝑥 1 ∴ ∫ 2𝑥 sin 2𝑥 2 𝑑𝑥 = ∫ 4𝑥 sin 2𝑥 2 . 𝑑𝑥 2 1 = − cos 2𝑥 2 + 𝐶 2 4.2 Trigonometric functions 4.2.1𝐬𝐢𝐧𝟐 𝒂𝒙; 𝐜𝐨𝐬 𝟐 𝒂𝒙 ;𝐭𝐚𝐧𝟐 𝒂𝒙 ; and 𝐜𝐨𝐭 𝟐 𝒂𝒙 We use the following identities: 1 1  sin2 𝑥 = − cos 2𝑥 2 2 1 1  cos 2 𝑥 = + cos 2𝑥 2 2  tan2 𝑥 = sec 2 𝑥 − 1  cot 2 𝑥 = cosec 2 𝑥 − 1 Downloaded from hlayiso.com Page 31 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Example 1. Find ∫ 𝑠𝑖𝑛2 𝑥𝑑𝑥 1 1 sin 2𝑥 𝑥 sin 2𝑥 ∫ 𝑠𝑖𝑛2 𝑥𝑑𝑥 = ∫(1 − cos 2𝑥) 𝑑𝑥 = (𝑥 − )+𝑐 = − +𝑐 2 2 2 2 4 2. Find ∫ tan2 2𝑥 𝑑𝑥 ∫ tan2 2𝑥 𝑑𝑥 = ∫(sec 2 2𝑥 − 1) 𝑑𝑥 = ∫ sec 2 2𝑥 𝑑𝑥 − ∫ 𝑑𝑥 tan 2𝑥 = −𝑥+𝐶 2 4.2.2 sin ax and cos bx We use the following identities 1  sin 𝐴 cos 𝐵 = [sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵)] 2 1  cos 𝐴 sin 𝐵 = [sin(𝐴 + 𝐵) − sin(𝐴 − 𝐵)] 2 1  cos 𝐴 cos 𝐵 = [cos(𝐴 + 𝐵) + cos(𝐴 − 𝐵)] 2 1  sin 𝐴 sin 𝐵 = [cos(𝐴 − 𝐵) − cos(𝐴 − 𝐵)] 2 Example Find  sin 4x cos 3x dx 1 ∫ sin 4𝑥 cos 3𝑥 𝑑𝑥 = ∫ sin(4𝑥 + 3𝑥) + sin(4𝑥 − 3𝑥) 𝑑𝑥 2 1 = ∫ sin(7𝑥) + sin(𝑥) 𝑑𝑥 2 1 cos 7𝑥 = ( + cos 𝑥) + 𝑐 2 7 cos 7𝑥 = + cos 𝑥 + 𝑐 14 Downloaded from hlayiso.com Page 32 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

4.2.3 𝐬𝐢𝐧𝒎 𝒙. 𝐜𝐨𝐬 𝒏 𝒙 Where m and n are uneven and ≤ 𝟓 Examples 2. Find  cos x dx 3 . ∫ 𝑐𝑜𝑠 3 𝑥𝑑𝑥 = ∫ 𝑐𝑜𝑠 2 𝑥. cos 𝑥 𝑑𝑥 = ∫(1 − 𝑠𝑖𝑛2 𝑥) cos 𝑥 𝑑𝑥 𝑙𝑒𝑡 𝑢 = sin 𝑥 𝑑𝑢 = cos 𝑥 𝑑𝑥 𝑢3 𝑠𝑖𝑛3 𝑥 = ∫(1 − 𝑢2 ) 𝑑𝑢 = 𝑢 − + 𝑐 = sin 𝑥 − +𝑐 3 3 4.2.4 sin and tan substitution We use the following substitutions: 𝑑𝑥 𝑎  For ∫ , 𝑥 = 𝑏 sin 𝛼 √𝑎 2 −𝑏2 𝑥 2 𝑑𝑥 𝑎  For ∫ , 𝑥 = tan 𝛼 √𝑎 2 +𝑏2 𝑥 2 𝑏 𝑎  For ∫ √𝑎2 − 𝑏 2 𝑥 2 𝑑𝑥 , 𝑏 sin 𝛼 4.3 Integration by parts Formula used: ∫ 𝑢𝑣′ = 𝑢𝑣 − ∫ 𝑢′𝑣 As we have u, to get u’ we differentiate u. As we have v’, to get v, we integrate v’. Priority List for u You choose u depending on this priority list: 1. ln 𝑥 2. 𝑥 𝑛 3. 𝑒 𝑥 /𝑎 𝑥 4. 𝑡𝑟𝑖𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 Example 1 Find  x ln xdx . Solution 1 1 𝑥2 ∫ 𝑥 ln 𝑥 𝑑𝑥 𝑙𝑒𝑡 𝑢 = ln 𝑥 𝑎𝑛𝑑 𝑣 ′ = 𝑥 ∴ 𝑢′ = 𝑎𝑛𝑑 𝑣 = ∫ 𝑢𝑣 ′ 𝑥 2 Downloaded from hlayiso.com Page 33 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

= 𝑢𝑣 − ∫ 𝑢′ 𝑣 𝑥2 1 𝑥2 = (ln 𝑥) () − ∫ ( ) ( ) 𝑑𝑥 2 𝑥 2 𝑥2 𝑥 = ln 𝑥 − ∫ 𝑑𝑥 2 2 𝑥2 𝑥2 = ln 𝑥 − + 𝑐 2 4 Example 2: Find  x sin x dx . Solution 2 ∫ 𝑥 sin 𝑥 𝑑𝑥 𝑙𝑒𝑡 𝑢 = 𝑥 𝑎𝑛𝑑 𝑣 ′ = sin 𝑥 ∴ 𝑢′ = 1 𝑎𝑛𝑑 𝑣 = − cos 𝑥 ∫ 𝑢𝑣 ′ = 𝑢𝑣 − ∫ 𝑢′ 𝑣 = (𝑥)(− cos 𝑥) − ∫(1)(− cos 𝑥) 𝑑𝑥 = −𝑥. cos 𝑥 + ∫ cos 𝑥 𝑑𝑥 = −𝑥𝑐𝑜𝑠 𝑥 + sin 𝑥 + 𝑐 Sometimes it is necessary to introduce a v’ when it is not there. That is let v’=1 Example 3 will show this. Example 3: Find  ln x dx . Solution 3 1 ∫ ln 𝑥 𝑑𝑥 = ∫ 1 ln 𝑥 𝑑𝑥 𝑙𝑒𝑡 𝑢 = ln 𝑥 𝑣 ′ = 1 ∴ 𝑢′ = 𝑎𝑛𝑑 𝑣 = 𝑥 𝑥 ∫ 𝑢𝑣 ′ = 𝑢𝑣 − ∫ 𝑢′ 𝑣 1 = (ln 𝑥)(𝑥) − ∫ ( ) (𝑥) 𝑑𝑥 𝑥 = −𝑥𝑙𝑛 𝑥 − ∫ 1 𝑑𝑥 = −𝑥. ln 𝑥 + 𝑥 + 𝑐 Sometimes, it is necessary to use integration by parts more than once. Example 4: Downloaded from hlayiso.com Page 34 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Find  x cos x dx 2 . Solution 4 ∫ 𝑥 2 cos 𝑥 𝑑𝑥 𝑙𝑒𝑡 𝑢 = 𝑥 2 𝑎𝑛𝑑 𝑣 ′ = cos 𝑥 ∴ 𝑢′ = 2𝑥 𝑎𝑛𝑑 𝑣 = sin 𝑥 ∫ 𝑢𝑣 ′ = 𝑢𝑣 − ∫ 𝑢′ 𝑣 = (𝑥 2 )(sin 𝑥) − ∫(2𝑥)(sin 𝑥) 𝑑𝑥 𝑙𝑒𝑡 𝑢 = 2𝑥 𝑎𝑛𝑑 𝑣 ′ = sin 𝑥 ∴ 𝑢′ = 2 𝑎𝑛𝑑 𝑣 = − cos 𝑥 = 𝑥 2 sin 𝑥 − [𝑢𝑣 − ∫ 𝑢′ 𝑣] [Integration by parts is applied again for the 2nd time] = 𝑥 2 sin 𝑥 − [(2𝑥)(− cos 𝑥) − ∫ 2. −cos 𝑥 𝑑𝑥] = 𝑥 2 sin 𝑥 − [(−2𝑥 cos 𝑥) + 2 ∫ cos 𝑥 𝑑𝑥] = 𝑥 2 sin 𝑥 − [(−2𝑥 cos 𝑥) + 2(sin 𝑥)] + 𝑐 = 𝑥 2 sin 𝑥 + 2𝑥 cos 𝑥 − 2 sin 𝑥) + 𝑐 When the two functions are an exponential function and a sin or cos function, the “I-method” Example 5 Find  e sin x dx x . Solution 5 𝐿𝑒𝑡 𝐼 = ∫ 𝑒 𝑥 sin 𝑥 𝑑𝑥 𝐼 = ∫ 𝑒 𝑥 sin 𝑥 𝑑𝑥 𝑢 = 𝑒 𝑥 𝑣 ′ = sin 𝑥 ∴ 𝑢′ = 𝑒 𝑥 ; 𝑣 = − cos 𝑥 = 𝑢𝑣 − ∫ 𝑢′ 𝑣 = (𝑒 𝑥 )(− cos 𝑥) − ∫(𝑒 𝑥 )(− cos 𝑥) 𝑑𝑥 = −𝑒 𝑥 cos 𝑥 + ∫(𝑒 𝑥 )(cos 𝑥) 𝑑𝑥 𝑙𝑒𝑡 𝑢 = 𝑒 𝑥 𝑎𝑛𝑑 𝑣 ′ = cos 𝑥 ∴ 𝑢′ = 𝑒 𝑥 𝑎𝑛𝑑 𝑣 = sin 𝑥 = −𝑒 𝑥 cos 𝑥 + [𝑢𝑣 − ∫ 𝑢′ 𝑣] = −𝑒 𝑥 cos 𝑥 + [(𝑒 𝑥 )(sin 𝑥) − ∫ 𝑒 𝑥 sin 𝑥 𝑑𝑥] = −𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 − ∫ 𝑒 𝑥 sin 𝑥 𝑑𝑥 = −𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 − 𝐼 2𝐼 = −𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 −𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 𝐼 = 2 𝑥 −𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 ∫ 𝑒 sin 𝑥 𝑑𝑥 = 2 Downloaded from hlayiso.com Page 35 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

TEST YOUR KNOWLEDGE 1.  3 x ln x dx = 2.  arcsin x dx =  ln x 3. dx = x3 4.  x arcsec x dx = 5.  e cos x dx = x  4 6.  0 cos 2 (2 x) dx =  8 7.  0 sin(5 x) cos(3x) dx =  tan x dx = 3 8.  sin x cos x dx = 3 9. 10. Determine the integrals in each of the following cases: Downloaded from hlayiso.com Page 36 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

10.1 x y 2x 3  6 10.2 y  cot 4 x 10.3 y  cos 3 x 10.4 1 y 4  16 x 2 11 Determine  cos 2 x. cos x.dx by using the following techniques: Integration by parts 11.1 11.2 u -substitution Downloaded from hlayiso.com Page 37 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

MODULE 5: PARTIAL FRACTIONS LEARNING OUTCOMES On completion of this module the students should be able to: Integrate integrals of the following forms: 𝑓(𝑥) 1. ∫ (𝑎𝑥±𝑏)(𝑐𝑥±𝑑) 𝑑𝑥 𝑓(𝑥) 2. ∫ (𝑎𝑥±𝑏)2 𝑑𝑥 𝑓(𝑥) 3. ∫ (𝑎𝑥±𝑏)3 𝑑𝑥 5.1 Introduction When we add fractions, they are reduced to a single fraction. For example 𝑎 𝑏 + 𝑥 𝑦 𝑎𝑦 + 𝑏𝑥 = 𝑥𝑦 When we break up a fraction into two or more fractions, we get partial fractions. It is very important to remember that if the highest power in the numerator is greater than or equal to the highest power in the denominator, we must first divide. 5.2 Different linear factors in the denominator Integrating Proper Rational Functions 𝑓(𝑥) 𝐴 𝐵 = + (𝑎𝑥 ± 𝑏)(𝑐𝑥 ± 𝑑) 𝑎𝑥 ± 𝑏 𝑐𝑥 ± 𝑑 Example 1  3x  17 Find dx . x  2x  3 2 Solution 1 3𝑥 − 17 3𝑥 − 17 𝐴 𝐵 = = + 𝑥 2 − 2𝑥 − 3 (𝑥 − 3)(𝑥 + 1) 𝑥 − 3 𝑥 + 1 3𝑥 − 17 𝐴 𝐵 ∴ = + (𝑥 − 3)(𝑥 + 1) 𝑥 − 3 𝑥 + 1 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑏𝑦 (𝑥 − 3)(𝑥 + 1) Downloaded from hlayiso.com Page 38 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

3𝑥 − 17 = 𝐴(𝑥 + 1) + 𝐵(𝑥 − 3) Let x = 3 3(3) − 17 = 𝐴(3 + 1) + 𝐵(3 − 3) −8 = 𝐴(4) ≫ 𝐴 = −2 Let x = -1 3(−1) − 17 = 𝐴(−1 + 1) + 𝐵(−1 − 3) −20 = 𝐵(−4) ≫𝐵=5 3𝑥 − 17 −2 5 ∴∫ 𝑑𝑥 = ∫ + 𝑑𝑥 (𝑥 − 3)(𝑥 + 1) 𝑥−3 𝑥+1 = −2 ln(𝑥 − 3) + 5 ln(𝑥 + 1) + 𝑐 Repeated Linear Factors 𝑓(𝑥) 𝐴 𝐵 𝐶 = + + (𝑎𝑥 ± 𝑏)3 𝑎𝑥 ± 𝑏 (𝑎𝑥 ± 𝑏)2 (𝑎𝑥 ± 𝑏)3 Example 2  3x  4 Find dx . x  4x  4 2 Solution 2 3𝑥 − 4 3𝑥 − 4 𝐴 𝐵 = = + 𝑥 2 − 4𝑥 + 4 (𝑥 − 2) 2 𝑥 − 2 (𝑥 − 2)2 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑏𝑦 (𝑥 − 2)2 3𝑥 − 4 = 𝐴(𝑥 − 2) + 𝐵 Let x = 2 3(2) − 4 = 𝐴(2 − 2) + 𝐵 ∴𝐵=2 Let x = 3 (or any other number) 3(3) − 4 = 𝐴(3 − 2) + 𝐵 ∴𝐵=2 5=𝐴+𝐵 ∴𝐴=3 3𝑥 − 4 3 2 ∫ 𝑑𝑥 = ∫ + 𝑑𝑥 (𝑥 − 2) 2 𝑥 − 2 (𝑥 − 2)2 3 =∫ + 2(𝑥 − 2)−2 𝑑𝑥 𝑥−2 2(𝑥 − 2)−1 = 3 ln(𝑥 − 2) + +𝑐 −1 2 = 3 ln(𝑥 − 2) − +𝑐 𝑥−2 Quadratic Factors 𝑓(𝑥) 𝐴𝑥 + 𝐵 𝐶 𝐷 = + + +⋯ (𝑎𝑥 2 + 𝑏𝑥 + 𝑐)(𝑑𝑥 ± 𝑒)𝑛 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 (𝑑𝑥 ± 𝑒)𝑛 (𝑑𝑥 ± 𝑒)𝑛−1 Example 3  7x2  x  2 Find dx ( x  1)( x 2  1) Downloaded from hlayiso.com Page 39 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

Solution 3 7𝑥 2 + 𝑥 + 2 𝐴𝑥 + 𝐵 𝐶 = + (𝑥 2 + 1)(𝑥 − 1) (𝑥 2 + 1) (𝑥 − 1) 𝑀𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡ℎ𝑟𝑜𝑢𝑔ℎ𝑜𝑢𝑡 𝑏𝑦 (𝑥 2 + 1)(𝑥 − 1) 7𝑥 2 + 𝑥 + 2 = (𝐴𝑥 + 𝐵)(𝑥 − 1) + 𝐶(𝑥 2 + 1) Let x = 1 7(1)2 + 1 + 2 = (𝐴(1) + 𝐵)(1 − 1) + 𝐶(12 + 1) ∴𝐶=5 Let x = 0 7(0)2 + 0 + 2 = (𝐴(0) + 𝐵)(0 − 1) + 𝐶(02 + 1) 𝑏𝑢𝑡 𝐶 = 5 ∴ 2 = −𝐵 + 5 ∴𝐵=3 Let x = 2 (or any other number) 7(2)2 + 2 + 2 = (𝐴(2) + 𝐵)(2 − 1) + 𝐶(22 + 1) 𝑏𝑢𝑡 𝐶 = 5 𝑎𝑛𝑑 𝐵 = 3 32 = 2𝐴 + 𝐵 + 5𝐶 ∴𝐴=2 7𝑥 2 + 𝑥 + 2 2𝑥 + 3 5 ∫ 2 𝑑𝑥 = ∫ 2 + 𝑑𝑥 (𝑥 + 1)(𝑥 − 1) (𝑥 + 1) (𝑥 − 1) 2𝑥 3 5 =∫ + + 𝑑𝑥 (𝑥 2 + 1) (𝑥 2 + 1) (𝑥 − 1) = ln(𝑥 2 + 1) + 3 tan−1 𝑥 + 5 ln(𝑥 − 1) + 𝑐 Downloaded from hlayiso.com Page 40 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

TEST YOUR KNOWLEDGE   x 2  3x  4 (1) dx x( x  2) 2  4x  2 (2) dx ( x  1)( x 2  1)  x6 (3) dx 2 x  2x  3x 2  x  1 (4) dx ( x  1)( x 2  4) 5. Determine  y.dx by resolving the integrand into partial fractions: x 1 y .dx x  10 x  25 2 Downloaded from hlayiso.com Page 41 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

MODULE 6: APPLICATIONS OF DEFINITE INTEGRALS LEARNING OUTCOMES On completion of this module the students should be able to: Evaluate the following definite integrals: 𝑏 1. ∫𝑎 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎) 𝑥 𝑢 2. ∫𝑥 2 𝑓(𝑥) 𝑑𝑥 = ∫𝑢 2 𝑔(𝑢)𝑑𝑢 1 1 ∞ 3. ∫0 𝑒 −𝑠𝑡 . 𝑓(𝑡)𝑑𝑡 6.1 Solving definite integrals 6.1.1 Introduction Definition of a definite integral 𝑏 ∫ 𝑔 (𝑥)𝑑𝑥 = 𝑔(𝑏) − 𝑔(𝑎) 𝑎 𝑎 is called the lower limit and 𝑏 the upper limit. When integrating with respect to 𝑥, 𝑎 will always be the limit on the left hand side of a graphical representation of 𝑔 ′(𝑥). We integrate with respect to 𝑦 from the bottom to the top, so 𝑎 will always be the lowest limit on the graph. Example 2 ∫ 𝑥 𝑑𝑥 1 𝑥2 = + 𝐶12 2 22 12 = + 𝐶 − ( + 𝐶) 2 2 1 =2− 2 = 1.5 Downloaded from hlayiso.com Page 42 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

6.2 Laplace transforms Definition of Laplace transform The Laplace transform 𝐿[𝑓(𝑡)] of a function 𝑓(𝑡) is: ∞ 𝐿[𝑓(𝑡)] = ∫ 𝑒 −𝑠𝑡 𝑓(𝑡)𝑑𝑡 = 𝐹(𝑠), whenever the integral exits 0 ∞ 𝐿(𝑡) = ∫ 𝑒 −𝑠𝑡 . 𝑡 𝑑𝑡 0 ∞ 𝐿(𝐴) = ∫ 𝑒 −𝑠𝑡 . 𝐴 𝑑𝑡 0 Example ∞ 𝐿(𝐴) = ∫ 𝑒 −𝑠𝑡 . (𝐴) 𝑑𝑡 0 𝑘 = lim 𝐴 ∫ 𝑒 −𝑠𝑡 𝑑𝑡 𝑘→∞ 0 𝑘 𝑒 −𝑠𝑡 = 𝐴 lim ( ) 𝑘→∞ −𝑠 0 𝑒 −𝑠𝑘 𝑒 −0 = 𝐴 lim ( − ) 𝑘→∞ −𝑠 −𝑠 −1 1 = 𝐴 lim ( 𝑠𝑘 + ) 𝑘→∞ 𝑠𝑒 𝑠 1 = 𝐴 [0 + ] 𝑠 𝐴 = 𝑠 5 ∴ 𝐿(5) = 𝑠 Downloaded from hlayiso.com Page 43 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

TEST YOUR KNOWLEDGE 11 1. 0 x dx 1 1 2.  3 dx 1 x 2 𝑥 3. ∫0 𝑑𝑥 (6+𝑥 2 )2 𝜋 4. ∫04 tan 𝑥 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 5. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑙𝑎𝑝𝑙𝑎𝑐𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑎)𝑓(𝑡) = 𝑡 𝑏) 𝑓(𝑡) = −6 3 𝑥 6. ∫1 1+𝑥 2 𝑑𝑥 3 4 1 7. Determine  1 x2. 5x 3 .dx Downloaded from hlayiso.com Page 44 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

MODULE 7: AREAS AND VOLUMES LEARNING OUTCOMES On completion of this module the students should be able to: Use integration to calculate 1. The area bounded by a curve and one of the reference axes. 2. The area bounded by two curves 3. The volume generated when an area bounded by a curve and a reference axis is rotated about that axis. 4. The volume generated when the area bounded by two curves is rotated about a reference axis. The reference strip is perpendicular to the reference axis. 7.1 Areas 7.1.1 The area bounded by a curve and a reference axis Area in the x-axis y ∆𝐴 = 𝑦∆𝑥 y = f (x) 𝑏 𝐴 = ∫ 𝑦𝑑𝑥 𝑎 0 a Δx b x Area in the y-axis y x = f (y) d ∆𝐴 = 𝑥∆𝑦 𝑏 Δy 𝐴 = ∫ 𝑥𝑑𝑦 𝑎 c x 0 Downloaded from hlayiso.com Page 45 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

7.1.2 The area bounded by two curves Area in the x-axis y ∆𝐴 = (𝑦1 − 𝑦2 )∆𝑥 y1 𝑏 y2 𝐴 = ∫ (𝑦1 − 𝑦2 )𝑑𝑥 𝑎 0 a Δx b x Area in the y-axis y d ∆𝐴 = (𝑥1 − 𝑥2 )∆𝑦 x2 x1 Δy 𝑏 𝐴 = ∫ (𝑥1 − 𝑥2 )𝑑𝑦 𝑎 c x 0 Example Calculate the area bounded by 𝑦 = 𝑥 + 2 and 𝑦 = 𝑥 2 Solution Points of intersection: 𝑥2 = 𝑥 + 2 ∴ 𝑥2 − 𝑥 − 2 = 0 ∴ (𝑥 − 2)(𝑥 + 1) = 0 ∴ 𝑥 = 2 𝑎𝑛𝑑 𝑥 = −1 𝑓(𝑥) = 𝑥 + 2 ∴ 𝑓(2) = 2 + 2 = 4 And 𝑓(−1) = −1 + 2 = 1 ∴ The points of intersection are (2; 4) and (−1; 1) Downloaded from hlayiso.com Page 46 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

𝑑 = (𝑦2 − 𝑦1 )𝑑𝑥 = [𝑥 + 2 − 𝑥 2 ]𝑑𝑥 2 ∴ 𝐴 = ∫ (𝑥 + 2 − 𝑥 2 ) −1 2 2 𝑥 𝑥3 = [ + 2𝑥 − ] 2 3 −1 22 23 −12 −13 = + 2(2) − − ( + 2(−1) − ) 2 3 2 3 8 1 1 = 2+4− − +2− 3 2 3 = 4.5 units 2 Downloaded from hlayiso.com Page 47 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

7.2 Volume 7.2.1 The volume of the area bounded by a curve and a reference axis Area rotating in the x-axis ∆𝑉 = 𝜋𝑦 2 ∆𝑥 𝑏 ∴ 𝑉 = 𝜋 ∫ 𝑦 2 𝑑𝑥 𝑎 Area rotating in the y-axis ∆𝑉 = 𝜋𝑥 2 ∆𝑦 𝑏 ∴ 𝑉 = 𝜋 ∫ 𝑥 2 𝑑𝑦 𝑎 Downloaded from hlayiso.com Page 48 of 60 CTC: STUDY GUIDE MATHEMATICS N5 VERSION 5 2019

7.2.1 The volume of the area bounded by two curves and a reference axisArea rotating in the x-axis ∆𝑉 = 𝜋(𝑦1 2 − 𝑦2 2 )∆𝑥 𝑏 ∴ 𝑉 = 𝜋 ∫ (𝑦1 2 − 𝑦2 2 )𝑑𝑥 𝑎 Area rotating in the y-axis ∆𝑉 = 𝜋(𝑥1 2 − 𝑥2 2 )∆𝑦 𝑏 ∴ 𝑉 = 𝜋 ∫ (𝑥1 2 − 𝑥2 2 )𝑑𝑦 𝑎 Example 𝑥 Determine the volume generated when the area bounded by 𝑦 = ln , 𝑦 = 0 and 𝑦 = 2 is rotated about the 3 𝑦-axis. Solution y 𝑥 𝑦 = ln 2 3 Δy x 0 𝑥 𝑦 = ln ∴ 𝑥 = 3𝑒 𝑦 3 ∆𝑉 = 𝜋𝑥 2 ∆𝑦 2 ∴ 𝑉 = 𝜋 ∫ 𝑥 2 𝑑𝑦 0 2 2 9 2 9 = 𝜋 ∫ (3𝑒 𝑦 )2 𝑑𝑦 = 𝜋 ∫ 9𝑒 2𝑦 𝑑𝑦 = [ 𝑒 2𝑦 ] = (54,598 − 1) = 241,192 0 0 2 0 2 Downloaded from hlayiso.com

TEST YOUR KNOWLEDGE 1. Calculate the area bound by a) 𝑦 = 𝑥 2 + 𝑥 − 6, 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠 𝑎𝑛𝑑 𝑥 = 4 b) 𝑥 = ln 𝑦 𝑡ℎ𝑒 𝑦 𝑎𝑥𝑖𝑠 𝑎𝑛𝑑 𝑦 = 4 c) 𝑦 = 1 − 𝑥 2 𝑎𝑛𝑑 𝑦 = 2𝑥 − 2 𝑟𝑜𝑡𝑎𝑡𝑒𝑑 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠 2. Calculate the volume generated when the area bound by a) 𝑦 = 2𝑥 + 1 , 𝑥 = 0 , 𝑦 = 0 𝑎𝑛𝑑 𝑥 = 3 𝑖𝑠 𝑟𝑜𝑡𝑎𝑡𝑒𝑑 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑥 − 𝑎𝑥𝑖𝑠 2 b) 𝑦 = −𝑥 + 3 𝑎𝑛𝑑 𝑦 = 𝑖𝑠 𝑟𝑜𝑡𝑎𝑡𝑒𝑑 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑦 𝑎𝑥𝑖𝑠 𝑥 3. Given the curves: x  y  49 and xy  4 2 2 a) Draw the curves in the ANSWER BOOK and show on the sketch the enclosed area and the representative strip, as well as the lower and upper limits. b) Calculate the magnitude of the bounded area in the first quadrant of QUESTION 5.2.1 above. Downloaded from hlayiso.com Page 50 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

MODULE 8: THE SECOUND MOMENT OF AREA LEARNING OUTCOMES On completion of this module the students should be able to: Calculate the second moment of area of: 1. A rectangular lamina with respect to a reference axis in the same plane as the lamina and parallel to one side of the lamina. (through the centroid or a side, or outside the lamina) 2. A circular lamina with respect to a reference axis perpendicular to the lamina and through its centre. 8.1 Introduction Definition of the second moment of area The second moment of area is the product of the area and the square of the perpendicular distance of its centroid from the axis of rotation. 8.2 Laminas A lamina is a thin plate where the thickness is negligible, that is we can ignore it 8.2.1 The second moment of area of a rectangular lamina a) Determine the second moment area of a rectangular lamina about an axis parallel to one side of the lamina y b x 0 Δx a x ∆𝐴 = 𝑏. ∆𝑥 ∆𝐼𝑦 = 𝑎𝑟𝑒𝑎 × (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2 = 𝑏∆𝑥. 𝑥 2 = 𝑏𝑥 2 ∆𝑥 𝑎 𝑏𝑥 3 𝑎 𝑎3 𝑏 𝐼𝑦 = ∫ 𝑏𝑥 2 𝑑𝑥 = [ ] = 0 3 0 3 Downloaded from hlayiso.com Page 51 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

b) Determine the second moment area of a rectangular lamina with respect to a reference axis parallel to one side of the lamina that bisects the lamina y b x 𝑎 0 Δx 𝑎 x − 2 2 ∆𝐴 = 𝑏. ∆𝑥 ∆𝐼𝑦 = 𝑎𝑟𝑒𝑎 × (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2 = 𝑏∆𝑥. 𝑥 2 = 𝑏𝑥 2 ∆𝑥 𝑎 𝑎 2 2 𝑏𝑥 3 2 𝑎3 𝑏 𝐼𝑦 = ∫ 𝑏𝑥 𝑑𝑥 = [ ] 𝑎= − 𝑎 3 − 12 2 2 c) Determine the second moment area of a uniform circular lamina with a radius R about an axis through its centre and perpendicular to the plane of the lamina. ∆𝐴 = 2𝜋𝑟∆𝑟 ∆𝐼𝑦 = 𝑎𝑟𝑒𝑎 × (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2 = 2𝜋𝑟∆𝑟. 𝑟 2 = 2𝜋𝑟 3 ∆𝑟 𝑅 𝜋𝑟 4 𝑅 𝜋𝑅4 𝐼𝑦 = ∫ 2𝜋𝑟 3 𝑑𝑟 = [ ] = 0 2 0 2 TEST YOUR KNOWLEDGE 1. Calculate the second moment of area of a washer with inside diameter 20mm and outside diameter 30mm if the washer rotates about and axis through its centre and perpendicular to the washer. 2. Calculate the second moment of area of a 6 cm by 4 cm rectangular lamina about an axis parallel to, and 2cm away from a 4cm side. The axis is outside the lamina. 3. Determine the second moment of mass of a rectangular lamina of mass m about an axis parallel to ONE side of the lamina Downloaded from hlayiso.com Page 52 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

MODULE 9: THE MOMENT OF INERTIA LEARNING OUTCOMES On completion of this module the students should be able to: Calculate the moment of inertia of: A uniform circular disc of mass 𝑚 with respect to a reference axis through the centre of the disc perpendicular to the plane of the disc. A uniform rectangular prism of mass, 𝑚 with respect to reference axis parallel to one side of the prism. The axis could be drawn through the centroid or outside the prism. 9.1 The relationship between the moment of inertia and energy Definition of the moment of inertia (second moment of mass) The moment of inertia is the product of the mass and the square of its rotation radius. 𝐼 = 𝑟 2 𝑚 , where 𝐼 = moment of inertia; 𝑚 = mass and 𝑟 = rotation radius 9.2 The parallel axis theorem of laminas 9.2.1 The second moment of area of a rectangular lamina a) Determine the moment of inertia of a rectangular lamina of mass, m, about an axis parallel to one side of the lamina y b x 0 Δx a x 𝑚 𝑚𝑎𝑠𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎 = 𝑎𝑏 𝑚 𝑚 ∆𝑚 = 𝑏. ∆𝑥 × = ∆𝑥 𝑎𝑏 𝑎 𝑚 𝑚𝑥 2 ∆𝐼𝑦 = 𝑚𝑎𝑠𝑠 × (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2 = ∆𝑥. 𝑥 2 = ∆𝑥 𝑎 𝑎 Downloaded from hlayiso.com Page 53 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

𝑚𝑥 2 𝑎 𝑚𝑥 3 𝑎 𝑚𝑎2 𝐼𝑦 = ∫ 𝑑𝑥 = [ ] = 0 𝑎 3𝑎 0 3 b) Determine the second moment area of a rectangular lamina with respect to a reference axis parallel to one side of the lamina that bisects the lamina y b x 𝑎 0 Δx 𝑎 x − 2 2 𝑚 𝑚𝑎𝑠𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎 = 𝑎𝑏 𝑚 𝑚 ∆𝑚 = 𝑏. ∆𝑥 × = ∆𝑥 𝑎𝑏 𝑎 𝑚 𝑚𝑥 2 ∆𝐼𝑦 = 𝑚𝑎𝑠𝑠 × (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2 = ∆𝑥. 𝑥 2 = ∆𝑥 𝑎 𝑎 𝑎 𝑎 2 2 𝑚𝑥 𝑚𝑥 3 2 𝑚𝑎2 𝐼𝑦 = ∫ 𝑑𝑥 = [ ] 𝑎= − 𝑎 𝑎 3𝑎 − 12 2 2 c) Determine the second moment area of a uniform circular lamina with a radius R about an axis through its centre and perpendicular to the plane of the lamina. 𝑚 𝑚𝑎𝑠𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎 = 𝜋𝑅2 𝑚 2𝑚𝑟∆𝑟 ∆𝑚 = 2𝜋𝑟∆𝑟 × 2 = 𝜋𝑅 𝑅2 2𝑚𝑟∆𝑟 2 2𝑚𝑟 3 ∆𝐼𝑦 = 𝑎𝑟𝑒𝑎 × (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2 = .𝑟 = ∆𝑟 𝑅2 𝑅2 2𝑚𝑟 3 𝑅 𝑚𝑟 4 𝑅 𝑚𝑅2 𝐼𝑦 = ∫ 𝑑𝑟 = [ ] = 0 𝑅2 2𝑅2 0 2 Downloaded from hlayiso.com Page 54 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

TEST YOUR KNOWLEDGE 1. Calculate the moment of inertia of a rectangular lamina of dimensions 12cm by 8 cm about an axis 2cm from a 8cm side outside the lamina and parallel to the 8cm side 2. Calculate the moment of inertia of a flywheel of radius 50cm and a thick ness of 10cm about an axis through its centre and perpendicular to the flywheel. The mass of the flywheel is 8 kg. Downloaded from hlayiso.com Page 55 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

MODULE 10: DIFFERENTIAL EQUATIONS LEARNING OUTCOMES On completion of this module the students should be able to: The general and particular solutions of: 1. First order differential equations by direct integration or by separating the variables 2. Second order differential equations of the form: 𝑑2𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑑𝑥 2 10.1 Introduction A differential equation is an equation involving derivatives or differentials. The order of a differential equation is determined by the order of the highest derivative. 𝑑𝑦  = 𝑎𝑥 [first order] 𝑑𝑥 𝑑2𝑦  = 𝑎𝑥 + 𝑏 [second order] 𝑑𝑥 2 𝑑3𝑦 𝑑2𝑦  + = 𝑓(𝑥) [third order] 𝑑𝑥 3 𝑑𝑥 2 A general solution has a constant of integration as part of the solution. In a particular solution, more information is given, which will enable the constant of integration to be calculated. 10.2 First Order Differential Equations a) Solve by direct integration Example 𝑑𝑦 Solve the differential equation = 3𝑥 2 − 4𝑥 − 2, 𝑦 = 4, 𝑥 = 2 𝑑𝑥 Solution 𝑑𝑦 = 3𝑥 2 − 4𝑥 − 2 𝑑𝑥 ∴ 𝑑𝑦 = 3𝑥 2 − 4𝑥 − 2 𝑑𝑥 ∴ ∫ 𝑑𝑦 = ∫ 3𝑥 2 − 4𝑥 − 2 𝑑𝑥 ∴ 𝑦 = 𝑥 3 − 2𝑥 2 − 2𝑥 + 𝑐 (general solution) But since we have more info (y = 4 when x = 2), we can find the particular solution. ∴ 4 = 23 − 2(2)2 − 2(2) + 𝑐 ≫𝑐=8 Downloaded from hlayiso.com Page 56 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

∴ 𝑦 = 𝑥 3 − 2𝑥 2 − 2𝑥 + 8 This is the particular solution b) Solve by separating the variables This is done by taking all x variables to one side that has dx and all y variables to another side with dy and then integrate both sides. Example Solve the following differential equations 𝑑𝑦 a. 𝑥 = 𝑦 + 𝑥𝑦 𝑑𝑥 b. 𝑦 ln 𝑥 ln 𝑦 𝑑𝑥 + 𝑥𝑑𝑦 = 0 𝑥 = 2, 𝑦 = 2 Solution 𝑑𝑦 a. 𝑥 = 𝑦 + 𝑥𝑦 𝑑𝑥 𝑑𝑦 𝑥 = 𝑦(1 + 𝑥) 𝑑𝑥 𝑑𝑦 1 + 𝑥 = 𝑑𝑥 𝑦 𝑥 𝑑𝑦 1 = ( + 1) 𝑑𝑥 𝑦 𝑥 𝑑𝑦 1 ∫ = ∫ ( + 1) 𝑑𝑥 𝑦 𝑥 ln 𝑦 = ln 𝑥 + 𝑥 + 𝑐 b. 𝑦 ln 𝑥 ln 𝑦 𝑑𝑥 + 𝑥𝑑𝑦 = 0 𝑥 = 2, 𝑦 = 2 𝑥𝑑𝑦 = −𝑦 ln 𝑥 ln 𝑦 𝑑𝑥 𝑑𝑦 ln 𝑥 =− 𝑑𝑥 𝑦 ln 𝑦 𝑥 𝑑𝑦 ln 𝑥 ∫ = ∫− 𝑑𝑥 𝑦 ln 𝑦 𝑥 2 𝑥 ln(ln 𝑦) = − + 𝑐 2 𝑏𝑢𝑡 𝑥 = 2 𝑤ℎ𝑒𝑛 𝑦 = 2 22 ln(ln 2) = − + 𝑐 ∴ 𝑐 = 1.633 2 2 𝑥 ln(ln 𝑦) = − + 1.633 2 10.3 Second order differential equations Second-order linear differential equation has the following basic equation (for N5): 𝑑2 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 𝑑𝑥 2 Downloaded from hlayiso.com Page 57 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

Example Solve the following: 𝑑2 𝑦 𝑑𝑦 = 3𝑥 2 − 2𝑥 + 1; = 2; 𝑦 = 1; 𝑥 = 0 𝑑𝑥 2 𝑑𝑥 Solution 𝑑2 𝑦 = 3𝑥 2 − 2𝑥 + 1 𝑑𝑥 2 Integrate both sides with respect to x 𝑑𝑦 = 𝑥3 − 𝑥2 + 𝑥 + 𝑐 𝑑𝑥 𝑑𝑦 But = 2; 𝑥 = 0; 𝑑𝑥 2 = 03 − 02 + 0 + 𝑐 ∴𝑐=2 𝑑𝑦 = 𝑥3 − 𝑥2 + 𝑥 + 2 𝑑𝑥 Again, integrate both sides with respect to x 𝑥4 𝑥3 𝑥2 𝑦= − + + 2𝑥 + 𝑑 4 3 2 𝐵𝑢𝑡 𝑦 = 1; 𝑥 = 0 04 03 02 1= − + + 2(0) + 𝑑 ∴𝑑=1 4 3 2 4 3 2 𝑥 𝑥 𝑥 𝑦= − + + 2𝑥 + 1 4 3 2 Downloaded from hlayiso.com Page 58 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

TEST YOUR KNOWLEDGE 1. Calculate the general solution of the following 𝑑𝑦 2𝑥 2 1.1 = 3 𝑑𝑥 5𝑦 𝑑2 𝑦 1.2 𝑑𝑥 2 = 𝑥2 + 4 2. Calculate the particular solution of 𝑑2 𝑦 𝑑𝑦 2.1 𝑑𝑥 2 = 𝑥 2 − 𝑥, 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡 𝑑𝑥 = 2 𝑖𝑓 𝑥 = 1 𝑎𝑛𝑑 𝑦 = 1 𝑖𝑓 𝑥 = 0 𝑑𝑦 2.2 𝑑𝑥 = 𝑥 2 − 2𝑥 − 4 𝑎𝑛𝑑 𝑦 = −4 𝑖𝑓 𝑥 = 2 3. Solve the differential equation: dy  e yx dx 1 d2y dy 4 Determine the particular solution of  . 2  5  18 x , given that  1, y  4 and x = 1. 2 dx dx Downloaded from hlayiso.com Page 59 of 60 CTC STUDY GUIDE MATHEMATICS N5 VERSION 3 2018

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