MMTH011/MAH101M:
MATHS MADE EASY
EXAM PRACTICE BOOK
PREPARED BY:
MR E. CHAUKE
It is illegal to photocopy any pages from this book without the
written permission of the copyright holder.
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
TABLE OF CONTENTS:
Compiled by Mr. E. Chauke:
OVERVIEW:…………….…………………………………………...……… 1





Algebra Refresher………………………………………………………………………………………………..……..
About the Author………………………………………………………………………………………….…………....
Acknowledgements…………………………………………………………………………………….…………......
Purpose of the book…………………………………….…………………………………………….……...……….
Massage to the students……………………………………………………………….…………………………….
5
8
9
10
10
CHAPTER 1: ………………………...………………..………..…..….. 11
ABSOLUTE VALUES/MODULUS.
CHAPTER 2: ……………………………………………………………… 17
THE RELTIONSHIP BETWEEN RADIANS AND DEGREES.
CHAPTER 3: ……………………………………………………………… 21
SOLVING LOGARITHMIC AND TRIGONOMETRIC EQUATIONS.
CHAPTER 4: …………………….……………………………………….. 34
FUNCTIONS.
CHAPTER 5: ………………………………………………………….….. 53
THE LIMIT OF A FUNCTIONS.
CHAPTER 6: ……………………………………….………………….…. 94
DERIVATIVES OF ORDINARY FUNCTIONS.
2|Page
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER 7: …………………………………………………...………. 120
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS.
CHAPTER 8: ……………………………………………………….…… 128
DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS.
CHAPTER 9: ……………………………………………………………. 140
IMPLICIT AND HIGHER ORDER DIFFERENTIATION.
CHAPTER 10: …………………………………………………………. 148
DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS.
CHAPTER 11…………………………………………………………… 153
L’HOPITAL’S RULE.
CHAPTER 12: …………………………………………………………. 159
ROLLE’S & MEAN THEOREM.
CHAPTER 13: ……………………………………………………….… 168
APPLICATION OF CALCULUS.
CHAPTER 14: ………………………………………………………… 178
INTEGRATION.
CHAPTER 15: ………………………………………………………… 244
THE AREAS BETWEEN THE CURVES.
3|Page
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER 16: …..………………………………………….…………. 249
THE ARC LENGTH.
SUMMERY AND THE FORMULAS: ….…………………………..
256
 FORMULAS.
 PROPERTIES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS.
 TRIG-INVERSE FUNCTIONS.
 DIFFEENTIATION TOOL-BOX.
 TOOL-BOX FOR INTEGRATION.
CHECK YOURSELF QUESTIONS: ……………………………….
TESTBANKS & EXAMS PAPAERS: ………………………….…
263
281
MEMORANDUMS: …………………………………………… 306-324
4|Page
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Trigonometry Tool-Box
Trigonometric ratios in right-angled triangle:
1. sin 𝑥 =
𝑦
𝑟
𝑥
2. cos 𝑥 = 𝑟
𝑦
3. tan 𝑥 = 𝑥
𝑟
4. cosec 𝑥 = 𝑦
𝑟
𝑦
𝑟
5. sec 𝑥 = 𝑥
𝜃
𝑥
6. cot 𝑥 = 𝑦
𝑥
Using special angles to get exact values:
𝜽
0°
30°
45°
60°
90°
Radians
0
𝜋⁄6
𝜋⁄4
𝜋⁄3
𝜋⁄2
𝐬𝐢𝐧 𝜽
0
1⁄2
1⁄√2
√3⁄2
1
𝐜𝐨𝐬 𝜽
1
√3⁄2
1⁄√2
1⁄2
0
𝐭𝐚𝐧 𝜽
0
1⁄√3
1
Measured in Radians
𝝅
𝟔
𝝅
𝟒
𝟐
√𝟐
𝟏
√𝟑
𝝅
𝟑
√3
−
𝝅
𝟒
𝟏
Pythagorean Identities:
1. sin2 𝜃 + cos2 𝜃 = 1
2. 1 + tan2 𝜃 = sec 2 𝜃
3. 1 + cot 2 𝜃 = cosec 2 𝜃
Double Angle Identities:
1. sin 2𝜃 = 2 sin 𝜃 cos 𝜃
2. cos 2𝜃 = cos2 𝜃 − sin2 𝜃
= 1 − 2sin2 𝜃
= 2cos2 𝜃 − 1
3. tan 2𝜃 =
Compound Angle Identities:
1.
2.
3.
4.
5.
sin(𝜃 + 𝛽) = sin 𝜃 cos 𝛽 + cos 𝜃 sin 𝛽
sin(𝜃 − 𝛽) = sin 𝜃 cos 𝛽 − cos 𝜃 sin 𝛽
cos(𝜃 + 𝛽) = cos 𝜃 cos 𝛽 − sin 𝜃 sin 𝛽
cos(𝜃 − 𝛽) = cos 𝜃 cos 𝛽 + sin 𝜃 sin 𝛽
tan 𝜃+tan 𝛽
tan(𝜃 + 𝛽) = 1−tan 𝜃 tan 𝛽
tan 𝜃−tan 𝛽
6. tan(𝜃 − 𝛽) = 1+tan 𝜃 tan 𝛽
𝟏
2 tan 𝜃
1−tan2 𝜃
Half Angle Identities:
1. sin2 𝜃 =
1−cos 2𝜃
2. cos 2 𝜃 =
1+cos 2𝜃
2
2
1−cos 2𝜃
3. tan2 𝜃 = 1+cos 2𝜃
𝜃
1−cos 𝜃
4. sin 2 = √
𝜃
2
1+cos 𝜃
5. cos 2 = √
𝜃
2
sin 𝜃
6. tan 2 = 1+cos 𝜃 =
5|Page
1−cos 𝜃
sin 𝜃
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Fundamental Identities / Negative Angles:
1.
2.
3.
4.
5.
6.
Co-functions Identities:
𝜋
1. sin ( 2 − 𝜃) = cos 𝜃
sin(−𝜃) = − sin 𝜃
cos(−𝜃) = cos 𝜃
tan(−𝜃) = − tan 𝜃
cosec(−𝜃) = − cosec 𝜃
sec(−𝜃) = sec 𝜃
cot(−𝜃) = − cot 𝜃
𝜋
2. sin ( 2 + 𝜃) = cos 𝜃
𝜋
3. cos (2 − 𝜃) = sin 𝜃
𝜋
4. cos (2 + 𝜃) = − sin 𝜃
𝜋
5. tan ( 2 − 𝜃) = cot 𝜃
𝜋
6. cot (2 − 𝜃) = tan 𝜃
𝜋
7. sec ( 2 − 𝜃) = cosec 𝜃
𝜋
8. cosec ( 2 − 𝜃) = sec 𝜃
Tangents and Cotangents Formulas:
1
sin 𝜃
1. cosec 𝜃 = sin 𝜃
cos 𝜃
2. sec 𝜃 = cos 𝜃
1. tan 𝜃 = cos 𝜃
2. cot 𝜃 =
Reciprocal Identities:
1
sin 𝜃
1
3. cot 𝜃 = tan 𝜃
Sum to product formulas:
1. sin 𝜃 + sin 𝛽 = 2 sin (
𝜃+𝛽
2
𝜃−𝛽
) cos (
𝜃+𝛽
2. sin 𝜃 − sin 𝛽 = 2 cos (
2
Product to sum formulas:
2
𝜃−𝛽
) sin (
𝜃+𝛽
3. cos 𝜃 + cos 𝛽 = 2 cos (
2
2
1. sin 𝜃 cos 𝛽 = 2 [ sin(𝜃 + 𝛽) + sin(𝜃 − 𝛽) ]
)
2. cos 𝜃 sin 𝛽 = 2 [ sin(𝜃 + 𝛽) − sin(𝜃 − 𝛽) ]
2
2
1
)
𝜃−𝛽
) sin (
1
)
𝜃−𝛽
) cos (
𝜃+𝛽
4. cos 𝜃 − cos 𝛽 = −2 sin (
2
1
3. cos 𝜃 cos 𝛽 = 2 [ cos(𝜃 + 𝛽) + cos(𝜃 − 𝛽) ]
1
)
4. sin 𝜃 sin 𝛽 = 2 [ cos(𝜃 − 𝛽) − cos(𝜃 + 𝛽) ]
6|Page
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Periodic Formulas:
If 𝒏 is an integer.
1.
2.
3.
4.
5.
6.
Trigonometric Quadrants:
sin(𝜃 + 2𝜋𝑛) = sin 𝜃
cos(𝜃 + 2𝜋𝑛) = cos 𝜃
tan(𝜃 + 2𝜋𝑛) = tan 𝜃
cosec(𝜃 + 2𝜋𝑛) = cosec 𝜃
sec(𝜃 + 2𝜋𝑛) = sec 𝜃
cot(𝜃 + 2𝜋𝑛) = cot 𝜃
Quadrant ll:
S
A
T
C
Quadrant lll:
1. sin(𝜋 − 𝜃) = + sin 𝜃
2. cos(𝜋 − 𝜃) = − cos 𝜃
3. tan(𝜋 − 𝜃) = − tan 𝜃
Quadrant lV:
1. sin(𝜋 + 𝜃) = − sin 𝜃
2. cos(𝜋 + 𝜃) = − cos 𝜃
3. tan(𝜋 + 𝜃) = + tan 𝜃
Negative Angles:
1. sin(2𝜋 − 𝜃) = − sin 𝜃
2. cos(2𝜋 − 𝜃) = + cos 𝜃
3. tan(2𝜋 − 𝜃) = − tan 𝜃
Area rule:
1.
2.
3.
4.
5.
6.
sin(−𝜋 − 𝜃) = + sin 𝜃
sin(−𝜋 + 𝜃) = − sin 𝜃
cos(−𝜋 − 𝜃) = − cos 𝜃
cos(−𝜋 + 𝜃) = − cos 𝜃
tan(−𝜋 − 𝜃) = − tan 𝜃
tan(−𝜋 + 𝜃) = + tan 𝜃
Sine Rule:
1
1. ∆𝐴𝐵𝐶 = 2 𝑎𝑏 sin 𝐶̂
1
2. ∆𝐴𝐵𝐶 = 2 𝑏𝑐 sin 𝐴̂
1.
2.
𝑎
sin 𝐴̂
sin 𝐴̂
𝑎
𝑏
𝑐
= sin 𝐵̂ = sin 𝐶̂
=
sin 𝐵̂
𝑏
=
sin 𝐶̂
𝑐
1
3. ∆𝐴𝐵𝐶 = 2 𝑎𝑐 sin 𝐵̂
7|Page
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
ABOUT THE AUTHOR:
CHAUKE EMMANUEL
EMMANUEL CHAUKE is a young Post Graduate in BSc (Mathematical Science) who is
passionate about education, more especially in Maths, Statistics and Applied Maths. He has
been giving extra lessons in both Calculus and Algebra for the past 3 years. He is a very
informative person who helps students who love math and logic, but struggle to have real
world, concise explanations of these subjects.
As an extra lesson Tutor, Emmanuel has established the critical areas where students struggle
and has written the Comprehensive guide for Introductory Algebra with these areas in mind.
The explanation are very clear to comprehend and it makes one to love the subjects at hand.
The books are written in a clear, simple, visual and logical manner. The colour coding
facilitates explanations, definition, formulas, and recaps of previous work, hints and ideas.
They are easy to read, easy to understand and easy to apply what has been learnt. They work
in conjunction with all other Calculus Books.
Emmanuel’s objective is for the Maths Handbook and Study Guides to demystify Maths and
help students to reach their potential in this challenging module/subject. The subtitles of the
books are ‘Maths made Easy’ and this is what he aims to do. Emmanuel ensures that his work
is up to date at all times and that it is suitable for any Calculus Courses and National
Curriculum students.
8|Page
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
ACKNOWLEDGEMENTS:
Words have no power to express and convery my heart full thanks to the Almighty God, the
one whom nothing before is impossible for giving me strength in fulfilling this work.
It gives me great pleasure to express my deep sense of gratitude and respect to my beloved Mentor,
Mr. Chauke T.E for infusing confidence and a sense of excitement and inspiring me in my work during
the course of study, through his constant encouragement and guidance. My sincere and heartful
thanks to him for his valuable suggestions. It is with great pride and pleasure that I submit this
dissertation work as his student.The preparation of this book has involved much time spent reading
the reasoned (but sometimes contradictory) advice from a large number of astute reviewers. I greatly
appreciate the time they spent to understand my motivation for the approach taken. I have learned
something from each of them.

I would like to extend my deep gratitude’s to my past and present lecturers for making me
the better person that I am today. I highly appreciate the knowledge they have imparted in
me, Namely: Prof Gopal Raja , Dr K. Adem , Dr PWN. Chin , Mrs. D Vijayasenan and
Mr. J.L Thabane.
My respectable gratitude to my past grade 11 and 12 teachers, Mrs. Mthamayendza and Mr. Shirinda
for making Mathematics come alive in their outstanding Maths classes. I am thankful to my Principal
Mr. Khosa for his kind of support and encouragement during course of study.my sincere and heart
ful thanks to him for his valuable suggestions.
I express my thanks to all my teachers who have taught me since my childhood: My high school
teachers, my undergraduate teachers, my graduate teachers and post graduate teachers.
Lastly but not least I would like to extend my deep gratitude to my very own blood ( Wombmate )
Chauke Welma, for her Love, affection and blessing and moral support without which this
work would remain Unfinished.
I WOULD LIKE TO DEDICATE THIS BOOK TO ALL MY PAST AND PRESENT STUDENTS, WISHING THEM
THE BEST OF LUCK IN MATHEMATICS AND IN LIFE.
9|Page
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
PURPOSE OF THE BOOK:
CALCULUS:
Written to improve algebra and problem solving skills of students taking a Calculus course, every
chapter in this companion is keyed to a calculus topic, providing conceptual background and specific
algebra techniques needed to understand and solve calculus problems related to that topic. It is
designed for calculus courses that integrate the review of precalculus concepts or individual use.
ALGEBRA:
This comprehensive book, designed to supplement the calculus course, provides an introduction to
and review of the basic ideas of Linear algebra.
TO THE STUDENTS:
Reading a calculus textbook is different from reading a newspaper or novel, or even a physics book.
Don’t be discouraged if you have to read a passage more than once in order to understand it. You
should have a pencil and paper and calculator at hand to sketch a diagram or make a calculations.
Some students start by trying their homework problems and read the text only if they get stuck on an
exercise. I suggest that a far better plan is to read and understand a section of the text before
attempting the exercises. In particular, you should look at the definition to see the exact meanings of
the terms. And before you read the example, I suggest that you cover up the solution and try solving
the problem yourself. You’ll get a lot more from looking at the solution if you do so.
The solutions to the exercises and the previous tests and exams are provided at the back of the book.
Some exercises ask for a verbal expression or interpretation or description. In such cases there is no
single correct way of expressing the answer, so don’t worry that you haven’t found the definitive
answer. In addition, there are often several different forms in which to express a numerical or
algebraic answer, so if your answer differs from mine, don’t immediately assume you’re wrong. From
example, if the answer given in the back of the book is √2 − 1 and you obtain 1⁄(1 + √2) , then
you’re right and rationalizing the denominator will show that the answers are equivalent.
I recommend that you keep this book for reference purpose after you finish the course. Because you
will likely forget some of the specific details of calculus, the book will serve as a useful reminder when
you need to use calculus in subsequent courses. And, because this book contains more materials than
can be covered in any one course, it can also serve as a valuable resource for working in scientist or
engineer.
Calculus is an exciting subject, justly considered to be one of the greatest achievements of the human
intellect. I hope you will discover that it is not only useful but also intrinsically beautiful.
10 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER ONE:
ABSOLUTE VALUES / MODULUS:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS:
11 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter one: Absolute Values / Modulus.
Definition 1.1:
Absolute Value: is the distance between the number and the origin.
Or
Absolute Value: describes the distance of a number on the number line from the origin without
considering which direction from zero the number lies.
About the Absolute Value:
 The absolute value of a number is never negative.
 The absolute value makes a negative number positive.
 We denote the absolute value by, | | which means modulus.
Definition 1.2.
The Absolute value or Magnitude of a real number 𝑎 is denoted by |𝑎| and is defined by
|𝑎| = { 𝑎
−𝑎
if 𝑎 ≥ 0
if 𝑎 < 0
Properties of an Absolute Values/Modulus.





|𝑥| = 0, if and only if 𝑥 = 0.
|𝑥| ≥ 0
|−𝑥| = |𝑥|, A number and its negative have the same
absolute value
|𝑥|2 = 𝑥 2
√𝑥 2 = |𝑥| , For any real number 𝑎
|𝑥𝑦| = |𝑥||𝑦| , The absolute value of a product is the
product of the absolute values
|𝑥 ± 𝑦|2 = (𝑥 ± 𝑦)2
|𝑥|
𝑥
|𝑦| = |𝑦| , The absolute value of a ratio is the ratio of the


absolute values.
|𝑥 − 𝑦| = |𝑦 − 𝑥|
|𝑥 + 𝑦| ≤ |𝑥| + |𝑦| , (Triangle Inequality)



Inequalities and Absolute
Values/Modulus.






|𝑥| = 𝑎
|𝑥| < 𝑎
|𝑥| > 𝑎
|𝑥| = |𝑥|
|𝑥| > |𝑥|
|𝑥| < |𝑥|
means
means
means
means
means
means
𝑥 = 𝑎 𝑜𝑟 𝑥 = −𝑎
−𝑎 < 𝑥 < 𝑎
𝑥 < −𝑎 𝑜𝑟 𝑥 > 𝑎
(𝑥)2 = (𝑥)2
(𝑥)2 > (𝑥)2
(𝑥)2 < (𝑥)2
12 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter one: Absolute Values / Modulus.
Worked Examples:
Example 1:
Evaluate the following Modulus:
1. |7|
2. |−11|
3. |8 − 12|
4. |2(−5) + 1|
Solution:
1. |7| = 7
2. |−11| = 11 , The absolute value makes a negative number positive.
3. |8 − 12| = |−4|
=4
4. |2(−5) + 1| = |−10 + 1|
= |−9|
=9
NB: Note that the effect of taking the absolute value of a number is to strip away the minus sign if the
number is negative and to leave the number unchanged if it is nonnegative.
13 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter one: Absolute Values / Modulus.
Theorem: Triangle Inequality.
Prove the Triangle Inequality: |𝑥 + 𝑦| ≤ |𝑥| + |𝑦|
Proof:
𝐿𝐻𝑆 = |𝑥 + 𝑦| 2
= (𝑥 + 𝑦)2
= 𝑥 2 + 2𝑥𝑦 + 𝑦 2
= |𝑥|2 + 2𝑥𝑦 + |𝑦|2
≤ |𝑥|2 + 2|𝑥||𝑦| + |𝑦|2
2
≤ ||𝑥| + |𝑦||
∴ |𝑥 + 𝑦| ≤ |𝑥| + |𝑦|
SOLVING MODULUS INEQUALITIES:
Worked Examples:
Example 2:
Solve for 𝑥 in the following modulus inequality:
1. |4𝑥 − 3| ≥ 5
2. |3𝑥 + 2| < 4
3. |3𝑥 + 9| = |2𝑥 + 1|
4. |2𝑥 − 3| ≥ |𝑥 + 3|
5. 𝑥 + 6 > |3𝑥 + 2|
6. 4 − 2|2𝑥 + 1| < 5
14 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter one: Absolute Values / Modulus.
Solution:
1. |4𝑥 − 3| ≥ 5
Using the properties of inequalities and absolute value, we obtain the following results,
∴ 4𝑥 − 3 ≤ −5 𝑜𝑟 4𝑥 − 3 ≥ 5
∴ 4𝑥 ≤ −2 𝑜𝑟 4𝑥 ≥ 8
Simplify the inequalities.
1
Thus, 𝑥 ≤ − 2 𝑜𝑟 𝑥 ≥ 2
Inequality Notation.
2. |3𝑥 + 2| < 4
Using the same procedure above, we have the following,
∴ −4 ≤ 3𝑥 + 2 < 4
∴ −6 < 3𝑥 < 2
Adding −2 to both sides of the equations.
2
Thus, −2 < 𝑥 < 3
Inequality Notation.
3. |3𝑥 + 9| = |2𝑥 + 1|
This is slightly different from the above problems, but it’s very easy if one can recall the
properties very well,
And hence we going to square both sides to remove the modulus sign.
∴ (3𝑥 + 9)2 = (2𝑥 + 1)2
∴ 9𝑥 2 + 54𝑥 + 81 = 4𝑥 2 + 4𝑥 + 1
Removing the brackets in both sides.
2
∴ 5𝑥 + 50𝑥 + 80 = 0
Rearranging the equation.
∴ 𝑥 2 + 10𝑥 + 16 = 0
Divide both sides by 5.
∴ (𝑥 + 2)(𝑥 + 8) = 0
Factorize the quadratic equation.
Thus, 𝑥 = −2 𝑜𝑟 𝑥 = −8
4. |2𝑥 − 3| ≥ |𝑥 + 3|
Remember modulus signs are removed by squaring both sides of the equation, and thus we
have the following results,
∴ (2𝑥 − 3)2 ≥ (𝑥 + 3)2
∴ 4𝑥 2 − 12𝑥 + 9 ≥ 𝑥 2 + 6𝑥 + 9 Removing the brackets in both sides of the equations.
∴ 3𝑥 2 − 18𝑥 ≥ 0
Rearranging the equation.
∴ 3𝑥(𝑥 − 6) ≥ 0
Taking out the common factor of the equation.
∴ 𝐶𝑉: 𝑥 = 0 𝑜𝑟 𝑥 = 6
Critical Values.
Thus, 𝑥 ≤ 0 𝑜𝑟 𝑥 ≥ 6
Inequality Notation.
15 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter one: Absolute Values / Modulus.
5.
𝑥 + 6 > |3𝑥 + 2|
The best way to solve this kind of a problem is to square both sides of the equations to deal with
the modulus sign, and thus obtain the following results,
∴ (𝑥 + 6)2 > (3𝑥 + 2)2
∴ −8𝑥 2 > −32
Removing the brackets in both sides of the equation.
∴ −8𝑥 2 + 32 > 0
Rearranging the equation.
2
∴ 𝑥 −4 < 0
Divide both sides by −8 , taking into an account the minus sign.
∴ (𝑥 + 2)(𝑥 − 2) < 0 This is a different of two squares factorization.
∴ 𝐶𝑉: 𝑥 = ±2
Critical Values.
Thus, −2 < 𝑥 < 2
Inequality Notation.
6. 4 − 2|2𝑥 + 1| < 5
This is tricky to most of the most students, but let’s now think of Algebra manipulation,
∴ −2|2𝑥 + 1| < 1
Transposing 4 to the right side of the equation.
1
∴ |2𝑥 + 1| > − 2
Divide both sides by −2, taking into an account the minus sign.
1
1
∴ 2𝑥 + 1 < − (− 2) 𝑜𝑟 2𝑥 + 1 > − 2 Applying Modulus properties introduced earlier.
∴ 2𝑥 < −
1
2
Thus, 𝑥 <
1
−4
𝑜𝑟 2𝑥 > −
𝑜𝑟 𝑥 >
3
2
3
−4
Adding −1 to both sides of the equation.
Inequality Notation.
16 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER TWO:
THE RELATIONSHIP BETWEEN RADIANS AND
DEGREES:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS:
17 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Two: The relationship between radians and degrees.
THE RELATIONSHIP BETWEEN RADIANS AND DEGREES:
At high school we usually lean to measure an angle in degrees. However, there are other ways of
measuring an angle. One that we are going to have a look at here is measuring angles in units called
Radians. In many scientific and engineering calculators radians are used in preference to degrees.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
Conversions between Degrees and Radians:

𝜋 𝑟𝑎𝑑 = 180°
𝑎𝑛𝑑
2𝜋 𝑟𝑎𝑑 = 360°
1. To convert degrees to radians, multiply degrees by
2. To convert radians to degrees, multiply radians by
3. 1° =
𝜋
180°
𝑎𝑛𝑑
1 𝑟𝑎𝑑 =
𝜋
180°
180°
𝜋
180°
𝜋
1°
4. 1′ = 𝑜𝑛𝑒 𝑚𝑖𝑛𝑢𝑡𝑒 = 60
1°
5. 1′′ = 𝑜𝑛𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 = 3600
18 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Two: The relationship between radians and degrees.
Worked Examples:
Example 3:
Change the following angle sizes to Radians:
a)
b)
c)
d)
e)
270°
120°
540°
315°
135°
Solution:
Change the following angle sizes to Radians:
a) 270°
𝜋
= 270° × 180°
=
3𝜋
2
b) 120°
𝜋
= 120° × 180°
=
2𝜋
3
c) 540°
𝜋
= 540° × 180°
= 3𝜋
d) 315°
𝜋
= 315° × 180°
=
7𝜋
4
e) 135°
𝜋
= 135° × 180°
=
3𝜋
4
19 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Two: The relationship between radians and degrees.
Worked Examples:
Example 4:
Change the following angle sizes to Degrees:
a)
b)
c)
d)
e)
− 𝜋 ⁄2
9𝜋⁄2
3𝜋
4𝜋⁄5
− 5𝜋⁄6
Solution:
a) −
𝜋
2
𝜋
2
=− ×
180°
𝜋
= −90°
b)
9𝜋
2
=
9𝜋
2
×
180°
𝜋
= 810°
c) 3𝜋
= 3𝜋 ×
180°
𝜋
= 540°
d)
4𝜋
5
4𝜋 180°
×
5
𝜋
= 144°
=
e) −
5𝜋
6
5𝜋 180°
×
6
𝜋
= −150°
=−
20 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER THREE:
SOLVING LOGARITHMIC AND TRIGNOMETRIC
EQUATIONS:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS:
21 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Logarithmic Equations.
Properties of the Logarithmic and Exponential functions:
2. PROPERTIES OF LOGARITHMIC
WITH A BASE "𝒂"
1. LOGARITHMIC LAWS:
Definition:
𝑦 = log 𝑎 𝑥 if and only if 𝑥 = 𝑎 𝑦 where 𝑎 > 0.
In other words, Logarithmic are exponents.
Remarks:
 log 𝑥 always refers to log base 10,
i.e. log 𝑥 = log10 𝑥
 ln 𝑥 Is called the natural logarithmic and is used to
represent log 𝑒 𝑥 , where the irrational number
𝑒 ≈ 2.71828128. Therefore, ln 𝑥 = 𝑦 if and only if
𝑥 = 𝑒𝑦.
 Change of base formulas:
ln 𝑎
a)
b)
c)
d)
e)
f)
g)
h)
i)
log 𝑎 𝑎 = 1 for all 𝑎 > 0 and
log 10 = 1
log 𝑎 1 = 0 for all 𝑎 > 0
log 𝑎 𝑥𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦
𝑥
log 𝑎 𝑦 = log 𝑎 𝑥 − log 𝑎 𝑦
log 𝑎 𝑥 𝑦 = 𝑦 log 𝑎 𝑥
log 𝑎 𝑎 𝑦 = 𝑦 log 𝑎 𝑎 = 𝑦(1) = 𝑦
𝑎log𝑎 𝑥 = 𝑥
log 𝑒 𝑥 = ln 𝑥
log 𝑎
log 𝑏 𝑎 = ln 𝑏 = log 𝑏
3. PROPERTIES OF NATURAL LOGARITHMIC:
a)
b)
c)
d)
e)
ln 𝑒 = 1
ln 1 = 0
ln 𝑥 = 𝑦 ⟺ 𝑒 𝑦 = 𝑥
ln 𝑥𝑦 = ln 𝑥 + ln 𝑦
𝑥
ln 𝑦 = ln 𝑥 − ln 𝑦
f) ln 𝑥 𝑦 = 𝑦 ln 𝑥
g) ln 𝑒 𝑥 = 𝑥 ln 𝑒 = 𝑥(1) = 𝑥 ,
h) 𝑒 ln 𝑥 = 𝑥 , 𝑥 > 0
4. PROPERTIES OF EXPONENTS:
If 𝑚 and 𝑛 are integers then the following
holds:
a) 𝑎𝑚 𝑎𝑛 = 𝑎𝑚+𝑛
b) (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛
c) (𝑎𝑏)𝑚 = 𝑎𝑚 𝑏 𝑚
d)
𝑥∈ℝ
e)
f)
g)
𝑎𝑚
=
𝑎𝑛
𝑎 𝑚
(𝑏 )
−𝑚
𝑎
1
𝑎𝑛
𝑚
𝑛
𝑎𝑚−𝑛
𝑎𝑚
= 𝑏𝑚 , where 𝑏 ≠ 0
1
= 𝑎𝑚 , where 𝑎 ≠ 0
𝑛
= √𝑎
𝑛
, where 𝑎 ≠ 0
𝑛
𝑚
h) 𝑎 = √𝑎𝑚 = ( √𝑎)
i) 𝑎0 = 1
, where 𝑎 ≠ 0 .
22 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Logarithmic Equations.
Worked Examples:
Example 5:
Solve for 𝑥 in the following examples:
a) 3𝑥+4 = 32𝑥−1
b) 3𝑥+4 = 5𝑥−6
c) 3𝑥 = 25
d) 𝑥 2 2𝑥 − 2𝑥 = 0
e) 𝑥 2 𝑒 𝑥 − 5𝑥𝑒 𝑥 − 6𝑒 𝑥 = 0
Solution:
a) 3𝑥+4 = 32𝑥−1
∴ 𝑥 + 4 = 2𝑥 − 1
∴𝑥=5
Copy the original equation.
Equate the powers of the exponents
b) 3𝑥+4 = 5𝑥−6
∴ (𝑥 + 4) log 3 = (𝑥 − 6) log 5
∴ 𝑥 log 3 + 4 log 3 = 𝑥 log 5 − 6 log 5
∴ 𝑥(log 3 − log 5) = −6 log 5 − 4 log 3
Thus, 𝑥 = − [
6 log 5+4 log 3
log 3−log 5
c) 3𝑥 = 25
∴ 𝑥 ln 3 = ln 25
∴𝑥=
ln 25
ln 3
]
𝑜𝑟
𝑥=
Copy the original equation.
Logarithmic property
Remove the brackets
−(6 log 5+4 log 3)
log 3−log 5
Copy the original equation.
Notice that we could have used the Log property if we wanted to.
≈ 2.930
23 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Logarithmic Equations.
d) 𝑥 2 2𝑥 − 2𝑥 = 0
∴ 2𝑥 (𝑥 2 − 1) = 0
∴ 2𝑥 = 0 𝑜𝑟 𝑥 2 − 1 = 0
∴ (𝑥 − 1)(𝑥 + 1) = 0 𝑏𝑢𝑡 2𝑥 ≠ 0
Thus, 𝑥 = ±1
Copy the original equation.
Take out the common factor of the equation.
e) 𝑥 2 𝑒 𝑥 − 5𝑥𝑒 𝑥 − 6𝑒 𝑥 = 0
∴ 𝑥 2 − 5𝑥 − 6 = 0
∴ (𝑥 + 1)(𝑥 − 6) = 0
∴ 𝑥 = −1 𝑜𝑟 𝑥 = 6
Copy the original equation.
Divide throughout by 𝑒 𝑥
Factorize the quadratic equation.
Factorize the quadratic equation.
Worked Examples:
Example 6:
Solve for 𝑥 in the following examples:
a) 𝑒 2𝑥+3 − 7 = 0
b) ln(5 − 2𝑥) = −3
c)
10
1+𝑒 −𝑥
=2
d) 4 + 3𝑥+1 = 8
e) 𝑒 2𝑥 − 3𝑒 𝑥 + 2 = 0
24 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Logarithmic Equations.
Solution:
a) 𝑒 2𝑥+3 − 7 = 0
∴ 𝑒 2𝑥+3 = 7
∴ ln(𝑒 2𝑥+3 ) = ln 7
∴ (2𝑥 + 3) ln 𝑒 = ln 7
∴ 2𝑥 + 3 = ln 7
Copy the original equation.
Transpose 7 to the right side of the equation.
Take the ln in both sides of the equation.
Applying the Natural Log property.
Note: ln 𝑒 = 1
1
2
∴ 𝑥 = (ln 7 − 3)
b) ln(5 − 2𝑥) = −3
∴ 𝑒 ln(5−2𝑥) = 𝑒 −3
∴ (5 − 2𝑥)𝑒 ln = 𝑒 −3
∴ 5 − 2𝑥 = 𝑒 −3
Copy the original equation.
Exponentiate both sides of the equation.
From the properties.
Note: 𝑒 ln = 1
1
2
Thus, 𝑥 = − (𝑒 −3 − 5) ≈ 2.475
c)
10
1+𝑒 −𝑥
=2
∴ 2 + 2𝑒 −𝑥 = 10
∴ 𝑒 −𝑥 = 4
∴ −𝑥 = ln 4
∴ 𝑥 = − ln 4 ≈ −1.386
d) 4 + 3𝑥+1 = 8
∴ 3𝑥+1 = 4
∴ (𝑥 + 1) ln 3 = ln 4
ln 4
∴ 𝑥 = (ln 3) − 1
Copy the original equation.
Cross Multiplication.
Simplify the equation.
Take the ln in both sides of the equation.
Copy the original equation.
Transpose 7 to the right side of the equation.
Take the ln in both sides of the equation.
Remove the brackets and solve the equation.
Thus, 𝑥 = 0.262
e)
𝑒 2𝑥 − 3𝑒 𝑥 + 2 = 0
Copy the original equation.
Let 𝑘 = 𝑒 𝑥 and 𝑘 2 = 𝑒 2𝑥 = 𝑒 𝑥 𝑒 𝑥
∴ 𝑘 2 − 3𝑘 + 2 = 0
Replace, 𝑘 = 𝑒 𝑥 and 𝑘 2 = 𝑒 2𝑥
∴ (𝑘 − 1)(𝑘 − 2) = 0
Factorize the equation.
∴ 𝑘 = 1 𝑜𝑟 𝑘 = 2
Hence, since 𝑘 = 𝑒 𝑥 = 1 Or 𝑘 = 𝑒 𝑥 = 2
∴ 𝑒 𝑥 = 1 Or 𝑒 𝑥 = 2
∴ 𝑥 = ln 1 = 0
𝑜𝑟
𝑥 = ln 2 ≈ 0.693
25 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Logarithmic Equations.
Worked Examples:
Example 6:
Solve for 𝑥 in the following examples:
a)
b)
c)
d)
e)
log 3 (7𝑥 + 3) = log 3(5𝑥 + 9)
log 2 (5𝑥 + 7) = 5
log 4 𝑥 + log 4 (𝑥 − 12) = 3
log(𝑥 − 3) + log(𝑥) = log 18
log 4 (2𝑥 + 1) = log 4(𝑥 + 2) − log 4 3
Solution:
a) log 3 (7𝑥 + 3) = log 3(5𝑥 + 9)
∴ 7𝑥 + 3 = 5𝑥 + 9
∴𝑥=3
Copy the original equation.
Drop the Logs since the base are the same.
b) log 2 (5𝑥 + 7) = 5
∴ 5𝑥 + 7 = 25
∴ 5𝑥 + 7 = 32
∴𝑥=5
Copy the original equation.
By the Log definition.
c) log 4 𝑥 + log 4 (𝑥 − 12) = 3
∴ log 4 [ 𝑥(𝑥 − 12)] = 3
∴ 𝑥(𝑥 − 12) = 43
∴ 𝑥 2 − 12𝑥 = 64
∴ 𝑥 2 − 12𝑥 − 64 = 0
∴ (𝑥 − 16)(𝑥 + 4) = 0
∴ 𝑥 = 16 𝑜𝑟 𝑥 = −4
Thus, 𝑥 = 16 𝑏𝑢𝑡 𝑥 ≠ −4
Copy the original equation.
By the Log property.
d) log(𝑥 − 3) + log(𝑥) = log 18
∴ log(𝑥 − 3)(𝑥) = log 18
∴ 𝑥 2 − 3𝑥 = 18
∴ (𝑥 − 6)(𝑥 + 3) = 0
∴ 𝑥 = 6 𝑜𝑟 𝑥 = −3
Thus, 𝑥 = 6 𝑏𝑢𝑡 𝑥 ≠ −3
Copy the original equation.
By the Log property.
Drop the Logarithmic.
Factorize the quadratic equation.
Factorize the quadratic equation.
26 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Logarithmic Equations.
e) log 4 (2𝑥 + 1) = log 4(𝑥 + 2) − log 4 3
𝑥+2
)
3
∴ log 4 (2𝑥 + 1) = log 4 (
∴ 2𝑥 + 1 =
𝑥+2
3
∴ 6𝑥 + 3 = 𝑥 + 2
∴𝑥=−
Copy the original equation.
By the Log property.
Drop the Logs since the base are the same.
Multiply both sides by 3
1
5
Thus, there is no solution for this problem. Since we can’t take the logarithm of a negative
number.
27 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Solving equations involving Trigonometry:
Recap:
When solving trig equations you have to find a reference angle (𝑅𝐴) and remember the
sin 𝜃 and cos 𝜃 functions have a period of 2𝜋 while tan 𝜃 has a period of 𝜋.
1. If sin 𝜃 = 𝑚
𝜃 = 𝑅𝐴 + 2𝜋𝑘
And
Or
−1 ≤ 𝑚 ≤ 1 then:
𝜃 = (𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ
2. If cos 𝜃 = 𝑚
𝜃 = 𝑅𝐴 + 2𝜋𝑘
And
Or
−1 ≤ 𝑚 ≤ 1 then:
𝜃 = (2𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ
And
−1 ≤ 𝑚 ≤ 1 then:
Alternative method:
If cos 𝜃 = 𝑚
𝜃 = ±𝑅𝐴 + 2𝜋𝑘 , 𝑘 ∈ ℤ
3. If tan 𝜃 = 𝑚
And
𝜃 = 𝑅𝐴 + 𝜋𝑘 , 𝑘 ∈ ℤ
Or
𝑚 ∈ ℝ then:
𝜃 = (𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ
NOTE:
The general solution must be used when no interval is given. The general solution is used to find the
values of the angles that falls within the given interval [𝑎, 𝑏].
4.
5.
6.
7.
8.
When sin 𝜃 = 0 , write 𝜃 = 0 + 𝜋𝑘 (𝑁𝑂𝑇 2𝜋).
𝜋
When cos 𝜃 = 0 , write 𝜃 = + 𝜋𝑘 (𝑁𝑂𝑇 2𝜋).
2
For all other values of sin 𝜃 and cos 𝜃 add 2𝜋𝑘 to the solution.
For all the solution of tan 𝜃 including tan 𝜃 = 0 add 𝜋𝑘.
For sin 𝜃 and cos 𝜃 , any value greater than 1 or less than −1 has NO solution.
Solving negative angles:
9. If sin 𝜃 = −𝑚 , then 𝜃 = sin−1 (𝑚) + 2𝜋𝑘 = 𝑅𝐴 + 2𝜋𝑘 , 𝑘 ∈ ℤ
Or 𝜃 = (𝜋 + 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (2𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ
10. If cos 𝜃 = −𝑚 , then 𝜃 = cos −1 (−𝑚) + 2𝜋𝑘 = 𝑅𝐴 + 2𝜋𝑘 , 𝑘 ∈ ℤ
Or 𝜃 = (𝜋 − 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (𝜋 + 𝑅𝐴) + 2𝜋𝑘 , 𝑘 ∈ ℤ
Alternative method: If cos 𝜃 = −𝑚 , then 𝜃 = cos−1(−𝑚) + 2𝜋𝑘 = ±𝑅𝐴 + 2𝜋𝑘
11. If tan 𝜃 = −𝑚 , then 𝜃 = tan−1 (𝑚) + 𝜋𝑘 = 𝑅𝐴 + 𝜋𝑘 , 𝑘 ∈ ℤ
Or 𝜃 = (𝜋 − 𝑅𝐴) + 𝜋𝑘 , 𝑘 ∈ ℤ Or 𝜃 = (2𝜋 − 𝑅𝐴) + 𝜋𝑘 , 𝑘 ∈ ℤ
28 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Trigonometric Equations.
Worked Examples:
Example 6:
Solve for 𝑥 in the following examples:
a) √3 tan 𝑥 = 1
𝜋
b) √2 sin (𝑥 − ) − 1 = 0
2
𝜋
𝜋
c) sin 2𝑥 cos 3 − sin 3 cos 2𝑥 =
d) 2sin2 𝑥 + 3 sin 𝑥 − 2 = 0
e) cos 2𝑥 + 3 cos 𝑥 − 1 = 0
√3
2
Solution:
a) √3 tan 𝑥 = 1
Copy the original equation.
1
√3
1
tan−1 ( 3
√
𝜋
6
∴ tan 𝑥 =
∴𝑥=
∴𝑥=
Divide throughout by √3.
)
Take the arctangent in both sides.
𝜋
b) √2 sin (𝑥 − ) − 1 = 0
2
Copy the original equation.
𝜋
∴ √2 sin (𝑥 − 2 ) = 1
𝜋
2
1
√2
𝜋
1
𝑥 − = sin−1 (
2
√2
𝜋
𝜋
𝑥−2=4
∴ sin (𝑥 − ) =
∴
∴
∴𝑥=
Transpose 1 to the right side of the equation
Divide throughout by √2.
)
Take the arsine in both sides.
Then solve the equation.
3𝜋
4
29 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Trigonometric Equations.
𝜋
𝜋
c) sin 2𝑥 cos 3 − sin 3 cos 2𝑥 =
𝜋
3
∴ sin (2𝑥 − ) =
𝜋
√3
2
√3
∴ 2𝑥 − 3 = sin−1 ( 2 )
𝜋
∴ 2𝑥 − 3 =
∴ 2𝑥 =
∴ 𝑥=
𝜋
3
√3
2
Copy the original equation.
Recall: sin(𝜃 − 𝛽) = sin 𝜃 cos 𝛽 − cos 𝜃 sin 𝛽
Take the arsine in both sides.
Then solve the equation.
2𝜋
3
𝜋
3
d) 2sin2 𝑥 + 3 sin 𝑥 − 2 = 0
Copy the original equation.
∴ (2 sin 𝑥 − 1)(sin 𝑥 + 2) = 0
Factorize the quadratic equation.
∴ 2 sin 𝑥 − 1 = 0 𝑜𝑟
sin 𝑥 + 2 = 0
1
∴ sin 𝑥 = 2
𝑜𝑟
sin 𝑥 = −2 (No solution)
∴𝑥=
𝜋
6
e) cos 2𝑥 + 3 cos 𝑥 − 1 = 0
Copy the original equation.
∴ (2cos2 𝑥 − 1) − 3 cos 𝑥 − 1 = 0
Double angle identity: cos 2𝑥 = 2cos 2 𝑥 − 1
2
∴ 2cos 𝑥 − 3 cos 𝑥 − 2 = 0
∴ (2 cos 𝑥 + 1)(cos 𝑥 − 2) = 0
Factorize the quadratic equation.
∴ 2 cos 𝑥 + 1 = 0 𝑜𝑟
cos 𝑥 − 2 = 0
1
∴ cos 𝑥 = − 2
𝑜𝑟
cos 𝑥 = 2 (No solution)
∴𝑥=±
2𝜋
3
30 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Trigonometric Equations.
Worked Examples:
Example 6:
Solve for 𝑥 in the following examples:
a) cos 2𝑥 − 3 sin 𝑥 = −1
−2𝜋 ≤ 𝑥 ≤ 2𝜋
b) sin 2𝑥 + sin 𝑥 = 6 cos 𝑥 + 3
5𝜋
−𝜋 ≤ 𝑥 ≤ 𝜋
𝜋
c) sin (3𝑥 + 18 ) + cos (2𝑥 − 18) = 0
−𝜋 ≤ 𝑥 ≤ 𝜋
d) 4cos2 𝑥 + sin 2𝑥 − 1 = 0
0 ≤ 𝑥 ≤ 2𝜋
e) sin 2𝑥 + 2 sin 𝑥 + cos 2 𝑥 + cos 𝑥 = 0
0 ≤ 𝑥 ≤ 2𝜋
Solution:
a) cos 2𝑥 − 3 sin 𝑥 = −1
−2𝜋 ≤ 𝑥 ≤ 2𝜋
∴ (1 − 2sin2 𝑥) − 3 sin 𝑥 + 1 = 0
∴ 2sin2 𝑥 + 3 sin 𝑥 − 2 = 0
Double angle identity: cos 2𝑥 = 1 − 2sin2 𝑥
Rearrange the equation in standard form.
∴ (2 sin 𝑥 − 1)(sin 𝑥 + 2) = 0
Factorize the quadratic equation.
∴ 2 sin 𝑥 − 1 = 0
1
∴ sin 𝑥 = 2
𝑜𝑟
𝑜𝑟
sin 𝑥 + 2 = 0
sin 𝑥 = −2 (No solution)
𝜋
∴ 𝑥 = 6 + 2𝜋𝑘 , 𝑘 ∈ ℤ
𝜋
𝑥 = (𝜋 − 6 ) + 2𝜋𝑘 , 𝑘 ∈ ℤ
OR
=
Thus, 𝑥 =
5𝜋
−
3
;−
7𝜋
6
;
𝜋
6
;
5𝜋
6
5𝜋
6
+ 2𝜋𝑘 , 𝑘 ∈ ℤ
31 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Trigonometric Equations.
b) sin 2𝑥 + sin 𝑥 = 6 cos 𝑥 + 3
−𝜋 ≤ 𝑥 ≤ 𝜋
∴ 2 sin 𝑥 cos 𝑥 + sin 𝑥 = 6 cos 𝑥 + 3
∴ 2 sin 𝑥 cos 𝑥 + sin 𝑥 − 6 cos 𝑥 − 3 = 0
∴ sin 𝑥 (2 cos 𝑥 + 1) − 3(2 cos 𝑥 + 1) = 0
∴ (2 cos 𝑥 + 1)(sin 𝑥 − 3) = 0
∴ 2 cos 𝑥 + 1 = 0 𝑜𝑟
cos 𝑥 − 3 = 0
1
∴ cos 𝑥 = − 2
𝑜𝑟
cos 𝑥 = 3 (No solution)
𝜋
∴ 𝑥 = 3 + 2𝜋𝑘 , 𝑘 ∈ ℤ
𝜋
𝑥 = (𝜋 − 3 ) + 2𝜋𝑘 , 𝑘 ∈ ℤ
Or
𝜋
Or 𝑥 = (𝜋 + 3 ) + 2𝜋𝑘 , 𝑘 ∈ ℤ
=
4𝜋
3
=
2𝜋
3
, ±
2𝜋
3
5𝜋
𝜋
c) sin (3𝑥 + 18 ) + cos (2𝑥 − 18) = 0
5𝜋
−𝜋 ≤ 𝑥 ≤ 𝜋
𝜋
𝜋
∴ sin (3𝑥 + 18 ) = − cos (2𝑥 − 18)
Transpose cos (2𝑥 − 18) to the right side
5𝜋
𝜋
𝜋
) = − sin [ − (2𝑥 − )]
18
2
18
5𝜋
5𝜋
sin (3𝑥 + 18 ) = − sin ( 9 − 2𝑥)
5𝜋
5𝜋
sin (3𝑥 + 18 ) = sin (2𝑥 − 9 )
5𝜋
5𝜋
3𝑥 + 18 = 2𝑥 − 9 + 2𝜋𝑘
5𝜋
𝑥 = − 6 + 2𝜋𝑘 , 𝑘 ∈ ℤ
∴ sin (3𝑥 +
∴
∴
∴
∴
Thus, 𝒙 =
23𝜋
90
+ 2𝜋𝑘 , 𝑘 ∈ ℤ
+ 2𝜋𝑘 , 𝑘 ∈ ℤ
𝜋
3
Thus, 𝑥 =
Change: sin 2𝑥 = 2 sin 𝑥 cos 𝑥
Subtract 6 cos 𝑥 and 3 in both sides.
Take out the common factor.
Factorize the quadratic equation.
;
59𝜋
90
;−
13𝜋
90
;−
49𝜋
90
;−
17𝜋
18
Co-functions identities.
OR
5𝜋
3𝑥 + 18 = 𝜋 − (2𝑥 −
𝑥=
; −
23𝜋
90
+
2𝜋
𝑘
5
5𝜋
)+
9
2𝜋𝑘
, 𝑘∈ℤ
5𝜋
6
32 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Three: Solving Trigonometric Equations.
d) 4cos2 𝑥 + sin 2𝑥 − 1 = 0
0 ≤ 𝑥 ≤ 2𝜋
[cos2 𝑥 + sin2 𝑥 = 1].
∴ 4cos2 𝑥 + 2 sin 𝑥 cos 𝑥 − (cos 2 𝑥 + sin2 𝑥) = 0
∴ 3cos2 𝑥 + 2 sin 𝑥 cos 𝑥 − sin2 𝑥 = 0
Simplify the equation.
∴ (3 cos 𝑥 − sin 𝑥)(sin 𝑥 + cos 𝑥) = 0
Factorize the quadratic equation.
∴ 3 cos 𝑥 − sin 𝑥 = 0 𝑜𝑟 sin 𝑥 + cos 𝑥 = 0
∴ 3 cos 𝑥 = sin 𝑥 𝑜𝑟 sin 𝑥 = − cos 𝑥
∴ tan 𝑥 = 3 𝑜𝑟 tan 𝑥 = −1
Divide by sin 𝑥 in both sides of the equation
𝜋
∴ 𝑥 = 1.23 𝑟𝑎𝑑 + 𝜋𝑘 , 𝑘 ∈ ℤ
OR 𝑥 = − + 𝜋𝑘 , 𝑘 ∈ ℤ
∴𝑥=
3𝜋
4
+ 𝜋𝑘 , 𝑘 ∈ ℤ
Thus, 𝑥 = 1.23 𝑟𝑎𝑑 ,
OR 𝑥 =
3𝜋
4
,
7𝜋
7𝜋
4
4
+ 𝜋𝑘 , 𝑘 ∈ ℤ
4
e) sin 2𝑥 + 2 sin 𝑥 + cos2 𝑥 + cos 𝑥 = 0
∴ 2 sin 𝑥 cos 𝑥 + 2 sin 𝑥 + cos2 𝑥 + cos 𝑥 = 0
∴ 2 sin 𝑥 (cos 𝑥 + 1) + cos 𝑥 (cos 𝑥 + 1) = 0
∴ (cos 𝑥 + 1)(2 sin 𝑥 + cos 𝑥) = 0
∴ cos 𝑥 + 1 = 0
𝑜𝑟 2 sin 𝑥 + cos 𝑥 = 0
1
∴ cos 𝑥 = −1
𝑜𝑟
tan 𝑥 = − 2
∴ 𝑥 = 0 + 2𝜋𝑘 , 𝑘 ∈ ℤ
Or 𝑥 = 𝜋 + 2𝜋𝑘 , 𝑘 ∈ ℤ
Or
Or
Or
0 ≤ 𝑥 ≤ 2𝜋
Change: sin 2𝑥 = 2 sin 𝑥 cos 𝑥
Take out the common factor.
Factorize the quadratic equation.
𝑥 = 0.464 𝑟𝑎𝑑 + 𝜋𝑘 , 𝑘 ∈ ℤ
𝑥 = 2.678 𝑟𝑎𝑑 + 𝜋𝑘 , 𝑘 ∈ ℤ
𝑥 = 5.819 + 𝜋𝑘 , 𝑘 ∈ ℤ
Thus, 𝑥 = 0 ; 0.464 ; 2.678 ; 5.819 ; 𝜋
33 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER FOUR:
FUNCTIONS:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS:
34 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
Definition:
A function 𝑓 is a rule that assigns to each element 𝑥 in a set 𝐷 exactly one element, called 𝑓(𝑥)
in a set 𝐸.
The properties of a function:





Domain
Range
Asymptotes
Period
Amplitudes
Types of functions and its Format:
Types of
functions:
1.
2.
3.
4.
5.
6.
Linear function
Quadratic Function
Cubic function
Exponential function
Power function
Hyperbolic function
Format to which it takes:
𝑓(𝑥) = 𝑚𝑥 + 𝑐
𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
𝑓(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑
𝑓(𝑥) = 𝑏 𝑥
𝑓(𝑥) = 𝑥 𝑛
𝑓(𝑥) =
𝑎
+𝑞
𝑥±𝑝
7. Rational function
𝑓(𝑥) =
8. Radical function
9. Trigonometric function
10. Composite function.
11. Piecewise
𝑓(𝑥) = √𝑔(𝑥)
𝑓(𝑥) = 𝑎 sin 𝑏𝑥 , 𝑎 cos 𝑏𝑥 and
tan 𝑏𝑥
ℎ(𝑥) = 𝑓 ∘ 𝑔(𝑥) = 𝑓(𝑔(𝑥))
𝑓(𝑥) = {
12. Inverse function
13. Absolute function
𝑃(𝑥)
𝑄(𝑥)
𝑥
−𝑥
𝑖𝑓 𝑥 ≥ 0
𝑖𝑓 𝑥 < 0
𝑦 = 𝑓 −1 (𝑥)
𝑓(𝑥) = |𝑥|
35 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
Graphs:
Define the following terms:


Domain: is the set of all inputs over which the function has defined outputs.
Range: is the set of all values that the function takes when 𝑥 takes values in the domain.
Worked Examples:
Example 7:
Sketch and find the domain and the range of the following functions:
1. 𝑓(𝑥) = 2𝑥 − 1
; −1 ≤ 𝑥 ≤ 1
2. 𝑓(𝑥) = 𝑥 2
;
−1 ≤ 𝑥 ≤ 3
3. 𝑓(𝑥) = √𝑥 + 4
1
4. 𝑓(𝑥) = 2𝑥 + 2𝑥
;
1≤𝑥≤4
Solution:
1. 𝑓(𝑥) = 2𝑥 − 1
; −1 ≤ 𝑥 ≤ 1
𝑦
Note:
𝑥
i)
ii)
This is a linear
function, where
our domain is the
restricted domains
and the range, we
just substituted
the values of the
domain into the
original function.
𝐷𝑓 = [−1,1]
𝑅𝑓 = [−3,1]
36 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
2. 𝑓(𝑥) = 𝑥 2
;
−1 ≤ 𝑥 ≤ 3
𝑦
𝑥
i)
ii)
𝐷𝑓 = [−1,3]
𝑅𝑓 = [1,9]
3. 𝑓(𝑥) = √𝑥 + 4
∴𝑥+4≥0
Thus, 𝑥 ≥ −4 , this will give us a clear picture of how our grah will look like.
i)
ii)
𝐷𝑓 = [−4, ∞)
𝑅𝑓 = [0, ∞)
37 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
4. 𝑓(𝑥) = 2𝑥 +
1
2𝑥
;
1≤𝑥≤4
𝐷𝑓 = [1,4]
𝑅𝑓 = [2.5 , 8.125]
i)
ii)
Worked Examples:
Example 8:
Sketch and find the domain and the asymptotes of the following functions:
𝑥
1. 𝑓(𝑥) = 𝑥−2
𝑥+1
2. 𝑓(𝑥) = 𝑥−2
𝑥+1
3. 𝑓(𝑥) = 𝑥−1
𝑥 2 +1
4. 𝑓(𝑥) = 𝑥 2 −1
5. 𝑓(𝑥) =
𝑥 2 +𝑥−6
𝑥−3
38 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
Solution:
𝑥
1. 𝑓(𝑥) = 𝑥−2
∴ 𝑓(𝑥) =
(𝑥−2)+2
𝑥−2
2
= 𝑥−2 + 1
𝑦=1
𝑥=2
i)
ii)
iii)
iv)
v)
𝐷𝑓 = ℝ − {2}
𝐷𝑓 = ℝ − {1}
Vertical asymptotes:
∴𝑥=2
Horizontal asymptotes:
∴𝑦=1
Oblique asymptotes: No Oblique asymptotes.
There are no oblique/slant asymptotes because the degree of the numerator and the
denominator are the same, note that if we have the horizontal asymptotes we won
have the oblique asymptotes.
39 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
2. 𝑓(𝑥) =
𝑥+1
𝑥−2
∴ 𝑓(𝑥) =
(𝑥−2)+3
𝑥−2
=
3
𝑥−2
+1
𝑦=1
𝑥=2
i)
ii)
iii)
iv)
v)
𝐷𝑓 = ℝ − {2}
𝐷𝑓 = ℝ − {1}
Vertical asymptotes:
∴𝑥=2
Horizontal asymptotes:
∴𝑦=1
Oblique asymptotes: No Oblique asymptotes.
𝑥+1
3. 𝑓(𝑥) = 𝑥−1
∴ 𝑓(𝑥) =
(𝑥−1)+2
𝑥−1
2
= 𝑥−1 + 1
𝑦=1
𝑥=1
i)
ii)
iii)
iv)
v)
𝐷𝑓 = ℝ − {1}
𝐷𝑓 = ℝ − {1}
Vertical asymptotes: 𝑥 = 2
Horizontal asymptotes: 𝑦 = 1
Oblique asymptotes: No Oblique asymptotes.
40 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
𝑥 2 +1
4. 𝑓(𝑥) = 𝑥 2 −1
∴ 𝑓(𝑥) =
(𝑥 2 −1)+2
𝑥 2 −1
2
= 𝑥 2 −1 + 1
Vertical asymptotes:
Oblique asymptotes: No Oblique asymptotes.
∴ 𝑥 2 − 1 = 0 ⟹ 𝑥 = ±1
Horizontal asymptotes:
∴𝑦=1
𝑦=1
𝑥 = −1
𝑥=1
41 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
5. 𝑓(𝑥) =
𝑥 2 +𝑥−6
𝑥−3
This is slightly different from the previous example because here, the degree of the numerator is
higher than the degree of the denominator by 1. Which tells us that we are going to have the
Oblique asymptotes, of which we going to use long division method to find the Oblique
asymptotes, and if we have the oblique then we won’t have the horizontal asymptotes.
Vertical asymptotes:
∴𝑥=3
Oblique asymptotes:
𝑥+4
𝑥2 + 𝑥 − 6
−(𝑥 2 − 3𝑥)
4𝑥 − 6
−(4𝑥 − 12)
6
Thus, the Oblique asymptotes by, 𝑦 = 𝑥 + 4
𝑥−3
𝑦 = 𝑥+4
𝑥=3
42 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
Piecewise Function:
Definition:
Piecewise function: is a function that is defined on a sequence of intervals.
Or
Piecewise function: Is a pieces of different functions all on one graph.
Worked Examples:
Example 9:
Find the domain and the range of the following functions:
𝑥+3
1. 𝑓(𝑥) = { 3
2𝑥 + 1
; 𝑥≤0
; 0<𝑥≤2
; 𝑥>2
2. 𝑓(𝑥) = {
2𝑥 + 1
𝑥2 − 2
; 𝑥 ≤ −1
; 𝑥 > −1
3. 𝑓(𝑥) = {
1−𝑥
𝑥2
;
;
𝑥≤1
𝑥>1
43 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
Solution:
𝑥+3
1. 𝑓(𝑥) = { 3
2𝑥 + 1
; 𝑥≤0
; 0<𝑥≤2
; 𝑥>2
Note:
Label each function as
follows 𝑦 = 𝑥 + 3 ,
𝑦=3
And 𝑦 = 2𝑥 + 1 , the
domains are given
corresponding to each
function. To get the
range of 𝑓, substitute
the values of the
restricted domains
independtely.
i)
Domain: 𝐷𝑓 = (−∞, 0] ∪ (0,2] ∪ (2, ∞)
ii)
Range: 𝑅𝑓 = (−∞, 3] ∪ {3} ∪ (5, ∞)
44 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
2. 𝑓(𝑥) = {
2𝑥 + 1
𝑥2 − 2
;
;
𝑥 ≤ −1
𝑥 > −1
Here we notice that we only have two pieces of functions drawn in one function, which is called
Piecewise function. So to draw the function 𝑓, we let 𝑦 = 2𝑥 + 1 which is a linear function with
the gradient of 2. And 𝑦 = 𝑥 2 − 2 is a parabolic function with the turning point 𝑦 = −2, but the
graph is greater than −1, which is our domain for the parabola function.
i)
Domain: 𝐷𝑓 = (−∞, −1] ∪ (−1, ∞)
ii)
Range: 𝑅𝑓 = (−∞, −1] ∪ (−1, ∞)
45 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
3. 𝑓(𝑥) = {
1−𝑥
𝑥2
;
;
𝑥≤1
𝑥>1
Let 𝑦 = 1 − 𝑥 with 𝑥 ≤ 1 , and 𝑦 = 𝑥 2 where 𝑥 > 1 , to get the range of the original
function 𝑓, Substitute the restricted domains in each function respectively and write them as an
interval notation. But the sketched graph will also give you the domains and the range.
i)
Domain: 𝐷𝑓 = (−∞, 1] ∪ (1, ∞)
ii)
Range: 𝑅𝑓 = (−∞, 0] ∪ (1, ∞)
46 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
EVALUATION OF A PIECEWISE FUNCTIONS:
Worked Examples:
Example 10:
Consider the given piecewise functions and answer the below.
1. Let 𝑓(𝑥) = {
𝑥−2
𝑥2
;
;
𝑥≥2
𝑥<2
Evaluate the following:
𝑓(1) , 𝑓(2) And 𝑓(3)
1−𝑥
𝑥 2 − 2𝑥
2. 𝑓(𝑥) = { 3
𝑥 −2
3 − 𝑥2
;
𝑥 < −1
; −1 ≤ 𝑥 ≤ 1
; 1<𝑥<2
;
𝑥≥2
Evaluate the following:
𝑓(−1) , 𝑓(2) , 𝑓(−5) And 𝑓(7)
tan−1 𝑥
3. 𝑓(𝑥) = { 2cos 2 𝑥 − 1
1 − sin 2𝑥
;
;
;
−16 < 𝑥 ≤ 0
𝜋⁄2 ≤ 𝑥 ≤ 5𝜋⁄6
5𝜋⁄4 ≤ 𝑥 < 2𝜋
Evaluate the following:
𝑓(−1)
,
2𝜋
3
𝑓( )
And
7𝜋
4
𝑓( )
47 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
Solution:
1. Let 𝑓(𝑥) = {
𝑥−2
𝑥2
;
;
𝑥≥2
𝑥<2
The challenge to most of the students is to pick up the correct function for each evaluation. Now
let’s have a look at our given domains in each function respectively, to evaluate 𝑓(1) we need to
check at which interval does 1 lies, for example we have [2, ∞) and (−∞, 2) as an interval from
the restricted domains , so it’s clear that 𝑓(1) lies within (−∞, 2) and 𝑓(2) and 𝑓(3) lies
within [2, ∞).
∴ 𝑓(1) = 𝑥 2
= 12
=1
,
1−𝑥
𝑥 2 − 2𝑥
2. 𝑓(𝑥) = { 3
𝑥 −2
3 − 𝑥2
∴ 𝑓(2) = 𝑥 − 2
=2−2
=0
And
∴ 𝑓(3) = 𝑥 − 2
=3−2
=1
;
𝑥 < −1
; −1 ≤ 𝑥 ≤ 1
;
1<𝑥<2
;
𝑥≥2
First step is to write out our interval to see which function to pick for each evaluation.
Now we have, (−∞, −1) , [−1,1] , (1,2) and [2, ∞) as an interval which becomes easy for us
now to see where does each evaluation lies using the restricted domains as an interval notation.
∴ 𝑓(−1) = 𝑥 2 − 2𝑥
= (−1)2 − 2(−1)
=3
∴ 𝑓(−5) = 1 − 𝑥
= 1 − (−5)
=6
∴ 𝑓(2) = 3 − 𝑥 2
= 3 − 22
= −1
∴ 𝑓(7) = 3 − 𝑥 2
= 3 − 72
= −46
48 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
tan−1 𝑥
3. 𝑓(𝑥) = { 2cos 2 𝑥 − 1
1 − sin 2𝑥
; −16 < 𝑥 ≤ 0
; 𝜋⁄2 ≤ 𝑥 ≤ 5𝜋⁄6
; 5𝜋⁄4 ≤ 𝑥 < 2𝜋
Do not be intimidated by Trig functions, we still use the same procedure we used in the above
examples. Remember that writing out the intervals will remove all your fears against the
problem, the intervals are as follows, (−16 , 0 ] , [ 𝜋⁄2 , 5𝜋⁄6 ] and [5𝜋⁄4 , 2𝜋).
Remember to switch off your calculator in Radians to avoid mistakes.
∴ 𝑓(−1) = tan−1 𝑥
= tan−1 (−1)
=−
𝜋
4
2𝜋
∴ 𝑓 ( 3 ) = 2cos2 𝑥 − 1
2𝜋
7𝜋
∴ 𝑓 ( 4 ) = 1 − sin 2𝑥
7𝜋
= 2cos2 ( 3 ) − 1
= 1 − sin 2 ( 4 )
=−
=2
1
2
49 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
COMPOSITE FUNCTIONS:
Definition:
Composite function: is when one function is inside of another function.
I.e. (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) → The function is inside of the other.
Or
And (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥)) → The function is inside of the other.
Worked Examples:
Example 10:
Consider the following given function below and answer the questions below.
Let 𝑓(𝑥) = √4 − 𝑥 , 𝑔(𝑥) = 𝑥 − 1 and ℎ(𝑥) = 𝑥 2 + 2𝑥 − 3
Evaluate the following:
i.
(𝑓 ∘ 𝑔)(𝑥)
ii.
(𝑔 ∘ 𝑓)(𝑥)
iii.
(ℎ ∘ 𝑔)(𝑥)
iv.
(𝑓 ∘ ℎ)(𝑥)
v.
(𝑓 ∘ 𝑔 ∘ ℎ)(𝑥)
50 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
Solution:
i.
(𝑓 ∘ 𝑔)(𝑥)
∴ (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
= 𝑓(𝑥 − 1)
= √4 − (𝑥 − 1 )
= √5 − 𝑥
By the definition.
Substitute 𝑥 − 1 into the function 𝑓.
Simplify under the square root.
1
= (√5 − 𝑥 )2
ii.
(𝑔 ∘ 𝑓)(𝑥)
∴ (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥))
= 𝑔(√4 − 𝑥)
= √4 − 𝑥 − 1
By the definition.
Substitute √4 − 𝑥 into the function 𝑔.
1
2
= (√4 − 𝑥) − 1
iii.
iv.
(ℎ ∘ 𝑔)(𝑥)
∴ (ℎ ∘ 𝑔)(𝑥) = ℎ(𝑔(𝑥))
= ℎ(𝑥 − 1)
= (𝑥 − 1)2 + 2(𝑥 − 1) − 3
= 𝑥 2 − 2𝑥 + 1 + 2𝑥 − 2 − 3
= 𝑥2 − 4
(𝑓 ∘ ℎ)(𝑥)
∴ (𝑓 ∘ ℎ)(𝑥) = 𝑓(ℎ(𝑥))
= 𝑓(𝑥 2 + 2𝑥 − 3)
= √4 − (𝑥 2 + 2𝑥 − 3)
= √4 − 𝑥 2 − 2𝑥 + 3
By the definition.
Substitute 𝑥 − 1 into the function ℎ.
Remove the brackets.
By the definition.
Substitute 𝑥 2 + 2𝑥 − 3 into 𝑓.
Simplify under the radical sign.
= √7 − 𝑥 2 − 2𝑥
v.
(𝑓 ∘ 𝑔 ∘ ℎ)(𝑥)
∴ (𝑓 ∘ 𝑔 ∘ ℎ)(𝑥) = 𝑓(𝑔 ∘ ℎ(𝑥))
By the definition.
= 𝑓 (𝑔(ℎ(𝑥)))
Evaluate 𝑔(ℎ(𝑥)) first.
= 𝑓((𝑥 2 + 2𝑥 − 3) − 1)
= 𝑓(𝑥 2 + 2𝑥 − 4)
Simplify the inside part first.
− (𝑥 2
= √4
+ 2𝑥 − 4)
2
= √8 − 𝑥 − 2𝑥
Simplify under the radical sign.
51 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Four: Functions.
Worked Examples:
Example 11:
Consider the following functions,
𝑓(𝑥) = 𝑒 𝑥 , 𝑔(𝑥) = ln 𝑥 and ℎ(𝑥) = tan 𝑥 and 𝑘(𝑥) = tan−1 𝑥
i.
ii.
iii.
Show that (𝑓 ∘ 𝑔)(−1) = (𝑔 ∘ 𝑓)(−1) = −1
(𝑔 ∘ ℎ)(𝑥)
Prove that (ℎ ∘ 𝑘)(𝑥) = 𝑥
Solution:
i.
(𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
= 𝑓(ln 𝑥)
= 𝑒 ln 𝑥
=𝑥
and
∴ (𝑓 ∘ 𝑔)(−1) = 𝑥 = −1
Thus, (𝑓 ∘ 𝑔)(−1) = (𝑔 ∘ 𝑓)(−1) = −1
(𝑔 ∘ 𝑓) = 𝑔(𝑓(𝑥))
= 𝑔(𝑒 𝑥 )
= ln 𝑒 𝑥
=𝑥
∴ (𝑔 ∘ 𝑓)(−1) = 𝑥 = −1
ii.
(𝑔 ∘ ℎ)(𝑥) = 𝑔(ℎ(𝑥))
= 𝑔(tan 𝑥)
= ln tan 𝑥
By the composite definition.
Substitute tan 𝑥 into the function of 𝑔.
iii.
(ℎ ∘ 𝑘)(𝑥) = ℎ(𝑘(𝑥))
= ℎ(tan−1 𝑥)
= tan(tan−1 𝑥)
=𝑥
By the composite definition.
Substitute tan−1 𝑥 into the function of ℎ.
By the Inverse Trig equivalency.
52 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER FIVE:
THE LIMIT OF A FUNCTION:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS:
53 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Definition:
Suppose that 𝑓(𝑥) is defined when 𝑥 is near the number 𝑎. (This means that 𝑓 is defined on some
open interval that contains 𝑎, except possibly at 𝑎 itself.)
Then we write
𝐥𝐢𝐦 𝒇(𝒙) = 𝑳
𝒙→𝒂
And say, “The limit of 𝑓(𝑥), as 𝑥 approaches 𝑎, equals 𝐿 ”
If we can make the values of 𝑓(𝑥) arbitrary close to 𝐿 (as close to 𝐿 as we like) by taking 𝑥 to be
sufficiently close to 𝑎 (on either side of 𝑎) but not equal to 𝑎.
Direct Substitute Property:
lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
Worked Examples:
Example 12:
Evaluate the following limit using Direct Substitute property of Limits.
1. lim (5)
𝑥→2
2.
lim (𝑥 2 + 2𝑥 − 1)
𝑥→−2
3. lim (𝑥 3 − 3)(𝑥 + 2)
𝑥→1
4.
5.
𝑥 3 −3
lim ( 𝑥+2 )
𝑥→−1
𝑥 3 +2𝑥 2 −1
)
5−3𝑥
𝑥→−2
lim (
54 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
1. lim (5) = 5 , the limit of a constant, it’s a constant.
𝑥→2
2.
lim (𝑥 2 + 2𝑥 − 1)
𝑥→−2
= (−2)2 + 2(−2) − 1
Substitute 𝑥 = −2 into the function.
= −1
3. lim (𝑥 3 − 3)(𝑥 + 2)
𝑥→1
= (13 − 3)(1 + 2)
Substitute 𝑥 = 1 into the function.
= −6
4.
𝑥 3 −3
)
𝑥→−1 𝑥+2
lim (
=
(−1)3 −3
−1+2
Substitute 𝑥 = −1 into the function.
= −4
5.
𝑥 3 +2𝑥 2 −1
)
5−3𝑥
𝑥→−2
lim (
=
(−2)3 +2(−2)2 −1
5−3(−2)
Substitute 𝑥 = −2 into the function.
1
= − 11
NB: Note that in the above examples, the limits exists because when we substituted our 𝑥 approach,
the limits was defined. But sometimes you will counteract a lot of problem where the limit is not
defined, whereby you will be required to factorize and simplify where is applicable and rationalization is
of great recall at some problem that will be studied later in this chapter.
55 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Calculating Limits using the Limit Laws:
In this section we use the following properties of limits, called Limit Laws, to calculate
limits.
Limit Laws: Suppose that 𝑐 is a constant and the limits lim 𝑓(𝑥) and lim 𝑔(𝑥) exists.
𝑥→𝑎
𝑥→𝑎
Then:
1. lim 𝐶 = 𝐶
Constant Law
𝑥→𝑎
2. lim [𝑐 ∙ 𝑔(𝑥)] = 𝑐lim 𝑓(𝑥)
𝑥→𝑎
Constant multiple Law
𝑥→𝑎
3. lim [𝑓(𝑥) + 𝑔(𝑥)] = lim 𝑓(𝑥) + lim 𝑔(𝑥)
Sum Law
4. lim [𝑓(𝑥) − 𝑔(𝑥)] = lim 𝑓(𝑥) − lim 𝑔(𝑥)
Difference Law
5. lim [𝑓(𝑥) ∙ 𝑔(𝑥)] = lim 𝑓(𝑥) ∙ lim 𝑔(𝑥)
Product Law
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
𝑓(𝑥)
6. lim [ 𝑔(𝑥) ] =
𝑥→𝑎
lim 𝑓(𝑥)
𝑥→𝑎
lim 𝑔(𝑥)
𝑥→𝑎
Where 𝑔(𝑥) ≠ 0
𝑥→𝑎
𝑛
8. lim √𝑓 (𝑥) = 𝑛√ lim 𝑓(𝑥)
𝑥→𝑎
𝑥→𝑎
𝑛
𝑛
9. lim √𝑥 = √𝑎
𝑥→𝑎
Quotient Law
𝑥→𝑎
7. lim [ 𝑓(𝑥) ]𝑛 = [ lim 𝑓(𝑥) ]
𝑥→𝑎
𝑥→𝑎
𝑛
Where 𝑛 is a positive integer, Power Law.
Where 𝑛 is a positive integer, Radical Law.
Where 𝑎 > 0
56 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Worked Examples:
Example 13:
Evaluate the following limits:
𝑥 2 +𝑥−6
)
𝑥−2
𝑥→2
1. lim (
𝑥 4 −1
2. lim (𝑥 3 −1)
𝑥→1
𝑥−3
)
𝑥→3 𝑥 3 −27
3. lim (
√9+𝑥 −3
)
𝑥
4. lim (
𝑥→0
5.
𝑥 2 −9
lim (2𝑥 2 +7𝑥+3)
𝑥→−3
6. lim
(−5+𝑥)2 −25
𝑥
𝑥→0
7.
𝑥 2 +2𝑥+1
𝑥→−1 𝑥 4 −1
lim
𝑥 4 −𝑎4
8. lim 𝑥 2 −𝑎2
𝑥→𝑎
√4𝑥+1−3
𝑥→2 𝑥−2
9. lim
𝑥−2
10. lim (𝑥−2 + 3𝑥)
𝑥→2
57 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
𝑥 2 +𝑥−6
)
𝑥−2
𝑥→2
1. lim (
= lim
(𝑥+3)(𝑥−2)
𝑥−2
𝑥→2
Factorizing the numerator.
= lim (𝑥 + 3)
𝑥→2
=2+3
Substitute 𝑥 = 2 , taking away the limit
=5
𝑥 4 −1
)
𝑥→1 𝑥 3 −1
2. lim (
(𝑥 2 −1)(𝑥 2 +1)
= lim (𝑥−1)(𝑥2 +𝑥+1)
𝑥→1
= lim
(𝑥+1)(𝑥−1)(𝑥 2 +1)
𝑥→1 (𝑥−1)(𝑥 2 +𝑥+1)
= lim
(𝑥+1)(𝑥 2 +1)
𝑥→1 (𝑥 2 +𝑥+1)
=
(1+1)(1+1)
1+1+1
Factorize the numerator and the denominator.
Different of two squares factorization on the numerator
Simplification
Substitute 𝑥 = 1 , taking away the limit
4
=3
𝑥−3
3. lim (𝑥 3 −27)
𝑥→3
𝑥−3
= lim (𝑥−3)(𝑥2 +3𝑥+9)
𝑥→3
1
= lim 𝑥 2 +3𝑥+9
𝑥→3
=
1
9+9+9
Factorize the denominator (different of cubes).
Simplify before substituting the limit approach.
Substitute 𝑥 = 3 , taking away the limit
1
= 27
58 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
√9+𝑥 −3
)
𝑥
𝑥→0
4. lim (
= lim [
𝑥→0
√9+𝑥 −3
√9+𝑥+3
× 9+𝑥+3
𝑥
√
]
Rationalizing the numerator.
9+𝑥−9
]
√9+𝑥+3)
= lim [ 𝑥(
𝑥→0
𝑥
= lim
𝑥→0 𝑥(√9+𝑥+3)
1
= lim
𝑥→0 (√9+𝑥+3)
=
1
√9+0+3
Simplification.
You can now substitute the limit approach since it is defined.
Substitute 𝑥 = 0 , taking away the limit.
1
=6
5.
𝑥 2 −9
lim (2𝑥 2 +7𝑥+3)
𝑥→−3
= lim
(𝑥+3)(𝑥−3)
𝑥→−3 (2𝑥+1)(𝑥+3)
= lim
𝑥−3
𝑥→−3 2𝑥+1
=
−3−3
2(−3)+1
Different of two squares at the numerator.
You can now substitute the limit value since the limit is defined.
Substitute 𝑥 = −3 , taking away the limit.
6
=5
59 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
(−5+𝑥)2 −25
6. lim
𝑥
𝑥→0
25−10𝑥+𝑥 2 −25
𝑥
𝑥→0
𝑥 2 −10𝑥
lim 𝑥
𝑥→0
= lim
=
𝑥(𝑥−10)
𝑥
𝑥→0
Remove the brackets and simplify the numerator.
= lim
Take out the common factor at the numerator.
= lim (𝑥 − 10)
Simplify.
= 0 − 10
Substitute 𝑥 = 0 , taking away the limit.
𝑥→0
= −10
7.
𝑥 2 +2𝑥+1
𝑥→−1 𝑥 4 −1
lim
(𝑥+1)2
𝑥→−1 (𝑥+1)(𝑥−1)(𝑥 2 +1)
= lim
𝑥+1
= lim
𝑥→−1 (𝑥−1)(𝑥 2 +1)
0
= −4
Factorize and simplify.
Simplify.
Substitute 𝑥 = −1 , taking away the limit.
=0
𝑥 4 −𝑎4
8. lim 𝑥 2 −𝑎2
𝑥→𝑎
= lim
(𝑥 2 −𝑎 2 )(𝑥 2 +𝑎2 )
𝑥→𝑎
𝑥 2 −𝑎 2
Different of two squares at the numerator.
= lim (𝑥 2 + 𝑎2 )
𝑥→𝑎
= 𝑎2 + 𝑎2
Substitute 𝑥 = 𝑎 , taking away the limit.
= 2𝑎2
60 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
9. lim
𝑥→2
√4𝑥+1−3
𝑥−2
= lim [
𝑥→2
√4𝑥+1−3
√4𝑥+1+3
× 4𝑥+1+3 ]
𝑥−2
√
Rationalize the numerator.
4𝑥+1−9
]
√4𝑥+1+3)
= lim [ (𝑥−2)(
𝑥→2
4𝑥−8
]
√4𝑥+1+3)
= lim [ (𝑥−2)(
𝑥→2
Simplification.
4(𝑥−2)
]
√4𝑥+1+3)
= lim [ (𝑥−2)(
𝑥→2
= lim [
𝑥→2
=
4
(√4𝑥+1+3)
4
9+3
√
]
Simplification.
Substitute 𝑥 = 2 , taking away the limit.
2
=3
𝑥−2
10. lim (𝑥−2 + 3𝑥)
𝑥→2
= lim (1 + 3𝑥)
Simplify inside the brackets first.
= 1 + 3(2)
Substitute 𝑥 = 2 , taking away the limit.
𝑥→2
=7
61 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
INFINITE LIMITS AND LIMITS AT INFINITY:
Definition 1:
Let 𝑓 be a function defined on some interval (𝑎, ∞). Then
lim 𝑓(𝑥) = 𝐿
𝑥→∞
Means that the value of 𝑓(𝑥) can be made arbitrarily close to 𝐿 by taking 𝑥 sufficiently large.
Definition 2:
Let 𝑓 be a function defined on some interval (−∞, 𝑎). Then
lim 𝑓(𝑥) = 𝐿
𝑥→−∞
Means that the value of 𝑓(𝑥) can be made arbitrarily close to 𝐿 by taking 𝑥 sufficiently large negative.
Theorem: If 𝑟 > 0 is a rational number, then
lim
1
𝑥→∞ 𝑥 𝑟
=0
And If 𝑟 > 0 is a rational number such that 𝑥 𝑟 is defined for all 𝑥, then
lim
1
𝑥→−∞ 𝑥 𝑟
=0
NB: To evaluate the limit at infinity of any rational function, we first divide both the numerator and
denominator by the highest power of 𝑥 that occurs in the denominator.
We may assume that 𝑥 ≠ 0, since we are interested only in large values of 𝑥.
62 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Worked Examples:
Example 14:
Evaluate the following limits:
1.
√2𝑥 2 +1
𝑥→∞ 3𝑥−5
2.
√1+𝑥 2
𝑥→∞ 2𝑥+3
3.
lim
4.
5.
lim
lim
𝑥−√𝑥 2 −𝑥+1
𝑥→∞ 2𝑥−√4𝑥 2 +𝑥
lim [ √4𝑥 2 − 𝑥 + 1 − 2𝑥 ]
𝑥→∞
𝑥 2 +𝑥
𝑥→∞ 3−𝑥
lim
Solution:
NB: To evaluate the limit at infinity of any rational function, we first divide both the numerator and
denominator by the highest power of 𝑥 that occurs in the denominator.
We may assume that 𝑥 ≠ 0, since we are interested only in large values of 𝑥.
In some cases, the limits may be given in a form of radical, it is advisable that students should factor out
the highest degree and do some preliminary algebra.
63 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
1.
√2𝑥 2 +1
𝑥→∞ 3𝑥−5
lim
1
= lim
√𝑥 2 (2+ 2 )
𝑥
5
𝑥
𝑥→∞
𝑥(3− )
Take out the highest degree in both denominator and numerator.
1
= lim
𝑥√(2+ 2 )
𝑥
5
𝑥
𝑥→∞ 𝑥(3− )
Recall: √𝑎𝑏 = √𝑎 ∙ √𝑏 at the numerator.
1
= lim
√(2+ 2)
𝑥
5
𝑥
𝑥→∞ (3− )
2.
=
√2+0
3−0
=
√2
3
The 𝑥 undo each other and now the limit is defined.
Since
1
𝑥
→ 0 as 𝑥 → ∞
√1+𝑥 2
𝑥→∞ 2𝑥+3
lim
1
= lim
√𝑥 2 ( 2 +1)
𝑥
3
𝑥
𝑥→∞
𝑥(2+ )
Take out the highest degree in both denominator and numerator.
1
= lim
𝑥√( 2 +1)
𝑥
3
𝑥
𝑥→∞ 𝑥(2+ )
Recall: √𝑎𝑏 = √𝑎 ∙ √𝑏 at the numerator.
1
= lim
√( 2 +1)
𝑥
3
𝑥
𝑥→∞ (2+ )
=
√0+1
2+0
The 𝑥 undo each other and now the limit is defined.
Since
1
𝑥
→ 0 as 𝑥 → ∞
1
=2
64 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
3.
lim
𝑥−√𝑥 2 −𝑥+1
𝑥→∞ 2𝑥−√4𝑥 2 +𝑥
1
= lim
1
𝑥−√𝑥 2 (1− + 2)
𝑥 𝑥
𝑥→∞ 2𝑥−√𝑥 2 (4+1)
Take out 𝑥 2 as the highest degree under the radical sign.
𝑥
1
= lim
1
𝑥−𝑥√(1− + 2 )
𝑥 𝑥
𝑥→∞ 2𝑥−𝑥√(4+1)
Recall: √𝑎𝑏 = √𝑎 ∙ √𝑏 at the numerator.
𝑥
1
= lim
𝑥→∞
1
𝑥
𝑥→∞
=
Factor out 𝑥 in both denominator and numerator.
𝑥(2−√(4+ ))
1
= lim
1
𝑥(1−√(1− + 2 ))
𝑥 𝑥
1
(1−√(1−𝑥+ 2 ))
𝑥
1
The 𝑥 undo each other.
(2−√(4+𝑥))
1−√1−0+0
2−√4+0
Since
1
𝑥
→ 0 as 𝑥 → ∞.
0
0
= !
Thus the limit doesn’t exist.
65 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
4.
lim [ √4𝑥 2 − 𝑥 + 1 − 2𝑥 ]
𝑥→∞
= lim [ √4𝑥 2 − 𝑥 + 1 − 2𝑥 ×
𝑥→∞
= lim [
𝑥→∞
4𝑥 2 −𝑥+1−4𝑥 2
√4𝑥 2 −𝑥+1+2𝑥
𝑥→∞
1
𝑥→∞
1
𝑥
𝑥(−1+ )
1
𝑥→∞
5.
1
1
𝑥
(−1+ )
Rationalize the function.
]
Simplify.
]
Factor out 𝑥 in both denominator and numerator.
1
1
]
The 𝑥 undo each other.
(√(4−𝑥+ 2 )+2)
𝑥
(−1+0)
(√(4−0+0)+2)
=−
]
𝑥(√(4− + 2 )+2)
𝑥 𝑥
= lim [
=
1
√𝑥 2 (4− + 2 )+2𝑥
𝑥 𝑥
= lim [
√4𝑥 2 −𝑥+1+2𝑥
]
−𝑥+1
= lim [
√4𝑥 2 −𝑥+1+2𝑥
Since
1
𝑥
→ 0 as 𝑥 → ∞.
1
4
𝑥 2 +𝑥
𝑥→∞ 3−𝑥
lim
= lim
𝑥(𝑥+1)
3
𝑥
𝑥→∞ 𝑥( −1)
= lim
𝑥+1
3
𝑥→∞ 𝑥−1
∞
= −1
Factor out 𝑥 in both denominator and numerator.
The 𝑥 undo each other.
Because, 𝑥 + 1 → ∞
and
3
𝑥
− 1 → −1
= −∞
66 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Theorem of Limits/ Definition of Limits.
Definition:
lim 𝑓(𝑥) = 𝐿 Means for every 𝜀 > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 𝑎| < 𝛿 ,
𝑥→𝑎
Then |𝑓(𝑥) − 𝐿| < 𝜀.
Worked Examples:
Example 15:
Show from a definition that:
1. lim (2𝑥 + 1) = 3
𝑥→1
2. lim (4 − 3𝑥) = −2
𝑥→2
3. lim (5 − 2𝑥) = 1
𝑥→2
4. lim (3𝑥 − 1) = 2
𝑥→1
5. lim (1 − 3𝑥) = −5
𝑥→2
6.
lim |𝑥 + 4| = 1
𝑥→−3
67 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
1. lim (2𝑥 + 1) = 3
𝑥→1
For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 1| < 𝛿 , then |2𝑥 + 1 − 3| < 𝜀.
∴ Compare |𝑥 − 1| and |2𝑥 + 1 − 3|.
Simply: |2𝑥 + 1 − 3| = |2𝑥 − 2|
= |2(𝑥 − 1)|
= |2||𝑥 − 1|
∴ 2|𝑥 − 1| < 𝜀
𝜀
∴ |𝑥 − 1| < 2
Thus, 𝛿 =
𝜀
2
2. lim (4 − 3𝑥) = −2
𝑥→2
For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 2| < 𝛿 , then |4 − 3𝑥 + 2| < 𝜀.
∴ Compare |𝑥 − 2| and |4 − 3𝑥 + 2|.
Simply: |4 − 3𝑥 + 2| = |6 − 3𝑥|
= |−3(𝑥 − 2)|
= |−3||𝑥 − 2|
∴ 3|𝑥 − 2| < 𝜀
𝜀
∴ |𝑥 − 2| <
3
𝜀
Thus, 𝛿 = 3
3. lim (5 − 2𝑥) = 1
𝑥→2
For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 2| < 𝛿 , then |5 − 2𝑥 − 1| < 𝜀.
∴ Compare |𝑥 − 2| and |5 − 2𝑥 − 1|.
Simply: |5 − 2𝑥 − 1| = |4 − 2𝑥|
= |−2(𝑥 − 2)|
= |−2||𝑥 − 2|
∴ 2|𝑥 − 2| < 𝜀
𝜀
∴ |𝑥 − 2| <
Thus, 𝛿 =
𝜀
2
2
68 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
4. lim (3𝑥 − 1) = 2
𝑥→1
For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 1| < 𝛿 , then |3𝑥 − 1 − 2| < 𝜀.
∴ Compare |𝑥 − 1| and |3𝑥 − 1 − 2|.
Simply: |3𝑥 − 1 − 2| = |3𝑥 − 3|
= |3(𝑥 − 1)|
= |3||𝑥 − 1|
∴ 3|𝑥 − 1| < 𝜀
𝜀
∴ |𝑥 − 1| < 3
𝜀
Thus, 𝛿 = 3
5. lim (1 − 3𝑥) = −5
𝑥→2
For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 − 2| < 𝛿 , then |1 − 3𝑥 + 5| < 𝜀.
∴ Compare |𝑥 − 2| and |1 − 3𝑥 + 5|.
Simply: |1 − 3𝑥 + 5| = |6 − 3𝑥|
= |−3(𝑥 − 2)|
= |−3||𝑥 − 2|
∴ 3|𝑥 − 2| < 𝜀
𝜀
∴ |𝑥 − 2| < 3
𝜀
Thus, 𝛿 = 3
6.
lim |𝑥 + 4| = 1
𝑥→−3
For every > 0 , there exist a 𝛿 > 0 such that if 0 < |𝑥 + 3| < 𝛿 , then ||𝑥 + 4| − 1| < 𝜀.
∴ Compare |𝑥 + 3| and ||𝑥 + 4| − 1|.
Simply: ||𝑥 + 4| − 1| < 𝜀.
⟹ |𝑥 + 4| − |1| < ||𝑥 + 4| − 1| < 𝜀.
⟹ |𝑥 + 4| − |1| < 𝜀.
⟹ |𝑥 + 4| − 1 < 𝜀.
⟹ |𝑥 + 4| < 𝜀 + 1
⟹ −𝜀 − 1 < 𝑥 + 4 < 𝜀 + 1
From the modulus inequality property.
⟹ −𝜀 − 2 < 𝑥 + 3 < 𝜀 < 𝜀 + 2
Subtract 1 in both sides of the equations.
⟹ −𝜀 − 2 < 𝑥 + 3 < 𝜀 + 2
⟹ |𝑥 + 3| < 𝜀 + 2
Thus, 𝛿 = 𝜀 + 2
69 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
ONE SIDED LIMIT:
THEOREM: lim 𝑓(𝑥) = 𝐿
if and only if
𝑥→𝑎
lim 𝑓(𝑥) =
𝑥→𝑎 −
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎 +
In other words, the limit Exist if and only if the limit from the right is equal to the limit from the left.
Note:


When computing one-sided limits, we use the fact that the Limit Laws also holds for one-sided
limits.
If the limit computed from the right and the left are not equal, it means that the limit doesn’t
exist.
Worked Examples:

Example 16:
1. Determine whether the lim |𝑥| exists or not.
𝑥→0
2. Show that, lim
|𝑥|
𝑥→0 𝑥
does not exist
3. Consider the piecewise function and answer the questions below.
−1 − 𝑥
𝑓(𝑥) = {3𝑥 2 + 1
(𝑥 − 3)2
a) Compute lim− 𝑓(𝑥) and
𝑥→1
; 𝑥 ≤ −7
; 𝑥<1
; 𝑥≥1
lim 𝑓(𝑥)
𝑥→1+
b) Does the lim 𝑓(𝑥) exist?
𝑥→1
70 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
1. Determine whether the lim |𝑥| exists or not.
𝑥
Recall that, |𝑥| = {
−𝑥
𝑥→0
;
;
𝑥≥0
𝑥<0
from modulus definition.
Since, |𝑥| = 𝑥 for 𝑥 > 0, we have
∴ lim+|𝑥| = lim+ 𝑥
𝑥→0
𝑥→0
=0
For 𝑥 < 0 , we have |𝑥| = −𝑥 and so,
∴ lim−|𝑥| = lim+(−𝑥)
𝑥→0
𝑥→0
=0
Thus, the given limit, lim |𝑥| exist since lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 0
𝑥→0
2. Show that, lim
|𝑥|
𝑥→0 𝑥
𝑥→0
|𝑥|
𝑥
𝑥→0
does not exist.
Recall that, |𝑥| = {
∴ lim+
𝑥→0
𝑥
−𝑥
;
;
𝑥≥0
𝑥<0
from modulus definition.
𝑥
= lim+ 𝑥 , apply the same procedure used in the previous example.
𝑥→0
= lim+ 1 ,
𝑥→0
=1
∴ lim−
𝑥→0
|𝑥|
𝑥
= lim−
𝑥→0
−𝑥
𝑥
, from the definition of modulus.
= lim−(−1)
𝑥→0
= −1
Since the limit tend to have different limit values from the left and the right, then we have
shown that the limit doesn’t exist at 𝑥 = 0.
71 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
3. Consider the piecewise function and answer the questions below.
−1 − 𝑥
𝑓(𝑥) = {3𝑥 2 + 1
(𝑥 − 3)2
a)
; 𝑥 ≤ −7
; 𝑥<1
; 𝑥≥1
lim 𝑓(𝑥) = lim−(3𝑥 2 + 1) , note we only look where 𝑥 is less than 1.
𝑥→1−
𝑥→1
= 3(1)2 + 1
=4
∴ lim+ 𝑓(𝑥) = lim+(𝑥 − 3)2 , and here we only look where 𝑥 is greater than 1.
𝑥→1
𝑥→1
= (1 − 3)2
=4
b) Yes, the limit exist since lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 1.
𝑥→1
𝑥→1
Worked Examples:

Example 17:
cos2 𝑥 − sin2 𝑥
1. Let 𝑓(𝑥) = { 3
1 − 2 sin 4𝑥
2𝜋
; 0<𝑥≤ 3
;
𝑥=2
2𝜋
;
𝑥> 3
Discuss the existence of the limit, lim
𝑓(𝑥).
2𝜋
𝑥→
3
𝑎2 𝑒 2𝑥 − 8
2. Consider the function, 𝑓(𝑥) = {
𝑥
1
;
;
;
−2 < 𝑥 ≤ 0
𝑥=6
𝑥>0
Find the value of 𝑎 if it is given that, the limit exist at 𝑥 = 0.
72 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
(cos2 𝑥 − sin2 𝑥)
1. ∴ lim
𝑓(𝑥) = lim
2𝜋−
2𝜋−
𝑥→
𝑥→
3
3
2𝜋
2𝜋
= cos 2 ( 3 ) − sin2 ( 3 ) , switch your calculator in radians
=−
1
2
∴ lim+ 𝑓(𝑥) = lim+(1 − 2 sin 4𝑥 )
2𝜋
3
𝑥→
2𝜋
3
𝑥→
2𝜋
3
= 1 − 2 sin 4 ( )
= 1 − √3
Thus, the limit doesn’t exist since lim
𝑓(𝑥) ≠ lim+ 𝑓(𝑥)
2𝜋−
𝑥→
3
2𝜋
3
𝑥→
2. To find the value of 𝑎, one should remember the conditions on which the limit exists at a given
point, of which in our case is 𝑥 = 0.
We have to check, lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→0
𝑥→0
∴ lim− 𝑓(𝑥) = lim− (𝑎2 𝑒 2𝑥 − 8 )
𝑥→0
𝑥→0
= 𝑎2 𝑒 2(0) − 8
= 𝑎2 − 8
∴ lim+ 𝑓(𝑥) = lim+ 1
𝑥→0
𝑥→0
=1
Hence, 𝑎2 − 8 = 1
The limit from the left should be equal to the limit from the right.
∴ 𝑎2 − 9 = 0
∴ (𝑎 − 3)(𝑎 + 3) = 0
∴ 𝑎 = ±3
73 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
THEOREMS OF CONTINUITY AND PROOFS:
THEOREMS: If 𝑓 and 𝑔 are continuous functions at 𝑎 and 𝑐 is a constant, then the following
functions are also continuous at 𝑎.
1.
2.
3.
4.
5.
𝑓+𝑔
𝑓−𝑔
𝑐𝑓
𝑓𝑔
𝑓
where 𝑔 ≠ 0
𝑔
PROOFS:
1. lim(𝑓 + 𝑔)(𝑥) = lim [ 𝑓(𝑥) + 𝑔(𝑥) ]
𝑥→𝑐
𝑥→𝑐
= lim 𝑓(𝑥) + lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
= 𝑓(𝑐) + 𝑔(𝑐)
= (𝑓 + 𝑔)(𝑐)
Thus, this shows that 𝑓 + 𝑔 is continuous at 𝑐.
2. lim(𝑓 − 𝑔)(𝑥) = lim [ 𝑓(𝑥) − 𝑔(𝑥) ]
𝑥→𝑐
𝑥→𝑐
= lim 𝑓(𝑥) − lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
= 𝑓(𝑐) − 𝑔(𝑐)
= (𝑓 − 𝑔)(𝑐)
Thus, this shows that 𝑓 − 𝑔 is continuous at 𝑐.
3. lim(𝑐 ∙ 𝑓)(𝑥) = lim[ 𝑐 ∙ 𝑓(𝑥) ]
𝑥→𝑐
𝑥→𝑐
= 𝑐 ∙ lim 𝑓(𝑥)
𝑥→𝑐
= 𝑐 ∙ 𝑓(𝑐)
= (𝑐𝑓)(𝑐)
Thus, this shows that 𝑐𝑓 is continuous at 𝑐.
74 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
4.
lim(𝑓 ∙ 𝑔)(𝑥) = lim [ 𝑓(𝑥) ∙ 𝑔(𝑥) ]
𝑥→𝑐
𝑥→𝑐
= lim 𝑓(𝑥) ∙ lim 𝑔(𝑥)
𝑥→𝑐
𝑥→𝑐
= 𝑓(𝑐) ∙ 𝑔(𝑐)
= (𝑓 ∙ 𝑔)(𝑐)
Thus, this shows that 𝑓𝑔 is continuous at 𝑐.
𝑓
𝑔
𝑥→𝑐
𝑓(𝑥)
𝑔(𝑥)
𝑥→𝑐
5. lim ( ) (𝑥) = lim
lim 𝑓(𝑥)
= 𝑥→𝑐
lim 𝑔(𝑥)
𝑥→𝑐
𝑓(𝑐)
= 𝑔(𝑐)
𝑓
𝑔
= ( ) (𝑐) Provided that 𝑔(𝑐) ≠ 0.
𝑓
Thus, this shows that 𝑔 is continuous at 𝑐.
75 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
CONTINUITY AT A POINT:
THEOREM 1: A function 𝑓 is continuous at a number 𝑎 if lim 𝑓(𝑥) = 𝑓(𝑎)
𝑥→𝑎
Three conditions about continuity:
a) 𝑓(𝑎) Is defined/exists (that is 𝑎 is in the domain of 𝑓).
b) lim 𝑓(𝑥) Exists.
𝑥→𝑎
a) lim 𝑓(𝑥) = 𝑓(𝑎).
𝑥→𝑎
THEOREM 2: A function 𝑓 is continuous in an open interval (𝑎, 𝑏) if it is continuous at every
point at this interval.
THEOREM 3: A function 𝑓 is continuous in the closed interval [𝑎, 𝑏] if the following conditions
hold:
a) 𝑓 Is continuous at every 𝑥 in the interval (𝑎, 𝑏).
b) 𝑓(𝑎) And 𝑓(𝑏) both exist.
c)
lim 𝑓(𝑥) = 𝑓(𝑎) and
𝑥→𝑎 +
lim 𝑓(𝑥) = 𝑓(𝑏).
𝑥→𝑏−
76 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Worked Examples:

Example 18:
𝑥+1
1. Is the function 𝑓(𝑥) = {
2𝑥 − 1
; 𝑥<2
continuous at the point 𝑥 = 2?
; 𝑥≥2
2. determine whether the function 𝑓 is continuous at 𝑥 = 1
If 𝑓(𝑥) = {
𝑥−1
1
2𝑥 − 2
;
0≤𝑥<1
𝑥=1
;
𝑥>1
;
3. Show that the function 𝑓 is not continuous at 𝑥 = 2,
If 𝑓(𝑥) = {
𝑥+1
2𝑥 + 1
; 𝑥<2
; 𝑥≥2
4. Show that the function 𝑓, is continuous at a point 𝑥 = 2
−1 + 2𝑥
𝑥7
If 𝑓(𝑥) = {
1−𝑥
3 − 𝑥2
; 𝑥 < −1
; −1 ≤ 𝑥 ≤ 1
; 1<𝑥<2
; 𝑥≥2
5. Determine whether the function is continuous at 𝑥 = 2
2𝑥 2 − 4
𝑥2
If 𝑓(𝑥) = 6 − 𝑥
𝑥+3
{ √
;
;
;
0≤𝑥<3
𝑥=2
2<𝑥<5
;
𝑥<2
3
77 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
𝑥+1
1. Is the function 𝑓(𝑥) = {
2𝑥 − 1
i)
ii)
; 𝑥<2
continuous at the point 𝑥 = 2?
; 𝑥≥2
𝑓(2) = 2𝑥 − 1
= 2(2) − 1
=3
lim 𝑓(𝑥) = lim−(𝑥 + 1)
𝑥→2−
𝑥→2
And
lim 𝑓(𝑥) = lim+(2𝑥 − 1)
𝑥→2+
=2+1
=3
𝑥→2
= 2(2) − 1
=3
Since the two limits are the same.
Thus, the lim 𝑓(𝑥) = 3 and it exists.
𝑥→2
iii) lim 𝑓(𝑥) = 𝑓(2) = 3
𝑥→2
Hence, the function 𝑓 is continuous at 𝑥 = 2.
2. determine whether the function 𝑓 is continuous at 𝑥 = 1
𝑥−1
1
2𝑥 − 2
If 𝑓(𝑥) = {
i)
ii)
;
;
;
0≤𝑥<1
𝑥=1
𝑥>1
𝑓(1) = 1
lim 𝑓(𝑥) = lim−(𝑥 − 1)
𝑥→1−
𝑥→1
and
= 1−1
=0
lim 𝑓(𝑥) = lim+(2𝑥 − 2)
𝑥→1+
𝑥→1
= 2(1) − 2
=0
Because the two limits are the same.
Thus, the lim 𝑓(𝑥) = 0 and it exists.
𝑥→1
iii)
But lim 𝑓(𝑥) ≠ 𝑓(1) , i.e. 0 ≠ 1
𝑥→1
Thus 𝑓 is not continuous at 𝑥 = 1.
78 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
3. Show that the function 𝑓 is not continuous at 𝑥 = 2,
𝑓(𝑥) = {
i)
ii)
𝑥+1
2𝑥 + 1
; 𝑥<2
; 𝑥≥2
𝑓(2) = 2𝑥 + 1
= 2(2) + 1
=5
lim 𝑓(𝑥) = lim−(𝑥 + 1)
𝑥→2−
𝑥→2
And
lim 𝑓(𝑥) = lim+(2𝑥 + 1)
𝑥→2+
=2+1
=3
𝑥→2
= 2(2) + 1
=5
Because the two limits are not the same.
Thus, the lim 𝑓(𝑥) doesn’t exists.
𝑥→2
iii)
lim 𝑓(𝑥) ≠ 𝑓(2) ,
𝑥→2
Thus, the function 𝑓 is not continuous at 𝑥 = 2.
79 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
4. Show that the function 𝑓, is continuous at a point 𝑥 = 2
−1 + 2𝑥
𝑥7
𝑓(𝑥) = {
1−𝑥
3 − 𝑥2
i)
ii)
; 𝑥 < −1
; −1 ≤ 𝑥 ≤ 1
; 1<𝑥<2
; 𝑥≥2
𝑓(2) = 3 − 𝑥 2
= 3 − (2)2
= −1
lim 𝑓(𝑥) = lim−(1 − 𝑥)
𝑥→2−
𝑥→2
And
lim 𝑓(𝑥) = lim+(3 − 𝑥 2 )
𝑥→2+
=1−2
= −1
𝑥→2
= 3 − (2)2
= −1
Because the two limits are the same.
Thus, the lim 𝑓(𝑥) = −1 and it exists.
𝑥→2
iii)
lim 𝑓(𝑥) = 𝑓(2) = −1
𝑥→2
Hence, the function 𝑓 is continuous at 𝑥 = 2.
80 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
5. Determine whether the function is continuous at 𝑥 = 2
2𝑥 2 − 4
𝑥2
𝑓(𝑥) = 6 − 𝑥
𝑥+3
{ √
i)
ii)
;
;
;
0≤𝑥<3
𝑥=2
2<𝑥<5
;
𝑥<2
3
𝑓(2) = 𝑥 2
= 22
=4
lim 𝑓(𝑥) = lim−(2𝑥 2 − 4)
𝑥→2−
𝑥→2
And
= 2(2)2 − 4
=4
lim 𝑓(𝑥) = lim+(6 − 𝑥)
𝑥→2+
𝑥→2
= 6−2
=4
Because the two limits are the same.
Thus, the lim 𝑓(𝑥) = 4 and it exists.
𝑥→2
iii)
lim 𝑓(𝑥) = 𝑓(2) = 4
𝑥→2
Hence, the function 𝑓 is continuous at 𝑥 = 2.
81 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Worked Examples:

Example 19:
1. Determine the value(s) of 𝑚 if the following function is continuous at 𝑥 = 3
17
𝑓(𝑥) = { 𝑥 + 2𝑚
3𝑥 + 8
2. Let 𝑓(𝑥) = {
𝑥 2 −4
𝑥−2
;
2𝑘
;
;
;
;
𝑥<4
2<𝑥≤8
𝑥=3
𝑥≠2
𝑥=2
Find the value(s) of 𝑘 such that the function 𝑓 is continuous at 𝑥 = 2
3. Is the function 𝑓, continuous at 𝑥 = 2?
𝑥 3 − 2𝑥
𝑓(𝑥) = {
3−𝑥
;
;
𝑥>2
𝑥<2
4. For what value(s) of 𝑎 is the function continuous at 𝑥 = 4?
𝑎𝑥 + 5
−11
𝑓(𝑥) = {
𝑥(𝑥 − 1)
;
𝑥<4
; −2 ≤ 𝑥 ≤ 0
;
𝑥≥4
82 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
1. Determine the value of 𝑚 if the following function is continuous at 𝑥 = 3
17
𝑓(𝑥) = { 𝑥 + 2𝑚
3𝑥 + 8
i)
ii)
𝑓(3) = 3𝑥 + 8
= 3(3) + 8
= 17
lim 𝑓(𝑥) = lim−(17)
𝑥→3−
And
𝑥→3
= 17
iii)
; 𝑥<4
; 2<𝑥≤8
;
𝑥=3
lim 𝑓(𝑥) = lim+(𝑥 + 2𝑚)
𝑥→3+
𝑥→3
= 3 + 2𝑚
Since it is given that the function 𝑓 is continuous at 𝑥 = 3
Thus, lim 𝑓(𝑥) = 𝑓(3),
𝑥→3
∴ 𝑓(3) = lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→3
𝑥→3
∴ 17 = 17 = 3 + 2𝑚
∴ 3 + 2𝑚 = 17
∴𝑚=7
The limit will exists at 𝑥 = 7 and the other two conditions will be fulfilled. Therefore
the value of 𝑚 = 7.
83 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
2. Let 𝑓(𝑥) = {
𝑥 2 −4
𝑥−2
;
2𝑘
;
𝑥≠2
𝑥=2
Find the value(s) of 𝑘 such that the function 𝑓 is continuous at 𝑥 = 2
i)
𝑓(2) = 2𝑘
ii)
lim 𝑓(𝑥) = lim
𝑥 2 −4
𝑥→2 𝑥−2
(𝑥−2)(𝑥+2)
𝑥→2
= lim
𝑥−2
𝑥→2
= lim (𝑥 + 2)
𝑥→2
=4
iii)
𝑓(2) = lim 𝑓(𝑥) (the condition for continuity holds)
𝑥→2
∴ 2𝑘 = 4
∴𝑘=2
3. Is the function 𝑓, continuous at 𝑥 = 2?
𝑓(𝑥) = {
i)
𝑥 3 − 2𝑥
3−𝑥
;
;
𝑥>2
𝑥<2
Condition 1:
Does 𝑓(2) 𝐸𝑥𝑖𝑠𝑡𝑠?
No. The function of 𝑥 is greater than or less than 2, but not if 𝑥 is equal to 2. Therefore,
the function is not continuous at 𝑥 = 2. Notice that we don’t have to bother with the
other two conditions. Once you find a problem of this kind, the function is automatically
not continuous, and you can stop.
84 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
4. For what value(s) of 𝑎 is the function continuous at 𝑥 = 4?
𝑎𝑥 + 5
𝑓(𝑥) = { −11
𝑥(𝑥 − 1)
;
𝑥<4
; −2 ≤ 𝑥 ≤ 0
;
𝑥≥4
i)
𝑓(4) = 𝑥(𝑥 − 1)
= 4(4 − 1)
= 12 , the function passes the first condition.
ii)
For condition 2, lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→4
𝑥→4
∴ lim− 𝑓(𝑥) = lim−(𝑎𝑥 + 5)
𝑥→4
𝑥→4
= 4𝑎 + 5
iii)
And
lim 𝑓(𝑥) = lim+ 𝑥(𝑥 − 1)
𝑥→4 +
𝑥→4
= 4(4 − 1) = 12
Since it is given that the function 𝑓 is continuous at 𝑥 = 4
Thus, lim 𝑓(𝑥) = 𝑓(4),
𝑥→4
∴ 𝑓(4) = lim− 𝑓(𝑥) = lim+ 𝑓(𝑥)
𝑥→4
𝑥→4
∴ 12 = 4𝑎 + 5 = 12
∴ 4𝑎 + 5 = 12
7
∴𝑎=
4
7
The limit will exists at 𝑥 = 4 and the other two conditions will be fulfilled. Therefore
7
the value of 𝑎 = 4.
85 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
TRIGONOMETRIC LIMITS:
THEOREM:
lim
𝑥→0
sin 𝑥
𝑥
=1
,
𝑥
lim sin 𝑥 = 1
𝑥→0
and lim
𝑥→0
cos 𝑥−1
𝑥
=0
We use a geometric argument to prove the above theorems.
NB: To evaluate trigonometric limits, students should not temper with the angle but rather do some
preliminary algebra such as factorization, manipulation and rationalization.
Worked Examples:

Example 20:
2022
Evaluate the following limits.
1. lim
𝑥→0
sin 5𝑥
𝑥
1−cos 𝑥
𝑥→0 sin 𝑥
2. lim
sin(𝑥−1)
𝑥→1 𝑥 2 +𝑥−2
3. lim
4. lim (1 +
𝑥→0
5. lim [
𝑥→0
sin(
sin 𝑥
)
𝑥
𝑥−2
)
2
𝑥−2
]
1−cos2 2𝑥
)
𝑥2
𝑥→0
6. lim (
86 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
1. lim
𝑥→0
sin 5𝑥
𝑥
= lim
𝑥→0
= lim
𝑥→0
sin 5𝑥 5
∙5
𝑥
Always multiply by the coefficient of the angle but do not change the problem.
5sin 5𝑥
5𝑥
= 5 lim
𝑥→0
sin 5𝑥
5𝑥
= 5(1)
=1
1−cos 𝑥
𝑥→0 sin 𝑥
2. lim
= lim [
𝑥→0
1−cos 𝑥
sin 𝑥
1+cos 𝑥
× 1+cos 𝑥 ]
Rationalize the numerator and simplify.
1−cos2 𝑥
= lim [ sin 𝑥(1+cos 𝑥)]
𝑥→0
sin2 𝑥
= lim [ sin 𝑥(1+cos 𝑥)]
𝑥→0
Pythagorean identity: sin2 𝑥 = 1 − cos 2 𝑥
sin 𝑥
= lim [ (1+cos 𝑥)]
𝑥→0
=
sin 0
1+cos 𝑥
=
0
2
=0
87 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
sin(𝑥−1)
𝑥→1 𝑥 2 +𝑥−2
3. lim
sin(𝑥−1)
= lim (𝑥−1)(𝑥+2)
Remember not to temper with the angle.
𝑥→1
sin(𝑥−1)
1
× (𝑥+2)
(𝑥−1)
𝑥→1
= lim
Notice that we didn’t change the formality of the question.
sin(𝑥−1)
1
× lim (𝑥+2)
(𝑥−1)
𝑥→1
𝑥→1
= lim
By Limit Laws (Product of limit)
1
)
1+2
= (1) (
1
=3
4. lim (1 +
𝑥→0
sin 𝑥
)
𝑥
= lim 1 + lim
𝑥→0
𝑥→0
sin 𝑥
𝑥
By Limit Laws (Sum of the limit)
=1+1
=2
5. lim [
sin(
𝑥→0
𝑥−2
)
2
𝑥−2
= lim [
sin(
𝑥−2
)
2
𝑥−2
𝑥→0
1
𝑥→0
= 2 lim [
𝑥→0
1
× 21 ]
𝑥−2
)
2
𝑥−2
2
sin(
= lim [ 2
1
]
] Now the angle and the denominator are the same which holds the Theorem.
𝑥−2
)
2
𝑥−2
2
sin(
Preliminary Algebra
2
]
1
=2
88 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
1−cos2 2𝑥
)
𝑥2
𝑥→0
6. lim (
sin2 2𝑥
)
𝑥2
𝑥→0
Pythagorean identity: sin2 2𝑥 = 1 − cos2 2𝑥.
= lim (
sin 2𝑥
sin 2𝑥
)( 𝑥 )
𝑥
Manipulation.
sin 2𝑥
sin 2𝑥
) ∙ lim (
)
𝑥
𝑥
𝑥→0
By limit Laws.
= lim (
𝑥→0
= lim (
𝑥→0
sin 2𝑥
sin 2𝑥
) ∙ 2 lim ( 2𝑥 )
2𝑥
𝑥→0
= 2 lim (
𝑥→0
Look the procedure in example 1.
= (2)(1)(2)(1)
=4
Worked Examples:

Example 21:
2022
Evaluate the following limits.
cos 𝑥−1
)
𝑥
𝑥→0
1. lim (
2. lim
sin(𝑥2 −4)
𝑥−2
𝑥→2
3. lim
𝑥→2
sin(𝑥−2)
𝑥 2 −4
4. lim𝜋 [
𝑥→
sin 𝑥−cos 𝑥
cos 2𝑥
]
4
3 tan 3𝑥−2 sin 𝑥
]
𝑥
𝑥→0
5. lim [
sin 7𝑥
6. lim sin 3𝑥
𝑥→0
tan 𝑥
)]
𝑥
7. lim [(𝑥 2 + 1) (
𝑥→0
89 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
Solution:
cos 𝑥−1
)
𝑥
𝑥→0
1. lim (
cos 𝑥−1
cos 𝑥+1
× cos 𝑥+1)
𝑥
𝑥→0
= lim (
Rationalize the numerator and simplify.
cos2 𝑥−1
= lim (𝑥(cos 𝑥+1))
𝑥→0
−(1−cos2 𝑥)
= lim (
𝑥→0
𝑥(cos 𝑥+1)
)
−sin2 𝑥
= lim (𝑥(cos 𝑥+1))
𝑥→0
sin 𝑥
sin 𝑥
∙
𝑥
cos 𝑥+1
Manipulation
sin 𝑥
sin 𝑥
∙ lim cos 𝑥+1
𝑥
𝑥→0
𝑥→0
By limit Laws.
= − lim
𝑥→0
= − lim
0
)
1+1
= −(1) (
=0
2. lim
sin(𝑥 2 −4)
𝑥→2
= lim
𝑥→2
= lim
𝑥→2
= lim
𝑥→2
= lim
𝑥→2
𝑥−2
sin(𝑥 2 −4)
𝑥 2 −4
sin(𝑥 2 −4)
𝑥 2 −4
sin(𝑥 2 −4)
𝑥 2 −4
sin(𝑥 2 −4)
𝑥 2 −4
×
𝑥 2 −4
𝑥−2
Preliminary algebra.
𝑥 2 −4
𝑥→2 𝑥−2
∙ lim
∙ lim
𝑥→2
By limit Laws.
(𝑥−2)(𝑥+2)
𝑥−2
∙ lim (𝑥 + 2)
𝑥→2
= (1)(2 + 2)
=4
90 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
sin(𝑥−2)
𝑥→2 𝑥 2 −4
3. lim
sin(𝑥−2)
= lim (𝑥−2)(𝑥+2)
Remember not to temper with the angle.
𝑥→2
sin(𝑥−2)
1
× 𝑥+2
𝑥−2
𝑥→2
Manipulation.
sin(𝑥−2)
1
∙ lim 𝑥+2
𝑥−2
𝑥→2
𝑥→2
By Limit Laws.
= lim
= lim
1
= (1) (2+2)
=
1
4
4. lim𝜋 [
𝑥→
sin 𝑥−cos 𝑥
cos 2𝑥
]
4
= lim𝜋 [
𝑥→
sin 𝑥−cos 𝑥
cos2 𝑥−sin2 𝑥
]
4
sin 𝑥−cos 𝑥
= lim𝜋 [ (cos 𝑥−sin 𝑥)(cos 𝑥+sin 𝑥) ]
𝑥→
Double angle identity: cos 2𝑥 = cos2 𝑥 − sin2 𝑥.
Different of two squares factorization.
4
−(cos 𝑥−sin 𝑥)
= lim𝜋 [ (cos 𝑥−sin 𝑥)(cos 𝑥+sin 𝑥) ]
𝑥→
4
−1
= lim𝜋 [ (cos 𝑥+sin 𝑥) ]
𝑥→
=
4
−1
𝜋
4
𝜋
4
cos( )+sin( )
=−
√2
2
=−
1
√2
91 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
5. lim [
𝑥→0
3 tan 3𝑥−2 sin 𝑥
]
𝑥
= lim [
3 tan 3𝑥
𝑥
−
2 sin 𝑥
𝑥
]
= lim [
3 sin 3𝑥
𝑥 cos 3𝑥
−
2 sin 𝑥
𝑥
]
= lim [
3 sin 3𝑥
1
∙ cos 3𝑥
𝑥
𝑥→0
𝑥→0
𝑥→0
−
2 sin 𝑥
𝑥
3 sin 3𝑥
1
∙ lim cos 3𝑥
𝑥
𝑥→0
𝑥→0
𝑥→0
]
Manipulation.
2 sin 𝑥
𝑥→0 𝑥
= lim
= 9 lim
Preliminary algebra.
− lim
sin 3𝑥
1
∙ lim cos 3𝑥
3𝑥
𝑥→0
− 2 lim
𝑥→0
By Limit Laws.
sin 𝑥
𝑥
= (9)(1)(1) − 2
=7
6. lim
sin 7𝑥
𝑥→0 sin 3𝑥
= lim [
𝑥→0
sin 7𝑥 7𝑥
3𝑥
∙ 3𝑥 ∙ sin 3𝑥
7𝑥
]
sin 7𝑥
7𝑥
3𝑥
∙ lim 3𝑥 ∙ lim sin 3𝑥
7𝑥
𝑥→0
𝑥→0
𝑥→0
= lim
Notice that we didn’t change the formality of the question.
By Limit Laws.
sin 7𝑥
7
3𝑥
∙ lim 3 ∙ lim sin 3𝑥
7𝑥
𝑥→0
𝑥→0
𝑥→0
= lim
7
= (1) (3) (1)
=
By the theorem of trig limits.
7
3
92 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Five: The Limit of a functions.
tan 𝑥
)]
𝑥
7. lim [(𝑥 2 + 1) (
𝑥→0
tan 𝑥
)
𝑥
= lim (𝑥 2 + 1) ∙ lim (
𝑥→0
𝑥→0
sin 𝑥
𝑥
= lim (𝑥 2 + 1) ∙ lim (
𝑥→0
𝑥→0
= lim (𝑥 2 + 1) ∙ lim
𝑥→0
𝑥→0
By Limit Law.
1
× cos 𝑥)
sin 𝑥
1
∙ lim
𝑥
𝑥→0 cos 𝑥
Manipulation.
By Limit Law.
= (0 + 1)(1)(1)
=1
93 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER SIX:
DERIVATIVES OF ORDINARY FUNCTIONS:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS:
94 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
THE FIRST PRINCIPLE OF DIFFERENTIATION:
In high school we studied differentiation of ordinary function whereby we were introduced to the first
principle of differentiation, even though some texts books called it a definition of derivation. In this
chapter we will start from there. This chapter is very interesting provided that you have a passion for
mathematics. We will also prove some theorems that will help us to solve some complicated problems
for differentiation.
RECAP:
First Principle or Definition of derivatives.
𝒇(𝒙 + 𝒉) − 𝒇(𝒙)
𝒉→𝟎
𝒉
𝒇′(𝒙) = 𝐥𝐢𝐦
Worked Examples:

Example 22:
2022
Find the derivative of the following functions using the Definition or the first Principle.
a) 𝑓(𝑥) = 𝑥 2 − 4
b) 𝑓(𝑥) = 𝑥 3
1
c) 𝑓(𝑥) = 𝑥+1
d) 𝑓(𝑥) = (2𝑥 − 3)2
e) 𝑓(𝑥) =
1
√𝑥−1
95 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Solution:
a) 𝑓(𝑥) = 𝑥 2 − 4
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
ℎ→0
((𝑥+ℎ)2 −4)−(𝑥 2 −4)
∴ 𝑓′(𝑥) = lim
By the definition.
= lim
ℎ
𝑥 2 +2𝑥ℎ+ℎ2 −4−𝑥 2 +4
lim
ℎ
ℎ→0
2𝑥ℎ+ℎ2
lim ℎ
ℎ→0
ℎ(2𝑥+ℎ)
lim ℎ
ℎ→0
ℎ→0
=
=
=
= lim (2𝑥 + ℎ)
ℎ→0
= 2𝑥 + 0
= 2𝑥
as ℎ → 0
b) 𝑓(𝑥) = 𝑥 3
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
(𝑥+ℎ)3 −𝑥 3
lim
ℎ
ℎ→0
𝑥 3 +3𝑥 2 ℎ+3𝑥ℎ 2 +ℎ3 −𝑥 3
lim
ℎ
ℎ→0
3𝑥 2 ℎ+3𝑥ℎ 2 +ℎ3
lim
ℎ
ℎ→0
ℎ(3𝑥 2 +3𝑥ℎ+ℎ 2 )
∴ 𝑓′(𝑥) = lim
ℎ→0
=
=
=
= lim
ℎ→0
By the definition.
ℎ
= lim (3𝑥 2 + 3𝑥ℎ + ℎ2 )
ℎ→0
= 3𝑥 2 + 0 + 0 as ℎ → 0
= 3𝑥 2
96 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
1
c) 𝑓(𝑥) = 𝑥+1
By the definition.
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
ℎ→0
∴ 𝑓′(𝑥) = lim
= lim
1
1
−
𝑥+ℎ+1 𝑥+1
ℎ
1
1
1
lim ℎ (𝑥+ℎ+1 − 𝑥+1)
ℎ→0
1 𝑥+1−(𝑥+ℎ+1)
lim (
)
ℎ→0 ℎ (𝑥+ℎ+1)(𝑥+1)
1
−ℎ
lim ℎ ((𝑥+ℎ+1)(𝑥+1))
ℎ→0
−1
lim (
)
ℎ→0 (𝑥+ℎ+1)(𝑥+1)
−1
as ℎ
(𝑥+0+1)(𝑥+1)
1
− (𝑥+1)2
ℎ→0
=
=
=
=
=
=
→ 0.
d) 𝑓(𝑥) = (2𝑥 − 3)2
∴ 𝑓′(𝑥) = lim
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
By the definition.
∴ 𝑓(𝑥 + ℎ) = (2(𝑥 + ℎ) − 3)2
= (2𝑥 + 2ℎ − 3)2
= 4𝑥 2 + 8𝑥ℎ − 12𝑥 − 12ℎ + 9 + 4ℎ2
∴ 𝑓(𝑥) = (2𝑥 − 3)2
= 4𝑥 2 − 12𝑥 + 9
∴ 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = (4𝑥 2 + 8𝑥ℎ − 12𝑥 − 12ℎ + 9 + 4ℎ2 ) − (4𝑥 2 − 12𝑥 + 9 )
= 8𝑥ℎ − 12ℎ + 4ℎ2
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
8𝑥ℎ−12ℎ+4ℎ2
lim (
)
ℎ
ℎ→0
Thus, 𝑓′(𝑥) = lim
ℎ→0
=
= lim (8𝑥 − 12 + 4ℎ)
ℎ→0
= 8𝑥 − 12 + 0
= 8𝑥 − 12
as ℎ → 0.
97 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
e) 𝑓(𝑥) =
1
√𝑥−1
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
∴ 𝑓′(𝑥) = lim
ℎ→0
∴ 𝑓(𝑥 + ℎ) =
1
By the definition.
And 𝑓(𝑥) =
√𝑥+ℎ−1
1
√𝑥−1
1
1
−
√𝑥+ℎ−1 √𝑥−1
= lim
ℎ
ℎ→0
1
1
= lim ℎ (
√𝑥+ℎ−1
ℎ→0
−
1
)
√𝑥−1
1
√𝑥−1−√𝑥+ℎ−1
)
(√𝑥+ℎ−1)(√𝑥−1)
1
√𝑥−1−√𝑥+ℎ−1
√𝑥−1+√𝑥+ℎ−1
) ( 𝑥−1+√𝑥+ℎ−1)
(√𝑥+ℎ−1)(√𝑥−1)
√
= lim ℎ (
ℎ→0
= lim ℎ (
ℎ→0
1
𝑥−1−(𝑥+ℎ+1)
)
ℎ
(√𝑥+ℎ−1)(√𝑥−1)(√𝑥−1)+(√𝑥−1)(√𝑥+ℎ−1)(√𝑥+ℎ−1)
ℎ→0
= lim (
1
−ℎ
= lim ℎ (
)
(√𝑥+ℎ−1)(√𝑥−1)(√𝑥−1)+(√𝑥−1)(√𝑥+ℎ−1)(√𝑥+ℎ−1)
ℎ→0
−1
= lim (
)
ℎ→0 (√𝑥+ℎ−1)(√𝑥−1)(√𝑥−1)+(√𝑥−1)(√𝑥+ℎ−1)(√𝑥+ℎ−1)
=
=
−1
(√𝑥+0−1)(√𝑥−1)(√𝑥−1)+(√𝑥−1)(√𝑥+0−1)(√𝑥+0−1)
as ℎ → 0.
−1
(√𝑥−1)(√𝑥−1)(√𝑥−1)+(√𝑥−1)(√𝑥−1)(√𝑥−1)
=−
1
3
3
(𝑥−1)2 +(𝑥−1)2
1
= −3
√(𝑥−1)2
=−
3
+ √(𝑥−1)2
1
3
2 √(𝑥−1)2
98 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
THE EXISTENCE OF THE DERIVATIVES:
For the derivative to exists at a given point or a number 𝑐, the following property should holds:
𝒇(𝒄 + 𝒉) − 𝒇(𝒄)
𝒉→𝟎
𝒉
𝒇′− (𝒄) = 𝒇′+ (𝒄) = 𝐥𝐢𝐦
Worked Examples:

Example 23:
2022
Find 𝑓′(𝑐) if it exists:
a) 𝑓(𝑥) = {
3𝑥 2
2𝑥 3 + 1
b) 𝑓(𝑥) = {
3𝑥 2 + 1
𝑥3 + 1
c) 𝑓(𝑥) = {
− 2 𝑥2
−3
1
;
;
𝑥≤1
𝑥≥1
at 𝑐 = 1
;
;
𝑥≤0
0<𝑥<1
at 𝑐 = 0
;
;
𝑥<3
𝑥≥0
at 𝑐 = 3
99 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Solution:
a) 𝑓(𝑥) = {
3𝑥 2
2𝑥 3 + 1
;
;
𝑥≤1
𝑥≥1
at 𝑐 = 1
We want to show that, 𝑓′− (𝑐) = 𝑓′+ (𝑐) = lim
ℎ→0
∴ 𝑓′− (𝑐) = lim−
ℎ→0
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
𝑓(1+ℎ)−𝑓(1)
ℎ
3(1+ℎ)2 −3(1)2
lim
ℎ
ℎ→0−
3+6ℎ+3ℎ2 −3
lim
ℎ
ℎ→0−
6ℎ+3ℎ2
lim
ℎ
ℎ→0−
∴ 𝑓′− (1) = lim−
ℎ→0
=
=
=
= lim−(6 + 3ℎ)
ℎ→0
=6
∴ 𝑓′+ (𝑐) = lim+
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
∴ 𝑓′+ (1) = lim+
𝑓(1+ℎ)−𝑓(1)
ℎ
2(1+ℎ)3 +1−(2(1)3 +1)
ℎ→0
ℎ→0
= lim+
ℎ
2+6ℎ+6ℎ2 +2ℎ 3+1−3
lim
ℎ
ℎ→0+
6ℎ+6ℎ2 +2ℎ3
lim
ℎ
ℎ→0+
lim (6 + 6ℎ + 2ℎ2 )
ℎ→0+
ℎ→0
=
=
=
=6
Thus, 𝑓′(1) = 6 and it exists
100 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
b) 𝑓(𝑥) = {
3𝑥 2 + 1
𝑥3 + 1
;
;
𝑥≤0
0<𝑥<1
at 𝑐 = 0
We want to show that, 𝑓′− (𝑐) = 𝑓′+ (𝑐) = lim
ℎ→0
∴ 𝑓′− (𝑐) = lim−
ℎ→0
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
𝑓(0+ℎ)−𝑓(0)
ℎ
3(0+ℎ)2 +1−3(0)2 −1
lim
ℎ
ℎ→0−
3ℎ 2 +1−1
lim
ℎ
ℎ→0−
3ℎ 2
lim
ℎ→0− ℎ
∴ 𝑓′− (0) = lim−
ℎ→0
=
=
=
= lim−(3ℎ)
ℎ→0
=0
∴ 𝑓′+ (0) = lim+
ℎ→0
= lim+
ℎ→0
=
=
=
𝑓(0+ℎ)−𝑓(0)
ℎ
(0+ℎ)3 +1−(0)3 −1
ℎ
ℎ 3 +1−1
lim+ ℎ
ℎ→0
ℎ3
lim+ ℎ
ℎ→0
lim (ℎ2 )
ℎ→0+
=0
Thus, 𝑓′(0) = 0 and it exists
101 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
1
c) 𝑓(𝑥) = {
− 𝑥2
2
−3
;
;
𝑥<3
𝑥≥0
at 𝑐 = 3
We want to show that, 𝑓′− (𝑐) = 𝑓′+ (𝑐) = lim
ℎ→0
∴ 𝑓′− (𝑐) = lim−
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
∴ 𝑓′− (3) = lim−
𝑓(3+ℎ)−𝑓(3)
ℎ
ℎ→0
ℎ→0
= lim−
ℎ→0
= lim−
1
2
1
2
− (3+ℎ)2 −(− (3)2 )
ℎ
9
ℎ2 9
− −3ℎ− +
2
2 2
ℎ
ℎ→0
= lim−
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
−3ℎ−
ℎ2
2
ℎ
ℎ→0
ℎ
2
= lim− (3 − )
ℎ→0
=0
∴ 𝑓′+ (𝑐) = lim+
ℎ→0
𝑓(𝑐+ℎ)−𝑓(𝑐)
ℎ
𝑓(3+ℎ)−𝑓(3)
ℎ
−3−(−3)
lim
ℎ
ℎ→0+
0
lim
ℎ→0+ ℎ
∴ 𝑓′+ (3) = lim+
ℎ→0
=
=
= lim+ 0
ℎ→0
=0
Thus, 𝑓′(3) = 0 and it exists.
102 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
RULES FOR DIFFERENTITATION:
1. Derivatives of a constant function:
If 𝑓(𝑥) = 𝑐 where 𝑐 is a constant, then 𝑓′(𝑥) = 0.
Proof:
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
𝑐−𝑐
lim
ℎ→0 ℎ
0
lim ℎ
ℎ→0
𝑓′(𝑥) = lim
ℎ→0
=
=
= lim 0
ℎ→0
=0
2. The constant multiple rule:
If 𝑐 is a constant and 𝑓 is a differential function then,
𝑑
(𝑐
𝑑𝑥
∙ 𝑓(𝑥)) = 𝑐 ∙ 𝑓′(𝑥)
Proof:
Let 𝑔(𝑥) = 𝑐𝑓(𝑥) then 𝑔′(𝑥) = 𝑐𝑓′(𝑥)
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ
𝑐𝑓(𝑥+ℎ)−𝑐𝑓(𝑥)
lim
ℎ
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
lim 𝑐 (
)
ℎ
𝑥→𝑐
𝑓(𝑥+ℎ)−𝑓(𝑥)
𝑐 lim
ℎ
ℎ→0
𝐿𝐻𝑆 = 𝑔′(𝑥) = lim
ℎ→0
=
=
=
= 𝑐𝑓′(𝑥)
= 𝑅𝐻𝑆.
3. The power rule:
𝑑
(𝑥 𝑛 )
𝑑𝑥
= 𝑛𝑥 𝑛−1
(No proof is required, one should be very careful when using BINOMIAL THEOREM).
103 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
4. The Sum Rule:
If 𝑓 and 𝑔 are differentiable functions, then
𝑑
(𝑓(𝑥) +
𝑑𝑥
𝑑
𝑑
𝑔(𝑥)) = 𝑑𝑥 𝑓(𝑥) + 𝑑𝑥 𝑔(𝑥)
Proof:
Let 𝑟(𝑥) = 𝑓(𝑥) + 𝑔(𝑥) , then 𝑟′(𝑥) = 𝑓′(𝑥) + 𝑔′(𝑥)
𝑟(𝑥+ℎ)−𝑟(𝑥)
ℎ
ℎ→0
𝐿𝐻𝑆 = 𝑟′(𝑥) = lim
𝑓(𝑥+ℎ)+𝑔(𝑥+ℎ)−(𝑓(𝑥)+𝑔(𝑥))
= lim (
ℎ
ℎ→0
)
𝑓(𝑥+ℎ)+𝑔(𝑥+ℎ)−𝑓(𝑥)−𝑔(𝑥)
)
ℎ
= lim (
ℎ→0
(𝑓(𝑥+ℎ)−𝑓(𝑥))+(𝑔(𝑥+ℎ)−𝑔(𝑥))
= lim (
ℎ→0
ℎ
)
(𝑔(𝑥+ℎ)−𝑔(𝑥))
𝑓(𝑥+ℎ)−𝑓(𝑥)
+
)
ℎ
ℎ
= lim (
ℎ→0
= lim
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
𝑔(𝑥+ℎ)−𝑔(𝑥)
+ lim
ℎ
ℎ
ℎ→0
= 𝑓′(𝑥) + 𝑔′(𝑥)
= 𝑅𝐻𝑆.
104 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Worked Examples:

Example 24:
2022
Sum and Power rule:
Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 = 𝑥 7 − 2𝑥 5 + 10
b) 𝑦 =
2𝑥 2 −3𝑥
√𝑥
− 5𝑥 −3
c) 𝑦 = √𝑥(2 − 5𝑥)
3
2
d) 𝑦 = (√𝑥 − 2𝑥 2 )
4
2
e) 𝑦 = √𝑥 − 𝑥 2 + 3𝑥 − 7𝑥 −3
105 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Solution:
a) 𝑦 = 𝑥 7 − 2𝑥 5 + 10
𝑑𝑦
∴ 𝑑𝑥 = 7𝑥 6 − 10𝑥 4
b) 𝑦 =
2𝑥 2 −3𝑥
√𝑥
Power and Sum rule differentiation
− 5𝑥 −3
3
1
𝑦 = 2𝑥 2 − 3𝑥 2 − 5𝑥 −3
∴
𝑑𝑦
𝑑𝑥
1
2
1
3
= 3𝑥 − 𝑥 −2 + 15𝑥 −4
2
3
15
= 3√𝑥 − 2 𝑥 + 𝑥 4
√
Simplify the problem first, using the exponents law.
Apply the derivatives
c) 𝑦 = √𝑥(2 − 5𝑥)
3
1
𝑦 = 2𝑥 2 − 5𝑥 2
1
Multiply throughout by 𝑥 2
1
𝑑𝑦
15 1
𝑥2
2
15√𝑥
2
∴ 𝑑𝑥 = 𝑥 −2 −
=
1
√𝑥
−
3
Apply the derivatives
2
d) 𝑦 = (√𝑥 − 2𝑥 2 )
𝑦 = 𝑥 − 4𝑥 2 + 4𝑥 3
∴
𝑑𝑦
𝑑𝑥
Remove the brackets.
= 1 − 8𝑥 + 12𝑥
2
2
e) 𝑦 = 4√𝑥 − 𝑥 2 + 3𝑥 − 7𝑥 −3
1
𝑦 = 𝑥 4 − 2𝑥 −2 + 3𝑥 − 7𝑥 −3
∴
𝑑𝑦
𝑑𝑥
=
=
1 −3
𝑥 4
4
1
4
4 √𝑥 3
Rewrite the equation in the simplest form.
+ 4𝑥 −3 + 3 + 21𝑥 −4
+
4
𝑥3
+
21
𝑥4
+3
106 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
5. The Product Rule:
If 𝑓 and 𝑔 are differentiable functions,
𝑑
(𝑔(𝑥) ∙
𝑑𝑥
𝑟(𝑥)) = 𝑔(𝑥) ∙ 𝑟′(𝑥) + 𝑟(𝑥) ∙ 𝑔′(𝑥)
Proof:
Let 𝑓(𝑥) = 𝑔(𝑥) ∙ 𝑟(𝑥) then, 𝑓′(𝑥) = 𝑔(𝑥) ∙ 𝑟′(𝑥) + 𝑟(𝑥) ∙ 𝑔′(𝑥)
𝑓′(𝑥) = lim
ℎ→0
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
= lim
𝑔(𝑥+ℎ)∙𝑟(𝑥+ℎ)−𝑔(𝑥)∙𝑟(𝑥)
ℎ
= lim
𝑔(𝑥+ℎ)∙𝑟(𝑥+ℎ)−𝑔(𝑥+ℎ)∙𝑟(𝑥)+𝑔(𝑥+ℎ)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟(𝑥)
ℎ
ℎ→0
ℎ→0
= lim
[𝑔(𝑥+ℎ)∙𝑟(𝑥+ℎ)−𝑔(𝑥+ℎ)∙𝑟(𝑥)]+[𝑔(𝑥+ℎ)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟(𝑥)]
ℎ
ℎ→0
[𝑔(𝑥+ℎ)∙𝑟(𝑥+ℎ)−𝑔(𝑥+ℎ)∙𝑟(𝑥)]
= lim (
ℎ
ℎ→0
= lim
[𝑔(𝑥+ℎ)∙𝑟(𝑥+ℎ)−𝑔(𝑥+ℎ)∙𝑟(𝑥)]
ℎ→0
ℎ
+
[𝑔(𝑥+ℎ)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟(𝑥)]
+ lim
ℎ→0
)
ℎ
[𝑔(𝑥+ℎ)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟(𝑥)]
ℎ
𝑟(𝑥+ℎ)−𝑟(𝑥)
𝑔(𝑥+ℎ)−𝑔(𝑥)
) + lim (𝑟(𝑥)) (
)
ℎ
ℎ
ℎ→0
= lim (𝑔(𝑥 + ℎ)) (
ℎ→0
= lim (𝑔(𝑥 + ℎ)) ∙ lim
ℎ→0
ℎ→0
𝑟(𝑥+ℎ)−𝑟(𝑥)
𝑔(𝑥+ℎ)−𝑔(𝑥)
+ lim 𝑟(𝑥) ∙ lim
ℎ
ℎ
ℎ→0
ℎ→0
= 𝑔(𝑥) ∙ 𝑟′(𝑥) + 𝑟(𝑥) ∙ 𝑔′(𝑥)
= 𝑅𝐻𝑆.
107 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Worked Examples:

Example 25:
2022
Product Rule:
Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 = (𝑥 + 2)(𝑥 − 3)
b) 𝑦 = (𝑥 − 1)2 (2𝑥 + 5)
c) 𝑦 = (2𝑥 3 − 4𝑥)(𝑥 5 − 5𝑥)
d) 𝑦 = (√𝑥(2 − 5𝑥)) (2𝑥 + 5)
e) 𝑦 = ( 4√𝑥 −
2
𝑥2
3
2
+ 3𝑥 − 7𝑥 −3 ) ((√𝑥 − 2𝑥 2 ) )
Solution:
a) 𝑦 = (𝑥 + 2)(𝑥 − 3)
∴
𝑑𝑦
𝑑
𝑑
= (𝑥 − 3) (𝑥 + 2) + (𝑥 + 2) (𝑥 − 3)
𝑑𝑥
𝑑𝑥
𝑑𝑥
= (𝑥 − 3)(1) + (𝑥 + 2)(1)
= 2𝑥 − 1
b) 𝑦 = (𝑥 − 1)2 (2𝑥 + 5)
𝑦 = (𝑥 2 − 2𝑥 + 1)(2𝑥 + 5)
∴
𝑑𝑦
𝑑
𝑑
= (2𝑥 + 5) (𝑥 2 − 2𝑥 + 1) + (𝑥 2 − 2𝑥 + 1) (2𝑥 + 5)
𝑑𝑥
𝑑𝑥
𝑑𝑥
= (2𝑥 + 5)(2𝑥 − 2) + (𝑥 2 − 2𝑥 + 1)(2)
108 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
c) 𝑦 = (2𝑥 3 − 4𝑥)(𝑥 5 − 5𝑥)
𝑑𝑦
𝑑
𝑑
∴ 𝑑𝑥 = (𝑥 5 − 5𝑥) 𝑑𝑥 (2𝑥 3 − 4𝑥) + (2𝑥 3 − 4𝑥) 𝑑𝑥 (𝑥 5 − 5𝑥)
= (𝑥 5 − 5𝑥)(6𝑥 2 − 4) + (2𝑥 3 − 4𝑥)(5𝑥 4 − 5)
d) 𝑦 = (√𝑥(2 − 5𝑥)) (2𝑥 + 5)
∴
𝑑𝑦
𝑑
𝑑
= (2𝑥 + 5) (√𝑥(2 − 5𝑥)) + (√𝑥(2 − 5𝑥)) (2𝑥 + 5)
𝑑𝑥
𝑑𝑥
𝑑𝑥
= (2𝑥 + 5) (
1
√𝑥
4
15√𝑥
)+
2
−
(√𝑥(2 − 5𝑥)) (2)
3
2
2
e) 𝑦 = ( √𝑥 − 𝑥 2 + 3𝑥 − 7𝑥 −3 ) ((√𝑥 − 2𝑥 2 ) )
∴
3 2 𝑑
3 2
𝑑𝑦
2
2
𝑑
4
4
= (√𝑥 − 2𝑥 2 )
( √𝑥 − 2 + 3𝑥 − 7𝑥 −3 ) + ( √𝑥 − 2 + 3𝑥 − 7𝑥 −3 ) (√𝑥 − 2𝑥 2 )
𝑑𝑥
𝑑𝑥
𝑥
𝑥
𝑑𝑥
3
2
= (√𝑥 − 2𝑥 2 ) (
1
4
4 √𝑥 3
4
21
2
+ 𝑥 3 + 𝑥 4 + 3 ) + ( 4√𝑥 − 𝑥 2 + 3𝑥 − 7𝑥 −3 ) (1 − 8𝑥 + 12𝑥 2 )
109 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
6. Quotient Rule:
If 𝑓 and 𝑔 are differentiable functions then,
𝑑 𝑔(𝑥)
[
]
𝑑𝑥 𝑟(𝑥)
=
𝑔′(𝑥)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟′(𝑥)
[𝑟(𝑥)]2
Proof:
Let 𝑓(𝑥) =
𝑔(𝑥)
𝑓′(𝑥) = lim
such that 𝑟(𝑥) ≠ 0, then 𝑓′(𝑥) =
𝑟(𝑥)
ℎ
𝑔(𝑥+ℎ) 𝑔(𝑥)
−
𝑟(𝑥+ℎ) 𝑟(𝑥)
ℎ
ℎ→0
= lim
𝑟(𝑥)𝑔(𝑥+ℎ)−𝑔(𝑥)𝑟(𝑥+ℎ)
ℎ 𝑟(𝑥) 𝑟(𝑥+ℎ)
ℎ→0
= lim
𝑟(𝑥)𝑔(𝑥+ℎ)−𝑔(𝑥)𝑟(𝑥+ℎ)+𝑟(𝑥)𝑔(𝑥)−𝑟(𝑥)𝑔(𝑥)
ℎ 𝑟(𝑥) 𝑟(𝑥+ℎ)
ℎ→0
= lim
𝑟(𝑥)[𝑔(𝑥+ℎ)−𝑔(𝑥)]−𝑔(𝑥)[𝑟(𝑥+ℎ)−𝑟(𝑥)]
ℎ 𝑟(𝑥) 𝑟(𝑥+ℎ)
ℎ→0
= lim
𝑟(𝑥)(
𝑔(𝑥+ℎ)−𝑔(𝑥)
𝑟(𝑥+ℎ)−𝑟(𝑥)
)−𝑔(𝑥)(
)
ℎ
ℎ
𝑟(𝑥) 𝑟(𝑥+ℎ)
ℎ→0
=
[𝑟(𝑥)]2
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ→0
= lim
𝑔′(𝑥)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟′(𝑥)
𝑟(𝑥) lim
ℎ→0
𝑔(𝑥+ℎ)−𝑔(𝑥)
𝑟(𝑥+ℎ)−𝑟(𝑥)
−𝑔(𝑥) lim
ℎ
ℎ
ℎ→0
𝑟(𝑥) lim 𝑟(𝑥+ℎ)
ℎ→0
=
=
𝑔′(𝑥)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟′(𝑥)
𝑟(𝑥)∙𝑟(𝑥)
𝑔′(𝑥)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟′(𝑥)
[𝑟(𝑥)]2
110 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Worked Examples:

Example 26:
2022
Quotient Rule:
Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 =
𝑥−3
𝑥+9
b) 𝑦 =
𝑥 2 +6𝑥−2
𝑥 4 +2𝑥 2 +8
c) 𝑦 =
d) 𝑦 =
(𝑥−1)2
2𝑥−7
√𝑥(2−5𝑥)
3 2
(√𝑥−2𝑥 2 )
2
4
e) 𝑦 =
√𝑥 −𝑥2 +3𝑥−7𝑥 −3
2𝑥2 −3𝑥
−5𝑥 −3
√𝑥
Solution:
𝑥−3
a) 𝑦 = 𝑥+9
∴
𝑑𝑦
𝑑𝑥
=
=
(𝑥+9)
𝑑
𝑑
(𝑥−3)−(𝑥−3) (𝑥+9)
𝑑𝑥
𝑑𝑥
(𝑥+3)2
(𝑥+9)(1)−(𝑥−3)(1)
(𝑥+3)2
111 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
𝑥 2 +6𝑥−2
b) 𝑦 = 𝑥 4 +2𝑥2 +8
𝑑
𝑑𝑦
∴ 𝑑𝑥 =
=
c) 𝑦 =
(𝑥 4 +2𝑥 2 +8)2
(𝑥 4 +2𝑥 2 +8)(2𝑥+6)−(𝑥 2 +6𝑥−2)(4𝑥 3 +4𝑥)
(𝑥 4 +2𝑥 2 +8)2
(𝑥−1)2
2𝑥−7
𝑑𝑦
∴ 𝑑𝑥 =
=
(2𝑥−7)
𝑑
𝑑
(𝑥−1)2 −(𝑥−1)2 (2𝑥−7)
𝑑𝑥
𝑑𝑥
(2𝑥−7)2
(2𝑥−7)(2𝑥−2)−(𝑥−1)2 (2)
(2𝑥−7)2
√𝑥(2−5𝑥)
𝑦=
d)
𝑑
(𝑥 4 +2𝑥 2 +8)𝑑𝑥(𝑥 2 +6𝑥−2)−(𝑥 2 +6𝑥−2)𝑑𝑥(𝑥 4 +2𝑥 2 +8)
3 2
(√𝑥−2𝑥 2 )
3 2
∴
𝑑𝑦
𝑑𝑥
𝑑
𝑑
3 2
(√𝑥−2𝑥 2 ) 𝑑𝑥(√𝑥(2−5𝑥))−(√𝑥(2−5𝑥))𝑑𝑥(√𝑥−2𝑥 2 )
=
2
3 2
((√𝑥−2𝑥 2 ) )
3 2
1 15√𝑥
−
)−(√𝑥(2−5𝑥))(1−8𝑥+12𝑥 2 )
2
√𝑥
2
3 2
2
((√𝑥−2𝑥 ) )
(√𝑥−2𝑥 2 ) (
=
2
4
e) 𝑦 =
∴
𝑑𝑦
𝑑𝑥
√𝑥 −𝑥2 +3𝑥−7𝑥 −3
2𝑥2 −3𝑥
−5𝑥 −3
√𝑥
=
2𝑥2 −3𝑥
𝑑
2
2
𝑑 2𝑥2 −3𝑥
−5𝑥 −3 ) ( 4√𝑥 − 2 +3𝑥−7𝑥 −3 )−( 4√𝑥− 2 +3𝑥−7𝑥 −3 ) (
−5𝑥 −3 )
𝑑𝑥
𝑑𝑥
𝑥
𝑥
√𝑥
√𝑥
2
2𝑥2 −3𝑥
(
−5𝑥 −3 )
√𝑥
(
2𝑥2 −3𝑥
1
4 21
2
3
15
−5𝑥 −3 )( 4 + 3 + 4 +3 )−( 4√𝑥 − 2+3𝑥−7𝑥 −3 )(3√𝑥−
+ )
𝑥
𝑥
2√𝑥 𝑥4
√𝑥
4 √𝑥3 𝑥
2
2𝑥2 −3𝑥
(
−5𝑥 −3 )
√𝑥
(
=
112 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
7. Chain Rule:
if 𝑓 and 𝑔 are differentiable functions then 𝑦 = (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) is a
𝑑𝑦
𝑑𝑥
function of 𝑥 and
differentiable
= 𝑓′(𝑔(𝑥)) ⋅ 𝑔′(𝑥)
Proof:
𝑑𝑦
𝑑𝑥
= lim
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
ℎ
ℎ→0
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]] 𝑔(𝑥+ℎ)−𝑔(𝑥)
∙
ℎ
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ→0
= lim
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]] 𝑔(𝑥+ℎ)−𝑔(𝑥)
∙
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ
ℎ→0
= lim
= lim
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ→0
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ
ℎ→0
∙ lim
As ℎ → 0 , then 𝑔(𝑥 + ℎ) → 0 because 𝑔 is differentiable and continuous at 𝑥
Let ∆𝑔 = 𝑔(𝑥 + ℎ) − 𝑔(𝑥) , then 𝑔(𝑥 + ℎ) = 𝑔(𝑥) + ∆𝑔 as ℎ → 0 , then ∆𝑔 → 0
𝑑𝑦
𝑑𝑥
= lim
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
ℎ→0
= lim
∆𝑔→0
= lim
∆𝑔→0
𝑔(𝑥+ℎ)−𝑔(𝑥)
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
𝑔(𝑥)+∆𝑔−𝑔(𝑥)
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
∆𝑔
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ
ℎ→0
∙ lim
∙ 𝑔′(𝑥)
∙ 𝑔′(𝑥)
= 𝑓′[𝑔(𝑥)] ∙ 𝑔′(𝑥)
113 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Worked Examples:

Example 27:
2022
Chain Rule:
Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 = (𝑥 3 + 2𝑥 2 − 3𝑥 + 7)2
b) 𝑦 = √2𝑥 10 − 5𝑥
7
3
c) 𝑦 = √(√𝑥 + 3)
d) 𝑦 = √𝑥 + √1 + √𝑥
e) 𝑦 = 4[(3𝑥 − 8)(3𝑥 2 − 2𝑥)3 ]4
114 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Solution:
a) 𝑦 = (𝑥 3 + 2𝑥 2 − 3𝑥 + 7)2
𝑑𝑦
𝑑
∴ 𝑑𝑥 = 2(𝑥 3 + 2𝑥 2 − 3𝑥 + 7)1 𝑑𝑥 (𝑥 3 + 2𝑥 2 − 3𝑥 + 7)
= 2(𝑥 3 + 2𝑥 2 − 3𝑥 + 7)(3𝑥 2 + 4𝑥 − 3)
b) 𝑦 = √2𝑥 10 − 5𝑥
1
𝑦 = (2𝑥 10 − 5𝑥)2
𝑑𝑦
Write the function with the exponent of half.
1
1
1
2
1
𝑑
∴ 𝑑𝑥 = 2 (2𝑥 10 − 5𝑥)−2 𝑑𝑥 (2𝑥 10 − 5𝑥)
= (2𝑥 10 − 5𝑥)−2 (20𝑥 9 − 5)
=
(20𝑥 9 −5)
Note: Simplification is not necessary.
2√2𝑥 10 −5𝑥
7
3
c) 𝑦 = √(√𝑥 + 3)
3
𝑦 = (√𝑥 + 3)7
𝑑𝑦
3
−
3
−
∴ 𝑑𝑥 = 7 (√𝑥 + 3)
= 7 (√𝑥 + 3)
=
4
7
4
7
𝑑
(√𝑥
𝑑𝑥
+ 3)
1
(2 𝑥)
√
3
7
4
14 √(√𝑥+3) ∙√𝑥
115 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
d) 𝑦 = √𝑥 + √1 + √𝑥
1
1 2
1 2
2
𝑦 = {𝑥 + (1 + 𝑥 ) }
∴
𝑑𝑦
𝑑𝑥
=
=
=
1
{𝑥
2
1
{𝑥
2
1
{𝑥
2
=(
1
1 2
2
1
−2
+ (1 + 𝑥 ) }
−2
1
1 2
2
−2
(1 +
1
(1 +
2
(1 +
1
(1 +
2
) (1 +
1
1
1
2
−2
1
2
−2
1
𝑑
𝑥 ) ) 𝑑𝑥 (1 + 𝑥 2 )
1
+ (1 + 𝑥 ) }
2√𝑥+√1+√𝑥
+ (1 + 𝑥 ) )
1
1
1 2
2
+ (1 + 𝑥 ) }
1
1
1 2
2
𝑑
(𝑥
𝑑𝑥
1
1
𝑥 ) ) (2 𝑥)
√
1
2√1+√𝑥
) (2 𝑥)
√
e) 𝑦 = 4[(3𝑥 − 8)(3𝑥 2 − 2𝑥)3 ]4
𝑑𝑦
𝑑𝑥
𝑑
= 16[(3𝑥 − 8)(3𝑥 2 − 2𝑥)3 ]3 𝑑𝑥 {(3𝑥 − 8)(3𝑥 2 − 2𝑥)3 }
= 16[(3𝑥 − 8)(3𝑥 2 − 2𝑥)3 ]3 {3(3𝑥 2 − 2𝑥)3 + (3𝑥 − 8)(3)(3𝑥 2 − 2𝑥)2 (6𝑥 − 2)}
Notice that, we used product rule and chain rule at the same time for derivative of the
function, 𝑦 = (3𝑥 − 8)(3𝑥 2 − 2𝑥)3 which appears to be the inside function.
116 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Worked Examples:

Example 28:
2022
Mixed Differentiations:
Find
𝑑𝑦
𝑑𝑥
in each of the following:
a) 𝑦 = (𝑥 2 + 3)2 (𝑥 2 − 𝑥 + 1)
(8𝑥−1)5
b) 𝑦 = (3𝑥−1)3
3𝑥+1
c) 𝑦 = √2𝑥+5
(𝑥+3)(𝑥−2)
d) 𝑦 = √
√𝑥−1
(8𝑥−1)5
3𝑥+1
e) 𝑦 = ((3𝑥−1)3 ) (√2𝑥+5)
117 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
Solution:
a) 𝑦 = (𝑥 2 + 3)2 (𝑥 2 − 𝑥 + 1)
𝑑𝑦
𝑑
𝑑
∴ 𝑑𝑥 = (𝑥 2 − 𝑥 + 1) 𝑑𝑥 (𝑥 2 + 3)2 + (𝑥 2 + 3)2 𝑑𝑥 (𝑥 2 − 𝑥 + 1)
= 2(𝑥 2 + 3)(2𝑥)(𝑥 2 − 𝑥 + 1) + (𝑥 2 + 3)2 (2𝑥 − 1)
(8𝑥−1)5
b) 𝑦 = (3𝑥−1)3
∴
𝑑𝑦
𝑑𝑥
=
=
(3𝑥−1)3
𝑑
𝑑
(8𝑥−1)5 −(8𝑥−1)5 (3𝑥−1)3
𝑑𝑥
𝑑𝑥
((3𝑥−1)3 )2
(3𝑥−1)3 (5)(8𝑥−1)4 (8)−(8𝑥−1)5 (3)(3𝑥−1)2 (3)
((3𝑥−1)3 )2
Do not simplify!
3𝑥+1
2𝑥+5
c) 𝑦 = √
1
⟹ 𝑦=
3𝑥+1 2
(2𝑥+5)
1
1 3𝑥+1 −2 𝑑 3𝑥+1
(
)
𝑑𝑥 2𝑥+5
𝑑𝑦
∴ 𝑑𝑥 = 2 (2𝑥+5)
1
=
1 3𝑥+1 −2 (3)(2𝑥+)−(2)(3𝑥+1)
(
) (
)
(2𝑥+5)2
2 2𝑥+5
118 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Six: Derivatives of ordinary functions.
(𝑥+3)(𝑥−2)
d) 𝑦 = √
√𝑥−1
1
⟹ 𝑦=
(𝑥+3)(𝑥−2) 2
(
)
√𝑥−1
1
∴
𝑑𝑦
𝑑𝑥
=
𝑑
𝑑
1 (𝑥+3)(𝑥−2) −2 (√𝑥−1)𝑑𝑥(𝑥+3)(𝑥−2)−(𝑥+3)(𝑥−2)𝑑𝑥√𝑥−1
(
)
(
)
2
2
√𝑥−1
(√𝑥−1)
=
1 (𝑥+3)(𝑥−2) −2 (√𝑥−1){(1)(𝑥−2)+(𝑥+3)(1)}−{(𝑥+3)(𝑥−2)}(2√𝑥−1)
(
) {
}
2
2
√𝑥−1
(√𝑥−1)
1
(8𝑥−1)5
1
3𝑥+1
e) 𝑦 = ((3𝑥−1)3 ) (√2𝑥+5)
𝑑𝑦
3𝑥+1
𝑑
(8𝑥−1)5
(8𝑥−1)5
𝑑
3𝑥+1
∴ 𝑑𝑥 = (√2𝑥+5) 𝑑𝑥 ((3𝑥−1)3 ) + ((3𝑥−1)3 ) 𝑑𝑥 (√2𝑥+5)
(3𝑥−1)3 (5)(8𝑥−1)4 (8)−(8𝑥−1)5 (3)(3𝑥−1)2 (3)
3𝑥+1
(
)
)
((3𝑥−1)3 )2
2𝑥+5
= (√
1
(8𝑥−1)5 1 3𝑥+1 −2 (3)(2𝑥+)−(2)(3𝑥+1)
+ ((3𝑥−1)3 ) (
) (
)
(2𝑥+5)2
2 2𝑥+5
119 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER SEVEN:
DERIVATIVES OF TRIGONOMETRIC
FUNCTIONS:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS:
120 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Seven: Derivatives of Trigonometric functions.
DERIVATIVES OF TRIGONOMETRIC FUNCTION:
1. Let 𝒇(𝒙) = 𝐬𝐢𝐧 𝒙 , then 𝒇′(𝒙) = 𝐜𝐨𝐬 𝒙
Proof:
𝑓(𝑥+ℎ)−𝑓(𝑓)
ℎ
ℎ→0
sin(𝑥+ℎ)−sin 𝑥
lim
ℎ
ℎ→0
sin 𝑥 cos ℎ+cos 𝑥 sin ℎ−sin 𝑥
lim
ℎ
ℎ→0
sin 𝑥 cos ℎ−sin 𝑥 + cos 𝑥 sin ℎ
lim
ℎ
ℎ→0
sin 𝑥(cos ℎ−1)+cos 𝑥 sin ℎ
lim
ℎ
ℎ→0
sin 𝑥(cos ℎ−1)
cos 𝑥 sin ℎ
lim (
) + lim (
)
ℎ
ℎ
ℎ→0
ℎ→0
cos ℎ−1
sin ℎ
sin 𝑥 lim ( ℎ ) + cos 𝑥 lim ( ℎ )
ℎ→0
ℎ→0
𝑓′(𝑥) = lim
=
=
=
=
=
=
= sin 𝑥 (0) + cos 𝑥 (1)
= cos 𝑥
2. Let 𝒇(𝒙) = 𝐜𝐨𝐬 𝒙 , then 𝒇′(𝒙) = − 𝐬𝐢𝐧 𝒙
Proof:
𝑓(𝑥+ℎ)−𝑓(𝑓)
ℎ
cos(𝑥+ℎ)−cos 𝑥
lim
ℎ
ℎ→0
cos 𝑥 cos ℎ−sin 𝑥 sin ℎ−cos 𝑥
lim
ℎ
ℎ→0
cos 𝑥 cos ℎ−cos 𝑥−sin 𝑥 sin ℎ
lim
ℎ
ℎ→0
cos 𝑥(cos ℎ−1)−sin 𝑥 sin ℎ
lim
ℎ
ℎ→0
cos 𝑥(cos ℎ−1)
sin 𝑥 sin ℎ
lim (
) − lim ( ℎ )
ℎ
ℎ→0
ℎ→0
cos ℎ−1
sin ℎ
cos 𝑥 lim ( ℎ ) − sin 𝑥 lim ( ℎ )
ℎ→0
ℎ→0
𝑓′(𝑥) = lim
ℎ→0
=
=
=
=
=
=
= cos 𝑥 (0) − sin 𝑥 (1)
= − sin 𝑥
121 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Seven: Derivatives of Trigonometric functions.
𝐬𝐢𝐧 𝒙
then 𝒇′(𝒙) = 𝐬𝐞𝐜 𝟐 𝒙
3. Let 𝒇(𝒙) = 𝐭𝐚𝐧 𝒙 = 𝐜𝐨𝐬 𝒙 ,
Proof:
𝑓′(𝑥) =
cos 𝑥
𝑑
𝑑
(sin 𝑥)−sin 𝑥 (cos 𝑥)
𝑑𝑥
𝑑𝑥
(cos 𝑥)2
=
cos 𝑥 cos 𝑥−sin 𝑥(− sin 𝑥)
cos2 𝑥
=
cos2 𝑥+sin2 𝑥
cos2 𝑥
=
1
cos2 𝑥
= sec 2 𝑥
4. Let 𝒇(𝒙) = 𝐜𝐨𝐬𝐞𝐜 𝒙 =
𝟏
𝐬𝐢𝐧 𝒙
,
then
𝒇′ (𝒙) = − 𝐜𝐨𝐬𝐞𝐜 𝒙 𝐜𝐨𝐭 𝒙
Proof:
𝑓′(𝑥) =
=
sin 𝑥
𝑑
𝑑
(1)−(1) (sin 𝑥)
𝑑𝑥
𝑑𝑥
(sin 𝑥)2
0−cos 𝑥
sin2 𝑥
cos 𝑥
= − sin2 𝑥
cos 𝑥
1
= − sin 𝑥 × sin 𝑥
= − cosec 𝑥 cot 𝑥
122 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Seven: Derivatives of Trigonometric functions.
5. Let 𝒇(𝒙) = 𝐬𝐞𝐜 𝒙 =
𝟏
𝐜𝐨𝐬 𝒙
,
then 𝒇′(𝒙) = 𝐬𝐞𝐜 𝒙 𝐭𝐚𝐧 𝒙
Proof:
𝑓′(𝑥) =
=
cos 𝑥
𝑑
𝑑
(1)−(1) (cos 𝑥)
𝑑𝑥
𝑑𝑥
(cos 𝑥)2
0−(− sin 𝑥)
cos2 𝑥
sin 𝑥
= cos2 𝑥
=
sin 𝑥
cos 𝑥
×
1
cos 𝑥
= sec 𝑥 tan 𝑥
6. Let 𝒇(𝒙) = 𝐜𝐨𝐭 𝒙 =
𝐜𝐨𝐬 𝒙
𝐬𝐢𝐧 𝒙
,
then
𝒇′(𝒙) = −𝐜𝐨𝐬𝐞𝐜 𝟐 𝒙
Proof:
𝑓′(𝑥) =
sin 𝑥
𝑑
𝑑
(cos 𝑥)−cos 𝑥 (sin 𝑥)
𝑑𝑥
𝑑𝑥
(sin 𝑥)2
=
sin 𝑥(−sin 𝑥)−cos 𝑥(cos 𝑥)
sin2 𝑥
=
−sin2 𝑥−cos2 𝑥
sin2 𝑥
=
−(sin2 𝑥+cos2 𝑥)
sin2 𝑥
−1
= sin2 𝑥
= −cosec 2 𝑥
123 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Seven: Derivatives of Trigonometric functions.
Worked Examples:
Example 29:
2022
1. Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 = sin 𝑥 cos 𝑥
b) 𝑦 = sin(cos 𝑥)
c) 𝑦 = sin2 𝑥
d) 𝑦 = sin 𝑥 2
e) 𝑦 = cos 𝑥 tan 𝑥 2
2. Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 = sec(3𝑥 − 1)
b) 𝑦 = sec 8 (3𝑥 − 1)
c) 𝑦 = tan(𝑥 2 + 1)
d) 𝑦 = cos(tan 𝑥)
cos 2𝑥
e) 𝑦 = 2+sin 𝑥
f)
𝑦 = cot 3 (2𝑥)
124 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Seven: Derivatives of Trigonometric functions.
Solution:
1. Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 = sin 𝑥 cos 𝑥
𝑑𝑦
𝑑
𝑑
∴ 𝑑𝑥 = cos 𝑥 𝑑𝑥 (sin 𝑥) + sin 𝑥 𝑑𝑥 (cos 𝑥)
Product rule.
= (cos 𝑥)(cos 𝑥) + (sin 𝑥)(− sin 𝑥)
= cos 2 𝑥 − sin2 𝑥
= cos 2𝑥
b) 𝑦 = sin(cos 𝑥)
𝑑𝑦
𝑑𝑥
∴
𝑑
= cos(cos 𝑥) 𝑑𝑥 (cos 𝑥) , Differentiate the outside function first, and derive the inside.
= cos(cos 𝑥)(− sin 𝑥)
= − sin 𝑥 cos(cos 𝑥)
c) 𝑦 = sin2 𝑥
⟹ 𝑦 = (sin 𝑥)2
𝑑𝑦
𝑑
∴ 𝑑𝑥 = 2 sin 𝑥 𝑑𝑥 (sin 𝑥)
Using the power rule.
= 2 sin 𝑥 cos 𝑥
= sin 2𝑥
d) 𝑦 = sin 𝑥 2
Do not be fooled by saying this is the same as sin 𝑥 2 = sin 𝑥 sin 𝑥, it’s not like that, rather
treat the function as, 𝑦 = sin(𝑥 2 ).
𝑑𝑦
𝑑𝑥
𝑑
= cos(𝑥 2 ) 𝑑𝑥 (𝑥 2 )
= 2𝑥 cos(𝑥 2 )
e) 𝑦 = cos 𝑥 tan 𝑥 2
∴
𝑑𝑦
𝑑𝑥
= tan 𝑥 2
𝑑
(cos 𝑥) +
𝑑𝑥
= tan 𝑥 2 (− sin 𝑥) +
𝑑
(tan 𝑥 2 ) Product
𝑑𝑥
𝑑
cos 𝑥 (sec 2 𝑥 2 ) 𝑑𝑥 (𝑥 2 )
2 2
cos 𝑥
and Chain rule
= − sin 𝑥 tan 𝑥 2 + 2𝑥 cos 𝑥 sec 𝑥
125 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Seven: Derivatives of Trigonometric functions.
2. Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 = sec(3𝑥 − 1)
𝑑𝑦
𝑑
∴ 𝑑𝑥 = sec(3𝑥 − 1) tan(3𝑥 − 1) 𝑑𝑥 (3𝑥 − 1)
Using Chain rule.
= sec(3𝑥 − 1) tan(3𝑥 − 1) (3)
= 3 sec(3𝑥 − 1) tan(3𝑥 − 1)
b) 𝑦 = sec 8 (3𝑥 − 1)
⟹ 𝑦 = (sec(3𝑥 − 1))8
∴
𝑑𝑦
𝑑𝑥
= 8(sec(3𝑥 − 1))7
𝑑
(sec(3𝑥
𝑑𝑥
− 1))
Using Chain rule.
7 (3
= 8(sec(3𝑥 − 1))
sec(3𝑥 − 1) tan(3𝑥 − 1))
7 (sec(3𝑥
= 24(sec(3𝑥 − 1))
− 1) tan(3𝑥 − 1))
c) 𝑦 = tan(𝑥 2 + 1)
𝑑𝑦
𝑑
∴ 𝑑𝑥 = sec 2 (𝑥 2 + 1) 𝑑𝑥 (𝑥 2 + 1)
= 2𝑥 sec
2 (𝑥 2
Using Chain rule.
+ 1)
d) 𝑦 = cos(tan 𝑥)
∴
𝑑𝑦
𝑑𝑥
= − sin(tan 𝑥)
= − sin(tan 𝑥)
𝑑
(tan 𝑥)
𝑑𝑥
(sec 2
Using Chain rule.
𝑥)
cos 2𝑥
e) 𝑦 = 2+sin 𝑥
𝑑𝑦
∴ 𝑑𝑥 =
=
f)
(2+sin 𝑥)
𝑑
𝑑
(cos 2𝑥)−(cos 2𝑥) (2+sin 𝑥)
𝑑𝑥
𝑑𝑥
(2+sin 𝑥)2
By Quotient rule.
(2+sin 𝑥)(−2 sin 2𝑥)−(cos 2𝑥)(cos 𝑥)
(2+sin 𝑥)2
𝑦 = cot 3 (2𝑥)
⟹ 𝑦 = (cot 2𝑥)3
𝑑𝑦
𝑑
∴ 𝑑𝑥 = 3(cot 2𝑥)2 𝑑𝑥 (cot 2𝑥)
=
Using Chain rule.
𝑑
3(cot 2𝑥)2 (−cosec 2 2𝑥) 𝑑𝑥 (2𝑥)
2 (cosec 2
= −6(cot 2𝑥)
2𝑥)
126 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Seven: Derivatives of Trigonometric functions.
Worked Examples:
Example 29:
2022
More Complex
Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 = sec 5 𝑥 sin(cos 𝑥)
b) 𝑦 = (cosec(tan 𝑥))5
𝑥
c) 𝑦 = √sin 𝑥 cos 2 + cot 5𝑥
Solution:
a) 𝑦 = sec 5 𝑥 sin(cos 𝑥)
∴
𝑑𝑦
𝑑𝑥
= sin(cos 𝑥)
𝑑
(sec 5 𝑥) +
𝑑𝑥
4
(sec 5 𝑥)
𝑑
sin(cos 𝑥)
𝑑𝑥
(sec 5
= (sin(cos 𝑥))(5sec 𝑥 ∙ sec 𝑥 tan 𝑥) +
𝑥)(− sin 𝑥 cos(cos 𝑥))
5
5
= 5(sin(cos 𝑥))(sec 𝑥 ∙ tan 𝑥) − (sec 𝑥)(sin 𝑥 cos(cos 𝑥))
b) 𝑦 = (cosec(tan 𝑥))5
∴
𝑑𝑦
𝑑𝑥
= 5(cosec(tan 𝑥))4
𝑑
(cosec(tan 𝑥))
𝑑𝑥
𝑑
= 5(cosec(tan 𝑥))4 ∙ (− cosec(tan 𝑥) cot(tan 𝑥)) 𝑑𝑥 (tan 𝑥)
= 5(cosec(tan 𝑥))4 ∙ (− cosec(tan 𝑥) cot(tan 𝑥))(sec 2 𝑥)
𝑥
c) 𝑦 = √sin 𝑥 cos 2 + cot 5𝑥
⟹
𝑑𝑦
𝑦=
1
𝑥
(sin 𝑥 cos 2
𝑥
1
2
+ cot 5𝑥)
−
∴ 𝑑𝑥 = 2 (sin 𝑥 cos 2 + cot 5𝑥)
1
𝑥
1
2
1
−
2
= 2 (sin 𝑥 cos 2 + cot 5𝑥)
𝑑
𝑥
(sin 𝑥 cos 2
𝑑𝑥
𝑥
+ cot 5𝑥)
1
𝑥
(cos 𝑥 cos 2 + 2 sin 𝑥 sin 2 − 5cosec 2 5𝑥)
127 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER EIGHT:
DERIVATIVES OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS:
128 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Exponential functions.
DERIVATIVES OF EXPONENTIAL FUNCTION:
Special function:
𝑑
THEOREM: States that if 𝑓(𝑥) = 𝑒 𝑔(𝑥) then 𝑓′(𝑥) = 𝑒 𝑔(𝑥) 𝑑𝑥 (𝑔(𝑥))
For example:
Let 𝑓(𝑥) = 𝑒 𝑥 ,
then
𝑓′(𝑥) = 𝑒 𝑥
Proof:
ln 𝑓(𝑥) = ln 𝑒 𝑥
ln 𝑓(𝑥) = 𝑥
1
∙
𝑓(𝑥)
Remember that, ln 𝑒 𝑥 = 𝑥 ln 𝑒 = 𝑥
𝑓′(𝑥) = 1
𝑓′(𝑥) = 𝑓(𝑥)
𝑓′(𝑥) = 𝑒 𝑥
Substitute 𝑓(𝑥) = 𝑒 𝑥
Worked Examples:
Example 30:
2022
Find
𝑑𝑦
𝑑𝑥
in each of the following function.
a) 𝑦 = 𝑒 −3𝑥
2
b) 𝑦 = 𝑥 2 𝑒 2𝑥
c) 𝑦 = 𝑒 √sec 3𝑥
d) 𝑦 = 𝑒 sin(𝑥
2 𝑠in2 𝑥)
𝑒 𝑥 −1
e) 𝑦 = 𝑒 𝑥 +1
129 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Exponential functions.
Solution:
a) 𝑦 = 𝑒 −3𝑥
2
𝑑𝑦
∴ 𝑑𝑥 = 𝑒 −3𝑥
𝑑
(−3𝑥 2 )
𝑑𝑥
2
2
= 𝑒 −3𝑥 (−6𝑥)
2
= −6𝑥𝑒 −3𝑥
b) 𝑦 = 𝑥 2 𝑒 2𝑥
𝑑𝑦
𝑑
𝑑
∴ 𝑑𝑥 = 𝑒 2𝑥 𝑑𝑥 (𝑥 2 ) + 𝑥 2 𝑑𝑥 (𝑒 2𝑥 )
= 𝑒 2𝑥 (2𝑥) + 𝑥 2 (𝑒 2𝑥 ∙ 2)
= 2𝑥𝑒 2𝑥 + 2𝑥 2 𝑒 2𝑥
= 2𝑥𝑒 2𝑥 (1 + 𝑥)
c) 𝑦 = 𝑒 √sec 3𝑥
𝑑𝑦
𝑑
∴ 𝑑𝑥 = 𝑒 √sec 3𝑥 𝑑𝑥 (√sec 3𝑥)
1
1
2
= 𝑒 √sec 3𝑥 ( (sec 3𝑥)−2 ) (sec 3𝑥 tan 3𝑥)
1
1
𝑑
(3𝑥)
𝑑𝑥
= 3𝑒 √sec 3𝑥 (2 (sec 3𝑥)−2 ) (sec 3𝑥 tan 3𝑥)
d) 𝑦 = 𝑒 sin(𝑥
∴
𝑑𝑦
𝑑𝑥
2 𝑠in2 𝑥)
𝑑
(sin(𝑥 2 𝑠in2 𝑥))
𝑑𝑥
2
2
𝑑
𝑒 sin(𝑥 𝑠in 𝑥) (cos(𝑥 2 𝑠in2 𝑥)) (𝑥 2 𝑠in2 𝑥)
𝑑𝑥
sin(𝑥 2 𝑠in2 𝑥) (cos(𝑥 2
2
2
2
= 𝑒 sin(𝑥
=
2 𝑠in2 𝑥)
=𝑒
𝑠in 𝑥))(2𝑥𝑠in 𝑥 + 2𝑥 sin 𝑥 cos 𝑥)
𝑒 𝑥 −1
e) 𝑦 = 𝑒 𝑥 +1
𝑑𝑦
∴ 𝑑𝑥 =
=
=
=
𝑑
𝑑
(𝑒 𝑥 −1)−(𝑒 𝑥 −1) (𝑒 𝑥 +1)
𝑑𝑥
𝑑𝑥
(𝑒 𝑥 +1)2
(𝑒 𝑥 +1)(𝑒 𝑥 )−(𝑒 𝑥 −1)(𝑒 𝑥 )
(𝑒 𝑥 +1)
(𝑒 𝑥 +1)2
𝑒 2𝑥 +𝑒 𝑥 −𝑒 2𝑥 +𝑒 𝑥
(𝑒 𝑥 +1)2
2𝑒 𝑥
𝑥
(𝑒 +1)2
130 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
DERIVATIVES OF LOGARITHMIC FUNCTION:
Let 𝒇(𝒙) = 𝐥𝐨𝐠 𝒃 𝒙 ,
𝟏
𝒇′(𝒙) = 𝒙 𝐥𝐨𝐠 𝒃 𝒆
then
Proof:
𝑓′(𝑥) = lim
ℎ→0
= lim
𝑓(𝑥+ℎ)−𝑓(𝑓)
ℎ
log𝑏 (𝑥+ℎ)−log𝑏 𝑥
ℎ
ℎ→0
= lim
ℎ→0
log𝑏 (
𝑥+ℎ
)
𝑥
ℎ
1
𝑥+ℎ
)
𝑥
= lim ℎ log 𝑏 (
ℎ→0
1 𝑥
ℎ
= lim 𝑥 ∙ ℎ log 𝑏 (1 + 𝑥 )
ℎ→0
𝑥
=
1
lim log 𝑏
𝑥
ℎ→0
=
1
lim log 𝑏
𝑥 ℎ→0
(1 +
ℎ ℎ
)
𝑥
(1 +
ℎ ℎ
)
𝑥
𝑥
ℎ
𝑥
Let 𝑝 = , if ℎ → 0 , then
ℎ
𝑥
→0
𝑥
∴ 𝑓′(𝑥) =
1
lim log 𝑏
𝑥 ℎ→0
(1 +
ℎ ℎ
)
𝑥
1
1
𝑥 ℎ→0
= lim log 𝑏 (1 + 𝑝)𝑝
1
𝑥
1
= log 𝑏 (lim (1 + 𝑝)𝑝 )
ℎ→0
1
𝑥
= log 𝑏 𝑒
Hence the theorem.
131 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
Worked Examples:
Example 30:
2022
1. Find
𝑑𝑦
𝑑𝑥
in each of the following Logarithmic functions.
a) 𝑦 = log(8𝑥 2 − 5𝑥 + 1)
b) 𝑦 = log 3 (𝑥 6 + 2𝑥 3 + 𝑒 √𝑥 )
c) 𝑦 = log10 (3𝑥 5 + 2𝑎𝑥 2 )
d) 𝑦 = log 𝑒 [(𝑥 − 1)7 (3𝑥 + 2)3 ]
(𝑥−1)7
e) 𝑦 = log 6 ((3𝑥+2)3 )
2. Find
𝑑𝑦
𝑑𝑥
in each of the following Natural logarithmic functions.
a) 𝑦 = ln(sec 2𝑥 8 )
b) 𝑦 = ln(𝑥 2 ∙ √3𝑥 − 2)
8−sin 11𝑥
c) 𝑦 = ln (
2
𝑒𝑥
)
d) 𝑦 = ln(sec 7 (ln tan 3𝑥))
e) 𝑦 = 5𝑥
f)
2
𝑦 = 18𝑒
g) 𝑦 = 18𝑒
𝑥2
𝑥2
h) 𝑦 = 7tan(cot 𝑥)
i)
𝑦 = 𝑥𝑥
132 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
Solution:
1. Find
𝑑𝑦
𝑑𝑥
in each of the following Logarithmic functions.
a) 𝑦 = log(8𝑥 2 − 5𝑥 + 1)
𝑑𝑦
1
∴ 𝑑𝑥 = (8𝑥2 −5𝑥+1)
𝑑
(8𝑥 2
𝑑𝑥
− 5𝑥 + 1)
1
= (8𝑥 2 −5𝑥+1) ∙ (16𝑥 − 5)
16𝑥−5
= 8𝑥 2 −5𝑥+1
b) 𝑦 = log 3 (𝑥 6 + 2𝑥 3 + 𝑒 √𝑥 )
𝑑𝑦
∴ 𝑑𝑥 =
=
1
𝑑
(𝑥 6 +2𝑥 3 +𝑒 √𝑥 )
1
(𝑥 6 +2𝑥 3 +𝑒 √𝑥 )
∙ log 3 𝑒 𝑑𝑥 (𝑥 6 + 2𝑥 3 + 𝑒 √𝑥 )
∙ log 3 𝑒 (6𝑥 5 + 6𝑥 2 +
𝑒 √𝑥
)
2√𝑥
c) 𝑦 = log10 (3𝑥 5 + 2𝑎𝑥 2 )
𝑑𝑦
1
𝑑
∴ 𝑑𝑥 = (3𝑥5 +2𝑎𝑥 2 ) log 10 ∙ 𝑑𝑥 (3𝑥 5 + 2𝑎𝑥 2 )
1
= (3𝑥 5 +2𝑎𝑥 2 ) log 10 ∙ (15𝑥 4 + 4𝑎𝑥)
15𝑥 4 +4𝑎𝑥
= (3𝑥 5 +2𝑎𝑥 2 ) log 10
133 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
d) 𝑦 = log 𝑒 [(𝑥 − 1)7 (3𝑥 + 2)3 ]
𝑦 = log 𝑒 (𝑥 − 1)7 + log 𝑒 (3𝑥 + 2)3
𝑑𝑦
1
𝑑
1
𝑑
∴ 𝑑𝑥 = (𝑥−1)7 ∙ log 𝑒 𝑒 ∙ 𝑑𝑥 (𝑥 − 1)7 + (3𝑥+2)3 ∙ log 𝑒 𝑒 ∙ 𝑑𝑥 (3𝑥 + 2)3
1
1
= (𝑥−1)7 ∙ 7(𝑥 − 1)6 (1) + (3𝑥+2)3 ∙ 3(3𝑥 + 2)2 (2)
7(𝑥−1)6
(𝑥−1)7
=
7
+
6(3𝑥+2)2
(3𝑥+2)3
6
= 𝑥−1 + 3𝑥+2
(𝑥−1)7
e) 𝑦 = log 6 ((3𝑥+2)3 )
𝑦 = log 6 (𝑥 − 1)7 − log 6 (3𝑥 + 2)3
∴
𝑑𝑦
𝑑𝑥
1
= (𝑥−1)7 ∙ log 6 𝑒 ∙
𝑑
(𝑥
𝑑𝑥
1
− 1)7 − (3𝑥+2)3 ∙ log 6 𝑒 ∙
1
𝑑
(3𝑥
𝑑𝑥
+ 2)3
1
= (𝑥−1)7 ∙ log 6 𝑒 ∙ 7(𝑥 − 1)6 (1) − (3𝑥+2)3 ∙ log 6 𝑒 ∙ 3(3𝑥 + 2)2 (2)
=
7(𝑥−1)6
(𝑥−1)7
7
∙ log 6 𝑒 −
6(3𝑥+2)2
(3𝑥+2)3
∙ log 6 𝑒
6
= 𝑥−1 ∙ log 6 𝑒 − 3𝑥+2 ∙ log 6 𝑒
134 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
2. Find
𝑑𝑦
𝑑𝑥
in each of the following Natural logarithmic functions.
a) 𝑦 = ln(sec 2𝑥 8 )
𝑑𝑦
1
𝑑
∴ 𝑑𝑥 = sec 2𝑥 8 ∙ 𝑑𝑥 (sec 2𝑥 8 )
=
1
sec 2𝑥 8
∙ sec(2𝑥 8 ) tan(2𝑥 8 ) ∙
𝑑
(2𝑥 8 )
𝑑𝑥
1
= sec 2𝑥 8 ∙ sec(2𝑥 8 ) tan(2𝑥 8 ) ∙ (16𝑥 7 )
b) 𝑦 = ln(𝑥 2 ∙ √3𝑥 − 2)
𝑑𝑦
∴ 𝑑𝑥 =
=
=
1
𝑑
∙ (𝑥 2
(𝑥 2 ∙√3𝑥−2) 𝑑𝑥
1
∙
(𝑥 2 ∙√3𝑥−2)
1
2
𝑥
1
1
(2𝑥√3𝑥 − 2 + 𝑥 2 (2 (3𝑥 − 2)−2 ) (3))
1
(𝑥 2 ∙√3𝑥−2)
= +
∙ √3𝑥 − 2)
1
∙ (2𝑥√3𝑥 − 2 + 3𝑥 2 (2 (3𝑥 − 2)−2 ))
3
2(√3𝑥−2)(√3𝑥−2)
OR
⟹ 𝑦 = ln 𝑥 2 + ln √3𝑥 − 2
𝑑𝑦
1
𝑑
∴ 𝑑𝑥 = 𝑥 2 ∙ 𝑑𝑥 (𝑥 2 ) +
1
= 𝑥 2 (2𝑥) +
2
1
𝑑
∙ (√3𝑥
√3𝑥−2 𝑑𝑥
1
1
((2 (3𝑥
3𝑥−2
√
− 2)
1
− 2)−2 ) (3))
3
3𝑥−2)(
√
√3𝑥−2)
= 𝑥 + 2(
135 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
8−sin 11𝑥
c) 𝑦 = ln (
2
𝑒𝑥
)
2
⟹ 𝑦 = ln(8 − sin 11𝑥) − ln(𝑒 𝑥 )
𝑑𝑦
1
𝑑
∴ 𝑑𝑥 = (8−sin 11𝑥) ∙ 𝑑𝑥 (8 − sin 11𝑥) −
1
= (8−sin 11𝑥) (−11 cos 11𝑥) −
1
2
𝑒𝑥
1
2
𝑒𝑥
𝑑
2
∙ 𝑑𝑥 (𝑒 𝑥 )
2
(2𝑥𝑒 𝑥 )
11 cos 11𝑥
= − 8−sin 11𝑥 − 2𝑥
d) 𝑦 = ln(sec 7 (ln tan 3𝑥))
∴
𝑑𝑦
𝑑𝑥
=
=
=
=
e) 𝑦 = 5𝑥
∴
1
𝑑
1
∙ (
)
(sec7 (ln tan 3𝑥)) 𝑑𝑥 (sec7 (ln tan 3𝑥))
1
(sec7 (ln tan 3𝑥))
1
(sec7 (ln tan 3𝑥))
1
(sec (ln tan 3𝑥))
∙ (7sec 6 (ln tan 3𝑥)(sec(ln tan 3𝑥) tan(ln tan 3𝑥)))
𝑑
(ln tan 3𝑥)
𝑑𝑥
3sec2 3𝑥
)
tan 3𝑥
∙ (7sec 6 (ln tan 3𝑥)(sec(ln tan 3𝑥) tan(ln tan 3𝑥))) (
3sec2 3𝑥
)
tan 3𝑥
∙ ((sec(ln tan 3𝑥) ∙ tan(ln tan 3𝑥))) (
2
𝑑𝑦
𝑑𝑥
2
= 5𝑥 ln 5 ∙
𝑑
(𝑥 2 )
𝑑𝑥
2
= 5𝑥 ln 5 (2𝑥)
f)
𝑦 = 18𝑒
𝑥2
𝑑𝑦
∴ 𝑑𝑥 = 18𝑒
= 18𝑒
𝑥2
𝑥2
= 2𝑥18𝑒
𝑑
2
∙ ln 18 ∙ 𝑑𝑥 (𝑒 𝑥 )
2
∙ ln 18 ∙ (𝑒 𝑥 (2𝑥))
𝑥2
∙ ln 18 ∙ 𝑒 𝑥
2
136 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
g) 𝑦 = 7tan(cot 𝑥)
𝑑𝑦
𝑑
∴ 𝑑𝑥 = 7tan(cot 𝑥) ∙ ln 7 ∙ 𝑑𝑥 (tan(cot 𝑥))
𝑑
= 7tan(cot 𝑥) ∙ ln 7 (sec 2 (cot 𝑥)) 𝑑𝑥 (cot 𝑥)
= 7tan(cot 𝑥) ∙ ln 7 (sec 2 (cot 𝑥))(−cosec 2 𝑥)
= −7tan(cot 𝑥) ∙ ln 7 (sec 2 (cot 𝑥))(cosec 2 𝑥)
h) 𝑦 = 𝑥 𝑥
⟹
ln 𝑦 = ln 𝑥 𝑥
⟹
ln 𝑦 = 𝑥 ln 𝑥
∴
1 𝑑𝑦
∙
𝑦 𝑑𝑥
𝑑
𝑑
= ln 𝑥 𝑑𝑥 (𝑥) + 𝑥 𝑑𝑥 (ln 𝑥)
1
𝑥
= ln 𝑥 + 𝑥 ( )
= ln 𝑥 + 1
𝑑𝑦
∴ 𝑑𝑥 = 𝑦(ln 𝑥 + 1 )
= 𝑥 𝑥 (ln 𝑥 + 1 )
137 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
Worked Examples:
Example 31:
2022
More Complex:
Find
𝑑𝑦
𝑑𝑥
in each of the following functions.
a) 𝑦 = log 𝑒 𝑥 𝑥
5
b) 𝑦 = (𝑒 ln cot 𝑥 )
c) 𝑦 = 𝑥 cos 𝑥
d) 𝑦 =
1
√cos √𝑥
e) 𝑦 = 𝑥 𝑥
𝑥
Solution:
a) 𝑦 = log 𝑒 𝑥 𝑥
𝑑𝑦
1
𝑑
∴ 𝑑𝑥 = 𝑥 𝑥 ∙ log 𝑒 𝑒 ∙ 𝑑𝑥 (𝑥 𝑥 )
1
= 𝑥 𝑥 ∙ log 𝑒 𝑒 ∙ (𝑥 𝑥 (ln 𝑥 + 1 ))
= ln 𝑥 + 1
= log 𝑒 𝑥 + 1
The relationship between the two Logarithmic.
Or
⟹ 𝑦 = 𝑥 log 𝑒 𝑥
𝑑𝑦
Logarithmic property.
𝑑
𝑑
∴ 𝑑𝑥 = log 𝑒 𝑥 ∙ 𝑑𝑥 (𝑥) + 𝑥 𝑑𝑥 (log 𝑒 𝑥)
Product rule.
1
= log 𝑒 𝑥 + 𝑥 (𝑥 ∙ log 𝑒 𝑒 (1))
= log 𝑒 𝑥 + 1
138 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eight: Derivatives of Logarithmic functions.
5
b) 𝑦 = (𝑒 ln cot 𝑥 )
∴
𝑑𝑦
𝑑𝑥
4
𝑑
(𝑒 ln cot 𝑥 )
𝑑𝑥
4
𝑑
5(𝑒 ln cot 𝑥 ) (𝑒 ln cot 𝑥 ) 𝑑𝑥 (ln cot 𝑥)
4
1
5(𝑒 ln cot 𝑥 ) (𝑒 ln cot 𝑥 ) (cot 𝑥) (−cose2 𝑥)
= 5(𝑒 ln cot 𝑥 ) ∙
=
=
c) 𝑦 = 𝑥 cos 𝑥
⟹ ln 𝑦 = ln 𝑥 cos 𝑥 = cos 𝑥 ln 𝑥
1 𝑑𝑦
𝑦 𝑑𝑥
∴ ∙
𝑑𝑦
∴ 𝑑𝑥 =
=
d) 𝑦 =
𝑑
𝑑
(cos 𝑥) + cos 𝑥 (ln 𝑥)
𝑑𝑥
𝑑𝑥
cos 𝑥
= − sin 𝑥 ln 𝑥 +
𝑥
cos 𝑥
𝑦 (− sin 𝑥 ln 𝑥 + 𝑥 )
cos 𝑥
𝑥 cos 𝑥 (− sin 𝑥 ln 𝑥 + 𝑥 )
= ln 𝑥
1
√cos √𝑥
−
⟹ 𝑦 = (cos √𝑥)
𝑑𝑦
1
2
1
−
1
−
∴ 𝑑𝑥 = − 2 (cos √𝑥)
3
2
3
𝑑
∙ 𝑑𝑥 (cos √𝑥)
1
= − 2 (cos √𝑥) 2 (− sin √𝑥) (2 𝑥)
√
sin √𝑥
=
2
3
4√𝑥∙ √(cos √𝑥)
e) 𝑦 = 𝑥 𝑥
𝑥
⟹ ln 𝑦 = 𝑥 𝑥 ln 𝑥
1 𝑑𝑦
𝑦 𝑑𝑥
∴ ∙
= ln 𝑥 ∙
𝑑
(𝑥 𝑥 ) +
𝑑𝑥
𝑥𝑥
𝑑
(ln 𝑥)
𝑑𝑥
1
𝑥
= (ln 𝑥)(𝑥 𝑥 (ln 𝑥 + 1 )) + (𝑥 𝑥 ) ( )
𝑑𝑦
1
∴ 𝑑𝑥 = 𝑦 {(ln 𝑥)(𝑥 𝑥 (ln 𝑥 + 1 )) + (𝑥 𝑥 ) (𝑥) }
𝑥
1
= (𝑥 𝑥 ) {(ln 𝑥)(𝑥 𝑥 (ln 𝑥 + 1 )) + (𝑥 𝑥 ) (𝑥) }
139 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER NINE:
IMPLICIT AND HIHER ORDER
DIFFERENTIATION:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS
140 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Nine: Implicit Differentiation.
IMPLICT DIFFERENTIATION:
Sometimes functions are given not in the form 𝑦 = 𝑓(𝑥) but in a more complicated form in which it is
difficult or impossible to express 𝑦 explicitly in terms of 𝑥. Such functions are called implicit functions. In
this unit we explain how these can be differentiated using implicit differentiation.
In order to master the techniques explained here it is vital that you undertake plenty of practice
exercises so that they become second nature.
Worked Examples:
Example 32:
2022
Find
𝑑𝑦
𝑑𝑥
in each of the following functions.
a) 𝑦 + 2𝑥𝑦 2 − ln 2𝑦 = 2𝑥 2 + 3𝑦 3
b) 𝑥 4 + 𝑦 4 = 81
c) 𝑒 𝑥𝑦 = 𝑥 + 𝑦
d) 𝑥 + tan(𝑥𝑦) = 0
e) 𝑥𝑦 2 + √𝑥𝑦 = 0
f)
log 𝑦 = sin(𝑥 + 𝑦)
g) 𝑒 𝑦 cos 𝑥 = 1 + sin(𝑥𝑦 2 )
141 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Nine: Implicit Differentiation.
Solution:
a) 𝑦 + 2𝑥𝑦 2 − ln 2𝑦 = 2𝑥 2 + 3𝑦 3
∴
∴
∴
∴
∴
𝑑𝑦
𝑑
𝑑
1
𝑑
𝑑
+ 2𝑦 2 𝑑𝑥 (𝑥) + 2𝑥 𝑑𝑥 (𝑦 2 ) − ln 2𝑦 ∙ 𝑑𝑥 (2𝑦) = 𝑑𝑥 (2𝑥 2 ) +
𝑑𝑥
𝑑𝑦
𝑑𝑦
2
𝑑𝑦
𝑑𝑦
+ 2𝑦 2 + 4𝑥𝑦 ∙ 𝑑𝑥 − ln 2𝑦 ∙ 𝑑𝑥 = 4𝑥 + 9𝑦 2 𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑦
2
𝑑𝑦
𝑑𝑦
+ 4𝑥𝑦 ∙ 𝑑𝑥 − ln 2𝑦 ∙ 𝑑𝑥 − 9𝑦 2 𝑑𝑥 = 4𝑥 − 2𝑦 2
𝑑𝑥
𝑑𝑦
2
(1 + 4𝑥𝑦 −
− 9𝑦 2 ) = 4𝑥 − 2𝑦 2
𝑑𝑥
ln 2𝑦
𝑑𝑦
𝑑𝑥
=
𝑑
3 𝑑𝑥 (𝑦 3 )
4𝑥−2𝑦 2
1+4𝑥𝑦−
2
−9𝑦 2
ln 2𝑦
b) 𝑥 4 + 𝑦 4 = 81
𝑑𝑦
∴ 4𝑥 3 + 4𝑦 3 𝑑𝑥 = 0
∴ 4𝑦 3
∴
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=−
= −4𝑥 3
𝑥3
𝑦3
c) 𝑒 𝑥𝑦 = 𝑥 + 𝑦
𝑑
𝑑𝑦
∴ 𝑒 𝑥𝑦 𝑑𝑥 (𝑥𝑦) = 1 + 𝑑𝑥
𝑑𝑦
𝑑𝑦
∴ 𝑒 𝑥𝑦 (𝑦 + 𝑥 𝑑𝑥 ) = 1 + 𝑑𝑥
𝑑𝑦
𝑑𝑦
∴ 𝑦𝑒 𝑥𝑦 + 𝑥𝑒 𝑥𝑦 𝑑𝑥 = 1 + 𝑑𝑥
𝑑𝑦
𝑑𝑦
∴ 𝑥𝑒 𝑥𝑦 𝑑𝑥 − 𝑑𝑥 = 1 − 𝑦𝑒 𝑥𝑦
𝑑𝑦
∴ 𝑑𝑥 (𝑥𝑒 𝑥𝑦 − 1) = 1 − 𝑦𝑒 𝑥𝑦
𝑑𝑦
1−𝑦𝑒 𝑥𝑦
∴ 𝑑𝑥 = 𝑥𝑒 𝑥𝑦 −1
142 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Nine: Implicit Differentiation.
d) 𝑥 + tan(𝑥𝑦) = 0
𝑑
∴ 1 + sec 2 (𝑥𝑦) 𝑑𝑥 (𝑥𝑦) = 0
𝑑𝑦
∴ 1 + sec 2 (𝑥𝑦) (𝑦 + 𝑥 𝑑𝑥 ) = 0
𝑑𝑦
∴ 1 + 𝑦sec 2 (𝑥𝑦) + 𝑥sec 2 (𝑥𝑦) 𝑑𝑥 = 0
𝑑𝑦
∴ 𝑥sec 2 (𝑥𝑦) 𝑑𝑥 = −1 − 𝑦sec 2 (𝑥𝑦)
𝑑𝑦
∴ 𝑑𝑥 =
−1−𝑦sec2 (𝑥𝑦)
𝑥sec2 (𝑥𝑦)
e) 𝑥𝑦 2 + √𝑥𝑦 = 0
1
𝑑
𝑑
1
𝑑
(𝑥) + 𝑥 (𝑦 2 ) + (𝑥𝑦)−2 (𝑥𝑦)
𝑑𝑥
𝑑𝑥
2
𝑑𝑥
𝑑𝑦
1
𝑑𝑦
𝑦 2 + 2𝑥𝑦 +
(𝑦 + 𝑥 ) = 0
𝑑𝑥
2√𝑥𝑦
𝑑𝑥
𝑑𝑦
𝑦
𝑥 𝑑𝑦
2
𝑦 + 2𝑥𝑦 +
+
=0
𝑑𝑥
2√𝑥𝑦
2√𝑥𝑦 𝑑𝑥
𝑑𝑦
𝑥
𝑦
(2𝑥𝑦 + 2 𝑥𝑦) = −𝑦 2 − 2 𝑥𝑦
𝑑𝑥
√
√
∴ 𝑦2
∴
∴
∴
∴
f)
𝑑𝑦
𝑑𝑥
=0
𝑦
2√𝑥𝑦
𝑥
2𝑥𝑦+
2√𝑥𝑦
−𝑦 2 −
=
log 𝑦 = sin(𝑥 + 𝑦)
1 𝑑𝑦
𝑑
∴ 𝑦 ∙ 𝑑𝑥 = cos(𝑥 + 𝑦) 𝑑𝑥 (𝑥 + 𝑦)
1 𝑑𝑦
𝑑𝑦
∴ 𝑦 ∙ 𝑑𝑥 = cos(𝑥 + 𝑦) (1 + 𝑑𝑥 )
1 𝑑𝑦
𝑑𝑦
∴ 𝑦 ∙ 𝑑𝑥 = cos(𝑥 + 𝑦) + cos(𝑥 + 𝑦) 𝑑𝑥
1 𝑑𝑦
𝑑𝑦
∴ 𝑦 ∙ 𝑑𝑥 − cos(𝑥 + 𝑦) 𝑑𝑥 = − cos(𝑥 + 𝑦)
𝑑𝑦 1
∴ 𝑑𝑥 (𝑦 − cos(𝑥 + 𝑦)) = − cos(𝑥 + 𝑦)
𝑑𝑦
− cos(𝑥+𝑦)
∴ 𝑑𝑥 = 1
−cos(𝑥+𝑦)
𝑦
143 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Nine: Implicit Differentiation.
g) 𝑒 𝑦 cos 𝑥 = 1 + sin(𝑥𝑦 2 )
𝑑
𝑑
𝑑
(𝑒 𝑦 ) + 𝑒 𝑦 (cos 𝑥) = 0 + cos(𝑥𝑦 2 ) (𝑥𝑦 2 )
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑦
𝑒 𝑦 cos 𝑥 𝑑𝑥 − 𝑒 𝑦 sin 𝑥 = cos(𝑥𝑦 2 ) (𝑦 2 + 2𝑥𝑦 𝑑𝑥 )
𝑑𝑦
𝑑𝑦
𝑒 𝑦 cos 𝑥 − 𝑒 𝑦 sin 𝑥 = 𝑦 2 cos(𝑥𝑦 2 ) + 2𝑥𝑦 cos(𝑥𝑦 2 )
𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑦
2
2
2
𝑦
(𝑒 cos 𝑥 − 2𝑥𝑦 cos(𝑥𝑦 )) = 𝑦 cos(𝑥𝑦 ) + 𝑒 sin 𝑥
𝑑𝑥
∴ cos 𝑥
∴
∴
∴
𝑑𝑦
𝑦 2 cos(𝑥𝑦 2 )+𝑒 𝑦 sin 𝑥
∴ 𝑑𝑥 = 𝑒 𝑦 cos 𝑥−2𝑥𝑦 cos(𝑥𝑦2 )
144 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Nine: Higher Order Differentiation.
HIGHER ORDER DIFFERENTITAION:
Notation for the Higher Order differentiation:
;
𝑓 ′ (𝑥)
;
𝑑𝑦
𝑑𝑥
;
𝑑
(𝑓(𝑥))
𝑑𝑥
and 𝐷𝑥 𝑦
 Second Derivative: 𝑦 ′′ ;
𝑓 ′′ (𝑥)
;
𝑑2 𝑦
𝑑𝑥 2
;
𝑑2
(𝑓(𝑥))
𝑑𝑥 2
and 𝐷 2 𝑥 𝑦
 First Derivative:
𝑦′
 Third Derivative:
𝑦 ′′′ ;
𝑓 ′′′ (𝑥) ;
𝑑3 𝑦
𝑑𝑥 3
;
𝑑3
(𝑓(𝑥))
𝑑𝑥 3
and 𝐷 3 𝑥 𝑦
 Forth Derivative:
𝑦4 ;
𝑓 4 (𝑥) ;
𝑑4 𝑦
𝑑𝑥 4
;
𝑑4
(𝑓(𝑥))
𝑑𝑥 4
and 𝐷 4 𝑥 𝑦
Worked Examples:
Example 33:
\
2022
𝑑2𝑥
Find 𝑑𝑥 2 of the following problem:
a) 𝑥 4 + 𝑦 4 = 81
b) 𝑥 2 + 4𝑦 2 = 10
c) 𝑥𝑦 + 𝑦 − 𝑥 = 8
d) sin2 𝑥 + cos 2 𝑦 = 1
e) 𝑒 𝑥 − 𝑒 𝑦 = 𝑥 2 + 𝑦 2
145 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Nine: Higher Order Differentiation.
Solution:
a) 𝑥 4 + 𝑦 4 = 81
𝑑𝑦
∴ 4𝑥 3 + 4𝑦 3 𝑑𝑥 = 0
𝑑𝑦
∴ 4𝑦 3 𝑑𝑥 = −4𝑥 3
𝑥3
𝑑𝑦
∴ 𝑑𝑥 = − 𝑦3
Now let’s find the second derivative.
∴
𝑑2 𝑥
𝑑𝑥 2
= −(
𝑑𝑦
𝑑𝑥
3𝑥 2 𝑦 3 −(𝑥 3 )(3𝑦 2 )
(𝑦 3 )2
)
𝒙𝟑
= −(
3𝑥 2 𝑦 3 −(𝑥 3 )(3𝑦 2 )(− 𝟑 )
𝒚
(𝑦 3 )2
)
b) 𝑥 2 + 4𝑦 2 = 10
𝑑𝑦
∴ 2𝑥 + 8𝑦 𝑑𝑥 = 0
𝑑𝑦
𝑥
∴ 𝑑𝑥 = − 4𝑦
Finding the second derivative it’s easy, we just need to find the derivative of the previous
one.
𝑑2 𝑥
∴ 𝑑𝑥 2 = − (
= −(
=
=
𝑑𝑦
𝑑𝑥
(4𝑦)2
4𝑦−4𝑥
4𝑦−4𝑥(−
(4𝑦)2
)
𝑥
)
4𝑦
) , Note: it is advisable to stop here. DO NOT simplify.
𝑥2
𝑦
16𝑦 2
−4𝑦−
−4𝑦 2 −𝑥 2
16𝑦 2
146 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Nine: Higher Order Differentiation.
c) 𝑥𝑦 + 𝑦 − 𝑥 = 8
𝑑𝑦
𝑑𝑦
∴ 𝑦 + 𝑥 𝑑𝑥 + 𝑑𝑥 − 1 = 0
∴
∴
𝑑𝑦
(𝑥 + 1)
𝑑𝑥
𝑑𝑦
1−𝑦
= 𝑥+1
𝑑𝑥
=1−𝑦
Thus, the result is in quotient form, let find the second derivative.
𝑑2 𝑥
∴ 𝑑𝑥 2 =
=
𝑑𝑦
)−(1−𝑦)(1)
𝑑𝑥
(𝑥+1)2
(𝑥+1)(−
1−𝑦
)−(1−𝑦)(1)
𝑥+1
(𝑥+1)2
−(𝑥+1)(
d) sin2 𝑥 + cos 2 𝑦 = 1
𝑑𝑦
∴ 2 sin 𝑥 cos 𝑥 − 2 cos 𝑥 sin 𝑥 𝑑𝑥 = 0
𝑑𝑦
∴ − sin 2𝑥 𝑑𝑥 = − sin 2𝑥
∴
𝑑𝑦
𝑑𝑥
=1
Thus, our first derivative it’s a constant, which tells us that our second derivative will also be
a constant. i.e.
𝑑2 𝑦
𝑑𝑥 2
=0
e) 𝑒 𝑥 − 𝑒 𝑦 = 𝑥 2 + 𝑦 2
𝑑𝑦
𝑑𝑦
∴ 𝑒 𝑥 − 𝑒 𝑦 𝑑𝑥 = 2𝑥 + 2𝑦 𝑑𝑥
∴
∴
𝑑𝑦
(−𝑒 𝑦 − 2𝑦)
𝑑𝑥
𝑑𝑦
2𝑥−𝑒 𝑥
=
𝑑𝑥
−𝑒 𝑦 −2𝑦
= 2𝑥 − 𝑒 𝑥
One should be very careful when taking the second derivative of this quotient first
derivative, and one should have the following results.
Thus,
𝑑2 𝑦
𝑑𝑥 2
=
(2−𝑒 𝑥 )(−𝑒 𝑦 −2𝑦)−(2𝑥−𝑒 𝑥 )(−𝑒 𝑦
𝑑𝑦
𝑑𝑦
−2 )
𝑑𝑥
𝑑𝑥
(−𝑒 𝑦 −2𝑦)2
2𝑥−𝑒𝑥
2𝑥−𝑒𝑥
)−2( 𝑦
))
−𝑒𝑦 −2𝑦
−𝑒 −2𝑦
(2−𝑒 𝑥 )(−𝑒 𝑦 −2𝑦)−(2𝑥−𝑒 𝑥 )(−𝑒 𝑦 (
=
(−𝑒 𝑦 −2𝑦)2
147 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER TEN:
DERIVATIVES OF INVERSE TRIGONOMETRIC
FUNCTIONS:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS
148 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Ten: Derivatives of Inverse Trigonometric functions.
Derivatives of Inverse Trig. Function:
1. Let 𝒚 = 𝐬𝐢𝐧−𝟏 𝒙 in the open interval (−𝟏, 𝟏) and be differentiable, then
𝒅𝒚
𝒅𝒙
=
𝟏
𝒅𝒚
𝒅𝒙
= −
√𝟏−𝒙𝟐
Proof:
If 𝑦 = sin−1 𝑥 with 𝑥 ∈ (−1,1) ⟹ sin 𝑦 = 𝑥 if and only sin2 𝑦 = 𝑥 2
𝑑
𝑑
⟹ 𝑑𝑥 sin 𝑦 = 𝑑𝑥 𝑥
⟹
⟹
𝑑𝑦
cos 𝑦 𝑑𝑥 = 1
𝑑𝑦
𝑑𝑥
1
= cos 𝑦
But we know that, cos 2 𝑦 = 1 − sin2 𝑦
=
Now substitute, cos 𝑦 = √1 − sin2 𝑦
=
𝑑𝑦
∴ 𝑑𝑥 =
1
√1−sin2 𝑦
1
√1−𝑥 2
1
Since sin 𝑦 = 𝑥 if and only sin2 𝑦 = 𝑥 2
; −1 < 𝑥 < 1
√1−𝑥 2
2. Let 𝒚 = 𝐜𝐨𝐬 −𝟏 𝒙 in the open interval (−𝟏, 𝟏) and be differentiable, then
𝟏
√𝟏−𝒙𝟐
Proof:
If 𝑦 = cos−1 𝑥 with 𝑥 ∈ (−1,1) ⟹ cos 𝑦 = 𝑥 if and only cos2 𝑦 = 𝑥 2
𝑑
𝑑
⟹
cos 𝑦 = 𝑥
𝑑𝑥
⟹
⟹
𝑑𝑦
𝑑𝑦
𝑑𝑥
1
=−
𝑑𝑦
𝑑𝑥
But we know that, sin2 𝑦 = 1 − cos2 𝑦
= − sin 𝑦
=−
∴
𝑑𝑥
−sin 𝑦 𝑑𝑥 = 1
= −
1
√1−cos2 𝑦
1
√1−𝑥 2
1
√1−𝑥 2
Now substitute, sin 𝑦 = √1 − cos 2 𝑦
Since cos 𝑦 = 𝑥 if and only cos 2 𝑦 = 𝑥 2
; −1 < 𝑥 < 1
3. Let 𝒚 = 𝐭𝐚𝐧−𝟏 𝒙 in the open interval (−∞, ∞) and be differentiable, then
Proof:
If 𝑦 = tan−1 𝑥 with 𝑥 ∈ (−∞, ∞) ⟹ tan 𝑦 = 𝑥 if and only tan2 𝑦 = 𝑥 2
𝑑
𝑑
⟹ 𝑑𝑥 tan 𝑦 = 𝑑𝑥 𝑥
⟹
⟹
=
𝟏
𝟏+𝒙𝟐
𝑑𝑦
sec 2 𝑦 𝑑𝑥 = 1
𝑑𝑦
𝑑𝑥
1
= sec2 𝑦
1
= 1+tan2 𝑦
1
𝑑𝑦
𝒅𝒚
𝒅𝒙
∴ 𝑑𝑥 =
= 1+𝑥2
1
1+𝑥 2
By Pythagorean Theorem we have, 1 + tan2 𝑦 = sec 2 𝑦
Now substitute, 1 + tan2 𝑦 = sec 2 𝑦
Since tan2 𝑦 = 𝑥 2
; −∞ < 𝑥 < ∞
149 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Ten: Derivatives of Inverse Trigonometric functions.
Worked Examples:
Example 34:
2022
Find
𝑑𝑦
𝑑𝑥
in each of the following functions.
a) 𝑦 = arc cos 4𝑥 2
b) 𝑦 = cos −1 (𝑥 + 𝑥 3 )
c) 𝑦 = tan−1 (sin−1 √𝑥)
d) 𝑦 = cos −1 (𝑒 √tan 3𝑥 )
1
2
e) 𝑦 = 𝑥 tan−1 (ln 𝑥) − ln(1 + 𝑥 2 )4
f)
𝑦 = sin−1(𝑥 2 ) − 𝑥𝑒 𝑥
2
g) 𝑦 = ln 2 cos−1 (sin−1 (tan−1 𝑥))
150 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Ten: Derivatives of Inverse Trigonometric functions.
Solution:
a) 𝑦 = arc cos 4𝑥 2
𝑑𝑦
∴ 𝑑𝑥 = −
=−
1
𝑑
∙ (4𝑥 2 )
√1−(4𝑥 2 )2 𝑑𝑥
8𝑥
√1−(4𝑥 2 )2
b) 𝑦 = cos −1 (𝑥 + 𝑥 3 )
𝑑𝑦
∴ 𝑑𝑥 = −
=−
1
𝑑
√1−(𝑥+𝑥 3 )2
∙ 𝑑𝑥 (𝑥 + 𝑥 3 )
1+3𝑥 2
√1−(𝑥+𝑥 3 )2
c) 𝑦 = tan−1 (sin−1 √𝑥)
∴𝑦=
=
=
1
𝑑
2
1+(sin−1 √𝑥)
1
2
1+(sin−1 √𝑥)
1
2
1+(sin−1 √𝑥)
∙ 𝑑𝑥 (sin−1 √𝑥)
∙(
1
2
√1−(√𝑥)
∙(
1
2
√1−(√𝑥)
)∙
𝑑
(√𝑥)
𝑑𝑥
1
)
2√𝑥
)∙(
151 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Ten: Derivatives of Inverse Trigonometric functions.
d) 𝑦 = cos −1 (𝑒 √tan 3𝑥 )
𝑑𝑦
∴ 𝑑𝑥 = −
1
𝑑
2
∙ 𝑑𝑥 (𝑒 √tan 3𝑥 )
2
∙ (𝑒 √tan 3𝑥 ) ∙ 𝑑𝑥 (√tan 3𝑥)
√1−(𝑒 √tan 3𝑥 )
1
=−
𝑑
√1−(𝑒 √tan 3𝑥 )
1
=−
1
1
√1−(𝑒 √tan 3𝑥 )
∙ (𝑒 √tan 3𝑥 ) ∙ (2 (tan 3𝑥)−2 (sec 2 3𝑥) (3))
2
1
e) 𝑦 = 𝑥 tan−1 (ln 𝑥) − 2 ln(1 + 𝑥 2 )4
𝑑𝑦
𝑑
𝑑
1
1
∴ 𝑑𝑥 = tan−1 (ln 𝑥) ∙ 𝑑𝑥 (𝑥) + 𝑥 𝑑𝑥 (tan−1 (ln 𝑥)) − 2 ∙ (1+𝑥2 )4 ∙ (4(1 + 𝑥 2 )3 (2𝑥))
1
3
1
= tan−1 (ln 𝑥) + (𝑥) (1+(ln 𝑥)2 ) (𝑥) −
1
4𝑥(1+𝑥 2 )
(1+𝑥 2 )4
4𝑥
= tan−1 (ln 𝑥) + 1+(ln 𝑥)2 − 1+𝑥 2
f)
𝑦 = sin−1(𝑥 2 ) − 𝑥𝑒 𝑥
∴
𝑑𝑦
𝑑𝑥
=
=
1
√1−(𝑥 2 )2
2𝑥
√1−(𝑥 2 )2
2
𝑑
(𝑥 2 ) −
𝑑𝑥
∙
2
𝑒𝑥 ∙
2
− 𝑒 𝑥 + 2𝑥 2 𝑒 𝑥
𝑑
(𝑥) +
𝑑𝑥
𝑥
2
𝑑
(𝑒 𝑥 )
𝑑𝑥
2
g) 𝑦 = ln 2 cos−1 (sin−1 (tan−1 𝑥))
𝑑𝑦
1
𝑑
∴ 𝑑𝑥 = 2 cos−1 (sin−1(tan−1 𝑥)) ∙ 𝑑𝑥 (2 cos −1(sin−1(tan−1 𝑥)))
1
−2
= 2 cos−1 (sin−1(tan−1 𝑥)) ∙ (
√1−(sin−1(tan−1 𝑥))2
1
−2
= 2 cos−1 (sin−1(tan−1 𝑥)) ∙ (
√1−(sin−1(tan−1 𝑥))2
1
= − cos−1 (sin−1(tan−1 𝑥)) ∙ (
𝑑
) ∙ 𝑑𝑥 (sin−1(tan−1 𝑥))
1
𝑑
) (tan−1 𝑥)
√1−(tan−1 𝑥)2 𝑑𝑥
)∙(
1
√1−(sin−1(tan−1 𝑥))2
)∙(
1
√1−(tan−1 𝑥)2
1
) ∙ (1+𝑥2 )
152 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER ELEVEN:
L’HOPITAL’S RULE:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS
153 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eleven: L’Hopital’s Rule.
Indeterminate Forms and L’Hopital’s Rule:
THEOREM: L’Hopital’s Rule.
Suppose that 𝑓 and 𝑔 are differentiable functions and 𝑔(𝑥) ≠ 0 near (except possible at 𝑎).
Suppose that, lim 𝑓(𝑎) = 0 and lim 𝑔(𝑎) = 0
𝑥→𝑎
𝑥→𝑎
lim 𝑓(𝑎) = ±∞ and lim 𝑔(𝑎) = ±∞
Or that,
𝑥→𝑎
𝑥→𝑎
( In other words, we have an indeterminate form of type
0
0
𝑜𝑟
∞
∞
) Then,
𝑓(𝑥)
𝑓′(𝑥)
= lim
𝑥→𝑎 𝑔(𝑥)
𝑥→𝑎 𝑔′(𝑥)
lim
Types of Indeterminate form:
0
 0
L’Hopital’s Rule.








∞
∞
0
=
∞
∞
L’Hopital’s Rule.
0
0 =0
1
=∞
Not Indeterminate form.
Not Indeterminate form.
Not Indeterminate form.
∞0
∞±∞
∞∙∞
∞∙0
L’Hopital’s Rule.
L’Hopital’s Rule.
L’Hopital’s Rule.
L’Hopital’s Rule.
0
PITFALL:
 In applying L’Hopital’s Rule to
𝑓(𝑎)
𝑔(𝑎)
do not differentiate the quotient
𝑓(𝑎)
.
𝑔(𝑎)
𝑓′(𝑥)
 Rather, differentiate the numerator and denominator independently, obtaining 𝑔′(𝑥) .
 Also, do not apply L’Hopital’s Rule to forms which are not considered to be Indeterminate.
The approach for L’Hopital’s Rule.
o 𝑥→𝑎
o 𝑥 → 𝑎+
o 𝑥 → 𝑎−
o 𝑥→∞
o 𝑥 → −∞ , Where 𝑎 is any real number and ∞ is an abstract concept describing
something without any bound or larger than any natural number.
 lim [ 𝑓(𝑥) ] 𝑔(𝑥) = 𝑒 ln[ 𝑓(𝑥) ]
𝑥→𝑎
 lim 𝑓(𝑥) ∙ 𝑔(𝑥) = lim (
𝑥→𝑎
𝑥→𝑎
𝑔(𝑥)
𝑔(𝑥)
1
𝑓(𝑥)
= 𝑒 𝑔(𝑥) ln 𝑓(𝑥)
)
154 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eleven: L’Hopital’s Rule.
Worked Examples:
Example 35:
2022
Evaluate the following limits:
1.
cos 𝑥
𝑥
𝑥→0
lim+
2. lim
𝑥→0
sin 𝑥
𝑥
𝑒 𝑥 −1
𝑥→0 2𝑥
3. lim
2sin2 𝑥
4. lim 1+cos 𝑥
𝑥→𝜋
5. lim 𝑥 ln 𝑥
𝑥→0
6. lim [ cosec𝑥 ∙ ln(1 − sin 𝑥) ]
𝑥→0
6𝑥 2 +5𝑥−4
7. lim1 4𝑥2 +16−9
𝑥→
2
𝑒 2𝑥 −1
𝑥→0 sin 𝑥
8. lim
cos 𝑥
9. lim𝜋 1−sin 𝑥
𝑥→
2
𝑥 3 −2𝑥 2 +1
𝑥→1 𝑥 3 −1
10. lim
11. lim+(𝑥 𝑥 )
𝑥→0
12. lim+(1 + sin 4𝑥)cot 𝑥
𝑥→0
155 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eleven: L’Hopital’s Rule.
Solution:
1.
lim
cos 𝑥
𝑥→0+ 𝑥
cos(0)
= 0
1
=
0
This is not L’Hopital’s rule because the numerator is not 0.
=∞
sin 𝑥
𝑥→0 𝑥
cos 𝑥
= lim 1
𝑥→0
cos 0
= 1
0
0
2. lim
(L’Hopital’s Rule)
Applying L’Hopital’s rule since we have Indeterminate form.
Apply the limit.
=1
𝑒 𝑥 −1
𝑥→0 2𝑥
𝑒𝑥
= lim 2
𝑥→0
1
=2
3. lim
2sin2 𝑥
4. lim 1+𝑐𝑜𝑠 𝑥
𝑥→𝜋
= lim
𝑥→𝜋
0
0
(L’Hopital’s Rule)
0
0
(L’Hopital’s Rule)
4 sin 𝑥 cos 𝑥
− sin 𝑥
= −4 lim cos 𝑥
𝑥→𝜋
= −4 cos 𝜋
=4
5. lim 𝑥 ln 𝑥
𝑥→0
= lim
=
ln 𝑥
Writing the limit in Quotient form
1
𝑥→∞ 𝑥
1
lim (𝑥
𝑥→∞
1
÷ (−) 𝑥 2 )
Applying L’Hopital’s Rule
1
= − lim (𝑥)
𝑥→∞
= −(0)
=0
156 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eleven: L’Hopital’s Rule.
6. lim [ cosec𝑥 ∙ ln(1 − sin 𝑥) ]
𝑥→0
Remember you can only apply L’Hopital’s Rule if the limit is in Quotient form,
ln(1−sin 𝑥)
sin 𝑥
− cos 𝑥
lim [
÷
𝑥→0+ 1−sin 𝑥
1
− lim+ (1−sin 𝑥)
𝑥→0
1
−( )
1−0
∴ lim [ cosec𝑥 ∙ ln(1 − sin 𝑥) ] = lim+ [
𝑥→0
𝑥→0
=
=
=
]
Recall: cosec𝑥 =
cos 𝑥 ]
L’Hopital’s Rule
1
sin 𝑥
= −1
6𝑥 2 +5𝑥−4
2
𝑥→ 4𝑥 +16−9
2
12𝑥+5
= lim1 8𝑥+16
𝑥→
7. lim1
=
0
0
(L’Hopital’s Rule)
2
1
2
1
8( )+16
2
12( )+5
11
= 20
𝑒 2𝑥 −1
𝑥→0 sin 𝑥
2𝑒 2𝑥
= lim cos 𝑥
𝑥→0
2𝑒 0
= cos(0)
8. lim
0
0
(L’Hopital’s Rule)
=2
cos 𝑥
9. lim𝜋 1−sin 𝑥
𝑥→
0
0
(L’Hopital’s Rule)
2
− sin 𝑥
= lim𝜋 − cos 𝑥
𝑥→
=
2
𝜋
2
𝜋
cos( )
2
sin( )
1
=0
=∞
157 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Eleven: L’Hopital’s Rule.
𝑥 3 −2𝑥 2 +1
𝑥→1 𝑥 3 −1
3𝑥 2 −4𝑥
= lim 3𝑥 2
𝑥→1
3(1)2 −4(1)
= 3(1)2
1
= −3
0
0
10. lim
(L’Hopital’s Rule)
11. lim+(𝑥 𝑥 )
𝑥→0
One can be tempted to say the limit doesn’t exist, whereas the limit does exist,
Note that lim+ 𝑥 𝑥 is 00 type of an indeterminate form.
𝑥→0
Apply the properties of limits stated above, we have the following results,
∴ lim+ 𝑥 𝑥 = lim+ 𝑒 𝑥 ln 𝑥
𝑥→0
𝑥→0
lim (𝑥 ln 𝑥)
= lim+ 𝑒 𝑥→0+
See the previous example.
𝑥→0
= lim+ 𝑒 0
𝑥→0
=1
The limit of a constant, it’s constant.
12. lim+(1 + sin 4𝑥)cot 𝑥
𝑥→0
Note that is lim+(1 + sin 4𝑥)cot 𝑥 1∞ type of an indeterminate form.
𝑥→0
Recalling the theory of L’Hopital’s Rule behind the problem, we obtain the following,
cot 𝑥 )
∴ lim+(1 + sin 4𝑥)cot 𝑥 = lim+ 𝑒 ln((1+sin 4𝑥)
𝑥→0
𝑥→0
= lim+ 𝑒 cot 𝑥 ln(1+sin 4𝑥)
𝑥→0
= lim+ 𝑒
ln(1+sin 4𝑥)
tan 𝑥
𝑥→0
= lim+ 𝑒
𝑥→0
(
4 cos 4𝑥
×sec2 𝑥)
1+sin 4𝑥
1
Recall: cot 𝑥 = tan 𝑥
Applying L’Hopital’s Rule
= lim+ 𝑒 4
𝑥→0
= 𝑒4
The limit of a constant, it’s a constant.
158 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER TWELVE:
ROLLE’S & MEAN VALUE THEOREM:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS
159 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Twelve: Rolle’s Theorem.
ROLLE’S THEOREM:
Rolle’s Theorem: States that let 𝑓 be a function that satisfies the following three hypotheses:
1. 𝑓 is continuous on the closed interval [𝑎, 𝑏]
2. 𝑓 is differentiable on the open interval (𝑎, 𝑏)
3. 𝑓(𝑎) = 𝑓(𝑏)
Then there is a number 𝑐 in (𝑎, 𝑏) such that 𝑓′(𝑐) = 0
ABSTRACT:
Rolle's Theorem can be used to show that a function has a horizontal tangent line inside an interval. If
you can show that a function is continuous over an interval, differentiable over the same interval, and
that the function has the same value at the endpoints of the interval, then you can use Rolle's Theorem.
Without each of these three conditions being met, Rolle's Theorem cannot be applied. If however you
can show that these three conditions are all met, then Rolle's Theorem proves that there is a horizontal
tangent line somewhere in the interval. Remember, a function has a horizontal tangent line wherever it
changes direction from increasing to decreasing or vice versa. Therefore, if you can prove the existence
of a horizontal tangent line, you've also proven the existence of a local extrema, which means you've got
a local maximum or a local minimum.
Worked Examples:
Example 36:
2022
Exhibit the validity of the Rolle’s Theorem:
1. 𝑓(𝑥) = 𝑥 2 − 4𝑥 + 3
𝑥 ∈ [ 1, 3 ]
2. 𝑓(𝑥) = (𝑥 2 − 2𝑥)𝑒 𝑥
𝑥 ∈ [ 0, 2]
3. 𝑓(𝑥) = sin(2𝑥)
𝑥 ∈ [ 𝜋 ⁄6 , 𝜋 ⁄3 ]
4. 𝑓(𝑥) = 𝑥 4 − 2𝑥 2
𝑥 ∈ [−2, 2 ]
5. 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4
𝑥 ∈ [1, 4]
160 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Twelve: Rolle’s Theorem.
Solution:
1. 𝑓(𝑥) = 𝑥 2 − 4𝑥 + 3
[ 1, 3 ]
First thing to do in this problem is to check if 𝑓(𝑎) = 𝑓(𝑏), where 𝑎 = 1 and 𝑏 = 3 in this
problem, if 𝑓(𝑎) ≠ 𝑓(𝑏) then the Rolle’s Theorem cannot be applied.
Thus, 𝑓(𝑎) = 𝑓(1) = (1)2 − 4(1) + 3 = 0
And 𝑓(𝑏) = 𝑓(3) = (3)2 − 4(3) + 3 = 0 , which satisfies the Rolle’s Theorem.
∴ 𝑓′(𝑐) = 0
∴ 𝑓′(𝑥) = 2𝑥 − 4
∴ 𝑓′(𝑐) = 2𝑐 − 4
∴ 2𝑐 − 4 = 0
∴𝑐=2
Since 𝑓′(𝑐) = 0 from the theorem.
Which lies in the given closed interval [ 1,3 ].
Thus, 𝐶 = 2 ∈ (1,3)
2. 𝑓(𝑥) = (𝑥 2 − 2𝑥)𝑒 𝑥
[ 0, 2]
So first step is to determine whether 𝑓(0) = 𝑓(2), and if it doesn’t don’t apply the Rolle’s
Theorem in this problem.
𝑓(𝑎) = 𝑓(0) = (02 − 2(0))𝑒 0
= 0 And
𝑓(𝑏) = 𝑓(2) = (22 − 2(2))𝑒 2
= (0)𝑒 2 = 0
Which tells us that the Rolle’s Theorem holds in this problem.
∴ 𝑓′(𝑐) = 0
∴ 𝑓 ′ (𝑥) = (𝑥 2 − 2𝑥)𝑒 𝑥 + (2𝑥 − 2)𝑒 𝑥
∴ 𝑓 ′ (𝑐) = (𝑐 2 − 2𝑐)𝑒 𝑐 + (2𝑐 − 2)𝑒 𝑐
Replacing 𝑥 with 𝑐 and solve the equation.
𝑐 (𝑐 2
∴ 𝑒
− 2𝑐 + 2𝑐 − 2) = 0
Note: 𝑓′(𝑐) = 0 from the theorem.
∴ 𝑒 𝑐 (𝑐 2 − 2) = 0
Simplify what’s inside the brackets.
𝑐
2
Which implies that, 𝑒 ≠ 0 and 𝑐 − 2 = 0
Hence 𝑐 = ±√2 which of cause −√2 doesn’t lie within the given interval.
Thus 𝐶 = √2 ∈ (1, 2)
161 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Twelve: Rolle’s Theorem.
𝜋 𝜋
3. 𝑓(𝑥) = sin(2𝑥)
[6,3]
Just to show that the Rolle’s Theorem holds and make sense in this particular problem, one may
𝜋
𝜋
𝜋
𝜋
not need to show that ( 6 ) ≠ 𝑓 ( 3 ) , but rather show that 𝑓 ( 6 ) ≠ 𝑓 ( 3 ), doing that we have,
𝜋
𝜋
2𝜋
3
And
𝜋
𝜋
2𝜋
3
Which tells us that the Rolle’s Theorem holds in this problem.
𝑓 ( 6 ) = sin (2 ∙ 6 ) =
𝑓 ( 3 ) = sin (2 ∙ 3 ) =
∴ 𝑓′(𝑐) = 0
∴ 𝑓′(𝑥) = 2 cos(2𝑥)
∴ 𝑓′(𝑐) = 2 cos(2𝑐)
∴ 2 cos(2𝑐) = 0
∴ cos(2𝑐) = 0
𝜋
∴ 2𝑐 = cos −1(0) = 2
𝜋
𝜋 𝜋
∴𝑐=4
Thus, 𝐶 =
Chain rule differentiation studied earlier.
Replacing 𝑥 with 𝑐 and solve the equation.
Note: 𝑓′(𝑐) = 0 from the theorem.
Which lies in the given closed interval [ 6 , 3 ]
𝜋
4
𝜋 𝜋
6 3
∈( , )
4. 𝑓(𝑥) = 𝑥 4 − 2𝑥 2
[−2, 2 ]
Just to show that the Rolle’s Theorem holds and make sense in this particular problem, one may
not need to show that (−2) ≠ 𝑓(2) , but rather show that 𝑓(−2) = 𝑓(2), doing that we have,
𝑓(−2) = (−2)4 − 2(2)2
= 16 − 8 = 8 And
𝑓(2) = (2)4 − 2(2)2
= 16 − 8 = 8
Which tells us that the Rolle’s Theorem holds in this problem.
∴ 𝑓′(𝑐) = 0
∴ 𝑓 ′ (𝑥) = 4𝑥 3 − 4𝑥
Differentiate the polynomial 𝑓.
′ (𝑐)
3
∴𝑓
= 4𝑐 − 4𝑐
Replacing 𝑥 with 𝑐 and solve the equation.
∴ 4𝑐 3 − 4𝑐 = 0
Note: 𝑓′(𝑐) = 0 from the theorem.
∴ 4𝑐(𝑐 − 1)(𝑐 + 1) = 0
Factorize the cubic polynomial.
∴ 𝑐 = 0 𝑜𝑟 ± 1
Thus this shows us that the Rolle’s Theorem does work provided that 𝑓 is continuous on the
closed interval [−2, 2 ] and differentiable on the open interval (−2, 2)
162 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Twelve: Rolle’s Theorem.
5. 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4
[1, 4]
∴ 𝑓(1) = 12 − 5(1) + 4 = 0
∴ 𝑓(4) = 42 − 5(4) + 4 = 0
But 𝑓′(𝑐) = 0
∴ 𝑓 ′ (𝑥) = 2𝑥 − 5
∴ 𝑓 ′ (𝑐) = 2𝑐 − 5
∴ 2𝑐 − 5 = 0
∴𝑐=
5
2
𝑜𝑟 2
1
2
Which lies within the given closed interval [1, 4]
1
Thus, 𝐶 = 2 2 ∈ (1 , 4)
163 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Mean Value Theorem: States that let 𝑓 be a function that satisfies the following three hypotheses:
1. 𝑓 is continuous on the closed interval [𝑎, 𝑏]
2. 𝑓 is differentiable on the open interval (𝑎, 𝑏)
𝑓(𝑏)−𝑓(𝑎)
Then there is a number 𝑐 in (𝑎, 𝑏) such that, 𝑓′(𝑐) = 𝑏−𝑎
Or Equivalently 𝑓(𝑏) − 𝑓(𝑎) = 𝑓′(𝑐)(𝑏 − 𝑎)
𝒇′(𝒄) =
𝒇(𝒃)−𝒇(𝒂)
𝒃−𝒂
𝑦 = 𝑓(𝑥)
ℎ(𝑥)
𝐴(𝑎, 𝑓(𝑎)) ●
𝐿 = 𝑓(𝑎) +
𝑓(𝑏)−𝑓(𝑎)
(𝑥
𝑏−𝑎
− 𝑎)
𝑓(𝑥)
● 𝐵(𝑏, 𝑓(𝑏))
𝐴
𝐵
𝑷𝒓𝒐𝒐𝒇:
We apply Rolle’s Theorem to a new function ℎ defined as the difference between 𝑓 and the function
whose graph is the secant line 𝐴𝐵. The equation of the secant line can be written as,
𝑦 − 𝑓(𝑎) =
𝑓(𝑏)−𝑓(𝑎)
(𝑥
𝑏−𝑎
∴ 𝑦 = 𝑓(𝑎) +
− 𝑎)
𝑓(𝑏)−𝑓(𝑎)
(𝑥
𝑏−𝑎
− 𝑎)
First we must verify that ℎ satisfies the Three Hypotheses of Rolle’s Theorem.
1. The formula ℎ is continuous on the closed interval [𝑎, 𝑏]
2. The formula ℎ differentiable on the open interval (𝑎, 𝑏)
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
𝑓(𝑏)−𝑓(𝑎)
(𝑏 − 𝑎) = 0
3. ℎ(𝑎) = 𝑓(𝑎) −
𝑏−𝑎
𝑓(𝑏)−𝑓(𝑎)
∴ ℎ(𝑏) = 𝑓(𝑏) − 𝑓(𝑎) − 𝑏−𝑎 (𝑏 − 𝑎) =
ℎ′(𝑥) = 𝑓′(𝑥) −
0
Thus ℎ(𝑎) = ℎ(𝑏)
Since ℎ satisfies the Three Hypotheses of Rolle’s Theorem, ℎ′(𝑐) = 0
∴ 0 = ℎ′(𝑐) = 𝑓′(𝑐) −
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
Hence the theorem, 𝑓′(𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
164 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Twelve: Mean Value Theorem.
Worked Examples:
Example 37:
2022
Exhibit the validity of the Mean Value Theorem:
1. 𝑓(𝑥) = 𝑥 2 + 3𝑥 + 2
𝑥 ∈ [ 1, 2]
4
2. 𝑓(𝑥) = 5 − 𝑥
𝑥 ∈ [1, 4]
3. 𝑓(𝑥) = 𝑒 −2𝑥
𝑥 ∈ [ 0, 3 ]
4. 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 1
𝑥 ∈ [1,4]
5. 𝑓(𝑥) = 𝑥 3 − 5𝑥 2 − 3𝑥
1≤𝑥≤3
Solution:
1. 𝑓(𝑥) = 𝑥 2 + 3𝑥 + 2
[ 1, 2]
Mean Value theorem it is similar to the Rolle’s Theorem, but slightly different, here we just need
find the gradient which is the average change of rate in the given problem, now applying the
Mean value theorem, have the following results.
𝑓 ′ (𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
2
, where 𝑎 = 1 and 𝑏 = 2 given as an interval.
Now 𝑓(1) = 1 + 3(1) + 2 = 6 and
𝑓(2) = 22 + 3(2) + 2 = 12
∴ 𝑓 ′ (𝑐) =
𝑓(2)−𝑓(1)
2−1
=
=6
∴ 𝑓′(𝑥) = 2𝑥 + 3
∴ 𝑓′(𝑐) = 2𝑐 + 3
∴ 2𝑐 + 3 = 6
∴𝑐=
3
2
12−6
1
Note: 𝑓′(𝑐) = 6
Which lies within the given closed interval [ 1, 2].
3
2
Thus, 𝐶 = ∈ (1,2)
165 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Twelve: Mean Value Theorem.
4
𝑥
2. 𝑓(𝑥) = 5 −
[1, 4]
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
𝑓 ′ (𝑐) =
, where 𝑎 = 1 and 𝑏 = 4 given as an interval.
4
Now 𝑓(1) = 5 − 1 = 1 and
4
𝑓(4) = 5 − 4 = 4
∴ 𝑓 ′ (𝑐) =
𝑓(4)−𝑓(1)
4−1
=
4−1
3
=1
∴𝑓
′ (𝑥)
4
= − (− 𝑥 2 )
4
∴ 𝑓 ′ (𝑐) = − (− 𝑐 2 )
4
∴ 𝑐2 = 1
Note: 𝑓′(𝑐) = 1
2
∴ 𝑐 =4
∴𝑐=2
but
(since it is outside the given interval)
𝑐 ≠ −2
Thus, 𝐶 = 2 ∈ (1 , 4)
3. 𝑓(𝑥) = 𝑒 −2𝑥
𝑓 ′ (𝑐) =
[ 0, 3 ]
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
0
, where 𝑎 = 0 and 𝑏 = 3 given as an interval.
Now 𝑓(0) = 𝑒 = 1 and
𝑓(3) = 𝑒 −2(3) = 𝑒 −6
∴ 𝑓 ′ (𝑐) =
=
𝑓(2)−𝑓(0)
3−0
=
𝑒 −6 −1
3
𝑒 −6 −1
3
∴ 𝑓 ′ (𝑥) = −2𝑒 −2𝑥
∴ 𝑓 ′ (𝑐) = −2𝑒 −2𝑐
𝑒 −6 −1
3
𝑒 −6 −1
−2𝑐
𝑒
=− 6
𝑒 −6 −1
−2𝑐 = − ln ( 6 )
1
𝑒 −6 −1
𝑐 = 2 ln ( 6 )
1
1
𝑐 = − 2 ln [6 (1 − 𝑒 −6 )]
∴ −2𝑒 −2𝑐 =
∴
∴
∴
∴
1
1
Thus, 𝐶 = − 2 ln [6 (1 − 𝑒 −6 )] ∈ (1,3)
166 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Twelve: Mean Value Theorem.
4. 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 1
∴ 𝑓′(𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
=
[1,4]
17−(−1)
3
=
18
3
=6
∴ 𝑓 ′ (𝑥) = 3𝑥 2 − 6𝑥
∴ 𝑓 ′ (𝑐) = 3𝑐 2 − 6𝑐
∴ 6 = 3𝑐 2 − 6𝑐
∴ 𝑐 2 − 2𝑐 − 2 = 0
Thus 𝑐 =
=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
−(−2)±√(−2)2 −4(1)(−2)
2
= 1 ± √3
Hence 𝑐 = 1 + √3 and 𝑐 ≠ 1 + √3
∴ 𝑐 = 1 + √3𝜖 (1,4)
5.
𝑓(𝑥) = 𝑥 3 − 5𝑥 2 − 3𝑥
1≤𝑥≤3
∴ 𝑓′(𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
=
𝑓(3)−𝑓(1)
3−1
=
−27−(−7)
2
= −10
∴ 𝑓 ′ (𝑐) = 3𝑐 2 − 10𝑐 − 3
∴ 3𝑐 2 − 10𝑐 − 3 = −10
∴ 3𝑐 2 − 10𝑐 + 7 = 0
∴ (3𝑐 − 7)(𝑐 − 1) = 0
∴ 𝑐 = 1 𝑜𝑟 𝑐 =
7
3
7
3
Thus 𝑐 = ∈ (1,3) where 𝑐 ≠ 1 because 𝑓 is differentiable on the open (1,3).
167 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER THIRTEEN:
APPLICATION OF CALCULUS:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS
168 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
THEOREMS:
The following theorems are provided and summarized so that the following examples becomes easy to
solve and make sense, students should practice and have interest in this section, most of the works is
done in high school grades if not in high school calculus class.
1. Theorem: if 𝑓 has a local maximum or minimum at 𝑐 , then 𝑐 is a critical number of 𝑓.
2. Fermat’s Theorem: if 𝑓 has a local maximum or minimum at 𝑐 and 𝑓′(𝑐) = 0 exists,
then 𝑓′(𝑐) = 0
3. Critical Number: is a number 𝑐 in the domain of 𝑓 such that either 𝑓′(𝑐) = 0 or 𝑓′(𝑐)
does not exists.
4. The Closed Interval Method: to find the absolute maximum and minimum values of
a continuous function 𝑓 on a closed interval [𝑎, 𝑏]:
a) Find the values of 𝑓 at the critical numbers of 𝑓 in (𝑎, 𝑏).
b) Find the values of 𝑓 at the endpoint of the interval.
c) The largest of the values from Step 1 and 2 is the absolute maximum value;
The smallest of these values is the absolute minimum value.
5. The Extreme Value Theorem: if 𝑓 is continuos on a closed interval [𝑎, 𝑏], then 𝑓
attains an absolute maximum value 𝑓(𝑐) and absolute minimum value 𝑓(𝑑) at some number 𝑐
and 𝑑 in [𝑎, 𝑏].
6. The Concavity Test:
a) If 𝑓′′(𝑥) > 0 for all 𝑥 𝑖𝑛 𝐼, then then graph of 𝑓 is concave Upward on 𝐼
b) If 𝑓′′(𝑥) < 0 for all 𝑥 𝑖𝑛 𝐼, then then graph of 𝑓 is concave Downward on 𝐼
7. The Second Derivative Test: Suppose 𝑓′′ is continuous near 𝑐
a) If 𝑓′(𝑐) = 0 and 𝑓′′(𝑐) > 0, then 𝑓 has a local minimum at 𝑐
b) If 𝑓′(𝑐) = 0 and 𝑓′′(𝑐) < 0, then 𝑓 has a local maximum at 𝑐
8. Increasing and Decreasing Test (Ascending and Descending):
a) If 𝑓′(𝑥) > 0 On an interval, then 𝑓 is increasing on that interval.
b) If 𝑓′(𝑥) < 0 On an interval, then 𝑓 is decreasing on that interval.
169 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
Worked Examples:
Example 38:
2022
Determine the interval on which the function 𝑓 is Increasing or Decreasing (Ascending/Descending).
1. 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2
2. 𝑓(𝑥) = 3𝑥 4 − 4𝑥 3 + 12𝑥 2 + 5
3. 𝑓(𝑥) = 𝑥 + 2 sin 𝑥
0 ≤ 𝑥 ≤ 2𝜋
4. 𝑓(𝑥) = 2𝑥 3 + 9𝑥 2 − 24𝑥 − 10
5. 𝑓(𝑥) = 𝑥 2 𝑒 −3𝑥
Solution:
1. 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2
Let’s now first find the gradient function that will help us to find the critical number of the
function, and remember the gradient function is always equals to zero.
∴ 𝑓 ′ (𝑥) = 3𝑥 2 − 3
∴ ∴ 𝑓 ′ (𝑐) = 3𝑐 2 − 3
∴ 3𝑐 2 − 3 = 0
∴ 3(𝑐 − 1)(𝑐 + 1) = 0
∴ 𝑐 = ±1
Recall: 𝑓′(𝑐) = 0
Let’s represent these values on the number line,
−1
1
Then it is clear now that we have the interval as, (−∞, −1) , (−1,1) and (1, ∞)
But we not done, remember we need to check where is the function increasing/decreasing.
And the function 𝑓 is increasing if 𝑓′(𝑥) > 0 and decreasing if 𝑓′(𝑥) < 0.
170 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
So at (−∞, −1)
∴ 𝑓′(𝑥) > 0
Substitute any values of 𝑥 that lie within the intervals (−∞, −1).
Thus, 𝒇 is increasing at (−∞, −𝟏).
And let’s also check at (−1,1)
∴ 𝑓′(𝑥) < 0
Substitute any values of 𝑥 that lie within the intervals (−1,1)
Thus, 𝒇 is decreasing at (−𝟏, 𝟏).
At (1, ∞)
∴ 𝑓′(𝑥) > 0
Substitute any values of 𝑥 that lie within the intervals (∞, 1).
Thus, 𝒇 is increasing at (∞, 𝟏).
2. 𝑓(𝑥) = 3𝑥 4 − 4𝑥 3 + 12𝑥 2 + 5
∴ 𝑓 ′ (𝑥) = 12𝑥 3 − 12𝑥 2 + 24𝑥
∴ 𝑓 ′ (𝑐) = 12𝑐 3 − 12𝑐 2 + 24𝑐
∴ 12𝑐 3 − 12𝑐 2 + 24𝑐 = 0
∴ 12𝑐(𝑐 − 2)(𝑐 + 1) = 0
∴ 𝑐 = 0 𝑜𝑟 − 1 𝑜𝑟 2
Gradient function 𝑓 ′ (𝑥)
Recall: 𝑓′(𝑐) = 0
Always remember to display your critical values on the number line.
−1
0
2
Intervals are, (−∞, −1) , (−1, 0 ) , ( 0, 2 ) 𝑎𝑛𝑑 ( 2, ∞ )
At (−∞, −1)
∴ 𝑓′(𝑥) < 0
Substitute any values of 𝑥 that lie within the intervals (−∞, −1)
Thus, 𝒇 is decreasing at (−∞, −𝟏).
At (−1, 0 )
∴ 𝑓′(𝑥) > 0
Substitute any values of 𝑥 that lie within the intervals (−1, 0).
Thus, 𝒇 is increasing at (−𝟏, 𝟎).
At ( 0, 2 )
∴ 𝑓′(𝑥) < 0
Substitute any values of 𝑥 that lie within the intervals ( 0 , 2 )
Thus, 𝒇 is decreasing at ( 0 , 2 ).
At ( 2, ∞ )
∴ 𝑓′(𝑥) > 0
Substitute any values of 𝑥 that lie within the intervals (2, ∞).
Thus, 𝒇 is increasing at (𝟐, ∞).
171 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
3. 𝑓(𝑥) = 𝑥 + 2 sin 𝑥
0 ≤ 𝑥 ≤ 2𝜋
Let’s first find the gradient function,
∴ 𝑓 ′ (𝑥) = 1 + 2 cos 𝑥
∴ 𝑓 ′ (𝑐) = 1 + 2 cos 𝑐
∴ 1 + 2 cos 𝑐 = 0
1
∴ cos 𝑐 = − 2
∴ 𝑅𝑒𝑓 𝑎𝑛𝑔𝑙𝑒: 𝐶 =
𝜋
3
𝜋
3
Thus, 𝐶 = 𝜋 − =
∴𝑪=
𝟐𝝅
𝟑
𝑶𝒓
At (0 ,
2𝜋
3
𝑜𝑟
𝟒𝝅
𝟑
2𝜋
3
), (
4𝜋
3
2𝜋 4𝜋
,
3 3
2𝜋 4𝜋
,
3 3
) 𝑎𝑛𝑑 (
2𝜋
4𝜋
3
, 2𝜋 )
2𝜋
3
).
𝟐𝝅
).
𝟑
)
∴ 𝑓′(𝑥) < 0
Substitute any values of 𝑥 that lie within the intervals (
Thus, 𝒇 is decreasing at (
4𝜋
3
(Check where cosine is negative)
Substitute any values of 𝑥 that lie within the intervals (0 ,
Thus, 𝒇 is increasing at (𝟎 ,
At (
4𝜋
3
)
∴ 𝑓′(𝑥) > 0
At (
𝜋
3
𝐶=𝜋+ =
, 𝒘𝒉𝒊𝒄𝒉 𝒍𝒊𝒆𝒔 𝒘𝒊𝒕𝒉𝒊𝒏 𝒕𝒉𝒆 𝒈𝒊𝒗𝒆𝒏 𝒅𝒐𝒎𝒂𝒊𝒏 𝟎 ≤ 𝒙 ≤ 𝟐𝝅
2𝜋
3
0
Intervals are, (0 ,
2𝜋
3
2𝜋 4𝜋
,
3 3
2𝜋 4𝜋
,
3 3
).
).
, 2𝜋 )
∴ 𝑓′(𝑥) > 0
Substitute any values of 𝑥 that lie within the intervals (
Thus, 𝒇 is increasing at(
4𝜋
3
4𝜋
3
, 2𝜋 )
.
, 2𝜋 ).
172 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
4. 𝑓(𝑥) = 2𝑥 3 + 9𝑥 2 − 24𝑥 − 10
∴ 𝑓 ′ (𝑥) = 6𝑥 2 + 18𝑥 − 24
∴ 𝑓 ′ (𝑐) = 6𝑐 2 + 18𝑐 − 24
∴ 6𝑐 2 + 18𝑐 − 24 = 0
∴ 6(𝑐 − 1)(𝑐 + 4) = 0
∴ 𝑐 = −4 𝑜𝑟 𝑐 = 1
Always remember to display your critical values on the number line.
−4
1
Intervals are, (−∞ , −4 ) , (−4 , 1 ) 𝑎𝑛𝑑 ( 1 , ∞ )
At (−∞ , −4 )
∴ 𝑓′(𝑥) > 0
Substitute any values of 𝑥 that lie within the intervals(−∞ , −4 ).
Thus, 𝒇 is increasing at (−∞ , −4 ).
At (−4 , 1 )
∴ 𝑓′(𝑥) < 0
Substitute any values of 𝑥 that lie within the intervals(−4 , 1 ).
Thus, 𝒇 is decreasing at (−4 , 1 ).
At ( 1 , ∞ )
∴ 𝑓′(𝑥) > 0
Substitute any values of 𝑥 that lie within the intervals( 1 , ∞ ).
Thus, 𝒇 is increasing at ( 1 , ∞ ).
173 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
5. 𝑓(𝑥) = 𝑥 2 𝑒 −3𝑥
Compute 𝑓′(𝑥) and find all critical numbers:
∴ 𝑓 ′ (𝑥) = 2𝑥𝑒 −3𝑥 − 3𝑥 2 𝑒 −3𝑥
∴ 𝑓 ′ (𝑐) = 2𝑐𝑒 −3𝑐 − 3𝑐 2 𝑒 −3𝑐
∴ 2𝑐𝑒 −3𝑐 − 3𝑐 2 𝑒 −3𝑐 = 0
∴ 𝑐𝑒 −3𝑐 (2 − 3𝑐) = 0
2
∴ 𝑐 = 0 ( 𝑖𝑓 𝑒 −3𝑐 ≠ 0 ) 𝑜𝑟 𝑐 = 3
Always remember to display your critical values on the number line.
2
3
0
2
2
Intervals are, (−∞ , 0 ) , (0 , 3 ) 𝑎𝑛𝑑 ( 3 , ∞ )
At (−∞ , 0 ) ,
∴ 𝑓′(𝑥) < 0
Substitute any values of 𝑥 that lie within the intervals(−∞ , 0 ).
Thus, 𝒇 is decreasing at (−∞ , 0 ).
2
At (0 , 3 )
∴ 𝑓′(𝑥) > 0
2
Substitute any values of 𝑥 that lie within the intervals(0 , 3 ) .
2
Thus, 𝒇 is increasing at (0 , 3 ).
2
At ( 3 , ∞ )
∴ 𝑓′(𝑥) < 0
2
Substitute any values of 𝑥 that lie within the intervals ( 3 , ∞ ).
Thus, 𝒇 is decreasing at (
2
3
, ∞ ).
174 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
Worked Examples:
Example 39:
2022
Determine the Global extreme values.
1. 𝑓(𝑥) =
𝑥3
3
−𝑥
2. 𝑓(𝑥) = 2𝑥 3 − 15𝑥 2 + 36𝑥
1≤𝑥≤5
𝜋
5𝜋
2
3. 𝑓(𝑥) = cos 𝑥
−2 ≤ 𝑥 ≤
4. 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 1
−2 ≤ 𝑥 ≤ 4
5. 𝑓(𝑥) = 𝑥 − 2 sin 𝑥
0 ≤ 𝑥 ≤ 2𝜋
1
Solution:
1. 𝑓(𝑥) =
𝑥3
3
−𝑥
∴ 𝑓′(𝑥) = 𝑥 2 − 1
∴ 𝑓′(𝑐) = 𝑐 2 − 1
∴ 𝑐2 − 1 = 0
∴ 𝑐 = ±1
∴ 𝑓(−1) =
∴ 𝑓(1) =
(−1)3
(1)3
3
3
2
− (−1) = 3
2
− (1) = − 3
We now compare the two values which are
2
3
2
and − 3 , the smaller one will have the global
minimum value at that point and the lager value will have the global maximum value at that
point.
Thus, 𝑓 has a global maximum value at −1.
And 𝑓 has a global minimum value at 1.
175 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
2. 𝑓(𝑥) = 2𝑥 3 − 15𝑥 2 + 36𝑥
∴ 𝑓′(𝑥) = 6𝑥 2 − 30𝑥 + 36
∴ 𝑓′(𝑐) = 6𝑐 2 − 30𝑐 + 36
∴ 6𝑐 2 − 30𝑐 + 36 = 0
∴ 6(𝑐 − 2)(𝑐 − 3) = 0
∴ 𝑐 = 2 𝑜𝑟 𝑐 = 3
1≤𝑥≤5
Hence, 𝑓(1) = 2(1)3 − 15(1)2 + 36(1) = 23
𝑓(2) = 2(2)3 − 15(2)2 + 36(2) = 28
𝑓(3) = 2(3)3 − 15(3)2 + 36(3) = 27
𝑓(5) = 2(5)3 − 15(5)2 + 36(5) = 55
The largest value is 55 and the smallest value is 23.
Thus, 𝑓 has a global maximum value at 5
And 𝑓 has a global minimum value at 1
𝜋
3. 𝑓(𝑥) = cos 𝑥
∴ 𝑓 ′ (𝑥) = − sin 𝑥
∴ 𝑓 ′ (𝑐) = − sin 𝑐
∴ sin 𝑐 = 0
∴ 𝑐 = 0 , 𝜋, 2𝜋
−2 ≤ 𝑥 ≤
5𝜋
2
Hence, 𝑓(0) = cos 0 = 1
𝑓(𝜋) = cos 𝜋 = −1
𝑓(2𝜋) = cos 2𝜋 = 1
𝜋
𝜋
𝑓 (− 2 ) = cos (− 2 ) = 0
5𝜋
5𝜋
𝑓 ( 2 ) = cos ( 2 ) = 0
The largest value is 1 and the smallest value is −1
Thus, 𝑓 has a global maximum value at 0 𝑎𝑛𝑑 2𝜋
And 𝑓 has a global minimum value at 𝜋
176 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Thirteen: Application of Calculus.
4. 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 1
1
2
− ≤𝑥≤4
∴ 𝑓 ′ (𝑥) = 3𝑥 2 − 6𝑥
∴ 𝑓 ′ (𝑐) = 3𝑐 2 − 6𝑐
∴ 3𝑐 2 − 6𝑐 = 0
∴ 3𝑐(𝑐 − 2) = 0
∴ 𝑐 = 0 𝑜𝑟 2
Hence, 𝑓(0) = 1
1
1
𝑓 (− 2) = 8
𝑓(2) = −3
𝑓(4) = 17
The largest value is 17 and the smallest value is −3
Thus, 𝑓 has a global maximum value at 4
And 𝑓 has a global minimum value at 2
5. 𝑓(𝑥) = 𝑥 − 2 sin 𝑥
∴ 𝑓 ′ (𝑥) = 1 − 2 cos 𝑥
∴ 𝑓 ′ (𝑐) = 1 − 2 cos 𝑐
∴ 1 − 2 cos 𝑐 = 0
1
∴ cos 𝑐 = 2
𝜋
∴𝑐=3
0 ≤ 𝑥 ≤ 2𝜋
5𝜋
3
𝑜𝑟
Hence, 𝑓(0) = 0
𝜋
𝜋
5𝜋
5𝜋
3
𝑓 ( 3 ) = 3 − √3 ≈ −0.68453
𝑓(3) =
+ √3 ≈ 6.968039
𝑓(2𝜋) = 2𝜋 ≈ 6.283185
The largest value is 6.968039 and the smallest value is −0.68453
Thus, 𝑓 has a global maximum value at
And 𝑓 has a global minimum value at
𝜋
3
5𝜋
3
177 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER FOURTEEN:
INTEGRATION:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS
178 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
INTEGRATION:
Definition:
 INTEGRATION: is the reversal process of Differentiation.
About the Integration:
 Integration is the most powerful section in Calculus.
 This section requires a solid understanding of Differentiation.
The notation 𝑓(𝑥)𝑑𝑥 is traditionally used for an antiderivative of 𝑓 and is called an Indefinite
Integral. It is advisable that one should go back into differentiation and make sure that chapter is well
understood because this section is very easy if and only if differentiation is not a problem.
NB: 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑥) Means 𝐹′(𝑥) = 𝑓(𝑥)
Properties of Integration:
1.
𝑐 𝑓(𝑥) 𝑑𝑥 = 𝑐 𝑓(𝑥) 𝑑𝑥
2.
[𝑓(𝑥) + 𝑔(𝑥)] 𝑑𝑥 =
𝑓(𝑥) 𝑑𝑥 + 𝑔(𝑥) 𝑑𝑥
3.
[𝑓(𝑥) − 𝑔(𝑥)] 𝑑𝑥 =
𝑓(𝑥) 𝑑𝑥 − 𝑔(𝑥) 𝑑𝑥
NOTE: Sometimes the use of algebraic manipulation or Trigonometric Identities will simplify the
integrand and make the method of integration obvious.
179 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
TOOL-BOX FOR INTEGRATION:
Basic Integrals:
Integrals with Trigonometric
Functions:
1.
𝑑𝑥 = 𝑥 + 𝐶
2.
𝑘 𝑑𝑥 = 𝑘𝑥 + 𝐶 , Where 𝑘 and 𝑐 are Constants.
3.
𝑥 𝑛 𝑑𝑥 =
𝑥 𝑛+1
𝑛+1
+ 𝐶 , Where 𝑛 ≠ −1
1.
2.
sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶
cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶
3.
sin(𝑛𝑥) 𝑑𝑥 = −
4.
cos(𝑛𝑥) 𝑑𝑥 =
5.
6.
7.
8.
9.
10.
11.
12.
tan 𝑥 𝑑𝑥 = − ln|cos 𝑥| + 𝐶 = ln|sec 𝑥| + 𝐶
cot 𝑥 𝑑𝑥 = ln|sin 𝑥| + 𝐶
sec 𝑥 𝑑𝑥 = ln|sec 𝑥 + tan 𝑥| + 𝐶
csc 𝑥 𝑑𝑥 = − ln|csc 𝑥 + cot 𝑥| + 𝐶
sec 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶
csc 2 𝑥 𝑑𝑥 = − cot 𝑥 + 𝐶
sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝐶
csc 𝑥 cot 𝑥 𝑑𝑥 = − cot 𝑥 + 𝐶
Integrals with Natural Logarithmics:
1.
1
𝑑𝑥
𝑥
2.
cos(𝑛𝑥)
+𝐶
𝑛
sin(𝑛𝑥)
+𝐶
𝑛
Integrals with Exponents:
1.
𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶
ln 𝑥 𝑑𝑥 = 𝑥 ln 𝑥 − 𝑥 + 𝐶
2.
𝑒 𝑛𝑥 𝑑𝑥 = 𝑛 𝑒 𝑛𝑥 + 𝐶
3.
1
𝑑𝑥
𝑎𝑥+𝑏
3.
𝑥 𝑒 𝑥 𝑑𝑥 = (𝑥 − 1)𝑒 𝑥 + 𝐶
4.
𝑎 𝑥 𝑑𝑥 = ln 𝑎 𝑎 𝑥 + 𝐶
1
4.
𝑥𝑒 𝑎𝑥 𝑑𝑥 = (𝑎 − 𝑎2 ) 𝑒 𝑎𝑥 + 𝐶
5.
ln 𝑎𝑥
𝑑𝑥
𝑥
1
5.
𝑥 2 𝑒 𝑥 𝑑𝑥 = (𝑥 2 − 2𝑥 + 2)𝑒 𝑥 + 𝐶
6.
𝑥 2 𝑒 𝑎𝑥 𝑑𝑥 = ( 𝑎 − 𝑎2 + 𝑎3 ) 𝑒 𝑎𝑥 + 𝐶
7.
𝑥 3 𝑒 𝑥 𝑑𝑥 = (𝑥 3 − 3𝑥 2 + 6𝑥 − 6)𝑒 𝑥 + 𝐶
8.
𝑥 𝑛 𝑒 𝑎𝑥 𝑑𝑥 =
6.
7.
8.
= ln|𝑥| + 𝐶
1
𝑎
= ln|𝑎𝑥 + 𝑏| + 𝐶
= 2 (ln 𝑎𝑥)2 + 𝐶
ln(𝑎𝑥 + 𝑏) 𝑑𝑥 = (𝑥 +
2
ln(𝑥 + 𝑎
2
ln(𝑥 − 𝑎
2)
2)
𝑏
) ln(𝑎𝑥
𝑎
2
𝑑𝑥 = 𝑥 ln(𝑥 + 𝑎
2
𝑑𝑥 = 𝑥 ln(𝑥 − 𝑎
+ 𝑏) − 𝑥 + 𝐶 , 𝑥 ≠ 0
2)
2)
+
+
𝑥
2𝑎 tan−1
𝑎
𝑥+𝑎
𝑎 ln 𝑥−𝑎
− 2𝑥 + 𝐶
− 2𝑥 + 𝐶
1
𝑥
1
𝑥2
𝑥 𝑛 𝑒 𝑎𝑥
𝑎
2𝑥
2
𝑛
− 𝑎 𝑥 𝑛−1 𝑒 𝑎𝑥 𝑑𝑥 + 𝐶
180 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
INTEGRALS WITH ROOTS:
INTEGRALS WITH RATIONAL
FRACTIONS:
1.
2.
3.
4.
5.
6.
7.
√𝑥 − 𝑎 𝑑𝑥 =
1
𝑑𝑥
√𝑥±𝑎
1
𝑑𝑥
𝑎−𝑥
√
2
(𝑥
3
3
2
− 𝑎) + 𝐶
3
2
2
3
2
5
2𝑏
√𝑎𝑥 + 𝑏 𝑑𝑥 = (3𝑎 +
2𝑥
) √𝑎𝑥
3
14.
15.
3.
1
𝑑𝑥
1+𝑥 2
4.
1
𝑑𝑥
𝑎 2 +𝑥 2
= 3 tan−1 𝑎 + 𝐶
5.
𝑥
𝑑𝑥
𝑎 2 +𝑥 2
= 2 ln|𝑎2 + 𝑥 2 | + 𝐶
+𝑏+𝐶
𝑥
√𝑎−𝑥 𝑑𝑥 = −√𝑥(𝑎 − 𝑥) − 𝑎 tan−1
6.
√𝑥(𝑎−𝑥)
+
𝑥−𝑎
𝐶
7.
𝑥
1
1
√𝑥 2 ±𝑎2
1
𝑥
√𝑥 2 ±𝑎
1
𝑥
√𝑎 2 −𝑥 2
+𝐶
𝑑𝑥 = ln |𝑥 + √𝑥 2 ± 𝑎2 | + 𝐶
𝑥
√𝑎 2 −𝑥 2
𝑑𝑥 = sin−1 𝑎 + 𝐶
𝑥
𝑥2
√𝑥 2 ±𝑎2
𝑛+1
+ 𝐶 , 𝑛 ≠ −1
= tan−1 𝑥 + 𝐶
1
𝑥
1
𝑥2
𝑎 2 +𝑥 2
𝑥3
𝑎 2 +𝑥 2
𝑥
𝑑𝑥 = 𝑥 − 𝑎 tan−1 𝑎 + 𝐶
1
1
𝑑𝑥 = 2 𝑥 2 − 2 𝑎2 ln|𝑎2 + 𝑥 2 | + 𝐶
8.
1
𝑑𝑥
𝑎𝑥 2 +𝑏𝑥+𝑐
9.
1
𝑑𝑥
(𝑥+𝑎)(𝑥+𝑏)
10.
𝑥
𝑑𝑥
(𝑥+𝑎)2
11.
𝑥
𝑑𝑥
𝑎𝑥 2 +𝑏𝑥+𝑐
=
=
2
√4𝑎𝑐−𝑏2
1
tan−1
2𝑎𝑥+𝑏
√4𝑎𝑐−𝑏2
𝑎+𝑥
𝑎
𝑎+𝑥
+ ln|𝑎 + 𝑥| + 𝐶
1
= 2𝑎 ln|𝑎𝑥 2 + 𝑏𝑥 + 𝑐|
𝑏
𝑎√4𝑎𝑐−𝑏2
tan−1
2𝑎𝑥+𝑏
√4𝑎𝑐−𝑏2
𝑑𝑥 = −√𝑥 2 ± 𝑎2 + 𝐶
1
1
𝑑𝑥 = 2 𝑥 √𝑥 2 ± 𝑎2 ± 2 𝑎2 ln |𝑥 + √𝑥 2 ± 𝑎2 | + 𝐶
𝑑𝑥
(𝑎 2 +𝑥 2 )3⁄2
=
𝑥
𝑎 2 √𝑎2 +𝑥 2
+𝐶
= 𝑏−𝑎 ln 𝑏+𝑥 + 𝐶 , 𝑎 ≠ 𝑏
−
𝑑𝑥 = √𝑥 2 ± 𝑎2 + 𝐶
2
√𝑎 2 ±𝑥 2
(𝑥+𝑎)𝑛+1
2
= 3 (𝑥 ± 2𝑎)√𝑥 ± 𝑎 + 𝐶
√𝑎2 − 𝑥 2 𝑑𝑥 = 2 𝑥√𝑎2 − 𝑥 2 + 𝑎2 tan−1
13.
(𝑥 + 𝑎)𝑛 𝑑𝑥 =
5
2
9.
12.
2.
𝑥 √𝑥 − 𝑎 𝑑𝑥 = 𝑎(𝑥 − 𝑎) + (𝑥 − 𝑎) + 𝐶
𝑥
𝑑𝑥
√𝑥±𝑎
1
= − 𝑥+𝑎 + 𝐶
= −2√𝑎 − 𝑥 + 𝐶
√𝑎+𝑥 𝑑𝑥 = √𝑥(𝑎 + 𝑥) − 𝑎 ln[√𝑥 + √𝑥 + 𝑎] + 𝐶
11.
1
𝑑𝑥
(𝑥+𝑎)2
= 2√𝑥 ± 𝑎 + 𝐶
8.
10.
1.
+𝐶
181 | P a g e
+𝐶
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
BASIC INTEGRALS:
Worked Examples:
Example 40:
2022
Evaluate the following Integrals:
a)
𝑥 7 𝑑𝑥
b)
(3𝑥 4 − 5𝑥 + 4)𝑑𝑥
c)
(𝑦 − 3)2 𝑑𝑦
d)
(𝑦 − 3)3 𝑑𝑦
e)
(𝑥−2)(𝑥+3)
5
𝑑𝑥
Solution:
a)
b)
𝑥 7 𝑑𝑥 , Note: 𝑛 = 7 so 𝑛 + 1 = 8
⟹
𝑥 7+1
7+1
⟹
𝑥8
8
+𝐶
Basic integrals from the table.
+𝐶
(3𝑥 4 − 5𝑥 + 4)𝑑𝑥
⟹ 3 𝑥 4 𝑑𝑥 − 5 𝑥 𝑑𝑥 + 4 𝑑𝑥
⟹
3𝑥 5
5
−
5𝑥 2
2
+ 4𝑥 + 𝐶
Basic integrals.
182 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
c)
(𝑦 − 3)2 𝑑𝑦
⟹ (𝑦 2 − 6𝑦 + 9)𝑑𝑦
𝑦 2 𝑑𝑦 − 6 𝑦𝑑𝑦 + 9 𝑑𝑦
⟹
d)
⟹
𝑦3
3
− 6 ( ) + 9𝑦 + 𝐶
𝑦2
2
⟹
𝑦3
3
− 3𝑦 2 + 9𝑦 + 𝐶
Remove the brackets.
Integrate each function within the integrand.
Basic integrals.
(𝑦 − 3)3 𝑑𝑦
⟹ (𝑦 2 − 6𝑦 + 9)(𝑦 − 3)𝑑𝑦
⟹ (𝑦 3 − 6𝑦 2 + 9𝑦 − 3𝑦 2 + 18𝑦 − 27)𝑑𝑦
⟹ (𝑦 3 − 9𝑦 2 + 27𝑦 − 27)𝑑𝑦
⟹
e)
𝑦4
4
− 3𝑦 3 +
(𝑥−2)(𝑥+3)
5
⟹
1
5
27𝑦 3
2
− 27𝑦 + 𝐶
Basic integrals.
𝑑𝑥
(𝑥 − 2)(𝑥 + 3) 𝑑𝑥
1
⟹ 5 (𝑥 2 + 𝑥 − 6)𝑑𝑥
1 𝑥3
⟹ 5( 3 +
𝑥3
𝑥2
𝑥2
2
⟹ 15 + 10 −
− 6𝑥) + 𝐶
6𝑥
5
Basic integrals.
+𝐶
183 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 41:
2022
Evaluate the following Integrals leave your answer in terms of ln where necessary:
a)
1
𝑑𝑥
3𝑥−5
b)
𝑒 8𝑥 𝑑𝑥
c)
𝑒 −5𝑦 𝑑𝑦
d)
31−𝑦 𝑑𝑦
e)
3
√𝑦(6𝑦 2
4
1
− 𝑦 + 4)𝑑𝑦
Solution:
a)
1
𝑑𝑥
3𝑥−5
1
3
⟹ ln|3𝑥 − 5| + 𝐶 From the table of integrals with natural logs.
b)
𝑒 8𝑥 𝑑𝑥
1
⟹ 8 𝑒 8𝑥 + 𝐶
From the table of integrals with Exponents.
184 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
1
c)
𝑒 −5𝑦 𝑑𝑦
1
⟹ −5𝑒 −5𝑦 + 𝐶
d)
31−𝑦 𝑑𝑦
1
⟹ − ln 3 ∙ 31−𝑦 + 𝐶
e)
3
√𝑦(6𝑦 2
4
− 𝑦 + 4)𝑑𝑦
5
3
3
1
⟹ 4 (6𝑦 2 − 𝑦 2 + 4𝑦 2 ) 𝑑𝑦
7
3
⟹ 4(
6𝑦 2
7
2
5
−
𝑦2
5
2
3
+
7
5
⟹
3 12𝑦 2
( 7
4
2𝑦 2
5
7
5
⟹
9𝑦 2
7
3𝑦 2
10
−
−
4𝑦 2
3
2
)+𝐶
3
+
8𝑦 2
)+
3
𝐶
3
+ 2𝑦 2 + 𝐶
185 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 42:
2022
Evaluate the following Integrals leave your answer in terms of ln where necessary:
a)
2+𝑥 2
𝑑𝑥
√𝑥
b)
𝑥 3 +3𝑥 6
𝑑𝑥
𝑥4
c)
𝑦 3 −2𝑦 2 −𝑦
𝑑𝑦
𝑦2
d)
(2𝑥 − 𝑥 2 +
e)
(
1
2
𝑒 3𝑦
5
4
) 𝑑𝑥
√𝑥
3
𝑦
− 2𝑦3 + 𝑒 7 + √2 ) 𝑑𝑦
Solution:
a)
2+𝑥 2
√𝑥
⟹
𝑑𝑥
(
2
𝑥2
√
√𝑥
⟹2
+
𝑥
) 𝑑𝑥
3
1
√
𝑑𝑥 + 𝑥 2 𝑑𝑥
𝑥
1
3
⟹ 2 𝑥 −2 𝑑𝑥 + 𝑥 2 𝑑𝑥
⟹ 2(
1
5
𝑥2
𝑥2
1
2
)+
5
2
+𝐶
5
2𝑥 2
⟹ 4√𝑥 +
+𝐶
5
186 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
b)
𝑥 3 +3𝑥 6
𝑑𝑥
𝑥4
3
𝑥
3𝑥 6
⟹ (𝑥 3 + 𝑥 4 ) 𝑑𝑥
1
⟹ (𝑥 + 3𝑥 2 ) 𝑑𝑥
3
⟹ ln|𝑥| + 𝑥 + 𝐶
c)
𝑦 3 −2𝑦 2 −𝑦
𝑑𝑦
𝑦2
𝑦3
2𝑦 2
⟹ (𝑦2 − 𝑦2
1
𝑦
⟹
⟹
d)
𝑦
− 𝑦2 ) 𝑑𝑦
(𝑦 − 2 − ) 𝑑𝑦
𝑦2
2
− 2𝑦 − ln|𝑦| + 𝐶
1
2
4
) 𝑑𝑥
√𝑥
1
2
4
𝑑𝑥 − 2 𝑑𝑥 +
𝑑𝑥
2𝑥
𝑥
√𝑥
1 1
1
1
𝑑𝑥 − 2 2 𝑑𝑥 + 4
𝑑𝑥
2 𝑥
𝑥
√𝑥
(2𝑥 − 𝑥 2 +
⟹
⟹
1
⟹
1
ln|𝑥| −
2
2(−𝑥
1
2
−1 )
+4(
𝑥2
1
2
)+𝐶
⟹ 2 ln|𝑥| + 𝑥 + 8√𝑥 + 𝐶
e)
𝑒 3𝑦
5
1
⟹5
(
−
3
2𝑦 3
𝑦
2
1
𝑑𝑦
𝑦3
+ 𝑒 7 + √ ) 𝑑𝑦
3
𝑒 3𝑦 𝑑𝑦 − 2
1
+ 𝑒 7 𝑑𝑦 + 2 √𝑦𝑑𝑦
3
⟹
1 1 3𝑦
3 𝑦 −2
(
𝑒
)
−
( )+
5 3
2 −2
𝑒 𝑦+
⟹
1 3𝑦
𝑒
15
1 3
𝑦2
3
+
3
4𝑦 2
7
+𝑒 𝑦+
7
1 𝑦2
( 3 )+𝐶
2
2
+𝐶
187 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 43:
2022
Evaluate the following integral for the Trigonometric functions:
a)
tan2 𝜃 𝑑𝜃
b)
(cot 2 𝛽 − 5𝛽 3 ) 𝑑𝛽
c)
(cos 5𝑥 − sin 3𝑥)𝑑𝑥
d)
(1+𝑥2 − tan 𝑥) 𝑑𝑥
e)
(ln 𝑦 +
f)
tan 𝜃
𝑑𝜃
sec2 𝜃
1
−1
√1−𝑦 2
− 𝑒 3 𝑦) 𝑑𝑦
Solution:
a)
tan2 𝜃 𝑑𝜃
⟹ (sec 2 𝜃 − 1) 𝑑𝜃
⟹
Pythagorean Identity: tan2 𝜃 = sec 2 𝜃 − 1
sec 2 𝜃 𝑑𝜃 − 𝑑𝜃
⟹ tan 𝜃 − 𝜃 + 𝐶
188 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
b)
(cot 2 𝛽 − 5𝛽 3 ) 𝑑𝛽
⟹ cot 2 𝛽𝑑𝛽 − 5 𝛽 3 𝑑𝛽
⟹ (cosec 2 𝛽 − 1) 𝑑𝛽 − 5 𝛽 3 𝑑𝛽
⟹ − cot 𝛽 − 𝛽 −
c)
⟹
cos 5𝑥 𝑑𝑥 − sin 3𝑥 𝑑𝑥
sin 5𝑥
5
+
cos 3𝑥
3
+𝐶
1
(1+𝑥2 − tan 𝑥) 𝑑𝑥
1
𝑑𝑥
1+𝑥 2
−1
⟹
⟹ tan
e)
+𝐶
(cos 5𝑥 − sin 3𝑥)𝑑𝑥
⟹
d)
5𝛽 4
4
(ln 𝑦 +
⟹
− tan 𝑥 𝑑𝑥
𝑥 + ln|cos 𝑥| + 𝐶
−1
√1−𝑦 2
− 𝑒 3 𝑦) 𝑑𝑦
ln 𝑦 𝑑𝑦 +
−1
√1−𝑦 2
𝑑𝑦 − 𝑒 3 𝑦𝑑𝑦
⟹ 𝑦 ln 𝑦 − 𝑦 + cos −1 𝑦 −
f)
𝑒3 2
𝑦
2
+𝐶
tan 𝜃
𝑑𝜃
sec2 𝜃
sin 𝜃
÷ sec 2 𝜃) 𝑑𝜃
cos 𝜃
sin 𝜃
(cos 𝜃 ÷ cos2 𝜃) 𝑑𝜃
⟹ (
⟹
Simply the identity
⟹ sin 𝜃 cos 𝜃 𝑑𝜃
1
sin 2𝜃 𝑑𝜃
2
⟹
Algebraic Manipulation.
1
⟹ 2 sin 2𝜃 𝑑𝜃
1
4
⟹ − cos(2𝜃) + 𝐶
189 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
TECHNIQUES OF INTEGRATION:
In this section we are going to study techniques of integration. In order to master the techniques
explained here it is vital that you undertake plenty of practice exercises so that they become second
nature. It is of much important that students should understand and master each technique of the
integral before attempting any examples and problems, integration might tempt you to use wrong
method if differentiation is not well understood in the previous sections. Hence it is advisable to look at
differentiation again.
TECHNIQUES OF INTEGRATION
1.
2.
3.
4.
5.
U-substitution Integration:
Change of Variable Integration:
Integration By-Parts:
Fractional Integration:
Trigonometric Substitution Integration
1. U-Substitution Integration:
Try to find some function 𝑢 = 𝑔(𝑥) in the Integrand whose differential 𝑑𝑢 = 𝑔′(𝑥)𝑑𝑥 does
occurs, apart from a constant factor.
Worked Examples:
Example 44:
2022
Evaluate the following integrals:
a)
(𝑥 2 + 3𝑥 − 105)(2𝑥 + 3) 𝑑𝑥
b)
𝑥3
𝑑𝑥
𝑥 4 −2
c)
𝑥 2 cos(𝑥 3 ) 𝑑𝑥
d)
𝑦 3 𝑒 𝑦 𝑑𝑦
e)
1
𝑑𝑥
𝑥 ln 𝑥
f)
sin √𝑥
𝑑𝑥
√𝑥
4
190 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
a)
(𝑥 2 + 3𝑥 − 105)(2𝑥 + 3) 𝑑𝑥
Do not be intimidated by this kind of integral, we need to be very careful in letting 𝑢
substitution. Let 𝑢 be that function whose differential is the other function within the integrand,
for example if we let 𝑢 = 𝑥 2 + 3𝑥 − 105 , then 𝑑𝑢 = (2𝑥 + 3)𝑑𝑥, now substitute those
𝑢 and 𝑑𝑢 and integrate with respect to 𝑢 and substitute the values of 𝑢 in terms of 𝑥.
⟹ (𝑥 2 + 3𝑥 − 105)(2𝑥 + 3) 𝑑𝑥
⟹ 𝑢 𝑑𝑢
⟹
⟹
b)
1 2
𝑢 +𝐶
2
1
(𝑥 2 + 3𝑥
2
𝑥3
𝑑𝑥
𝑥 4 −2
− 105)2 + 𝐶
1
𝑥 3 𝑑𝑥
𝑥 4 −2
=
Then the integrand becomes easy now because if we choose 𝑢 = 𝑥 4 − 2, then
𝑑𝑢 = 4𝑥 3 𝑑𝑥, and what we want is 𝑥 3 𝑑𝑥, so if we solve for 𝑥 3 𝑑𝑥 because it is appearing as the
1
function in the integrand, then 𝑥 3 𝑑𝑥 = 4 𝑑𝑢
1 1
∙ 𝑑𝑢
𝑢 4
⟹
⟹
⟹
c)
1
ln|𝑢| + 𝐶
4
1
ln|𝑥 4 − 2| +
4
1
Substitute 𝑢 = 𝑥 4 − 2 and 𝑥 3 𝑑𝑥 = 4 𝑑𝑢
𝐶
𝑥 2 cos(𝑥 3 ) 𝑑𝑥 = cos(𝑥 3 ) 𝑥 2 𝑑𝑥
Let us now think of of U-substitution, and we will have 𝑢 = 𝑥 3 ⟹ 𝑑𝑢 = 3𝑥 2 𝑑𝑥 , and we want
1
to solve for 𝑥 2 𝑑𝑥 because it is appearing as the other function, thus 𝑥 2 𝑑𝑥 = 3 𝑑𝑢
1
⟹
cos(𝑢) 3 𝑑𝑢
1
⟹ 3 cos 𝑢
1
⟹ 3 sin(𝑢) + 𝐶
1
⟹ 3 sin(𝑥 3 ) + 𝐶
191 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
d)
4
4
𝑦 3 𝑒 𝑦 𝑑𝑦 = 𝑒 𝑦 𝑦 3 𝑑𝑦
The problem becomes easy now if we can just think of U-Substitution and we will then
1
have, 𝑢 = 𝑦 4 ⟹ 𝑑𝑢 = 4𝑦 3 𝑑𝑦 then 𝑦 3 𝑑𝑦 = 4 𝑑𝑢 now let’s plug into the integral
problem.
4
⟹
𝑒 𝑦 𝑦 3 𝑑𝑦
⟹
𝑒 𝑢 4 𝑑𝑢
1
1
⟹ 4 𝑒𝑢
1
⟹ 4 𝑒𝑢 + 𝐶
1
4
⟹ 4 𝑒𝑦 + 𝐶
e)
1
𝑑𝑥 =
𝑥 ln 𝑥
1 1
ln 𝑥 𝑥
𝑑𝑥
1
It’s now clear that we will let 𝑢 = ln 𝑥 ⟹ 𝑑𝑢 = 𝑥 𝑑𝑥 , no we have everything to
substitute in the given integral in terms of 𝒖 and not 𝒙.
1 1
⟹
ln 𝑥 𝑥
1
⟹
𝑢
𝑑𝑥
𝑑𝑢
⟹ ln|𝑢| + 𝐶
⟹ ln|ln 𝑥| + 𝐶
f)
sin √𝑥
√𝑥
𝑑𝑥 =
sin √𝑥 .
1
√𝑥
𝑑𝑥 , Note that we didn’t change the formality of the given
problem, we have just noticed that its U-sub, just think of 𝑢 = √𝑥 ⟹ 𝑑𝑢 = 2
1
√𝑥
Note we looking for
⟹
sin √𝑥 .
1
√𝑥
1
√𝑥
𝑑𝑥 , then if solve the equataion we will have 2𝑑𝑢 =
1
√𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
⟹ sin 𝑢 2𝑑𝑢
⟹ 2 sin 𝑢 𝑑𝑢
⟹ −2 cos 𝑢 + 𝐶
⟹ −2 cos √𝑥 + 𝐶
192 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 45:
2022
More Complex Problem:
Evaluate the following integrals:
3 −𝑥
a)
(3𝑥 2 − 1)𝑒 𝑥
b)
3
(𝑥
7
c)
sin 𝑥 √cos 𝑥 𝑑𝑥
d)
sin 𝑥 ln(cos 𝑥) 𝑑𝑥
e)
3 −5 tan 2𝑥
𝑒
sec 2 2𝑥 𝑑𝑥
2
f)
1
𝑑𝑥
𝑥(ln 𝑥)5
− 2)𝑒 2+𝑥
𝑑𝑥
2 −4𝑥
𝑑𝑥
5
Solution:
a)
(3𝑥 2 − 1)𝑒 𝑥
3 −𝑥
𝑑𝑥 =
𝑒𝑥
3 −𝑥
(3𝑥 2 − 1)𝑑𝑥
Let 𝑢 = 𝑥 3 − 𝑥 , 𝑑𝑢 = (3𝑥 2 − 1)𝑑𝑥
⟹ ∫ 𝑒 𝑢 𝑑𝑢
⟹ 𝑒𝑢 + 𝐶
⟹ 𝑒𝑥
3 −𝑥
+𝐶
193 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
b)
3
(𝑥
7
− 2)𝑒 2+𝑥
2 −4𝑥
3
𝑑𝑥 = 7 𝑒 2+𝑥
2 −4𝑥
(𝑥 − 2) 𝑑𝑥
Let 𝑢 = 2 + 𝑥 2 − 4𝑥 , 𝑑𝑢 = (2𝑥 − 4)𝑑𝑥 = 2(𝑥 − 2)𝑑𝑥 ,
= (𝑥 − 2)𝑑𝑥
1
𝑒 𝑢 2 𝑑𝑢
⟹
c)
1
𝑑𝑢
2
⟹
1 𝑢
𝑒
2
⟹
1 2+𝑥 2 −4𝑥
𝑒
2
+𝐶
+𝐶
5
sin 𝑥 √cos 𝑥 𝑑𝑥 =
5
√cos 𝑥 sin 𝑥 𝑑𝑥
Let 𝑢 = cos 𝑥 , 𝑑𝑢 = − sin 𝑥 𝑑𝑥 thus −𝑑𝑢 = sin 𝑥 𝑑𝑥
5
⟹−
√𝑢 𝑑𝑢
5
6
6
⟹ − 𝑢5 + 𝐶
5
6
6
⟹ − (cos 𝑥)5 + 𝐶
d)
sin 𝑥 ln(cos 𝑥) 𝑑𝑥 =
ln(cos 𝑥) sin 𝑥 𝑑𝑥
Let 𝑢 = cos 𝑥 , 𝑑𝑢 = − sin 𝑥 𝑑𝑥 thus −𝑑𝑢 = sin 𝑥 𝑑𝑥
⟹ − ln 𝑢 𝑑𝑢
⟹ −(𝑢 ln 𝑢 − 𝑢) + 𝐶
⟹ −𝑢 ln 𝑢 + 𝑢 + 𝐶
⟹ − cos 𝑥 ln cos 𝑥 + cos 𝑥 + 𝐶
194 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
e)
3 −5 tan 2𝑥
𝑒
sec 2 2𝑥 𝑑𝑥
2
Let 𝑢 = −5 tan 2𝑥
3
3
= 2 𝑒 −5 tan 2𝑥 sec 2 2𝑥 𝑑𝑥
, 𝑑𝑢 = −10sec 2 2𝑥 𝑑𝑥
1
thus − 10 𝑑𝑢 = sec 2 2𝑥 𝑑𝑥
1
⟹ (2) (− 10) 𝑒 𝑢 𝑑𝑢
3
⟹ − 20 𝑒 𝑢 + 𝐶
3
⟹ − 20 𝑒 −5 tan 2𝑥 + 𝐶
f)
1
𝑑𝑥
𝑥(ln 𝑥)5
=
Let 𝑢 = ln 𝑥
⟹
1
𝑑𝑢
𝑢5
⟹
𝑢−5 𝑑𝑢
1
1
∙ 𝑑𝑥
(ln 𝑥)5 𝑥
1
, 𝑑𝑢 = 𝑑𝑥
𝑥
1
4
⟹ − 𝑢−4 + 𝐶
1
⟹ − 4 (ln 𝑥)−4 + 𝐶
195 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
2. Change of Variable Integration:
Try to find some function 𝑢 = 𝑔(𝑥) in the Integrand whose differential 𝑑𝑢 = 𝑔′(𝑥)𝑑𝑥 does not
occurs, apart from a constant factor.
Worked Examples:
Example 46:
2022
Evaluate the following integrals:
a)
b)
𝑥
4
√𝑥+4
𝑥
2𝑥+1
𝑑𝑥
𝑑𝑥
c)
𝑥+3
𝑑𝑥
(𝑥−4)2
d)
𝑥
𝑑𝑥
(𝑥−5)6
e)
𝑥 √𝑥 + 1 𝑑𝑥
f)
(2𝑥 + 3) √2𝑥 − 1 𝑑𝑥
196 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
a)
𝑥
4
√𝑥+4
𝑑𝑥
Let 𝑢 = 𝑥 + 4
𝑢−4
⟹
1
⟹ 𝑥 = 𝑢 − 4 and so 𝑑𝑥 = 𝑑𝑢
𝑑𝑢
𝑢4
3
−1
⟹
(𝑢4 − 2𝑢 4 ) 𝑑𝑢
⟹
𝑢4 𝑑𝑢 − 2𝑢 4 𝑑𝑢
3
⟹
−1
7
3
4𝑢4
𝑢4
7
4
− 2(
3
4
7
4
7
)+𝐶
8
3
3
⟹ (𝑥 + 4 )4 − (𝑥 + 4 )4 + 𝐶
b)
𝑥
𝑑𝑥
2𝑥+1
1
Let 𝑢 = 2𝑥 + 1
1
1
𝑢−
2
2
⟹
𝑢
1
1
⟹ 𝑥 = 2 𝑢 − 2 and so 𝑑𝑥 = 2 𝑑𝑢
1
∙ 2 𝑑𝑢
1
2
⟹
1
2
1
(2
⟹
1
4
(1 − ) 𝑑𝑢
− 𝑢 ) 𝑑𝑢
1
𝑢
1
4
⟹ (𝑢 − ln|𝑢|) + 𝐶
1
1
⟹ 4 (2𝑥 + 1) − 4 ln|2𝑥 + 1| + 𝐶
197 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
c)
𝑥+3
𝑑𝑥
(𝑥−4)2
Let 𝑢 = 𝑥 − 4 , ⟹ 𝑥 = 𝑢 + 4 and so 𝑑𝑥 = 𝑑𝑢
⟹
𝑢+7
𝑑𝑢
𝑢2
⟹
(𝑢 + 7𝑢−2 ) 𝑑𝑢
⟹
1
𝑑𝑢
𝑢
1
+ 7 𝑢−2 𝑑𝑢
7
𝑢
⟹ ln|𝑢| − + 𝐶
⟹ ln|𝑥 − 4| −
d)
7
+
𝑥−4
𝐶
𝑥
𝑑𝑥
(𝑥−5)6
Let 𝑢 = 𝑥 − 5 , ⟹ 𝑥 = 𝑢 + 5 and so 𝑑𝑥 = 𝑑𝑢
⟹
𝑢+5
𝑑𝑢
𝑢6
⟹ (𝑢−5 + 5𝑢−6 ) 𝑑𝑢
1
4
1
5
⟹ − 𝑢−4 + 5 (− 𝑢−5 ) + 𝐶
1
⟹ − 4 (𝑥 − 5)−4 − (𝑥 − 5)−5 + 𝐶
1
1
⟹ − 4(𝑥−5)4 − (𝑥−5)5 + 𝐶
198 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
1
e)
𝑥 √𝑥 + 1 𝑑𝑥 = 𝑥(𝑥 + 1)2 𝑑𝑥
Let 𝑢 = 𝑥 + 1 , 𝑥 = 𝑢 − 1 and so 𝑑𝑥 = 𝑑𝑢
⟹ (𝑢 − 1) √𝑢 𝑑𝑢
3
⟹
𝑢2 𝑑𝑢 − √𝑢 𝑑𝑢
5
⟹
2𝑢2
5
2
3
− 3 𝑢2 + 𝐶
5
2
3
2
⟹ 5 (𝑥 + 1)2 − 3 (𝑥 + 1)2 + 𝐶
1
f)
(2𝑥 + 3) √2𝑥 − 1 𝑑𝑥 = (2𝑥 + 3)(2𝑥 − 1)2 𝑑𝑥
1
1
1
1
Let 𝑢 = 2𝑥 − 1 , ⟹ 𝑥 = 2 𝑢 + 2 but 2𝑥 + 3 = 2 (2 𝑢 + 2) + 3 = 𝑢 + 4
1
And thus 𝑑𝑥 = 2 𝑑𝑢
⟹
1
2
(𝑢 + 4) √𝑢𝑑𝑢
⟹
1
2
(𝑢2 + 4√𝑢) 𝑑𝑢
3
5
1
⟹ 2(
⟹
2𝑢2
5
1
(2𝑥
5
2
3
+ 4 (3 𝑢2 )) + 𝐶
5
2
− 1) +
4
((𝑥
3
3
3 2
2
+ 1) ) + 𝐶
199 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
3. Integration by Parts:
This technique is particularly useful for integrating the products of two functions containing
continuity characteristics. Remember the Product Rule states that if 𝑓 and 𝑔 are differentiable
functions, then
𝑑
[𝑓(𝑥)𝑔(𝑥)]
𝑑𝑥
= 𝑓(𝑥)𝑔′(𝑥) + 𝑔′(𝑥)𝑓(𝑥) if we think of integrating both sides we
will then have,
𝑑
[𝑓(𝑥)𝑔(𝑥)] 𝑑𝑥
𝑑𝑥
=
𝑓(𝑥)𝑔′(𝑥)𝑑𝑥 + 𝑔(𝑥)𝑓′(𝑥)𝑑𝑥
⟹ 𝑓(𝑥)𝑔(𝑥) = 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥 + 𝑔(𝑥)𝑓′(𝑥)𝑑𝑥
⟹ 𝑓(𝑥)𝑔′(𝑥)𝑑𝑥 = 𝑓(𝑥)𝑔(𝑥) − 𝑔(𝑥)𝑓′(𝑥)𝑑𝑥
Equation
1
𝒇(𝒙)𝒈′(𝒙)𝒅𝒙 = 𝒇(𝒙)𝒈(𝒙) − 𝒈(𝒙)𝒇′(𝒙)𝒅𝒙
Hence the formula is called the The Formula for Integration by Parts. It is perhaps
easier to remember in the following notation. Let 𝑢 = 𝑓(𝑥) and 𝑣 = 𝑔(𝑥). Then the
differentials are, 𝑑𝑢 = 𝑓′(𝑥)𝑑𝑥 and 𝑑𝑣 = 𝑔′(𝑥)𝑑𝑥 , so by the Substitution Rule, the formula
for integration by parts becomes.
Equation
2
∫ 𝒖𝒅𝒗 = 𝒖𝒗 − ∫ 𝒗 𝒅𝒖
Guidelines for Selecting 𝒖 and 𝒅𝒗
(There are always exceptions, but these are generally helpful.)
“ L-I-A-T-E ” Choose "𝒖" to be the function that comes first in this list:
1.
2.
3.
4.
5.
Logarithmic Function
Inverse Trig Function
Algebraic Function
Trig Function
Exponential Function
Alternative General Guidelines for Choosing 𝒖 𝐚𝐧𝐝 𝒅𝒖.
1. Let 𝒅𝒗 be the most complicated portion of the integrand that can be “easily” integrated
2. Let 𝒖 be that portion of the integrand whose derivative 𝒅𝒖 is the “simpler” function that 𝒖
itself.
200 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 46:
2022
Evaluate the following integrals:
a)
𝑥 𝑒 −𝑥 𝑑𝑥
b)
𝑥 cos 𝑥 𝑑𝑥
c)
ln 𝑥 𝑑𝑥
d)
𝑥 7 ln 𝑥 𝑑𝑥
e)
cos 𝑥 ln(sin 𝑥) 𝑑𝑥
f)
cos −1 𝑥 𝑑𝑥
g)
𝑦 3 √4 − 𝑦 2 𝑑𝑦
h)
𝑥 sin 𝑥 cos 𝑥 𝑑𝑥
201 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
a)
𝑥 𝑒 −𝑥 𝑑𝑥, by mere looking at the problem one should be able to see that this problem is not a
U-substitution either change of variable integrals. So let’s now think of Integration by-parts since
we see two functions being multiplied.
𝑥 𝑒 −𝑥 𝑑𝑥
Let 𝑢 = 𝑥 and 𝑑𝑣 = 𝑒 −𝑥 𝑑𝑥 (Algebraic comes before the exponential function)
⟹ 𝑑𝑢 = 𝑑𝑥 And 𝑣 = 𝑒 −𝑥 𝑑𝑥 = −𝑒 −𝑥
Thus
b)
𝑥 cos 𝑥 𝑑𝑥 , it’s clear by now that the problem is not a U-sub, but by Parts Integrand.
Let 𝑢 = 𝑥 and 𝑑𝑣 = cos 𝑥 𝑑𝑥 (Algebraic comes before Trig functions using LIATE)
⟹ 𝑑𝑢 = 𝑑𝑥 And 𝑣 = cos 𝑥 𝑑𝑥 = sin 𝑥
Thus
c)
𝑥 𝑒 −𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= −𝑥𝑒 −𝑥 − (−𝑒 −𝑥 ) 𝑑𝑥
= −𝑥𝑒 −𝑥 + 𝑒 −𝑥 𝑑𝑥
= −𝑥𝑒 −𝑥 − 𝑒 −𝑥 + 𝐶
= 𝑒 −𝑥 (−𝑥 − 1) + 𝐶
𝑥 cos 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= 𝑥 sin 𝑥 − sin 𝑥 𝑑𝑥
= 𝑥 sin 𝑥 − (− cos 𝑥) + 𝐶
= 𝑥 sin 𝑥 + cos 𝑥 + 𝐶
ln 𝑥 𝑑𝑥 , In this problem for to be tempted to say the integral for this problem is 1⁄𝑥 and is
totally wrong! It’s clear that the problem is By-Parts.
Let 𝑢 = ln 𝑥 and 𝑑𝑣 = 𝑑𝑥 (Logarithmic comes before algebraic function)
1
⟹ 𝑑𝑢 = 𝑥 𝑑𝑢 And 𝑣 =
Thus
𝑑𝑥 = 𝑥
ln 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
1
= 𝑥 ln 𝑥 − 𝑥 ∙ 𝑥 𝑑𝑥
= 𝑥 ln 𝑥 − 𝑑𝑥
= 𝑥 ln 𝑥 − 𝑥 + 𝐶
202 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
d)
𝑥 7 ln 𝑥 𝑑𝑥 , now let 𝑢 = ln 𝑥 and 𝑑𝑣 = 𝑥 7 𝑑𝑥
1
⟹ 𝑑𝑢 = 𝑥 𝑑𝑥 𝑎𝑛𝑑 𝑣 =
∴
𝑥 7 𝑑𝑥 =
𝑥8
8
𝐥𝐧 𝒙 is a natural
logarithmic and 𝒙𝟕 is an
Algebraic function so
using the sequence we
can see that
𝑳 𝑐𝑜𝑚𝑒 𝑏𝑒𝑓𝑜𝑟𝑒 𝑨
𝑥 7 ln 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
𝑥8
𝑥8
1
= (ln 𝑥) ( 8 ) − ( 8 ) (𝑥 𝑑𝑥)
𝑥8
1 7
𝑥 𝑑𝑥
8
= (ln 𝑥) ( 8 ) −
𝑥8
1
= (ln 𝑥) ( 8 ) − 64 𝑥 8 + 𝐶
e)
cos 𝑥 ln(sin 𝑥) 𝑑𝑥 , This type of integral can be integrated using both techniques, Usubstitution and Integration by parts. So in this case we are going to use Integration by parts.
You can also verify your answer by using U-substitution.
Thus 𝑢 = ln(sin 𝑥) and 𝑑𝑣 = cos 𝑥 𝑑𝑥
1
⟹ 𝑑𝑢 = sin 𝑥 (cos 𝑥)𝑑𝑥 = cot 𝑥 𝑑𝑥
⟹ 𝑣=
∴
cos 𝑥 𝑑𝑥 = sin 𝑥
cos 𝑥 ln(sin 𝑥) 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= sin 𝑥 ln(sin 𝑥) − sin 𝑥 (cot 𝑥 𝑑𝑥 )
cos 𝑥
= sin 𝑥 ln(sin 𝑥) − (sin 𝑥) ( sin 𝑥 ) 𝑑𝑥
= sin 𝑥 ln(sin 𝑥) − cos 𝑥 𝑑𝑥
= sin 𝑥 ln(sin 𝑥) − sin 𝑥 + 𝐶
203 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
f)
cos−1 𝑥 𝑑𝑥 , This is the most trick question in the exam, it’s hard for students to see that this is
By-Parts integration even though the other function which is 1 does not appear clearly but it
reflect as 1𝑑𝑥.
So following the sequence we see that Trig Inverse function comes before the algebraic
function.
So 𝑢 = cos −1 𝑥
⟹ 𝑑𝑢 = −
∴
and 𝑑𝑣 = 1𝑑𝑥
1
√1−𝑥 2
𝑑𝑥
And 𝑣 =
1𝑑𝑥 = 𝑥
cos −1 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= 𝑥 cos−1 𝑥 − (𝑥) (−
= 𝑥 cos−1 𝑥 +
But we notice that
𝑥
√1−𝑥 2
) 𝑑𝑥
𝑥𝑑𝑥
𝑑𝑥 can be integrated by U-Substitutions.
1
So 𝑢 = 1 − 𝑥 2
⟹
𝑥
√1−𝑥 2
1
√1−𝑥 2
1
√1−𝑥 2
⟹ 𝑑𝑢 = −2𝑥𝑑𝑥 thus − 2 𝑑𝑢 = 𝑥𝑑𝑥
𝑑𝑥 =
1
1
(− 2 𝑑𝑢)
√𝑢
1
1
2
= (− ) (
𝑢2
1
2
)+𝐶
= −√1 − 𝑥 2 + 𝐶
∴
cos −1 𝑥 𝑑𝑥 = 𝑥 cos−1 𝑥 +
1
√1−𝑥 2
𝑥𝑑𝑥
= 𝑥 cos−1 𝑥 − √1 − 𝑥 2 + 𝐶
204 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
g)
𝑦 3 √4 − 𝑦 2 𝑑𝑦 = 𝑦 2 𝑦√4 − 𝑦 2 𝑑𝑦
#Since both of these are algebraic functions, the LIATE Rule of Thumb is not helpful. Applying
Case ll of the alternative guidelines above, we see that 𝑦√4 − 𝑦 2 is the “most complicated part
of the integrand that can easily be integrated.” Therefore:
𝑦 3 √4 − 𝑦 2 𝑑𝑦 =
⟹
𝑦 2 𝑦√4 − 𝑦 2 𝑑𝑦
Let 𝑢 = 𝑦 2 And 𝑑𝑣 = 𝑦√4 − 𝑦 2 𝑑𝑦
⟹ 𝑑𝑢 = 2𝑦𝑑𝑦
And 𝑣 =
𝑦√1 − 𝑦 2 𝑑𝑦 this is U-Sub so let’s suppose that we have
1
2
𝑢 = 4 − 𝑦 2 And 𝑑𝑢 = −2𝑦𝑑𝑦 𝑡ℎ𝑢𝑠 − 𝑑𝑢 = 𝑦𝑑𝑦
1
∴
1
1
1
3
1
𝑢2 (− 2 𝑑𝑢) = − 2 𝑢2 𝑑𝑢 = − 3 (4 − 𝑦 2 )2
So now we have 𝑢 = 𝑦 2 𝑎𝑛𝑑 𝑑𝑢 = 2𝑦𝑑𝑦
And
Hence,
1
3
3
1
3
1
1
3
𝑑𝑣 = 𝑦√4 − 𝑦 2 𝑑𝑦 𝑎𝑛𝑑 𝑣 = − (4 − 𝑦 2 )2
𝑦 2 𝑦√4 − 𝑦 2 𝑑𝑦 = 𝑢𝑣 − 𝑣 𝑑𝑢
3
= (𝑦 2 ) (− 3 (4 − 𝑦 2 )2 ) − (− 3 (4 − 𝑦 2 )2 ) (2𝑦)𝑑𝑦
3
1
= (𝑦 2 ) (− 3 (4 − 𝑦 2 )2 ) + 3 ((4 − 𝑦 2 )2 ) (2𝑦)𝑑𝑦
Again we see that
1
3
3
5
1
2
((4 − 𝑦 2 )2 ) (2𝑦)𝑑𝑦 = − 3 (4 − 𝑦 2 )2 (5) by U-Substitution:
3
1
3
1
∴
𝑦 2 𝑦√4 − 𝑦 2 𝑑𝑦 = (𝑦 2 ) (− 3 (4 − 𝑦 2 )2 ) + 3 ((4 − 𝑦 2 )2 ) (2𝑦)𝑑𝑦
∴
𝑦 2 𝑦√4 − 𝑦 2 𝑑𝑦 = −
𝑦2
(4 −
3
3
2
5
𝑦 2 )2 − 15 (4 − 𝑦 2 )2 + 𝐶
205 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
h)
1
𝑥 sin 𝑥 cos 𝑥 𝑑𝑥 = 2 𝑥 sin 2𝑥 𝑑𝑥
1
Remember that sin 𝑥 cos 𝑥 = 2 sin 2𝑥 now 𝑢 = 𝑥 𝐴𝑛𝑑 𝑑𝑣 = sin 2𝑥 𝑑𝑥
⟹ 𝑑𝑢 = 𝑑𝑥
Thus,
𝐻𝑒𝑛𝑐𝑒,
𝑣=
1
sin 2𝑥 𝑑𝑥 = − 2 cos 2𝑥
𝑥 sin 2𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
1
2
=−
𝑥 cos 2𝑥
2
− (− cos 2𝑥) 𝑑𝑥
=−
𝑥 cos 2𝑥
2
+ 2 cos 2𝑥 𝑑𝑥
=−
𝑥 cos 2𝑥
2
+ 4 sin 2𝑥 + 𝐶
𝑥 sin 2𝑥 𝑑𝑥 = −
1
2
1
1
𝑥 cos 2𝑥
4
1
+ 8 sin 2𝑥 + 𝐶
206 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Using Repeated Application of Integration by Parts:
Sometimes integration by parts must be repeated to obtain an answer.
Note: DO NOT switch choices for " 𝒖 𝒂𝒏𝒅 𝒅𝒗 " in successive applications.
Worked Examples:
Example 47:
2022
Evaluate the following integrals:
a)
𝑥 2 sin 𝑥 𝑑𝑥
b)
(ln 𝑥)2 𝑑𝑥
c)
𝑒 𝑥 cos 𝑥 𝑑𝑥
d)
𝑒 𝑥 sin 𝑥 𝑑𝑥
e)
𝑒 2𝑥 sin 2𝑥 𝑑𝑥
207 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
a)
𝑥 2 sin 𝑥 𝑑𝑥 , remember for us to be able to choose the correct substitution we need to
understand sequence so A comes before T,
Thus 𝑢 = 𝑥 2 and 𝑑𝑣 = sin 𝑥 𝑑𝑥
⟹ 𝑑𝑢 = 2𝑥𝑑𝑥
𝑎𝑛𝑑
𝑣=
sin 𝑥 𝑑𝑥 = − cos 𝑥
𝑥 2 sin 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= −𝑥 2 cos 𝑥 − (− cos 𝑥) (2𝑥𝑑𝑥)
= −𝑥 2 cos 𝑥 + 2 𝑥 cos 𝑥 𝑑𝑥
So 𝑥 cos 𝑥 𝑑𝑥 it’s by part again so remember we don’t change our choice for selecting
" 𝑢 𝑎𝑛𝑑 𝑑𝑣 ".
Now let 𝑢 = 𝑥 and 𝑑𝑣 = cos 𝑥 𝑑𝑥
⟹ 𝑑𝑢 = 𝑑𝑥 𝑎𝑛𝑑 𝑣 =
⟹
cos 𝑥 𝑑𝑥 = sin 𝑥
𝑥 cos 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= 𝑥 sin 𝑥 − sin 𝑥 𝑑𝑥
= 𝑥 sin 𝑥 + sin 𝑥
But 2 𝑥 cos 𝑥 𝑑𝑥 = 2𝑥 sin 𝑥 + 2 cos 𝑥
So
𝑥 2 sin 𝑥 𝑑𝑥 = −𝑥 2 cos 𝑥 + 2 𝑥 cos 𝑥 𝑑𝑥
= −𝑥 2 cos 𝑥 + 2𝑥 sin 𝑥 + 2 cos 𝑥.
208 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
b)
(ln 𝑥)2 𝑑𝑥
Since (ln 𝑥)2 is a natural logarithmic function and 1𝑑𝑥 is an Algebraic function
Then let 𝑢 = (ln 𝑥)2 and 𝑑𝑣 = 1𝑑𝑥
1
⟹ 𝑑𝑢 = 2(ln 𝑥) 𝑥 𝑑𝑥 And 𝑣 = 𝑥
⟹ (ln 𝑥)2 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
1
𝑥
= 𝑥(ln 𝑥)2 − 𝑥 (2(ln 𝑥) ) 𝑑𝑥
= 𝑥(ln 𝑥)2 − 2 ln 𝑥 𝑑𝑥
But we notice that,
ln 𝑥 𝑑𝑥 it’s By-Part again,
Thus, Let 𝑢 = ln 𝑥 and 𝑑𝑣 = 𝑑𝑥
1
𝑥
⟹ 𝑑𝑢 = 𝑑𝑢 And 𝑣 =
Thus
𝑑𝑥 = 𝑥
ln 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
1
= 𝑥 ln 𝑥 − 𝑥 ∙ 𝑥 𝑑𝑥
= 𝑥 ln 𝑥 − 𝑑𝑥
= 𝑥 ln 𝑥 − 𝑥 + 𝐶
∴ (ln 𝑥)2 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= 𝑥(ln 𝑥)2 − 2 ln 𝑥 𝑑𝑥
= 𝑥(ln 𝑥)2 − 2(𝑥 ln 𝑥 − 𝑥) + 𝐶
= 𝑥(ln 𝑥)2 − 2𝑥 ln 𝑥 + 2𝑥 + 𝐶
209 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
𝑒 𝑥 cos 𝑥 𝑑𝑥
c)
Using LIATE Rule we notice that 𝑇 𝑐𝑜𝑚𝑒𝑠 𝑏𝑒𝑓𝑜𝑟𝑒 𝐸
𝑢 = cos 𝑥 (𝑇𝑟𝑖𝑔 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛) and 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 (𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛)
⟹ 𝑑𝑢 = − sin 𝑥 𝑑𝑥 𝑎𝑛𝑑 𝑣 = 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥
𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= 𝑒 𝑥 cos 𝑥 − 𝑒 𝑥 (− sin 𝑥)𝑑𝑥
= 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 𝑑𝑥
𝑺𝒆𝒄𝒐𝒏𝒅 𝒂𝒑𝒑𝒍𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒊𝒐𝒏 𝒃𝒚 𝒑𝒂𝒓𝒕𝒔:
𝑢 = sin 𝑥 (𝑀𝑎𝑘𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 Choices 𝑓𝑜𝑟 𝑢 𝑎𝑛𝑑 𝑑𝑣 )
𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 ⟹ 𝑑𝑢 = cos 𝑥 𝑎𝑛𝑑 𝑣 = 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥
⟹
𝑒 𝑥 sin 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
⟹
𝑒 𝑥 sin 𝑥 𝑑𝑥 = 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 𝑑𝑥
But, 𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 𝑑𝑥
= 𝑒 𝑥 cos 𝑥 𝑒 𝑥 + sin 𝑥 − 𝑒 𝑥 cos 𝑥 𝑑𝑥 + 𝐶
Note: the appearance of original integral on right side of equation. Move to left side and
solve the integral as follows:
⟹
𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑒 𝑥 cos 𝑥 𝑒 𝑥 + sin 𝑥 − 𝑒 𝑥 cos 𝑥 𝑑𝑥 + 𝐶
⟹
𝑒 𝑥 cos 𝑥 𝑑𝑥 + 𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 + 𝐶
⟹ 2 𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 + 𝐶
∴
1
2
𝑒 𝑥 cos 𝑥 𝑑𝑥 = (𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥) + 𝐶
210 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
𝑒 𝑥 sin 𝑥 𝑑𝑥
d)
Using LIATE Rule we notice that 𝑇 𝑐𝑜𝑚𝑒𝑠 𝑏𝑒𝑓𝑜𝑟𝑒 𝐸
𝑢 = sin 𝑥 (𝑇𝑟𝑖𝑔 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛) And 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥 (𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛
⟹ 𝑑𝑢 = cos 𝑥 𝑑𝑥 𝑎𝑛𝑑 𝑣 = 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥
𝑒 𝑥 sin 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 𝑑𝑥
= 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 𝑑𝑥
𝑺𝒆𝒄𝒐𝒏𝒅 𝒂𝒑𝒑𝒍𝒊𝒄𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒊𝒏𝒕𝒆𝒈𝒓𝒂𝒕𝒊𝒐𝒏 𝒃𝒚 𝒑𝒂𝒓𝒕𝒔:
𝑢 = cos 𝑥 (𝑀𝑎𝑘𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 Choices 𝑓𝑜𝑟 𝑢 𝑎𝑛𝑑 𝑑𝑣 ) And 𝑑𝑣 = 𝑒 𝑥 𝑑𝑥
⟹ 𝑑𝑢 = − sin 𝑥 𝑎𝑛𝑑 𝑣 = 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥
𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
𝑒 𝑥 cos 𝑥 𝑑𝑥 = 𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 (sin 𝑥)𝑑𝑥
But,
𝑒 𝑥 sin 𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
= 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 𝑑𝑥
= 𝑒 𝑥 sin 𝑥 − (𝑒 𝑥 cos 𝑥 + 𝑒 𝑥 sin 𝑥 𝑑𝑥)
= 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 − 𝑒 𝑥 sin 𝑥 𝑑𝑥 + 𝐶
Note: the appearance of original integral on right side of equation. Move to left side and
solve the integral as follows:
⟹
𝑒 𝑥 sin 𝑥 𝑑𝑥 = 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 − 𝑒 𝑥 sin 𝑥 𝑑𝑥 + 𝐶
⟹
𝑒 𝑥 sin 𝑥 𝑑𝑥 + 𝑒 𝑥 sin 𝑥 𝑑𝑥 = 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 + 𝐶
⟹ 2 𝑒 𝑥 sin 𝑥 𝑑𝑥 = 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 + 𝐶
∴
1
𝑒 𝑥 sin 𝑥 𝑑𝑥 = 2 (𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥) + 𝐶
211 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
e)
𝑒 2𝑥 sin 2𝑥 𝑑𝑥
Let 𝑢 = 𝑒 2𝑥 𝐴𝑛𝑑 𝑑𝑣 = sin 2𝑥 𝑑𝑥
⟹ 𝑑𝑢 = 2𝑒 2𝑥 𝑑𝑥 𝐴𝑛𝑑 𝑣 =
1
sin 2𝑥 𝑑𝑥 = − 2 cos 2𝑥
𝑒 2𝑥 sin 2𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
=−
𝑒 2𝑥 cos 2𝑥
2
− (− 2 cos 2𝑥) (2𝑒 2𝑥 )𝑑𝑥
=−
𝑒 2𝑥 cos 2𝑥
2
+ 𝑒 2𝑥 cos 2𝑥 𝑑𝑥
1
Let 𝑢 = 𝑒 2𝑥 𝐴𝑛𝑑 𝑑𝑣 = cos 2𝑥 𝑑𝑥
1
2
𝑑𝑢 = 2𝑒 2𝑥 𝑑𝑥 𝐴𝑛𝑑 𝑣 =
cos 2𝑥 𝑑𝑥 = sin 2𝑥
𝑒 2𝑥 cos 2𝑥 𝑑𝑥 =
𝑒 2𝑥 sin 2𝑥
2
− ( sin 2𝑥) (2𝑒 2𝑥 )𝑑𝑥
=
𝑒 2𝑥 sin 2𝑥
2
− 𝑒 2𝑥 sin 2𝑥 𝑑𝑥
𝐵𝑢𝑡,
𝑒 2𝑥 sin 2𝑥 𝑑𝑥 = −
𝑒 2𝑥 cos 2𝑥
2
+ 𝑒 2𝑥 cos 2𝑥 𝑑𝑥
=−
𝑒 2𝑥 cos 2𝑥
2
+
⟹ 2 𝑒 2𝑥 sin 2𝑥 𝑑𝑥 = −
⟹
∴
1
2
1
2
𝑒 2𝑥 sin 2𝑥 𝑑𝑥 = (−
𝑒 2𝑥 sin 2𝑥 =
𝑒 2𝑥 sin 2𝑥
4
𝑒 2𝑥 sin 2𝑥
2
− 𝑒 2𝑥 sin 2𝑥 𝑑𝑥 + 𝐶
𝑒 2𝑥 cos 2𝑥
2
+
𝑒 2𝑥 sin 2𝑥
2
𝑒 2𝑥 cos 2𝑥
2
+
𝑒 2𝑥 sin 2𝑥
)+
2
−
𝑒 2𝑥 cos 2𝑥
4
𝐶
+𝐶
212 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
4. Partial Fractions Integration:
In this section we show how to integrate any Rational function (a ratio of polynomials) by
expressing it as a sum of simpler fractions, called 𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠, that we already know
how to integrate.
1. Case l: Denominator with only Linear factors.
𝒇(𝒙)
𝑨
𝑩
= (𝒂𝒙+𝒃) + (𝒄𝒙+𝒅) +
(𝒂𝒙+𝒃)(𝒄𝒙+𝒅)(𝒆𝒙+𝒇)
𝑪
(𝒆𝒙+𝒇)
2. Case ll: Denominator with a Quadratic factors.
𝒇(𝒙)
(𝒂𝒙𝟐 +𝒃𝒙+𝒄)(𝒅𝒙+𝒆)
=
𝑨𝒙+𝑩
𝑪
(𝒂𝒙𝟐 +𝒃𝒙+𝒄)
+ (𝒅𝒙+𝒆)
3. Case lll: Denominator with a Repeated Linear factors.
𝒇(𝒙)
𝑨
𝑩
𝑪
= (𝒂𝒙+𝒃) + (𝒂𝒙+𝒃)𝟐 + (𝒂𝒙+𝒃)𝟑 +
(𝒂𝒙+𝒃)𝟑 (𝒄𝒙+𝒅)
𝑫
𝒄𝒙+𝒅
213 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 48:
2022
Evaluate the following integrals:
a)
𝑥−9
𝑑𝑥
(𝑥+5)(𝑥−2)
b)
𝑥−4
𝑑𝑥
𝑥 2 −5𝑥+6
c)
2𝑥 2 −𝑥+4
𝑑𝑥
𝑥 3 +4𝑥
d)
10
𝑑𝑥
(𝑥−1)(𝑥 2 +9)
e)
4𝑥 2 −7𝑥−12
𝑑𝑥
𝑥(𝑥+2)(𝑥−3)
f)
1
𝑑𝑥
𝑥 3 −1
g)
𝑥 3 +2𝑥
𝑑𝑥
𝑥 4 +4𝑥 2 +3
h)
𝑥 3 +𝑥 2 +2𝑥+1
(𝑥 2 +1)(𝑥 2 +2)
214 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
𝑥−9
𝑑𝑥
(𝑥+5)(𝑥−2)
a)
𝑥−9
(𝑥+5)(𝑥−2)
𝐴
𝐵
= 𝑥+5 + 𝑥−2 Clear fractions and solve for 𝐴 & 𝐵
⟹ 𝑥 − 9 = 𝐴(𝑥 − 2) + 𝐵(𝑥 + 5)
Now let 𝑥 = 2 , thus −7 = 7𝐵 ⟹ 𝐵 = −1
Now let 𝑥 = −5 Thus −14 = −7𝐴
b)
So
𝑥−9
(𝑥+5)(𝑥−2)
2
∴
𝑥−9
𝑑𝑥
(𝑥+5)(𝑥−2)
−1
= 𝑥+5 + 𝑥−2
∴
2
𝑑𝑥
𝑥+5
∴
𝑥−9
𝑑𝑥
(𝑥+5)(𝑥−2)
−
⟹ 𝐴=2
=
2
−1
(𝑥+5 + 𝑥−2) 𝑑𝑥
1
𝑑𝑥
𝑥−2
= 2 ln|𝑥 + 5| − ln|𝑥 − 2| + 𝐶
𝑥−4
𝑑𝑥
𝑥 2 −5𝑥+6
⟹
𝑥−4
𝑥 2 −5𝑥+6
𝑥−4
= (𝑥−3)(𝑥−2) =
𝐴
𝐵
+
𝑥−3
𝑥−2
∴ 𝑥 − 4 = 𝐴(𝑥 − 2) + 𝐵(𝑥 − 3)
Let 𝑥 = 2 to eliminate 𝐴 and solve for 𝐵.
∴ 2 − 4 = 𝐴(2 − 2) + 𝐵(2 − 3)
∴ −2 = −𝐵
Thus, 𝐵 = 2
and
let 𝑥 = 3 to eliminate 𝐵 and solve for 𝐴.
∴ 3 − 4 = 𝐴(3 − 2) + 𝐵(3 − 3)
∴ −1 = 𝐴
Thus, 𝐴 = −1
𝑥−4
−1
2
+
𝑥−3
𝑥−2
−1
2
(𝑥−3 + 𝑥−2) 𝑑𝑥
∴ (𝑥−3)(𝑥−2) =
⟹
⟹ − ln|𝑥 − 3| + 2 ln|𝑥 − 2| + 𝐶
215 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
2𝑥 2 −𝑥+4
c)
𝑥 3 +4𝑥
∴
2𝑥 2 −𝑥+4
𝑥(𝑥 2 +4)
𝑑𝑥
𝐴
𝐵𝑥+𝐶
= 𝑥 + 𝑥 2 +4 ,
Clear fractions and solve for 𝐴, 𝐵& 𝐶
∴ 2𝑥 2 − 𝑥 + 4 = 𝐴(𝑥 2 + 4) + (𝐵𝑥 + 𝐶)𝑥
∴ 2𝑥 2 − 𝑥 + 4 = (𝐴 + 𝐵)𝑥 2 + 𝐶𝑥 + 4𝐴
Equating the coefficients, we obtain
𝐴 + 𝐵 = 2 , 𝐶 = −1 , 4𝐴 = 4
Therefore, 𝐴 = 1 , 𝐵 = 1 , 𝐶 = −1
∴
2𝑥 2 −𝑥+4
𝑑𝑥
𝑥 3 +4𝑥
1
𝑥−1
=
(𝑥 + 𝑥 2 +4) 𝑑𝑥
=
( +
𝑥
1
− 2 ) 𝑑𝑥
𝑥 2 +4
𝑥 +4
=
1
𝑑𝑥
𝑥
+
1
𝑥
1
𝑥
𝑑𝑥
𝑥 2 +4
−
1
𝑑𝑥
𝑥 2 +4
1
𝑥
= ln|𝑥| + 2 ln|𝑥 2 + 4| − 2 tan−1 (2) + 𝐶
216 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
d)
10
𝑑𝑥
(𝑥−1)(𝑥 2 +9)
This is case ll from the table of partial integration, so it becomes easy now to integrate if we can
see which techniques of integral to follow. Once you can be able to solve the values of
𝐴, 𝐵 𝑎𝑛𝑑 𝐶 then you are done.
10
𝐴
𝐵𝑥+𝐶
∴ (𝑥−1)(𝑥2 +9) = 𝑥−1 + 𝑥 2 +9
⟹ 10 = 𝐴(𝑥 2 + 9) + (𝐵𝑥 + 𝐶)(𝑥 − 1)
⟹Let 𝑥 = 1 then 10 = 𝐴(1 + 9)
⟹Let 𝑥 = 0
⟹ 𝐴=1
then 10 = 9𝐴 − 𝐶 but 𝐴 = 1 ⟹ 𝐶 = −1
⟹ 10 = 𝐴𝑥 2 + 9𝐴 + 𝐵𝑥 2 − 𝐵𝑥 + 𝐶𝑥 − 𝐶
⟹ 10 = 𝑥 2 (𝐴 + 𝐵) + 𝑥(𝐶 − 𝐵) + (9𝐴 − 𝐶)
⟹ 𝐴 + 𝐵 = 0 , 𝐶 − 𝐵 = 0 , 9𝐴 − 𝐶 = 10
𝐻𝑒𝑛𝑐𝑒 𝐵 = −1
⟹
1
𝑑𝑥
𝑥−1
+
−𝑥−1
𝑑𝑥
𝑥 2 +9
⟹
1
𝑑𝑥
𝑥−1
−
−𝑥
𝑑𝑥
𝑥 2 +9
1
+
−1
𝑑𝑥
𝑥 2 +9
1
𝑥
⟹ ln|𝑥 − 1| − 2 ln|𝑥 2 + 9| − 3 tan−1 (3) + 𝐶
217 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
4𝑥 2 −7𝑥−12
𝑑𝑥
𝑥(𝑥+2)(𝑥−3)
e)
⟹
4𝑥 2 −7𝑥−12
𝑥(𝑥+2)(𝑥−3)
𝐴
𝑥
= +
𝐵
𝑥+2
+
𝐶
𝑥−3
∴ 4𝑥 2 − 7𝑥 − 12 = 𝐴(𝑥 + 2)(𝑥 − 3) + 𝐵(𝑥)(𝑥 − 3) + 𝐶(𝑥)(𝑥 + 2)
Let 𝑥 = 0 to eliminate 𝐵 and 𝐶.
∴ 4(0)2 − 7(0) − 12 = 𝐴(0 + 2)(0 − 3) + 𝐵(0)(0 − 3) + 𝐶(0)(0 + 2)
∴ −12 = −6𝐴
Thus, 𝐴 = 2
Now let 𝑥 = 3 to eliminate 𝐴 and 𝐵.
∴ 4(3)2 − 7(3) − 12 = 𝐴(3 + 2)(3 − 3) + 𝐵(3)(3 − 3) + 𝐶(3)(3 + 2)
∴ 3 = 15𝐶
1
Thus, 𝐶 = 5
Now let 𝑥 = −2 to eliminate 𝐴 and 𝐶.
∴ 4(−2)2 − 7(−2) − 12 = 𝐴(−2 + 2)(−2 − 3) + 𝐵(−2)(−2 − 3) + 𝐶(−2)(−2 + 2)
∴ 18 = 10𝐵
9
Thus, 𝐵 = 5
∴
4𝑥 2 −7𝑥−12
𝑥(𝑥+2)(𝑥−3)
⟹
2
(𝑥 +
2
𝑥
= +
9
5
𝑥+2
9
+
9
5
𝑥+2
1
5
𝑥−3
+
1
5
𝑥−3
) 𝑑𝑥
1
⟹ 2 ln|𝑥| + 5 ln|𝑥 + 2| + 5 ln|𝑥 − 3| + 𝐶
218 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
f)
1
𝑑𝑥
𝑥 3 −1
1
1
⟹ 𝑥 3 −1 = (𝑥−1)(𝑥2 +𝑥+1)
𝐴
𝐵𝑥+𝐶
= 𝑥−1 + 𝑥 2 +𝑥+1
∴ 1 = 𝐴(𝑥 2 + 𝑥 + 1) + (𝐵𝑥 + 𝐶)(𝑥 − 1)
∴ 1 = 𝐴𝑥 2 + 𝐴𝑥 + 𝐴 + 𝐵𝑥 2 − 𝐵𝑥 + 𝐶𝑥 − 𝐶
∴ 1 = 𝑥 2 (𝐴 + 𝐵) + 𝑥(𝐴 − 𝐵 + 𝐶) + (𝐴 − 𝐵)
∴ 𝐴 + 𝐵 = 0 ……………….………… (1)
∴ 𝐴 − 𝐵 + 𝐶 = 0 ..………………… (2)
∴ 𝐴 − 𝐵 = 1 …..………………………. (3)
From (1) 𝐴 = −𝐵……………………. (4)
Substitute (4) into (3) ⟹ −2𝐵 = 1
1
Thus, 𝐵 = − 2 and
1
1
𝐴=2
1
Hence, 2 + 2 + 𝐶 = 0 ⟹ 𝐶 = −1
1
𝑥 3 −1
⟹
1
=
1
(𝑥−1)(𝑥 2 +𝑥+1)
=
1
2
𝑥−1
+
1
2
𝑥 2 +𝑥+1
− 𝑥−1
1
− 𝑥−1
2
+ 𝑥 22+𝑥+1) 𝑑𝑥
( 𝑥−1
⟹
1
3
1
6
⟹ ln|𝑥 − 1| − ln|𝑥 2 + 𝑥 + 1| −
1
2𝑥+1
tan−1 (
)+
√3
√3
𝐶
219 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
g)
𝑥 3 +2𝑥
𝑑𝑥
𝑥 4 +4𝑥 2 +3
⟹
𝑥 3 +2𝑥
𝑥 4 +4𝑥 2 +3
𝑥 3 +2𝑥
= (𝑥 2 +1)(𝑥2 +3) =
𝐴𝑥+𝐵
𝑥 2 +1
𝐶𝑥+𝐷
+ 𝑥 2 +3
⟹ 𝑥 3 + 2𝑥 = (𝐴𝑥 + 𝐵)(𝑥 2 + 3) + (𝐶𝑥 + 𝐷)(𝑥 2 + 1)
⟹ 𝑥 3 + 2𝑥 = 𝐴𝑥 3 + 3𝑥𝐴 + 𝐵𝑥 2 + 3𝐵 + 𝐶𝑥 3 + 𝐶𝑥 + 𝐷𝑥 2 + 𝐷
⟹ 𝑥 3 + 2𝑥 = 𝑥 3 (𝐴 + 𝐶) + 𝑥 2 (𝐵 + 𝐷) + 𝑥(3𝐴 + 𝐶) + (3𝐵 + 𝐷)
∴ 𝐴 + 𝐶 = 1 ………………………... (1)
∴ 𝐵 + 𝐷 = 0 ………………………… (2)
∴ 3𝐴 + 𝐶 = 2 ………………………. (3)
∴ 3𝐵 + 𝐶 = 2 ………………………. (4)
From (1) 𝐴 = 1 − 𝐶 ……………. (5)
Substitute (5) into (3) ⟹ 3 − 3𝐶 + 𝐶 = 2
1
⟹ 𝐶=
1
2
1
1
And 𝐴 = 2 , Substitute 𝐶 into (4) ⟹ 3𝐵 + 2 = 2 ⟹
⟹
𝑥 3 +2𝑥
𝑥 4 +4𝑥 2 +3
1
⟹
⟹
𝑥+
1
𝑥 3 +2𝑥
1
𝑥+
1
1
𝑥−
𝐵=2
1
and 𝐷 = − 2
1
= (𝑥 2 +1)(𝑥2 +3) = 𝑥2 2 +12 + 𝑥2 2 +32
1
𝑥−
1
(𝑥2 2 +12 + 𝑥2 2 +32 ) 𝑑𝑥
1
ln|𝑥 2
4
1
+ 1| + 2 tan−1 𝑥 +
1
ln|𝑥 2
4
+ 3| − 2
1
√3
𝑥
tan−1 ( 3) + 𝐶
√
220 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
h)
𝑥 3 +𝑥 2 +2𝑥+1
(𝑥 2 +1)(𝑥 2 +2)
𝑥 3 +𝑥 2 +2𝑥+1
⟹ (𝑥2 +1)(𝑥2 +2) =
𝐴𝑥+𝐵
𝑥 2 +1
𝐶𝑥+𝐷
+ 𝑥 2 +2
⟹ 𝑥 3 + 𝑥 2 + 2𝑥 + 1 = (𝐴𝑥 + 𝐵)(𝑥 2 + 2) + (𝐶𝑥 + 𝐷)(𝑥 2 + 1)
= 𝐴𝑥 3 + 2𝑥𝐴 + 𝐵𝑥 2 + 2𝐵 + 𝐶𝑥 3 + 𝐶𝑥 + 𝐷𝑥 2 + 𝐷
= 𝑥 3 (𝐴 + 𝐶) + 𝑥 2 (𝐵 + 𝐷) + 𝑥(2𝐴 + 𝐶) + (2𝐵 + 𝐷)
∴ 𝐴 + 𝐶 = 1 ………………………... (1)
∴ 𝐵 + 𝐷 = 1 ………………………… (2)
∴ 2𝐴 + 𝐶 = 2 ………………………. (3)
∴ 2𝐵 + 𝐶 = 1 ………………………. (4)
From (1) 𝐴 = 1 − 𝐶 …………….. (5)
Substitute (5) into (3) ⟹ 2 − 2𝐶 + 𝐶 = 2
⟹ 𝐶=0
And 𝐴 = 1 , Substitute 𝐶 into (4) ⟹ 2𝐵 + 0 = 1 ⟹
𝑥 3 +𝑥 2 +2𝑥+1
𝑥+
1
2
⟹ (𝑥2 +1)(𝑥2 +2) = 𝑥 2 +1
+
⟹
⟹
𝑥+
1
2
+
(𝑥2 +1
1
ln|𝑥 2
2
1
2
𝑥 2 +2
𝐵=
1
2
and 𝐷 =
1
2
1
2
𝑥 2 +2
) 𝑑𝑥
1
+ 1| + 2 tan−1 𝑥 +
1
𝑥
tan−1 ( 2) +
√2
√
𝐶
221 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
EVALUATING PRODUCT TO SUM INTEGRALS:
To evaluate the integrals:
a)
sin 𝑚𝑥 cos 𝑛𝑥 𝑑𝑥 (b) cos 𝑚𝑥 sin 𝑛𝑥 𝑑𝑥 (c)
Where 𝑚 and 𝑛 are real numbers.
cos 𝑚𝑥 cos 𝑛𝑥 𝑑𝑥 and d) sin 𝑚𝑥 sin 𝑛𝑥 𝑑𝑥
Such integrals could be integrated or evaluated by using integration by parts, but it’s easier to use the
preceding identities.
Make use of the corresponding identity:
1
5. sin 𝜃 cos 𝛽 = 2 [ sin(𝜃 + 𝛽) + sin(𝜃 − 𝛽) ]
1
6. cos 𝜃 sin 𝛽 = 2 [ sin(𝜃 + 𝛽) − sin(𝜃 − 𝛽) ]
1
7. cos 𝜃 cos 𝛽 = 2 [ cos(𝜃 + 𝛽) + cos(𝜃 − 𝛽) ]
1
8. sin 𝜃 sin 𝛽 = 2 [ cos(𝜃 − 𝛽) − cos(𝜃 + 𝛽) ]
Worked Examples:
Example 49:
2022
Evaluate the following integrals:
1.
sin 3𝑥 cos 6𝑥 𝑑𝑥
2.
cos 4𝑥 sin 2𝑥 𝑑𝑥
3.
cos 5𝑥 cos 2𝑥 𝑑𝑥
4.
sin 3𝑥 sin 8𝑥 𝑑𝑥
222 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
sin 3𝑥 cos 6𝑥 𝑑𝑥
1.
∴
sin 3𝑥 cos 6𝑥 𝑑𝑥 =
=
=
=
=
1
[sin(3𝑥
2
+ 6𝑥) + sin(3𝑥 − 6𝑥)]𝑑𝑥
1
[sin(4𝑥
2
+ 2𝑥) − sin(4𝑥 − 2𝑥)]𝑑𝑥
1
(sin 9𝑥 + sin(−3𝑥)) 𝑑𝑥
2
1
(sin 9𝑥 − sin 3𝑥) 𝑑𝑥
2
1
1
1
(− cos 9𝑥 + cos 3𝑥)
2
9
3
1
1
cos 3𝑥 − 18 cos 9𝑥 + 𝐶
6
cos 4𝑥 sin 2𝑥 𝑑𝑥
2.
∴
cos 4𝑥 sin 2𝑥 𝑑𝑥 =
=
=
=
1
(sin 6𝑥 + sin 2𝑥) 𝑑𝑥
2
1
1
1
(− cos 6𝑥 − cos 2𝑥) +
2
6
2
1
1
− 12 cos 6𝑥 − 4 cos 2𝑥 + 𝐶
𝐶
cos 5𝑥 cos 2𝑥 𝑑𝑥
3.
∴
cos 5𝑥 cos 2𝑥 𝑑𝑥 =
1
1
[
2
cos(5𝑥 + 2𝑥) + cos(5𝑥 − 2𝑥) ] 𝑑𝑥
= 2 (cos(7𝑥) + cos(3𝑥)) 𝑑𝑥
1 1
1
= 2 (7 sin 7𝑥 + 3 sin 3𝑥) + 𝐶
1
1
= 14 sin 7𝑥 + 6 sin 𝑥 + 𝐶
sin 3𝑥 sin 8𝑥 𝑑𝑥
4.
∴
sin 3𝑥 sin 8𝑥 𝑑𝑥 =
1
1
[
2
cos(3𝑥 − 8𝑥) − cos(3𝑥 + 8𝑥) ] 𝑑𝑥
= 2 (cos(−5𝑥) + cos(11𝑥)) 𝑑𝑥
1
= 2 (cos(5𝑥) + cos(11𝑥)) 𝑑𝑥
1 1
2 5
1
sin 5𝑥
10
= ( sin 5𝑥 +
=
1
1
sin 11𝑥) +
11
𝐶
+ 22 sin 11𝑥 + 𝐶
223 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
INTEGRATION OF TRIGONOMETRIC FUNCTIONS INVOLVING POWERES
ABSTRACT
In this section we use Trigonometric Identities to integrate a certain combination of Trigonometric
functions. Students may want to review some basic trigonometric identities before reading further.
 cos2 𝑥 + sin2 𝑥 = 1
 1 + tan2 𝑥 = sec 2 𝑥
1−cos 2𝑥
 sin2 𝑥 =
2
 cos 𝑥 =
2
1+cos 2𝑥
2
 sin 2𝑥 = 2 sin 𝑥 cos 𝑥
1
 sin 𝑥 cos 𝑥 = 2 sin 2𝑥
The Powers of Sine , Cosine , Secant and Tangent:
1. Strategies for Evaluating
𝐬𝐢𝐧𝒏 𝒙 𝒅𝒙 ,
𝐜𝐨𝐬 𝒏 𝒙 𝒅𝒙 &
𝐬𝐢𝐧𝒎 𝒙 𝐜𝐨𝐬𝒏 𝒙 𝒅𝒙
a) If the powers of both Sine and Cosine are even, use the half angle identities.
sin2 𝑥 =
1−cos 2𝑥
2
And
cos 2 𝑥 =
1+cos 2𝑥
2
b) If the power of Cosine is Odd, save one Cosine factor and use the Identity:
cos 2 𝑥 = 1 − sin2 𝑥 to express the remaining factors in terms of Sine and let 𝑢 = sin 𝑥
c) If the power of Sine is Odd , save one Sine factor factors and use the Identity:
sin2 𝑥 = 1 − cos2 𝑥 , to express the remaining factors in terms of Cosine and let 𝑢 = cos 𝑥
1
d) It is helpful to use the identity sin 𝑥 cos 𝑥 = 2 sin 2𝑥
2. Strategies for Evaluating
𝐭𝐚𝐧𝒎 𝒙 𝐬𝐞𝐜 𝒏 𝒙 𝒅𝒙
a) If the power of the Secant is Even, save one factor and use of 𝐬𝐞𝐜 𝟐 𝒙 and use 𝐬𝐞𝐜 𝟐 𝒙 = 𝟏 + 𝐭𝐚𝐧𝟐 𝒙 to express
the remaining factors in terms of tan 𝑥 And let 𝑢 = tan 𝑥
b) If the power of the tangent is Odd, save one factor for sec 𝑥 tan 𝑥 and use tan2 𝑥 = sec 2 − 1 to express the
remaining factors in terms of sec 𝑥 And let 𝑢 = sec 𝑥
224 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 50:
2022
Evaluate the following integrals:
1.
sin2 𝑥 𝑑𝑥
2.
cos2 𝑥 𝑑𝑥
3.
tan2 𝑥 𝑑𝑥
4.
sin3 𝑥 𝑑𝑥
5.
cos3 𝑥 𝑑𝑥
6.
tan3 𝑥 𝑑𝑥
Solution:
1.
sin2 𝑥 𝑑𝑥 =
1−cos 2𝑥
) 𝑑𝑥
2
(
1
Half angle identity: sin2 𝑥 =
1−cos 2𝑥
2
= 2 (1 − cos 2𝑥) 𝑑𝑥
1
1
= 2 (𝑥 − 2 sin 2𝑥) + 𝐶
1
1
= 2 𝑥 − 4 sin 2𝑥 + 𝐶
2.
cos2 𝑥 𝑑𝑥 =
1+cos 2𝑥
) 𝑑𝑥
2
(
1
Half angle identity: cos2 𝑥 =
1+cos 2𝑥
2
= 2 (1 + cos 2𝑥) 𝑑𝑥
1
2
1
2
= (𝑥 + sin 2𝑥) + 𝐶
1
1
= 2 𝑥 + 4 sin 2𝑥 + 𝐶
225 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
3.
tan2 𝑥 𝑑𝑥 = (sec 2 𝜃 − 1) 𝑑𝜃
= sec 2 𝜃 𝑑𝜃 − 𝑑𝜃
Pythagorean Identity: tan2 𝑥 = sec 2 𝜃 − 1
= tan 𝜃 − 𝜃 + 𝐶
4.
sin3 𝑥 𝑑𝑥 = sin2 𝑥 sin 𝑥 𝑑𝑥
= (1 − cos 2 𝑥) sin 𝑥 𝑑𝑥
Save one Sine factor factors and use the Identity:
Pythagorean Identity: [ sin2 𝑥 = 1 − cos2 𝑥 ]
Then let 𝑢 = cos 𝑥 , 𝑑𝑢 = − sin 𝑥 𝑑𝑥 𝑡ℎ𝑢𝑠 − 𝑑𝑢 = sin 𝑥 𝑑𝑥
= − (1 − 𝑢2 ) 𝑑𝑢
U-Substitution.
1
3
= − (𝑢 − 𝑢3 ) + 𝐶
1
= − cos 𝑥 + 3 cos3 𝑥 + 𝐶
5.
cos3 𝑥 𝑑𝑥 = cos 2 𝑥 cos 𝑥 𝑑𝑥
= (1 − sin2 𝑥) cos 𝑥 𝑑𝑥
Save one Cosine factor factors and use the Identity:
Pythagorean Identity: [ cos2 𝑥 = 1 − sin2 𝑥 ]
Then let 𝑢 = sin 𝑥 , 𝑑𝑢 = cos 𝑥 𝑑𝑥
= (1 − 𝑢2 ) 𝑑𝑢
U-Substitution.
1
3
= 𝑢 − 𝑢3 + 𝐶
1
= sin 𝑥 − 3 sin3 𝑥 + 𝐶
6.
tan3 𝑥 𝑑𝑥 = tan2 𝑥 tan 𝑥 𝑑𝑥
= (sec 2 𝑥 − 1) tan 𝑥 𝑑𝑥
tan 𝑥 sec 2 𝑥 𝑑𝑥 − tan 𝑥 𝑑𝑥
=
Then let 𝑢 = tan 𝑥
=
Save one factor for tan 𝑥
Pythagorean Identity: tan2 𝑥 = sec 2 𝜃 − 1
⟹ 𝑑𝑢 = sec 2 𝑥 𝑑𝑥
𝑢 𝑑𝑢 − tan 𝑥 𝑑𝑥
=
𝑢2
2
=
tan2 𝑥
2
U-Substitution.
+ ln|cos 𝑥| + 𝐶
− ln|sec 𝑥| + 𝐶
226 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 51:
2022
Evaluate the following integrals:
1.
(1 + cos 𝜃)2 𝑑𝜃
2.
cos4 𝑥 𝑑𝑥
3.
sin5 𝑥 𝑑𝑥
4.
tan5 𝑥 𝑑𝑥
5.
sec 6 𝑥 𝑑𝑥
6.
sin3 𝑥 cos 2 𝑥𝑑𝑥
Solution:
(1 + cos 𝜃)2 𝑑𝜃
1.
∴
(1 + cos 𝜃)2 𝑑𝜃 = (1 + 2 cos 𝜃 + cos2 𝜃) 𝑑𝜃
=
𝑑𝜃 + 2 cos 𝜃 𝑑𝜃 + cos 2 𝜃 𝑑𝜃
=
𝑑𝜃 + 2 cos 𝜃 𝑑𝜃 + 2 (1 + cos 2𝜃) 𝑑𝜃
1
1
1
= 𝜃 + 2 sin 𝜃 + 2 𝜃 + 4 sin 2𝜃 + 𝐶
3
1
= 2 𝜃 + 2 sin 𝜃 + 4 sin 2𝜃 + 𝐶
227 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
2.
cos4 𝑥 𝑑𝑥 = (cos 2 𝑥)2 𝑑𝑥 = (cos 2 𝑥) (cos2 𝑥)𝑑𝑥
1+cos 2𝑥
1+cos 2𝑥
) ( 2 ) 𝑑𝑥
2
=
(
1
= 4 (1 + cos 2𝑥) (1 + cos 2𝑥)𝑑𝑥
1
= 4 (1 + 2 cos 2𝑥 + cos 2 2𝑥)
1
1
1
= 4 𝑑𝑥 + 2 cos 2𝑥 𝑑𝑥 + 4 cos2 2𝑥 𝑑𝑥
1
1
1
1+cos 4𝑥
)+𝐶
2
1
4
1
4
1
8
(1 + cos 4𝑥) 𝑑𝑥
1
1
1
= 4 𝑥 + 4 sin 2𝑥 + 4 (
= 𝑥 + sin 2𝑥 +
1
= 4 𝑥 + 4 sin 2𝑥 + 8 𝑥 + 32 sin 4𝑥 + 𝐶
3
1
1
1
= 8 𝑥 + 4 sin 2𝑥 + 8 𝑥 + 32 sin 4𝑥 + 𝐶
3.
sin5 𝑥 𝑑𝑥 =
=
sin3 𝑥 sin2 𝑥 𝑑𝑥
sin3 𝑥 (1 − cos2 𝑥)𝑑𝑥
=
sin3 𝑥 𝑑𝑥 − sin3 𝑥 cos2 𝑥𝑑𝑥
=
sin2 𝑥 sin 𝑥 𝑑𝑥 − sin2 𝑥cos2 𝑥 sin 𝑥 𝑑𝑥
= (1 − cos2 𝑥) sin 𝑥 𝑑𝑥 − (1 − cos 2 𝑥) cos2 𝑥 sin 𝑥 𝑑𝑥
Then let 𝑢 = cos 𝑥 𝑑𝑢 = − sin 𝑥 𝑑𝑥 𝑡ℎ𝑢𝑠 − 𝑑𝑢 = sin 𝑥 𝑑𝑥
= − (1 − 𝑢2 ) 𝑑𝑢 − (−1) (1 − 𝑢2 ) 𝑢2 𝑑𝑢
= − (1 − 𝑢2 ) 𝑑𝑢 + (𝑢2 − 𝑢4 ) 𝑑𝑢
= −𝑢 +
𝑢3
3
+
= − cos 𝑥 +
𝑢3
3
−
𝑢5
5
2cos3 𝑥
3
+𝐶
−
cos5 𝑥
5
+𝐶
228 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
4.
tan5 𝑥 𝑑𝑥
∴
tan4 𝑥 tan 𝑥 𝑑𝑥 = (sec 2 𝑥 − 1)2 tan 𝑥 𝑑𝑥
= sec 4 𝑥 tan 𝑥 𝑑𝑥 − 2 sec 2 𝑥 tan 𝑥 𝑑𝑥 + tan 𝑥 𝑑𝑥
= sec 3 𝑥 sec 𝑥 tan 𝑥 𝑑𝑥 − 2 sec 2 𝑥 tan 𝑥 𝑑𝑥 + tan 𝑥 𝑑𝑥
notice here we le𝑡 𝑢 = sec 𝑥 for sec 3 𝑥 sec 𝑥 tan 𝑥 𝑑𝑥
And we let 𝑢 = tan 𝑥 for 2 sec 2 𝑥 tan 𝑥 𝑑𝑥
1
4
1
sec 4 𝑥
4
= sec 4 𝑥 − tan2 𝑥 − ln|cos 𝑥| + 𝐶
=
5.
− tan2 𝑥 + ln|sec 𝑥| + 𝐶
sec 6 𝑥 𝑑𝑥
∴
sec 6 𝑥 𝑑𝑥 = sec 4 𝑥 sec 2 𝑥 𝑑𝑥
= (1 + tan2 𝑥)2 sec 2 𝑥 𝑑𝑥
[ where 𝑢 = tan 𝑥 , 𝑑𝑢 = sec 2 𝑥 𝑑𝑥 ]
= (1 + 𝑢2 )2 𝑑𝑢
4
2
= (𝑢 + 2𝑢 + 1) 𝑑𝑢
1
2
= 5 𝑢5 + 3 𝑢3 + 𝑢 + 𝐶
1
2
= 5 tan5 𝑥 + 3 tan3 𝑥 + tan 𝑥 + 𝐶
6.
sin3 𝑥 cos 2 𝑥𝑑𝑥
sin3 𝑥 cos 2 𝑥𝑑𝑥 = sin2 𝑥cos2 𝑥 sin 𝑥 𝑑𝑥
= (1 − cos2 𝑥) cos2 𝑥 sin 𝑥 𝑑𝑥
Then let 𝑢 = cos 𝑥 𝑑𝑢 = − sin 𝑥 𝑑𝑥 𝑡ℎ𝑢𝑠 − 𝑑𝑢 = sin 𝑥 𝑑𝑥
= − (1 − 𝑢2 ) 𝑢2 𝑑𝑢
= − (𝑢2 − 𝑢4 ) 𝑑𝑢
=−
𝑢3
3
+
𝑢5
5
+𝐶
=−
cos3 𝑥
3
+
cos5 𝑥
5
+𝐶
229 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Worked Examples:
Example 52:
2022
More Complex Problems:
Evaluate the following integrals:
1.
tan3 𝑥 sec 𝑥 𝑑𝑥
2.
sin3 𝑥 √cos 𝑥 𝑑𝑥
3.
cos2 𝑥 tan3 𝑥 𝑑𝑥
4.
cos5 𝑥 sin4 𝑥 𝑑𝑥
5.
tan6 𝑥 sec 4 𝑥𝑑𝑥
6.
tan5 𝑥 sec 7 𝑥𝑑𝑥
7.
cot 3 𝑥 cosec 3 𝑥 𝑑𝑥
230 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
tan3 𝑥 sec 𝑥 𝑑𝑥
1.
∴
tan3 𝑥 sec 𝑥 𝑑𝑥 = tan2 𝑥 sec 𝑥 tan 𝑥 𝑑𝑥
= (sec 2 𝑥 − 1) sec 𝑥 tan 𝑥 𝑑𝑥
= (𝑢2 − 1)𝑑𝑢
where 𝑢 = sec 𝑥 , 𝑑𝑢 = sec 𝑥 tan 𝑥 𝑑𝑥
1
3
1
sec 3 𝑥
3
= 𝑢3 − 𝑢 + 𝐶
=
− sec 𝑥 + 𝐶
Sin3 𝑥 √cos 𝑥 𝑑𝑥
2.
∴
sin3 𝑥 √cos 𝑥 𝑑𝑥 = sin2 𝑥 √cos 𝑥 sin 𝑥 𝑑𝑥
= (1 − cos2 𝑥) √cos 𝑥 sin 𝑥 𝑑𝑥
= (1 − 𝑢2 )√𝑢 (−𝑑𝑢) [ where 𝑢 = cos 𝑥 , −𝑑𝑢 = sin 𝑥 𝑑𝑥 ]
5
=
1
(𝑢2 − 𝑢2 ) 𝑑𝑢
2
7
3
2
= 7 𝑢2 − 3 𝑢2 + 𝐶
7
2
2
3
= 7 (cos 𝑥)2 − 3 (cos 𝑥)2 + 𝐶
cos2 𝑥 tan3 𝑥 𝑑𝑥
3.
∴
sin3 𝑥
𝑑𝑥
cos 𝑥
(1−cos2 𝑥)
cos2 𝑥 tan3 𝑥 𝑑𝑥 =
=
cos 𝑥
(1−𝑢2 )
=
𝑢
sin 𝑥 𝑑𝑥
(−𝑑𝑢) [ where 𝑢 = cos 𝑥 , −𝑑𝑢 = sin 𝑥 𝑑𝑥 ]
1
2
= − ln|𝑢| + 𝑢2 + 𝐶
1
= 2 cos2 𝑥 − ln|cos 𝑥| + 𝐶
231 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
cos5 𝑥 sin4 𝑥 𝑑𝑥
4.
∴
cos 5 𝑥 sin4 𝑥 𝑑𝑥 = cos 4 𝑥 sin4 𝑥 cos 𝑥 𝑑𝑥
= (1 − sin2 𝑥)2 sin4 𝑥 cos 𝑥 𝑑𝑥
[ where 𝑢 = sin 𝑥 , 𝑑𝑢 = cos 𝑥 𝑑𝑥 ]
= (1 − 𝑢2 )2 𝑢4 𝑑𝑢
4
6
8
= (𝑢 − 2𝑢 + 𝑢 ) 𝑑𝑢
1
2
1
= 5 𝑢5 − 7 𝑢7 + 9 𝑢9 + 𝐶
1
5
2
7
1
9
= sin5 𝑥 − sin7 𝑥 + sin9 𝑥 + 𝐶
tan5 𝑥 sec 4 𝑥 𝑑𝑥
5.
∴
tan5 𝑥 sec 4 𝑥 𝑑𝑥 =
=
tan6 𝑥 sec 2 𝑥sec 2 𝑥 𝑑𝑥
tan6 𝑥 (1 + tan2 𝑥)sec 2 𝑥 𝑑𝑥
Then let 𝑢 = tan 𝑥 , 𝑑𝑢 = sec 2 𝑥 𝑑𝑥
𝑢6 (1 + 𝑢2 )𝑑𝑥
=
= (𝑢6 + 𝑢8 ) 𝑑𝑥
1
7
1
9
= 𝑢7 + 𝑢9 + 𝐶
1
1
= 7 tan7 𝑥 + 9 tan9 𝑥 + 𝐶
tan5 𝑥 sec 7 𝑥 𝑑𝑥
6.
∴
tan5 𝑥 sec 7 𝑥𝑑𝑥 = tan4 𝑥 sec 6 𝑥 tan 𝑥 sec 𝑥 𝑑𝑥
= (sec 2 𝑥 − 1)2 sec 6 𝑥 tan 𝑥 sec 𝑥 𝑑𝑥
Then let 𝑢 = sec 𝑥 , 𝑑𝑢 = tan 𝑥 sec 𝑥 𝑑𝑥
= (𝑢2 − 1)2 𝑢6 𝑑𝑥
= (𝑢10 + 2𝑢8 + 𝑢6 ) 𝑑𝑢
1
2
1
= 11 𝑢11 + 9 𝑢9 + 7 𝑢7 + 𝐶
1
2
1
= 11 sec11 𝑥 + 9 sec 9 𝑥 + 7 sec 7 𝑥 + 𝐶
232 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
cot 3 𝑥 cosec 3 𝑥 𝑑𝑥
7.
∴
cot 3 𝑥 cosec 3 𝑥 𝑑𝑥 =
cot 2 𝑥cosec 2 𝑥 cot 𝑥 cosec 𝑥 𝑑𝑥
= (cosec 2 𝑥 − 1)cosec 2 𝑥 cot 𝑥 cosec 𝑥 𝑑𝑥
Then applying U-substitution, we have 𝑢 = cosec 𝑥
and − 𝑑𝑢 = cot 𝑥 cosec 𝑥 𝑑𝑥
= (𝑢2 − 1) 𝑢2 (−𝑑𝑢)
(𝑢2 − 𝑢4 ) 𝑑𝑢
=
1
1
= 3 𝑢3 − 5 𝑢5 + 𝐶
1
1
= 3 cosec 3 𝑥 − 5 cosec 5 𝑥 + 𝐶
233 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Integration of Trigonometric Substitution:
Table of Trigonometric Substitution:
Expression
Substitution
Identity
√𝒂𝟐 − 𝒙𝟐
𝑥 = 𝑎 sin 𝜃 , − 𝜋⁄2 ≤ 𝜃 ≤ 𝜋⁄2
cos 2 𝜃 = 1 − sin2 𝜃
√𝒂𝟐 + 𝒙𝟐
𝑥 = 𝑎 tan 𝜃 , − 𝜋⁄2 ≤ 𝜃 ≤ 𝜋⁄2
1 + tan2 𝜃 = sec 2 𝜃
𝑥 = 𝑎 sec 𝜃 , 0 ≤ 𝜃 ≤ 𝜋⁄2
√𝒙𝟐 − 𝒂𝟐
or
𝜋 ≤ 𝜃 ≤ 3𝜋⁄2
sec 2 𝜃 − 1 = tan2 𝜃
Worked Examples:
Example 53:
2022
Evaluate the following integrals:
1.
2.
3.
4.
5.
√9−𝑥 2
𝑑𝑥
𝑥2
1
𝑥 2 √𝑥 2 +4
𝑥
√𝑥 2 +4
𝑑𝑥
𝑑𝑥
𝑑𝑥
√𝑥 2 −𝑎2
𝑥
√3−2𝑥−𝑥 2
𝑑𝑥
234 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
√9−𝑥 2
𝑑𝑥
𝑥2
1.
Let’s write express our integrand in this form,
√32 −𝑥 2
𝑑𝑥
𝑥2
, so that we can see it clearly from our
table of substitution, which one to let. Notice that we didn’t change the formality of the given
integral. Then,
Let 𝑥 = 3 sin 𝜃, where − 𝜋⁄2 ≤ 𝜃 ≤ 𝜋⁄2. Then 𝑑𝑥 = 3 cos 𝜃 𝑑𝜃
Let’s now substitute what we have and simply the integrand later.
∴
√32 −𝑥 2
𝑑𝑥
𝑥2
=
=
=
√9−(3 sin 𝜃)2
3 cos 𝜃 𝑑𝜃
(3 sin 𝜃)2
√9(1−sin2 𝜃)
9sin2 𝜃
√9(cos2 𝜃)
9sin2 𝜃
3 cos 𝜃 𝑑𝜃
3 cos 𝜃 𝑑𝜃
=
9cos2 𝜃
𝑑𝜃
9sin2 𝜃
=
cot 2 𝜃 𝑑𝜃
= (cosec 2 𝜃 − 1) 𝑑𝜃
= − cot 𝜃 − 𝜃 + 𝐶
Since this is an indefinite integral, we must return to the original variable 𝑥. This can be done by
either using trigonometric identities or ratios to express cot 𝜃 in terms of sin 𝜃 = 𝑥 ⁄3 or by
drawing a diagram below, where 𝜃 is interpreted as an angle of a right triangle.
Since sin 𝜃 = 𝑥 ⁄3, then the Pythagorean theorem gives the length of the adjacent side as
√9 − 𝑥 2 , so we can simply read the value of cot 𝜃 from the diagram below.
235 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
𝑥
3
𝜃
√9 − 𝑥 2
∴ sin 𝜃 = 𝑥 ⁄3 ⟹
∴
√32 −𝑥 2
𝑑𝑥
𝑥2
𝑥
3
𝜃 = sin−1 ( )
= − cot 𝜃 − 𝜃 + 𝐶
=−
√9−𝑥 2
𝑥
𝑥
− sin−1 (3) + 𝐶
236 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
1
2.
𝑥 2 √𝑥 2 +4
∴
𝑑𝑥
1
𝑥 2 √𝑥 2 +4
1
𝑑𝑥 =
𝑥 2 √𝑥 2 +22
𝑑𝑥
Let 𝑥 = 2 tan 𝜃 − 𝜋⁄2 ≤ 𝜃 ≤ 𝜋⁄2 whihch implies that, 𝑑𝑥 = 2sec 2 𝜃 𝑑𝜃
∴
1
𝑥 2 √𝑥 2 +4
1
𝑑𝑥 =
(2 tan 𝜃)2 √(2 tan 𝜃)2 +4
1
=
4tan2 𝜃√4(1+tan2 𝜃)
=
2sec2 𝜃 𝑑𝜃
4tan2 𝜃∙2 sec 𝜃
=
cos 𝜃 𝑑𝜃
4sin2 𝜃
1
=4
=−
𝑑𝑢
𝑢2
1
4𝑢
2sec 2 𝜃 𝑑𝜃
2sec 2 𝜃 𝑑𝜃
[𝑤ℎ𝑒𝑟𝑒 , 𝑢 = sin 𝜃 , 𝑑𝑢 = cos 𝜃]
+𝐶
1
= − 4 sin 𝜃 + 𝐶
=−
cosec 𝜃
4
+𝐶
Since, 𝑥 = 2 tan 𝜃 , which implies that tan 𝜃 =
𝑥
2
𝑥
2
𝑎𝑛𝑑 𝜃 = tan−1 ( )
By Pythagorean Theorem, the length of the hypotenuse is given by, √𝑥 2 + 4 .
∴
1
𝑥 2 √𝑥 2 +4
𝑑𝑥 = −
=−
=−
cosec 𝜃
4
√𝑥2 +4
𝑥
4
+𝐶
+𝐶
√𝑥 2 +4
+
4𝑥
√𝑥 2 + 4
𝑥
𝐶
237 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
𝑥
3.
√𝑥 2 +4
∴
𝑑𝑥
𝑥
√𝑥 2 +4
𝑥
𝑑𝑥 =
√𝑥 2 +22
𝑑𝑥
Let 𝑥 = 2 tan 𝜃 − 𝜋⁄2 ≤ 𝜃 ≤ 𝜋⁄2 whihch implies that, 𝑑𝑥 = 2sec 2 𝜃 𝑑𝜃
∴
𝑥
√𝑥 2 +4
2 tan 𝜃
𝑑𝑥 =
√(2 tan 𝜃)2 +4
=
=
2sec 2 𝜃 𝑑𝜃
2 tan 𝜃
√4(1+tan2 𝜃)
2sec 2 𝜃 𝑑𝜃
2 tan 𝜃sec2 𝜃 𝑑𝜃
2 sec 𝜃
= 2 tan 𝜃 sec 𝜃 𝑑𝜃
= 2 sec 𝜃 + 𝐶
√𝑥 2 +4
+
2
= 2(
𝐶)
= √𝑥 2 + 4 + 𝐶
Alternative Method:
𝑥
√𝑥 2 +4
𝑑𝑥
Using U-Substitution we have the following results.
Let 𝑢 = 𝑥 2 + 4 , 𝑑𝑢 = 2𝑥𝑑𝑥
∴
𝑥
√𝑥 2 +4
1
𝑑𝑥 = 2
𝑑𝑢
√𝑢
= √𝑢 + 𝐶
= √𝑥 2 + 4 + 𝐶
238 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
𝑑𝑥
4.
√𝑥 2 −𝑎2
We let 𝑥 = 𝑎 sec 𝜃 where, 0 ≤ 𝜃 ≤ 𝜋⁄2
Then 𝑑𝑥 = 𝑎 sec 𝜃 tan 𝜃 𝑑𝜃
∴
𝑑𝑥
√𝑥 2 −𝑎2
=
=
=
𝑎 sec 𝜃 tan 𝜃𝑑𝜃
or
𝜋 ≤ 𝜃 ≤ 3𝜋⁄2
Replace 𝑑𝑥 with 𝑎 sec 𝜃 tan 𝜃 𝑑𝜃 and 𝑥 with 𝑎 sec 𝜃.
√𝑎 2 sec2 𝜃−𝑎 2
𝑎 sec 𝜃 tan 𝜃𝑑𝜃
Factor out 𝑎2 as a common factor at the denominator.
√𝑎2 (sec2 𝜃−1)
𝑎 sec 𝜃 tan 𝜃𝑑𝜃
√𝑎 2 (tam2 𝜃)
=
𝑎 sec 𝜃 tan 𝜃𝑑𝜃
𝑎 tan 𝜃
=
sec 𝜃 𝑑𝜃
= ln|sec 𝜃 + tan 𝜃| + 𝐶
∴
𝑑𝑥
√𝑥 2 −𝑎2
= ln|sec 𝜃 + tan 𝜃| + 𝐶
𝑥
𝑎
= ln | +
= ln |
√𝑥 2 −𝑎2
|+
𝑎
𝑥+√𝑥 2 −𝑎2
|+
𝑎
𝐶
√𝑥 2 − 𝑎2
𝑥
𝐶
= ln|𝑥 + √𝑥 2 − 𝑎2 | − ln 𝑎 + 𝐶
𝜃
𝑎
239 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
𝑥
5.
√3−2𝑥−𝑥 2
𝑑𝑥
We can transform the integrand into a function for which trigonometric substitution is
appropriate by first completing the square under the root sign:
∴ 3 − 2𝑥 − 𝑥 2 = (−𝑥 2 − 2𝑥 − 1) + 1 + 3
= 4 − (𝑥 + 1)2
This suggest that we make the substitution, 𝑢 = 𝑥 + 1 , 𝑥 = 𝑢 − 1
∴
𝑥
√4−(𝑥+1)2
𝑑𝑥 =
𝑢−1
√4−𝑢2
and 𝑑𝑥 = 𝑑𝑢.
𝑑𝑢
√4 − 𝑢2 = √4 − sin2 𝜃 = 2 cos 𝜃
Let 𝑢 = 2 sin 𝜃 , 𝑑𝑢 = 2 cos 𝜃 𝑑𝜃 and
𝑢
And 𝜃 = sin−1 (2 ) , from Pythagorean Theorem the length of the adjacent is given by √4 − 𝑢2 .
∴
𝑥
√4−(𝑥+1)2
𝑑𝑥 =
2 sin 𝜃−1
2 cos 𝜃 𝑑𝜃
2 cos 𝜃
= (2 sin 𝜃 − 1) 𝑑𝜃
= −2 cos 𝜃 − 𝜃 + 𝐶
𝑢
2
= −√4 − 𝑢2 − sin−1 ( ) + 𝐶
From, √4 − 𝑢2 = √4 − sin2 𝜃 = 2 cos 𝜃
𝑥+1
)+𝐶
2
= −√4 − (𝑥 + 1)2 − sin−1 (
𝑥+1
)+
2
= √3 − 2𝑥 − 𝑥 2 − sin−1 (
𝐶
240 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
THE FUNDAMENTAL THEOREM OF CALCULUS:
The Fundamental Theorem of Calculus:
States that if 𝑓 is continuous on a closed interval [𝑎, 𝑏], then
𝑏
∫ 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)
𝑎
Where 𝐹 is any antiderivative of 𝑓, that is, a function such that 𝐹 ′ = 𝑓.
Worked Examples:
Example 54:
2022
Evaluate the following integrals:
1.
2
(1
1
2.
3
(2 sin 𝑥
0
3.
𝑒𝑥
ln 𝑥 𝑑𝑥
1
𝜋
4
+ 2𝑦)2 𝑑𝑦
− 𝑒 𝑥 ) 𝑑𝑥
sec 𝜃 tan 𝜃 𝑑𝜃
4.
0
5.
4 𝑥+𝑒 𝑥
𝑒
2
6.
𝜋⁄2
cos2 𝑥 𝑑𝑥
0
𝑑𝑥
241 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
Solution:
1.
2
(1
1
⟹
+ 2𝑦)2 𝑑𝑦
2
(1 +
1
4𝑦 + 4𝑦 2 ) 𝑑𝑦
2
4
⟹ [𝑦 + 2𝑦 2 + 3 𝑦 3 ]
2
⟹ (2 + 2(2) +
⟹
⟹
2.
3
(2 sin 𝑥
0
⟹
⟹
⟹
⟹
⟹
3.
62
13
− 3
3
49
𝑜𝑟
3
1
4
(2)3 ) −
3
4
(1 + 2(1)2 + 3 (1)3 )
1
16 3
− 𝑒 𝑥 ) 𝑑𝑥
3
3
2 sin 𝑥 𝑑𝑥 − 0 𝑒 𝑥 𝑑𝑥
0
[−2 cos 𝑥]30 − [𝑒 𝑥 ]30
(−2 cos 3 + 2 cos 0) − (𝑒 3 − 𝑒 0 )
−2 cos 3 + 2 − 𝑒 3 + 1
3 − 2 cos 3 − 𝑒 3
𝑒𝑥
ln 𝑥 𝑑𝑥
1
𝑥
⟹ [𝑥 ln 𝑥 − 𝑥]1𝑒
⟹ (𝑒 𝑥 ln 𝑒 𝑥 − 𝑒 𝑥 ) − ((1) ln 1 − 1)
⟹ 𝑥𝑒 𝑥 − 𝑒 𝑥 − 0 + 1
⟹ 𝑥(𝑒 𝑥 − 1) + 1
4.
𝜋
4
0
sec 𝜃 tan 𝜃 𝑑𝜃
𝜋
⟹ [sec 𝜃]04
𝜋
⟹ sec 4 − sec 0
= −1 + √2
242 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fourteen: Integration.
5.
4 𝑥+𝑒 𝑥
𝑒
2
⟹
𝑑𝑥
4 𝑒𝑥 𝑥
𝑒 𝑒 𝑑𝑥
2
⟹ Let 𝑢 = 𝑒 𝑥
⟹
by U−substitution we have,
, 𝑑𝑢 = 𝑒 𝑥 𝑑𝑥
4 𝑢
𝑒 𝑑𝑢
2
⟹ [𝑒 𝑢 ]42
𝑥
4
⟹ [𝑒 𝑒 ]2
4
⟹ 𝑒𝑒 − 𝑒𝑒
6.
2
𝜋⁄2
cos2 𝑥 𝑑𝑥
0
⟹
𝜋⁄2 1+cos 2𝑥
(
) 𝑑𝑥
0
2
⟹
1 𝜋⁄2
(
2 0
Half angle identity
1 + cos 2𝑥) 𝑑𝑥
1
1
𝜋⁄2
⟹ (2 𝑥 + 4 sin 2𝑥)
0
⟹
1 𝜋
[(
2 2
⟹
𝜋
4
+ 0) − (0 + 0)]
243 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER FIFTEEN:
THE AREAS BETWEEN THE CURVES:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS
244 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fifteen: The Areas between the curves.
THE AREA BETWEEN THE CURVES:
The area between the curves 𝑦 = 𝑓(𝑥) and 𝑦 = 𝑔(𝑥) and between 𝑥 = 𝑎 and 𝑥 = 𝑏 is given by,
𝑏
𝐴 = ∫ |𝑓(𝑥) − 𝑔(𝑥)| 𝑑𝑥
𝑎
Worked Examples:
Example 54:
2022
Find the area of the region bounded or enclosed by the curves:
1. 𝑦 = 𝑥 2 and 𝑦 = 2𝑥 − 𝑥 2
2. 𝑦 = 𝑥 − 1
and 𝑦 2 = 2𝑥 + 6
3. 𝑦 = 𝑥 2 − 𝑥 − 2 above the 𝑥 −axis from 𝑥 = −2 to 𝑥 = 2.
4. 𝑥 = 𝑦 4
and
5. 𝑦 = 12 − 𝑥 2
𝑦 = √2 − 𝑥
and
6. 𝑦 = cos 𝑥 and 𝑦 = sin 𝑥
,
𝑦=0
𝑦 = 𝑥2 − 6
𝜋
𝑥 ∈ [0, 2 ]
245 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fifteen: The Areas between the curves.
Solution:
1. 𝑦 = 𝑥 2 and 𝑦 = 2𝑥 − 𝑥 2
∴ 𝑥 2 = 2𝑥 − 𝑥 2 , the point of intersection
∴ 2𝑥 2 − 2𝑥 = 0
∴ 2𝑥(𝑥 − 1) = 0
Thus, 𝑥 = 0 𝑜𝑟 𝑥 = 1
1
(2𝑥 2
0
∴ 𝐴=
− 2𝑥) 𝑑𝑥
1
2
= [3 𝑥 3 − 𝑥 2 ]
0
2
2
= (3 (1)3 − (1)2 ) − (3 (0)3 − (0)2 )
=−
1
3
1
= 3 square units
2. 𝑦 = 𝑥 − 1
and 𝑦 2 = 2𝑥 + 6
1
∴ 𝑥 = 2 𝑦2 − 3
∴
1 2
𝑦
2
2
𝑎𝑛𝑑 𝑥 = 𝑦 + 1
−3 = 𝑦+1
∴ 𝑦 − 2𝑦 − 8 = 0
∴ (𝑦 − 4)(𝑦 + 2) = 0
Thus, 𝑦 = 4 𝑜𝑟 𝑦 = −2
∴𝐴=
=
4
(𝑦 2 − 2𝑦 − 8) 𝑑𝑦
−2
4
1
[3 𝑦 3 − 𝑦 2 − 8𝑦]
−2
1
3
80
−3
1
3
= ( (4)3 − (4)2 − 8(4)) − ( (−2)3 − (−2)2 − 8(−2))
=
−
28
3
= 36 square units
246 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fifteen: The Areas between the curves.
3. 𝑦 = 𝑥 2 − 𝑥 − 2 above the 𝑥 −axis from 𝑥 = −2 to 𝑥 = 2.
∴ 𝑦 = 𝑥 2 − 𝑥 − 2 and 𝑦 = 0 (𝑥 − axis)
∴ 𝑥2 − 𝑥 − 2 = 0
∴ (𝑥 + 1)(𝑥 − 2) = 0
Thus, 𝑥 = −1 𝑜𝑟 𝑥 = 2 , but we also have 𝑥 values from 𝑥 = −2 to 𝑥 = 2.
Hence, 𝑥 = −2, −1, 2
So −2 ≤ 𝑥 ≤ −1
and −1 ≤ 𝑥 ≤ 2
∴𝐴=
−1 2
(𝑥
−2
− 𝑥 − 2) 𝑑𝑥 +
1
3
1
2
−1
= ( 𝑥 3 − 𝑥 2 − 2𝑥)
1
3
−2
1
2
2
(𝑥 2
−1
1
3
− 𝑥 − 2) 𝑑𝑥
2
1
2
+ ( 𝑥 3 − 𝑥 2 − 2𝑥)
8
3
−1
8
3
1
3
1
2
= [(− − + 2) − (− − 2 + 4)] + [( + 2 − 4) − (− − + 2)]
8
= 3 square units
4. 𝑥 = 𝑦 4
and
𝑦 = √2 − 𝑥
,
𝑦=0
∴ 𝑦2 = 2 − 𝑥
⟹ 𝑥 = 2 − 𝑦2
4
2
∴ 𝑦 =2−𝑦
∴ 𝑦4 + 𝑦2 − 2 = 0
∴ (𝑦 2 + 2)(𝑦 2 − 1) = 0
Thus, 𝑦 = ±1 and 𝑦 2 + 2 = 0 No solution.
So, −1 ≤ 𝑥 ≤ 0 and 0 ≤ 𝑥 ≤ 1
∴𝐴=
0
(𝑦 4
−1
+ 𝑦 2 − 2) 𝑑𝑦 +
1
1
= (5 𝑦 5 + 3 𝑦 3 − 2𝑦)
0
−1
1
1
1 4
(𝑦
0
1
+ 𝑦 2 − 2) 𝑑𝑦
1
+ (5 𝑦 5 + 3 𝑦 3 − 2𝑦)
1
0
1
1
= {(0) − (5 (−1)5 + 3 (−1)3 − 2(−1))} + {(5 (1)5 + 3 (1)3 − 2(1)) − (0)}
22
22
= − 15 + 15
= 0 square units
247 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Fifteen: The Areas between the curves.
5. 𝑦 = 12 − 𝑥 2
𝑦 = 𝑥2 − 6
and
∴ 12 − 𝑥 2 = 𝑥 2 − 6 The point of intersection.
∴ 𝑥2 − 9 = 0
Thus, 𝑥 = ±3
∴𝐴=
=
3 2
(𝑥 − 9)𝑑𝑥
−3
3
1
(3 𝑥 3 − 9𝑦)
−3
1
1
= {(3 (3)3 − 9(3)) − (3 (−3)3 − 9(−3))}
= −18 − 18
= −36
= 36 square units
6. 𝑦 = cos 𝑥 and 𝑦 = sin 𝑥
𝜋
𝑥 ∈ [0, 2 ]
∴ cos 𝑥 = sin 𝑥 , the point of intersection
∴ tan 𝑥 = 1
𝜋
4
Thus, 𝑥 =
So 0 ≤ 𝑥 ≤
but we also have 𝑥 values from 0 to
𝜋
4
𝜋
4
and
≤𝑥≤
𝜋
4
∴ 𝐴 = 0 (cos 𝑥 − sin 𝑥) 𝑑𝑥 +
= [sin 𝑥 + cos 𝑥]𝜋0
𝜋
⁄4
𝜋
2
𝜋
4
𝜋
2
𝜋
2
(cos 𝑥 − sin 𝑥) 𝑑𝑥
⁄
+ [sin 𝑥 + cos 𝑥]𝜋𝜋⁄24
𝜋
𝜋
𝜋
𝜋
𝜋
= {(sin ( 4 ) + cos ( 4 )) − (sin(0) + cos(0))} − {(sin ( 2 ) + cos ( 2 )) − (sin ( 4 ) + cos ( 4 ))}
1
√2
=(
=
1
√2
+
+
1
−
√2
1) − (1 + 0 −
1
1
+ 2
√2
√
+
1
√2
1
√2
−
1
)
√2
−2
= 2√2 − 2 square units.
248 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER SIXTEEN:
THE ARC LENGTH:
MMTH011 / MAH101M
DIFFERENTIAL AND INTEGRAL CALCULUS
249 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Sixteen: The Areas between the curves.
The Arc Length Formula:
The Arc Length Formula:
States that if 𝑓′ is continuous on a closed interval [𝑎, 𝑏], then the length of the curve
𝑦 = 𝑓(𝑥) , 𝑎 ≤ 𝑥 ≤ 𝑏 is given by,
𝒃
𝑳 = ∫ √𝟏 + (
𝒂
𝒅𝒚 𝟐
) 𝒅𝒙
𝒅𝒙
Worked Examples:
Example 55:
2022
Find the arc length of the following functions given the interval.
3
1. 𝑦 = 1 + 6𝑥 2
1
0≤𝑥≤1
2. 𝑦 = 𝑥 2 − 8 ln 𝑥
1≤𝑥≤2
3. 𝑦 = ln(sec 𝑥)
0 ≤ 𝑥 ≤ 𝜋 ⁄4
4. 𝑦 =
𝑥5
6
1
+ 10𝑥3
5. 𝑦 = ln(1 − 𝑥 2 )
1≤𝑥≤2
1
0≤𝑥≤2
250 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Sixteen: The Arc Length.
Solution:
3
1. 𝑦 = 1 + 6𝑥 2
∴
𝑑𝑦
𝑑𝑥
0≤𝑥≤1
,
= 9√𝑥
𝑑𝑦 2
𝑑𝑥
∴ 1 + ( ) = 1 + 81𝑥
2
𝑑𝑦
∴ √1 + ( ) = √1 + 81𝑥
𝑑𝑥
Thus, 𝐿 =
=
2
𝑏
√1 + (𝑑𝑦)
𝑎
𝑑𝑥
1
1+
0 √
1
= 81
1 1
𝑢2
0
2
(1
243
=(
𝑑𝑥
81𝑥 𝑑𝑥
𝑑𝑢
3
1
+ 81𝑥)2 )
0
2
= 243 (82√82 − 1)
251 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Sixteen: The Arc Length.
1
2. 𝑦 = 𝑥 2 − 8 ln 𝑥
∴
𝑑𝑦
𝑑𝑥
1≤𝑥≤2
,
1
= 2𝑥 − 8𝑥
𝑑𝑦 2
1 2
∴ 1 + (𝑑𝑥 ) = 1 + (2𝑥 − 8𝑥)
1
1
= 4𝑥 2 + 2 + 64𝑥2
2
1
= (2𝑥 + 8𝑥)
2
2
𝑑𝑦
1
∴ √1 + (𝑑𝑥 ) = √(2𝑥 + 8𝑥)
= 2𝑥 +
Thus, 𝐿 =
=
1
8𝑥
2
𝑏
√1 + (𝑑𝑦)
𝑎
𝑑𝑥
2
(2𝑥
1
+
1
8𝑥
1
𝑑𝑥
) 𝑑𝑥
2
= (𝑥 2 + 8 ln 𝑥)
1
1
1
= (22 + 8 ln 2) − (12 + 8 ln 1)
1
8
= 3 + ln 2
252 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Sixteen: The Arc Length.
3. 𝑦 = ln(sec 𝑥)
∴
𝑑𝑦
𝑑𝑥
0 ≤ 𝑥 ≤ 𝜋 ⁄4
,
1
= sec 𝑥 ∙ sec 𝑥 tan 𝑥
= tan 𝑥
𝑑𝑦 2
∴ 1 + (𝑑𝑥 ) = 1 + tan2 𝑥
2
𝑑𝑦
∴ √1 + ( ) = √1 + tan2 𝑥
𝑑𝑥
[ pythagorean identiy 1 + tan2 𝑥 = sec 2 𝑥 ]
= √sec 2 𝑥
= sec 𝑥
Thus, 𝐿 =
=
2
𝑏
√1 + (𝑑𝑦)
𝑎
𝑑𝑥
𝑑𝑥
𝜋⁄4
sec 𝑥 𝑑𝑥
0
= (ln|sec 𝑥 + tan 𝑥|)𝜋0
⁄4
= ln(1 + √2) − ln(1)
= ln(1 + √2)
253 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Sixteen: The Arc Length.
4. 𝑦 =
∴
𝑥5
6
𝑑𝑦
𝑑𝑥
1
+ 10𝑥3
1≤𝑥≤2
,
5
3
= 6 𝑥 4 − 10𝑥 4
𝑑𝑦 2
5
2
3
∴ 1 + (𝑑𝑥 ) = 1 + (6 𝑥 4 − 10𝑥 4 )
25
3
9
= 36 𝑥 8 + 2 + 100𝑥8
5
6
= ( 𝑥4 +
2
3
)
4
10𝑥
2
2
𝑑𝑦
5
3
∴ √1 + (𝑑𝑥 ) = √(6 𝑥 4 + 10𝑥 4 )
5
3
= 6 𝑥 4 + 10𝑥4
Thus, 𝐿 =
=
2
𝑏
√1 + (𝑑𝑦)
𝑎
𝑑𝑥
2 5 4
( 𝑥
1 6
+
1
𝑑𝑥
3
) 𝑑𝑥
10𝑥 4
2
1
= (6 𝑥 5 − 10𝑥3 )
1
1
1
1
1
= (6 (2)5 − 10(2)3 ) − (6 (1)5 − 10(1)3 )
=
1277
1
− 15
240
=
1261
240
254 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Chapter Sixteen: The Arc Length.
1
5. 𝑦 = ln(1 − 𝑥 2 ) ,
∴
𝑑𝑦
𝑑𝑥
0≤𝑥≤2
−2𝑥
= 1−𝑥2
𝑑𝑦 2
4𝑥 2
∴ 1 + (𝑑𝑥 ) = 1 + 𝑥 4 −2𝑥2 +1
=
𝑥 4 −2𝑥 2 +1+4𝑥 2
𝑥 4 −2𝑥 2 +1
=
𝑥 4 +2𝑥 2 +1
𝑥 4 −2𝑥 2 +1
2
=
(𝑥 2 +1)
(𝑥 2 −1)2
𝑥 2 +1
2
= (𝑥 2 −1)
2
2
2
𝑑𝑦
𝑥 +1
∴ √1 + (𝑑𝑥 ) = √(𝑥 2 −1)
𝑥 2 +1
= 𝑥 2 −1
Thus, 𝐿 =
2
𝑏
√1 + (𝑑𝑦)
𝑎
𝑑𝑥
=
1⁄2 𝑥 2 +1
(𝑥 2 −1) 𝑑𝑥
0
𝑑𝑥
1
= (ln(𝑥 − 1) − ln(𝑥 + 1))20
1
1
= {(ln (2 − 1) − ln (2 + 1))} − {(ln(0 − 1) − ln(0 + 1))}
1
3
= ln (− 2) − ln (2) − ln(−1) + ln(1)
1
3
= ln (− 2 ÷ 2) − ln(−1)(1)
= ln(−3−1 ) − ln(−1)
= − ln(3)
255 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
256 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
SUMMERY AND THE FORMULAS:
1. Quadratic formula:
If 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
Then, 𝑥 =
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
2. Factors:
a) 𝑎2 − 𝑏 2 = (𝑎 − 𝑏)(𝑎 + 𝑏)
b) 𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 )
c) 𝑎3 + 𝑏 3 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 )
3. Factor Theorem:
If 𝑦 = 𝑓(𝑥) and 𝑓(𝑎) = 0 , then (𝑥 − 𝑎) is a factor of 𝑓(𝑥).
4. Partial Fractions:
a)
𝑓(𝑥)
(𝑎𝑥+𝑏)(𝑐𝑥+𝑑)(𝑒𝑥+𝑓)
b)
𝑓(𝑥)
(𝑎𝑥 2 +𝑏𝑥+𝑐)(𝑑𝑥+𝑒)
c)
𝑓(𝑥)
(𝑎𝑥+𝑏)3 (𝑐𝑥+𝑑)
𝐴
𝐵
= (𝑎𝑥+𝑏) + (𝑐𝑥+𝑑) +
𝐴𝑥+𝐵
𝐶
(𝑒𝑥+𝑓)
𝐶
= (𝑎𝑥 2 +𝑏𝑥+𝑐) + (𝑑𝑥+𝑒)
𝐴
𝐵
𝐶
𝐷
= (𝑎𝑥+𝑏) + (𝑎𝑥+𝑏)2 + (𝑎𝑥+𝑏)3 + 𝑐𝑥+𝑑
5. Parabola:
If 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
or
𝑥 = 𝑎𝑦 2 + 𝑏𝑦 + 𝑐
𝑏
𝑏
Then the axis of symmetry at 𝑥 = − 2𝑎 𝑜𝑟 𝑦 = − 2𝑎
257 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Properties of the Logarithmic and Exponential functions:
6. PROPERTIES OF LOGARITHMIC WITH A BASE "𝒂"
5. LOGARITHMIC LAWS:
Definition:
𝑦 = log 𝑎 𝑥 if and only if 𝑥 = 𝑎 𝑦 where 𝑎 > 0.
In other words, Logarithmic are exponents.
Remarks:
 log 𝑥 always refers to log base 10, i.e.
log 𝑥 = log10 𝑥
 ln 𝑥 is called the natural logarithmic and is
used to represent log 𝑒 𝑥 , where the
irrational number 𝑒 ≈ 2.71828128.
Therefore, ln 𝑥 = 𝑦 if and only if 𝑥 = 𝑒 𝑦 .
 Change of base formulas:
ln 𝑎
j)
k)
l)
m)
n)
o)
p)
q)
log 𝑎 𝑎 = 1 for all 𝑎 > 0 and log 10 = 1
log 𝑎 1 = 0 for all 𝑎 > 0
log 𝑎 𝑥𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦
𝑥
log 𝑎 𝑦 = log 𝑎 𝑥 − log 𝑎 𝑦
log 𝑎 𝑥 𝑦 = 𝑦 log 𝑎 𝑥
log 𝑎 𝑎 𝑦 = 𝑦 log 𝑎 𝑎 = 𝑦(1) = 𝑦
𝑎log𝑎 𝑥 = 𝑥
log 𝑒 𝑥 = ln 𝑥
log 𝑎
log 𝑏 𝑎 = ln 𝑏 = log 𝑏
7. PROPERTIES OF NATURAL LOGARITHMIC:
i)
j)
k)
l)
m)
ln 𝑒 = 1
ln 1 = 0
ln 𝑥 = 𝑦
⟺ 𝑒𝑦 = 𝑥
ln 𝑥𝑦 = ln 𝑥 + ln 𝑦
𝑥
ln 𝑦 = ln 𝑥 − ln 𝑦
n) ln 𝑥 𝑦 = 𝑦 ln 𝑥
o) ln 𝑒 𝑥 = 𝑥 ln 𝑒 = 𝑥(1) = 𝑥 ,
p) 𝑒 ln 𝑥 = 𝑥
, 𝑥>0
8. PROPERTIES OF EXPONENTS:
If 𝑚 and 𝑛 are integers then the following holds:
j) 𝑎𝑚 𝑎𝑛 = 𝑎𝑚+𝑛
k) (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛
l) (𝑎𝑏)𝑚 = 𝑎𝑚 𝑏 𝑚
m)
𝑥∈ℝ
n)
𝑎𝑚
=
𝑎𝑛
𝑎 𝑚
(𝑏 )
−𝑚
o) 𝑎
p)
1
𝑎𝑛
𝑚
𝑛
𝑎𝑚−𝑛
𝑎𝑚
= 𝑏𝑚 , where 𝑏 ≠ 0
1
= 𝑎𝑚 , where 𝑎 ≠ 0
𝑛
= √𝑎
𝑛
, where 𝑎 ≠ 0
𝑛
𝑚
q) 𝑎 = √𝑎𝑚 = ( √𝑎)
r) 𝑎0 = 1
, where 𝑎 ≠ 0 .
258 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
TRIG-INVERSE FUNCTION:
1. Definition:
a) 𝑦 = sin−1 𝑥 is equivalent to sin 𝑦 = 𝑥
b) 𝑦 = cos −1 𝑥 is equivalent to cos 𝑦 = 𝑥
c) 𝑦 = tan−1 𝑥 is equivalent to tan 𝑦 = 𝑥
2. Inverse property:
a) sin(sin−1 𝑥) = 𝑥
b) cos(cos−1 𝑥) = 𝑥
c) tan(tan−1 𝑥) = 𝑥
3. Domain and Range:
Functions
Domain
𝑦 = sin−1 𝑥
−1 ≤ 𝑥 ≤ 1
𝑦 = cos −1 𝑥
−1 ≤ 𝑥 ≤ 1
𝑦 = tan−1 𝑥
−∞ ≤ 𝑥 ≤ ∞
Range
−
Alternative notation
𝜋
𝜋
≤𝑦≤
2
2
0≤𝑦≤𝜋
−
𝜋
𝜋
≤𝑦≤
2
2
sin−1 𝑥 = 𝑎𝑟𝑐 sin 𝑥
cos−1 𝑥 = 𝑎𝑟𝑐 cos 𝑥
tan−1 𝑥 = 𝑎𝑟𝑐 tan 𝑥
259 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
Differentiation Tool-Box
Functions: 𝑓(𝑥)
Differential functions: 𝑓′(𝑥)
1. 𝑓(𝑥) = 𝑐
𝑓′(𝑥) = 0
2. 𝑓(𝑥) = 𝑐 ∙ 𝑔(𝑥)
3. 𝑓(𝑥) = [ 𝑔(𝑥) + 𝑟(𝑥) ]
𝑓′(𝑥) = 𝑐 ∙ 𝑔′(𝑥)
𝑓′(𝑥) = [ 𝑔′(𝑥) + 𝑟′(𝑥) ]
4. 𝑓(𝑥) = 𝑥 𝑛
𝑓′(𝑥) = 𝑛𝑥 𝑛−1
5. 𝑓(𝑥) = [ 𝑔(𝑥) ∙ 𝑟(𝑥) ]
6. 𝑓(𝑥) =
𝑓′(𝑥) =
𝑔(𝑥)
𝑟(𝑥)
𝑓(𝑥) = sin 𝑥
𝑓(𝑥) = cos 𝑥
𝑓(𝑥) = tan 𝑥
𝑓(𝑥) = cosec 𝑥
𝑓(𝑥) = sec 𝑥
𝑓(𝑥) = cot 𝑥
𝑓(𝑥) = sin−1 𝑥
15. 𝑓(𝑥) = cos −1 𝑥
−1
16. 𝑓(𝑥) = tan
𝑔′(𝑥) ∙ 𝑟(𝑥) − 𝑔(𝑥) ∙ 𝑟′(𝑥)
2
(𝑟(𝑥))
7. 𝑓(𝑥) = 𝑔(𝑟(𝑥))
8.
9.
10.
11.
12.
13.
14.
𝑓′(𝑥) = 𝑔′(𝑥) ∙ 𝑟(𝑥) + 𝑔(𝑥) ∙ 𝑟′(𝑥)
𝑥
17. 𝑓(𝑥) = 𝑒 𝑔(𝑥)
𝑓′(𝑥) = 𝑔′(𝑟(𝑥)) ∙ 𝑟′(𝑥)
𝑓′(𝑥) = cos 𝑥
𝑓′(𝑥) = − sin 𝑥
𝑓′(𝑥) = sec 2 𝑥
𝑓′(𝑥) = − cosec 𝑥 ∙ cot 𝑥
𝑓′(𝑥) = sec 𝑥 ∙ tan 𝑥
𝑓′(𝑥) = −cosec 2 𝑥
1
𝑓′(𝑥) =
√1 − 𝑥 2
1
𝑓′(𝑥) = −
√1 − 𝑥 2
1
𝑓′(𝑥) =
1 + 𝑥2
𝑓′(𝑥) = 𝑒 𝑔(𝑥) ∙ 𝑔′(𝑥)
18. 𝑓(𝑥) = ln(𝑔(𝑥))
19. 𝑓(𝑥) = 𝑎 𝑔(𝑥)
𝑓′(𝑥) =
1
∙ 𝑔′(𝑥)
𝑔(𝑥)
𝑓′(𝑥) = 𝑎 𝑔(𝑥) ∙ ln 𝑎 ∙ 𝑔′(𝑥)
20. 𝑓(𝑥) = log 𝑏 𝑥
𝑓′(𝑥) =
21. 𝑓(𝑥) = log 𝑎 𝑔(𝑥)
𝑛
22. 𝑓(𝑥) = (𝑔(𝑥))
𝑓′(𝑥) =
1
log 𝑏 𝑒
𝑥
1
∙ log 𝑎 𝑒 ∙ 𝑔′(𝑥)
𝑔(𝑥)
𝑛−1
𝑓′(𝑥) = 𝑛(𝑔(𝑥))
∙ 𝑔′(𝑥)
260 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
TOOL-BOX FOR INTEGRATION:
Basic Integrals:
Integrals with Trigonometric
Functions:
𝑑𝑥 = 𝑥 + 𝐶
1.
𝑘 𝑑𝑥 = 𝑘𝑥 + 𝐶 , Where 𝑘 and 𝑐 are Constants.
2.
𝑥 𝑛 𝑑𝑥 =
3.
𝑥 𝑛+1
𝑛+1
+ 𝐶 , Where 𝑛 ≠ −1
13.
14.
sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶
cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶
15.
sin(𝑛𝑥) 𝑑𝑥 = −
16.
cos(𝑛𝑥) 𝑑𝑥 =
17.
18.
19.
20.
21.
22.
23.
24.
tan 𝑥 𝑑𝑥 = − ln|cos 𝑥| + 𝐶 = ln|sec 𝑥| + 𝐶
cot 𝑥 𝑑𝑥 = ln|sin 𝑥| + 𝐶
sec 𝑥 𝑑𝑥 = ln|sec 𝑥 + tan 𝑥| + 𝐶
csc 𝑥 𝑑𝑥 = − ln|csc 𝑥 + cot 𝑥| + 𝐶
sec 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝐶
csc 2 𝑥 𝑑𝑥 = − cot 𝑥 + 𝐶
sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝐶
csc 𝑥 cot 𝑥 𝑑𝑥 = − cot 𝑥 + 𝐶
Integrals with Natural Logarithmics:
9.
1
𝑑𝑥
𝑥
10.
cos(𝑛𝑥)
+𝐶
𝑛
sin(𝑛𝑥)
+𝐶
𝑛
Integrals with Exponents:
9.
𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶
ln 𝑥 𝑑𝑥 = 𝑥 ln 𝑥 − 𝑥 + 𝐶
10.
𝑒 𝑛𝑥 𝑑𝑥 = 𝑛 𝑒 𝑛𝑥 + 𝐶
11.
1
𝑑𝑥
𝑎𝑥+𝑏
11.
𝑥 𝑒 𝑥 𝑑𝑥 = (𝑥 − 1)𝑒 𝑥 + 𝐶
12.
𝑎 𝑥 𝑑𝑥 = ln 𝑎 𝑎 𝑥 + 𝐶
1
12.
𝑥𝑒 𝑎𝑥 𝑑𝑥 = (𝑎 − 𝑎2 ) 𝑒 𝑎𝑥 + 𝐶
13.
ln 𝑎𝑥
𝑑𝑥
𝑥
1
13.
𝑥 2 𝑒 𝑥 𝑑𝑥 = (𝑥 2 − 2𝑥 + 2)𝑒 𝑥 + 𝐶
14.
𝑥 2 𝑒 𝑎𝑥 𝑑𝑥 = ( 𝑎 − 𝑎2 + 𝑎3 ) 𝑒 𝑎𝑥 + 𝐶
15.
𝑥 3 𝑒 𝑥 𝑑𝑥 = (𝑥 3 − 3𝑥 2 + 6𝑥 − 6)𝑒 𝑥 + 𝐶
16.
𝑥 𝑛 𝑒 𝑎𝑥 𝑑𝑥 =
= ln|𝑥| + 𝐶
1
= 𝑎 ln|𝑎𝑥 + 𝑏| + 𝐶
= 2 (ln 𝑎𝑥)2 + 𝐶
𝑏
) ln(𝑎𝑥
𝑎
14.
ln(𝑎𝑥 + 𝑏) 𝑑𝑥 = (𝑥 +
15.
ln(𝑥 2 + 𝑎2 ) 𝑑𝑥 = 𝑥 ln(𝑥 2 + 𝑎2 ) + 2𝑎 tan−1 𝑎 − 2𝑥 + 𝐶
16.
+ 𝑏) − 𝑥 + 𝐶 , 𝑥 ≠ 0
𝑥
ln(𝑥 2 − 𝑎2 ) 𝑑𝑥 = 𝑥 ln(𝑥 2 − 𝑎2 ) +
𝑥+𝑎
𝑎 ln 𝑥−𝑎
− 2𝑥 + 𝐶
1
𝑥
1
𝑥2
𝑥 𝑛 𝑒 𝑎𝑥
𝑎
2𝑥
2
𝑛
− 𝑎 𝑥 𝑛−1 𝑒 𝑎𝑥 𝑑𝑥 + 𝐶
261 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
INTEGRALS WITH ROOTS:
INTEGRALS WITH RATIONAL
FRACTIONS:
16.
√𝑥 − 𝑎 𝑑𝑥 =
2
(𝑥
3
3
2
− 𝑎) + 𝐶
12.
17.
1
𝑑𝑥
√𝑥±𝑎
1
𝑑𝑥
(𝑥+𝑎)2
1
= − 𝑥+𝑎 + 𝐶
= 2√𝑥 ± 𝑎 + 𝐶
13. (𝑥 + 𝑎)𝑛 𝑑𝑥 =
18.
19.
20.
21.
22.
1
𝑑𝑥
𝑎−𝑥
√
3
2
2
3
2𝑏
√𝑎𝑥 + 𝑏 𝑑𝑥 = (3𝑎 +
2
5
2𝑥
) √𝑎𝑥
3
𝑥
√𝑥(𝑎−𝑥)
+
𝑥−𝑎
29.
30.
= 3 tan−1 𝑎 + 𝐶
16.
𝑥
𝑑𝑥
𝑎 2 +𝑥 2
= 2 ln|𝑎2 + 𝑥 2 | + 𝐶
𝐶
18.
𝑥
√𝑎2 − 𝑥 2 𝑑𝑥 = 2 𝑥√𝑎2 − 𝑥 2 + 𝑎2 tan−1
28.
1
𝑑𝑥
𝑎 2 +𝑥 2
17.
24.
27.
15.
1
1
1
𝑥
√𝑥 2 ±𝑎
1
𝑥
√𝑎 2 −𝑥 2
+𝐶
𝑑𝑥 = ln |𝑥 + √𝑥 2 ± 𝑎2 | + 𝐶
𝑥
√𝑎 2 −𝑥 2
𝑑𝑥 = sin−1 𝑎 + 𝐶
𝑥
𝑥2
√𝑥 2 ±𝑎2
1
𝑥
1
𝑥2
𝑎 2 +𝑥 2
𝑥3
𝑎 2 +𝑥 2
𝑥
𝑑𝑥 = 𝑥 − 𝑎 tan−1 𝑎 + 𝐶
1
1
𝑑𝑥 = 2 𝑥 2 − 2 𝑎2 ln|𝑎2 + 𝑥 2 | + 𝐶
19.
1
𝑑𝑥
𝑎𝑥 2 +𝑏𝑥+𝑐
20.
1
𝑑𝑥
(𝑥+𝑎)(𝑥+𝑏)
21.
𝑥
𝑑𝑥
(𝑥+𝑎)2
22.
𝑥
𝑑𝑥
𝑎𝑥 2 +𝑏𝑥+𝑐
=
=
2
√4𝑎𝑐−𝑏2
1
tan−1
2𝑎𝑥+𝑏
√4𝑎𝑐−𝑏2
𝑎+𝑥
𝑎
𝑎+𝑥
+ ln|𝑎 + 𝑥| + 𝐶
1
= 2𝑎 ln|𝑎𝑥 2 + 𝑏𝑥 + 𝑐|
𝑏
𝑎√4𝑎𝑐−𝑏2
tan−1
2𝑎𝑥+𝑏
√4𝑎𝑐−𝑏2
𝑑𝑥 = −√𝑥 2 ± 𝑎2 + 𝐶
1
1
𝑑𝑥 = 2 𝑥 √𝑥 2 ± 𝑎2 ± 2 𝑎2 ln |𝑥 + √𝑥 2 ± 𝑎2 | + 𝐶
𝑑𝑥
(𝑎 2 +𝑥 2 )3⁄2
=
𝑥
𝑎 2 √𝑎2 +𝑥 2
+𝐶
= 𝑏−𝑎 ln 𝑏+𝑥 + 𝐶 , 𝑎 ≠ 𝑏
−
𝑑𝑥 = √𝑥 2 ± 𝑎2 + 𝐶
2
√𝑎 2 ±𝑥 2
= tan−1 𝑥 + 𝐶
2
= 3 (𝑥 ± 2𝑎)√𝑥 ± 𝑎 + 𝐶
√𝑎+𝑥 𝑑𝑥 = √𝑥(𝑎 + 𝑥) − 𝑎 ln[√𝑥 + √𝑥 + 𝑎] + 𝐶
26.
1
𝑑𝑥
1+𝑥 2
+𝑏+𝐶
√𝑎−𝑥 𝑑𝑥 = −√𝑥(𝑎 − 𝑥) − 𝑎 tan−1
√𝑥 2 ±𝑎2
+ 𝐶 , 𝑛 ≠ −1
14.
5
2
23.
25.
𝑛+1
= −2√𝑎 − 𝑥 + 𝐶
𝑥 √𝑥 − 𝑎 𝑑𝑥 = 𝑎(𝑥 − 𝑎) + (𝑥 − 𝑎) + 𝐶
𝑥
𝑑𝑥
√𝑥±𝑎
(𝑥+𝑎)𝑛+1
+𝐶
262 | P a g e
+𝐶
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
263 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
1. Rewrite the expression without using the absolute value symbol.
a) |−2𝜋|
b) |6| − |−36|
c) |3 − 2𝑥 3 |
d) 2||−2| − |−5|| + 11|−3|
e) |𝑥 − 3| if 𝑥 < 3.
2. Show that ||𝑥| − |𝑦|| ≤ |𝑥 − 𝑦|
(ℎ𝑖𝑛𝑡: 𝑥 = 𝑥 − 𝑦 + 𝑦)
3. Solve for 𝑥 in the following equation:
a) 3|4𝑥 − 1| ≤ 9
b) |𝑥 2 + 5𝑥 + 4| = 0
c) |𝑥 2 + 1| = 2𝑥
d) |𝑥 + 3| = 𝑥 2 − 4𝑥 − 3
1
e) 3 |2 𝑥 + 2| + 6 < 15
4. Convert the following degrees to radians.
a) 900°
b) −315°
c) 10°
d) 270°
e) −5°
264 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
5. Convert the Radians to degrees.
a) −3𝜋
b)
c)
3𝜋
4
5𝜋
12
d) −
e)
11𝜋
4
5𝜋
6
6. Solve for 𝑥 in the following exponential equations:
a) 103𝑥 = 1000
1
b) 82𝑥−3 = (16)
𝑥−2
c) 𝑒 3𝑥−7 = 5𝑒 𝑥−1
d) 𝑒 2𝑥 =
𝑒𝑥
2
𝑒2
2
e) 𝑒 −𝑥 = 𝑒 𝑥+𝑥 𝑒 −11
7. Solve for 𝑥 in the following Logarithmic equations:
a) 6 + ln 𝑥 = 10
b) ln(𝑥 2 ) = ln(2𝑥 + 3)
c) log13 (2𝑥 2 − 4𝑥) = log13 (45 + 𝑥 2 )
d) log(𝑥 − 2) − log(2𝑥 − 3) = log 2
e) log 3 (𝑥 2 − 6𝑥) = 3
265 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
8. Solve the following trigonometric equations:
a) tan 𝑥 = tan2 𝑥
b) cos 𝑥 sin 𝑥 = cos 𝑥
c) cos 𝑥 = 1 + sin2 𝑥
d) 7 sin 𝑥 + 5 = 2cos2 𝑥
e) 2 cos 𝑥 − sin 𝑥 + 2 cos 𝑥 sin 𝑥 = 1
f) 3(1 − sin 𝑥) = 2cos2 𝑥
g)
1+sin 𝑥
cos 𝑥
cos 𝑥
+ 1+sin 𝑥 = 4
h) cos 3 𝑥 = cos2 𝑥
9. Sketch the following functions showing all the necessary information.
3
a) 𝑓(𝑥) = 2 − 𝑥+1
b) 𝑓(𝑥) =
2𝑥−1
𝑥+1
2𝑥
c) 𝑓(𝑥) = 𝑥 2 −16
d) 𝑓(𝑥) =
e) 𝑓(𝑥) =
𝑥 2 −𝑥−6
𝑥 2 −9
𝑥 2 −4
𝑥−2
𝑥 3 +1
f) 𝑓(𝑥) = 𝑥 2 −4
g) 𝑓(𝑥) =
𝑥 2 −4𝑥+2
1−𝑥
266 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
10.
𝑥+3
Let 𝑓 be defined by 𝑓(𝑥) = { 3
2𝑥 − 1
;
𝑥≤0
; 0<𝑥<2
;
𝑥>2
a) Evaluate 𝑓(0) ; 𝑓(1) ; 𝑓(2) and 𝑓(3).
b) Fins the domain and the range of the function 𝑓.
c) Sketch the function 𝑓.
11.
Let 𝑓 be defined by 𝑓(𝑥) = {
3𝑥 − 1
−𝑥 + 3
𝑥<2
𝑥>4
a) Evaluate 𝑓(0) ; 𝑓(4) ; and 𝑓(5).
b) Fins the domain and the range of the function 𝑓.
c) Sketch the function 𝑓.
𝑥2 − 1
12.
Let 𝑓 be defined by 𝑓(𝑥) = { 𝑥 − 1
3𝑥
3
a) Evaluate 𝑓(−5) ; 𝑓 (2) ; and 𝑓(88).
;
;
;
𝑥≤0
0≤𝑥≤4
𝑥≥2
b) Fins the domain and the range of the function 𝑓.
c) Sketch the function 𝑓.
13.
Consider the given functions below:
𝑓(𝑥) = 𝑒 2𝑥−8 ; 𝑔(𝑥) = log 𝑒 𝑒 3𝑥
amd
ℎ(𝑥) = 5 ln 𝑥 3 − 4
Find the following:
a) ℎ ∘ 𝑔
b) 𝑓 ∘ ℎ
c) 𝑔 ∘ 𝑓
d) ℎ ∘ 𝑓 ∘ 𝑔
e) 𝑓 ∘ 𝑔 ∘ ℎ
267 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
14.
Evaluate the following limits:
(𝑥+3)(𝑥 2 +1)
a) lim
𝑥+3
𝑥→3
1
𝑥 2 −3
b) lim
𝑥→9 𝑥−9
2𝑥 2 +1
c) lim √ 3𝑥−2
𝑥→2
𝑥 2 −2
d) lim (
2
)
𝑥→2 𝑥 3 −3𝑥+5
8𝑥 2 −8𝑥−6
e) lim1 (
𝑥→1
2𝑥−3
)
2
f) lim
√𝑥 2 +9−3
𝑥2
𝑥→0
(3+𝑥)−1 −3−1
g) lim (
𝑥
𝑥→0
1
1
−
(𝑥+ℎ)2 𝑥2
h) lim (
ℎ
ℎ→0
)
)
1 1
i)
j)
+
4 𝑥
lim (4+𝑥
)
𝑥→−4
lim (
4−√16
𝑥→16 16𝑥−𝑥 2
)
𝑥 3 +3𝑥 2 +𝑥+3
k) lim (
)
𝑥+3
3
1
1
l) lim (𝑥√1+𝑥 − 𝑥)
𝑥→0
268 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
15.
Evaluate the following limits to infinity:
1
a) lim (2𝑥+3)
𝑥→∞
2𝑥 2 +7𝑥−8
b) lim (
𝑥 2 +3
𝑥→∞
)
2𝑥
c) lim (√𝑥 2 )
+8
𝑥→∞
d) lim (
𝑥→∞
e) lim (
𝑥→∞
f)
√9𝑥 6 −𝑥
𝑥 3 +1
√9𝑥 6 −𝑥
2−𝑥 3
2𝑥−𝑥 2
lim (
𝑥→∞ √𝑥+𝑥 2
)
)
)
𝑥2
g) lim (√𝑥 4 )
+1
𝑥→∞
h) lim (√𝑥 2 + 4𝑥 − 𝑥)
𝑥→∞
i)
j)
lim (√9𝑥 2 + 𝑥 − 3𝑥)
𝑥→∞
𝑥 4 −3𝑥 2 +𝑥
lim (
𝑥→∞
k) lim (
𝑥 3 −𝑥+2
𝑥−𝑥√𝑥
3
𝑥→∞ 2𝑥 2 +3𝑥−5
l)
)
)
lim (√𝑥 2 + 1)
𝑥→∞
269 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
16.
Show from a definition of limits that:
a) lim(14 − 5𝑥) = 4
𝑥→2
1
b) lim (3 − 2 𝑥) = −5
𝑥→16
c) lim(4𝑥 − 5) = 7
𝑥→3
4
d) lim (3 − 𝑥) = −5
5
𝑥→10
2+4𝑥
e) lim (
3
𝑥→1
f)
)=2
lim (1 − 4𝑥) = 13
𝑥→−3
g) lim(2𝑥 + 3) = 5
𝑥→1
1
h) lim (2 𝑥 + 3) = 2
𝑥→−2
i)
lim (3𝑥 + 5) = −1
𝑥→−2
j) lim(𝑥 2 + 2𝑥 − 7) = 1
𝑥→2
𝑥 2 +𝑥−6
k) lim (
𝑥→2
𝑥−2
)=5
l) lim(𝑥 2 − 4𝑥 + 5) = 1
𝑥→2
270 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
17.
4𝑥 2 − 2𝑥
Is the function 𝑓(𝑥) = { 10𝑥 − 1
30
18.
Is the function 𝑓(𝑥) = {
19.
𝑎𝑥 + 2𝑏
Let 𝑓 be defined by 𝑓(𝑥) = { 𝑥 + 3𝑎 − 𝑏
3𝑥 − 5
5𝑥 + 7
4𝑥 − 5
;
;
;
𝑥<2
𝑥=2
𝑥>2
;
;
2
continuous at 𝑥 = 3?
𝑥<2
continuous at 𝑥 = 2
𝑥>2
;
𝑥≤0
; 0<𝑥≤2
;
𝑥>2
Find the values of 𝑎 and 𝑏 for which the function 𝑓 is continuous at every 𝑥.
20.
Find the values of 𝑎 and 𝑏 that makes the function 𝑓 continuous everywhere.
𝑥 2 −4
;
𝑥−2
𝑓(𝑥) = {𝑎𝑥 2 − 𝑏𝑥 + 3
;
;
2𝑥 − 𝑎 + 𝑏
21.
2≤𝑥<3
𝑥≥2
Explain why the function is discontinuous at the given number 𝑐.
cos 𝑥
a) 𝑓(𝑥) = { 0
1 − 𝑥2
b) 𝑓(𝑥) = {
𝑥(𝑥−1)
(𝑥−1)(𝑥+1)
1
22.
𝑥<2
; 𝑥<0
; 𝑥=0
; 𝑥>0
;
𝑥≠1
;
𝑥=1
For what value(s) of 𝑘 is the function continuous at 𝑥 = −3
−6𝑥 − 12
If 𝑓(𝑥) = { 𝑘 2 − 5𝑘
6
;
;
;
𝑥 < −3
𝑥 = −3
𝑥 > −3
271 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
23.
Evaluate the following trigonometric limits:
a) lim
sin 7𝑥
𝑥
𝑥→0
b) lim
sin2 𝑥
𝑥→0 3𝑥 2
𝑥
c) lim sin 8𝑥
𝑥→0
sin 8𝑥
d) lim sin 3𝑥
𝑥→0
e) lim
tan 4𝑥
5𝑥
𝑥→0
f)
lim
sin(𝑥+3)
𝑥→−4 𝑥 2 +5𝑥+6
g) lim
sin(𝑥 2 )
𝑥→0 𝑥 tan 𝑥
2𝜃2
h) lim
𝜃→0 1−cos 𝜃
i) lim
sin 2𝑥 tan 3𝑥
𝑥2
𝑥→0
j) lim
sin 2𝑥+sin 4𝑥
𝑥
𝑥→0
𝑥 sin 2𝑥
k) lim 2−2cos2𝑥
𝑥→0
l)
lim
cos2 𝑥
𝑥→𝜋⁄2 1−sin 𝑥
sin 𝑥
m) lim 𝑥+tan 𝑥
𝑥→0
n) lim
sin 3𝑥 sin 5𝑥
𝑥→0
o)
lim
𝑥2
1−tan 𝑥
𝑥→𝜋⁄4 sin 𝑥−cos 𝑥
272 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
24.
Use the first principle of derivative to find the derivatives of the following
Functions:
a) 𝑓(𝑥) = 2𝑥 − 6
b) 𝑓(𝑥) = 𝑥 4
c) 𝑓(𝑥) = 𝑎𝑥 + 𝑏
d) 𝑓(𝑥) = sin 𝑥
e) 𝑓(𝑥) = cos 𝑥
25.
Find 𝑓′(𝑐) if it exists:
1
− 𝑥2
𝑓(𝑥) = { 2
−3
26.
;
;
𝑥<3
𝑥≥0
at 𝑐 = 3
Find the derivatives of the following functions with respect to 𝑥:
a) 𝑦 = √𝑥(𝑥 − 2)
b) 𝑦 = (2𝑥 7 − 5)(2 − 3𝑥)
𝐴
c) 𝑦 = 𝑥 5 + 𝐵𝑒 3𝑥
d) 𝑦 = (2√𝑤 − 3
1
√2𝑥
e) 𝑦 =
3
)
3−𝑥𝑒 𝑥
𝑥+𝑒 𝑥
f) 𝑦 + 2𝑒 2𝑥 − 2𝑥 = 0
1
5
g) 𝑦 = (𝑥 5 − 2𝑥) (√𝑥 + 5𝑥 7 )
273 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
27.
Find the derivatives of the following functions with respect to 𝑥:
a) 𝑦 = sin2 𝑥 + cos 𝑥 2
(2𝑥−8)5
b) 𝑦 = (𝑒 3𝑥 +tan 2𝑥)3
c) 𝑦 = 𝑒 𝑎 sec √4𝑥
d) 𝑦 = [3𝑥 − (2𝑥 2 − cos3 𝑥)3 ]7
15
8√𝑥
)
𝑥 4 −2𝑐
e) 𝑦 = (
f) 𝑦 = 3𝑥 log10 √𝑥
g) 𝑦 = − cos[ln(3𝑥 + 𝑏 −1 )]
h) 𝑦 = ln(𝑒 −𝑥 + 𝑥𝑒 −𝑥 )
i) 𝑦 = log 3 (𝑒 −𝑥 cos(𝜋𝑥))
j) 𝑦 = √𝑥
𝑥
k) 𝑦 = 𝑥 √𝑥
l) 𝑦 = (sin 𝑥)cos 𝑥
m) 𝑦 = 15−2 ln tan 2𝑥
n) 𝑦 = (ln 𝑥)tan 𝑥
1
o) 𝑦 = (tan 𝑥)𝑥
p) 𝑦 = ln
(2𝑥−1)5
√6𝑥−8
q) 𝑦 = ln ln ln 7𝑥
r) 𝑥 𝑦 = 𝑦 𝑥
274 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
28.
Find
𝑑𝑦
𝑑𝑥
in each of the following functions:
a) 𝑦 5 + 𝑦 5 = 25
b) 2𝑥𝑦 − 𝑦 2 + cos 𝑦𝑥 = 1
c) 𝑒 𝑥 sin 𝑦 = 2 + tan 2𝑥𝑦
d)
e) 𝑥 tan 𝑦 + 𝑦 tan 𝑥 = 5𝑒
f) (𝑥 + 𝑦)𝑥 3 = 𝑦 2 (2𝑥 − 𝑦)
g) tan−1(𝑦 2 𝑥) = 𝑦 + 𝑦𝑥 2
h) 4 sin 𝑦 cos 𝑥 = −3
i) 𝑒 𝑥⁄𝑦 = 𝑦 − 𝑥 2
j) 𝑥 sin 𝑦 + 𝑒 𝑦 − ln 3𝑥𝑦 − 5−3 = 2𝑥
29.
Find
𝑑2 𝑦
𝑑𝑥 2
in each of the following functions:
a) 𝑥10 + 𝑦10 = 10𝑒
b) 𝑦 = ln(sec 𝑥 + 𝑒 𝑥 )
c) 5𝑥 2 + 𝑦 2 = −5
d) 𝑥𝑦 + 𝑒 𝑦 = 𝑒
e) 𝑥 2 + 𝑥𝑦 + 𝑦 3 = 23
30.
Find
𝑑8 𝑥
𝑑𝑥 8
(𝑥 7 ln 𝑥).
275 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
31.
Evaluate the following limits using L’Hopital’s Rule.
6𝑥 2 +5𝑥−4
a) lim1 (4𝑥 2 +16𝑥−9)
𝑥→
2
𝑥 3 −2𝑥 2 +1
b) lim (
)
𝑥 3 −1
𝑥→1
𝑒 2𝑥 −1
c) lim ( sin 𝑥 )
𝑥→0
𝑥2
d) lim (1−cos 𝑥)
𝑥→0
ln 𝑥
e) lim (sin 𝜋𝑥)
𝑥→1
f)
ln(ln 𝑥)
lim
𝑥
𝑥→∞
g) lim
sin−1 𝑥
𝑥
𝑥→0
√1+2𝑥−√1−4𝑥
)
𝑥
𝑥→0
h) lim (
i) lim(cosec 𝑥 − cot 𝑥)
𝑥→0
j)
k)
l)
lim [ ln(𝑥 7 − 1) − ln(𝑥 5 − 1) ]
𝑥→1+
lim+
𝑒 𝑥 −𝑒 −𝑥 −2𝑥
1−sin 𝑥
𝑥→0
3
lim ( √𝑥)
𝑥→∞
−1
ln 𝑥
m) lim+ sin 𝑥 ln 𝑥
𝑥→0
n) lim+ 𝑥 √𝑥
𝑥→0
276 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
32.
Exhibit the validity of the Rolle’s Theorem:
a) 𝑓(𝑥) = 𝑥 2 − 8𝑥 + 12
𝑥 ∈ [2,6]
b) 𝑓(𝑥) = 5 − 12𝑥 + 3𝑥 2
𝑥 ∈ [1,3]
c) 𝑓(𝑥) = 𝑥 3 − 𝑥
𝑥 ∈ [−1,1]
3
3
d) 𝑓(𝑥) = √𝑥 2 − √𝑥
𝑥 ∈ [0,1]
e) 𝑓(𝑥) = 𝑥 3 − 𝑥 2 − 6𝑥 + 2
𝑥 ∈ [0,3]
33.
Exhibit the validity of the Mean Value Theorem:
a) 𝑓(𝑥) = 𝑥 3 + 24𝑥 − 16
6
𝑥 ∈ [0,4]
𝑥 ∈ [1,2]
b) 𝑓(𝑥) = 𝑥 − 3
𝑥
c) 𝑓(𝑥) = 𝑥+2
𝑥 ∈ [1,4]
d) 𝑓(𝑥) = 𝑥 3 + 12𝑥 2 + 7𝑥
𝑥 ∈ [−4,4]
e) 𝑓(𝑥) = 𝑒 −2𝑥
𝑥 ∈ [0,3]
34.
Determine the intervals on which the function 𝑓 is Ascending or Descending:
a) 𝑓(𝑥) = 𝑥 4 − 2𝑥 2 + 3
b) 𝑓(𝑥) = 𝑒 2𝑥 + 𝑒 −𝑥
c) 𝑓(𝑥) = 4cos2 − 2 sin 𝑥
0 ≤ 𝑥 ≤ 2𝜋
5
d) 𝑓(𝑥) = 5𝑥 3 − 2𝑥 3
e) 𝑓(𝑥) = 2𝑥 3 − 3𝑥 2 − 12𝑥
277 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
35.
Determine the Global extreme values:
a) 𝑓(𝑥) = 2𝑥 3 + 3𝑥 2 − 36𝑥
b) 𝑓(𝑥) = 4𝑥 3 + 3𝑥 2 − 6𝑥 + 1
f) 𝑓(𝑥) = 2 cos 𝑥 + cos2 𝑥
0 ≤ 𝑥 ≤ 2𝜋
c) 𝑓(𝑥) = sin 𝑥 + cos 𝑥
0 ≤ 𝑥 ≤ 2𝜋
d) 𝑓(𝑥) = 𝑒 tan
36.
−1 𝑥
−1 ≤ 𝑥 ≤ √3
Evaluate the following integrals:
a)
(2𝑥 − 𝑥 7 + 3 sin 𝑥) 𝑑𝑥
b)
3𝑥 cos 𝑥 2 𝑑𝑥
c)
d)
e)
37.
a)
b)
12−𝜃4
𝜃2
𝑑𝜃
𝑥2
2𝑥 3 −6
𝑑𝑥
𝑥 sin 5𝑥 𝑑𝑥
Evaluate the following integrals:
6𝑥 arc tan 3𝑥 2 𝑑𝑥
(6𝑦−1) sin(√3𝑦 2 −𝑦+8)
√3𝑦 2 −𝑦+8
c)
sec 3 𝑥 𝑑𝑥
d)
𝑥+1
𝑑𝑥
(𝑥−2)(𝑥 2 +4)
e)
𝑑𝑥
(81+𝑥 2 )2
𝑑𝑦
278 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
38.
Evaluate the following integrals:
a)
𝜋⁄2
(3 −
0
b)
3𝜋⁄4
sin5 𝑥 cos 3 𝑥
𝜋⁄2
c)
𝜋
sin4 (3θ) 𝑑𝜃
0
d)
𝜋⁄3
tan5 𝑥 sec 4 𝑥
0
e)
39.
a)
b)
c)
d)
e)
40.
a)
b)
c)
cos 𝑥)2
𝑑𝑥
𝑑𝑥
cos 𝜃 cos 5 (sin 𝜃) 𝑑𝜃
Evaluate the following integrals:
cos(−3𝑦) sin(7𝑦) 𝑑𝑦
𝑥
√1−𝑥 4
𝑑𝑥
1
1−cos 𝑥
1−sin 𝑥
cos 𝑥
𝑑𝑥
𝑑𝑥
𝑥+4
𝑥 2 +2𝑥+5
Evaluate the following integrals:
𝑒 arc tan 𝜃
𝜃2 +1
𝑑𝜃
𝑥 3 sin 𝑥 𝑑𝑥
cos−1 (𝑥 −2 )
𝑥3
𝑑𝑥
d)
cos3 𝑥 sin 2𝑥 𝑑𝑥
e)
sin2 𝑥 cos 𝑥 ln(sin 𝑥) 𝑑𝑥
279 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
41.
Find the area of the region bounded or enclosed by the curves:
a) 𝑦 = 𝑥 3 and
𝑦=𝑥
b) 𝑦 = 2𝑥 and
𝑦 = 𝑥 √𝑥 + 1
c) 𝑥 = 𝑦 2
and
𝑥 = 𝑦+2
d) 𝑦 = 𝑥 2
and
𝑦 = 𝑥 given that 𝑦 = 1
8
e) 𝑦 = tan−1 𝑥 in the 𝑥 −axis such that 𝑥 = 1
42.
Compute the arc length of the following function given the intervals.
a) 𝑦 = 𝑥 3⁄2
0≤𝑥≤1
b) 𝑦 = 2(𝑥 − 1)3⁄2
1≤𝑥≤5
c) 𝑦 =
𝑥3
6
1
+ 2𝑥
2
1≤𝑥≤3
d) 𝑦 = 3 (𝑥 2 + 1)3⁄2
1≤𝑥≤4
e) 𝑦 = ln(cos 𝑥)
0 ≤ 𝑥 ≤ 𝜋⁄4
280 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
281 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
TEST 1A BSc
MODULE CODE: MMTH011
MODULE NAME: DIFFERENTIAL AND INTEGRAL CALCULUS
TOTAL MARKS: 70
DURATION: 2 HRS
SEMESTER TEST:
INTERNAL EXAMINER: MR CHAUKE E.
282 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 1:
Find 𝐷𝑓 , 𝑅𝑓 and sketch the following:
1.1.
𝑥+1
𝑓(𝑥) = { 2
1
1.2.
𝑖𝑓
𝑥<1
𝑖𝑓 1 ≤ 𝑥 < 2
𝑖𝑓
𝑥=2
Determine all the asymptotes and sketch the following:
𝑥 2 +4
(12)
𝑓(𝑥) = 𝑥 2 −4
If 𝑓(𝑥) = 1 − 𝑥 and 𝑔(𝑥) = √𝑥 , determine the following:
1.3.
i)
ii)
iii)
1.4.
(8)
𝑔∘𝑓
𝑓∘𝑔
𝑓∘𝑓
(12)
Show that ||𝑥| − |𝑦|| ≤ |𝑥 − 𝑦| if 𝑥, 𝑦 ∈ 𝑅 Hint: 𝑥 = 𝑥 − 𝑦 + 𝑦
(8)
QUESTION 2:
2.1
Evaluate the following limits:
a) lim
𝑥 3 +2𝑥 2 −1
𝑥→−2
b) lim
5−3𝑥
√2𝑥 2 +1
𝑥→∞ 3𝑥−5
c) lim
sin(𝑥−1)
𝑥→1 𝑥 2 +𝑥−2
(4)
(4)
(4)
283 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
2.2. Prove that if 𝑓 and 𝑔 are continuous at 𝑥 = 𝑐 then 𝑓𝑔 is continuous at 𝑥 = 𝑐.
(9)
2.3. Determine whether the function 𝑓 is continuous at 𝑐 = 2.
1
𝑓(𝑥) = { 0
−1
𝑖𝑓 𝑥 > 0
𝑖𝑓 𝑥 = 0
𝑖𝑓 𝑥 < 0
(9)
284 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
TEST 1B BSc
MODULE CODE: MMTH011
MODULE NAME: DIFFERENTIAL AND INTEGRAL CALCULUS
TOTAL MARKS: 100
DURATION: 2 HRS
SEMESTER TEST:
INTERNAL EXAMINER: MR CHAUKE E.
285 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 1 [50]:
1.1.
If 𝑓(𝑥) = 𝑥 2 + 2𝑥 − 1 and 𝑔(𝑥) = 2𝑥 − 3 Then find the following:
1.1.1. (𝑓 ∘ 𝑔)(𝑥)
(2)
1.1.2. (𝑔 ∘ 𝑓)(𝑥)
(2)
1.1.3. (𝑔 ∘ 𝑔 ∘ 𝑔)(𝑥)
(3)
1.2.
𝑥+2
3
A function 𝑓 is defined by 𝑓(𝑥) = {
2 − 𝑥2
𝑥−3
1
1.2.1. Evaluate: 𝑓 (2) ;
3
𝑓 (2)
;
𝑥<1
; 𝑥=1
; 1<𝑥≤2
;
𝑥>2
7
(6)
and 𝑓 (3).
(14)
1.2.2. Sketch the function 𝑓(𝑥) and find 𝐷𝑓 & 𝑅𝑓 .
1.2.3. Convert
5
7
(2)
𝜋 to degrees and 330° to Radians.
1.2.4. Show that |𝑥 + 𝑦| ≤ |𝑥| + |𝑦| (𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝐼𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦)
𝑥+1
(8)
1.2.5. Sketch the function 𝑓(𝑥) = 𝑥−1 and show all the asymptotes and intercepts.
1.2.6. Prove that if 𝑥, 𝑦 ∈ 𝑅 then ||2| − |𝑦|| ≤ |𝑥 − 𝑦|
(10)
(3)
286 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 2 [39]:
2.1. Complete the following:
(3)
lim 𝑓(𝑥) = 𝐿 Means…………………………..
𝑥→𝑎
2.2. Use 2.1. To show that: lim(14 − 5𝑥) = 4
(6)
2.3. Prove that if 𝑓 and 𝑔 are continuous at 𝑥 = 𝑐 then 𝑓𝑔 is continuous at 𝑥 = 𝑐.
(8)
𝑥→2
2.4. Evaluate the following limits:
1 1
−
3 𝑥
(2)
2.4.1. lim (𝑥−3)
𝑥→3
4−√𝑥
(5)
2.4.2. lim (16𝑥−𝑥 2 )
𝑥→16
1
1
(5)
2.4.3. lim (𝑥√1+𝑥 − 𝑥)
𝑥→0
√9𝑥 6 −𝑥
2.4.4. lim (
𝑥→∞
2.4.5. lim (
2−𝑥 3
2𝑥−𝑥 2
𝑥→∞ √𝑥+𝑥 2
(5)
)
(5)
)
QUESTION 3 [11]:
2𝑥 2 − 4
𝑥2
3.1. Consider the given function 𝑓(𝑥) = 6 − 𝑥
𝑥+3
{√
; 0≤𝑥<3
;
𝑥=2
; 2<𝑥<5
3
;
𝑥<2
3.1.1. Determine whether the function 𝑓 is continuous at 𝑥 = 2.
(11)
287 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
TEST 2A BSc
MODULE CODE: MMTH011
MODULE NAME: DIFFERENTIAL AND INTEGRAL CALCULUS
TOTAL MARKS: 120
DURATION: 2 HRS
SEMESTER TEST:
INTERNAL EXAMINER: MR CHAUKE E.
288 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 1:
1.1.
Evaluate the following limits.
sin 3𝑥
(4)
1.1.1. lim 5𝑥 3 −4𝑥
𝑥→0
1.1.2. lim
𝑥→0
sin(𝑥 2 )
(4)
𝑥
𝑥 3 −1
1.1.3. lim 𝑒 1−𝑥 −1
(4)
ln(𝑥−3)
(3)
𝑥→1
1.1.4. lim
𝑥→4
𝑥−4
1
1.2.
Show from a definition that if 𝑓(𝑥) = log 𝑏 𝑥 then the 𝑓′(𝑥) = 𝑥 log 𝑏 𝑒
(7)
1.3.
State the Rolle’s and Mean Value Theorem.
(3)
1.4.
Prove, using the definition of derivatives , that if 𝑓(𝑥) = cos 𝑥 then the 𝑓 ′ (𝑥) = − sin 𝑥
(7)
1.5.
Use the Chain Rule to show that if 𝑦 = (𝑥 6 − 1)2 , then
𝑑𝑦
𝑑𝑥
= 12𝑥 5 √𝑦 where 𝑦 > 0
(4)
QUESTION 2:
2.1. Exhibit the validity of the Mean Value Theorem in the following functions.
𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 1
; 𝑥𝜖[1,4]
2.2. Determine the Extreme values for then function: 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2
(7)
(6)
2.3. Prove that if 𝑓 and 𝑔 are differentiable function the 𝑓𝑔 is also differentiable at 𝑥 and
(𝑓𝑔)′ = 𝑓𝑔′ + 𝑓′𝑔
(7)
289 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
2.4. Determine 𝑦′ in the following differential problem.
a) tan−1(𝑥 2 𝑦) = 𝑥 + 𝑥𝑦 2
(7)
b) 𝑦 = 𝑥 𝑥
(7)
(𝑥+3)(𝑥−2)
c) 𝑦 = √
√𝑥−1
d) 𝑦 = (𝑒 ln cot 𝑥 )
5
(7)
(7)
e) 𝑦 = (ln 𝑥)cos 𝑥
(7)
f) 𝑦 = log 2 (𝑥 log 5 𝑥)
(7)
g) 𝑥 𝑦 = 𝑦 𝑥
(7)
2.5. Determine the intervals on which the function 𝑓 is Increasing/Decreasing.
If 𝑓(𝑥) = 𝑥 4 − 4𝑥 2 + 2
2.6. Find
𝑑2 𝑦
𝑑𝑥 2
: 𝑒 𝑥 − 𝑒𝑦 = 𝑥2 + 𝑦2
(8)
(7)
290 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
TEST 2B BSc
MODULE CODE: MMTH011
MODULE NAME: DIFFERENTIAL AND INTEGRAL CALCULUS
TOTAL MARKS: 130
DURATION: 2 HRS
SEMESTER TEST:
INTERNAL EXAMINER: MR CHAUKE E.
291 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 1 [52]:
1. Find the derivative using the Definition of the following function:
(6)
If 𝑓(𝜃) = √𝜃
2. Evaluate the limit: lim
𝑒 𝑥 −𝑒 3
(4)
𝑥→3 𝑥−3
3. Prove that if 𝑦 = tan−1 𝑥 in the open interval (−∞, ∞) then
4. Consider the given function 𝑓 if 𝑓(𝑥) = sin 𝑥 + cos 𝑥
𝒅𝒚
𝒅𝒙
=
𝟏
(7)
𝟏+𝒙𝟐
where 𝑥 ∈ [0,2𝜋]
a) Find the critical number of the function 𝑓.
b) Find the intervals on which 𝑓 is increasing or decreasing.
c) Find the local maximum and minimum values of 𝑓.
(4)
(5)
(5)
5. Find 𝑓′(𝑐) if it exists:
𝑓(𝑥) = {
3𝑥 2
2𝑥 3 + 1
;
;
𝑥≤1
𝑥≥1
at 𝑐 = 1
(8)
6. Exhibit the validity of the Rolle’s Theorem:
If 𝑓(𝜃) = cos 2𝜃
(6)
𝜋⁄8 ≤ 𝜃 ≤ 7𝜋⁄8
𝑔(𝑥)
7. If 𝑓 and 𝑔 are differentiable functions and 𝑓(𝑥) = 𝑟(𝑥) such that 𝑟(𝑥) ≠ 0,
Then 𝑓′(𝑥) =
𝑔′(𝑥)∙𝑟(𝑥)−𝑔(𝑥)∙𝑟′(𝑥)
[𝑟(𝑥)]2
(7)
292 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 2 [78]:
1. Find the derivatives of the following functions:
a) 𝑓(𝑥) = 𝑥 2 sin 𝑥 tan 𝑥
𝑥 2 +𝑎3
(5)
3
(5)
b) 𝑓(𝑥) = (𝑥 2 −𝑏3 )
c) 𝑓(𝑥) = 𝑒 𝑒
𝑥
(5)
2
d) 𝑓(𝑥) = sin2 (𝑒 sin 𝑥 )
(5)
e) 𝑓(𝑥) = 2sin 𝜋𝑥
(5)
2. Find the equation of the tangent to the circle 𝑥 2 + 𝑦 2 = 25 at the point (3,4).
(6)
3. Find 𝑦′′ by implicit differentiation if, √𝑥 + √𝑦 = 0
(6)
4. Evaluate the following integrals:
𝑠𝑒𝑐 2 (1⁄𝑥)
𝑑𝑥
(5)
b)
sin−1 𝑥 𝑑𝑥
(7)
c)
sin6 𝑥 cos3 𝑥 𝑑𝑥
(7)
a)
d)
e)
𝑥2
𝑥
√𝑥 2 +𝑥+1
𝑑𝑥
𝑒 2𝑥
𝑒 2𝑥 +3𝑒 𝑥 +1
𝑑𝑥
(8)
(7)
5. Find the area of the region bounded or enclosed by the curves:
If 4𝑥 + 𝑦 2 = 12 and 𝑥 = 𝑦
(7)
293 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
294 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION PAPER 2016
SEFAKO MAKGATHO HEALTH SCIENCES UNIVERSITY
ASSESSMENT AND CERTIFICATION MANAGEMET
FACULTY OF SCIENCES & TECHNOLOGY
SCHOOL OF PATHOLOGY AND PRE-CLINICAL SCIENCES
SUBJECT NAME: Differential and Integral Calculus
TYPE OF EXAMINATION: STANDARD
X
:RE-EXAMINATION
SUBJECT CODE : MMTH011
COURSE
jdk
..
: MATHEMATICS 1
PAPER NUMBER : 1
DATE OF EXAMINATION:
24th MAY 2016
NUMBER OF STUDENTS:
372
H
k
INTERNAL EXAMINERS PARTICULARS: 1. Mr J.L Thabane
2.
EXTERNAL EXAMINERS PARTICULARS: 1. Mrs. A vd Merwe
2.
…
DURATION
: 3HRS
O
TOTAL MARKS
: 150
H
3.
o
4.
p
3.
i
4.
o
THIS QUESTIO PAPER CONSIST OF 3 PAGES INCLUDING COVER PAGE:
Specified Instructions to
Students.
1
2
3
4
Stationary.
4 PAGE BOOK
8 PAGE BOOK
MCQ’ s ( must be supplied by
the department on day of
submission )
Double Folios
Please confirm what
you need.
Please indicate the quality
per Student.
x
1
295 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 1 [20]:
1.1.
A function 𝑓 is defined by:
𝑓(𝑥) = {
1−𝑥
𝑥2
𝑖𝑓
𝑖𝑓
𝑥≤1
𝑥>1
(10)
Evaluate 𝑓(0) ; 𝑓(1) ; 𝑓(2) and sketch the function 𝑓.
1.2.
i)
ii)
1.3.
i)
ii)
If 𝑓(𝑥) = 1 − 𝑥 and 𝑔(𝑥), then determine the following:
(3)
(3)
𝑔∘𝑓
𝑓∘𝑔
Change the following angle sizes to radians:
(2)
(2)
270°
120°
QUESTION 2 [40]:
2.1. Determine whether the following function is continuous at the indicated points.
𝑓(𝑥) =
𝑥3
;
𝑥 ≤ −1
𝑥2 − 2
;
−1 < 𝑥 < 0
3−𝑥
;
0≤𝑥<2
;
2≤𝑥<4
;
4<𝑥<7
;
𝑥≥7
4𝑥−1
𝑥−1
15
7−𝑥
{ 5𝑥
i)
𝑥=1
ii)
𝑥=7
iii)
𝑥=2
at
(21)
296 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
2.2. Evaluate the following limits:
a) lim
√2𝑥 2 +1
(4)
𝑥→∞ 3𝑥−5
b) lim
sin(𝑥−2)
(4)
𝑥 2 −4
𝑥→2
𝑥 3 −1
(5)
c) lim 𝑒 1−𝑥 −1
𝑥→1
1
d) lim
𝑥 2 −3
(3)
𝑥→9 𝑥−9
2.3. Complete the following:
The function 𝑓 is continuous on [𝑎, 𝑏] if……………
(3)
QUESTION 3 [38]:
3.1. Show from definition that if 𝑓(𝑥) = cos 𝑥 then 𝑓′(𝑥) = − sin 𝑥
3.2. Determine
𝑑𝑦
𝑑𝑥
of the following:
I.
𝑦 = 𝑥 2 sin 𝑥
II.
𝑦 = (1−𝑥 2 )1⁄2
3.3. Find
(4)
𝑥
𝑑2 𝑥
𝑑𝑥 2
(10)
(5)
given that 𝑥 4 + 𝑦 4 = 1
(7)
3.4. Prove that if 𝑓 and 𝑔 are differentiable functions then 𝑦 = (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) is a
𝑑𝑦
(12)
differentiable function of 𝑥 and
= 𝑓′(𝑔(𝑥)) ⋅ 𝑔′(𝑥)
𝑑𝑥
297 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 4 [52]:
4.1. Prove that if 1) 𝑓 is continuous on the closed interval [𝑎, 𝑏] 2) 𝑓 is differentiable
on (𝑎, 𝑏)then there exists a 𝑐 ∈ (𝑎, 𝑏) such that,
𝑓 ′ (𝑐) =
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
(12)
(The First Mean Value Theorem for Derivatives)
4.2. Find where the function 𝑓(𝑥) = 3𝑥 4 − 4𝑥 3 − 12𝑥 2 + 5 is increasing and where it is
(9)
Decreasing.
4.3. Evaluate the following integrals:
a)
sin(ln 𝑥)
𝑥
𝑑𝑥
(4)
b)
cos 3 𝑥 𝑑𝑥
(5)
c)
𝑒 𝑥 sin 𝑥 𝑑𝑥
(9)
d)
1
1−𝑥 2
𝑑𝑥
(7)
298 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION PAPER 2017
SEFAKO MAKGATHO HEALTH SCIENCES UNIVERSITY
ASSESSMENT AND CERTIFICATION MANAGEMET
FACULTY OF SCIENCES & TECHNOLOGY
SCHOOL OF PATHOLOGY AND PRE-CLINICAL SCIENCES
SUBJECT NAME: PRE-CALCULUS AND DIFFERENTIAL CALCULUS.
TYPE OF EXAMINATION: STANDARD
X
: RE-EXAMINATION
SUBJECT CODE : MMTH000
COURSE
jdk
..
: MATHEMATICS 1
PAPER NUMBER : 1
DATE OF EXAMINATION:
23th OCTOBER 2017
NUMBER OF STUDENTS:
160
H
k
INTERNAL EXAMINERS PARTICULARS: 1. Dr PWN. Chin
2.
EXTERNAL EXAMINERS PARTICULARS: 1. Mrs. D Vijayasenan
2.
…
DURATION
: 3HRS
O
TOTAL MARKS
: 120
H
3.
o
4.
p
3.
i
4.
o
THIS QUESTIO PAPER CONSIST OF 3 PAGES INCLUDING COVER PAGE:
Specified Instructions to
Students.
1
2
3
4
Stationary.
4 PAGE BOOK
8 PAGE BOOK
MCQ’ s ( must be supplied by
the department on day of
submission )
Double Folios
Please confirm what
you need.
Please indicate the quality
per Student.
x
1
299 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 1 [30]:
1. Solve for the value(s) of 𝑥 in the following equations:
a) 𝑒 2𝑥+3 − 7 = 0
(5)
b) sin 2𝑥 = cos 𝑥 𝑥 ∈ [0,2𝜋]
(5)
c) ln(5 − 2𝑥) = −3
(5)
2. Find the following in the interval notation:
If 𝑓(𝑥) = {
𝑥+2
𝑥2
;
;
𝑥 ≤ −1
𝑥 > −1
a) The Domain of the function.
b) The Range of the function.
c) Sketch the above function 𝑓(𝑥).
(5)
(5)
(5)
QUESTION 2 [30]:
1. Evaluate the following limits:
√𝑥−1
𝑥→1 𝑥−1
(7)
a) lim
b) lim
√1+𝑥 2
(7)
𝑥→∞ 2𝑥+3
c) lim
sin 6𝑥
(7)
𝑥→0 sin 2𝑥
𝑥 2 −4
2. Give that 𝑓(𝑥) = { 𝑥−2
2𝑘
𝑖𝑓 𝑥 ≠ 2
𝑖𝑓 𝑥 = 2
Find the value of 𝑘 if 𝑓 is continuous at 𝑥 = 2.
(9)
300 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 3 [30]:
1. Find the derivative with respect to 𝑥 of the following functions:
𝑥2
(7)
a) 𝑓(𝑥) = 𝑥−2
b) 𝑓(𝑥) = 𝑥 2 𝑒 2𝑥
2 +2
(7)
c) 𝑓(𝑥) = ln(sin 𝑥 2 )
2. Find
𝑑𝑦
𝑑𝑥
(7)
from the equation: 𝑥𝑦 + 𝑦 2 = 1.
(9)
QUESTION 4 [30]:
1. Find the following:
a)
b)
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦
𝑑𝑥
of the function , 𝑦 = 𝑥 3 + 𝑒 2𝑥 at any point 𝑥.
2
of the function , 𝑦 = sin−1(𝑥 2 ) − 𝑒 𝑥 .
c) The turning points of the function, 𝑓(𝑥) = 2𝑥 2 − 𝑥 4 .
2. Use (4)(𝑐) above to find the equation of the tangent at the point 𝑥 = 1.
(8)
(8)
(8)
(6)
QUESTION 5 [30]:
Evaluate the following Integrals:
1.
(𝑥 2 + 5𝑥 + 3) 𝑑𝑥
(7)
2.
1 5𝑥
𝑒
0
(7)
𝑑𝑥
3.
sec 4 𝑥 tan 𝑥 𝑑𝑥
(9)
4.
cos 3𝑥 𝑒 sin 3𝑥 𝑑𝑥
(7)
301 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION PAPER 2017
SEFAKO MAKGATHO HEALTH SCIENCES UNIVERSITY
ASSESSMENT AND CERTIFICATION MANAGEMET
FACULTY OF HEALTH SCIENCES
SCHOOL OF PATHOLOGY AND PRE-CLINICAL SCIENCES
SUBJECT NAME: Differential and Integral Calculus
TYPE OF EXAMINATION: STANDARD
X
:RE-EXAMINATION
SUBJECT CODE : MAH101M/MMTH011
COURSE
jdk
..
: MATHEMATICS 1
PAPER NUMBER : 1
DATE OF EXAMINATION:
O
19 MAY 2017
NUMBER OF STUDENTS:
70
H
k
DURATION
: 3HRS
TOTAL MARKS : 150
INTERNAL EXAMINERS PARTICULARS: 1. MR E. CHAUKE
3.
2.
EXTERNAL EXAMINERS PARTICULARS: 1. MR THOKA S.J
2.
…
H
o
4.
p
3.
i
4.
o
THIS QUESTIO PAPER CONSIST OF 4 PAGES INCLUDING COVER PAGE:
Specified Instructions to
Students.
1
2
3
4
Stationary.
4 PAGE BOOK
8 PAGE BOOK
MCQ’ s ( must be supplied by
the department on day of
submission )
Double Folios
Please confirm what
you need.
Please indicate the quality
per Student.
x
1
302 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 1:
4𝜋
Convert
1.2.
Let 𝑓 be a function defined by:
3
(2)
to Degrees and 225° to Radians
1.1.
5
𝑓(𝑥) = {𝑎𝑥 + 𝑏
21
𝑖𝑓 𝑥 ≤ 2
𝑖𝑓 2 < 𝑥 < 10
𝑖𝑓 𝑥 ≥ 10
a) Determine the values of 𝒂 and 𝒃 such that the function 𝑓 is
Continuous at 𝑐 = 2
and at 𝑐 = 10
5
b) Evaluate 𝑓 ( ):
2
(2)
c) Evaluate lim− 𝑓(𝑥):
(2)
d) Sketch the graph 𝑓(𝑥):
e) Find the Domain and the Range of the function 𝑓(𝑥):
(5)
(2)
Complete the following:
lim 𝑓(𝑥) = 𝐿 Means…………………………
(3)
𝑥→10
1.3.
(7)
𝑥→𝑎
1
(5)
1.4.
Use 1.3. to show lim (3 − 𝑥) = −5
2
1.5.
Let 𝑓(𝑥) = 2 + cos 𝑥 and 𝑔(𝑥) = sin2 2𝑥 − 1 , Find the following:
𝑥→16
2𝜋
i)
(𝑓 ∘ 𝑔) ( )
3
ii)
(𝑔 ∘ 𝑓) (−
3𝜋
4
(3)
)
(3)
303 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
[𝟑𝟒]
QUESTION 2:
2.1. Show from definition that if 𝑓(𝑥) = cos 𝑥 then 𝑓′(𝑥) = − sin 𝑥
(6)
2.2. Evaluate the following limits:
√1+𝑥 2
a) lim
(4)
𝑥→∞ 3+2𝑥
b) lim
sin(𝑥 2 −4)
𝑥→2
(4)
𝑥−2
𝑥 3 +1
c) lim
(4)
𝑥→−1 −𝑥 2 +𝑥+2
d) lim 𝑥 3 𝑒 −𝑥
2
(4)
𝑥→∞
2.3. Prove that if 𝑓 and 𝑔 are differentiable functions then 𝑦 = (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) is a
differentiable function of 𝑥 and
2.4. Find
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 𝑓′(𝑔(𝑥)) ⋅ 𝑔′(𝑥) ( Chain Rule )
if sin2 𝑦 + cos(𝑥𝑦) = 𝜋
(8)
(5)
2.5. Exhibit the validity of Mean Value Theorem in the function 𝑓.
𝑓(𝑥) = 𝑥 3 − 5𝑥 2 − 3𝑥
,
1≤𝑥≤3
(5)
2.6. State the Rolle’s Theorem:
(3)
2.7. Prove the Mean Value Theorem for Differential Calculus.
(8)
[𝟓𝟐]
304 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 3:
3.1. Determine the extremes values for the following function:
𝑓(𝜃) = 𝜃 − 2 tan−1 𝜃
(4)
3.2. Find the Derivatives of the following functions:
a) 𝑦 = 𝑎𝑟𝑐 cos(𝑥 + 𝑥 3 )
b) 𝑦 𝑥 + 𝑥 𝑦 = 1
(4)
(5)
3.3. Show that if 𝑦 2 = 𝑒 √𝑥 , then
3.4. Find
𝑑2𝑦
𝑑𝑥 2
𝑑𝑦
𝑑𝑥
=
𝑒√𝑥
4√𝑥𝑒 √𝑥
(5)
, 𝑥>0
𝑦
, if
(6)
𝑥 = tan ( )
3
3.5. Evaluate the following Integrals:
a)
b)
c)
d)
e)
3 sec √𝑥 tan √𝑥
7√𝑥
10
(𝑥−1)(𝑥 2 +9)
2𝑥
(4)
𝑑𝑥
(5)
𝑑𝑥
(10)
(6)
𝑒 sin 2𝑥 𝑑𝑥
𝑥 2 𝑠𝑖𝑛 𝑥 3 cos 5 𝑥 3 𝑑𝑥
3𝑥 2 ln(𝑥 3 +5)
(𝑥 3 +5)
(4)
𝑑𝑥
𝜋
𝜋
2
2
3.6. Show that if 𝑥 = 𝑎 tan 𝜃 , with 𝑎 > 0 and − < 𝜃 <
Then
√𝑎2 + 𝑥 2 𝑑𝜃 = 𝑎 ln|sec 𝜃 + tan 𝜃| + 𝐶
(6)
3.7. Calculate the area of the region bounded by the given parabola with equation,
𝑦 2 = 4𝑥 and the line. 4𝑥 − 2𝑦 − 4 = 0
(5)
[64]
………………………………….BEST OF LUCK………………………………
305 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
306 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
MEMORANDUM 2A BSc
MODULE CODE: MMTH011
MODULE NAME: DIFFERENTIAL AND INTEGRAL CALCULUS
TOTAL MARKS: 120
DURATION: 2 HRS
SEMESTER TEST:
INTERNAL EXAMINER: MR CHAUKE E.
307 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
1.1.
sin 3𝑥
OR
1.1.1. lim 5𝑥 3 −4𝑥
𝑥→0
sin 3𝑥
= lim
𝑥→0
= lim
3𝑥
∙
3𝑥
−3
𝑥→0
3 cos 3𝑥
𝑥→0
3 cos 0
∙ lim 𝑥(5𝑥 2 −4)
= 15(0)−4
𝑥→0
3(1)
= (1) ( 4 )
=
= −4
= −4
3
1.1.2. lim
sin(𝑥 2 )
𝑥
sin(𝑥2 )
lim 𝑥
𝑥→0
sin(𝑥 2 )
OR
𝑥→0
=
= lim
𝑥→0
= lim
𝑥2
sin(𝑥 2 )
𝑥2
𝑥→0
=0
Using L’hopital’s Rule
= lim 15𝑥 2 −4
3𝑥
5𝑥 3 −4𝑥
sin 3𝑥
𝑥(3)
𝑥→0
sin 3𝑥
lim 5𝑥 3 −4𝑥
−4
3
lim
sin(𝑥 2 )
𝑥→0
= lim
𝑥
2𝑥 cos(𝑥 2 )
𝑥→0
Using L’hopital’s Rule
1
∙𝑥
= lim 2𝑥 cos(𝑥 2 )
∙ lim 𝑥 = (1)(0)
= 2(0) cos(0)
𝑥→0
𝑥→0
=0
𝑥 3 −1
1.1.3. lim 𝑒 1−𝑥 −1
𝑥→1
3𝑥 2
= lim −𝑒 1−𝑥
𝑥→1
=
3(1)
−𝑒 0
= −3
1.1.4. lim
ln(𝑥−3)
𝑥−4
𝑥→4
1
= lim 𝑥−3
1
𝑥→4
1
= lim 𝑥−3
𝑥→4
1
= 4−3
=1
308 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
1.2.
Let 𝑓(𝑥) = log 𝑏 𝑥 ,
Proof:
1
𝑓′(𝑥) = 𝑥 log 𝑏 𝑒
then
𝑓(𝑥+ℎ)−𝑓(𝑓)
𝑓′(𝑥) = lim
ℎ→0
= lim
ℎ
log𝑏 (𝑥+ℎ)−log𝑏 𝑥
ℎ
ℎ→0
log𝑏 (
= lim
ℎ→0
𝑥+ℎ
)
𝑥
ℎ
1
𝑥+ℎ
= lim ℎ log 𝑏 (
ℎ→0
1
𝑥
ℎ→0 𝑥
ℎ
𝑥
)
ℎ
= lim ∙ log 𝑏 (1 + )
𝑥
1
ℎ
= lim 𝑥 log 𝑏 (1 + 𝑥 )
ℎ→0
1
= 𝑥 lim log 𝑏 (1 +
ℎ→0
𝑥
ℎ
𝑥
ℎ ℎ
)
𝑥
ℎ
Let 𝑝 = 𝑥 , if ℎ → 0 , then
ℎ
→0
𝑥
𝑥
1
∴ 𝑓′(𝑥) = 𝑥 lim log 𝑏 (1 +
ℎ→0
1
ℎ ℎ
)
𝑥
1
= 𝑥 lim log 𝑏 (1 + 𝑝)𝑝
ℎ→0
1
1
= 𝑥 log 𝑏 (lim (1 + 𝑝)𝑝 )
1
ℎ→0
= 𝑥 log 𝑏 𝑒
Hence the theorem.
1.3.
Rolle’s Theorem: States that let 𝑓 be a function that satisfies the following three hypotheses:
1. 𝑓 is continuous on the closed interval [𝑎, 𝑏]
2. 𝑓 is differentiable on the open interval (𝑎, 𝑏)
3. 𝑓(𝑎) = 𝑓(𝑏)
Then there is a number 𝑐 in (𝑎, 𝑏) such that 𝑓′(𝑐) = 0
309 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
1.4. Let 𝑓(𝑥) = cos 𝑥 , then 𝑓′(𝑥) = − sin 𝑥
Proof:
𝑓(𝑥+ℎ)−𝑓(𝑓)
ℎ
cos(𝑥+ℎ)−cos 𝑥
lim
ℎ
ℎ→0
cos 𝑥 cos ℎ−sin 𝑥 sin ℎ−cos 𝑥
lim
ℎ
ℎ→0
cos 𝑥 cos ℎ−cos 𝑥−sin 𝑥 sin ℎ
lim
ℎ
ℎ→0
cos 𝑥(cos ℎ−1)−sin 𝑥 sin ℎ
lim
ℎ
ℎ→0
cos 𝑥(cos ℎ−1)
sin 𝑥 sin ℎ
lim (
) − lim (
)
ℎ
ℎ
ℎ→0
ℎ→0
cos ℎ−1
sin ℎ
cos 𝑥 lim (
) − sin 𝑥 lim (
)
ℎ
ℎ
ℎ→0
ℎ→0
𝑓′(𝑥) = lim
ℎ→0
=
=
=
=
=
=
= cos 𝑥 (0) − sin 𝑥 (1)
= − sin 𝑥
1.5.
𝑦 = (𝑥 6 − 1)2
𝑑𝑦
∴ 𝑑𝑥 = 2(𝑥 6 − 1)(6𝑥 5 )
= 12𝑥 5 (𝑥 6 − 1)
But 𝑦 = (𝑥 6 − 1)2
⟹ (𝑥 6 − 1) = √𝑦
∴
𝑑𝑦
𝑑𝑥
= 12𝑥 5 √𝑦
310 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
2.
2.1.
𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 1
; 𝑥𝜖[1,4]
𝑓(𝑏)−𝑓(𝑎)
∴ 𝑓′(𝑐) = 𝑏−𝑎
𝑓(4)−𝑓(1)
= 4−1
17−(−1)
=
3
18
= 3 =6
′ (𝑥)
2
𝑓
= 3𝑥 − 6𝑥
∴ 𝑓 ′ (𝑐) = 3𝑐 2 − 6𝑐
∴ 6 = 3𝑐 2 − 6𝑐
∴ 𝑐 2 − 2𝑐 − 2 = 0
Thus 𝑐 =
=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
−(−2)±√(−2)2 −4(1)(−2)
2
= 1 ± √3
Hence 𝑐 = 1 + √3 and 𝑐 ≠ 1 + √3
∴ 𝑐 = 1 + √3𝜖(1,4)
2.2. 𝑓(𝑥) = 𝑥 3 − 3𝑥 + 2
∴ 𝑓′(𝑥) = 3𝑥 2 − 3
∴ 𝑓′(𝑐) = 3𝑐 2 − 3
∴ 3𝑐 2 − 3 = 0
⟹ 𝑐2 − 1 = 0
⟹ 𝑐 = ±1
∴ 𝑓(−1) = (−1)3 − 3(−1) + 2 = 4
∴ 𝑓(1) = (1)3 − 3(1) + 2 = 0
Thus 𝑓 has a global maximum at −𝟏 and 𝑓 has a global minimum at 𝟏
2.4.
a) tan−1 (𝑥 2 𝑦) = 𝑥 + 𝑥𝑦 2
∴
∴
∴
1
{2𝑥𝑦
1+(𝑥 2 𝑦)2
𝑑𝑦
𝑑𝑦
+ 𝑥 2 𝑑𝑥 } = 1 + 𝑦 2 + 2𝑥𝑦 𝑑𝑥
2𝑥𝑦
𝑥2
𝑑𝑦
𝑑𝑦
+
= 1 + 𝑦 2 + 2𝑥𝑦
1+(𝑥 2 𝑦)2
1+(𝑥 2 𝑦)2 𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑥2
2𝑥𝑦
(
− 2𝑥𝑦) = 1 + 𝑦 2 − 1+(𝑥2 𝑦)2
𝑑𝑥 1+(𝑥 2 𝑦)2
Thus
𝑑𝑦
𝑑𝑥
1+𝑦 2 −
=
2𝑥𝑦
2
1+(𝑥2 𝑦)
𝑥2
2 −2𝑥𝑦
1+(𝑥2 𝑦)
311 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
b) 𝑦 = 𝑥 𝑥
⟹ ln 𝑦 = ln 𝑥 𝑥
⟹ ln 𝑦 = 𝑥 ln 𝑥
1 𝑑𝑦
𝑑
𝑑
∴ 𝑦 ∙ 𝑑𝑥 = ln 𝑥 𝑑𝑥 (𝑥) + 𝑥 𝑑𝑥 (ln 𝑥)
1
= ln 𝑥 + 𝑥 (𝑥)
= ln 𝑥 + 1
𝑑𝑦
∴ = 𝑦(ln 𝑥 + 1 )
𝑑𝑥
= 𝑥 𝑥 (ln 𝑥 + 1 )
1
(𝑥+3)(𝑥−2)
c) 𝑦 = √
∴
∴
=
𝑑𝑦 1 (𝑥 + 3)(𝑥 − 2)
= (
)
𝑑𝑥 2
√𝑥 − 1
d) 𝑦 = (𝑒
𝑑𝑦
𝑑𝑥
√𝑥−1
(𝑥+3)(𝑥−2) 2
(
)
√𝑥−1
1
1
[(1)(𝑥 − 2) + (𝑥 + 3)(1)](√𝑥 − 1) − [(𝑥 + 3)(𝑥 − 2)] ( (𝑥 − 1)−2 (1))
2
1
−
2
(√𝑥 − 1)
{
ln cot 𝑥 5
2
}
)
= 5(𝑒
ln cot 𝑥 4
1
) {(𝑒 ln cot 𝑥 ) (cot 𝑥) (−cosec 2 𝑥)}
e) 𝑦 = (ln 𝑥)cos 𝑥
∴ ln 𝑦 = cos 𝑥 ln(ln 𝑥)
1
∴ 𝑦 ′ = − sin 𝑥 ln(ln 𝑥) + cos 𝑥 (
𝑦
′
∴ 𝑦 = 𝑦 {− sin 𝑥 ln(ln 𝑥) +
1
1
)( )
ln 𝑥
𝑥
1
1
cos 𝑥 ( ) ( ) }
ln 𝑥
𝑥
1
1
)( ) }
ln 𝑥
𝑥
Thus 𝑦 ′ = (ln 𝑥)cos 𝑥 {− sin 𝑥 ln(ln 𝑥) + cos 𝑥 (
f)
𝑦 = log 2 (𝑥 log 5 𝑥)
1
∴ 𝑦 ′ = 𝑥 log
5𝑥
=
1
∙ log 2 𝑒 ∙ {log 5 𝑥 + 𝑥 ∙ 𝑥 ∙ log 5 𝑒 (1)}
1
𝑥 log5 𝑥
∙ log 2 𝑒 ∙ {log 5 𝑥 + log 5 𝑒}
g) 𝑥 𝑦 = 𝑦 𝑥
∴ 𝑦 ln 𝑥 = 𝑥 ln 𝑦
𝑑𝑦
1
1 𝑑𝑦
⟹ 𝑑𝑥 ln 𝑥 + 𝑦 ∙ 𝑥 = ln 𝑦 + 𝑥 ∙ 𝑦 ∙ 𝑑𝑥
⟹
⟹
⟹
𝑑𝑦
𝑥 𝑑𝑦
𝑦
ln 𝑥 − 𝑦 𝑑𝑥 = ln 𝑦 − 𝑥
𝑑𝑥
𝑑𝑦
𝑥
𝑦
(ln 𝑥 − 𝑦) = ln 𝑦 − 𝑥
𝑑𝑥
𝑦
ln 𝑦−
𝑑𝑦
𝑑𝑥
=
ln 𝑥−
𝑥
𝑥
𝑦
312 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
2.5. 𝑓(𝑥) = 𝑥 4 − 4𝑥 2 + 2
∴ 𝑓 ′ (𝑥) = 4𝑥 3 − 8𝑥
∴ 𝑓 ′ (𝑐) = 4𝑐 3 − 8𝑐
⟹ 4𝑐 3 − 8𝑐 = 0
⟹ 4𝑐(𝑐 2 − 2) = 0
⟹ 4𝑐 = 0 Or 𝑐 2 − 2 = 0
⟹ 𝑐 = 0 Or 𝑐 = ±√2
−√2
0
√2
Thus (−∞, −√2) , (−√2, 0) , (0, √2), (√2, ∞)
At (−∞, −√2) 𝑓′(𝑥) < 0 hence 𝑓 is Decreasing on (−∞, −√2)
At (−√2, 0)
𝑓′(𝑥) > 0 hence 𝑓 is Increasing on (−√2, 0)
At (0, √2)
𝑓′(𝑥) < 0 hence 𝑓 is Decreasing on (−√2, 0)
At (√2, ∞)
𝑓′(𝑥) > 0 hence 𝑓 is Increasing on (√2, ∞)
2.6. Find
𝑑2 𝑦
𝑑𝑥 2
: 𝑒 𝑥 − 𝑒𝑦 = 𝑥2 + 𝑦2
𝑑𝑦
𝑑𝑦
∴ 𝑒 𝑥 − 𝑒 𝑦 𝑑𝑥 = 2𝑥 + 2𝑦 𝑑𝑥
𝑑𝑦
∴
∴
(−𝑒 𝑦 − 2𝑦) = 2𝑥 − 𝑒 𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑2 𝑦
𝑑𝑥 2
2𝑥−𝑒 𝑥
= −𝑒 𝑦 −2𝑦
=
(2−𝑒 𝑥 )(−𝑒 𝑦 −2𝑦)−(2𝑥−𝑒 𝑥 )(−𝑒 𝑦
𝑑𝑦
𝑑𝑦
−2 )
𝑑𝑥
𝑑𝑥
(−𝑒 𝑦 −2𝑦)2
2𝑥−𝑒𝑥
2𝑥−𝑒𝑥
)−2( 𝑦
))
−𝑒𝑦 −2𝑦
−𝑒 −2𝑦
(2−𝑒 𝑥 )(−𝑒 𝑦 −2𝑦)−(2𝑥−𝑒 𝑥 )(−𝑒 𝑦 (
=
(−𝑒 𝑦 −2𝑦)2
313 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
MEMORANDUM 2017
SEFAKO MAKGATHO HEALTH SCIENCES UNIVERSITY
ASSESSMENT AND CERTIFICATION MANAGEMET
FACULTY OF HEALTH SCIENCES
SCHOOL OF PATHOLOGY AND PRE-CLINICAL SCIENCES
SUBJECT NAME: Differential and Integral Calculus
TYPE OF EXAMINATION: STANDARD
X
:RE-EXAMINATION
SUBJECT CODE : MAH101M/MMTH011
COURSE
jdk
..
: MATHEMATICS 1
PAPER NUMBER : 1
DATE OF EXAMINATION:
O
19 MAY 2017
NUMBER OF STUDENTS:
70
H
k
DURATION
: 3HRS
TOTAL MARKS : 150
INTERNAL EXAMINERS PARTICULARS: 1. MR E. CHAUKE
3.
2.
EXTERNAL EXAMINERS PARTICULARS: 1. MR THOKA S.J
2.
…
H
o
4.
p
3.
i
4.
o
THIS QUESTIO PAPER CONSIST OF 4 PAGES INCLUDING COVER PAGE:
Specified Instructions to
Students.
1
2
3
4
Stationary.
4 PAGE BOOK
8 PAGE BOOK
MCQ’ s ( must be supplied by
the department on day of
submission )
Double Folios
Please confirm what
you need.
Please indicate the quality
per Student.
x
1
314 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 1:
1.1.
4𝜋
3
= 240° & 225° =
5𝜋
4
(2)
1.2.
a) lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(2)
𝑥→2
𝑥→2
lim− 5 = lim+(𝑎𝑥 + 𝑏) = 5
𝑥→2
𝑥→2
5 = 2𝑎 + 𝑏 = 5
Thus 2𝑎 + 𝑏 = 5……………………………….(1)
lim 𝑓(𝑥) = lim+ 𝑓(𝑥) = 𝑓(10)
𝑥→10−
𝑥→10
lim−(𝑎𝑥 + 𝑏) = lim+(21) = 21
𝑥→10
𝑥→10
10𝑎 + 𝑏 = 21 = 21
10𝑎 + 𝑏 = 21……………………………………(2)
From (1) 𝑏 = 5 − 2𝑎……………………….(3)
Now sub 𝑏 into (2)
10𝑎 + 5 − 2𝑎 = 21
𝑎 = 2 and sub 𝑎 into (3)
Thus 𝑏 = 5 − 2(2) = 1
Hence 𝑎 = 2 𝑎𝑛𝑑 𝑏 = 1
5
(7)
(2)
5
b) 𝑓 (2) = 2 (2) + 1 = 6
(2)
(2𝑥 + 1) = 2(10) + 1 = 21
c) lim
+
10
d)
𝑦 = 21
21
𝑦 = 2𝑥 + 1
𝑦=5
(5)
5
2
10
e) 𝐷𝑓 = (−∞, 2] ∪ (2 , 10)[10 , ∞)
𝑅𝑓 = {5} ∪ (5, 21) ∪ {21}
(2)
315 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
lim 𝑓(𝑥) = 𝐿 Means that for every 𝜀 > 0 there exist a 𝛿 > 0 such that if 0 < |𝑥 − 𝑎| < 𝛿
1.3.
𝑥→𝑎
(3)
then |𝑓(𝑥) − 𝐿| < 𝜀
1
lim (3 − 2 𝑥) = −5
1.4.
𝑥→16
for every 𝜀 > 0 there exist a 𝛿 > 0 such that if 0 < |𝑥 − 16| < 𝛿 then
Compare |𝑥 − 16|
1
1
1
|3 − 2 𝑥 + 5| < 𝜀
(5)
and |3 − 2 𝑥 + 5|
1
Simply |3 − 𝑥 + 5| = |8 − 𝑥|
2
1
2
= |− 2 (𝑥 − 16)|
1
= |− 2| |𝑥 − 16|
1
∴ 2 |𝑥 − 16| < 𝜀
∴ |𝑥 − 16| < 2𝜀
Thus 𝛿 = 2𝜀
1.5.
𝑓(𝑥) = 2 + cos 𝑥 and 𝑔(𝑥) = sin2 2𝑥 − 1
(𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
i.
= 2 + cos(sin2 2𝑥 − 1)
2𝜋
(𝑓 ∘ 𝑔)(𝑥) = (2 + cos(sin2 2𝑥 − 1) ) ( )
2
2𝜋
(3)
3
= 2 + cos (sin 2 ( 3 ) − 1)
= 2.9689 𝑅𝑎𝑑
ii.
(𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥))
= sin2 2(2 + cos 𝑥) − 1
3𝜋
= sin2 2 (2 + cos (− 4 )) − 1
= −0.7216 𝑅𝑎𝑑
(3)
316 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
QUESTION 2:
2.1. Let 𝑓(𝑥) = cos 𝑥 , then 𝑓′(𝑥) = − sin 𝑥
Proof:
𝑓′(𝑥) = lim
ℎ→0
= lim
ℎ→0
= lim
ℎ→0
= lim
ℎ→0
= lim
𝑓(𝑥+ℎ)−𝑓(𝑓)
ℎ
cos(𝑥+ℎ)−cos 𝑥
(6)
ℎ
cos 𝑥 cos ℎ−sin 𝑥 sin ℎ−cos 𝑥
ℎ
cos 𝑥 cos ℎ−cos 𝑥−sin 𝑥 sin ℎ
ℎ
cos 𝑥(cos ℎ−1)−sin 𝑥 sin ℎ
ℎ
cos 𝑥(cos ℎ−1)
ℎ→0
= lim (
sin 𝑥 sin ℎ
) − lim (
ℎ
cos ℎ−1
ℎ→0
= cos 𝑥 lim (
ℎ→0
= cos 𝑥 (0) − sin 𝑥 (1)
= − sin 𝑥
)
sin ℎ
) − sin 𝑥 lim (
ℎ
ℎ→0
ℎ
ℎ→0
ℎ
)
2.2.
√1+𝑥 2
a) lim
𝑥→∞ 3+2𝑥
1
= lim
√𝑥 2 ( 2 +1)
𝑥
𝑥→∞
(4)
3
𝑥
𝑥( +2)
1
= lim
=
𝑥√( 2 +1)
𝑥
3
𝑥→∞ 𝑥(𝑥+2)
√0+1
lim 0+2
𝑥→∞
1
=2
b) lim
sin(𝑥 2 −4)
𝑥→2
𝑥−2
= lim
= lim
sin(𝑥 2 −4) 𝑥 2 −4
𝑥→2
sin(𝑥 2 −4)
𝑥 2 −4
∙ lim
∙
𝑥−2
𝑥 2 −4
OR
lim
sin(𝑥 2 −4)
𝑥→2
= (1) lim
𝑥→2
lim
𝑥→−1
𝑥→2
= lim(𝑥 + 2)
= 4(1)
=4
=4
= lim
−𝑥 2 +𝑥+2
= lim
(𝑥+1)(𝑥 2 −𝑥+1)
𝑥→−1 −(𝑥 2 −𝑥−2)
(𝑥+1)(𝑥 2 −𝑥+1)
𝑥→−1 −(𝑥+1)(𝑥−2)
𝑥 2 −𝑥+1
OR
lim
𝑥→−1
𝑥 3 +1
= lim
−𝑥 2 +𝑥+2
(4)
3𝑥 2
𝑥→−1 −2𝑥+1
3(−1)2
L’hopital’s
= −2(−1)+1
(4)
3
= lim
=3
= −(−3)
=1
𝑥→−1 −(𝑥−2)
3
L’hopital’s
= 2(2) cos(0)
𝑥−2
𝑥 3 +1
1
𝑥→2
𝑥→2
c)
𝑥−2
2𝑥 cos(𝑥 2 −4)
= lim 2𝑥 cos(𝑥 2 − 4)
𝑥 2 −4
𝑥→2 𝑥−2
(𝑥−2)(𝑥+2)
𝑥→2
= lim
=1
317 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
2
d) lim 𝑥 3 𝑒 −𝑥 = lim
𝑥→∞
= lim
𝑥→∞ 𝑒 𝑥
3𝑥 2
𝑥→∞ 2𝑥𝑒 𝑥
= lim
𝑥3
2
(4)
6𝑥
2
𝑥→∞ 2𝑒 𝑥 +4𝑥 2 𝑒 𝑥
6
= lim
Using L’Hopital’s Rule
2
2
2
2
𝑥→∞ 4𝑥𝑒 𝑥 +8𝑥𝑒 𝑥 +8𝑥 3 𝑒 𝑥
6
2
=∞
=0
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
𝑑𝑦
2.3. 𝑑𝑥 = lim
ℎ→0
= lim
ℎ→0
= lim
ℎ
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]] 𝑔(𝑥+ℎ)−𝑔(𝑥)
∙
ℎ
𝑔(𝑥+ℎ)−𝑔(𝑥)
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]] 𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ→0
= lim
𝑔(𝑥+ℎ)−𝑔(𝑥)
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
∙
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ→0
∙ lim
ℎ
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ
ℎ→0
As ℎ → 0 , then 𝑔(𝑥 + ℎ) → 0
because 𝑔 is differentiable and continuous at 𝑥
Let ∆𝑔 = 𝑔(𝑥 + ℎ) − 𝑔(𝑥) , then 𝑔(𝑥 + ℎ) = 𝑔(𝑥) + ∆𝑔 as ℎ → 0 , then ∆𝑔 → 0
𝑑𝑦
𝑑𝑥
= lim
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
ℎ→0
= lim
∆𝑔→0
= lim
∆𝑔→0
∙ lim
𝑔(𝑥+ℎ)−𝑔(𝑥)
𝑔(𝑥+ℎ)−𝑔(𝑥)
ℎ→0
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
𝑔(𝑥)+∆𝑔−𝑔(𝑥)
𝑓[𝑔(𝑥+ℎ)−𝑓[𝑔(𝑥)]]
∆𝑔
(8)
ℎ
∙ 𝑔′(𝑥)
∙ 𝑔′(𝑥)
= 𝑓′[𝑔(𝑥)] ∙ 𝑔′(𝑥)
2.4. sin2 𝑦 + cos(𝑥𝑦) = 𝜋
𝑑𝑦
𝑑𝑦
⟹ 2 sin 𝑦 cos 𝑦 𝑑𝑥 − sin(𝑥𝑦) {𝑦 + 𝑥 𝑑𝑥 } = 0
𝑑𝑦
(5)
𝑑𝑦
⟹ 2 sin 𝑦 cos 𝑦 𝑑𝑥 − 𝑦 sin(𝑥𝑦) − 𝑥 sin(𝑥𝑦) 𝑑𝑥 = 0
⟹
⟹
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
(2 sin 𝑦 cos 𝑦 − 𝑥 sin(𝑥𝑦)) = 𝑦 sin(𝑥𝑦)
𝑦 sin(𝑥𝑦)
= sin 2𝑦−𝑥 sin(𝑥𝑦)
318 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
2.5. 𝑓(𝑥) = 𝑥 3 − 5𝑥 2 − 3𝑥
𝑓′(𝑐) =
𝑓(𝑏)−𝑓(𝑎)
=
𝑏−𝑎
−27−(−7)
,
1≤𝑥≤3
𝑓(3)−𝑓(1)
3−1
=
= −10
2
∴ 𝑓 ′ (𝑐) = 3𝑐 2 − 10𝑐 − 3
∴ 3𝑐 2 − 10𝑐 − 3 = −10
∴ 3𝑐 2 − 10𝑐 + 7 = 0
∴ (3𝑐 − 7)(𝑐 − 1) = 0
7
∴ 𝑐 = 1 𝑜𝑟 𝑐 = 3
(5)
𝟕
Thus 𝒄 = 𝟑 ∈ (𝟏, 𝟑) where 𝑐 ≠ 1 because 𝑓 is differentiable on the open (1,3).
2.6. Rolle’s Theorem: States that let 𝑓 be a function that satisfies the following three hypotheses:
4. 𝑓 is continuous on the closed interval [𝑎, 𝑏]
5. 𝑓 is differentiable on the open interval (𝑎, 𝑏)
6. 𝑓(𝑎) = 𝑓(𝑏)
Then there is a number 𝑐 in (𝑎, 𝑏) such that 𝑓′(𝑐) = 0
(3)
2.7. Mean Value Theorem: States that let 𝑓 be a function that satisfies the following three hypotheses:
1. 𝑓 is continuous on the closed interval [𝑎, 𝑏]
2. 𝑓 is differentiable on the open interval (𝑎, 𝑏)
Then there is a number 𝑐 in (𝑎, 𝑏) such that, 𝑓′(𝑐) =
Or Equivalently 𝑓(𝑏) − 𝑓(𝑎) = 𝑓′(𝑐)(𝑏 − 𝑎)
𝑓′(𝑐) =
(8)
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
𝑦 = 𝑓(𝑥)
ℎ(𝑥)
𝐴(𝑎, 𝑓(𝑎))
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
𝐿 = 𝑓(𝑎) +
𝑓(𝑏)−𝑓(𝑎)
(𝑥
𝑏−𝑎
− 𝑎)
𝑓(𝑥)
𝐵(𝑏, 𝑓(𝑏))
𝐴
𝐵
319 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
𝑷𝒓𝒐𝒐𝒇:
We apply Rolle’s Theorem to a new function ℎ defined as the difference between 𝑓 and the function whose
graph is the secant line 𝐴𝐵. The equation of the secant line can be written as,
𝑦 − 𝑓(𝑎) =
𝑓(𝑏)−𝑓(𝑎)
(𝑥 − 𝑎)
𝑏−𝑎
𝑓(𝑏)−𝑓(𝑎)
(𝑥 − 𝑎)
+
𝑏−𝑎
∴ 𝑦 = 𝑓(𝑎)
First we must verify that ℎ satisfies the Three Hypotheses of Rolle’s Theorem.
The formula ℎ is continuous on the closed interval [𝑎, 𝑏]
The formula ℎ differentiable on the open interval (𝑎, 𝑏)
ℎ′(𝑥) = 𝑓′(𝑥) −
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
𝑓(𝑏) − 𝑓(𝑎)
(𝑏 − 𝑎) = 0
𝑏−𝑎
𝑓(𝑏)−𝑓(𝑎)
(𝑏 − 𝑎) = 0
∴ ℎ(𝑏) = 𝑓(𝑏) − 𝑓(𝑎) −
𝑏−𝑎
Thus ℎ(𝑎) = ℎ(𝑏)
Since ℎ satisfies the Three Hypotheses of Rolle’s Theorem, ℎ′(𝑐) = 0
ℎ(𝑎) = 𝑓(𝑎) −
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
𝑓(𝑏)−𝑓(𝑎)
𝑏−𝑎
∴ 0 = ℎ′(𝑐) = 𝑓′(𝑐) −
Hence 𝑓′(𝑐) =
QUESTION 3:
3.1.
𝑓(𝜃) = 𝜃 − 2 tan−1 𝜃
1
⟹ 𝑓 ′ (𝜃) = 1 − 2 (1+𝑥 2 )
2
⟹ 0 = 1 − 1+𝑥 2
2
(4)
⟹ −1 = − 1+𝑥 2
⟹ 𝑥2 − 1 = 0
𝑇ℎ𝑢𝑠 𝑥 = ±1
∴ 𝑓(1) = 1 − 2 tan−1(1) = −89
∴ 𝑓(1) = −1 − 2 tan−1(−1) = 89
Hence 𝑓 has a global maximum at 𝑥 = −1
And 𝑓 has a global minimum at 𝑥 = 1
3.2.
a) 𝑦 = 𝑎𝑟𝑐 cos(𝑥 + 𝑥 3 )
1
(1 + 3𝑥 2 )
𝑦′ = −
3 2
=−
√1−(𝑥+𝑥 )
1+3𝑥 2
(4)
√1−(𝑥+𝑥 3 )2
320 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
b) 𝑦 𝑥 + 𝑥 𝑦 = 1
∴ 𝑥 ln 𝑦 + 𝑦 ln 𝑥 = ln 1
1
1
∴ ln 𝑦 + 𝑥 ∙ 𝑦 𝑦 ′ + 𝑦 ′ ln 𝑥 + 𝑦 ∙ 𝑥 = 0
𝑥
(5)
𝑦
∴ 𝑦 ′ (𝑦 + ln 𝑥) = − 𝑥 − ln 𝑦
𝑦
𝑥
𝑥
+ln 𝑥
𝑦
− −ln 𝑦
∴ 𝑦′ =
𝑦 2 = 𝑒 √𝑥
3.3.
𝑑𝑦
1
⟹ 2𝑦 𝑑𝑥 = (𝑒 √𝑥 ) (2 𝑥)
√
𝑑𝑦
⟹
(5)
1
= [(𝑒 √𝑥 ) (2 𝑥)] ÷ 2𝑦
𝑑𝑥
√
𝑒 √𝑥
=2
1
∙
√𝑥 2𝑦
𝑒 √𝑥
= 4𝑦
=
=
√𝑥
𝑒 √𝑥
, but 𝑦 = √𝑒 √𝑥
4√𝑒 √𝑥 √𝑥
𝑒 √𝑥
𝑁𝑜𝑡𝑒:
4√𝑥𝑒 √𝑥
√𝑎𝑏 = √𝑎 ∙ √𝑏
𝑦
𝑥 = tan (3)
3.4.
𝑦
1 𝑑𝑦
1 = sec 2 ( 3) (3) 𝑑𝑥
𝑑𝑦
𝑇ℎ𝑢𝑠
⟹
=
𝑑𝑥
𝑑2 𝑦
𝑑𝑥 2
=
3
𝑦
3
sec2 ( )
𝑦
3
𝑦
3
𝑦
3
0−3[2 sec( ) sec( ) tan( )](
𝑦
3
2
(6)
1𝑑𝑦
)
3𝑑𝑥
(sec2 ( ))
𝑦
3
𝑦
3
𝑦
3
1
3
))
3 sec2 (𝑦)
3
−3[2 sec( ) sec( ) tan( )]( (
=
𝑦
3
2
(sec2 ( ))
321 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
3.5.
3 sec √𝑥 tan √𝑥
a)
7√𝑥
𝐿𝑒𝑡 𝑢 = √𝑥
3
⟹ 𝑑𝑢 = 2
1
√
3
sec √𝑥 tan √𝑥
𝑑𝑥 = 7
√𝑥
𝑑𝑥
𝑥
3
𝑑𝑥 = 7 sec √𝑥 tan √𝑥 ∙
⟹ 2𝑑𝑢 =
1
√𝑥
1
√𝑥
𝑑𝑥
𝑑𝑥
⟹ 7 sec 𝑢 tan 𝑢 ∙ (2𝑑𝑢)
(4)
6
⟹ 7 sec 𝑢 tan 𝑢 𝑑𝑢
⟹
⟹
6
7
6
7
sec 𝑢 + 𝐶
sec √𝑥 + 𝐶
10
𝑑𝑥
(𝑥−1)(𝑥 2 +9)
b)
10
(𝑥−1)(𝑥 2 +9)
𝐴
= 𝑥−1 +
𝐵𝑥+𝐶
𝑥 2 +9
(5)
10 = 𝐴(𝑥 2 + 9) + (𝐵𝑥 + 𝐶)(𝑥 − 1)
Let 𝑥 = 1 then 10 = 𝐴(1 + 9) ⟹ 𝐴 = 1
Let 𝑥 = 0 then 10 = 9𝐴 − 𝐶 but 𝐴 = 1 ⟹ 𝐶 = −1
10 = 𝐴𝑥 2 + 9𝐴 + 𝐵𝑥 2 − 𝐵𝑥 + 𝐶𝑥 − 𝐶
10 = 𝑥 2 (𝐴 + 𝐵) + 𝑥(𝐶 − 𝐵) + (9𝐴 − 𝐶)
𝐴 + 𝐵 = 0 , 𝐶 − 𝐵 = 0 , 9𝐴 − 𝐶 = 10
𝐻𝑒𝑛𝑐𝑒 𝐵 = −1
1
−𝑥−1
⟹ 𝑥−1 𝑑𝑥 + 𝑥 2 +9 𝑑𝑥
⟹
1
𝑥−1
−𝑥
𝑑𝑥 −
−1
𝑑𝑥 +
𝑥 2 +9
1
2
𝑥 2 +9
1
𝑑𝑥
𝑥
⟹ ln|𝑥 − 1| − 2 ln|𝑥 + 9| − 3 tan−1 (3) + 𝐶
c)
Let 𝑢 = 𝑒 2𝑥
𝑒 2𝑥 sin 2𝑥 𝑑𝑥
𝐴𝑛𝑑 𝑑𝑣 = sin 2𝑥 𝑑𝑥
2𝑥
⟹ 𝑑𝑢 = 2𝑒 𝑑𝑥 𝐴𝑛𝑑 𝑣 =
1
sin 2𝑥 𝑑𝑥 = − 2 cos 2𝑥
(10)
𝑒 2𝑥 sin 2𝑥 𝑑𝑥 = 𝑢𝑣 − 𝑣 𝑑𝑢
=−
𝑒 2𝑥 cos 2𝑥
=−
2
𝑒 2𝑥 cos 2𝑥
2
1
− (− 2 cos 2𝑥) (2𝑒 2𝑥 )𝑑𝑥
+ 𝑒 2𝑥 cos 2𝑥 𝑑𝑥
Let 𝑢 = 𝑒 2𝑥 𝐴𝑛𝑑 𝑑𝑣 = cos 2𝑥 𝑑𝑥
𝑑𝑢 = 2𝑒 2𝑥 𝑑𝑥 𝐴𝑛𝑑 𝑣 =
1
cos 2𝑥 𝑑𝑥 = 2 sin 2𝑥
322 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
𝑒 2𝑥 cos 2𝑥 𝑑𝑥 =
=
𝑒 2𝑥 sin 2𝑥
2
𝑒 2𝑥 sin 2𝑥
2
1
− (2 sin 2𝑥) (2𝑒 2𝑥 )𝑑𝑥
− 𝑒 2𝑥 sin 2𝑥 𝑑𝑥
𝑒 2𝑥 sin 2𝑥 𝑑𝑥 = −
𝐵𝑢𝑡,
𝑒 2𝑥 cos 2𝑥
2
𝑒 2𝑥 cos 2𝑥
=−
⟹ 2 𝑒 2𝑥 sin 2𝑥 𝑑𝑥 = −
1
𝑒 2𝑥 sin 2𝑥 𝑑𝑥 = 2 (−
⟹
𝑒 2𝑥 sin 2𝑥 =
∴
𝑒 2𝑥 sin 2𝑥
4
+
2
𝑒 2𝑥 cos 2𝑥
2
𝑒 2𝑥 cos 2𝑥
2
−
+ 𝑒 2𝑥 cos 2𝑥 𝑑𝑥
+
+
𝑒 2𝑥 cos 2𝑥
4
𝑒 2𝑥 sin 2𝑥
2
− 𝑒 2𝑥 sin 2𝑥 𝑑𝑥 + 𝐶
𝑒 2𝑥 sin 2𝑥
2
𝑒 2𝑥 sin 2𝑥
2
)+𝐶
+𝐶
d) 𝑥 2 𝑠𝑖𝑛 𝑥 3 cos 5 𝑥 3 𝑑𝑥 = 𝑠𝑖𝑛 𝑥 3 cos5 𝑥 3 𝑥 2 𝑑𝑥
1
𝐿𝑒𝑡 𝑢 = 𝑥 3 , 𝑑𝑢 = 3𝑥 2 𝑑𝑥 𝑡ℎ𝑢𝑠 3 𝑑𝑢 = 𝑥 2 𝑑𝑥
1
1
⟹ 3 sin 𝑢 cos5 𝑢𝑑𝑢 = 3 sin 𝑢 cos4 𝑢 cos 𝑢 𝑑𝑢
1
⟹ 3 sin 𝑢 (1 − sin2 𝑢)2 cos 𝑢 𝑑𝑢
𝑁𝑜𝑤 𝑙𝑒𝑡 𝑣 = sin 𝑢 𝑑𝑣 = cos 𝑢 𝑑𝑢
⟹ 𝑣 (1 − 𝑣)2 𝑑𝑣 = 𝑣(1 − 2𝑣 + 𝑣 2 ) 𝑑𝑣
⟹ (𝑣 − 2𝑣 2 + 𝑣 3 ) 𝑑𝑣
1
2
1
⟹ 2 𝑣2 − 3 𝑣3 + 4 𝑣4 + 𝐶
1
2
(6)
1
sin2 𝑢 − 3 sin3 𝑢 + 4 sin4 𝑢 + 𝐶
𝐵𝑢𝑡 𝑢 = 𝑥 3
1
2
1
⟹ 2 sin2 𝑥 3 − 3 sin3 𝑥 3 + 4 sin4 𝑥 3 + 𝐶
⟹
2
e)
3𝑥 2 ln(𝑥 3 +5)
(𝑥 3 +5)
𝐿𝑒𝑡 𝑢 = ln(𝑥 3 + 5)
1
2
,
1
𝑢𝑑𝑢 = 2 𝑢 + 𝐶 = 2
𝑑𝑥 =
3𝑥 2
ln(𝑥 3 + 5) 𝑥 3 +5 𝑑𝑥
3𝑥 2
1
𝑑𝑢 = 𝑥 3 +5 3𝑥 2 𝑑𝑥 = 𝑥 3 +5 𝑑𝑥
(ln(𝑥 3
2
+ 5)) + 𝐶
OR
3𝑥 2 ln(𝑥 3 +5)
(𝑥 3 +5)
𝑑𝑥 =
(ln(𝑥 3 +5))
2
2
+𝐶
(4)
323 | P a g e
Author: Emmanuel Chauke
DIFFERENTIAL AND INTEGRAL CALCULUS
if 𝑥 = 𝑎 tan 𝜃 then
3.6.
√𝑎2 + 𝑥 2 𝑑𝜃 = 𝑎 ln|sec 𝜃 + tan 𝜃| + 𝐶
(6)
𝑎2 + 𝑥 2 = 𝑎2 + 𝑎2 tan2 𝜃
= 𝑎2 (1 + tan2 𝜃)
= 𝑎2 sec 2 𝜃
√𝑎2 + 𝑥 2 = 𝑎 sec 𝜃
𝑇ℎ𝑢𝑠
Hence
√𝑎2 + 𝑥 2 𝑑𝜃 = 𝑎 sec 𝜃 𝑑𝜃
= 𝑎 ln|sec 𝜃 + tan 𝜃| + 𝐶
𝑦 2 = 4𝑥
3.7.
and
4𝑥 − 2𝑦 − 4 = 0
𝑦 = √4𝑥 𝑎𝑛𝑑 𝑦 = 2𝑥 − 2
2𝑥 − 2 = √4𝑥
⟹ 4𝑥 2 − 12𝑥 + 4 = 0
⟹ 𝑥 2 − 3𝑥 + 1 = 0
⟹ 𝑥=
⟹ 𝑥=
∴ 𝑥=
Thus
𝟑+√𝟓
𝟐
𝟑−√𝟓
𝟐
(5)
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
−(−3)±√(−3)2 −4(1)(1)
2
3±√5
2
𝟐
(𝒙 − 𝟑𝒙 + 𝟏) 𝒅𝒙 =
𝟑+√𝟓
𝟐
𝟑−√𝟓
𝟐
𝒙𝟑
(𝟑 −
𝟑𝒙𝟐
𝟐
𝟑+√𝟓
𝟐
+ 𝒙) ⃒𝟑−√𝟓 = −
𝟕√𝟓
𝟐
𝟐
(𝟏𝟓𝟎)
BEST OF LUCK
324 | P a g e