PURE MATHEMATICS II
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Table of Contents
DIFFERENTIATION ................................................................................................................................................................. 4
IMPLICIT DIFFERENTIATION ....................................................................................................................................... 5
PARAMETRIC DIFFERENTIATION .............................................................................................................................. 7
TRIGONOMETRIC DIFFERENTIATION ...................................................................................................................... 9
DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS................................................................ 10
DIFFERENTIATION OF EXPONENTIAL FUNCTIONS ......................................................................................... 11
DIFFERENTIATION OF NATURAL LOGARITHMS ............................................................................................... 13
PARTIAL DERIVATIVES ................................................................................................................................................. 15
First Partial Derivative .............................................................................................................................................. 15
Second PARTIAL DERIVATIVE ............................................................................................................................... 16
INTEGRATION RESULTS................................................................................................................................................ 17
Even powers of 𝐬𝐢𝐧𝒙 and 𝐜𝐨𝐬𝒙 ............................................................................................................................... 19
Odd powers on 𝐬𝐢𝐧𝒙 and 𝐜𝐨𝐬𝒙 ............................................................................................................................... 20
Even powers of 𝐭𝐚𝐧𝒙 .................................................................................................................................................. 21
INTEGRATION BY PARTS .............................................................................................................................................. 22
Reduction Formulae ................................................................................................................................................... 25
PARTIAL FRACTIONS ...................................................................................................................................................... 28
Denominator with Linear Factors ......................................................................................................................... 28
Denominator with unfactorizable quadratic factor. ...................................................................................... 29
Denominator with a repeated factor ................................................................................................................... 30
Improper Fractions (degree of numerator ≥ degree of denominator)................................................ 31
TRAPEZIUM RULE (NUMERICAL INTEGRATION).............................................................................................. 33
COMPLEX NUMBERS ....................................................................................................................................................... 34
SQUARE ROOT OF NEGATIVE NUMBERS .......................................................................................................... 34
Operations on Complex Numbers ......................................................................................................................... 34
Adding and Subtracting Complex Numbers........................................................................................................................34
Multiplying Complex Numbers .................................................................................................................................................34
Dividing Complex Numbers .......................................................................................................................................................34
Square Roots of Complex Numbers ...................................................................................................................... 35
Quadratic Equations ................................................................................................................................................... 35
Equations with Real Coefficients .............................................................................................................................................35
Equations with Complex Coefficients ....................................................................................................................................35
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Roots of Equations....................................................................................................................................................... 36
Argand Diagram............................................................................................................................................................ 37
Representing Sums and Differences on Argand Diagrams ..........................................................................................37
Modulus – Argument Form ...................................................................................................................................... 37
The Modulus of a Complex Number .......................................................................................................................................37
The Argument of Complex Number ........................................................................................................................................38
Modulus – Argument Form ........................................................................................................................................................39
De Moivre’s Theorem ................................................................................................................................................. 41
Multiples of Sine and Cosine ......................................................................................................................................................43
The Exponential Form of a Complex Number .................................................................................................. 43
Locus on the Argand diagram ................................................................................................................................. 44
SEQUENCES.............................................................................................................................................................................. 49
SEQUENCES ......................................................................................................................................................................... 50
Types of Sequences ..................................................................................................................................................... 51
Convergent Sequences .................................................................................................................................................................51
Divergent Sequences .....................................................................................................................................................................51
Convergence of a Sequence ..................................................................................................................................... 52
Recurrence Relations ................................................................................................................................................. 52
SERIES ........................................................................................................................................................................................ 54
SERIES ................................................................................................................................................................................... 55
Using Sigma Notation ................................................................................................................................................. 55
Sum of a Series .............................................................................................................................................................. 55
Mathematical Induction ............................................................................................................................................ 56
Method of Differences ................................................................................................................................................ 57
ARITHMETIC PROGRESSIONS ................................................................................................................................ 58
GEOMETRIC PROGRESSIONS .................................................................................................................................. 60
MACLAURIN’S SERIES .................................................................................................................................................... 63
TAYLOR SERIES ............................................................................................................................................................ 66
BINOMIAL THEOREM .......................................................................................................................................................... 68
PASCAL’S TRIANGLE ....................................................................................................................................................... 69
FACTORIALS ....................................................................................................................................................................... 69
THE BINOMIAL THEOREM ........................................................................................................................................... 71
Extension of the Binomial Expansion .................................................................................................................. 72
ROOTS OF EQUATIONS ....................................................................................................................................................... 76
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THE INTERMEDIATE VALUE THEOREM ................................................................................................................ 77
DETERMINING THE ROOTS OF AN EQUATION ................................................................................................... 78
BISECTION METHOD .................................................................................................................................................. 78
LINEAR INTERPOLATION ........................................................................................................................................ 78
NEWTON RAPHSON.................................................................................................................................................... 79
DERIVING AN ITERATIVE FORMULA .................................................................................................................. 80
MATRICES................................................................................................................................................................................. 81
MATRICES ............................................................................................................................................................................ 82
Matrix Multiplication .................................................................................................................................................. 82
THE DETERMINANT OF A 𝟑 × 𝟑 MATRIX ......................................................................................................... 83
The Transpose of a Matrix........................................................................................................................................ 84
Finding the inverse of A Matrix (Cofactor Method) ...................................................................................... 85
SYSTEMS OF EQUATIONS......................................................................................................................................... 86
ROW REDUCTION ........................................................................................................................................................ 87
Row Reduction and Systems of Equations ..........................................................................................................................87
DIFFERENTIAL EQUATIONS ............................................................................................................................................. 90
DIFFERENTIAL EQUATIONS ........................................................................................................................................ 91
Separable Differential Equations........................................................................................................................... 91
The Integrating Factor ............................................................................................................................................... 92
Linear Differential Equations with Constant Coefficients........................................................................... 93
Homogenous Differential Equations ......................................................................................................................................93
Non – Homogeneous Differential Equations ......................................................................................................................94
Differential Equations Requiring a Substitution ............................................................................................. 98
Mathematical Modelling......................................................................................................................................... 101
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DIFFERENTIATION
At the end of this section students should be able to:
1. find the derivative of 𝑒 𝑓(𝑥) , where 𝑓(𝑥) is a differentiable function of 𝑥;
2. find the derivative of ln 𝑓(𝑥) (to include functions of 𝑥 – polynomials or trigonometric);
3. apply the chain rule to obtain gradients and equations of tangents and normals to curves
given by their parametric equations;
4. use the concept of implicit differentiation, with the assumption that one of the variables is a
function of the other;
5. differentiate any combinations of polynomials, trigonometric, exponential and logarithmic
functions;
6. differentiate inverse trigonometric functions
7. obtain second derivatives, 𝑓 ′′ (𝑥), of the functions in 3, 4, 5 above;
8. find the first and second partial derivatives of 𝑢 = 𝑓(𝑥, 𝑦).
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IMPLICIT DIFFERENTIATION
A function which is written in the form 𝑦 = 𝑓(𝑥) is
called an explicit function: 𝑦 is stated explicitly in
terms of 𝑥. However, functions such as
1
𝑥 2 + 𝑦 2 = 0 or 𝑥𝑦 + = 5𝑥 2 are implicit
𝑥
functions.
Differentiate 𝑦 2 with respect to 𝑥.
LESSON 1
𝑑𝑦
10𝑥
=
−1
𝑑𝑥 4(𝑥 + 𝑦)3
LESSON 4
=
𝑑
𝑑𝑦
−3 =0
𝑑𝑥
𝑑𝑦
= 3 − 2𝑥𝑦
𝑑𝑥
𝑑𝑦 3 − 2𝑥𝑦
=
𝑑𝑥
𝑥2
𝑑𝑦
𝑑𝑥
LESSON 5
Find the equations of the
tangents at the points where 𝑥 = 6 on the curve
LESSON 2
Use implicit differentiation to
𝑑𝑦
𝑑𝑥
𝑥 2 + 𝑦 2 − 6𝑥 − 2𝑦 = 3.
SOLUTION
for 𝑥 2 + 𝑦 2 = 1.
𝑥 2 + 𝑦 2 − 6𝑥 − 2𝑦 = 3
SOLUTION
𝑑𝑦
𝑑𝑦
−6−2
=0
𝑑𝑥
𝑑𝑥
𝑑𝑦
𝑑𝑦
2𝑦
−2
= 6 − 2𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑦
(2𝑦 − 2)
= 6 − 2𝑥
𝑑𝑥
𝑑𝑦 6 − 2𝑥
=
𝑑𝑥 2𝑦 − 2
2𝑥 + 2𝑦
𝑥 2 + 𝑦2 = 1
𝑑 2 𝑑 2
𝑑
(1)
𝑥 +
𝑦 =
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑦
2𝑥 + 2𝑦
=0
𝑑𝑥
𝑑𝑦
2𝑦
= −2𝑥
𝑑𝑥
𝑑𝑦
2𝑥
=−
𝑑𝑥
2𝑦
When 𝑥 = 6
62 + 𝑦 2 − 6(6) − 2𝑦 = 3
𝑑𝑦
𝑥
=−
𝑑𝑥
𝑦
LESSON 3
for 𝑥 2 𝑦 − 3𝑥 = 5.
𝑥2
𝑑𝑥
𝑑 2 𝑑𝑦
=
𝑦
𝑑𝑦 𝑑𝑥
determine
𝑑𝑥
𝑥 2 𝑦 − 3𝑥 = 5
2𝑥𝑦 + 𝑥 2
𝑦2
= 2𝑦
𝑑𝑦
SOLUTION
SOLUTION
𝑑 2
𝑦
𝑑𝑥
Determine
𝑦 2 − 2𝑦 − 3 = 0
Determine
𝑑𝑦
𝑑𝑥
for
(𝑦 − 3)(𝑦 + 1) = 0
𝑦 = −1, 3
(𝑥 + 𝑦)4 − 5𝑥 2 = 0.
SOLUTION
(𝑥 + 𝑦)4 − 5𝑥 2 = 0
𝑑𝑦
4(𝑥 + 𝑦)3 (1 + ) − 10𝑥 = 0
𝑑𝑥
𝑑𝑦
10𝑥
1+
=
𝑑𝑥 4(𝑥 + 𝑦)3
Gradient at (6, −1)
𝑑𝑦
6 − 2(6)
3
=
=
𝑑𝑥 2(−1) − 2 2
Equation of line: 𝑦 = 𝑚𝑥 + 𝑐
Using (6, −1)
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3
−1 = (6) + 𝑐
2
We have already stated that
2𝑥 − 2𝑦 = 0
−10 = 𝑐
3
𝑦 = 𝑥 − 10
2
𝑑 2 𝑦 2(2𝑥 + 1)
2
=
=
2
2
(2𝑥
𝑑𝑥
+ 1)
2𝑥 + 1
Gradient at (6, 3)
For (−3, −3)
𝑑𝑦 6 − 2(6)
3
=
=−
𝑑𝑥 2(3) − 2
2
𝑑2 𝑦
2
=
2
𝑑𝑥
2(−3) + 1
Equation of line
𝑑2 𝑦
2
=−
2
𝑑𝑥
5
Using (6, 3)
→ Maximum
For (2, 2)
3
3 = − (6) + 𝑐
2
12 = 𝑐
𝑑2 𝑦
2
=
2
𝑑𝑥
2(2) + 1
3
𝑦 = − 𝑥 + 12
2
𝑑2 𝑦 2
=
𝑑𝑥 2 5
LESSON 6
Find and classify the stationary
points on the curve 2𝑥𝑦 + 𝑦 − 𝑥 2 = 6.
SOLUTION
2𝑥𝑦 + 𝑦 − 𝑥 2 = 6
𝑑𝑦 𝑑𝑦
+
− 2𝑥 = 0
𝑑𝑥 𝑑𝑥
𝑑𝑦
(2𝑥 + 1)
= 2𝑥 − 2𝑦
𝑑𝑥
𝑑𝑦 2𝑥 − 2𝑦
=
𝑑𝑥
2𝑥 + 1
2𝑦 + 2𝑥
Stationary points occur when
𝑑𝑦
𝑑𝑥
= 0.
2𝑥 − 2𝑦
=0
2𝑥 + 1
2𝑥 − 2𝑦 = 0
𝑥=𝑦
Sub. 𝑥 = 𝑦 into 2𝑥𝑦 + 𝑦 − 𝑥 2 = 6
2𝑥(𝑥) + 𝑥 − 𝑥 2 − 6 = 0
𝑥2 + 𝑥 − 6 = 0
(𝑥 + 3)(𝑥 − 2) = 0
𝑥 = −3, 2
𝑦 = −3, 2
(−3, −3) and (2, 2)
𝑑𝑦
𝑑 2 𝑦 (2 − 2 𝑑𝑥 ) (2𝑥 + 1) − (2𝑥 − 2𝑦)(2)
=
(2𝑥 + 1)2
𝑑𝑥 2
→ Minimum
𝑑𝑦
𝑑𝑥
= 0 and
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PARAMETRIC DIFFERENTIATION
Given that 𝑥 = 𝑓(𝑡) and 𝑦 = 𝑔(𝑡) where 𝑡 is called
a parameter, then
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
LESSON 1
Find the gradient of the stated
curve at the point defined.
𝑥 = 𝑡 + 5; 𝑦 = 𝑡 2 − 3𝑡 where 𝑡 = 2
𝑑𝑦 𝑑𝑦 𝑑𝑥
=
÷
𝑑𝑥 𝑑𝑡 𝑑𝑡
1
(2𝑡 + 7)−2
=
12𝑡 2 − 2𝑡
=
SOLUTION
1
2(6𝑡 2
− 𝑡)√2𝑡 + 7
when 𝑡 = 1
𝑥 =𝑡+5
𝑑𝑦
1
=
2
𝑑𝑥 2(6(1) − (1))√2(1) + 7
𝑑𝑥
=1
𝑑𝑡
𝑦 = 𝑡 2 − 2𝑡
=
𝑑𝑦
= 2𝑡 − 3
𝑑𝑡
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
= (2𝑡 − 3)(1)
= 2𝑡 − 3
when 𝑡 = 2
1
30
LESSON 3
Find the equation of the normal
8
to the curve 𝑥 = 3 , 𝑦 = 2𝑡 2 − 1 at the point where
𝑡
the curve crosses the line 𝑥 = 1.
SOLUTION
8
𝑥 = 3 = 8𝑡 −3
𝑡
𝑑𝑥
24
= −24𝑡 −4 = − 4
𝑑𝑡
𝑡
𝑑𝑦
= 2(2) − 3 = 1
𝑑𝑥
𝑦 = 2𝑡 2 − 1
𝑑𝑦
= 4𝑡
𝑑𝑡
LESSON 2
Find the gradient of the stated
curve at the point defined.
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
𝑥 = 𝑡 2 (4𝑡 − 1); 𝑦 = √2𝑡 + 7 when 𝑡 = 1
= 4𝑡 × (−
SOLUTION
=−
𝑥 = 𝑡 2 (4𝑡 − 1)
𝑥 = 4𝑡 3 − 𝑡 2
𝑑𝑥
= 12𝑡 2 − 2𝑡
𝑑𝑡
𝑦 = √2𝑡 + 7
1
𝑦 = (2𝑡 + 7)2
1
𝑑𝑦 1
= (2𝑡 + 7)−2 (2)
𝑑𝑡 2
1
= (2𝑡 + 7)−2
𝑡4
)
24
𝑡5
6
when 𝑥 = 1
8
𝑥= 3
𝑡
8
1= 3
𝑡
𝑡3 = 8
𝑡=2
when 𝑡 = 2
𝑦 = 2𝑡 2 − 1
𝑦 = 2(2)2 − 1
𝑦=7
(1, 7)
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At (1, 7):
𝑑𝑦
𝑡5
=−
𝑑𝑥
6
25
=−
6
16
=−
3
3
Gradient of normal is
16
𝑦 = 𝑚𝑥 + 𝑐
3
(1) + 𝑐
7=
16
109
=𝑐
16
3
109
𝑦=
𝑥+
16
16
𝑑𝑦
𝑑2𝑦
LESSON 4
Find and 2 for the
𝑑𝑥
𝑑𝑥
parametric equations
𝑥 = 4𝑡 − 1 and 𝑦 = 𝑡 3 + 5
SOLUTION
𝑥 = 4𝑡 − 1
𝑑𝑥
=4
𝑑𝑡
𝑦 = 𝑡3 + 5
𝑑𝑦
= 3𝑡 2
𝑑𝑡
𝑥 = 3(3)2 − 1 = 26
𝑦 = 32 − 6(3) − 3 = −12
(26, −12)
𝑑2 𝑦
1
1
=
=
2
3
𝑑𝑥
6(3)
162
Minimum point
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
2
3𝑡
=
4
𝑑 2 𝑦 3𝑡 𝑑𝑡
=
𝑑𝑥 2
2 𝑑𝑥
3𝑡 1
= ×
2 4
3𝑡
=
8
𝑑𝑦
𝑑2𝑦
LESSON 5
Find and 2 for the
𝑑𝑥
𝑑𝑥
parametric equations 𝑥 = 3𝑡 2 − 1 and
𝑦 = 𝑡 2 − 6𝑡 − 3.
Hence find and classify the stationary point(s).
SOLUTION
𝑥 = 3𝑡 2 − 1
𝑑𝑥
= 6𝑡
𝑑𝑡
𝑦 = 𝑡 2 − 6𝑡 − 3
𝑑𝑦
= 2𝑡 − 6
𝑑𝑡
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
2𝑡 − 6
=
6𝑡
𝑡−3
=
3𝑡
𝑑 2 𝑦 1(3𝑡) − (𝑡 − 3)(3) 𝑑𝑡
=
(3𝑡)2
𝑑𝑥 2
𝑑𝑥
3𝑡 − 3𝑡 + 9 1
=
×
9𝑡 2
6𝑡
1
= 3
6𝑡
Stationary points occur when
𝑑𝑦
=0
𝑑𝑥
𝑡−3
=0
3𝑡
𝑡−3=0
𝑡=3
Page |9
TRIGONOMETRIC DIFFERENTIATION
Function
1st Derivative
(e) 𝑦 =
𝑥2
sec 𝑥 3
sec 𝑥 3 (2𝑥) − (3𝑥 2 sec 𝑥 3 tan 𝑥 3 )(𝑥 2 )
(sec 𝑥 3 )2
3 (2𝑥
sec 𝑥
− 3𝑥 4 tan 𝑥 3 )
=
(sec 𝑥 3 )2
2𝑥 − 3𝑥 4 tan 𝑥 3
=
sec 𝑥 3
𝑦′ =
𝐬𝐢𝐧 𝒙
cos 𝑥
𝐬𝐢𝐧(𝒖(𝒙))
𝑢′ cos(𝑢(𝑥))
𝐜𝐨𝐬 𝒙
− sin 𝑥
′
𝐜𝐨𝐬(𝒖(𝒙))
−𝑢 sin(𝑢(𝑥))
𝐭𝐚𝐧 𝒙
sec 2 𝑥
′
2
𝐭𝐚𝐧(𝒖(𝒙))
𝑢 sec 𝑥
𝐜𝐬𝐜 𝒙
− csc 𝑥 cot 𝑥
𝐜𝐬𝐜(𝒖(𝒙))
−𝑢′ csc(𝑢(𝑥)) cot(𝑢(𝑥))
𝐬𝐞𝐜 𝒙
sec 𝑥 tan 𝑥
𝐬𝐞𝐜(𝒖(𝒙))
𝑢′ sec(𝑢(𝑥)) tan(𝑢(𝑥))
𝐜𝐨𝐭 𝒙
− csc 2 𝑥
𝐜𝐨𝐭(𝒖(𝒙))
−𝑢 ′ csc 2 (𝑢(𝑥))
LESSON 2
𝑑2𝑦
𝑥2 2
𝑑𝑥
Given that 𝑦 = 𝑥 tan 𝑥, show that
2
≡ 2(𝑥 + 𝑦 2 )(1 + 𝑦)
SOLUTION
𝑑𝑦
= (1) tan 𝑥 + 𝑥 sec 2 𝑥
𝑑𝑥
𝑑2 𝑦
= sec 2 𝑥 + (1) sec 2 𝑥 + 𝑥(2 sec 𝑥)(sec 𝑥 tan 𝑥)
𝑑𝑥 2
= 2 sec 2 𝑥 + (2 sec 2 𝑥)𝑥 tan 𝑥
= 2 sec 2 𝑥 (1 + 𝑥 tan 𝑥)
LESSON 1
Differentiate the following w.r.t 𝑥
(a) 𝑦 = sec 4𝑥
(b) 𝑦 = cot 5𝑥
(c) 𝑦 = 3 csc(1 − 2𝑥 3 )
(d) 𝑦 =
(e) 𝑦 =
1
2+csc(−4𝑥)
𝑥2
sec 𝑥 3
𝑑2 𝑦
= 2(1 + tan2 𝑥)(1 + 𝑥 tan 𝑥)
𝑑𝑥 2
𝑥2
𝑑2 𝑦
= 2(𝑥 2 + 𝑥 2 tan2 𝑥)(1 + 𝑥 tan 𝑥)
𝑑𝑥 2
= 2(𝑥 2 + (𝑥 tan 𝑥)2 )(1 + 𝑥 tan 𝑥)
= 2(𝑥 2 + 𝑦 2 )(1 + 𝑦)
SOLUTION
(a) 𝑦 = sec 4𝑥
𝑦 ′ = 4 sec 4𝑥 tan 4𝑥
(b) 𝑦 = cot 5𝑥
𝑦 ′ = −5 csc 2 5𝑥
(c) 𝑦 = 3 csc(1 − 2𝑥 3 )
𝑦 ′ = 3[(−6𝑥 2 ) csc(1 − 2𝑥 3 ) cot(1 − 2𝑥 3 )
= −18𝑥 2 csc(1 − 2𝑥 3 ) cot(1 − 2𝑥 3 )
(d) 𝑦 =
Recall: sec 2 𝑥 = 1 + tan2 𝑥
1
2+csc(−4𝑥)
= (2 + csc(−4𝑥))−1
𝑦 ′ = −1(2 + csc(−4𝑥))−2 (4 csc(−4𝑥) cot(−4𝑥))
4 csc(−4𝑥) cot(−4𝑥)
=−
(2 + csc(−4𝑥))2
P a g e | 10
DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS
LESSON 1
Differentiate sin−1 𝑥.
PROOF
Let 𝑦 = sin−1 𝑥
𝜋
𝜋
then sin 𝑦 = 𝑥, – < 𝑦 <
2
2
𝑑
𝑑
(sin 𝑦) =
(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑦
cos 𝑦
=1
𝑑𝑥
𝑑𝑦
1
=
𝑑𝑥 cos 𝑦
But cos 2 𝑦 + sin2 𝑦 = 1
cos 2 𝑦 = 1 − sin2 𝑦
cos 𝑦 = √1 − sin2 𝑦 = √1 − 𝑥 2
𝑑𝑦
1
=
𝑑𝑥 √1 − 𝑥 2
𝑑
1
(sin−1 𝑥) =
𝑑𝑥
√1 − 𝑥 2
LESSON 4
3
𝑦′ = −
LESSON 2
Differentiate cos −1 𝑥
PROOF
Let 𝑦 = cos −1 𝑥
then 𝑥 = cos 𝑦 , 0 ≤ 𝑦 ≤ 𝜋
𝑑
𝑑
(cos 𝑦) =
(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑦
− sin 𝑦
=1
𝑑𝑥
𝑑𝑦
1
=−
𝑑𝑥
sin 𝑦
But cos 2 𝑦 + sin2 𝑦 = 1
sin2 𝑦 = 1 − cos 2 𝑦
sin 𝑦 = √1 − cos 2 𝑦 = √1 − 𝑥 2
𝑑
1
(cos −1 𝑥) = −
𝑑𝑥
√1 − 𝑥 2
In general, given that 𝑢 is a function of 𝑥, we have
LESSON 3
Differentiate tan−1 𝑥.
PROOF
Let 𝑦 = tan−1 𝑥
𝜋
𝜋
then tan 𝑦 = 𝑥, − < 𝑦 <
2
2
𝑑
𝑑
(tan 𝑦) =
(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑦
sec 2 𝑦
=1
𝑑𝑥
𝑑𝑦
1
=
𝑑𝑥 sec 2 𝑦
But sec 2 𝑦 = 1 + tan2 𝑦
= 1 + 𝑥2
𝑑
1
(tan1 𝑥) =
𝑑𝑥
1 + 𝑥2
Show that if 𝑦 = cos −1 3𝑥, then
√1−9𝑥 2
SOLUTION
𝑦 = cos −1 3𝑥 iff 3𝑥 = cos 𝑦
𝑑
𝑑
(cos 𝑦) =
(3𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑦
− sin 𝑦
=3
𝑑𝑥
𝑑𝑦
3
=−
𝑑𝑥
sin 𝑦
sin 𝑦 = √1 − cos 2 𝑦
= √1 − (3𝑥)2
𝑑
3
(cos −1 3𝑥) = −
𝑑𝑥
√1 − (3𝑥)2
3
=−
√1 − 9𝑥 2
𝑦 = sin−1 𝑢
𝑦 = cos −1 𝑢
𝑦 = tan−1 𝑢
𝑦′ =
1
√1 − 𝑢2
𝑦′ = −
𝑦′ =
. 𝑢′
1
√1 − 𝑢2
. 𝑢′
1
. 𝑢′
1 + 𝑢2
P a g e | 11
DIFFERENTIATION OF EXPONENTIAL FUNCTIONS
FUNCTION
DERIVATIVE
𝒆𝒙
𝑒𝑥
𝒆𝒖(𝒙)
𝑢′ 𝑒 𝑢(𝑥)
(e) 𝑦 = (1 − 𝑒 5𝑥 )3
𝑑𝑦
= 3(1 − 𝑒 5𝑥 )2 (−5𝑒 5𝑥 )
𝑑𝑥
= −15(1 − 𝑒 5𝑥 )2 𝑒 5𝑥
LESSON 2
Determine
𝑑𝑦
𝑑𝑥
for
𝑥𝑒 3𝑦 + 𝑦 2 = 5𝑥 3
LESSON 1
Differentiate the following
(a) 𝑦 = 𝑒 2𝑥−3
(b) 𝑦 = 2𝑒
𝑥𝑒 3𝑦 + 𝑦 2 = 5𝑥 3
sin 3𝑥
(c) 𝑦 = 𝑥𝑒 𝑥
(d) 𝑦 =
SOLUTION
2
𝑑𝑦
𝑑𝑦
) + 2𝑦
= 15𝑥 2
𝑑𝑥
𝑑𝑥
1(𝑒 3𝑦 ) + 𝑥 (3𝑒 3𝑦
𝑒 −2𝑥
𝑥+1
𝑑𝑦
𝑑𝑦
+ 2𝑦
= 15𝑥 2
𝑑𝑥
𝑑𝑥
(e) 𝑦 = (1 − 𝑒 5𝑥 )3
𝑒 3𝑦 + 3𝑥𝑒 3𝑦
SOLUTION
(3𝑥𝑒 3𝑦 + 2𝑦)
(a) 𝑦 = 𝑒 2𝑥−3
𝑑𝑦
= 2𝑒 2𝑥−3
𝑑𝑥
𝑑𝑦 15𝑥 2 − 𝑒 3𝑦
=
𝑑𝑥 3𝑥𝑒 3𝑦 + 2𝑦
(b) 𝑦 = 2𝑒 sin 3𝑥
𝑑𝑦
= 2[3 cos 3𝑥]𝑒 sin 3𝑥
𝑑𝑥
= 6𝑒 sin 3𝑥 cos 3𝑥
2
(c) 𝑦 = 𝑥𝑒 𝑥
𝑑𝑦
2
2
= 1𝑒 𝑥 + 𝑥[2𝑥𝑒 𝑥 ]
𝑑𝑥
2
2
= 𝑒 𝑥 + 2𝑥 2 𝑒 𝑥
2
= 𝑒 𝑥 [1 + 2𝑥 2 ]
(d) 𝑦 =
𝑒 −2𝑥
LESSON 3
that
Given that 𝑦 = 𝑒 𝑥 sin 𝑥, prove
𝑑2𝑦
𝑑𝑦
−2
+ 2𝑦 = 0
2
𝑑𝑥
𝑑𝑥
SOLUTION
𝑑𝑦
= 𝑒 𝑥 sin 𝑥 + 𝑒 𝑥 cos 𝑥
𝑑𝑥
= 𝑒 𝑥 (sin 𝑥 + cos 𝑥)
𝑑2 𝑦
= 𝑒 𝑥 (sin 𝑥 + cos 𝑥) + 𝑒 𝑥 (cos 𝑥 − sin 𝑥)
𝑑𝑥 2
𝑥+1
𝑑𝑦 −2𝑒 −2𝑥 (𝑥 + 1) − 𝑒 −2𝑥 (1)
=
(𝑥 + 1)2
𝑑𝑥
−2𝑥
−2𝑥𝑒
− 2𝑒 −2𝑥 − 𝑒 −2𝑥
=
(𝑥 + 1)2
−2𝑥 (2𝑥
−𝑒
+ 2 + 1)
=
(𝑥 + 1)2
𝑒 −2𝑥 (2𝑥 + 3)
=−
(𝑥 + 1)2
𝑑𝑦
= 15𝑥 2 − 𝑒 3𝑦
𝑑𝑥
= 2𝑒 𝑥 cos 𝑥
𝑑2 𝑦
𝑑𝑦
−2
+ 2𝑦
𝑑𝑥 2
𝑑𝑥
= 2𝑒 𝑥 cos 𝑥 − 2𝑒 𝑥 (cos 𝑥 + sin 𝑥) + 2𝑒 𝑥 sin 𝑥
=0
P a g e | 12
LESSON 4
Determine
𝑑𝑦
𝑑𝑥
for the equation
defined parametrically by
𝑥 = 𝑒 2𝑡 − 1 and 𝑦 = sin−1 2𝑡.
SOLUTION
𝑥 = 𝑒 2𝑡 − 1
𝑑𝑥
= 2𝑒 2𝑡
𝑑𝑡
𝑦 = sin−1 2𝑡
𝑑𝑦
2
=
𝑑𝑡 √1 − (2𝑡)2
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
=
=
2
√1 −
4𝑡 2
×
1
𝑒 2𝑡 √1
− 4𝑡 2
1
2𝑒 2𝑡
P a g e | 13
DIFFERENTIATION OF NATURAL LOGARITHMS
FUNCTION
DERIVATIVE
𝐥𝐧 𝒙
1
𝑥
𝐥𝐧(𝒖(𝒙))
𝑢′ (𝑥)
𝑢(𝑥)
LESSON 1
following.
Differentiate each of the
(ii) 𝑦 = ln(3𝑥 + 1)
(iii) 𝑦 = 3 ln(7𝑥 − 2)
(iv) 𝑦 = ln(𝑥 2 − 𝑥 + 1)
(v) 𝑦 = ln(sin 4𝑥)
(vi) 𝑦 = 𝑥 2 ln 𝑥
2 ln 𝑥
(vii)
𝑦=
(viii)
𝑦 = ln[(2𝑥 − 1)(3𝑥 + 2)2 ]
1+𝑥
SOLUTION
(ii) 𝑦 = ln(3𝑥 + 1)
𝑑𝑦
3
=
𝑑𝑥 3𝑥 + 1
(iii) 𝑦 = 3 ln(7𝑥 − 2)
𝑑𝑦
7
= 3(
)
𝑑𝑥
7𝑥 − 2
𝑑𝑦
21
=
𝑑𝑥 7𝑥 − 2
(iv) 𝑦 = ln(𝑥 2 − 𝑥 + 1)
(vii)
𝑦=
2 ln 𝑥
1+𝑥
1
𝑑𝑦 (2 × 𝑥 ) (1 + 𝑥) − 2 ln 𝑥 (1)
=
(1 + 𝑥)2
𝑑𝑥
2(1 + 𝑥) − 2𝑥 ln 𝑥
=
𝑥(1 + 𝑥)2
2 + 2𝑥 − 2𝑥 ln 𝑥
=
𝑥(1 + 𝑥)2
2(1 + 𝑥 − 𝑥 ln 𝑥)
=
𝑥(1 + 𝑥)2
(viii) 𝑦 = ln[(2𝑥 − 1)(3𝑥 + 2)2 ]
𝑦 = ln(2𝑥 − 1) + ln(3𝑥 + 2)2
𝑦 = ln(2𝑥 − 1) + 2 ln(3𝑥 + 2)
𝑑𝑦
2
3
=
+ 2(
)
𝑑𝑥 2𝑥 − 1
3𝑥 + 2
2
6
=
+
2𝑥 − 1 3𝑥 + 2
2(3𝑥 + 2) + 6(2𝑥 − 1)
=
(2𝑥 − 1)(3𝑥 + 2)
6𝑥 + 4 + 12𝑥 − 6
=
(2𝑥 − 1)(3𝑥 + 2)
2(9𝑥 − 1)
=
(2𝑥 − 1)(3𝑥 + 2)
LESSON 2
that (1 −
𝑑𝑦
𝑥2)
𝑑𝑥
Given that 𝑦 = ln (
≡2
SOLUTION
𝑦 = ln(1 + 𝑥) − ln(1 − 𝑥)
𝑑𝑦
1
1
=
+
𝑑𝑥 1 + 𝑥 1 − 𝑥
=
(1 − 𝑥) + (1 + 𝑥)
1 − 𝑥2
(v) 𝑦 = ln(sin 4𝑥)
=
𝑑𝑦 4 cos 4𝑥
=
𝑑𝑥
sin 4𝑥
= 4 cot 4𝑥
2
𝑑𝑦
(1 − 𝑥 2 )
1 − 𝑥2
𝑑𝑥
=
2
. (1 − 𝑥 2 )
1 − 𝑥2
𝑑𝑦
2𝑥 − 1
= 2
𝑑𝑥 𝑥 − 𝑥 + 1
=2
2
(vi) 𝑦 = 𝑥 ln 𝑥
𝑑𝑦
1
= 2𝑥 ln 𝑥 + 𝑥 2 ( )
𝑑𝑥
𝑥
= 2𝑥 ln 𝑥 + 𝑥
= 𝑥(2 ln 𝑥 + 1)
1+𝑥
1−𝑥
), show
P a g e | 14
Differentiate 3𝑥 w.r.t 𝑥
LESSON 3
SOLUTION
𝑦 = 3𝑥
ln 𝑦 = ln 3𝑥
1 𝑑𝑦
= 𝑥 ln 3
𝑦 𝑑𝑥
𝑑𝑦
= 𝑦 ln 3
𝑑𝑥
= 3𝑥 ln 3
In general, if 𝑦 = 𝑎 𝑥 then
LESSON 4
If 𝑦 =
𝑑𝑦
𝑑𝑥
= 𝑎 𝑥 ln 𝑎
𝑥
√𝑥 2 −3
, find
SOLUTION
𝑥
ln 𝑦 = ln (
)
2
√𝑥 − 3
1
ln 𝑦 = ln 𝑥 − ln(𝑥 2 − 3)2
1 𝑑𝑦 1 1 2𝑥
= − (
)
𝑦 𝑑𝑥 𝑥 2 𝑥 2 − 3
=
1
𝑥
−
𝑥 𝑥2 − 3
=
𝑥2 − 3 − 𝑥2
𝑥(𝑥 2 − 3)
=−
3
𝑥(𝑥 2 − 3)
𝑑𝑦
3𝑦
=−
𝑑𝑥
𝑥(𝑥 2 − 3)
𝑥
Sub 𝑦 =
1
(𝑥 2 − 3)2
3𝑥
=−
1
𝑥(𝑥 2 − 3)2 (𝑥 2 − 3)
=−
3
3
(𝑥 2 − 3)2
𝑑𝑦
𝑑𝑥
P a g e | 15
PARTIAL DERIVATIVES
For partial derivatives we differentiate with
LESSON 3
Given that
respect to one variable and treat the other
𝑓(𝑟, 𝑠) = 4𝑟 ln(𝑟 + 𝑠 2 ), determine
variable(s) as constants.
(i) 𝑓𝑟
(ii) 𝑓𝑠
First Partial Derivative
LESSON 1
SOLUTION
Given that
3
2
𝑓(𝑥, 𝑦) = 𝑥 + 2𝑥𝑦 +
2𝑥
2
(i) 𝑓(𝑟, 𝑠) = 4𝑟 ln(𝑟 2 + 𝑠 2 )
, evaluate
(i) 𝑓𝑥
2𝑟
𝑓𝑟 = 4 ln(𝑟 2 + 𝑠 2 ) + 4𝑟 ( 2
)
𝑟 + 𝑠2
(ii) 𝑓𝑦
𝑓𝑟 = 4 ln(𝑟 2 + 𝑠 2 ) +
5𝑦
SOLUTION
(ii) 𝑓(𝑟, 𝑠) = 4𝑟 ln(𝑟 2 + 𝑠 2 )
(i) For 𝑓𝑥 we differentiate 𝑓(𝑥, 𝑦) with respect to
2𝑠
𝑓𝑠 = 4𝑟 ( 2
)
𝑟 + 𝑠2
8𝑟𝑠
𝑓𝑠 = 2
𝑟 + 𝑠2
𝑥, treating 𝑦 as a constant.
𝑓(𝑥, 𝑦) = 𝑥 3 + 2𝑥𝑦 2 +
𝑓𝑥 = 3𝑥 2 + 2𝑦 2 +
2
𝑥
5𝑦
2
5𝑦
(ii) 𝑓(𝑥, 𝑦) = 𝑥 3 + 2𝑥𝑦 2 +
LESSON 4
2𝑥
5
𝑦 −1
2𝑥 −2
𝑦
5
2𝑥
𝑓𝑦 = 4𝑥𝑦 − 2
5𝑦
𝑓𝑦 = 4𝑥𝑦 −
LESSON 2
Given that
𝑧 = (3𝑥 2 + 2𝑥𝑦 + 𝑦 2 )4 , determine
(i)
(ii)
8𝑟 2
𝑟2 + 𝑠2
𝜕𝑧
Given that 𝑤 = 𝑥𝑦 2 𝑧 3 − 𝑦 4 + 𝑒 𝑧 ,
determine
(i)
(ii)
(iii)
𝜕𝑤
𝜕𝑥
𝜕𝑤
𝜕𝑦
𝜕𝑤
𝜕𝑧
SOLUTION
(i) 𝑤 = 𝑥𝑦 2 𝑧 3 − 𝑦 4 + 𝑒 𝑧
𝜕𝑤
= 𝑦2𝑧3
𝜕𝑥
𝜕𝑥
𝜕𝑧
𝜕𝑦
SOLUTION
(i) 𝑧 = (3𝑥 2 + 2𝑥𝑦 + 𝑦 2 )4
𝜕𝑧
= 4(3𝑥 2 + 2𝑥𝑦 + 𝑦 2 )3 (6𝑥 + 2𝑦)
𝜕𝑥
= 8(3𝑥 + 𝑦)(3𝑥 2 + 2𝑥𝑦 + 𝑦 2 )3
(ii) 𝑧 = (3𝑥 2 + 2𝑥𝑦 + 𝑦 2 )4
𝜕𝑧
= 4(3𝑥 2 + 2𝑥𝑦 + 𝑦 2 )3 (2𝑥 + 2𝑦)
𝜕𝑦
= 8(𝑥 + 𝑦)(3𝑥 2 + 2𝑥𝑦 + 𝑦 2 )
(ii) 𝑤 = 𝑥𝑦 2 𝑧 3 − 𝑦 4 + 𝑒 𝑧
𝜕𝑤
= 2𝑥𝑦𝑧 3 − 4𝑦 3
𝜕𝑦
(iii) 𝑤 = 𝑥𝑦 2 𝑧 3 − 𝑦 4 + 𝑒 𝑧
𝜕𝑤
= 3𝑥𝑦 2 𝑧 2 + 𝑒 𝑧
𝜕𝑧
P a g e | 16
For 𝑓(𝑥, 𝑦) = 𝑥 3 𝑦 + 2𝑦,
LESSON 5
determine 𝑓𝑥 (2, 1).
𝑓(𝑥, 𝑦) = 𝑥 3 𝑦 + 2𝑦
(iv)
𝑓𝑥 = 3𝑥 2 𝑦
𝑓𝑥 (2, 1) = 3(2)
𝜕𝑥𝜕𝑦
(iii)
SOLUTION
2 (1)
𝜕2 𝑤
(ii)
Second PARTIAL DERIVATIVE
Given that
3
𝑦
2
𝑓(𝑥, 𝑦) = 𝑥 𝑦 − 4𝑥𝑦 +
𝑥2
, determine
2𝑦
3𝑧
𝜕2 𝑤
(ii)
𝜕𝑥𝜕𝑦
= 8𝑥 3 𝑦𝑧 2
2𝑦
(iii) 𝑤 = 𝑥 4 𝑦 2 𝑧 2 − ln(𝑥𝑦) + sin ( )
3𝑧
SOLUTION
𝜕𝑤
1 2
2𝑦
= 2𝑥 4 𝑦𝑧 2 − + cos ( )
𝜕𝑦
𝑦 3𝑧
3𝑧
(i) 𝑓(𝑥, 𝑦) = 𝑥 3 𝑦 − 4𝑥𝑦 2 + 𝑦𝑥 −2
𝑓𝑥 = 3𝑥 2 𝑦 − 4𝑦 2 − 2𝑦𝑥 −3
(iv)
6𝑦
𝑓𝑥𝑥 = 6𝑥𝑦 + 4
𝑥
(ii) 𝑓(𝑥, 𝑦) = 𝑥 3 𝑦 − 4𝑥𝑦 2 +
𝑦
1
2
2𝑦
𝑦
3
3
= 2𝑥 4 𝑦𝑧 2 − + 𝑧 −1 cos (
= 4𝑥 4 𝑦𝑧 −
1
𝑥2
𝑧 −1 )
2
2𝑦
4𝑦
2𝑦
cos ( ) + 3 sin ( )
3𝑧 2
3𝑧
9𝑧
3𝑧
(v) 𝑤 = 𝑥 4 𝑦 2 𝑧 2 − ln(𝑥𝑦) + sin (
𝑦
2
𝑥3
2𝑦
3
𝑧 −1 )
𝜕𝑤
2𝑦
2𝑦
= 2𝑥 4 𝑦 2 𝑧 + (− 2 ) cos ( )
𝜕𝑧
3𝑧
3𝑧
2𝑦
2𝑦
= 2𝑥 4 𝑦 2 𝑧 − 2 cos ( )
3𝑧
3𝑧
𝑓𝑦 = 𝑥 3 − 8𝑥𝑦 + 𝑥 −2
𝑓𝑦𝑥 = 3𝑥 2 − 8𝑦 −
𝜕𝑦
𝑥2
2
(iii) 𝑓(𝑥, 𝑦) = 𝑥 3 𝑦 − 4𝑥𝑦 2 +
𝜕𝑤
𝜕2𝑤
2
2𝑦
2
2𝑦
2𝑦
= 4𝑥4 𝑦𝑧 − 2 cos ( ) + [− (− 2 ) sin ( )]
𝜕𝑦𝜕𝑧
3𝑧
3𝑧
3𝑧
3𝑧
3𝑧
2
𝑓𝑥 = 3𝑥 𝑦 − 4𝑦 − 3 𝑦
𝑥
2
𝑓𝑥𝑦 = 3𝑥 2 − 8𝑦 − 3
𝑥
2
(vi)
𝜕𝑤
𝜕𝑧
= 2𝑥 4 𝑦 2 𝑧 −
2𝑦
3
𝑧 −2 cos (
2𝑦
3
𝑧 −1 )
𝜕2𝑤
4𝑦
2𝑦
2𝑦
2𝑦
2𝑦
= 2𝑥 4 𝑦 2 + 3 cos ( ) − 2 (− (− 2 ) sin ( ))
𝜕𝑧 2
3𝑧
3𝑧
3𝑧
3𝑧
3𝑧
NB:𝑓𝑥𝑦 = 𝑓𝑦𝑥
= 2𝑥 4 𝑦 2 +
Given that
2𝑦
𝑤 = 𝑥 4 𝑦 2 𝑧 2 − ln(𝑥𝑦) + sin ( ), determine
3𝑧
𝜕𝑥
𝜕𝑧 2
𝜕𝑤
1
= 4𝑥 3 𝑦 2 𝑧 2 −
𝜕𝑥
𝑥
(iii) 𝑓𝑦𝑥
(i)
𝜕2 𝑤
(i) 𝑤 = 𝑥 4 𝑦 2 𝑧 2 − ln(𝑥𝑦) + sin ( )
(ii) 𝑓𝑥𝑦
𝜕𝑤
𝜕𝑦𝜕𝑧
SOLUTION
(i) 𝑓𝑥𝑥
LESSON 2
𝜕2 𝑤
𝜕𝑧
(vi)
LESSON 1
𝜕𝑦
𝜕𝑤
(v)
= 12
𝜕𝑤
4𝑦
2𝑦
4𝑦 2
2𝑦
cos
(
)
−
sin ( )
3𝑧 3
3𝑧
9𝑧 4
3𝑧
P a g e | 17
INTEGRATION RESULTS
SOLUTION
FUNCTION
INTEGRAL
𝟏
𝒙
′ (𝒙)
𝒇
𝒇(𝒙)
𝒇
ln|𝑓(𝑥)| + 𝑐
(i) ∫ 𝑑𝑥
𝑥
= 2∫
1
𝑑𝑥
𝑥
= 2 ln|𝑥| + 𝑐
1
[𝑓(𝑥)]𝑛+1 + 𝑐
𝑛+1
′ (𝒙)[𝒇(𝒙)]𝒏
𝒆
2
ln|𝑥| + 𝑐
𝒙
𝑥
𝑒 +𝑐
1 𝑎𝑥+𝑏
𝑒
+𝑐
𝑎
𝒆𝒂𝒙+𝒃
𝒇′ (𝒙)𝒆𝒇(𝒙)
𝑒 𝑓(𝑥) + 𝑐
𝐭𝐚𝐧 𝒙
− ln|cos 𝑥| + 𝑐 or
ln|sec 𝑥| + 𝑐
1
ln sec(𝑎𝑥 + 𝑏) + 𝑐
𝑎
𝐭𝐚𝐧(𝒂𝒙 + 𝒃)
𝐬𝐞𝐜 𝒙
ln|sec 𝑥 + tan 𝑥| + 𝑐
𝟐
= ln 𝑥 2 + 𝑐
4
(ii) ∫
𝑑𝑥
4𝑥−1
= ln|4𝑥 − 1| + 𝑐
5
(iii) ∫
𝑑𝑥
1−2𝑥
5
−2
=− ∫
𝑑𝑥
2 1 − 2𝑥
5
= − ln|1 − 2𝑥| + 𝑐
2
(iv) ∫
2𝑥+3
2𝑥 2 +6𝑥−9
1
4𝑥 + 6
= ∫ 2
𝑑𝑥
2 2𝑥 + 6𝑥 − 9
𝐬𝐞𝐜 𝒙
tan 𝑥 + 𝑐
𝐜𝐬𝐜 𝒙
𝑥
ln |tan ( )| + 𝑐
2
𝐜𝐨𝐭 𝒙
(v) ∫ tan 𝑥 𝑑𝑥
√𝒂𝟐 − 𝒖𝟐
ln|sin 𝑥| + 𝑐
𝑢
sin−1 ( ) + 𝑐
𝑎
𝒖′
𝟐
𝒂 + 𝒖𝟐
1
𝑢
tan−1 ( ) + 𝑐
𝑎
𝑎
= −∫
𝒖′
𝑑𝑥
1
= ln|2𝑥 2 + 6𝑥 − 9| + 𝑐
2
=∫
sin 𝑥
𝑑𝑥
cos 𝑥
−sin 𝑥
𝑑𝑥
cos 𝑥
= − ln|cos 𝑥| + 𝑐
LESSON 1
Determine
=
2
(i) ∫ 𝑑𝑥
𝑥
(ii)
4
∫ 4𝑥−1
(iii)
(vii)
𝑑𝑥
5
∫ 1−2𝑥
𝑑𝑥
(v) ∫ tan 𝑥 𝑑𝑥
(vi)∫ 12𝑥 3 (3𝑥 4 + 5)2 𝑑𝑥
3
(3𝑥 4 + 5)3
+𝑐
3
3
∫ 𝑥 (2 + ln 𝑥)3 𝑑𝑥
1
= 3 ∫ (2 + ln 𝑥)3 𝑑𝑥
𝑥
𝑑𝑥
2𝑥+3
(iv)∫ 2𝑥 2 +6𝑥−9
(vii)
(vi) ∫ 12𝑥 3 (3𝑥 4 + 5)2 𝑑𝑥
∫ 𝑥 (2 + ln 𝑥)3 𝑑𝑥
=
3(2 + ln 𝑥)4
+𝑐
4
P a g e | 18
LESSON 2
Evaluate each of the following.
(v) ∫ 𝑒 4𝑥 √𝑒 4𝑥 + 6 𝑑𝑥
1
(i) ∫ 𝑒 2𝑥 𝑑𝑥
= ∫ 𝑒 4𝑥 (𝑒 4𝑥 + 6)2 𝑑𝑥
(ii) ∫ 2𝑒 3−𝑥 𝑑𝑥
1
1
= ∫ 4𝑒 4𝑥 (𝑒 4𝑥 + 6)2 𝑑𝑥
4
(iii) ∫ 𝑒 𝑥 (2 + 𝑒 𝑥 )3 𝑑𝑥
1
5𝑒 −2𝑥
(iv) ∫ (1+𝑒 −2𝑥)2 𝑑𝑥
(v) ∫ 𝑒 4𝑥 √𝑒 4𝑥 + 6 𝑑𝑥
3
(vi) ∫ 3𝑥 2 𝑒 𝑥 𝑑𝑥
𝑥
(vii)
∫ 𝑒 𝑥+𝑒 𝑑𝑥
(viii)
∫ − csc 2 𝑥 𝑒 cot 𝑥 𝑑𝑥
1 (𝑒 4𝑥 + 6)2+1
= [
]+𝑐
3
4
2
3
1 4𝑥
= (𝑒 + 6)2 + 𝑐
6
3
(vi) ∫ 3𝑥 2 𝑒 𝑥 𝑑𝑥
3
SOLUTION
(i) ∫ 𝑒 2𝑥 𝑑𝑥
1
= 𝑒 2𝑥 + 𝑐
2
(ii) ∫ 2𝑒 3−𝑥 𝑑𝑥
= 𝑒𝑥 + 𝑐
𝑥
∫ 𝑒 𝑥+𝑒 𝑑𝑥
(vii)
𝑥
= ∫ 𝑒 𝑥 𝑒 𝑒 𝑑𝑥
𝑥
= 𝑒𝑒 + 𝑐
= 2 ∫ 𝑒 3−𝑥 𝑑𝑥
1
= 2 ( 𝑒 3−𝑥 ) + 𝑐
−1
∫ − csc 2 𝑥 𝑒 cot 𝑥 𝑑𝑥
(viii)
= 𝑒 cot 𝑥 + 𝑐
= −2𝑒 3−𝑥 + 𝑐
LESSON 3
(iii) ∫ 𝑒 𝑥 (2 + 𝑒 𝑥 )3
(2 + 𝑒 𝑥 )4
=
+𝑐
4
Determine
(a) ∫ (𝑥 +
√
1
4−𝑥 2
1
) 𝑑𝑥
2
(b) ∫ ((2𝑥−1)3 −
) 𝑑𝑥
9+4𝑥 2
(c) ∫
√
1
1−𝑥 2
𝑒 sin
−1 𝑥
𝑑𝑥
5𝑒 −2𝑥
(iv) ∫ (1+𝑒 −2𝑥)2 𝑑𝑥
= ∫ 5𝑒 −2𝑥 (1 + 𝑒 −2𝑥 )−2 𝑑𝑥
SOLUTION
(a) ∫ (𝑥 +
√
1
4−𝑥 2
=
5
∫ −2𝑒 2𝑥 (1 + 𝑒 −2𝑥 )−2 𝑑𝑥
−2
5 (1 + 𝑒 −2𝑥 )−1
=− [
]+𝑐
2
−1
5
= (1 + 𝑒 −2𝑥 )−1 + 𝑐
2
5
=
+𝑐
2(1 + 𝑒 −2𝑥 )
= ∫ (𝑥 +
) 𝑑𝑥
1
) 𝑑𝑥
√22 − 𝑥 2
𝑥2
𝑥
=
+ sin−1 ( ) + 𝑐
2
2
1
3
(b) ∫ ((2𝑥−1)3 −
) 𝑑𝑥
9+4𝑥 2
1
3
2
𝑑𝑥 − ∫ 2
𝑑𝑥
(2𝑥 − 1)3
2 3 + (2𝑥)2
3
2
= ∫(2𝑥 − 1)−3 𝑑𝑥 − ∫ 2
𝑑𝑥
2 3 + (2𝑥)2
=∫
P a g e | 19
1 (2𝑥 − 1)−2
3 1
2𝑥
= [
] − [ tan−1 ( )] + 𝑐
2
−2
2 3
3
1
1
2𝑥
= − (2𝑥 − 1)−2 − tan−1 ( ) + 𝑐
4
2
3
Even powers of 𝐬𝐢𝐧 𝒙 and 𝐜𝐨𝐬 𝒙
LESSON 1
Determine
∫ sin2 𝑥 𝑑𝑥
SOLUTION
(c) ∫
√
1
1−𝑥 2
= 𝑒 sin
𝑒
sin−1 𝑥
−1 𝑥
+𝑐
𝑑𝑥
∫ sin2 𝑥 𝑑𝑥
Using the identity
cos 2𝑥 = 1 − 2 sin2 𝑥
1 cos 2𝑥
sin2 𝑥 = −
2
2
∫ sin2 𝑥 𝑑𝑥
1 cos 2𝑥
= ∫( −
) 𝑑𝑥
2
2
𝑥 sin 2𝑥
= −
+𝑐
2
4
LESSON 2
Determine
∫ cos 2 𝑥 𝑑𝑥
SOLUTION
∫ cos 2 𝑥 𝑑𝑥
Using the identity
cos 2𝑥 = 2 cos 2 𝑥 − 1
1
cos
2𝑥
cos 2 𝑥 = +
2
2
∫ cos 2 𝑥 𝑑𝑥
1 cos 2𝑥
= ∫( +
) 𝑑𝑥
2
2
𝑥 sin 2𝑥
= +
+𝑐
2
4
P a g e | 20
Odd powers on 𝐬𝐢𝐧 𝒙 and 𝐜𝐨𝐬 𝒙
LESSON 1
LESSON 4
Determine
∫ cos 3 𝑥 sin4 𝑥 𝑑𝑥
Determine
∫ sin3 𝑥 𝑑𝑥
SOLUTION
SOLUTION
∫ sin3 𝑥 𝑑𝑥 = ∫ sin 𝑥 sin2 𝑥 𝑑𝑥
Since sin2 𝑥 = 1 − cos 2 𝑥
∫ sin 𝑥 sin2 𝑥 𝑑𝑥
= ∫ sin 𝑥 𝑑𝑥 − ∫ sin 𝑥 cos 2 𝑥 𝑑𝑥
= ∫ sin 𝑥 𝑑𝑥 + ∫(− sin 𝑥) cos 2 𝑥 𝑑𝑥
cos 3 𝑥
+𝑐
3
LESSON 2
Determine
∫ cos 3 𝑥 𝑑𝑥
SOLUTION
∫ cos 3 𝑥 𝑑𝑥
= ∫ cos 𝑥 cos 2 𝑥 𝑑𝑥
= ∫ cos 𝑥 (1 − sin2 𝑥) 𝑑𝑥
= ∫ cos 𝑥 𝑑𝑥 − ∫ cos 𝑥 sin2 𝑥 𝑑𝑥
= sin 𝑥 −
sin3 𝑥
+𝑐
3
LESSON 3
= ∫ cos 𝑥 cos 2 𝑥 sin4 𝑥 𝑑𝑥
= ∫ cos 𝑥 (1 − sin2 𝑥) sin4 𝑥 𝑑𝑥
= ∫ sin 𝑥 (1 − cos 2 𝑥) 𝑑𝑥
= − cos 𝑥 +
∫ cos 3 𝑥 sin4 𝑥 𝑑𝑥
Determine
∫ sin3 2𝑥 𝑑𝑥
SOLUTION
∫ sin3 2𝑥 𝑑𝑥
= ∫ sin 2𝑥 sin2 2𝑥 𝑑𝑥
= ∫ sin 2𝑥 (1 − cos 2 2𝑥) 𝑑𝑥
= ∫ sin 2𝑥 − ∫ sin 2𝑥 cos 2 2𝑥 𝑑𝑥
1
cos 3 2𝑥
= cos 2𝑥 +
+𝑐
2
6
= ∫ cos 𝑥 sin4 𝑥 − ∫ cos 𝑥 sin6 𝑥 𝑑𝑥
=
sin5 𝑥 sin7 𝑥
−
+𝑐
5
7
P a g e | 21
Even powers of 𝐭𝐚𝐧 𝒙
LESSON 1
Determine
∫ tan2 𝑥 𝑑𝑥
SOLUTION
∫ tan2 𝑥 𝑑𝑥 = ∫ sec 2 𝑥 − 1 𝑑𝑥
Since sec 2 𝑥 = 1 + tan2 𝑥
= tan 𝑥 − 𝑥 + 𝑐
1 + sin 𝑥
∫
𝑑𝑥
cos 𝑥
1
sin 𝑥
=∫
+
𝑑𝑥
cos 𝑥 cos 𝑥
= ∫ sec 𝑥 + tan 𝑥 𝑑𝑥
= ln|sec 𝑥 + tan 𝑥| + ln|sec 𝑥| + 𝑐
= ln|sec 𝑥 (sec 𝑥 + tan 𝑥)| + 𝑐
LESSON 2
Determine
∫ tan2 2𝑥 𝑑𝑥
SOLUTION
∫ tan2 2𝑥 𝑑𝑥
= ∫ sec 2 2𝑥 − 1 𝑑𝑥
1
= tan 2𝑥 − 𝑥 + 𝑐
2
P a g e | 22
INTEGRATION BY PARTS
𝑑
𝑑𝑣
𝑑𝑢
(𝑢𝑣) = 𝑢
+𝑣
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑣
𝑑
𝑑𝑢
(𝑢𝑣) − 𝑣
𝑢
=
𝑑𝑥 𝑑𝑥
𝑑𝑥
∫𝑢
𝑥
1
(2𝑥 + 3)6 −
(2𝑥 + 3)7 + 𝑐
12
168
1
(2𝑥 + 3)6 (14𝑥 − 2𝑥 − 3) + 𝑐
=
168
1
(2𝑥 + 3)6 (12𝑥 − 3) + 𝑐
=
168
1
(2𝑥 + 3)6 (4𝑥 − 1) + 𝑐
=
56
=
𝑑𝑣
𝑑𝑢
𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣
𝑑𝑥
𝑑𝑥
𝑑𝑥
∫ 𝑢 𝑑𝑣 = 𝑢𝑣 − ∫ 𝑣 𝑑𝑢
When choosing 𝑢 we use the following acronym.
LESSON 3
Evaluate
L – Logarithms
∫
I – Inverses
3𝑥 + 1
√1 − 2𝑥
𝑑𝑥
SOLUTION
A – Algebra
T – Trigonometric Ratios
E – Exponentials
LESSON 1
∫
3𝑥 + 1
1
√1 − 2𝑥
𝑑𝑥 = ∫(3𝑥 + 1)(1 − 2𝑥)−2 𝑑𝑥
𝑢 = 3𝑥 + 1
Determine
∫ 𝑥𝑒 𝑥 𝑑𝑥
SOLUTION
𝑑𝑢 = 3
1
𝑑𝑣
= (1 − 2𝑥)−2
𝑑𝑥
3𝑥 + 1
∫
𝑑𝑥
√1 − 2𝑥
1
𝑣 = −(1 − 2𝑥)2
1
1
= −(3𝑥 + 1)(1 − 2𝑥)2 − ∫ −3(1 − 2𝑥)2 𝑑𝑥
∫ 𝑥𝑒 𝑥 𝑑𝑥
1
3
= −(3𝑥 + 1)(1 − 2𝑥)2 − (1 − 2𝑥)2
𝑢=𝑥
𝑑𝑣 = 𝑒
𝑑𝑢 = 1
𝑥
1
= (1 − 2𝑥)2 (−3𝑥 − 1 − 1 + 2𝑥)
𝑣 = 𝑒𝑥
1
𝑥
𝑥
𝑥
∫ 𝑥𝑒 𝑑𝑥 = 𝑥𝑒 − ∫ 𝑒 𝑑𝑥
= 𝑥𝑒 𝑥 − 𝑒 𝑥 + 𝑐
= −(1 − 2𝑥)2 (𝑥 + 2) + 𝑐
LESSON 4
Evaluate
∫ 𝑥 cos 𝑥 𝑑𝑥
Determine ∫ 𝑥(2𝑥 + 3)5 𝑑𝑥
LESSON 2
SOLUTION
SOLUTION
∫ 𝑥(2𝑥 + 3)5 𝑑𝑥
∫ 𝑥 cos 𝑥 𝑑𝑥
𝑢=𝑥
𝑢=𝑥
𝑑𝑢 = 1
𝑑𝑣 = (2𝑥 + 3)5
𝑣=
1
(2𝑥 + 3)6
12
∫ 𝑥(2𝑥 + 3)5 𝑑𝑥
= 𝑥[
1
1
(2𝑥 + 3)6 ] − ∫ (2𝑥 + 3)6 𝑑𝑥
12
12
𝑑𝑢 = 1
𝑑𝑣
= cos 𝑥
𝑑𝑥
𝑣 = sin 𝑥
∫ 𝑥 cos 𝑥 𝑑𝑥 = 𝑥 sin 𝑥 − ∫ sin 𝑥 𝑑𝑥
= 𝑥 sin 𝑥 + cos 𝑥 + 𝑐
P a g e | 23
LESSON 5
Determine
LESSON 7
Evaluate
𝜋
∫ ln 𝑥 𝑑𝑥
∫(𝑥 − 𝜋)2 sin 𝑥 𝑑𝑥
0
SOLUTION
SOLUTION
∫ ln 𝑥 𝑑𝑥 = ∫ 1. ln 𝑥 𝑑𝑥
𝑢 = ln 𝑥
𝑑𝑣
=1
𝑑𝑥
𝜋
∫(𝑥 − 𝜋)2 sin 𝑥 𝑑𝑥
1
𝑑𝑢 =
𝑥
0
𝑣=𝑥
∫ ln 𝑥 𝑑𝑥
𝑑𝑢 = 2(𝑥 − 𝜋)
𝑑𝑣
= sin 𝑥
𝑑𝑥
𝑣 = − cos 𝑥
𝜋
∫(𝑥 − 𝜋)2 sin 𝑥 𝑑𝑥
= 𝑥 ln 𝑥 − ∫ 1 𝑑𝑥
0
= 𝑥 ln 𝑥 − 𝑥 + 𝑐
LESSON 6
𝑢 = (𝑥 − 𝜋)2
= −(𝑥 − 𝜋)2 cos 𝑥 − ∫ −2(𝑥 − 𝜋) cos 𝑥 𝑑𝑥
Determine
2
∫ 𝑥 ln 𝑥 𝑑𝑥
𝑢 = −2(𝑥 − 𝜋)
𝑑𝑢 = −2
𝑑𝑣
= cos 𝑥
𝑑𝑥
𝑣 = sin 𝑥
SOLUTION
∫ −2(𝑥 − 𝜋) cos 𝑥 𝑑𝑥
∫ 𝑥 2 ln 𝑥 𝑑𝑥
= −2(𝑥 − 𝜋) sin 𝑥 — 2 sin 𝑥 𝑑𝑥
𝑢 = ln 𝑥
1
𝑑𝑢 =
𝑥
𝑑𝑣
= 𝑥2
𝑑𝑥
𝑥3
𝑣=
3
= −2(𝑥 − 𝜋) sin 𝑥 − 2 cos 𝑥
𝜋
∫(𝑥 − 𝜋)2 sin 𝑥 𝑑𝑥
0
= [−(𝑥 − 𝜋)2 cos 𝑥 + 2(𝑥 − 𝜋) sin 𝑥 + 2 cos 𝑥]0
= 2 cos 𝜋 — 𝜋 2 cos(0) + 2 cos(0)
2
∫ 𝑥 ln 𝑥 𝑑𝑥
= −2 + 𝜋 2 − 2
𝑥3
𝑥2
= ln 𝑥 − ∫
𝑑𝑥
3
3
= 𝜋2 − 4
𝑥3
1
ln 𝑥 − 𝑥 3 + 𝑐
3
9
1
= 𝑥 3 (3 ln 𝑥 − 1) + 𝑐
9
LESSON 8
=
Evaluate
∫ 𝑒 2𝑥 sin 𝑥 𝑑𝑥
SOLUTION
∫ 𝑒 2𝑥 sin 𝑥 𝑑𝑥
𝑑𝑣 = 𝑒 2𝑥
1
𝑑𝑢 = cos 𝑥
𝑣 = 𝑒 2𝑥
2
1 2𝑥
1
2𝑥
∫ 𝑒 sin 𝑥 𝑑𝑥 = 𝑒 sin 𝑥 − ∫ 𝑒 2𝑥 cos 𝑥 𝑑𝑥
2
2
𝑢 = cos 𝑥
𝑑𝑣 = 𝑒 2𝑥
1
𝑑𝑢 = − sin 𝑥
𝑣 = 𝑒 2𝑥
2
𝑢 = sin 𝑥
𝜋
P a g e | 24
∫ 𝑒 2𝑥 sin 𝑥 𝑑𝑥
1
1 1
1
= 𝑒 2𝑥 sin 𝑥 − [ 𝑒 2𝑥 cos 𝑥 + ∫ 𝑒 2𝑥 sin 𝑥 𝑑𝑥]
2
2 2
2
1 2𝑥
1 2𝑥
1
2𝑥
∫ 𝑒 sin 𝑥 𝑑𝑥 = 𝑒 sin 𝑥 − 𝑒 cos 𝑥 − ∫ 𝑒 2𝑥 sin 𝑥 𝑑𝑥
2
4
4
5
1 2𝑥
1 2𝑥
2𝑥
∫ 𝑒 sin 𝑥 𝑑𝑥 = 𝑒 sin 𝑥 − 𝑒 cos 𝑥
4
2
4
2 2𝑥
1 2𝑥
2𝑥
∫ 𝑒 sin 𝑥 𝑑𝑥 = 𝑒 sin 𝑥 − 𝑒 cos 𝑥 + 𝑐
5
5
P a g e | 25
Reduction Formulae
LESSON 9
Establish a reduction formula
that could be used to find ∫ 𝑥 𝑛 𝑒 𝑥 𝑑𝑥 and use it
𝑑𝑣
= cos 𝑥
𝑑𝑥
𝑣 = sin 𝑥
when 𝑛 = 4.
𝐼𝑛 = sin 𝑥 cosn−1 𝑥 + ∫(𝑛 − 1) cos𝑛−2 𝑥 sin2 𝑥 𝑑𝑥
SOLUTION
STEP 3: Simplify
𝐼𝑛 = ∫ 𝑥 𝑛 𝑒 𝑥 𝑑𝑥
𝐼𝑛 = sin 𝑥 cos 𝑛−1 𝑥 + ∫(𝑛 − 1) cos 𝑛−2 𝑥 (1 − cos 2 𝑥)𝑑𝑥
𝑢 = 𝑥𝑛
𝑑𝑢 = 𝑛𝑥 𝑛−1
𝑑𝑣
= 𝑒𝑥
𝑑𝑥
𝑣 = 𝑒𝑥
𝐼𝑛 = sin 𝑥 cos 𝑛−1 𝑥 + (𝑛 − 1) ∫ cos 𝑛−2 𝑥 𝑑𝑥 − (𝑛 − 1) ∫ cos 𝑛 𝑥 𝑑𝑥
𝐼𝑛 = 𝑥 𝑛 𝑒 𝑥 − ∫ 𝑛𝑥 𝑛−1 𝑒 𝑥 𝑑𝑥
𝑛 𝑥
𝐼𝑛 = 𝑥 𝑒 − 𝑛 ∫ 𝑥
𝐼𝑛 = sin 𝑥 cos𝑛−1 𝑥 + (𝑛 − 1) ∫ cos𝑛−2 𝑥 (1 − cos2 𝑥) 𝑑𝑥
𝐼𝑛 = sin 𝑥 cos 𝑛−1 𝑥 + (𝑛 − 1)𝐼𝑛−2 − (𝑛 − 1)𝐼𝑛
𝐼𝑛 + 𝑛𝐼𝑛 − 𝐼𝑛 = sin 𝑥 cos 𝑛−1 𝑥 + (𝑛 − 1)𝐼𝑛−2
𝑛−1 𝑥
𝑒 𝑑𝑥
𝐼𝑛 = 𝑥 𝑛 𝑒 𝑥 − 𝑛𝐼𝑛−1
𝑛𝐼𝑛 = sin 𝑥 cos 𝑛−1 𝑥 + (𝑛 − 1)𝐼𝑛−2
𝐼𝑛 =
𝑛=4
𝐼4 = 𝑥 4 𝑒 𝑥 − 4𝐼3
= 𝑥 4 𝑒 𝑥 − 4[𝑥 3 𝑒 𝑥 − 3𝐼2 ]
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝐼2
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12[𝑥 2 𝑒 𝑥 − 2𝐼1 ]
1
𝑛−1
sin 𝑥 cos 𝑛−1 𝑥 +
𝐼
𝑛
𝑛 𝑛−2
STEP 4: Apply the derived formula to the rest of
the question
𝑛=5
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝑥 2 𝑒 𝑥 − 24𝐼1
1
4
𝐼5 = sin 𝑥 cos 4 𝑥 + 𝐼3
5
5
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝑥 2 𝑒 𝑥 − 24[𝑥𝑒 𝑥 − 𝐼0 ]
𝑛=3
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝑥 2 𝑒 𝑥 − 24𝑥𝑒 𝑥 + 24𝐼0
1
2
𝐼3 = sin 𝑥 cos 2 𝑥 + 𝐼1
3
3
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝑥 2 𝑒 𝑥 − 24𝑥𝑒 𝑥 + 24 [∫ 𝑒 𝑥 𝑑𝑥]
STEP 5: When you have reduced your integral to
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝑥 2 𝑒 𝑥 − 24𝑥𝑒 𝑥 + 24[𝑒 𝑥 + 𝑐]
= 𝑥 4 𝑒 𝑥 − 4𝑥 3 𝑒 𝑥 + 12𝑥 2 𝑒 𝑥 − 24𝑥𝑒 𝑥 + 24𝑒 𝑥 + 𝐴
If 𝐼𝑛 ≡ ∫ cos 𝑛 𝑥 𝑑𝑥 show that
LESSON 10
1
𝑛−1
𝑛
𝑥
𝐼𝑛 = sin 𝑥 cos 𝑛−1 𝑥 +
𝐼𝑛−2 .
Hence, find ∫ cos 5 𝑥 𝑑𝑥.
its lowest form, go back to the original integral &
plug in the final value of 𝑛.
𝑛=1
𝐼1 = ∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐
1
4 1
2
𝐼5 = sin 𝑥 cos 4 𝑥 + ( sin 𝑥 cos 2 𝑥 + sin 𝑥) + 𝑐
5
5 3
3
SOLUTION
STEP 1: Write the integral in the product form
𝐼𝑛 = ∫ cos1 𝑥 cos 𝑛−1 𝑥 𝑑𝑥
LESSON 11
that
2𝐼𝑛 = (𝑥 + 1)𝑛 𝑒 2𝑥 − 𝑛𝐼𝑛−1
STEP 2: Integrate by parts or an appropriate
SOLUTION
method
𝑢 = cos 𝑛−1 𝑥
If 𝐼𝑛 = ∫(𝑥 + 1)𝑛 𝑒 2𝑥 𝑑𝑥, show
𝑑𝑢 = (𝑛 − 1) cos 𝑛−2 𝑥 (− sin 𝑥)
𝐼𝑛 = ∫(𝑥 + 1)𝑛 𝑒 2𝑥 𝑑𝑥
P a g e | 26
𝑢 = (𝑥 + 1)𝑛
𝑑𝑢 = 𝑛(𝑥 + 1)𝑛−1
𝑑𝑣
1
= 𝑒 2𝑥
𝑣 = 𝑒 2𝑥
𝑑𝑥
2
1
1
𝐼𝑛 = (𝑥 + 1)𝑛 𝑒 2𝑥 − ∫ 𝑛(𝑥 + 1)𝑛−1 . 𝑒 2𝑥 𝑑𝑥
2
2
𝐼𝑛 =
1
𝑛
(𝑥 + 1)𝑛 𝑒 2𝑥 − ∫(𝑥 + 1)𝑛−1 𝑒 2𝑥 𝑑𝑥
2
2
𝐼𝑛 =
1
𝑛
(𝑥 + 1)𝑛 𝑒 2𝑥 − 𝐼𝑛−1
2
2
2𝐼𝑛 = (𝑥 + 1)𝑛 𝑒 2𝑥 − 𝑛𝐼𝑛−1
LESSON 12
It is given that
𝑒
𝐼𝑛 = ∫1 𝑥(ln 𝑥)𝑛 𝑑𝑥 , 𝑛 ≥ 0. By considering
𝑑
[𝑥 2 (ln 𝑥)𝑛 ] or otherwise, show that for
1
1
𝑛 ≥ 1, 𝐼𝑛 = 𝑒 2 − 𝑛𝐼𝑛−1
2
2
Hence, find 𝐼3 in terms of 𝑒
SOLUTION
𝑑𝑥
METHOD 1
𝑑 2
1
[𝑥 (ln 𝑥)𝑛 ] = 2𝑥(ln 𝑥)𝑛 + 𝑥 2 𝑛(ln 𝑥)𝑛−1 .
𝑑𝑥
𝑥
𝑑 2
[𝑥 (ln 𝑥)𝑛 ] = 2𝑥(ln 𝑥)𝑛 + 𝑛𝑥(ln 𝑥)𝑛−1
𝑑𝑥
Rearrange the equation
2𝑥(ln 𝑥)𝑛 =
𝑑 2
[𝑥 (ln 𝑥)𝑛 ] − 𝑛𝑥(ln 𝑥)𝑛−1
𝑑𝑥
Take integrals of both sides 𝑤. 𝑟. 𝑡 𝑥
𝑒
𝑒
2 ∫ 𝑥(ln 𝑥)𝑛 = ∫
1
1
𝑑 2
[𝑥 (ln 𝑥)𝑛 𝑑𝑥
𝑑𝑥
𝑒
1
3
3
= − 𝑒 2 + 𝑒 2 − 𝐼0
4
4
4
𝑒
1
3
= 𝑒 2 − [∫ 𝑥(ln 𝑥)0 𝑑𝑥]
2
4
1
1 2 3 1 2 1
𝑒 − [ 𝑒 − ]
2
4 2
2
1
3
3
= 𝑒2 − 𝑒2 +
2
8
8
1 2 3
= 𝑒 +
8
8
=
METHOD 2
𝑒
𝐼𝑛 = ∫ 𝑥(ln 𝑥)𝑛 𝑑𝑥
1
𝑢 = (ln 𝑥)𝑛
𝑑𝑢 = 𝑛(ln 𝑥)𝑛−1 .
𝑑𝑣
=𝑥
𝑑𝑥
𝑣=
𝑒
1
1
𝑒 𝑛
𝐼𝑛 = [ 𝑥 2 (ln 𝑥)𝑛 ] − 𝐼𝑛−1
1 2
2
1
𝑛
𝐼𝑛 = 𝑒 2 − 𝐼𝑛−1
2
2
𝐼3 is found the same way
LESSON 13
By using the substitution
𝑥 = 4 sin 𝜃, find the exact value of
2
∫
1
1
𝑒 1
∫ 𝑥(ln 𝑥) 𝑑𝑥 = [𝑥 2 (ln 𝑥)𝑛 ] − 𝑛𝐼𝑛−1
1 2
2
𝑛
1
1
𝑛
𝐼𝑛 = 𝑒 2 − 𝐼𝑛−1
2
2
1
3
𝐼3 = 𝑒 2 − 𝐼2
2
2
1
3 1
1
3
3
= 𝑒 2 − [ 𝑒 2 − 𝐼1 ] = 𝑒 2 − 𝑒 2 + 𝐼1
2
2 2
2
4
2
1
3 1
1
= − 𝑒 2 + [ 𝑒 2 − 𝐼0 ]
4
2 2
2
1 2
𝑥
2
1
𝑒 𝑛
𝐼𝑛 = [ 𝑥 2 (ln 𝑥)𝑛 ] − ∫ 𝑥(ln 𝑥)𝑛−1 𝑑𝑥
1 2
2
− 𝑛 ∫ 𝑥(ln 𝑥)𝑛−1 𝑑𝑥
𝑒
1
𝑥
0
1
√16 − 𝑥 2
𝑑𝑥
Show that
𝑑 𝑛−1
16(𝑛 − 1)𝑥 𝑛−2
𝑛𝑥 𝑛
√16 − 𝑥 2 ] =
[𝑥
−
𝑑𝑥
√16 − 𝑥 2
√16 − 𝑥 2
Deduce, or prove otherwise, that if
2
𝐼𝑛 = ∫0
𝑥𝑛
√16−𝑥 2
𝑑𝑥 , for 𝑛 ≥ 2, then
𝑛𝐼𝑛 = 16(𝑛 − 1)𝐼𝑛−2 − 2𝑛 √3. Hence find 𝐼2 .
SOLUTION
𝑥 = 4 sin 𝜃
𝑑𝑥
= 4 cos 𝜃
𝑑𝜃
𝑑𝑥 = 4 cos 𝜃 𝑑𝜃
𝜋
𝑥 = 2; 𝜃 =
6
P a g e | 27
2
𝑥 = 0; 𝜃 = 0
𝑛∫
2
∫
0
1
0
√16 −
𝑥2
𝜋
6
=∫
0
𝜋
6
=∫
0
𝜋
6
=∫
0
𝜋
6
=∫
0
𝜋
6
2
= 16(𝑛 − 1) ∫
0
1
√16 − (4 sin 𝜃)2
4 cos 𝜃 𝑑𝜃
√16 − 16 sin2 𝜃
4 cos 𝜃
√16(1 − sin2 𝜃)
𝑑𝜃
2
𝐼0 = ∫
0
𝑑 𝑛−1
[𝑥 √16 − 𝑥 2 ]
𝑑𝑥
1
1
= 𝑥 𝑛−1 . (16 − 𝑥 2 )2 . −2𝑥 + √16 − 𝑥 2 (𝑛 − 1)𝑥 𝑛−2
2
=
2𝑥 𝑛−1 𝑥
2√16 − 𝑥 2
𝑥𝑛
√16 − 𝑥 2
+ (𝑛 − 1)√16 − 𝑥 2 𝑥 𝑛−2
+
+
+
(𝑛 − 1)(16 − 𝑥 2 )𝑥 𝑛−2
√16 − 𝑥 2
(𝑛 − 1)(16𝑥 𝑛−2 − 𝑥 𝑛 )
√16 − 𝑥 2
16(𝑛 − 1)𝑥 𝑛−2
16(𝑛 − 1)𝑥 𝑛−2
√16 − 𝑥 2
1
√16 −
𝑥2
8𝜋
− 4√3
3
4𝜋
𝐼2 =
− 2√3
3
2𝐼2 =
𝜋
𝜋
= [𝜃] 6 =
0 6
=−
2
0
2𝐼2 = 16𝐼0 − 22 √3
0
√16 − 𝑥 2
0
𝑑 𝑛−1
√16 − 𝑥 2 ] 𝑑𝑥
[𝑥
𝑑𝑥
𝑛𝐼𝑛 = 16(𝑛 − 1)𝐼𝑛−2 − 2𝑛 √2
𝑑𝜃
𝑑𝜃
√1 − sin2 𝜃
𝑥𝑛
𝑑𝑥 − ∫
1
𝑛𝐼𝑛 = 16(𝑛 − 1)𝐼𝑛−2 − 2𝑛 . . √4. √3
2
= ∫ 1 𝑑𝜃
=−
√16 −
𝑥2
2
2𝐼2 = 16𝐼0 − 4√3
cos 𝜃
√16 − 𝑥 2
𝑥 𝑛−2
𝑛𝐼𝑛 = 16(𝑛 − 1)𝐼𝑛−2 − 2𝑛−1 √12
4 cos 𝜃
𝑥𝑛
𝑑𝑥
𝑛𝐼𝑛 = 16(𝑛 − 1)𝐼𝑛−2 − [𝑥 𝑛−1 √16 − 𝑥 2 ]
0
𝜋
6
=−
√16 − 𝑥 2
𝑑𝑥
cos 𝜃
=∫
𝑑𝜃
cos 𝜃
=−
𝑥𝑛
√16 − 𝑥 2
−
−
(𝑛 − 1)𝑥 𝑛
√16 − 𝑥 2
𝑛𝑥 𝑛
√16 − 𝑥 2
𝑑 𝑛−1
16(𝑛 − 1)𝑥 𝑛−2
𝑛𝑥 𝑛
√16 − 𝑥 2 ] =
[𝑥
−
𝑑𝑥
√16 − 𝑥 2
√16 − 𝑥 2
𝑥 2 𝜋
𝑑𝑥 = [sin−1 ( )] =
4 0 6
P a g e | 28
PARTIAL FRACTIONS
LESSON 2
Denominator with Linear Factors
LESSON 1
Express
2𝑥−7
𝑥 2 −𝑥−2
in partial
(a) Express
13𝑥+19
in partial fractions
𝑥 3 +2𝑥 2 −5𝑥−6
(b) Hence determine ∫
fractions.
SOLUTION
2𝑥 − 7
2𝑥 − 7
𝐴
𝐵
=
≡
+
2
𝑥 − 𝑥 − 2 (𝑥 − 2)(𝑥 + 1) 𝑥 − 2 𝑥 + 1
2𝑥 − 7 = 𝐴(𝑥 + 1) + 𝐵(𝑥 − 2)
When 𝑥 = −1
When 𝑥 = 2
−3𝐵 = −9
3𝐴 = −3
𝐵=3
𝐴 = −1
2𝑥 − 7
1
3
=−
+
−𝑥−2
𝑥−2 𝑥+1
2𝑥 − 7
1
3
∫ 2
𝑑𝑥 = ∫ −
𝑑𝑥 + ∫
𝑑𝑥
𝑥 −𝑥−2
𝑥−2
𝑥+1
𝑥2
= − ln|𝑥 − 2| + 3 ln|𝑥 + 1| + 𝑐
= ln |
(𝑥 + 1)3
|+𝑐
𝑥−2
2𝑥 2 + 11𝑥 + 3
𝐴
𝐵
𝐶
= +
+
𝑥(3𝑥 + 1)(𝑥 + 3) 𝑥 3𝑥 + 1 𝑥 + 3
2𝑥 2 + 11𝑥 + 3 = 𝐴(3𝑥 + 1)(𝑥 + 3) + 𝐵𝑥(𝑥 + 3) + 𝐶𝑥(3𝑥 + 1)
When 𝑥 = −3
24𝐶 = −12
1
𝐶=−
2
When 𝑥 = 0
3𝐴 = 3
𝐴=1
Comparing coefficients of 𝑥 2
3𝐴 + 𝐵 + 3𝐶 = 2
3
3+𝐵− =2
2
1
𝐵=
2
2𝑥 2 + 11𝑥 + 3
1
1
1
= +
−
𝑥(3𝑥 + 1)(𝑥 + 3) 𝑥 2(3𝑥 + 1) 2(𝑥 + 3)
13𝑥+19
𝑥 3 +2𝑥 2 −5𝑥−6
𝑑𝑥
SOLUTION
(a)
13𝑥+19
𝑥 3 +2𝑥 2 −5𝑥−6
After factorizing the denominator we get
13𝑥 + 19
13𝑥 + 19
=
3
2
(𝑥
𝑥 + 2𝑥 − 5𝑥 − 6
+ 1)(𝑥 − 2)(𝑥 + 3)
13𝑥 + 19
𝐴
𝐵
𝐶
=
+
+
(𝑥 + 1)(𝑥 − 2)(𝑥 + 3) 𝑥 + 1 𝑥 − 2 𝑥 + 3
13𝑥 + 19 = 𝐴(𝑥 − 2)(𝑥 + 3) + 𝐵(𝑥 + 1)(𝑥 +
3) +𝐶(𝑥 + 1)(𝑥 − 2)
When 𝑥 = 2
45 = 15𝐵
3=𝐵
When 𝑥 = −3
−20 = 10𝐶
−2 = 𝐶
When 𝑥 = −1
6 = −6𝐴
−1 = 𝐴
13𝑥 + 19
1
3
2
=−
+
−
3
2
𝑥 + 2𝑥 − 5𝑥 − 6
𝑥+1 𝑥−2 𝑥+3
(13𝑥+19)
(b) ∫ 3 2
𝑥 +2𝑥 −5𝑥−6
1
3
2
𝑑𝑥 + ∫
𝑑𝑥 − ∫
𝑑𝑥
𝑥+1
𝑥−2
𝑥+3
= − ln|𝑥 + 1| + 3 ln|𝑥 − 2| − 2 ln|𝑥 + 3| + 𝑐
= −∫
P a g e | 29
Denominator with unfactorizable
quadratic factor.
LESSON 1
Express(𝑥 2
𝑥 2 +5𝑥+4
+3𝑥+1)(𝑥+3)
5𝑥 2 +6𝑥+2
LESSON 3
Express (𝑥+2)(𝑥 2
+2𝑥+5)
in partial
fractions and hence determine
in partial
∫
fractions.
5𝑥 2 + 6𝑥 + 2
𝑑𝑥
(𝑥 + 2)(𝑥 2 + 2𝑥 + 5)
SOLUTION
SOLUTION
𝑥 2 + 5𝑥 + 4
𝐴𝑥 + 𝐵
𝐶
=
+
(𝑥 2 + 3𝑥 + 1)(𝑥 + 3) 𝑥 2 + 3𝑥 + 1 𝑥 + 3
5𝑥 2 + 6𝑥 + 2
𝐴
𝐵𝑥 + 𝐶
=
+
(𝑥 + 2)(𝑥 2 + 2𝑥 + 5) 𝑥 + 2 𝑥 2 + 2𝑥 + 5
𝑥 2 + 5𝑥 + 4 = (𝐴𝑥 + 𝐵)(𝑥 + 3) + 𝐶(𝑥 2 + 3𝑥 + 1)
5𝑥 2 + 6𝑥 + 2 = 𝐴(𝑥 2 + 2𝑥 + 5) + (𝐵𝑥 + 𝐶)(𝑥 + 2)
= (𝐴 + 𝐵)𝑥 2 + (2𝐴 + 2𝐵 + 𝐶)𝑥 + 5𝐴 + 2𝐶
= (𝐴 + 𝐶)𝑥 2 + (𝐵 + 3𝐴 + 3𝐶)𝑥 + 3𝐵 + 𝐶
𝐴+𝐵 =5
2(𝐴 + 𝐵) + 𝐶 = 6
2(5) + 𝐶 = 6
𝐶 = −4
5𝐴 + 2𝐶 = 2
5𝐴 − 8 = 10
𝐴+𝐶 =1
𝐵 + 3(𝐴 + 𝐶) = 5
𝐵 + 3(1) = 5
𝐵=2
3𝐵 + 𝐶 = 4
6+𝐶 =4
𝐶 = −2
𝐴+𝐶 =1
𝐴=3
𝑥 2 + 5𝑥 + 4
3𝑥 + 2
2
=
−
(𝑥 2 + 3𝑥 + 1)(𝑥 + 3) 𝑥 2 + 3𝑥 + 1 𝑥 + 3
LESSON 2
Express
2𝑥 2 −5𝑥+2
𝑥 3 +𝑥
𝐴=2
𝐴+𝐵 =5
𝐵=3
∫
5𝑥 2 + 6𝑥 + 2
𝑑𝑥
(𝑥 + 2)(𝑥 2 + 2𝑥 + 5)
=∫
in partial fractions
2
3𝑥 − 4
+ 2
𝑑𝑥
𝑥 + 2 𝑥 + 2𝑥 + 5
= ln(𝑥 + 2)2 − ∫
3(𝑥 + 1) − 7
𝑑𝑥
𝑥 2 + 2𝑥 + 5
SOLUTION
= ln(𝑥 + 2)2 − ∫
2𝑥 2 − 5𝑥 + 2 2𝑥 2 − 5𝑥 + 2
=
𝑥3 + 𝑥
𝑥(𝑥 2 + 1)
3(𝑥 + 1)
7
𝑑𝑥 − ∫ 2
𝑑𝑥
𝑥 2 + 2𝑥 + 5
𝑥 + 2𝑥 + 5
3
2(𝑥 + 1)
7
= ln(𝑥 + 2)2 − ∫ 2
𝑑𝑥 − ∫ 2
𝑑𝑥
2 𝑥 + 2𝑥 + 5
𝑥 + 2𝑥 + 5
2𝑥 2 − 5𝑥 + 2 𝐴 𝐵𝑥 + 𝐶
= + 2
𝑥(𝑥 2 + 1)
𝑥 𝑥 +1
3
1
= ln(𝑥 + 2)2 − ln|𝑥 2 + 2𝑥 + 5| − 7 ∫
𝑑𝑥
(𝑥 + 1)2 + 4
2
2𝑥 2 − 5𝑥 + 2 = 𝐴(𝑥 2 + 1) + (𝐵𝑥 + 𝐶)𝑥
3
1
= ln(𝑥 + 2)2 − ln|𝑥 2 + 2𝑥 + 5| − 7 ∫
𝑑𝑥
(𝑥
2
+ 1)2 + 22
and hence determine ∫
2𝑥 2 −5𝑥+2
𝑥 3 +𝑥
𝑑𝑥.
= (𝐴 + 𝐵)𝑥 2 + 𝐶𝑥 + 𝐴
𝐴=2
𝐶 = −5
𝐴+𝐵 =2
2+𝐵 =2
𝐵=0
∫
2𝑥 2 − 5𝑥 + 2
2
5
𝑑𝑥 = ∫ − 2
𝑑𝑥
𝑥3 + 𝑥
𝑥 𝑥 +1
= ln 𝑥 2 − 5 tan−1 𝑥 + 𝑐
3
7
= ln(𝑥 + 2)2 − ln|𝑥 2 + 2𝑥 + 5| − tan−1 (𝑥 + 1) + 𝑐
2
2
P a g e | 30
Denominator with a repeated factor
LESSON 1
Express
𝑥 2 −3𝑥−9
𝑥 3 −6𝑥 2 +9𝑥
in partial
SOLUTION
𝑥 2 − 3𝑥 − 9
𝑥 2 − 3𝑥 − 9
𝑥 2 − 3𝑥 − 9
=
=
𝑥 3 − 6𝑥 2 + 9𝑥 𝑥(𝑥 2 − 6𝑥 + 9)
𝑥(𝑥 − 3)2
𝑥 2 − 3𝑥 − 9 𝐴
𝐵
𝐶
= +
+
2
𝑥(𝑥 − 3)
𝑥 𝑥 − 3 (𝑥 − 3)2
𝑥 2 − 3𝑥 − 9 = 𝐴(𝑥 − 3)2 + 𝐵𝑥(𝑥 − 3) + 𝐶𝑥
When 𝑥 = 3
When 𝑥 = 0
−9 = 3𝐶
−9 = 9𝐴
−3 = 𝐶
−1 = 𝐴
Equating coefficients of 𝑥 2
1=𝐴+𝐵
1 = −1 + 𝐵
2=𝐵
𝑥 2 − 3𝑥 − 9
1
2
3
=− +
−
𝑥 3 − 6𝑥 2 + 9𝑥
𝑥 𝑥 − 3 (𝑥 − 3)2
Express
2𝑥 2 +9𝑥+24
𝑥 3 +4𝑥 2 −3𝑥−18
in partial
fractions and hence determine
∫
5𝑥 2 + 6𝑥 + 2
𝑑𝑥
(𝑥 + 2)(𝑥 2 + 2𝑥 + 5)
SOLUTION
2𝑥 2 + 9𝑥 + 24
2𝑥 2 + 9𝑥 + 24
=
𝑥 3 + 4𝑥 2 − 3𝑥 − 18 (𝑥 − 2)(𝑥 + 3)2
2𝑥 2 + 9𝑥 + 2𝑥
𝐴
𝐵
𝐶
=
+
+
(𝑥 − 2)(𝑥 + 3)2 𝑥 − 2 𝑥 + 3 (𝑥 + 3)2
2𝑥 2 + 9𝑥 + 24 = 𝐴(𝑥 + 3)2 + 𝐵(𝑥 − 2)(𝑥 + 3) + 𝐶(𝑥 − 2)
When 𝑥 = −3
When 𝑥 = 2
15 = −5𝐶
50 = 25𝐴
−3 = 𝐶
2=𝐴
Equating coefficients of 𝑥 2
2=𝐴+𝐵
2= 2+𝐵
0=𝐵
2𝑥 2 + 9𝑥 + 24
2
3
=
−
3
2
(𝑥
𝑥 + 4𝑥 − 3𝑥 − 18 𝑥 − 2
+ 3)2
∫
2𝑥 2 + 9𝑥 + 24
𝑑𝑥
𝑥 3 + 4𝑥 2 − 3𝑥 − 18
2
𝑑𝑥 − ∫ 3(𝑥 + 3)−2 𝑑𝑥
𝑥−2
= 2 ln|𝑥 − 2| +
fractions
LESSON 2
=∫
3
+𝑐
𝑥+3
P a g e | 31
Improper Fractions (degree of
numerator ≥ degree of denominator)
If for
𝑃(𝑥)
𝑄(𝑥)
, 𝑃(𝑥) has degree 𝑛 and 𝑄(𝑥) has degree
𝑚, then quotient has degree 𝑛 − 𝑚
LESSON 1
Express
2𝑥 3 +3𝑥 2 −𝑥−4
𝑥 2 (𝑥+1)
in partial
fractions and hence determine
∫
2𝑥 3 + 3𝑥 2 − 𝑥 − 4
𝑑𝑥
𝑥 2 (𝑥 + 1)
SOLUTION
2𝑥 3 + 3𝑥 2 − 𝑥 − 4
𝐵 𝐶
𝐷
=𝐴+ + 2+
2
𝑥 (𝑥 + 1)
𝑥 𝑥
𝑥+1
2𝑥 3 + 3𝑥 2 − 𝑥 − 4 = 𝐴(𝑥 2 )(𝑥 + 1) + 𝐵𝑥(𝑥 + 1) + 𝐶(𝑥 + 1)
+ 𝐷𝑥 2
= 𝐴𝑥 3 + (𝐴 + 𝐵 + 𝐷)𝑥 2 + (𝐵 + 𝐶)𝑥 + 𝐶
𝐴=2
SOLUTION
3𝑥 2 + 2
𝐵
𝐶
=𝐴+
+
(2𝑥 + 1)(𝑥 − 2)
2𝑥 + 1 𝑥 − 2
3𝑥 2 + 2 = 𝐴(2𝑥 + 1)(𝑥 − 2) + 𝐵(𝑥 − 2) + 𝐶(2𝑥 + 1)
1
When 𝑥 = 2
14 = 5𝐶
14
=𝐶
5
When 𝑥 = −
2
11
5
=− 𝐵
4
2
11
−
=𝐵
10
3
Equating coefficients of 𝑥 2 3 = 2𝐴 → = 𝐴
2
3𝑥 2 + 2
3
11
14
= −
+
(2𝑥 + 1)(𝑥 − 2) 2 10(2𝑥 + 1) 5(𝑥 − 2)
∫
3𝑥 2 + 2
𝑑𝑥
(2𝑥 + 1)(𝑥 − 2)
3
11
2
14
1
= ∫ 𝑑𝑥 −
∫
𝑑𝑥 +
∫
𝑑𝑥
2
20 2𝑥 + 1
5 𝑥−2
=
3𝑥 11
14
−
ln|2𝑥 + 1| +
ln|𝑥 − 2| + 𝑐
2 20
5
𝐶 = −4
LESSON 3
𝐵 + 𝐶 = −1
𝐵=3
∫
𝐴+𝐵+𝐷 =3
(𝑥+3)(𝑥+2)(𝑥+1 )
in
𝑥 4 + 𝑥 3 − 19𝑥 2 − 44𝑥 − 21
𝑑𝑥
(𝑥 + 3)(𝑥 + 2)(𝑥 + 1 )
SOLUTION
2+3+𝐷 = 3
𝐷 = −2
2𝑥 3 + 3𝑥 2 − 𝑥 − 4
3 4
2
=2+ − 2−
2
𝑥 (𝑥 + 1)
𝑥 𝑥
𝑥+1
2𝑥 3 + 3𝑥 2 − 𝑥 − 4
𝑑𝑥
𝑥 2 (𝑥 + 1)
𝑥 4 + 𝑥 3 − 19𝑥 2 − 44𝑥 − 21
(𝑥 + 3)(𝑥 + 2)(𝑥 + 1 )
= 𝐴𝑥 + 𝐵 +
𝐶
𝐷
𝐸
+
+
𝑥+3 𝑥+2 𝑥+1
𝑥 4 + 𝑥 3 − 19𝑥 2 − 44𝑥 − 21
3
2
= ∫ 2 𝑑𝑥 + ∫ 𝑑𝑥 − ∫ 4𝑥 −2 𝑑𝑥 − ∫
𝑑𝑥
𝑥
𝑥+1
4
= 2𝑥 + ln 𝑥 + − 2 ln|𝑥 + 1| + 𝑐
𝑥
LESSON 2
𝑥 4 +𝑥 3 −19𝑥 2 −44𝑥−21
partial fractions and hence determine
𝐵 − 4 = −1
∫
Express
= 𝐴𝑥 + 𝐵 +
𝐶
𝐷
𝐸
+
+
𝑥+3 𝑥+2 𝑥+1
𝑥 4 + 𝑥 3 − 19𝑥 2 − 44𝑥 − 21
= (𝐴𝑥 + 𝐵)(𝑥 + 3)(𝑥 + 2)(𝑥 + 1) + 𝐶(𝑥 + 2)(𝑥 + 1)
+𝐷(𝑥 + 3)(𝑥 + 1) + 𝐸(𝑥 + 3)(𝑥 + 2)
3𝑥 2 +2
Express (2𝑥+1)(𝑥−2) in partial
3𝑥 2 +2
fractions and hence determine ∫ (2𝑥+1)(𝑥−2) 𝑑𝑥
When 𝑥 = −3
−6 = 2𝐶
−3 = 𝑐
When 𝑥 = −1
4 = 2𝐸
2=𝐸
When 𝑥 = −2
−1 = −𝐷
1=𝐷
P a g e | 32
Equating coefficients of 𝑥 4
1=𝐴
Equating coefficients of 𝑥 3
1 = 6𝐴 + 𝐵
1= 6+𝐵
−5 = 𝐵
𝑥 4 + 𝑥 3 − 19𝑥 2 − 44𝑥 − 21
(𝑥 + 3)(𝑥 + 2)(𝑥 + 1)
=𝑥−5−
∫
3
1
2
+
+
𝑥+3 𝑥−2 𝑥+1
𝑥 4 + 𝑥 3 − 19𝑥 2 − 44𝑥 − 21
𝑑𝑥
(𝑥 + 3)(𝑥 + 2)(𝑥 + 1)
= ∫(𝑥 − 5) 𝑑𝑥 − 3 ∫
=
1
1
1
𝑑𝑥 + ∫
𝑑𝑥 + 2 ∫
𝑑𝑥
𝑥+3
𝑥−2
𝑥+1
𝑥2
− 5𝑥 − 3 ln|𝑥 + 3| + ln|𝑥 − 2| + 2 ln|𝑥 + 1| + 𝑐
2
P a g e | 33
TRAPEZIUM RULE (NUMERICAL INTEGRATION)
Introduction
The area under the curve 𝑓(𝑥) = 𝑥 2 + 1 can be estimated by finding the sum of the areas of trapeziums of
equal width (as shown above)
ℎ
The area of a trapezium with parallel sides 𝑦0 and 𝑦1 and width ℎ is given by the formula (𝑦0 + 𝑦1 )
2
𝑏−𝑎
The width of each trapezium is
where 𝑛 is the number of trapeziums.
𝑛
Thus we have area under curve is
ℎ
ℎ
ℎ
(𝑦0 + 𝑦1 ) + (𝑦1 + 𝑦2 ) + ⋯ + (𝑦𝑛−1 + 𝑦𝑛 )
2
2
2
ℎ
= (𝑦0 + 𝑦1 + 𝑦1 + 𝑦2 + ⋯ + 𝑦𝑛−1 + 𝑦𝑛 )
2
ℎ
= [(𝑦0 + 𝑦𝑛 ) + 2(𝑦1 + 𝑦2 + ⋯ + 𝑦𝑛−1 )]
2
𝑏
∫ 𝑦 𝑑𝑥 ≈
𝑎
(𝑏 − 𝑎)
[(𝑦0 + 𝑦𝑛 ) + 2(𝑦1 + 𝑦2 + ⋯ + 𝑦𝑛−1 )]
2𝑛
LESSON 1
Using 5 trapeziums, estimate
2
∫ 𝑥 2 + 1 𝑑𝑥
SOLUTION
Width of each trapezium is
2
𝑦0 = 𝑓(0) = 0 + 1 = 1
𝑦1 = 𝑓(0.4) = 0.42 + 1 = 1.16
𝑦2 = 𝑓(0.8) = 0.82 + 1 = 1.64
𝑦3 = 𝑓(1.2) = 1.22 + 1 = 2.44
𝑦4 = 𝑓(1.6) = 1.62 + 1 = 3.56
𝑦5 = 𝑓(2) = 22 + 1 = 5
2
∫ 𝑥 2 + 1 𝑑𝑥 ≈
0
2−0
5
0
= 0.4
2−0
[(1 + 5) + 2(1.16 + 1.64 + 2.44 + 3.56)]
2(5)
≈ 4.72
P a g e | 34
COMPLEX NUMBERS
INTRODUCTION
Dividing Complex Numbers
Complex numbers are written in the form 𝑎 + 𝑖𝑏
where 𝑎 and 𝑏 are real numbers and 𝑖 is the
imaginary unit such that
If 𝑧 = 𝑎 + 𝑏𝑖 then 𝑧̅ = 𝑎 − 𝑏𝑖 is its conjugate and vice
versa. It is also important to note that the product of a
complex number and its conjugate is 𝑎2 + 𝑏 2 which is
always a Real number. The conjugate is also denoted
𝑧∗.
𝑖 = √−1
or 𝑖 2 = −1
Sometimes the letter 𝑧 is used to denote a complex
number, 𝑧: 𝑧 = 𝑎 + 𝑖𝑏. A complex number can also
be written as an ordered pair of its real numbers,
(𝑎, 𝑏).
𝑎 is also known as the real part i.e. 𝑅𝑒(𝑧) = 𝑎
𝑏 is also known as the imaginary part i.e. 𝐼𝑚(𝑧) = 𝑏
SQUARE ROOT OF NEGATIVE
NUMBERS
Thus complex numbers can be used to find the
square roots of negative numbers. Examples
(i)
3𝑖
2+𝑖
First we find the conjugate of the denominator
𝑧 = 2 + 𝑖 → 𝑧̅ = 2 − 𝑖
Multiply the numerator and the denominator by
the conjugate
3𝑖(2 − 𝑖)
(2 + 𝑖)(2 − 𝑖)
6𝑖 − 3𝑖 2
= 2
2 + 12
6𝑖 + 3
=
5
3 6𝑖
= +
5 5
√−16 = √(−1)16 = √−1√16 = 4𝑖
(ii) Express
√−18 = √(9)(2)(−1) = 3√2𝑖
With this extension of the number system we can
now solve equations which we once unsolvable.
Operations on Complex Numbers
Adding and Subtracting Complex Numbers
(𝑎 + 𝑏𝑖) ± (𝑐 + 𝑑𝑖) = (𝑎 ± 𝑐) + (𝑏 ± 𝑑)𝑖
For example,
(i) (6 + 2i) + (5 − 4i) = (6 + 5) + (2 − 4)𝑖
= 11 − 2𝑖
(ii) (−1 − 𝑖) − (8 − 2𝑖) = (−1 − 8) + (−1 + 2)𝑖
= −9 + 𝑖
Multiplying Complex Numbers
(i) 2𝑖(3𝑖) = 6𝑖 2 = 6(−1) = −6
(ii) −2𝑖(4 + 3𝑖) = −8𝑖 − 6𝑖 2 = 6 − 8𝑖
(iii) (3 + 2𝑖)(5 − 4𝑖) = 15 − 12𝑖 + 10𝑖 − 8𝑖 2
= 15 + 8 − 12𝑖 + 10𝑖
= 23 − 2𝑖
(iv) (𝑎 + 𝑏𝑖)(𝑎 − 𝑏𝑖) = 𝑎2 − 𝑏𝑖 2 = 𝑎2 + 𝑏 2
3+7𝑖
5−2𝑖
in the form 𝑎 + 𝑏𝑖
3 + 7𝑖
5 − 2𝑖
(3 + 7𝑖)(5 + 2𝑖)
=
(5 − 2𝑖)(5 + 2𝑖)
15 + 41𝑖 + 14𝑖 2
=
52 + 22
1 + 41𝑖
=
29
1 41
=
+
𝑖
29 29
P a g e | 35
SOLUTION
Square Roots of Complex Numbers
LESSON 1
(a) 𝑥 2 + 1 = 0
𝑥 2 = −1
𝑥 = ±√−1
𝑥 = ±𝑖
Find √15 + 8𝑖
SOLUTION
We assume that the square root of a
complex number is a complex number
√15 + 8𝑖 = 𝑎 + 𝑏𝑖
15 + 8𝑖 = (𝑎 + 𝑏𝑖)2
15 + 8𝑖 = 𝑎2 − 𝑏 2 + 2𝑎𝑏𝑖
𝑎2 − 𝑏 2 = 15
2𝑎𝑏 = 8
(b) 𝑥 2 + 3𝑥 + 3 = 0
−3 ± √32 − 4(1)(3)
2(1)
−3 ± √−3
𝑥=
2
−3 ± √3𝑖
𝑥=
2
(c) 4𝑥 2 − 2𝑥 = −1
4𝑥 2 − 2𝑥 + 1 = 0
𝑥=
−(−2) ± √(−2)2 − 4(4)(1)
2(4)
2 ± √−12
=
8
2 ± √4(3)(−1)
=
8
2 ± 2√3𝑖
=
8
1 √3
= ±
𝑖
4
4
4
𝑎=
𝑏
𝑥=
4 2
( ) − 𝑏 2 = 15
𝑏
𝑥
𝑥
16
− 𝑏 2 = 15
𝑏2
𝑥
𝑏 4 + 15𝑏 2 − 16 = 0
𝑥
(𝑏 2 + 16)(𝑏 2 − 1) = 0
𝑏 2 = −16 Invalid since 𝑏 is real
Equations with Complex Coefficients
𝑏2 = 1
LESSON 2
𝑏 = ±1
Determine 𝑧 such that
𝑧 2 + (2 + 2𝑖)𝑧 − (15 − 10𝑖) = 0
𝑎 = ±4
SOLUTION
√15 + 8𝑖 = 4 + 𝑖 or − 4 − 𝑖
𝑧 2 + (2 + 2𝑖)𝑧 − (15 − 10𝑖) = 0
𝑧 2 + (2 + 2𝑖)𝑧 − 15 + 10𝑖 = 0
= ±(4 + 𝑖)
Thus we see that a complex number has 2 square
roots, which are complex numbers.
Quadratic Equations
Equations with Real Coefficients
LESSON 1
Solve the following equations
(a) 𝑥 2 + 1 = 0
(b) 𝑥 2 + 3𝑥 + 3 = 0
(c) 4𝑥 2 − 2𝑥 = −1
𝑧=
−(2 + 2𝑖) ± √(2 + 2𝑖)2 − 4(1)(−15 + 10𝑖)
2(1)
−(2 + 2𝑖) ± √(2 + 2𝑖)(2 + 2𝑖) + 60 − 40𝑖
2
−(2 + 2𝑖) ± √8𝑖 + 60 − 40𝑖
𝑧=
2
−(2 + 2𝑖) ± √60 − 32𝑖
𝑧=
2
𝑧=
√60 − 32𝑖 = 𝑎 + 𝑏𝑖
60 − 32𝑖 = (𝑎 + 𝑏𝑖)2
P a g e | 36
60 − 32𝑖 = 𝑎2 − 𝑏 2 + 2𝑎𝑏𝑖
𝑎2 − 𝑏 2 = 60
16
𝑎𝑏 = −16 → 𝑎 = −
𝑏
If 𝑥 2 + 𝑥 + = 0, then 𝛼 + 𝛽 = − and 𝛼𝛽 =
𝑎
𝑎
𝑎
𝑎
where 𝛼 and 𝛽 are the roots of the equation
16 2
(− ) − 𝑏 2 = 60
𝑏
256
− 𝑏 2 = 60
𝑏2
𝑏 4 + 60𝑏 2 − 256 = 0
(𝑏 2 + 64)(𝑏 2 − 4) = 0
𝑏2 = 4
𝑏 = ±2
Also, if 𝑥 3 + 𝑥 2 + 𝑥 + = 0
𝑏
𝑐
𝑑
𝑎
𝑎
𝑏
𝑐
𝑎
𝑎
and
𝑑
LESSON 1
Given one root find the equation
(i) 5𝑖
(ii) 4 − 3𝑖
SOLUTION
(i) Let 𝛼 = 5𝑖 then 𝛽 = −5𝑖
𝛼 + 𝛽 = (5𝑖) + (−5𝑖) = 0
𝛼𝛽 = (5𝑖)(−5𝑖) = 25
Equation is 𝑥 2 + 25 = 0
(ii) Let 𝛼 = 4 − 3𝑖 then 𝛽 = 4 + 3𝑖
𝛼 + 𝛽 = (4 − 3𝑖) + (4 + 3𝑖) = 8
𝛼𝛽 = (4 − 3𝑖)(4 + 3𝑖) = 25
Equation is 𝑥 2 − 8𝑥 + 25 = 0
LESSON 2
Given that 1 − 2𝑖 is a root of the
equation 𝑥 3 + 𝑥 2 − 𝑥 + 15 = 0, find the other 2
roots.
Roots of Equations
SOLUTION
Since complex roots occur in
conjugate pairs and a cubic polynomial has 3 roots
one root must be real.
For the equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
−𝑏±√−𝑝
2𝑎
𝑏 √𝑝
±
𝑖
2𝑎 2𝑎
As a result we can conclude that if a quadratic
equation has complex roots they occur in conjugates.
In general, if a polynomial has complex roots they
occur in conjugate pairs. For example, if 2 + 𝑖 is the
root of a polynomial equation then 2 − 𝑖 is also a root
of the same equation.
Recall:
𝑐
𝛼𝛽𝛾 = − where 𝛼, 𝛽 and 𝛾 are the roots of the
𝑎
equation.
−(2 + 2𝑖) ± (−8 + 2𝑖)
2
−(2 + 2𝑖) + (−8 + 2𝑖)
𝑧=
= −5
2
−(2 + 2𝑖) − (−8 + 2𝑖)
𝑧=
= 3 − 2𝑖
2
𝑥=−
𝑏
𝑎
then 𝛼 + 𝛽 + 𝛾 = − , 𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾 =
𝑧=
Letting −𝑝 = 𝑏 2 − 4𝑎𝑐, 𝑝 > 0, we have 𝑥 =
𝑏
i.e. 𝑥 2 − (sum of roots)𝑥 + (product of roots) = 0
When 𝑏 = 2
16
𝑎=−
= −8
2
−8 + 2𝑖
When 𝑏 = −2
16
𝑎=−
=8
−2
8 − 2𝑖
𝑥=
𝑐
Let 𝛼 = 1 − 2𝑖, 𝛽 = 1 + 2𝑖, 𝛾 ∈ ℝ
𝛼 + 𝛽 + 𝛾 = −1
(1 − 2𝑖) + (1 + 2𝑖) + (𝛾) = −1
2 + 𝛾 = −1
𝛾 = −3
P a g e | 37
Argand Diagram
INTRODUCTION
A complex number 𝑧 = 𝑎 + 𝑏𝑖 can be represented
on a diagram called an Argand diagram as
(i) a point with coordinates (𝑎, 𝑏)
(ii) a vector
Modulus – Argument Form
Representing Sums and Differences on Argand
Diagrams
The Modulus of a Complex Number
LESSON 1
Find 𝑧1 + 𝑧2 and 𝑧1 − 𝑧2 for
𝑧1 = 5 + 8𝑖 and 𝑧2 = 2 + 𝑖. Hence, represent
𝑧1 + 𝑧2 and 𝑧1 − 𝑧2 on Argand diagrams.
SOLUTION
(i) 𝑧1 + 𝑧2 = (5 + 8𝑖) + (2 + 𝑖) = 7 + 9𝑖
(ii) 𝑧1 − 𝑧2 = (5 + 8𝑖) − (2 + 𝑖) = 3 + 7𝑖
The modulus of a complex number, 𝑧 = 𝑎 + 𝑏𝑖, is a
measure of the magnitude of 𝑧, and is written as
|𝑧|.
Thus modulus 𝑧 = |𝑧| = √𝑎2 + 𝑏 2 .
LESSON 1
(a)
(b)
(c)
(d)
𝑧1
𝑧2
𝑧3
𝑧4
Determine the modulus of
=1+𝑖
= −3 + 4𝑖
= −1 − √3𝑖
= −5
SOLUTION
(a) 𝑧1 = 1 + 𝑖
|𝑧1 | = √12 + 12 = √2
P a g e | 38
(b) 𝑧2 = −3 + 4𝑖
|𝑧2 | = |−3 + 4𝑖| =
SOLUTION
√(−3)2
+
42
=5
(a) 𝑧 = 1 + 𝑖
arg 𝑧1 = 𝜃
(c) 𝑧3 = −1 − √3𝑖
2
|𝑧3 | = |−1 − √3𝑖| = √(−1)2 + (−√3) = 2
(d) 𝑧4 = −5
|𝑧4 | = 5
LESSON 2
If 𝑧1 = −3 + 4𝑖 and 𝑧2 = 2 − 𝑖,
what is |𝑧1 − 𝑧2 |?
SOLUTION
We are trying to find the distance between 𝑧1 and
𝑧2 . In other words, what is the distance between
the points 𝑃(−3, 4) and 𝑄(2, −1) on the Argand
Diagram?
1
= tan−1 ( )
1
𝜋
=
4
(b) 𝑧 = −3 + 4𝑖
arg 𝑧2 = 𝜃
4
= 𝜋 − tan−1 ( )
3
= 2.21
(c) 𝑧 = −1 − √3𝑖
arg 𝑧3 = 𝜃
= −𝜋 + tan−1 (
=−
2𝜋
3
(d) 𝑧4 = −5
arg 𝑧4 = 𝜋
|𝑧1 − 𝑧2 | = |(−3 + 4𝑖) − (2 − 𝑖)|
= |(−3 − 2) − (−4 − 1)𝑖|
= √(−3 − 2)2 + (−4 − 1)2
= √50
The Argument of Complex Number
The angle 𝜃 is called the argument of 𝑧 (arg 𝑧)
where 𝜃 is the angle the vector representing the
complex number on the Argand diagram makes
𝑏
with the positive real axis. Thus tan 𝜃 = . To
𝑎
avoid complications we use – 𝜋 < 𝜃 ≤ 𝜋 and this
is known as the principal argument of 𝑧.
LESSON 1
(a)
(b)
(c)
(d)
𝑧1
𝑧2
𝑧3
𝑧4
Determine the argument of
=1+𝑖
= −3 + 4𝑖
= −1 − √3𝑖
= −5
√3
)
1
P a g e | 39
𝜋 𝜋
𝜋 𝜋
= 5 (cos ( − ) − 𝑖 sin ( − ))
2 3
2 3
𝜋
𝜋
= 5 (cos ( ) − 𝑖 sin ( ))
6
6
𝜋
𝜋
= 5 (cos (− ) + 𝑖 sin (− ))
6
6
|𝑧| = 5
𝜋
arg 𝑧 = −
6
Modulus – Argument Form
If 𝑧 = 𝑎 + 𝑏𝑖 has modulus 𝑟 and argument 𝜃 then
𝑎 = 𝑟 cos 𝜃 and 𝑏 = 𝑟 sin 𝜃
∴ 𝑎 + 𝑏𝑖 = 𝑟 cos 𝜃 + 𝑖 sin 𝜃
= 𝑟(cos 𝜃 + 𝑖 sin 𝜃)
Therefore, 𝑧 = 1 + 𝑖 in modulus – argument form
is
𝜋
𝜋
√2 (cos ( ) + 𝑖 sin ( ))
4
4
LESSON 1
Write the following in modulus –
argument form
LESSON 3
Prove that for 𝑧1 = cos 𝐴 + 𝑖 sin 𝐴
and 𝑧2 = cos 𝐵 + 𝑖 sin 𝐵
(a) 𝑧1 𝑧2 = 𝑟1 𝑟2 [cos(𝐴 + 𝐵) + 𝑖 sin(𝐴 + 𝐵)]
𝑧
𝑟
(b) 1 = 1 [cos(𝐴 − 𝐵) + 𝑖 sin(𝐴 − 𝐵)]
𝑧2
(a)
(b)
(c)
(d)
𝑧1
𝑧2
𝑧3
𝑧4
=1+𝑖
= −3 + 4𝑖
= −1 − √3𝑖
= −5
SOLUTION
SOLUTION
(i) 𝑧1 = 1 + 𝑖
𝑧1 = √2 (cos
𝜋
𝜋
+ 𝑖 sin )
4
4
(ii) 𝑧2 = −3 + 4𝑖
𝑧2 = 5(cos(2.21) + 𝑖 sin(2.21))
(iii) 𝑧3 = −1 − √3𝑖
2𝜋
2𝜋
𝑧3 = 2 (cos (− ) + 𝑖 sin (− ))
3
3
(iv) 𝑧4 = −5
𝑧4 = 5(cos 𝜋 + 𝑖 sin 𝜋)
LESSON 2
Find the modulus and argument
of the following
3𝜋
𝑟2
(a) 𝑧1 𝑧2 = 𝑟1 [cos 𝐴 + 𝑖 sin 𝐴]𝑟2 [cos 𝐵 + 𝑖 sin 𝐵]
= 𝑟1 𝑟2 [cos 𝐴 cos 𝐵 + 𝑖 cos 𝐴 sin 𝐵
+ 𝑖 sin 𝐴 cos 𝐵
+ 𝑖 2 sin 𝐴 sin 𝐵]
= 𝑟1 𝑟2 [cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵
+(cos 𝐴 sin 𝐵 + sin 𝐴 cos 𝐵)
= 𝑟1 𝑟2 [cos(𝐴 + 𝐵) + 𝑖 sin(𝐴 + 𝐵)]
SOLUTION
(b)
𝑟 [cos 𝐴+𝑖 sin 𝐴]
𝑧1
𝑧2
= 𝑟1[cos 𝐵+𝑖 sin 𝐵]
2
𝑟1 [cos 𝐴 + 𝑖 sin 𝐴][cos 𝐵 − 𝑖 sin 𝐵]
=
𝑟2 [cos 𝐵 + 𝑖 sin 𝐵][cos 𝐵 − 𝑖 sin 𝐵]
=
𝑟1 [cos 𝐴 cos 𝐵 − 𝑖 cos 𝐴 sin 𝐵 + 𝑖 sin 𝐴 cos 𝐵 − 𝑖 2 sin 𝐴 sin 𝐵]
𝑟2 [cos2 𝐵 + sin2 𝐵]
=
𝑟1 [cos 𝐴 cos 𝐵 + sin 𝐴 sin 𝐵 + (sin 𝐴 cos 𝐵 − cos 𝐴 sin 𝐵)𝑖
𝑟2
=
𝑟1
[cos(𝐴 − 𝐵) + 𝑖 sin(𝐴 − 𝐵)]
𝑟2
NB:
3𝜋
If 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃) then
(i) 𝑧 = cos ( ) + 𝑖 sin ( )
5
𝜋
5
𝑧̅ = 𝑟(cos 𝜃 − 𝑖 sin 𝜃)
𝜋
(ii) 𝑧 = 5 sin ( ) − 5𝑖 cos ( )
3
3
= 𝑟(cos(−𝜃) + 𝑖 sin(−𝜃) )
SOLUTION
3𝜋
3𝜋
5
5
(i) 𝑧 = cos ( ) + 𝑖 sin ( )
|𝑧| = 1
3𝜋
5
𝜋
𝜋
(ii) 𝑧 = 5 sin ( ) − 5𝑖 cos ( )
3
3
𝜋
𝜋
= 5 (sin ( ) − 𝑖 cos ( ))
3
3
arg 𝑧 =
Furthermore, we can conclude
(a) |𝑧1 𝑧2 | = 𝑟1 𝑟2 = |𝑧1 ||𝑧2 |
𝑧
𝑟1
𝑧2
𝑟2
(b) | 1| =
|𝑧 |
= |𝑧1|
2
(c) arg(𝑧1 𝑧2 ) = 𝐴 + 𝐵 = arg 𝑧1 + arg 𝑧2
𝑧
(d) arg ( 1) = 𝐴 − 𝐵 = arg 𝑧1 − arg 𝑧2
𝑧2
P a g e | 40
LESSON 4
Given that 𝑧1 = 1 + √3𝑖 and
𝑧2 = 1 − 𝑖, determine (i) 𝑧1 𝑧2 and (ii)
𝑧1
𝑧2
in modulus –
argument form.
SOLUTION
2
|𝑧1 | = √12 + (√3) = 2
arg 𝑧1 = tan−1 (
𝜋
√3
)=
1
3
|𝑧2 | = √12 + 12 = √2
1
𝜋
arg 𝑧2 = − tan−1 ( ) =
1
4
𝜋
𝜋
𝜋
𝜋
3
4
3
4
(i) 𝑧1 𝑧2 = 2√2 [cos ( + ) + 𝑖 sin ( + )]
(ii)
𝑧1
𝑧2
7𝜋
7𝜋
= 2√2 [cos
+ 𝑖 sin ]
12
12
2
𝜋
𝜋
𝜋
𝜋
= [cos ( − ) + 𝑖 sin ( − )]
3
4
3
4
√2
𝜋
𝜋
= √2 [cos
+ 𝑖 sin ]
12
12
P a g e | 41
De Moivre’s Theorem
INTRODUCTION
𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃) then
𝑧 2 = 𝑟 2 (cos 2 𝜃 − sin2 𝜃 + 2𝑖 sin 𝜃 cos 𝜃)
= 𝑟 2 (cos 2𝜃 + 𝑖 sin 2𝜃)
This can be extended to give
𝑧 𝑛 = 𝑟 𝑛 (cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃)
LESSON 1
Use De Moivre’s Theorem to prove the following identities
cos 4𝜃 ≡ 8 cos 4 𝜃 − 8 cos 2 𝜃 + 1
SOLUTION
When 𝑛 = 4
cos 4𝜃 + 𝑖 sin 4𝜃 = (cos 𝜃 + 𝑖 sin 𝜃)4
= cos 4 𝜃 + 4 cos 3 𝜃 𝑖 sin 𝜃 + 6 cos 2 𝜃 𝑖 2 sin2 𝜃 + 4 cos 𝜃 𝑖 3 sin3 𝜃 + 𝑖 4 sin4 𝜃
= cos 4 𝜃 + 4𝑖 cos 3 𝜃 sin 𝜃 − 6 cos 2 𝜃 sin2 𝜃 − 4𝑖 cos 𝜃 sin3 𝜃 + sin4 𝜃
Equating real parts
cos 4𝜃 = cos 4 𝜃 − 6 cos 2 sin2 𝜃 + sin4 𝜃
= cos 4 𝜃 − 6 cos 2 𝜃 (1 − cos 2 𝜃) + (1 − cos 2 𝜃)(1 − cos 2 𝜃)
= cos 4 𝜃 − 6 cos 2 𝜃 + 6 cos 4 𝜃 + cos 4 𝜃 − 2 cos 2 𝜃 + 1
= 8 cos 4 𝜃 − 8 cos 2 𝜃 + 1
LESSON 2
Use de Moivre’s theorem to show that
sin 5𝜃 = 𝑎 cos 4 𝜃 sin 𝜃 + 𝑏 cos 2 𝜃 sin3 𝜃 + 𝑐 sin5 𝜃
where 𝑎, 𝑏 and 𝑐 are integers determined.
SOLUTION
When 𝑛 = 5
cos 5𝜃 + 𝑖 sin 5𝜃 = (cos 𝜃 + 𝑖 sin 𝜃)5
= 1 cos 5 𝜃 + 5 cos 4 𝜃 (𝑖 sin 𝜃) + 10 cos 3 𝜃 (𝑖 sin 𝜃)2 + 10cos 2 𝜃 (𝑖 sin 𝜃)3 + 5cos 𝜃 (𝑖 sin 𝜃)4 + 1(𝑖 sin 𝜃)5
= cos5 𝜃 + 5𝑖 cos4 𝜃 sin 𝜃 − 10 cos3 𝜃 sin2 𝜃 − 10𝑖 cos2 𝜃 sin3 𝜃 + 5 cos 𝜃 sin4 𝜃 + 𝑖 sin5 𝜃
Equating Imaginary Parts
sin 5𝜃 = 5 cos 4 𝜃 sin 𝜃 − 10 cos 2 𝜃 sin3 𝜃 + sin5 𝜃
𝑎 = 5, 𝑏 = −10, 𝑐 = 1
P a g e | 42
LESSON 3
that
Use de Moivre’s theorem to show
3 tan 𝜃 − tan3 𝜃
tan 3𝜃 =
1 − 3 tan2 𝜃
SOLUTION
cos 3𝜃 + 𝑖 sin 3𝜃 = (cos 𝜃 + 𝑖 sin 𝜃)3
= cos 3 𝜃 + 3 cos 2 𝜃 (𝑖 sin 𝜃) + 3 cos 𝜃 (𝑖 sin 𝜃)2 + (𝑖 sin 𝜃)3
= cos3 𝜃 + 3𝑖 cos2 𝜃 sin 𝜃 − 3 cos 𝜃 sin2 𝜃 − 𝑖 sin3 𝜃
(ii)
1
3
(1−𝑖√3)
Rewriting 1 − 𝑖√3 in Modulus – Argument
form
2
|1 − √3𝑖| = √12 + (−√3) = 2
𝜃 = − tan−1 (√3)
𝜋
=−
3
By De Moivre’s Theorem
−3
(1 − √3𝑖)
𝜋
𝜋 −3
= [2 (cos (− ) + 𝑖 sin (− ))]
3
3
Equating Real Parts:
𝜋
𝜋
= 2−3 (cos (−3 (− )) + 𝑖 sin(−3 (− ))
3
3
cos 3𝜃 = cos 3 𝜃 − 3 cos 𝜃 sin2 𝜃
1
= (cos 𝜋 + 𝑖 sin 𝜋)
8
1
= (−1)
8
1
=−
8
Equating Imaginary Parts
sin 3𝜃 = 3 cos 2 𝜃 sin 𝜃 − sin3 𝜃
tan 3𝜃 =
sin 3𝜃
cos 3𝜃
LESSON 5
10
2
=
3
3 cos 𝜃 sin 𝜃 − sin 𝜃
cos 3 𝜃 − 3 cos 𝜃 sin2 𝜃
3 cos 2 𝜃 sin 𝜃 sin3 𝜃
−
cos 3 𝜃
cos 3 𝜃
=
cos 3 𝜃 3 cos 𝜃 sin2 𝜃
−
cos 3 𝜃
cos 3 𝜃
3
=
3 tan 𝜃 − tan 𝜃
1 − 3 tan2 𝜃
LESSON 4
Express √3 + 𝑖 in the modulus –
Find the value of
𝜋
𝜋
4
4
12
argument form. Hence, find (√3 + 𝑖)
𝑎 + 𝑏𝑖.
in the form
SOLUTION
2
|√3 + 𝑖| = √(√3) + 12 = 2
1
𝜋
𝜃 = tan−1 ( ) =
6
√3
𝜋
𝜋
√3 + 𝑖 = 2 (cos ( ) + 𝑖 sin ( ))
6
6
By De Moivre’s Theorem
(i) (cos ( ) + 𝑖 sin ( ))
(ii)
10
1
(√3 + 𝑖)
3
(1−𝑖√3)
𝜋
𝜋 10
= [2 (cos ( ) + 𝑖 sin ( ))]
6
6
𝜋
𝜋
= 210 (cos(10 ( ) + 𝑖 sin(10 ( ))
6
6
SOLUTION
𝜋
𝜋
4
4
12
(i) (cos ( ) + 𝑖 sin ( ))
By De Moivre’s Theorem
𝜋
𝜋 12
(cos ( ) + 𝑖 sin ( ))
4
4
𝜋
𝜋
= cos (12 × ) + 𝑖 sin (12 × )
4
4
= cos 3𝜋 + 𝑖 sin 3𝜋
= −1
5𝜋
5𝜋
= 1024 (cos ( ) + 𝑖 sin ( ))
3
3
1 √3
= 1024 ( −
𝑖)
2
2
= 512 − 512√3𝑖
P a g e | 43
LESSON 2
multiples of 𝜃.
Multiples of Sine and Cosine
INTRODUCTION
Expressions for powers of sin 𝜃 and cos 𝜃 in terms
of sines and cosines of multiples of 𝜃 can be
derived using the following results
Express sin3 𝜃 in terms of sines of
SOLUTION
1 3
1
1
1
(𝑧 − ) = 𝑧 3 − 3𝑧 2 . + 3𝑧. 2 − 3
𝑧
𝑧
𝑧
𝑧
If 𝑧 = cos 𝜃 + 𝑖 sin 𝜃 then
= 𝑧 3 − 3𝑧 +
1
1
cos 𝜃 − 𝑖 sin 𝜃
=
.
𝑧 (cos 𝜃 + 𝑖 sin 𝜃) cos 𝜃 − 𝑖 sin 𝜃
= cos 𝜃 − 𝑖 sin 𝜃
∴𝑧+
1
= (cos 𝜃 + 𝑖 sin 𝜃) + (cos 𝜃 − 𝑖 sin 𝜃)
𝑧
1
1
) − 3 (𝑧 − )
3
𝑧
𝑧
If 𝑧 = cos 𝜃 + 𝑖 sin 𝜃 , 𝑧 𝑛 −
1
𝑧𝑛
= 2𝑖 sin 𝑛𝜃
∴ (2𝑖 sin 𝜃)3 = 2𝑖 sin 3𝜃 − 3(2𝑖 sin 𝜃)
= 2 cos 𝜃
𝑧−
= (𝑧 3 −
3 1
−
𝑧 𝑧3
1
= (cos 𝜃 + 𝑖 sin 𝜃) − (cos 𝜃 − 𝑖 sin 𝜃)
𝑧
= 2𝑖 sin 𝜃
−8𝑖 sin3 𝜃 = 2𝑖 sin 3𝜃 − 6𝑖 sin 𝜃
1
3
sin3 𝜃 = − sin 3𝜃 + sin 𝜃
4
4
By De Moivre’s Theorem
𝑧 𝑛 = cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃 , so that
1
𝑧𝑛
= cos 𝑛𝜃 − 𝑖 sin 𝑛𝜃
𝑧𝑛 +
The Exponential Form of a Complex
Number
From Maclaurin’s Theorem
1
1
= 2 cos 𝑛𝜃 and 𝑧 𝑛 − 𝑛 = 2𝑖 sin 𝑛𝜃
𝑛
𝑧
𝑧
LESSON 1
Express cos 4 𝜃 in terms of
cosines of multiples of 𝜃.
cos 𝜃 = 1 −
𝜃2 𝜃4 𝜃6
+
−
+⋯
2! 4! 6!
sin 𝜃 = 𝜃 −
𝜃3 𝜃5
+
−⋯
3! 5!
𝜃2
This series cos 𝜃 + sin 𝜃 appears to be similar to
the expansion of 𝑒 𝜃
= 𝑧 4 + 4𝑧 2 + 6 +
= (𝑧 4 +
4
1
+ 4
2
𝑧
𝑧
1
1
) + 4 (𝑧 2 + 2 ) + 6
4
𝑧
𝑧
𝑛
If 𝑧 = cos 𝜃 + 𝑖 sin 𝜃 , 𝑧 +
1
𝑧𝑛
= 2 cos 𝑛𝜃
i.e. 𝑒 𝜃 = 1 + 𝜃 +
𝜃2
2!
+
𝜃3
3!
+
Looking at the powers of 𝑖
𝑖=𝑖
𝑖 2 = −1
∴ (2 cos 𝜃)4 = 2 cos 4𝜃 + 4(2 cos 2𝜃) + 6
𝑖 3 = 𝑖 2 (𝑖) = −𝑖
16 cos 4 𝜃 = 2 cos 4𝜃 + 8 cos 2𝜃 + 6
𝑖 4 = (𝑖 2 )2 = (−1)2 = 1
1
cos 4 𝜃 = (cos 4𝜃 + 4 cos 2𝜃 + 3)
8
𝑖 5 = (𝑖 4 )(𝑖) = 𝑖
𝜃4
4!
+
𝜃5
5!
+⋯
4!
+
𝜃5
1 4
1
1
1
1
(𝑧 + ) = 𝑧 4 + 4𝑧 3 . + 6𝑧 2 . 2 + 4𝑧. 3 + 4
𝑧
𝑧
𝑧
𝑧
𝑧
3!
+
𝜃4
SOLUTION
2!
−
𝜃3
Then cos 𝜃 + sin 𝜃 = 1 + 𝜃 −
⋯
5!
−
P a g e | 44
𝑖 6 = (𝑖 4 )(𝑖 2 ) = 1(−1) = −1
NB: If 𝑧 = 𝑟𝑒 𝑖𝜃 then 𝑧 ∗ = 𝑟𝑒 −𝑖𝜃
𝑖 7 = (𝑖 6 )(𝑖) = −𝑖
Now let’s try the expansion 𝑒 𝑖𝜃
𝑒 𝑖𝜃
(𝑖𝜃)2 (𝑖𝜃)3 (𝑖𝜃)4 (𝑖𝜃)5
= 1 + 𝑖𝜃 +
+
+
+
+⋯
2!
3!
4!
5!
𝜃 2 𝑖𝜃 3 𝜃 4 𝑖𝜃 5
= 1 + 𝑖𝜃 −
−
+
+
−⋯
2!
3!
4!
5!
Grouping Imaginary and Real terms
𝑒 𝑖𝜃 = (1 −
𝜃2 𝜃4
𝜃3 𝜃5
+
− ⋯ ) + 𝑖 (𝜃 −
+
− ⋯)
2
4
3! 5!
= cos 𝜃 + 𝑖 sin 𝜃
𝑧 = 𝑥 + 𝑦𝑖 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃) = 𝑟𝑒 𝑖𝜃
LESSON 1
Express the following complex
numbers in the form 𝑟𝑒 𝑖𝜃 .
(a) 𝑧1 = 1 + 𝑖
(b) 𝑧1 =
1+𝑖
Locus on the Argand diagram
We will be using the notation 𝑧 = 𝑥 + 𝑦𝑖
LESSON 1
If the point 𝑃 in the complex
plane corresponds to the complex number 𝑧, find
the locus of 𝑃 in each of the following situations.
(a) |𝑧| = 3
(b) |𝑧 − 2| = 4
(c) |𝑧 + 3 − 𝑖| = 2
SOLUTION
(a) |𝑧| = 3
The distance between the point (0, 0) and the
point 𝑃(𝑥, 𝑦) representing the complex
number 𝑧 = 𝑥 + 𝑦𝑖 is 3
∴ |(𝑥 + 𝑦𝑖) − (0 − 0𝑖)| = 3
CARTESIAN FORM
|𝑥 + 𝑦𝑖| = √𝑥 2 + 𝑦 2 = 3
𝑥 2 + 𝑦2 = 9
i.e. a circle with centre at (0, 0) and radius 3
√3−𝑖
SOLUTION
(a) 𝑧1 = 1 + 𝑖
𝑟 = |𝑧| = √12 + 12 = √2
𝜋
arg 𝑧 = tan−1 (1) =
4
𝜋
𝑧 = √2𝑒 4 𝑖
(b) 𝑧1 =
1+𝑖
√3−𝑖
Let 𝑧2 = 1 + 𝑖 and 𝑧3 = √3 − 𝑖
𝑟2 = |𝑧2 | = √12 + 12 = √2
𝜋
arg 𝑧2 =
4
2
𝑟3 = |𝑧3 | = √(√3) + (−1)2 = 2
1
𝜋
arg 𝑧3 = − tan−1 ( ) = −
6
√3
𝑧2
√2
|𝑧2 | = | | =
𝑧3
2
𝜋
𝜋
5𝜋
arg 𝑧1 = arg 𝑧2 − arg 𝑧3 = − (− ) =
4
6
12
√2 5𝜋𝑖
𝑧1 =
𝑒 12
2
(b) |𝑧 − 2| = 4
Circle with centre (2, 0) and radius 4
The distance between the point (2, 0) and the
point 𝑃(𝑥, 𝑦) representing the complex
number
𝑧 = 𝑥 + 𝑦𝑖 is 4
CARTESIAN FORM
|(𝑥 + 𝑦𝑖) − (2 − 0𝑖)| = 4
|𝑥 + 𝑦𝑖 − 2| = 4
|𝑥 − 2 + 𝑦𝑖| = 4
(𝑥 − 2)2 + 𝑦 2 = 42
(𝑥 − 2)2 + 𝑦 2 = 16
P a g e | 45
11
=0
2
7 2
27
(𝑥 + ) + 𝑦 2 =
2
4
𝑥 2 + 7𝑥 + 𝑦 2 +
7
√27
2
2
Circle with centre (− , 0) and radius
(c) |𝑧 + 3 − 𝑖| = 2
Circle with centre (−3, 1) and radius 2
The distance between the point (−3,1) and
the point 𝑃(𝑥, 𝑦) representing the complex
number 𝑧 = 𝑥 + 𝑦𝑖 is 2.
CARTESIAN FORM
|𝑥 + 𝑦𝑖 + 3 − 𝑖| = 2
|𝑥 + 3 + (𝑦 − 1)𝑖| = 2
(𝑥 + 3)2 + (𝑦 − 1)2 = 22
.
LESSON 3
Sketch the locus of the point
𝑃(𝑥, 𝑦) representing the complex number
𝑧 = 𝑥 + 𝑦𝑖, given that |𝑧 − 3𝑖| = |𝑧 + 2 + 5𝑖|.
Write down the Cartesian equation of the locus.
SOLUTION
|𝑧 − 3𝑖| = |𝑧 + 2 + 5𝑖|
Rewriting |𝑥 + 𝑦𝑖 − 3𝑖| = |𝑥 + 𝑦𝑖 − (−2 − 5𝑖)|
LESSON 2
Determine the Cartesian
equation of the locus of points satisfying the
following conditions.
The distance between the point 𝑃(𝑥, 𝑦),
representing the complex number 𝑧 = 𝑥 + 𝑦𝑖, and
the point 𝐴(0,3) is equal to the distance between
𝑃(𝑥, 𝑦) and the point 𝐵(−2, −5). Therefore, we are
finding the ⊥ bisector of 𝐴𝐵.
(a) 2|𝑧 − 3𝑖| = |𝑧|
|𝑥 + 𝑦𝑖 − (0 + 3𝑖)| = |𝑥 + 𝑦𝑖 − (−2 − 5𝑖)|
(b) |
𝑧−1
𝑧+2
| = √3
SOLUTION
(a) 2|𝑧 − 3𝑖| = |𝑧|
CARTESIAN FORM
2|𝑥 + 𝑦𝑖 − 3𝑖| = |𝑥 + 𝑦𝑖|
2|𝑥 + (𝑦 − 3)𝑖| = |𝑥 + 𝑦𝑖|
4[𝑥 2 + (𝑦 − 3)2 ] = 𝑥 2 + 𝑦 2
4𝑥 2 + 4𝑦 2 − 24𝑦 + 36 = 𝑥 2 + 𝑦 2
3𝑥 2 + 3𝑦 2 − 24𝑦 + 36 = 0
𝑥 2 + 𝑦 2 − 8𝑦 + 12 = 0
𝑥 2 + (𝑦 − 4)2 = 4
Circle with centre (0, 4) and radius 2
(b) |
𝑧−1
𝑧+2
|𝑥 + (𝑦 − 3)| = |(𝑥 + 2) + (𝑦 + 5)|
𝑥 2 + (𝑦 + 3)2 = (𝑥 + 2)2 + (𝑦 + 5)2
𝑥 2 + 𝑦 2 − 6𝑦 + 9 = 𝑥 2 + 4𝑥 + 4 + 𝑦 2 + 10𝑦 + 25
16𝑦 + 4𝑥 + 20 = 0
4𝑦 + 𝑥 + 5 = 0
LESSON 4
Describe and sketch the locus of
the points satisfying the following conditions.
(a) arg(𝑧 − 3) =
𝜋
4
(b) arg(𝑧 + 3 − 2𝑖) =
| = √3
CARTESIAN FORM
|𝑧 − 1| = √3|𝑧 + 2|
|𝑥 + 𝑦𝑖 − 1| = √3|𝑥 + 𝑦𝑖 + 2|
|𝑥 − 1 + 𝑦𝑖| = √3|𝑥 + 2 + 𝑦𝑖|
(𝑥 − 1)2 + 𝑦 2 = 3[(𝑥 + 2)2 + 𝑦 2 ]
𝑥 2 − 2𝑥 + 1 + 𝑦 2 = 3𝑥 2 + 12𝑥 + 12 + 3𝑦 2
2𝑥 2 + 2𝑦 2 + 14𝑥 + 11 = 0
SOLUTION
(a) arg(𝑧 − 3) =
𝜋
4
𝜋
3
P a g e | 46
𝜋
arg[(𝑥 + 𝑦𝑖) − (3 − 0𝑖)] =
4
𝜋
arg[(𝑥 − 3) + 𝑦𝑖] =
4
𝑦
𝜋
−1
tan (
)=
𝑥−3
4
𝑦
𝜋
= tan ( ) = 1
𝑥−3
4
𝑦 = 𝑥 − 3; 𝑥 > 3
This is the half line starting at (3, 0), not
𝜋
including (3, 0), making an angle of with the
LESSON 6
Shade on an Argand diagram the
region in which |𝑧 − 2𝑖| ≤ 1.
SOLUTION
|𝑧 − 2𝑖| = 1
|𝑧 − (0 + 2𝑖)| = 1
Circle with centre (0, 2) and radius 1.
4
positive real axis.
(b) arg(𝑧 + 3 − 2𝑖) =
𝜋
3
𝜋
arg[(𝑥 + 𝑦𝑖) − (−3 + 2𝑖)] =
3
𝜋
arg[(𝑥 + 3) + (𝑦 − 2)𝑖] =
3
𝑦−2
𝜋
= tan ( ) = √3
𝑥+3
3
𝑦 − 2 = √3𝑥 + 3√3
𝑦 = √3𝑥 + 2 + 3√3; 𝑥 > −3
The half – line starting at (−3, 2), exclusive,
𝜋
which makes an angle of with the positive
3
real axis.
LESSON 7
(a) Sketch on one Argand diagram:
(i) the locus of points satisfying
|𝑧 − 𝑖| = |𝑧 − 2|
(ii) the locus of points satisfying
𝜋
arg(𝑧 − 𝑖) =
4
(b) Shade on your diagram the region in which
𝜋
𝜋
|𝑧 − 𝑖| ≤ |𝑧 − 2| and – ≤ arg(𝑧 − 𝑖) ≤
2
4
SOLUTION
(a) (i) |𝑧 − 𝑖| = |𝑧 − 2|
|(𝑥 + 𝑦𝑖) − (0 + 𝑖)| = |(𝑥 + 𝑦𝑖) − (2 + 0𝑖)|
LESSON 5
Describe and sketch the locus of 𝑧
where 𝑧 = (2 + 𝑖) + 𝜆(1 − 𝑖)
SOLUTION
Using vectors
This is the perpendicular bisector of the
line segment joining the points (0, 1) and
(2, 0)
𝜋
(ii) arg(𝑧 − 𝑖) =
4
𝜋
arg[(𝑥 + 𝑦𝑖) − (0 + 𝑖)] =
4
Half – line starting at (0, 1), excluding,
𝜋
making an angle of with the positive 𝑥4
2
1
𝑧 = ( )+𝜆( )
1
−1
This is the line passing through the point (2, 1)
1
and parallel to the vector ( ), i.e (1 − 𝑖)
−1
axis.
P a g e | 47
(b)
(ii) arg(𝑧) =
𝜋
2
3
+ sin−1 (6) =
2𝜋
3
LESSON 8
LESSON 9
(a) Sketch on an Argand diagram the locus of
points satisfying the equation
|𝑧 − 6𝑖| = 3
(b) It is given that 𝑧 satisfies the equation
|𝑧 − 6𝑖| = 3.
(i) Write down the greatest possible value of
|𝑧|.
(ii) Find the greatest possible value of arg 𝑧,
giving your answer in the form 𝑝𝜋, where
−1 < 𝑝 ≤ 1.
(a) On the same Argand diagram, sketch the loci
of points satisfying:
(i) |𝑧 + 3 + 𝑖| = 5
(ii) arg(𝑧 + 3) = −
3𝜋
4
(b) (i) From your sketch, explain why there is
only one complex number satisfying
both equations.
(ii) Verify that this complex number is
−7 − 4𝑖
SOLUTION
SOLUTION
(a) Circle with centre (0, 6) and radius 3
(a) (i) Circle with centre (−3, −1) and radius 5
(ii) Half – line, starting at (−3, 0), exclusive,
making an angle of −
3𝜋
4
with the positive
real axis.
(b) (i) 9 is the largest possible value of |𝑧|.
(b) (i) There is only one complex number
satisfying both equations since there is
only one point of intersection due to the
P a g e | 48
half-line which starts within the circle.
(ii) If −7 − 3𝑖 is the point of intersection it
must satisfy both conditions.
|−7 − 4𝑖 + 3 + 𝑖|
= |−4 − 3𝑖|
= √(−4)2 + (−3)2
=5
arg(−7 − 4𝑖 + 3)
= arg(−4 − 4𝑖)
−4
3𝜋
= −𝜋 + tan−1 ( ) = −
−4
4
P a g e | 49
SEQUENCES
At the end of this section, students should be able to:
1. define the concept of a sequence {𝑎𝑛 } of terms 𝑎𝑛 as a function from the positive integers to
the real numbers;
2. write a specific term from the formula for the 𝑛th term, or from a recurrence relation;
3. describe the behaviour of convergent and divergent sequences, through simple examples;
4. apply mathematical induction to establish properties of sequences.
P a g e | 50
SEQUENCES
INTRODUCTION
A sequence is a list of numbers which obey a
particular pattern. Each number in the sequence is
called a term of the sequence. These are usually
denoted 𝑢1 , 𝑢2 , 𝑢3 , … , 𝑢𝑛−1 , 𝑢𝑛 where 𝑢1 is the first
term, 𝑢2 is the second term and 𝑢𝑛 is the 𝑛th term.
In some cases the sequence can be defined by a
formula – an expression for the 𝑛th term.
1
,
2
(d) 𝑢𝑛
(a) 𝑢𝑛 = 4𝑛 − 1
(c) 𝑢𝑛 =
𝑛+1
𝑛
1
2𝑛
(d) 𝑢𝑛 = (−1)𝑛+1 (
𝑛
𝑛+1
𝑛+1
1
1
𝑢1 =
(
)=
1+1
2
2
2
3
𝑢2 = (−1) (
)=−
2+1
3
3
3
𝑢3 = (−1)4 (
)=
3+1
4
4
4
𝑢4 = (−1)5 (
)=−
4+1
5
5
5
𝑢5 = (−1)6 (
)=
5+1
6
1
2 3
4 5
, − ,
, − ,
, …
2
3 4
5 6
(−1)2
LESSON 1
Write down the first 5 terms of
the following sequences:
(b) 𝑢𝑛 =
1
1
=
5
2
32
1 1 1
1
, ,
,
, …
4 8 16 32
𝑛
= (−1)𝑛+1 ( )
𝑢5 =
)
SOLUTION
(a) 𝑢𝑛 = 4𝑛 − 1
𝑢1 = 4(1) − 1 = 3
𝑢2 = 4(2) − 1 = 7
𝑢3 = 4(3) − 1 = 11
𝑢4 = 4(4) − 1 = 15
𝑢5 = 4(5) − 1 = 19
3, 7, 11, 15, 19,
(b) 𝑢𝑛 =
LESSON 2
For each of the following
sequences determine an expression for the 𝑛𝑡ℎ
term, 𝑢𝑛 .
(a) 5, 8, 11, 14, ….
(b) 8, 6, 4, 2,
0,
(c)
….
1
3
2
,
4
(d) 1,
𝑛
1
(e) 1,
(f)
1
1×2
(g) 2,
,
5
4
,
6
1
1
2
3
− ,
𝑛+1
1+1
𝑢1 =
=2
1
2+1 3
𝑢2 =
=
2
2
3+1 4
𝑢3 =
=
3
3
4+1 5
𝑢4 =
=
4
4
5+1 6
𝑢5 =
=
5
5
3 4 5 6
2,
,
,
,
, …
2 3 4 5
1
(c) 𝑢𝑛 = 𝑛
2
1
1
𝑢1 = 1 =
2
2
1
1
𝑢2 = 2 =
2
4
1
1
𝑢3 = 3 =
2
8
1
1
𝑢4 = 4 =
2
16
3
2
,
1
,
4
1
4
,
9
,
7
1
1
4
1
,
3×4
5
16
,
,
−2,
…
, …
, − ,
8
1
,
2×3
3
4
,
1
,
5
5
,
…
, …
16
1
4×5
,
1
5×6
,
….
…
SOLUTION
(a) Consecutive terms differ by 3 therefore we try
3𝑛. To create the right formula we add 2 i.e.
𝑢𝑛 = 3𝑛 + 2
(b) Consecutive terms differ by −2 therefore we
try −2𝑛. To create the correct expression we
need to add 10 i.e. 𝑢𝑛 = 10 − 2𝑛
(c) The numerators are the natural numbers 𝑛
and the denominators are two more than the
𝑛
numerator i.e. 𝑢𝑛 =
𝑛+2
(d) Ignoring the signs, each numerator is 1 and
the denominators are the natural numbers 𝑛.
Since the signs alternate between positive and
P a g e | 51
negative, starting with positive, we use
(−1)𝑛+1 . Therefore 𝑢𝑛 = (−1)𝑛+1 (
𝑛
𝑛+1
)
(e) Each numerator is 1 and the denominators
are powers of 2 i.e. 𝑢𝑛 =
1
2𝑛−1
(f) Each numerator is 1 and the first number of
the denominator is 𝑛 and the second is 𝑛 + 1.
Therefore 𝑢𝑛 =
1
𝑛(𝑛+1)
(g) The numerators are the natural numbers but
they begin with 2, i.e. 𝑛 + 1 and the
denominators are the square numbers.
Therefore 𝑢𝑛 =
𝑛+1
𝑛2
The sequence above diverges since it does not
converge to any specific value.
Types of Sequences
A sequence can be classified as being convergent,
divergent, oscillating or periodic.
Convergent Sequences
Convergent sequences as the name suggests
converge to a definite limit.
lim 𝑢𝑛 = 𝑙
𝑛→∞
This oscillating sequence above is divergent.
The sequence above is convergent because it is
tending to a value.
This divergent sequence is PERIODIC as it consists
of a set of values which are constantly repeated.
The repeating pattern of the sequence consists of
three values therefore the sequence is said to have
a period of 3.
The sequence above is OSCILLATING and
converges.
Divergent Sequences
Divergent sequences are sequences which are not
convergent.
P a g e | 52
Convergence of a Sequence
LESSON 1
Determine which of the following
functions is convergent or divergent. If the
sequence is convergent, determine the limit of the
sequence.
(a) 𝑢𝑛 =
(b) 𝑢𝑛 =
(c) 𝑢𝑛 =
(d) 𝑢𝑛 =
3𝑛
𝑛+1
𝑛3
𝑛4 −7
1−2𝑛
4𝑛2 4𝑛 1
2 − 𝑛2 + 𝑛2
= lim 𝑛
3𝑛 4
𝑛→∞
+
𝑛2 𝑛2
4 1
4− + 2
𝑛 𝑛
= lim
3 4
𝑛→∞
+
𝑛 𝑛2
DOES NOT EXIST
Not convergent
(d) lim
ln 𝑛
𝑛→∞ 𝑛3
√3𝑛+4
ln 𝑛
= lim
𝑛3
1
𝑛
By L’Hopital
𝑛→∞ 3𝑛2
SOLUTION
1
𝑛→∞ 3𝑛3
=0
Convergent and converges to 0.
= lim
(a) lim 𝑢𝑛
𝑛→∞
3𝑛
+1
3𝑛
= lim 𝑛
1
𝑛→∞ 𝑛
+
𝑛 𝑛
3
= lim
1
𝑛→∞
1+
𝑛
=3
𝑢𝑛 converges and it converges to 3
(b) lim 𝑢𝑛
= lim
𝑛→∞ 𝑛
𝑛→∞
3
𝑛
−7
𝑛3
4
= lim 4 𝑛
𝑛→∞ 𝑛
7
−
𝑛4 𝑛4
1
𝑛
= lim
7
𝑛→∞
1− 4
𝑛
=0
𝑢𝑛 is convergent and it converges to 0.
= lim
𝑛→∞ 𝑛4
(c) lim
Recurrence Relations
LESSON 1
following
A sequence is given by the
𝑢1 = 4
𝑢𝑛+1 = 𝑢𝑛 + 3
Write down the first four terms of the sequence.
SOLUTION
𝑢1 = 4
𝑢2 = 𝑢1+1 = 𝑢1 + 3 = 7
𝑢3 = 𝑢2+1 = 𝑢2 + 3 = 10
𝑢4 = 𝑢3+1 = 𝑢3 + 3 = 13
LESSON 2
defined by
A sequence of positive integers is
1−2𝑛
𝑢1 = 1,
𝑢𝑛+1 = 𝑢𝑛 + 𝑛(3𝑛 + 1), 𝑛 ∈ ℤ+
Prove by induction that 𝑢𝑛 = 𝑛2 (𝑛 − 1) + 1,
𝑛 ∈ ℤ+
𝑛→∞ √3𝑛+4
= lim
√(1 −
𝑛→∞
√3𝑛 + 4
= lim √
𝑛→∞
= lim (
𝑛→∞
2𝑛)2
4𝑛2 − 4𝑛 + 1
3𝑛 + 4
4𝑛2 − 4𝑛 + 1
)
3𝑛 + 4
SOLUTION
1
2
When 𝑛 = 1,
𝑢1 = 12 (1 − 1) + 1
𝑢1 = 1
P a g e | 53
Hence by mathematical oinduction 𝑢𝑛 = 2𝑛+1 + 1.
Therefore 𝑢1 is true
Assume true for 𝑛 = 𝑘
𝑢𝑘 = 𝑘 2 (𝑘 − 1) + 1
𝑢𝑘+1 = (𝑘 + 1)2 (𝑘) + 1
Now,
𝑢𝑘+1 = 𝑢𝑘 + 𝑘(3𝑘 + 1)
= 𝑘 2 (𝑘 − 1) + 1 + 𝑘(3𝑘 + 1)
= 𝑘 3 − 𝑘 2 + 1 + 3𝑘 2 + 𝑘
= 𝑘 3 + 2𝑘 2 + 𝑘 + 1
= 𝑘(𝑘 2 + 2𝑘 + 1) + 1
= 𝑘(𝑘 + 1)2 + 1
Therefore, 𝑢𝑘+1 is true when 𝑢𝑘 is true. Hence, by
Mathematical Induction
𝑢𝑛 = 𝑛2 (𝑛 − 1) + 1,
𝑛 ∈ ℤ+
LESSON 3
A sequence is defined by 𝑢1 = 5
and 𝑢𝑛+1 = 𝑢𝑛 + 2𝑛+1 . Prove by induction that
𝑢𝑛 = 2𝑛+1 + 1.
SOLUTION
𝑢1 = 21+1 + 1
𝑢1 = 22 + 1
𝑢1 = 5
Therefore 𝑢1 is true
Assume 𝑢𝑛 is true for 𝑛 = 𝑘
𝑢𝑘 = 2𝑘+1 + 1
𝑢𝑘+1 = 2𝑘+2 + 1
Now,
𝑢𝑘+1 = 𝑢𝑘 + 2𝑘+1
= 2𝑘+1 + 1 + 2𝑘+1
= 2(2𝑘+1 ) + 1
= 2𝑘+2 + 1
Therefore 𝑢𝑘+1 is ture whenever 𝑢𝑘 is true.
P a g e | 54
SERIES
At the end of this section, students should be able to:
1.
2.
3.
4.
use the summation (Ʃ) notation;
define a series, as the sum of the terms of a sequence;
identify the 𝑛th term of a series, in the summation notation;
define the 𝑚th partial sum 𝑆𝑚 as the sum of the first 𝑚 terms of the sequence, that is,
𝑚
𝑆𝑚 = ∑ 𝑎𝑟 ;
5.
6.
7.
8.
9.
𝑟=1
apply mathematical induction to establish properties of series;
find the sum to infinity of a convergent series;
apply the method of differences to appropriate series, and find their sums;
use the Maclaurin theorem for the expansion of series;
use the Taylor theorem for the expansion of series.
P a g e | 55
SERIES
∞
INTRODUCTION
∑
Given the sequence 𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 , … , 𝑢𝑛 , the
corresponding series is
𝑢1 + 𝑢2 + 𝑢3 + 𝑢4 + ⋯ + 𝑢𝑛
𝑆𝑛 is the 𝑛𝑡ℎ partial sum where:
𝑟=1
𝑛
𝑛+2
(iv) Ignoring the signs, each numerator is 1 and
the denominators are the natural numbers 𝑛.
Since the signs alternate between positive and
negative, starting with positive, we use
(−1)𝑛+1 . Therefore
∞
𝑆1 = 𝑢1
the first partial sum
𝑆2 = 𝑢1 + 𝑢2
the second partial sum
𝑆3 = 𝑢1 + 𝑢2 + 𝑢3
the third partial sum
LESSON 1
Write each of the following series
using sigma notation.
(a) 5 + 8 + 11 + 14 + ⋯
(b) 8 + 6 + 4 + 2 + 0 + (−2) + ⋯
3
2
3
4
4
5
1
6
5
+ + + + +⋯
7
1
1
1
2
1
3
1
4
5
4
1
8
1
(d) 1 + (− ) + + (− ) + + ⋯
1
(e) 1 + + +
2
(f)
1
1×2
+
3
2×3
4
4
9
+
(g) 2 + + +
+
3×4
5
16
1
16
+
4×5
+
2𝑛−1
∞
∑
1
5×6
1
2𝑛−1
(vi) Each numerator is 1 and the first number of
the denominator is 𝑛 and the second is 𝑛 + 1.
Therefore
∞
∑
𝑟=1
1
𝑛(𝑛 + 1)
(vii)
The numerators are the natural numbers
but they begin with 2, i.e. 𝑛 + 1 and the
denominators are the square numbers.
Therefore
∞
+⋯
1
1
are powers of 2 i.e.
𝑟=1
Using Sigma Notation
1
𝑟=1
𝑛
)
𝑛+1
(v) Each numerator is 1 and the denominators
the 𝑛𝑡ℎ partial
𝑆𝑛 = 𝑢1 + 𝑢2 + 𝑢3 + ⋯ + 𝑢𝑛
sum
(c)
∑(−1)𝑛+1 (
∑
+⋯
𝑟=1
𝑛+1
𝑛2
+⋯
SOLUTION
Sum of a Series
(i) Consecutive terms differ by 3 therefore we try
3𝑛. To create the right formula we add 2 i.e.
3𝑛 + 2
∑𝑟 =
∞
∑ 3𝑟 + 2
𝑟=1
(ii) Consecutive terms differ by −2 therefore we
try −2𝑛. To create the correct expression we
need to add 10 i.e. 10 − 2𝑛
∞
∑ 10 − 2𝑛
𝑟=1
(iii) The numerators are the natural numbers 𝑛
and the denominators are two more than the
𝑛
numerator i.e.
𝑛+2
The following standard results can be used to find
the sum of various series.
𝑛
𝑟=1
𝑛
(𝑛 + 1),
2
𝑛
∑ 𝑟3 =
𝑟=1
𝑛
∑ 𝑟2 =
𝑟=1
𝑛
(𝑛 + 1)(2𝑛 + 1),
6
𝑛2
(𝑛 + 1)2
4
LESSON 1
Find each of the following sums
(a) ∑4𝑟=1 𝑟(𝑟 + 1)
3
(b) ∑16
10 𝑟
SOLUTION
(a) ∑4𝑟=1 𝑟(𝑟 + 1)
P a g e | 56
4
= ∑(𝑟 + 𝑟)
𝑟=1
4
4
= ∑ 𝑟2 + ∑ 𝑟
𝑟=1
𝑟=1
4
4
= (4 + 1)(2(4) + 1) + (4 + 1)
6
2
= 40
3
(b) ∑16
𝑟=10 𝑟
16
3𝑛2 (𝑛 + 1)2 + 4𝑛(𝑛 + 1)(2𝑛 + 1)
12
𝑛(𝑛 + 1)[3𝑛(𝑛 + 1) + 4(2𝑛 + 1)]
=
12
𝑛(𝑛 + 1)[3𝑛2 + 3𝑛 + 8𝑛 + 4]
=
12
=
2
9
Mathematical Induction
LESSON 1
that
= ∑ 𝑟3 − ∑ 𝑟3
𝑟=1
2
Prove by mathematical induction
𝑛
∑ 𝑟 2 (𝑟 − 1) =
𝑟=1
2
16
9
(16 + 1)2 − (9 + 1)
4
4
= 16 471
𝑟=1
=
LESSON 2
Express each of the following in a
factorized form.
∑𝑛𝑟=1(𝑟
(a)
+ 1)(𝑟 − 1)
2 (𝑟
(b) ∑𝑛
𝑟
+ 2)
𝑟=1
for all positive integers 𝑛.
SOLUTION
𝑛
𝑃𝑛 : ∑ 𝑟 2 (𝑟 − 1) =
𝑟=1
𝑃1 : 12 (1 − 1) =
SOLUTION
1
𝑛(𝑛2 − 1)(3𝑛 + 2)
12
1
𝑛(𝑛2 − 1)(3𝑛 + 2)
12
1
(1)(12 − 1)(3(1) + 2)
12
0=0
(a) ∑𝑛𝑟=1(𝑟 + 1)(𝑟 − 1)
𝑛
= ∑(𝑟 2 − 1)
𝑟=1
𝑛
Assume 𝑃𝑛 is true for 𝑛 = 𝑘
𝑛
2
= ∑𝑟 − ∑1
𝑟=1
Therefore, 𝑃1 is true.
𝑟=1
𝑛
= (𝑛 + 1)(2𝑛 + 1) − 𝑛
6
𝑛(𝑛 + 1)(2𝑛 + 1) − 6𝑛
=
6
𝑛[(𝑛 + 1)(2𝑛 + 1) − 6]
=
6
𝑛[2𝑛2 + 3𝑛 + 1 − 6]
=
6
𝑛[2𝑛2 + 3𝑛 − 5]
=
6
𝑛(2𝑛 + 5)(𝑛 − 1)
=
6
(b) ∑𝑛𝑟=1 𝑟 2 (𝑟 + 2)
𝑛
𝑘
𝑃𝑘 : ∑ 𝑟 2 (𝑟 − 1) =
𝑟=1
𝑘+1
𝑃𝑘+1 : ∑ 𝑟 2 (𝑟 − 1) =
𝑟=1
𝑛
𝑃𝑘+1 =
=
𝑛
𝑛
(𝑛 + 1)2 + 2 [ (𝑛 + 1)(2𝑛 + 1)]
4
6
1
(𝑘 + 1)(𝑘 2 + 2𝑘)(3𝑘 + 5)
12
=
1
𝑘(𝑘 + 1)(𝑘 + 2)(3𝑘 + 5)
12
=
1
12(𝑘 + 1)2 𝑘
𝑘(𝑘 + 1)(𝑘 − 1)(3𝑘 + 2) +
12
12
=
1
𝑘(𝑘 + 1)[𝑘 − 1)(3𝑘 + 2) + 12(𝑘 + 1)]
12
= ∑ 𝑟 + 2 ∑ 𝑟2
𝑟=1
=
1
𝑘(𝑘 2 − 1)(3𝑘 + 2) + (𝑘 + 1)2 (𝑘 + 1 − 1)
12
3
𝑟=1
2
1
(𝑘 + 1)((𝑘 + 1)2 − 1)(3(𝑘 + 1) + 2)
12
Now, 𝑃𝑘+1 = 𝑃𝑘 + (𝑘 + 1) term
= ∑(𝑟 3 + 2𝑟 2 )
𝑟=1
𝑛
1
𝑘(𝑘 2 − 1)(3𝑘 + 2)
12
P a g e | 57
=
1
𝑘(𝑘 + 1)(3𝑘 2 − 𝑘 − 2 + 12𝑘 + 12)
12
1
=
𝑘(𝑘 + 1)(3𝑘 2 + 11𝑘 + 10)
12
=
1
𝑘(𝑘 + 1)(𝑘 + 2)(3𝑘 + 5)
12
Therefore 𝑃𝑘+1 is true whenever 𝑃𝑘 is true.
Hence by mathematical induction
𝑛
∑ 𝑟 2 (𝑟 − 1) =
𝑟=1
1
𝑛(𝑛2 − 1)(3𝑛 + 2)
12
for all positive integers 𝑛.
Method of Differences
If 𝑢𝑟 = 𝑓(𝑟 + 1) − 𝑓(𝑟), then
𝑛
𝑛
∑ 𝑢𝑟 = ∑{𝑓(𝑟 + 1) − 𝑓(𝑟)}
𝑟=1
𝑟=1
LESSON 1
𝑟 = 𝑛 − 1:
[𝑛(𝑛 + 1)(𝑛 + 2) − 𝑛(𝑛 − 1)(𝑛 + 1)]
𝑛
1
∑ 𝑟(𝑟 + 1) = 𝑛(𝑛 + 1)(𝑛 + 2)
3
𝑟=1
(iii) ∑𝑛𝑟=1 𝑟(𝑟 + 1) = ∑𝑛𝑟=1 𝑟 2 + ∑𝑛𝑟=1 𝑟
𝑛
𝑛
= (𝑛 + 1)(2𝑛 + 1) + (𝑛 + 1)
6
2
1
3
= 𝑛(𝑛 + 1)(2𝑛 + 1) + 𝑛(𝑛 + 1)
6
6
1
= 𝑛(𝑛 + 1)(2𝑛 + 1 + 3)
6
1
= 𝑛(𝑛 + 1)(2𝑛 + 4)
6
1
= 𝑛(𝑛 + 1)(𝑛 + 2)
3
LESSON 2
(i) Express
4+𝑟
𝑟(𝑟 + 1)(𝑟 + 2)
in partial fractions.
(ii) Use the method of differences to show that
𝑛
∑
(i) Show that
𝑟=1
𝑟(𝑟 + 1)(𝑟 + 2) − (𝑟 − 1)𝑟(𝑟 + 1) ≡ 3𝑟(𝑟 + 1)
[(𝑛 − 1)𝑛(𝑛 + 1) − (𝑛 − 1)(𝑛 − 2)𝑛] +
𝑟 = 𝑛:
4+𝑟
3
2
1
= −
+
𝑟(𝑟 + 1)(𝑟 + 2) 2 𝑛 + 1 𝑛 + 2
(iii) Write down the limit to which
𝑛
∑
(ii) Hence use the method of differences to find an
expression for
𝑛
∑ 𝑟(𝑟 + 1)
𝑟=1
converges as 𝑛 tends to infinity.
(iv) Find
100
∑
𝑟=1
(iii) Show that you can obtain the same expression
for ∑𝑛𝑟=1 𝑟(𝑟 + 1) using the standard results
for ∑𝑛𝑟=1 𝑟 and ∑𝑛𝑟=1 𝑟 2 .
𝑟=50
4+𝑟
𝑟(𝑟 + 1)(𝑟 + 2)
giving your answer to 3 significant figures.
SOLUTION
SOLUTION
(i)
(i) 𝑟(𝑟 + 1)(𝑟 + 2) − (𝑟 − 1)𝑟(𝑟 + 1)
= (𝑟 + 1)[𝑟(𝑟 + 2) − 𝑟(𝑟 − 1)]
= (𝑟 + 1)[𝑟 2 + 2𝑟 − 𝑟 2 + 𝑟]
= 3𝑟(𝑟 + 1)
(ii) ∑𝑛𝑟=1 𝑟(𝑟 + 1) = 13 [𝑟(𝑟 + 1)(𝑟 + 2) − 𝑟(𝑟 − 1)(𝑟 + 1)]
𝑟 = 1:
[1(2)(3) − 1(0)(2)] +
[2(3)(4) − 2(1)(3)] +
𝑟 = 2:
[3(4)(5) − 3(2)(4)] +
𝑟 = 3:
[4(5)(6) − 4(3)(5)] +
𝑟 = 4:
4+𝑟
𝑟(𝑟 + 1)(𝑟 + 2)
4+𝑟
𝑟(𝑟+1)(𝑟+2)
𝐴
𝐵
𝑟
𝑟+1
= +
+
𝐶
𝑟+2
4 + 𝑟 = 𝐴(𝑟 + 1)(𝑟 + 2) + 𝐵𝑟(𝑟 + 2)
+ 𝐶𝑟(𝑟 + 1)
When 𝑟 = 0
4 + 0 = 𝐴(1)(2)
4 = 2𝐴
2=𝐴
When 𝑟 = −1
4 + (−1) = 𝐵(−1)(−1 + 2)
P a g e | 58
3 = −𝐵
𝐵 = −3
When 𝑟 = −2
4 + (−2) = 𝐶(−2)(−2 + 1)
2 = 2𝐶
1=𝐶
4+𝑟
2
3
1
= −
+
𝑟(𝑟 + 1)(𝑟 + 2) 𝑟 𝑟 + 1 𝑟 + 2
4+𝑟
(ii) ∑𝑛𝑟=1
=
𝑟(𝑟+1)(𝑟+2)
4+𝑟
2
3
1
= −
+
𝑟(𝑟 + 1)(𝑟 + 2) 𝑟 𝑟 + 1 𝑟 + 2
2 3 1
𝑟 = 1:
( − + )+
1 2 3
2 3 1
𝑟 = 2:
( − + )+
2 3 4
2 3 1
𝑟 = 3:
( − + )+
3 4 5
2 3 1
𝑟 = 4:
( − + )+
4 5 6
2
3
1
(
− +
)
𝑛−1 𝑛 𝑛+1
2
3
1
𝑟 = 𝑛:
( −
+
)
𝑛 𝑛+1 𝑛+2
𝑛
4+𝑟
∑
𝑟(𝑟 + 1)(𝑟 + 2)
𝑟 = 𝑛 − 1:
𝑟=1
=
=
2 3 2
1
3
1
− + +
−
+
1 2 2 𝑛+1 𝑛+1 𝑛+2
3
2
1
−
+
2 𝑛+1 𝑛+2
2
1
4+𝑟
3
𝑛
𝑛
∑
= −
+
𝑟(𝑟 + 1)(𝑟 + 2) 2 𝑛 + 1 𝑛 + 2
𝑟=1
𝑛 𝑛 𝑛 𝑛
2
1
3
= − 𝑛 + 𝑛
2 1+1 1+2
𝑛
𝑛
3
=
2
4+𝑟
3
2
1
(iv) ∑100
= −
+
𝑟=50
𝑛
100
=∑
𝑟=1
2
𝑛+1
49
𝑛+2
4+𝑟
4+𝑟
−∑
𝑟(𝑟 + 1)(𝑟 + 2)
𝑟(𝑟 + 1)(𝑟 + 2)
𝑟=1
3
2
1
3
2
1
=( −
+
)−( −
+
)
2 100 + 1 100 + 2
2 49 + 1 49 + 2
= 0.104
INTRODUCTION
A sequence 𝑎1 , 𝑎2 , 𝑎3 , … 𝑎𝑛−1 , 𝑎𝑛 , … is called an
arithmetic sequence, or arithmetic progression, if
there exists a constant 𝑑, called the common
difference, such that
𝑎𝑛 − 𝑎𝑛−1 = 𝑑
That is
𝑎1
𝑎2 = 𝑎1 + 𝑑
𝑎3 = 𝑎2 + 𝑑
= 𝑎1 + 𝑑 + 𝑑
= 𝑎1 + 2𝑑
Therefore,
𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 for every 𝑛 > 1
LESSON 1
Find the common difference for
each of the following arithmetic progressions.
(a) 3, 5, 7, 9, 11, …
(b) 8, 3, −2, −7, …
(c) 2𝑏, 5𝑏, 8𝑏, 11𝑏, …
SOLUTION
(a) 𝑎1 = 3, 𝑎2 = 5, 𝑎3 = 7, …
𝑑 = 𝑎2 − 𝑎1 = 5 − 3 = 2
(b) 𝑎1 = 8, 𝑎2 = 3, 𝑎3 = −2, 𝑎4 = −5
𝑑 = 𝑎3 − 𝑎2 = −2 − 3 = −5
(c) 𝑎1 = 2𝑏, 𝑎2 = 5𝑏, 𝑎3 = 8𝑏, 𝑎4 = 11𝑏, …
𝑑 = 𝑎4 − 𝑎3 = 11𝑏 − 8𝑏 = 3𝑏
NB: Any pair of consecutive terms can be used.
(iii) As 𝑛 → ∞
𝑟(𝑟+1)(𝑟+2)
ARITHMETIC PROGRESSIONS
LESSON 2
Prove that the sequence 3, 7,
11, 15, … is an arithmetic progression.
SOLUTION
We need to show that 𝑎𝑛 − 𝑎𝑛−1 is a constant.
𝑎𝑛 = 4𝑛 − 1
𝑎𝑛−1 = 4(𝑛 − 1) − 1
= 4𝑛 − 5
𝑎𝑛 − 𝑎𝑛−1 = (4𝑛 − 1) − (4𝑛 − 5)
=4
Therefore, 𝑑 = 4
P a g e | 59
LESSON 3
The sum, 𝑆𝑛 , of the first 𝑛 terms
of a sequence is given by 𝑆𝑛 = 𝑛(5𝑛 − 2). Show
that the sequence is an arithmetic progression
with common difference 10.
SOLUTION
𝑎𝑛 = 𝑆𝑛 − 𝑆𝑛−1
= 𝑛(5𝑛 − 2) − [(𝑛 − 1)(5(𝑛 − 1) − 2)]
= 5𝑛2 − 2𝑛 − [(𝑛 − 1)(5𝑛 − 5 − 2)]
= 5𝑛2 − 2𝑛 − [(𝑛 − 1)(5𝑛 − 7)]
= 5𝑛2 − 2𝑛 − (5𝑛2 − 7𝑛 − 5𝑛 + 7)
= 5𝑛2 − 2𝑛 − (5𝑛2 − 12𝑛 + 7)
= 5𝑛2 − 2𝑛 − 5𝑛2 + 12𝑛 − 7
= 10𝑛 − 7
𝑎𝑛−1 = 10(𝑛 − 1) − 7
= 10𝑛 − 10 − 7
= 10𝑛 − 17
𝑑 = 𝑎𝑛 − 𝑎𝑛−1 = (10𝑛 − 7) − (10𝑛 − 17)
= 10
LESSON 4
If the first three terms of an
arithmetic progression are 5, 9, and 13, what is
the value of the 10th term?
SOLUTION
𝑎1 = 5,
𝑎2 = 9,
𝑎3 = 13
The common difference, 𝑑, is 4
𝑎10 = 𝑎1 + (𝑛 − 1)𝑑
= 5 + (10 − 1)(4)
= 41
Sum Formulae for Finite Arithmetic Sequence
If 𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 is a finite arithmetic sequence,
then the corresponding series 𝑎1 + 𝑎2 + 𝑎3 + ⋯ +
𝑎𝑛 is called a finite arithmetic series. The sum of
the first 𝑛 terms of the series, which we denote 𝑆𝑛 ,
would be stated as
𝑛
𝑆𝑛 = [2𝑎1 + (𝑛 − 1)𝑑]
2
LESSON 1
Find the sum of the even
numbers from 50 to 120 inclusive.
SOLUTION
𝑎1 = 50
𝑑=2
𝑎𝑛 = 120
𝑎1 + (𝑛 − 1)𝑑 = 120
50 + 2(𝑛 − 1) = 120
2(𝑛 − 1) = 70
𝑛 − 1 = 35
𝑛 = 36
36
[2(50) + (36 − 1)(2)]
𝑆36 =
2
𝑠36 = 3060
LESSON 2
The last term of an arithmetic
progression of 20 terms is 295 and the common
difference is 4. Calculate the sum of the
progression.
SOLUTION
𝑛 = 20, 𝑎20 = 295, 𝑑 = 4
We need to determine 𝑎1
𝑎20 = 𝑎1 + (𝑛 − 1)𝑑
𝑎1 + (20 − 1)(4) = 295
𝑎1 = 219
20
[2(219) + (20 − 1)(4)]
2
= 5140
𝑆20 =
𝑆20
LESSON 3
The sum of the first 6 terms of an
arithmetic progression is 54.75 and the sum of the
next 6 terms is 63.75. Find the common difference
and the first term.
SOLUTION
𝑆6 = 54.75
6
[2𝑎1 + (6 − 1)𝑑] = 54.75
2
6𝑎1 + 15𝑑 = 54.75
(1)
𝑆12 = 54.75 + 63.75 = 118.5
12
[2𝑎1 + (12 − 1)𝑑] = 118.5
2
12𝑎1 + 66𝑑 = 118.5
(2)
Solving (1) and (2) simultaneously
6𝑎1 + 15𝑑 = 54.75
12𝑎1 + 66𝑑 = 118.5
(1) × 2:
12𝑎1 + 30𝑑 = 109.5
12𝑎1 + 66𝑑 = 118.5
−36𝑑 = −9
1
𝑑=
4
𝑎 = 8.5
P a g e | 60
GEOMETRIC PROGRESSIONS
INTRODUCTION
A sequence 𝑎1 , 𝑎2 , 𝑎3 , … , 𝑎𝑛 , … is called a
geometric sequence, or geometric progression, if
there exists a nonzero constant 𝑟, called the
common ratio, such that
𝑎1
𝑎2 = 𝑎1 𝑟
𝑎3 = 𝑎2 𝑟
= 𝑎1 𝑟𝑟
= 𝑎1 𝑟 2
Therefore,
𝑎
𝑎𝑛 = 𝑎1 𝑟 𝑛−1 , 𝑛 > 1 or 𝑛 = 𝑟
𝑎𝑛−1
LESSON 1
Prove that the sequence 1, 3, 9,
27, … is a Geometric Progression.
𝑎
SOLUTION
We need to show that 𝑛 is a
𝑎𝑛−1
constant.
𝑎1 = 30 ,
𝑎2 = 31 ,
𝑛−1
𝑎𝑛 = 3
𝑎𝑛−1 = 3𝑛−1−1 = 3𝑛−2
𝑎𝑛
3𝑛−1
= 𝑛−2
𝑎𝑛−1 3
= 3𝑛−1−(𝑛−2)
=3
𝑎3 = 32 ,
𝑎4 = 33
LESSON 2
The first and fourth terms of a
geometric progression are 6 and 20.25
respectively. Determine the 8th term of the
progression.
SOLUTION
𝑎1 = 6
𝑎4 = 𝑎1 𝑟 3 = 20.25
20.25
𝑟3 =
= 3.375
6
3
𝑟 = √3.375 = 1.5
𝑎8 = 𝑎1 𝑟 7
3 7 6561
𝑎8 = 6 ( ) =
2
64
LESSON 3
The lengths of the sides of a
triangle are in geometric progression and the
longest side has a length of 36 cm. Given that the
perimeter is 76 cm, find the length of the shortest
side.
SOLUTION
Let longest side be 𝑎1 and shortest side be 𝑎3 .
𝑎1 = 36
𝑎1 + 𝑎2 + 𝑎3 = 76
𝑎1 + 𝑎1 𝑟 + 𝑎1 𝑟 2 = 76
36 + 36𝑟 + 36𝑟 2 = 76
36𝑟 2 + 36𝑟 = 40
9𝑟 2 + 9𝑟 − 10 = 0
(3𝑟 + 5)(3𝑟 − 2) = 0
2
𝑟=
3
Since length cannot be negative
2 2
𝑎1 𝑟 2 = 36 ( ) = 16
3
Sum of a Geometric Progression
The sum of the first 𝑛 terms of a G.P is given by
𝑎1 (1 − 𝑟 𝑛 )
𝑆𝑛 =
,
1−𝑟
𝑎1 (𝑟 𝑛 − 1)
𝑟 < 1 or 𝑆𝑛 =
,
𝑟−1
𝑟>1
LESSON 1
The fourth term of a geometric
progression is 6 and the seventh term is −48.
Calculate
(i)
the common ratio,
(ii)
the first term,
(iii)
the sum of the first eleven terms.
SOLUTION
(i)
𝑎4 = 𝑎1 𝑟 3 = 6
𝑎7 = 𝑎1 𝑟 6 = −48
𝑎7 𝑎1 𝑟 6
=
= 𝑟3
𝑎4 𝑎1 𝑟 3
48
𝑟3 = −
= −8
6
𝑟 = −2
(ii)
𝑎1 𝑟 3 = 6
𝑎1 (−8) = 6
3
𝑎1 = −
4
𝑎 (1−𝑟 𝑛 )
(iii)
𝑆𝑛 = 1
1−𝑟
3
− ((−2)11 − 1)
𝑆11 = 4
1 − (−2)
𝑆11 = −512.25
5
1 𝑛
LESSON 2
Given that 𝑆𝑛 = (1 − ( ) ), find
4
3
𝑎𝑛 and prove that this sequence is a Geometric
Progression.
SOLUTION
𝑎𝑛 = 𝑆𝑛 − 𝑆𝑛−1
5
1 𝑛
5
1 𝑛−1
= (1 − ( ) ) − (1 − ( ) )
4
3
4
3
5 5 1 𝑛 5 5 1 𝑛−1
= − ( ) − + ( )
4 4 3
4 4 3
5 1 𝑛 5 1 𝑛−1
=− ( ) + ( )
4 3
4 3
5 1 𝑛−1
1
= ( )( )
(1 − )
4 3
3
P a g e | 61
5 1 𝑛−1
= ( )
6 3
5 1 𝑛−2
𝑎𝑛−1 = ( )
6 3
5 1 𝑛−1
( )
𝑎𝑛
= 6 3 𝑛−2
𝑎𝑛−1 5 1
( )
6 3
1
=
3
Sum to Infinity
What would be the sum of the infinite series
1 1 1 1
1+ + + +
+⋯
2 4 8 16
If we think about it we should realise that the sum
appears to be 2. Since the sum appears to tend
towards a specific number as it goes on
indefinitely we refer to this series as a
CONVERGENT series. The sum of this series can be
given using the formula
𝑎1
𝑆∞ =
,
−1 < 𝑟 < 1
1−𝑟
1
For our series above we have 𝑎1 = 1 and 𝑟 = ,
2
therefore
1
𝑆∞ =
=2
1
1−
2
Thus we see that our intuitive answer is indeed
correct.
LESSON 1
The first and fourth terms of a
geometric progression are 500 and 32
respectively. Find
(i)
the values of second and third terms
(ii)
the sum to infinity of the progression
SOLUTION
(i)
𝑎1 = 500
𝑎4 = 32
𝑎1 𝑟 3 = 32
500𝑟 3 = 32
32
8
𝑟3 =
=
500 125
2
𝑟=
5
𝑎2 = 𝑎1 𝑟
2
= 500 ( )
5
= 200
𝑎3 = 𝑎1 𝑟 2
(ii)
2 2
= 500 ( )
5
= 80
𝑎
𝑆∞ =
1−𝑟
500
=
2
1−
5
2500
=
3
LESSON 2
The first term of a geometric
progression is 𝑎 and the common ratio is 𝑟. Given
that 𝑎 = 12𝑟 and that the sum to infinity is 4,
calculate the third term.
SOLUTION
𝑎1
𝑆∞ =
1−𝑟
12𝑟
4=
1−𝑟
4 − 4𝑟 = 12𝑟
4 = 16𝑟
1
𝑟=
4
𝑎1 = 12𝑟
1
𝑎1 = 12 ( )
4
𝑎1 = 3
𝑎3 = 𝑎1 𝑟 2
1 2
𝑎3 = 3 ( )
4
3
𝑎3 =
16
LESSON 3
The first term of a geometric
series is 120. The sum to infinity of the series is
480. Given that the sum of the first 𝑛 terms is
greater than 300, determine the smallest possible
value of 𝑛.
SOLUTION
𝑎
𝑆∞ =
1−𝑟
120
480 =
1−𝑟
480(1 − 𝑟) = 120
1
1−𝑟 =
4
3
𝑟=
4
𝑎(1 − 𝑟 𝑛 )
1−𝑟
𝑆𝑛 > 300
𝑆𝑛 =
P a g e | 62
3 𝑛
120 (1 − ( ) )
4
> 300
3
1−
4
3 𝑛
120 (1 − ( ) )
4
> 300
1
4
3 𝑛
480 (1 − ( ) ) > 300
4
3 𝑛 5
1−( ) >
4
8
3 𝑛 3
( ) <
4
8
3 𝑛
3
ln ( ) < ln ( )
4
8
3
3
𝑛 ln ( ) < ln ( )
4
8
3
ln ( )
3
8
𝑛>
ln ( ) is negative
3
4
ln ( )
4
𝑛 > 3.4
𝑛=4
LESSON 4
Determine whether the
geometric series
∞
1 𝑟
∑( )
2
𝑟=1
is convergent. If it converges, determine its sum.
SOLUTION
We need to show that
−1 < 𝑟 < 1.
∞
1 𝑟 1 1 1 1
∑( ) = + + +
+⋯
2
2 4 8 16
𝑟=1
1
1
𝑟=4=
1 2
2
Since −1 < 𝑟 < 1, the series converges.
1
( )
𝑆∞ = 2 = 1
1
1−
2
P a g e | 63
MACLAURIN’S SERIES
INTRODUCTION
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 ′ (0) +
𝑥 2 ′′
𝑥3
𝑥𝑛
𝑓 (0) + 𝑓 ′′′ (0) + ⋯ + 𝑓 𝑛 (0) + ⋯
2!
3!
𝑛!
𝑓(𝑥) must be differentiable
LESSON 2
Find the Maclaurin expansion for
(1 + 𝑥)2 cos 𝑥 up to and including the term in 𝑥 3 .
𝑓(𝑥) must exist at 𝑥 = 0
The derivatives of 𝑓(𝑥) must exist at 𝑥 = 0
Only within specific values of 𝑥 is the series valid.
SOLUTION
𝑓(𝑥) = (1 + 𝑥)2 cos 𝑥
= (1 + 2𝑥 + 𝑥 2 ) cos 𝑥
LESSON 1
Use Maclaurin’s Theorem to find
the first four non – zero terms for cos 𝑥, hence
determine an approximation for cos(0.2).
SOLUTION
𝑓(0) = cos(0) = 1
𝑓 ′ (𝑥) = − sin 𝑥
𝑓 ′ (0) = − sin(0) = 0
𝑓 ′′ (𝑥) = − cos 𝑥
𝑓 ′′ (0) = − cos(0) = −1
𝑓 ′′′ (𝑥) = sin 𝑥
𝑓 ′′′ (0) = sin(0) = 0
𝑓
𝑥2
+. . )
2!
using result Question 1
𝑓(𝑥) = cos 𝑥
′′′′ (𝑥)
= (1 + 2𝑥 + 𝑥 2 ) (1 −
= cos 𝑥
𝑓
′′′′ (0)
= cos(0) = 1
𝑓 ′′′′′ (𝑥) = − sin 𝑥
𝑓 ′′′′′ (0) = − sin(0) = 0
𝑓 ′′′′′′ (𝑥) = − cos 𝑥
𝑓 ′′′′′′ (0) = − cos(0) − 1
=1−
𝑥2
+ 2𝑥 − 𝑥 3 + 𝑥 2 + ⋯
2!
1
= 1 + 2𝑥 + 𝑥 2 − 𝑥 3
2
LESSON 3
tan 𝑥 up to 𝑥 3 .
Find the Maclaurin’s series for
SOLUTIONhttp://sirhunte.teachable.com/courses
/93027/lectures/2211764
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 ′ (0) +
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 ′ (0) +
+
𝑥 2 ′′
𝑥3
𝑓 (0) + 𝑓 ′′′ (0) + ⋯
2!
3!
𝑥𝑛 𝑛
𝑓 (0) + ⋯
𝑛!
𝑥2
𝑥3
𝑥4
+ (0) + (1)
2!
3!
4!
𝑥5
𝑥6
+ (0) + (−1)
5!
6!
𝑓(𝑥) = 1 + (0)𝑥 + (−1)
𝑥2 𝑥4 𝑥6
=1− + + +⋯
2! 4! 6!
+
𝑥 2 ′′
𝑥3
𝑓 (0) + 𝑓 ′′′ (0) + ⋯
2!
3!
𝑥𝑛 𝑛
𝑓 (0) + ⋯
𝑛!
𝑓(𝑥) = tan 𝑥
𝑓(0) = tan(0) = 0
𝑓 ′ (𝑥) = sec 2 𝑥
𝑓 ′ (0) = sec 2 (0) = 1
𝑓 ′′ (𝑥) = 2 tan 𝑥 sec 2 𝑥
𝑓 ′′ (0) = 2 tan(0) sec 2 (0)
=0
𝑓 ′′′ (𝑥)
= 2 sec 4 𝑥
+ 4 sec 2 𝑥 tan2 𝑥
𝑓 ′′′ (0)
= 2 sec 4 (0) + 4 sec 2 0 tan2 0
=2
(0.2)2 (0.2)4 (0.2)6
𝑓(0.2) = 1 −
+
−
2
24
720
= 0.98
𝑓(𝑥) = 0 + 𝑥 + (0)
𝑥2
𝑥3
+2 +⋯
2!
3!
P a g e | 64
=𝑥+
2𝑥 3
3!
LESSON 4
𝑔(𝑡) = 𝑒
𝑓 ′ (0) = 6
𝑓 ′′ (𝑥) = −18(1 − 3𝑥)−4 (−3) = 54(1 − 3𝑥)−4
A function is defined as
2𝑡+1
𝑓 ′′ (0) = 54
.
𝑓 ′′′ (𝑥) = −96(1 − 2𝑥)−5 (−3) = 288(1 − 3𝑥)−5
(a) Obtain the Maclaurin’s series expansion for
𝑔(𝑡) up to and including the term in 𝑡 4 .
(ii) Hence, estimate 𝑔(0.1) to four decimal places.
𝑓 ′′′ (0) = 288
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 ′ (0) +
SOLUTION
+
(i)
𝑔(𝑡) = 𝑒 2𝑡+1
𝑥𝑛 𝑛
𝑓 (0) + ⋯
𝑛!
𝑔(0) = 𝑒 1
𝑓(𝑥) = 1 + 6𝑥 + 36 (
𝑔′ (𝑡) = 2𝑒 2𝑡+1
𝑔′ (0) = 2𝑒 1
𝑔′′ (𝑡) = 4𝑒 2𝑡+1
𝑔′′ (0) = 4𝑒 1
𝑔′′′ (𝑡) = 8𝑒 2𝑡+1
𝑔′′′ (0) = 8𝑒 1
𝑔′′′′ (𝑡) = 16𝑒 2𝑡+1
𝑔′′′′ (0) = 16𝑒 1
𝑥 2 ′′
𝑥3
𝑓 (0) + 𝑓 ′′′ (0) + ⋯
2!
3!
𝑥2
𝑥3
) + 288 ( )
4
6
= 1 + 6𝑥 + 9𝑥 2 + 48𝑥 3
Valid for −1 < −3𝑥 < 1
→
1
1
3
3
− <𝑥<
LESSON 6
𝑡2
𝑡3
𝑔(𝑡) = 𝑔(0) + 𝑥𝑔
+ 𝑔′′ (0) + 𝑔′′′ (0)
2!
3!
𝑡𝑛 𝑛
+ ⋯ + 𝑔 (0) + ⋯
𝑛!
′ (0)
= 3.3201
Find 𝑓 ′ (𝑥), 𝑓′′(𝑥) and 𝑓 ′′′ (𝑥). Hence obtain the
Maclaurin series for 𝑓(𝑥) as far as the term in 𝑥 3 .
By considering the equivalent binomial expansion,
give the set of values of 𝑥 for which the Maclaurin
series is valid.
).
State the range of validity for this series.
1+𝑥
1−𝑥
= 3. Hence
SOLUTION
𝑓(𝑥) = ln(1 + 𝑥)
𝑓 ′ (𝑥) =
1
1+𝑥
𝑓 ′′ (𝑥) = −
𝑓 ′′′ (𝑥) =
1
(1 + 𝑥)2
2
(1 + 𝑥)3
𝑓(0) = ln(1) = 0
𝑓 ′ (0) =
1
=1
1+0
1
(1 + 0)2
= −1
𝑓 ′′ (0) = −
𝑓 ′′′ (0) =
2
(1 + 0)3
=2
𝑓(𝑥) = (1 − 3𝑥)−2
𝑓 ′ (𝑥) = −2(1 − 3𝑥)−3 (−3) = 6(1 − 3𝑥)−3
1−𝑥
(i)
1
You are given that 𝑓(𝑥) = (1−3𝑥)2.
𝑓(0) = 1
1+𝑥
find an approximation to ln 3, giving your
answer to three decimal places.
(ii) 𝑔(0.1) = 𝑒 + 2𝑒(0.1) + 2𝑒(0.1)2 + 4𝑒3 (0.1)3 + 2𝑒3 (0.1)4
SOLUTION
terms in the Maclaurin series for ln (
(ii) Find the value of 𝑥 for which
𝑡2
𝑡3
𝑡4
= 𝑒 + 2𝑒 𝑡 + 4𝑒 + 8𝑒 + 16𝑒
2!
3!
4!
4𝑒 3 2𝑒 4
2
= 𝑒 + 2𝑒 𝑡 + 2𝑒 𝑡 + 𝑡 + 𝑡
3
3
LESSON 5
(i) Use the Maclaurin series for ln(1 + 𝑥) and
ln(1 − 𝑥) to obtain the first three non – zero
𝑓 ′′′′ (𝑥) = −
6
(1 + 𝑥)4
6
(1 + 0)4
= −6
𝑓 ′′′′ (0) = −
P a g e | 65
𝑓 ′′′′′ (𝑥) =
24
(1 + 𝑥)5
24
(1 + 0)5
= 24
𝑓 ′′′′′ (0) =
1+𝑥
ln (
) = ln(1 + 𝑥) − ln(1 − 𝑥)
1−𝑥
𝑥2 𝑥3 𝑥4 𝑥5
+ − + + ⋯)
2
3
4
5
𝑥2 𝑥3 𝑥4 𝑥5
− (−𝑥 − − − − )
2
3
4
5
= (𝑥 −
𝑥2
𝑥3
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 ′ (0) + 𝑓 ′′ (0) + 𝑓 ′′′ (0)
2!
3!
𝑥𝑛 𝑛
+ ⋯ + 𝑓 (0) + ⋯
𝑛!
1
𝑥3
𝑥4
𝑓(𝑥) = 0 + 𝑥 − 𝑥 2 + (2) − (6)
2!
3!
4!
𝑥5
+ (24) + ⋯
5!
𝑥2 𝑥3 𝑥4 𝑥5
=𝑥− + − + +⋯
2
3
4
5
𝑓(𝑥) = ln(1 − 𝑥)
𝑓 ′ (𝑥) = −
1
1−𝑥
𝑓(0) = ln(1) = 0
𝑓 ′ (0) = −
1
1−0
= −1
𝑓 ′′ (𝑥) = −
1
(1 − 𝑥)2
𝑓 ′′ (0) = −
1
(1 − 0)2
= −1
𝑓 ′′′ (𝑥) = −
2
(1 − 𝑥)3
𝑓 ′′′ (0) = −
2
(1 − 0)3
= −2
𝑓 ′′′′ (𝑥)
=−
𝑓 ′′′′ (0)
6
(1 − 𝑥)4
=−
𝑓 ′′′′′ (𝑥)
24
=−
(1 − 𝑥)5
6
= −6
(1 − 0)4
𝑓 ′′′′′ (0)
24
=−
= −24
(1 − 0)5
𝑥 2 ′′
𝑥3
𝑓 (0) + 𝑓 ′′′ (0)
2!
3!
𝑥𝑛 𝑛
+ ⋯ + 𝑓 (0) + ⋯
𝑛!
𝑓(𝑥) = 𝑓(0) + 𝑥𝑓 ′ (0) +
𝑓(𝑥) = 0 − 𝑥 −
= −𝑥 −
𝑥2
𝑥3
𝑥4
𝑥5
− (2) − (6) − (24)
2!
3!
4!
5!
+⋯
𝑥2 𝑥3 𝑥4 𝑥5
− − −
2
3
4
5
= 2𝑥 +
(ii)
1+𝑥
1−𝑥
2𝑥 3 2𝑥 5
+
3
5
−1 <𝑥 < 1
=3
1 + 𝑥 = 3 − 3𝑥
4𝑥 = 2
1
𝑥=
2
1
1+
2)
ln 3 = ln (
1
1−
2
1
2 1 3 2 1 5
= 2( )+ ( ) + ( )
2
3 2
5 2
= 1.096
P a g e | 66
TAYLOR SERIES
INTRODUCTION
1.
2.
The function 𝑓(𝑥) has to be infinitely differentiable
The function 𝑓(𝑥) has to be defined in a region near the value 𝑥 = 𝑎.
Maclaurin’s Series is:
𝑔(𝑥) = 𝑔(0) + 𝑥𝑔′ (0) +
𝑥 2 ′′
𝑥3
𝑥𝑛
𝑔 (0) + 𝑔′′′ (0) + ⋯ + 𝑔𝑛 (0) + ⋯
2!
3!
𝑛!
For Taylor’s Series, we let 𝑓(𝑥 + 𝑎) = 𝑔(𝑥)
𝑓(𝑥 + 𝑎) = 𝑓(𝑎) + 𝑓 ′ (𝑎)𝑥 +
𝑓 ′′ (𝑎) 2 𝑓 ′′′ (𝑎) 3
𝑓 𝑛 (𝑎)
𝑥 +
𝑥 +⋯+
+⋯
2!
3!
𝑛!
Furthermore, replacing 𝑥 with 𝑥 − 𝑎 we get
𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) +
∞
𝑓(𝑥) = ∑
𝑘=0
𝑓 ′′ (𝑎)
𝑓 ′′′ (𝑎)
𝑓 (𝑛) (𝑎)
(𝑥 − 𝑎)2 +
(𝑥 − 𝑎)3 + ⋯ +
(𝑥 − 𝑎)𝑛 + ⋯
2!
3!
𝑛!
𝑓 𝑘 (𝑎)
(𝑥 − 𝑎)𝑘
𝑘!
LESSON 1
Find the first three non – zero
terms of the Taylor expansion of ln(𝑥 + 3).
SOLUTION
SOLUTION
𝑓(𝑥) = ln 𝑥
𝑓(𝑥 + 𝑎) = ln(𝑥 + 3)
𝑓(𝑥) = ln 𝑥
𝑓 ′ (𝑥) =
1
𝑥
𝑓 ′′ (𝑥) = −
LESSON 2
Find the first four non – zero
1
terms for the Taylor expansion of
with centre
𝑥+3
𝑎 = 1.
𝑓(3) = ln 3
𝑓 ′ (3) =
1
𝑥2
Let 𝑓(𝑥) =
→𝑎=3
1
3
𝑓 ′′ (3) = −
𝑓(𝑥) =
1
𝑥+3
1
𝑥+3
𝑓 ′ (𝑥) = −
1
9
𝑓 ′′ (𝑎) 2 𝑓 ′′′ (𝑎) 3
𝑥 +
𝑥
2!
3!
𝑛 (𝑎)
𝑓
+ ⋯+
+⋯
𝑛!
𝑓 ′′ (𝑥) =
1
(𝑥 + 3)2
2
(𝑥 + 3)3
𝑓(𝑥 + 𝑎) = 𝑓(𝑎) + 𝑓 ′ (𝑎)𝑥 +
1
1 1
ln(𝑥 + 3) = ln 3 + 𝑥 − ( ) 𝑥 2
3
9 2!
1
1 2
= ln 3 + 𝑥 −
𝑥
3
18
= (𝑥 + 3)−1
𝑓 ′′′ (𝑥) = −
6
(𝑥 + 3)4
𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) +
+
𝑓(1) =
1
4
𝑓 ′ (1) = −
𝑓 ′′ (1) =
1
16
1
32
𝑓 ′′′ (1) = −
3
128
𝑓 ′′ (𝑎)
(𝑥 − 𝑎)2
2!
𝑓 ′′′ (𝑎)
𝑓 (𝑛) (𝑎)
(𝑥 − 𝑎)3 + ⋯ +
(𝑥 − 𝑎)𝑛 + ⋯
3!
𝑛!
1
1 1
1 1
3 1
= − (𝑥 − 1) + ( ) (𝑥 − 1)2 −
( ) (𝑥 − 1)3
𝑥 + 3 4 16
32 2!
128 3!
P a g e | 67
=
1 1
1
3
(𝑥 − 1)2 −
(𝑥 − 1)3
− (𝑥 − 1) +
4 16
64
256
LESSON 3
(i) Obtain the first four non – zero terms of the
Taylor Series expansion of sin 𝑥 in ascending
𝜋
powers of (𝑥 − ).
4
𝜋
(ii) Hence, calculate an approximation to sin ( ).
16
SOLUTION
(i) let 𝑓(𝑥) = sin 𝑥
𝑓 ′′ (𝑎)
(𝑥 − 𝑎)2
2!
𝑓 ′′′ (𝑎)
𝑓 (𝑛) (𝑎)
(𝑥 − 𝑎)3 + ⋯ +
(𝑥 − 𝑎)𝑛
+
3!
𝑛!
𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) +
𝜋
4
𝑓(𝑥) = sin 𝑥
𝑎=
𝜋
𝜋
√2
𝑓 ( ) = sin ( ) =
4
4
2
𝑓 ′ (𝑥) = cos 𝑥
𝜋
𝜋
√2
𝑓 ′ ( ) = cos ( ) =
4
4
2
𝑓 ′′ (𝑥) = − sin 𝑥
𝜋
𝜋
𝑓 ′′ ( ) = − sin ( )
4
4
√2
=−
2
𝑓 ′′′ (𝑥)
= − cos 𝑥
𝜋
𝑓 ′′′ (𝑥) = − cos ( )
4
√2
=−
2
𝜋
𝜋 2
√2 √2
√2 1
+
(𝑥 − ) −
( ) (𝑥 − )
2
2
4
2 2!
4
𝜋 3
√2 1
−
( ) (𝑥 − )
2 3!
4
𝜋
𝜋 2
√2 √2
√2
=
+
(𝑥 − ) −
(𝑥 − )
2
2
4
4
4
𝜋 3
√2
−
(𝑥 − )
12
4
sin 𝑥 =
𝜋
16
(ii) sin ( ) → 𝑥 =
𝜋
16
𝜋
𝜋 𝜋
3𝜋
(𝑥 − ) = ( − ) = −
4
16 4
16
𝜋
3𝜋
3𝜋 2
√2 √2
√2
sin ( ) =
+
(− ) −
(− )
16
2
2
16
4
16
3
3𝜋
√2
−
(− )
12
= 0.1920
16
P a g e | 68
BINOMIAL THEOREM
At the end of this section, students should be able to:
𝑛
1. explain the meaning and use simple properties of 𝑛! and ( ), that is, 𝑛𝐶𝑟 where 𝑛, 𝑟 ∈ ℤ
𝑟
𝑛
2. recognise that 𝑛𝐶𝑟 that is, ( ), is the number of ways in which 𝑟 objects may be chosen
𝑟
from 𝑛 distinct objects;
3. expand (𝑎 + 𝑏)𝑛 for 𝑛 ∈ ℚ;
4. apply the Binomial Theorem to real-world problems, for example, in mathematics of
finance, science.
P a g e | 69
PASCAL’S TRIANGLE
INTRODUCTION
For any positive integer 𝑛:
PASCAL’S TRIANGLE
(𝑎 + 𝑏)0 = 1
1
(𝑎 + 𝑏)1 = 1𝑎 + 1𝑏
1
(𝑎 + 𝑏)2 = 1𝑎2 + 2𝑎𝑏 + 1𝑏
1
1
(𝑎 + 𝑏)3 = 1𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 1𝑏 3
(𝑎 + 𝑏)4 = 1𝑎4 + 4𝑎3 𝑏 + 6𝑎2 𝑏 2 + 3𝑎𝑏 3 + 1𝑏 4
1
1
2
1
3
4
3
1
6
4
1
FACTORIALS
𝑛2 − 3𝑛 + 2 = 72
𝑛! = 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) … (3)(2)(1)
𝑛2 − 3𝑛 − 70 = 0
For example, 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
(𝑛 − 10)(𝑛 + 7) = 0
0! Is defined as 1.
𝑛 = 10 since 𝑛 = −7 in invalid
LESSON 1
Simplify
9!
6!
LESSON 4
SOLUTION
9! 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
=
6!
6×5×4×3×2×1
= 9 × 8 × 7 = 504
LESSON 2
Simplify 𝑛! − (𝑛 − 2)!
Show that
1
𝑘+1
2
−
=
(𝑘 + 2)! (𝑘 + 3)! (𝑘 + 3)!
SOLUTION
SOLUTION
1
𝑘+1
−
(𝑘 + 2)! (𝑘 + 3)!
𝑛! − (𝑛 − 2)!
=
𝑘+3
𝑘+1
−
(𝑘 + 3)(𝑘 + 2)! (𝑘 + 3)!
=
𝑘+3
𝑘+1
−
(𝑘 + 3)! (𝑘 + 3)!
=
𝑘 + 3 − (𝑘 + 1)
(𝑘 + 3)!
=
2
(𝑘 + 3)!
= 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) … (3)(2)(1) − (𝑛 − 2)!
= 𝑛(𝑛 − 1)(𝑛 − 2)! − (𝑛 − 2)!
(𝑛 − 2)! [𝑛(𝑛 − 1) − 1]
= (𝑛 − 2)! (𝑛2 − 𝑛 − 1)
LESSON 3
(𝑛−1)!
Solve the equation (𝑛−3)! = 72.
SOLUTION
(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) … (3)(2)(1)
= 72
(𝑛 − 3)(𝑛 − 4)(𝑛 − 5) … (3)(2)(1)
(𝑛 − 1)(𝑛 − 2) = 72
P a g e | 70
LESSON 5
1
𝐴
1
1
−
]+
𝑛! (𝑛 + 1)!
𝑟 = 𝑛 − 1:
[
𝑟 = 𝑛:
[
𝐵
(a) Express (𝑟+2)𝑟! in the form (𝑟+1)! + (𝑟+2)!, where
𝐴 and 𝐵 are integers.
(b) Hence find
𝑛
𝑛
∑
𝑟=1
∑
1
(𝑟 + 2)𝑟!
𝑟=1
SOLUTION
(a)
1
𝐴
(𝑟+2)𝑟!
𝐵
= (𝑟+1)! + (𝑟+2)!
𝑟+1
1
(
)(
)
𝑟 + 1 (𝑟 + 2)𝑟!
=
𝐴(𝑟 + 2)
𝐵
+
(𝑟 + 2)(𝑟 + 1)! (𝑟 + 2)!
𝑟 + 1 = 𝐴(𝑟 + 2) + 𝐵
Equating coefficients of 𝑟:
𝐴=1
Equating constants:
2𝐴 + 𝐵 = 1
2+𝐵 =1
𝐵 = −1
1
1
1
=
−
(𝑟 + 2)𝑟! (𝑟 + 1)! (𝑟 + 2)!
(b)
𝑛
𝑛
𝑛
𝑟=1
𝑟=1
𝑟=1
1
1
1
∑
=∑
−∑
(𝑟 + 2)𝑟!
(𝑟 + 1)!
(𝑟 + 2)!
𝑟 = 1:
1 1
[ − ]+
2! 3!
𝑟 = 2:
1 1
[ − ]+
3! 4!
𝑟 = 3:
1 1
[ − ]+
4! 5!
1
1
−
]
(𝑛 + 1)! (𝑛 + 2)!
1
1
1
= −
(𝑟 + 2)𝑟! 2 (𝑛 + 2)!
P a g e | 71
THE BINOMIAL THEOREM
INTRODUCTION
We now look at an alternative to Pascal’s Triangle using Factorials.
𝑛𝐶𝑟 =
𝑛!
𝑛
=( )
𝑟
(𝑛 − 𝑟)! 𝑟!
For any positive integer 𝑛:
𝑛
(𝑎 + 𝑏)𝑛 = ∑ 𝑛𝐶𝑟 𝑎𝑛−𝑟 𝑏 𝑟
𝑟=0
𝑛
𝑛
𝑛
𝑛
𝑛
𝑛
= ( ) 𝑎𝑛 + ( ) 𝑎𝑛−1 𝑏1 + ( ) 𝑎𝑛−2 𝑏 2 + ( ) 𝑎𝑛−3 𝑏 3 + ⋯ + (
) 𝑎1 𝑏 𝑛−1 + ( ) 𝑎0 𝑏 𝑛
0
1
2
3
𝑛−1
𝑛
LESSON 1
Determine the expansion (3 + 2𝑥)5 .
SOLUTION
(3 + 2𝑥)5
5
5
5
5
5
5
= ( ) 35 + ( ) 34 (2𝑥)1 + ( ) 33 (2𝑥)2 + ( ) 32 (2𝑥)3 + ( ) 31 (2𝑥)4 + ( ) (2𝑥)5
0
1
2
3
4
5
= 243 + 5[81(2𝑥)] + 10[27(4𝑥 2 )] + 10[9(8𝑥 3 )] + 5[3(16𝑥 4 )] + 1(32𝑥 5 )
= 243 + 810𝑥 + 1080𝑥 2 + 720𝑥 3 + 240𝑥 4 + 32𝑥 5
LESSON 2
Find the 6th term of the expansion (2 − 3𝑥)10 .
SOLUTION
The 6th term of the expansion (2 − 3𝑥)10 occurs when 𝑟 = 5 since the summation index
begins with 𝑟 = 0.
=(
10 5
) 2 (−3𝑥)5
5
= 252(32)(−243𝑥 6 )
= −1959552𝑥 5
LESSON 3
Find the coefficient of 𝑥 6 in the expansion of (1 + 3𝑥)2 (2 − 𝑥)6 .
SOLUTION
(1 + 3𝑥)2 = 1 + 6𝑥 + 9𝑥 2
We only need the terms which will result in a 𝑥 6 term after multiplication.
6
6
6
(2 − 𝑥)6 = ( ) 22 (−𝑥)4 + ( ) 2(−𝑥)5 + ( ) (−𝑥)6
4
5
6
= 15(4)𝑥 4 + 6(2)(−𝑥 5 ) + 1(𝑥 6 )
= 60𝑥 4 − 12𝑥 5 + 𝑥 6
We isolate the multiplications which would create an 𝑥 6 term.
(1 + 6𝑥 + 9𝑥 2 )(… + 60𝑥 4 − 12𝑥 5 + 𝑥 6 ) = 1(𝑥 6 ) + 6𝑥(−12𝑥 5 ) + 9𝑥 2 (60𝑥 4 )
= 𝑥 6 − 72𝑥 6 + 540𝑥 6
= 469𝑥 6
P a g e | 72
LESSON 4
Find the term independent of 𝑥 in the expansion (3𝑥 3 +
1
2𝑥
8
) .
SOLUTION
The term independent of 𝑥:
1 6
8
= ( ) (3𝑥 3 )2 ( )
6
2𝑥
1
= 28(9𝑥 6 ) (
)
64𝑥 6
63
=
16
Extension of the Binomial Expansion
For any real number 𝑛
(1 + 𝑥)𝑛 = 1 + 𝑛𝑥 +
𝑛(𝑛 − 1) 2 𝑛(𝑛 − 1)(𝑛 − 2) 3
𝑥 +
𝑥 +⋯
2!
3!
provided that −1 < 𝑥 < 1
𝑥 𝑛
𝑥 𝑛
(𝑎 + 𝑥)𝑛 = [𝑎 (1 + )] = 𝑎𝑛 (1 + )
𝑎
𝑎
𝑥
𝑛(𝑛 − 1) 𝑥 2 𝑛(𝑛 − 1)(𝑛 − 2) 𝑥 3
= 𝑎𝑛 [1 + 𝑛 ( ) +
( ) +
( ) + ⋯]
𝑎
2!
𝑎
3!
𝑎
−1 <
LESSON 1
𝑥
<1
𝑎
→
−𝑎 < 𝑥 < 𝑎
Find the binomial expansion of √(1 + 2𝑥)3 up to and including the term in 𝑥 3 .
SOLUTION
3
√(1 + 2𝑥)3 = (1 + 2𝑥)2
3
3 3
1
3 3
3
1
= 1 + ( ) (2𝑥) + ( ) ( − 1) (
) (2𝑥)2 + ( ) ( − 1) ( − 2) (
) (2𝑥)3
2
2 2
2(1)
2 2
2
3(2)(1)
3 1 1
3 1
1
1
= 1 + 3𝑥 + ( ) ( ) ( ) 4𝑥 2 + ( ) ( ) (− ) (
) 8𝑥 3
2 2 2
2 2
2 (3)(2)
3
1
= 1 + 3𝑥 + 𝑥 2 − 𝑥 3
2
2
− 1 < 2𝑥 < 1
−
1
1
<𝑥<
2
2
P a g e | 73
LESSON 2
Find the first three terms of the expansion of
4
2−3𝑥
.
SOLUTION
4
3 −1
3 −1
= 4(2 − 3𝑥)−1 = 4 [2 (1 − 𝑥) ] = 4 [2−1 (1 − 𝑥) ]
2 − 3𝑥
2
2
3 −1
= 2 (1 − 𝑥)
2
3
1
3 2
= 2[1 + (−1) (− 𝑥) + (−1)(−1 − 1) (
) (− 𝑥)
(2)(1)
2
2
3
9
= 2 [1 + 𝑥 + 𝑥 2 + ⋯ ]
2
4
9
= 2 + 3𝑥 + 𝑥 2
2
LESSON 3
9𝑥
(i) Express (1+𝑥)(2+5𝑥) in the form
𝐴
1+𝑥
+
𝐵
2+5𝑥
where 𝐴 and 𝐵 are integers.
(ii) Hence, or otherwise, find the expansion of
9𝑥
(1+𝑥)(2+5𝑥)
as a power series in ascending order up to and
3
including the term in 𝑥 .
(iii) Find the range of values of 𝑥 for which the series expansion of
9𝑥
(1 + 𝑥)(2 + 5𝑥)
SOLUTION
(i)
9𝑥
(1+𝑥)(2+5𝑥)
=
𝐴
1+𝑥
+
𝐵
2+5𝑥
9𝑥 = 𝐴(2 + 5𝑥) + 𝐵(1 + 𝑥)
9𝑥 = 2𝐴 + 5𝐴𝑥 + 𝐵 + 𝐵𝑥
Equating constants:
2𝐴 + 𝐵 = 0
Equationg coefficients of 𝑥:
5𝐴 + 𝐵 = 9
𝐴=3
𝐵 = −6
9𝑥
3
6
=
−
(1 + 𝑥)(2 + 5𝑥) 1 + 𝑥 2 + 5𝑥
(ii)
3
1+𝑥
= 3(1 + 𝑥)−1
1
1
= 3 [1 + (−1)𝑥 + (−1)(−1 − 1) (
) 𝑥 2 + (−1)(−1 − 1)(−1 − 2) (
) 𝑥3]
(2)(1)
(3)(2)(1)
= 3[1 − 𝑥 + 𝑥 2 − 𝑥 3 ]
= 3 − 3𝑥 + 3𝑥 2 − 3𝑥 3
P a g e | 74
6
2 + 5𝑥
= 6(2 + 5𝑥)−1
5 −1
= 6 [2 (1 + 𝑥) ]
2
5 −1
= 6 [2−1 (1 + 𝑥) ]
2
5 −1
= 3 (1 + 𝑥)
2
5
1
5 2
1
5 3
= 3 [1 + (−1) ( 𝑥) + (−1)(−1 − 1) (
) ( 𝑥) + (−1)(−1 − 1)(−1 − 2) (
) ( 𝑥) ]
(2)(1) 2
(3)(2)(1) 2
2
5
25 2 125 3
= 3 [1 − 𝑥 +
𝑥 −
𝑥 ]
2
4
8
15
75 2 375 3
=3−
𝑥+
𝑥 −
𝑥
2
4
8
3
6
−
1 + 𝑥 2 + 5𝑥
= 3 − 3𝑥 + 3𝑥 2 − 3𝑥 3 − [3 −
15
75 2 375 3
𝑥+
𝑥 −
𝑥 ]
2
4
8
9
63 2 351 3
= 𝑥−
𝑥 +
𝑥
2
4
8
5
2
2
5
(iii) Valid for −1 < 𝑥 < 1 and −1 < 𝑥 < 1 → − < 𝑥 <
2
5
2
2
− <𝑥<
5
5
4
LESSON 4
4
Use the expansion of √1 + 𝑥 to setimate √82 to four decimal places.
SOLUTION
1
4
√1 + 𝑥 = (1 + 𝑥)4
1
1 1
1
1 1
1
1
= 1 + 𝑥 + ( ) ( − 1) (
) 𝑥 2 + ( ) ( − 1) ( − 2) (
) 𝑥3
(2)(1)
(3)(2)(1)
4
4 4
4 4
4
1
3 2
7 3
=1+ 𝑥−
𝑥 +
𝑥
4
32
128
This works for small 𝑥.
82 = 81 + 1
1
(1 +
1
81)4
1 4
= [81 (1 + )]
81
1
=
1
814
1 4
(1 + )
81
P a g e | 75
1
1 4
= 3 (1 + )
81
1 1
3 1 2
7
1 3
= 3[1 + ( ) −
( ) +
( )
4 81
32 81
128 81
= 3[1.003]
= 3.009
LESSON 5
2
(a) Find the binomial expansion of (1 + 6𝑥)3 up to and inclusing the term in 𝑥 2 .
2
(b) Find the binomial expansion of (8 + 6𝑥)3 up to and including the term in 𝑥 2 .
3
𝑎
(c) Hence, find an estimate for the the value of √144 in the form where 𝑎 and 𝑏 are integers.
𝑏
SOLUTION
2
2
2
2
3
3
3
1
(a) (1 + 6𝑥)3 = 1 + (6𝑥) + ( ) ( − 1) (2)(1) (6𝑥)2
= 1 + 4𝑥 − 4𝑥 2
2
2
6
3
(b) (8 + 6𝑥)3 = [8 (1 + 𝑥)]
8
2
=
2
83
6 3
(1 + 𝑥)
8
2
3
1
= 4 (1 + 6 ( 𝑥))
8
1
1
1 2
= 4 [1 + 4 ( 𝑥) − 4 ( ) ]
8
8
1 2
= 4 + 2𝑥 − 𝑥
4
1
2
(c) (144)3 = (122 )3 = 123
8 + 6𝑥 = 12
6𝑥 = 4
2
𝑥=
3
2
1 2 2 47
3
√144 ≅ 4 + 2 ( ) − ( ) ≅
3
4 3
9
P a g e | 76
ROOTS OF EQUATIONS
At the end of this section, students should be able to:
1. test for the existence of a root of 𝑓(𝑥) = 0 where f is continuous using the Intermediate
Value Theorem;
2. use interval bisection to find an approximation for a root in a given interval;
3. use linear interpolation to find an approximation for a root in a given interval;
4. explain, in geometrical terms, the working of the Newton-Raphson method;
5. use the Newton-Raphson method to find successive approximations to the roots of 𝑓(𝑥) =
0, where 𝑓 is differentiable;
6. use a given iteration to determine a root of an equation to a specified degree of accuracy.
P a g e | 77
THE INTERMEDIATE VALUE THEOREM
If 𝑓(𝑥) is a continuous function on the closed interval [𝑎, 𝑏] and the product 𝑓(𝑎)𝑓(𝑏) < 0 then there exists 𝑐
in [𝑎, 𝑏] such that 𝑓(𝑐) = 0.
LESSON 1
Use the Intermediate Value Theorem to show that
𝑓(𝑥) = 𝑥 3 − 2𝑥 2 + 𝑥 − 1 has a root between 1 and 2.
SOLUTION
𝑓(1) = 13 − 2(1)2 + 1 − 1 = −1
𝑓(2) = 23 − 2(2)2 + 2 − 1 = 1
𝑓(𝑥) is a polynomial and therefore continuous on the
interval [1, 2].
𝑓(1) × 𝑓(2) = −1
By the Intermediate Value Theorem there must be some 𝑐 ∈ [1, 2] such that 𝑓(𝑐) = 0. Therefore there is a
root between 1 and 2.
LESSON 2
Use the Intermediate Value Theorem to verify that there is a root of the equation 3𝑥 3 + 4𝑥 =
1 between 0 and 1.
SOLUTION
3𝑥 3 + 4𝑥 − 1 = 0
Let 𝑓(𝑥) = 3𝑥 3 + 4𝑥 − 1
𝑓(0) = −1
𝑓(1) = 3(1)3 + 4(1) − 1 = 6
𝑓(𝑥) is a polynomial and therefore continuous on the interval [0, 1].
𝑓(0) × 𝑓(1) = −6
By the Intermediate Value Theorem there must be some 𝑐 ∈ [0, 1] such that 𝑓(𝑐) = 0. Therefore there is a
root between 0 and 1.
P a g e | 78
DETERMINING THE ROOTS OF AN EQUATION
BISECTION METHOD
LESSON 1
Given that 𝑥 3 − 3𝑥 2 = 1 − 𝑥 has
a root between 2 and 3, find this root to 1 decimal
place using the bisection method.
LINEAR INTERPOLATION
LESSON 1
Use Linear Interpolation, twice
over, to determine the root of the equation
𝑥 3 − 3𝑥 2 = 1 − 𝑥 in the interval [2, 3].
SOLUTION
SOLUTION
𝑥 3 − 3𝑥 2 = 1 − 𝑥
3
2
𝑥 − 3𝑥 + 𝑥 − 1 = 0
𝑥 3 − 3𝑥 2 = 1 − 𝑥
𝑥 3 − 3𝑥 2 + 𝑥 − 1 = 0
Let 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 𝑥 − 1
𝑓(2) = 23 − 3(2)2 + 2 − 1 = −3
𝑓(3) = 33 − 3(3)2 + 3 − 1 = 2
Mid – point of interval is 2.5
𝑓(2.5) = 2.53 − 3(2.5)2 + 2.5 − 1 = −1.625
Since there is a sign change between 2.5 and 3 the
root occurs within this interval. We therefore now
bisect this interval. Mid – point is 2.75
Let 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 𝑥 − 1
𝑓(2.75) = 2.753 − 3(2.75)2 + 2.75 − 1
𝑓(2) = 23 − 3(2)2 + 2 − 1 = −3
= −0.140625
Due to sign changes the root must be in the
interval 2.75 and 3.
𝑓(3) = 33 − 3(3)2 + 3 − 1 = 2
Using similar triangles
Mid – point is 2.875
2 3−𝛼
=
3 𝛼−2
𝑓(2.875) = 2.8753 − 3(2.875)2 + 2.875 − 1
2(𝛼 − 2) = 3(3 − 𝛼)
= 0.8418
Due to sign changes the root is between 2.75 and
2.875
Mid – point is 2.8125
2𝛼 − 4 = 9 − 3𝛼
5𝛼 = 13
𝛼=
13
= 2.6
5
𝑓(2.8125) = 2.81253 − 3(2.8125)2 + 2.8125 − 1
= 0.33
Therefore root lies between 2.75 and 2.8125 and
since to 1 decimal place both limits are the same
the root is approximately 2.8.
13
13 3
13 2
13
𝑓 ( ) = ( ) − 3 ( ) + ( ) − 1 = −1.1
5
5
5
5
P a g e | 79
2
3−𝛼
=
1.1 (𝛼 − 2.6)
2(𝛼 − 2.6) = 1.1(3 − 𝛼)
2𝛼 − 5.2 = 3.3 − 1.1𝛼
3.1𝛼 = 8.5
𝛼 = 2.74
NEWTON RAPHSON
INTRODUCTION
LESSON 1
The equation
𝑥 3 − 𝑥 2 + 4𝑥 − 900 = 0 has exactly one real root,
𝛼. Taking 𝑥1 = 10 as the first approximation to 𝛼,
use the Newton – Raphson method to find a
second approximation, 𝑥2 , to 𝛼. Give your answer
to four significant figures.
SOLUTION
𝑓(𝑥) = 𝑥 3 − 𝑥 2 + 4𝑥 − 900
𝑓 ′ (𝑥) = 3𝑥 2 − 2𝑥 + 4
Let 𝑥1 = 10
𝑥2 = 𝑥1 −
𝑓(𝑥1 )
𝑓 ′ (𝑥1 )
𝑥2 = 10 −
103 − 102 + 4(10) − 900
3(10)2 − 2(10) + 4
𝑥2 = 10 −
40
284
𝑥2 = 9.859
LESSON 2
The equation
2𝑥 3 + 5𝑥 2 + 3𝑥 − 13 = 0 has exactly one real root
𝛼.
(i) Show that 𝛼 lies in the interval 1 < 𝛼 < 2.
(ii) Using the Newton – Raphson method with
initial estimate 𝑥1 = 1.5 to estimate the root
of the equation 2𝑥 3 + 5𝑥 2 + 3𝑥 − 13 = 0 in
the interval [1, 2], correct to 2 decimal places.
SOLUTION
(i) Let 𝑓(𝑥) = 2𝑥 3 + 5𝑥 2 + 3𝑥 − 13
𝑓(1) = 2(1)3 + 5(1)2 + 3(1) − 13 = −3
𝑓(2) = 23 + 5(2)2 + 3(2) − 13 = 29
Since 𝑓(1) × 𝑓(2) < 0, by the Intermediate
Value Theorem there is a root in the interval
[1, 2]
(ii) 𝑓(𝑥) = 2𝑥 3 + 5𝑥 2 + 3𝑥 − 13
𝑓 ′ (𝑥) = 6𝑥 2 + 10𝑥 + 3
𝑥1 = 1.5
2(1.5)3 + 5(1.5)2 + 3(1.5) − 13
𝑥2 = 1.5 −
6(1.5)2 + 10(1.5) + 3
19
𝑥2 = 1.5 −
63
P a g e | 80
𝑥2 = 1.1984
3
𝑥4 = √
𝑥3 = 1.1984 −
2(1.1984)3 + 5(1.1984)2 + 3(1.1984) − 13
6(1.1984)2 + 10(1.1984) + 3
Therefore, the approximation is −1.26 correct to 2
decimal places.
𝑥3 = 1.1984 − 0.0516
𝑥3 = 1.1467
𝑥3
= 1.1467 −
2(1.1467)3 + 5(1.1467)2 + 3(1.1467) − 13
6(1.1467)2 + 10(1.1467) + 3
𝑥3 = 1.1984 − 0.001357
𝑥3 = 1.1453
Since 𝑥2 and 𝑥3 correct to 2 decimal places are
both equal to 1.15, 𝛼 = 1.15.
DERIVING AN ITERATIVE FORMULA
EXAMPLE 1
3 4𝑥𝑛 +1
Show that 𝑥𝑛+1 = √
2
is an
approximate iterative formula for finding the root
of 𝑓(𝑥) = 2𝑥 3 − 4𝑥 − 1 = 0. Apply the iterative
formula with initial approximation 𝑥1 = −1.2, to
obtain an approximation of the root to 2 decimal
places.
SOLUTION
2𝑥 3 − 4𝑥 − 1 = 0
2𝑥 3 = 4𝑥 + 1
𝑥3 =
4𝑥 + 1
2
3
𝑥=√
4𝑥 + 1
2
3 4𝑥 + 1
𝑛
𝑥𝑛+1 = √
2
𝑥1 = −1
3
4(−1.2) + 1
= −1.2386
2
3
4(−1.2386) + 1
= −1.2551
2
𝑥2 = √
𝑥3 = √
4(−1.2551) + 1
= −1.2621
2
P a g e | 81
MATRICES
At the end of this section, students should be able to:
1.
2.
3.
4.
5.
6.
reduce a system of linear equations to echelon form;
row-reduce the augmented matrix of an 𝑛 × 𝑛 system of linear equations, 𝑛 = 2, 3;
determine whether the system is consistent, and if so, how many solutions it has;
find all solutions of a consistent system;
invert a non-singular 3 × 3 matrix;
solve a 3 × 3 system of linear equations, having a non-singular coefficient matrix, by using
its inverse.
P a g e | 82
MATRICES

A matrix is a rectangular array of elements enclosed in brackets. A matrix is defined: number of rows
× number of columns (this is the size/order of a matrix).

A square matrix contains the same number of rows and columns.

Two matrices are equal if they contain the same corresponding elements.

The addition and subtraction of matrices is only possible if the matrices are of the same size.
o

Matrix addition is commutative and associative.
Two matrices can be multiplied if the number of columns in the first matrix is equal to the number of
rows in the second.
o

Matrix multiplication is not commutative but it is associative.
The identity matrix 𝐼 for 2 × 2 matrices is (
1
0
1
0
) and for 3 × 3 matrices it is (0
1
0
Matrix Multiplication
1
If 𝐴 = (−1
1
LESSON 1
2
2
−2
−1
1 ) and
3
2
𝐵 = (1
0
−1
1
1
1
0)
1
(i) find 𝐴𝐵
(ii) deduce 𝐴−1
SOLUTION
1
2 −1 2 −1
2
1 ) (1 1
−2 3
0 1
(i) 𝐴𝐵 = (−1
1
1
0)
1
1(2) + 2(1) + (−1)(0)
= ( −1(2) + 2(1) + 1(0)
1(2) + (−2)(1) + 3(0)
4
= (0
0
0
4
0
0
0)
4
(ii) 𝐴𝐴−1 = 𝐼
𝐴𝐵 = 4𝐼
1
𝐴 ( 𝐵) = 𝐼
4
2
1
Therefore 𝐴−1 = (1
4
0
−1
1
1
1
0)
1
1(−1) + 2(1) + (−1)(1)
(−1)(−1) + 2(1) + 1(1)
1(−1) + (−2)(1) + 3(1)
1(1) + 2(0) + (−1)(1)
(−1)(1) + 2(0) + 1(1))
1(1) + (−2)(0) + 3(1)
0
1
0
0
0).
1
P a g e | 83
THE DETERMINANT OF A 𝟑 × 𝟑
SOLUTION
MATRIX
(i) |𝑀| = 3 |
INTRODUCTION
= 3[2(3) − 1(1)] − 1[1(3) − 1(1)] + 2[(1(1) − 1(2)]
𝑎
The determinant of 𝐴 = (𝑑
𝑔
𝑒
𝑎|
ℎ
o
𝑑
𝑓
|−𝑏|
𝑔
𝑖
𝑏
𝑒
ℎ
𝑐
𝑓) is
𝑖
𝑓
𝑑
|+𝑐|
𝑔
𝑖
𝑒
|.
ℎ
If the matrix 𝐴 contains a row with ALL
zeros then |𝐴| = 0.
o
For square matrices |𝐴𝑇 | = |𝐴|.
o
If 𝐴 contains two identical rows or
columns then |𝐴| = 0.
o
Interchanging two rows or columns of a
o
1
1
| + 2|
3
1
= 11
(ii) Since |𝑀| ≠ 0, 𝑀 is non – singular.
LESSON 3
Find the value of 𝑘 for which the
matrix
2
(𝑘
2
3
1
0
−1
4)
3
is singular.
SOLUTION
For 𝑘 ∈ ℝ, |𝑘𝐴| = 𝑘 𝑛 |𝐴| where 𝐴 is a 𝑛 ×
2|
𝑛 matrix.
2(3) − 3(3𝑘 − 8) − 1(−2) = 0
If a row or a column of a matrix is
6 − 9𝑘 + 24 + 2 = 0
multiplied by a constant, 𝑘, then the
9𝑘 = 32
𝑘.
Adding a multiple of one row to another
2
|
1
= 3(5) − 1(2) + 2(−1)
determinant.
determinant of the matrix is multiplied by
o
1
1
| − 1|
3
1
2
|𝑘
2
matrix 𝐴 changes the sign of the
o
2
1
3
1
0
1
0
𝑘=
−1
4 |=0
3
4
𝑘 4
𝑘
| −3|
| − 1|
3
2 3
2
1
|=0
0
32
9
LESSON 4
does not affect the determinant.
3
Determine |−6
9
6
1
2
15
0 | by
5
o
|𝐴𝐵| = |𝐴||𝐵|
factoring.
o
A square matrix is singular if |𝐴| = 0
SOLUTION
otherwise it is non – singular.
be obtained by factorizing the matrix. This is done
o
det 𝐴−1 =
1
The determinant of a matrix can
by factoring out the common factor from each row
det 𝐴
or column.
(i) Find the value of the determinant of 𝑀.
6 15
1 0 | factoring out 3 from 𝑐1
2 5
1 6 3
(3)(5) |−2 1 0| factoring out 5 from 𝑐3
3 2 1
(ii) State, giving a reason, whether 𝑀 is singular
To find the determinant it is easier to use the
LESSON 2
3
𝑀 = (1
1
The matrix 𝑀 is given by
1
2
1
2
1).
3
or non – singular.
1
3 |−2
3
elements of 𝑐3 because it contains a zero.
= 15 {3 |
−2
3
1
1
| + 1|
2
−2
6
|}
1
The Transpose of a Matrix
The transpose of a matrix is created by
= 15[3(−7) + 13] = −120
interchanging the rows and columns. The
LESSON 5
1
The matrix 𝐴 = (−3
−3
1
2
−1
1
2 ).
−2
transpose of a matrix 𝐴 is denoted 𝐴𝑇 .
o
(𝐴𝐵)𝑇 = 𝐵𝑇 𝐴𝑇
(i) Show that |𝐴| = −5.
o
A square matrix is symmetric if 𝐴𝑇 = 𝐴.
(ii) Matrix 𝐴 is changed to form new matrices 𝐵, 𝐶
o
A square matrix is called skew –
and 𝐷. Write down the determinant of EACH
symmetric is 𝐴𝑇 = −𝐴.
of the new matrices, giving a reason for your
For example, if
answers in EACH case.
(a) Matrix 𝐵 is formed by interchanging
column 1 and column 2 of matrix 𝐴 and
𝐴=(
4
6
6
9
) then 𝐴𝑇 = (3
5
5
−1
3
then interchanging row 1 and row 2 of the
LESSON 6
resulting matrix.
4 2
by 𝑋 = (−5 6
7 9
1 6
𝑌 = (5 2).
4 6
(b) Column 1 of matrix 𝐶 is formed by adding
column 2 to column 1 of matrix 𝐴. The
other columns remain unchanged.
4
−1)
9
The matrices 𝑋 and 𝑌 are given
1
8 ) and
−3
(c) Matrix 𝐷 is formed by multiplying each
element of matrix 𝐴 by 4.
SOLUTION
Calculate
(a) the determinant of 𝑋
2
−3
(i) |𝐴| = 1 | 2
|− 1|
−1 −2
−3
2
−3 2
|+ 1|
|
−2
−3 −1
= 1(−2) − 1(12) + 1(9)
= −5
(ii) (a) interchanging columns will change the
sign of the determinant to 5 and then
interchanging the rows will change the
sign of the determinant to −5.
(b) The determinant remains as −5 since
adding a multiple of a row does not affect
the determinant.
(c) Since each element is multiplied by 4 the
value of the determinant is
−5 × 43 = −320
(b) 𝑌 𝑇 𝑋
SOLUTION
(a) |𝑋| = 4 |
6
9
8
−5
| − 2|
−3
7
8
−5
| + 1|
−3
7
= 4(−90) − 2(−41) + 1(−87)
= −365
(b) 𝑌 𝑇 = (
1
6
𝑌𝑇 𝑋 = (
5
2
1
6
7
=(
56
4
)
6
4
4
) (−5
6
7
68 29
)
78 4
5
2
2
6
9
1
8)
−3
6
|
9
𝑎1
𝐵𝑇 = (𝑏1
𝑐1
Finding the inverse of A Matrix
(Cofactor Method)
𝑎2
𝑏2
𝑐2
𝑎3
𝑏3 )
𝑐3
1
Step 4: Use the relation 𝐴−1 = |𝐴| (𝐴 adj) to find
𝑎1
If 𝐴 = (𝑎2
𝑎3
|
𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑏
𝑐2 ) then | 2
𝑏3
𝑐3
𝑐2 𝑎2
|, |
𝑐3 𝑎3
𝑐2
𝑐3 | and
𝑏2
| are called the minors of 𝑎1 , 𝑏1 and 𝑐1
𝑏3
𝐴−1 .
LESSON 7
respectively. A minor of an element is obtained by
−1
(0
4
deleting the row and column containing that
SOLUTION
element and finding the determinant of the 2 × 2
−1 −1 1
Let 𝐴 = ( 0 −1 3)
4
1 2
−1 3
−1
|𝐴| = −1 |
| + 4|
1 2
−1
matrix which remains.
The cofactors of a matrix are determined by
multiplying the minor by ±1 in the following
−1
−1
1
Find the inverse of
1
3)
2
1
|
3
= −1(−5) + 4(−2)
order
+
(−
+
= −3
− +
+ −)
− +
Step 1: Find the determinant of the matrix
𝑎1
𝐴 = (𝑎2
𝑎3
𝑏1
𝑏2
𝑏3
𝑐1
𝑐2 )
𝑐3
Determinant of 𝐴, det 𝐴, |𝐴|, ∆
𝑏
|𝐴| = 𝑎1 | 2
𝑏3
𝑎2
𝑐2
| − 𝑏1 |𝑎
𝑐3
3
𝑐2
𝑎2
𝑐3 | + 𝑐1 |𝑎3
𝑏2
|
𝑏3
The elements of any row or column can their
corresponding cofactors can be used to determine
the determinant. i.e.
𝑏
|𝐴| = 𝑎1 | 2
𝑏3
𝑐2
𝑏
| − 𝑎2 | 1
𝑐3
𝑏3
𝑐1
𝑏
| + 𝑎3 | 1
𝑐3
𝑏2
𝑐1
| using
𝑐2
the first row and its corresponding cofactors.
Step 2: Write the matrix, say 𝐵, of cofactors.
𝑏2 𝑐2
|
𝑏3 𝑐3
𝑏 𝑐
𝐵 = − | 1 1|
𝑏3 𝑐3
𝑏 𝑐
+ | 1 1|
( 𝑏2 𝑐2
+|
𝑎2
− |𝑎
3
𝑎1
+ |𝑎
3
𝑎1
− |𝑎
2
𝑐2
𝑐3 |
𝑐1
𝑐3 |
𝑐2
𝑐2 |
𝑎2
𝑎3
𝑎
−| 1
𝑎3
𝑎
+| 1
𝑎2
+|
𝑏2
|
𝑏3
𝑏1
|
𝑏3
𝑏1
|
𝑏2 )
Step 3: Write the matrix, 𝐵𝑇 . This matrix is called
the adjoint of 𝐴 (𝐴 adj)
−1 3
0 3
| −|
|
1 2
4 2
−1 1
−1 1
𝐵 = −|
| +|
|
1 2
4 2
−1 1
−1 1
(+ |−1 3| − | 0 3|
−5 12 4
𝐵 = ( 3 −6 −3)
−2 3
1
−5 3 −2
𝐵𝑇 = ( 12 −6 3 )
4 −3 1
1 −5 3 −2
𝐴−1 =
( 12 −6 3 )
−3
4 −3 1
+|
0 −1
+|
|
4 1
−1 −1
−|
|
4
1
−1 −1
+|
|
0 −1 )
LESSON 9
SYSTEMS OF EQUATIONS
LESSON 8
−𝑥 − 5𝑦 − 5𝑧 = 2
𝑌 and 𝑋 are 3 × 1 matrices and
4𝑥 − 5𝑦 + 4𝑧 = 19
are related by the equation 𝑌 = 𝐴𝑋, where
1 0
𝐴 = (7 5
3 2
Solve the equations
𝑥 + 5𝑦 − 𝑧 = −20
3
0) is non – singular.
1
SOLUTION
Step 1: Write the system in the form 𝐴𝑋 = 𝐵
Find
where 𝐴, 𝑋 and 𝐵 are matrices
(a) 𝐴−1
−1
(4
1
10
(b) 𝑋, when 𝑌 = (12)
8
7
0
| −0|
3
1
2
−5 𝑥
4 ) (𝑦) = ( 19 )
−20
−1 𝑧
Step 2: Find 𝐴−1
SOLUTION
5
(a) |𝐴| = 1 |
2
−5
−5
5
0
7
| + 3|
1
3
5
|
2
𝐴−1 = −
1 −15
( 8
150
25
−30
6
0
−45
−16)
25
= 1(5) + 3(−1)
Step 3: Multiply both sides of equation by 𝐴−1
=2
𝑥
1 −15 −30
(𝑦 ) = −
( 8
6
150
𝑧
25
0
𝑥
300
1
(𝑦 ) = −
( 450 )
150
𝑧
−450
𝑥
−2
(𝑦) = (−3)
𝑧
3
7 0
0
7 5
| −|
| |
|
3 1
1
3 2
3
1 3
1 0
| |
| −|
|
1
3 1
3 2
3
1 3
1 0
| −|
| |
|
0
7 0
7 5 )
−7 −1
−8 −2)
21 5
6 −15
−8 21 )
−2
5
6 −15
1 5
= (−7 −8 21 )
2
−1 −2
5
5
|
2
0
𝐵 = −|
2
0
( |5
5
=( 6
−15
5
𝐵𝑇 = (−7
−1
𝐴−1
(b) 𝐴𝑋 = 𝑌
𝐴−1 𝐴𝑋 = 𝐴−1 𝑌
𝑋 = 𝐴−1 𝑌
1 5
𝑋 = (−7
2
−1
=
6
−8
−2
−15 10
21 ) (12)
8
5
5(10) + 6(12) + (−15)(8)
1
( (−7)(10) + (−8)(12) + 21(8) )
2
(−1)(10) + (−2)(12) + (5)(8)
1
1 2
= (2) = (1)
2 6
3
2
−45
−16) ( 19 )
−20
25
This equation is said to be consistent since it has a
unique solution.
ROW REDUCTION
The following elementary row operations can be
performed on a matrix.
-
The interchanging of rows
-
The multiplication of a row by a non –
zero scalar.
-
The adding of the multiple of a row to
another row.
These operations can convert a 3 × 3 matrix (or
1
any other matrix) to row echelon form (0
0
1 𝑎 𝑏
or (0 1 𝑐 )
0 0 0
𝑎
1
0
𝑏
𝑐)
1
This can be done by
1.
Beginning with the left – most column and
use row operations to make the first
element in this column a 1 and the
elements below it zeros.
𝑎
𝑑
(
𝑔
2.
𝑏
𝑒
ℎ
1
𝑐
𝑓 ) → (0
𝑖
0
𝑏1
𝑒1
ℎ1
𝑐1
𝑓1 )
𝑖1
Ignore the row and the column with the 1
1 −1 −1 0 0
−1 3 | 0 1 0)
1
2 0 0 1
1 1 −1 −1 0 0
𝑅3 − 4𝑅1 → (0 −1 3 | 0 1 0)
0 −3 6 4 0 1
1 1 −1 −1 0 0
−𝑅2 → (0 1 −3| 0 −1 0)
0 −3 6 4
0 1
1 0
2 −1 1 0
𝑅1 − 𝑅2 → (0 1 −3| 0 −1 0)
0 −3 6 4
0 1
1 0 2 −1 1 0
𝑅3 + 3𝑅2 → (0 1 −3| 0 −1 0)
0 0 −3 4 −3 1
0
1 0 2 −1 1
1
− 𝑅3 → (0 1 −3| 0 −1 0 )
4
1
3
0 0 1 −3 1 −3
1 0 2 −1 1 0
𝑅2 + 3𝑅3 → (0 1 0| −4 2 −1 )
4
1
0 0 1 −3 1 −3
5
2
1 0 0 3 −1 3
𝑅1 − 2𝑅3 → 0 1 0| −4 2 −1
0 0 1 −4 1 −1
(
3
3)
1
−𝑅1 → (0
4
1 5
𝐴−1 = (−12
3
−4
−3
6
3
2
−3)
−3
created from step 1 and repeat step 1 on
Row Reduction and Systems of Equations
the remaining matrix.
1
(0
0
3.
𝑏1
𝑒1
ℎ1
𝑐1
1
𝑓1 ) → (0
0
𝑖1
𝑏1
1
0
𝑐1
𝑓2 )
𝑖2
Repeat this process until the desired
matrix is obtained.
LESSON 10
−1
(0
4
−1
−1
1
Find the inverse of
To obtain the inverse of a matrix
augmented matrix (𝐴|𝐼) to (𝐴−1 |𝐼).
−1
−1
1
11
3| 0
20
A consistent system can have

a unique solution

infinitely many solutions.
The use of row reduction greatly assists in
we can use row operations to convert the
−1
(0
4
inconsistent.
An inconsistent system has no solution.
1
3).
2
SOLUTION
A system of equations can either be consistent or
0
1
0
0
0)
1
determining the consistency of a system of
equations.
1
An upper triangle matrix (0
0
unique solution.
𝑎
1
0
𝑏𝑑
𝑐 | 𝑒 ) indicates a
1𝑓
1
Matrices of the form (0
0
𝑎
1
0
𝑏𝑑
𝑐 | 𝑒 ) indicate
00
𝑎
1
0
𝑏𝑑
𝑐 | 𝑒 ) indicate no
0𝑓
solution.
LESSON 11
𝑥=2
Since the system has a unique solution it is said to
infinitely many solutions.
1
Matrices of the form (0
0
𝑥 + 2(−1) = 0
be consistent.
LESSON 12
of the system of equations.
𝑥 − 2𝑦 + 2𝑧 = −3
A system of three equations is
𝑥 + 2𝑦 − 𝑧 = 4
−𝑥 − 𝑦 + 𝑧 = −3
2𝑥 + 4𝑦 − 2𝑧 = 8
−𝑦 + 3𝑧 = −3
4𝑥 + 𝑦 + 2𝑧 = 6
(i) Write the augmented matrix for the system.
(ii) Use row reduction to solve the system of
equations.
SOLUTION
1 −3
3|−3)
2 6
1 −3
3|−3)
2 6
1 −1 3
−1 3 |−3)
1
2 6
1 1 −1 3
𝑅3 − 4𝑅1 → (0 −1 3 |−3)
0 −3 6 −6
1 1 −1 3
−𝑅2 → (0 1 −3| 3 )
0 −3 6 −6
1 0
2 0
𝑅1 − 𝑅2 → (0 1 −3| 3 )
0 −3 6 −6
1 0 2 0
𝑅3 + 3𝑅2 → (0 1 −3|3)
0 0 −3 3
1 0 2 0
1
𝑅3 → (0 1 −3| 3 )
3
0 0 1 −1
−1
(i) ( 0
4
−1
(ii) ( 0
4
−1
−1
1
−1
−1
1
1
−𝑅1 → (0
4
From 𝑅3 : 𝑧 = −1
From 𝑅2 : 𝑦 − 3𝑧 = 3
:𝑦 − 3(−1) = 3
:𝑦 = 0
From 𝑅3 : 𝑥 + 2𝑧 = 3
Determine the general solutions
SOLUTION
2 −3
−1| 4 )
−2 8
1 −2 2 −3
𝑅2 − 𝑅1 → (0 4 −3| 7 )
2 4 −2 8
1 −2 2 −3
𝑅3 − 2𝑅1 → (0 4 −3| 7 )
0 8 −6 14
1 −2 2 −3
1
3
𝑅2 → (0 1 − | 7 )
4 4
4
0 8 −6 14
1 −2 2 −3
3 7
𝑅3 − 8𝑅2 → (0 1 − | )
4 4
0 0
0 0
1
(1
2
−2
2
4
The row of zeros indicates that 0𝑥 + 0𝑦 + 0𝑧 = 0,
therefore we can use a parameter for one of the
variables and express the other variables in terms
of this parameter. Furthermore, the row of zeros
indicates that the system has infinitely many
solutions.
Let 𝑧 = 𝑡
From row 2:
3
7
𝑦− 𝑧=
4
4
3
7
𝑦− 𝑡=
4
4
3
7
𝑦= 𝑡+
4
4
From row 1:
This result indicates that when 𝑘 = 8, the
𝑥 − 2𝑦 + 2𝑧 = −3
system of equations has infinitely many
3
7
𝑥 − 2 ( 𝑡 + ) + 2𝑡 = −3
4
4
3
7
𝑥 − 𝑡 − + 2𝑡 = −3
2
2
1
1
𝑥=− 𝑡+
2
2
solutions.
Let 𝑧 = 𝑡
From 𝑅2 :
6𝑦 + 4𝑧 = 22
6𝑦 + 4𝑡 = 22
3
Therefore, if we say that 𝑧 = 1 then 𝑦 = (1) +
4
7
4
5
1
1
2
2
2
= and 𝑥 = − (1) + = 0.
𝑥
5
So (𝑦) = ( 2 ) would be one solution for the
𝑧
0
system.
Discuss the solutions of the given
equations when (i) 𝑘 = 4 (ii) 𝑘 = 8
3𝑥 + 3𝑦 + 𝑧 = 10
𝑥 − 𝑦 − 𝑧 = −4
−2𝑥 + 2𝑦 + 2𝑧 = 𝑘
SOLUTION
3
(1
−2
3
−1
2
1 10
−1|−4)
2 𝑘
1 −1
Interchange 𝑅2 and 𝑅1 → ( 3
3
−2 2
1 −1 −1 −4
𝑅2 − 3𝑅1
→ (0 6
4 | 22 )
𝑅3 + 2𝑅1
0 0
0 𝑘−8
11 2
− 𝑡
3 3
From 𝑅1 :
1
LESSON 13
𝑦=
−1 −4
1 | 10 )
2 𝑘
(i) When 𝑘 = 4, we have
0𝑥 + 0𝑦 + 0𝑧 = 𝑘 − 8 = −4
This result indicates that when 𝑘 = 4, the
system of equations is inconsistent and has no
solution.
(ii) When 𝑘 = 8, we have that
0𝑥 + 0𝑦 + 0𝑧 = 𝑘 − 8 = 0
𝑥 − 𝑦 − 𝑧 = −4
11 2
𝑥 − ( − 𝑡) − 𝑡 = −4
3 3
1 1
𝑥=− − 𝑡
3 3
DIFFERENTIAL EQUATIONS
At the end of this section, students should be able to:
1. solve first order linear differential equations 𝑦 ′ − 𝑘𝑦 = 𝑓(𝑥) using an integrating factor,
given that 𝑘 is a real constant or a function of 𝑥, and 𝑓 is a function;
2. solve first order linear differential equations given boundary conditions;
3. solve second order ordinary differential equations with constant coefficients of the form
𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0 = 𝑓(𝑥), where a, 𝑏, 𝑐 ∈ ℝ and 𝑓(𝑥) is:
(a) a polynomial,
(b) an exponential function,
(c) a trigonometric function;
and the complementary function may consist of
(a) real and distinct root,
(b) 2 equal roots,
(c) 2 complex roots;
4. solve second order ordinary differential equation given boundary conditions;
5. use substitution to reduce a second order ordinary differential equation to a suitable form.
DIFFERENTIAL EQUATIONS
A differential equation is an equation which contains derivatives of a function or functions.
For a first order differential equation the highest derivative is the first derivative.
For a second order differential equation the highest derivative is the second derivative.
Separable Differential Equations
sin 𝑦 = 𝐴𝑒 sin 𝑥 where 𝐴 = 𝑒 𝑐
The solution of this type of equation can be
These solutions are called general solutions of the
achieved by separating the variables and
differential equation because the value of the
integrating both sides of the equation with respect
constant is unknown.
to the relative variables.
LESSON 3
LESSON 1
Solve the differential equations
differential equation
𝑑𝑦 5𝑥 2 − 3
=
𝑑𝑥
𝑦
𝑑𝑦
𝜋
= csc 𝑦 ; 𝑦 =
when 𝑥 = 4
𝑑𝑥
3
1
1
𝑑𝑦 = 2 𝑑𝑥
csc 𝑦
𝑥
𝑥2
∫ 𝑦 𝑑𝑦 = ∫(5𝑥 2 − 3) 𝑑𝑥
𝑦 2 5𝑥 3
=
− 3𝑥 + 𝑐
2
3
∫
10𝑥 3
− 6𝑥 + 𝑐
3
Solve the differential equation
SOLUTION
𝑑𝑦
= cos 𝑥 tan 𝑦
𝑑𝑥
1
𝑑𝑦 = cos 𝑥 𝑑𝑥
tan 𝑦
cos 𝑦
𝑑𝑦 = ∫ cos 𝑥 𝑑𝑥
sin 𝑦
ln(sin 𝑦) = sin 𝑥 + 𝑐
𝑒 ln (sin 𝑦) = 𝑒 sin 𝑥+𝑐
1
𝑑𝑦 = ∫ 𝑥 −2 𝑑𝑥
csc 𝑦
∫ sin 𝑦 𝑑𝑦 = ∫ 𝑥 −2 𝑑𝑥
𝑑𝑦
= cos 𝑥 tan 𝑦
𝑑𝑥
sin 𝑦 = 𝑒 𝑐 𝑒 sin 𝑥
𝜋
SOLUTION
𝑦 𝑑𝑦 = (5𝑥 2 − 3) 𝑑𝑥
∫
𝑑𝑦
= csc 𝑦
𝑑𝑥
3
𝑑𝑦 5𝑥 2 − 3
=
𝑑𝑥
𝑦
LESSON 2
𝑥2
when 𝑥 = 4, 𝑦 = .
SOLUTION
𝑦2 =
Find the particular solution of the
∫
𝑓 ′ (𝑥)
𝑑𝑥 = ln 𝑓(𝑥) + 𝑐
𝑓(𝑥)
1
− cos 𝑦 = − + 𝑐
𝑥
𝜋
1
− cos ( ) = − + 𝑐
3
4
1
1
− = − +𝑐 →
2
4
1 1
cos 𝑦 = +
𝑥 4
1
− =𝑐
4
The Integrating Factor
𝑦=−
7𝑥
+ 𝑐𝑥 3
2
INTRODUCTION
Linear differential equations of the form
LESSON 5
𝑑𝑦
+ 𝑃(𝑥)𝑦 = 𝑄(𝑥)
𝑑𝑥
can be solved by multiplying throughout by the
Integrating Factor,
LESSON 4
Solve the differential equation
𝑑𝑦
+ (cot 𝑥)𝑦 = 2 cos 𝑥
𝑑𝑥
𝜋
given that 𝑦 = 2 when 𝑥 = .
2
𝑒 ∫ 𝑃(𝑥) .
SOLUTION
Solve the following differential
𝑑𝑦
+ (cot 𝑥)𝑦 = 2 cos 𝑥
𝑑𝑥
equations
𝑥
𝑑𝑦
− 3𝑦 = 7𝑥
𝑑𝑥
𝑃(𝑥) = cot 𝑥
cos 𝑥
𝑑𝑥 = ln sin 𝑥
sin 𝑥
SOLUTION
∫ cot 𝑥 𝑑𝑥 = ∫
Step 1: Write the DE in the form
𝐼. 𝐹 = 𝑒 ln sin 𝑥 = sin 𝑥
𝑑𝑦
+ 𝑃(𝑥)𝑦 = 𝑄(𝑥)
𝑑𝑥
𝑑𝑦 3
− 𝑦=7
𝑑𝑥 𝑥
𝑑𝑦
cos 𝑥
+ sin 𝑥 .
. 𝑦 = 2 sin 𝑥 cos 𝑥
𝑑𝑥
sin 𝑥
𝑑𝑦
∫ sin 𝑥
+ (cos 𝑥)𝑦 𝑑𝑥 = 2 ∫ cos 𝑥 sin 𝑥 𝑑𝑥
𝑑𝑥
Step 2: Calculate the I.F using
𝑦 sin 𝑥 = sin2 𝑥 + 𝐾
𝐼. 𝐹 = 𝑒 ∫ 𝑃(𝑥) 𝑑𝑥
𝑦 = sin 𝑥 + 𝐾 csc 𝑥
sin 𝑥
3
𝑃(𝑥) = −
𝑥
3
𝐼. 𝐹 = 𝑒 ∫ −𝑥 𝑑𝑥 = 𝑒 −3 ln 𝑥 = 𝑒 ln 𝑥
𝑦 = 2 when 𝑥 =
−3
= 𝑥 −3
Step 3: Multiply each term in the equation by I.F
𝑥 −3
𝑑𝑦
− 3𝑥 −4 𝑦 = 7𝑥 −3
𝑑𝑥
L.H.S resembles the product rule where I.F is
actually 𝑢
Step 4: Integrate both sides of the equation w.r.t.
𝑥
𝑑𝑦
∫ (𝑢
+ 𝑢 𝑃(𝑥)𝑦) 𝑑𝑥 = ∫ 𝑢𝑄(𝑥) 𝑑𝑥
𝑑𝑥
∫𝑥
−3
𝑑𝑦
− 3𝑥 −4 𝑦 𝑑𝑥 = ∫ 7𝑥 −3 𝑑𝑥
𝑑𝑥
𝑦
7
= − 2+𝑐
3
𝑥
2𝑥
𝜋
∫ 𝑓 ′ (𝑥)[𝑓(𝑥)]𝑛 𝑑𝑥 =
2
[𝑓(𝑥)]𝑛+1
+𝑐
𝑛+1
𝜋
𝐾
2 = sin ( ) +
𝜋
2
sin ( )
2
𝐾
2= 1+
1
𝐾=1
𝑦 = sin 𝑥 + csc 𝑥
LESSON 6
Determine the particular solution
of the differential equation
𝑑𝑦 + 4𝑦𝑑𝑥 = 𝑒 −3𝑥 𝑑𝑥
given that 𝑦 = 3 when 𝑥 = 0.
SOLUTION
𝑑𝑦 + 4𝑦𝑑𝑥 = 𝑒 −3𝑥 𝑑𝑥
𝑑𝑦
+ 4𝑦 = 𝑒 −3𝑥
𝑑𝑥
I. F = 𝑒 ∫ 4𝑑𝑥 = 𝑒 4𝑥
𝑑𝑦
𝑒 4𝑥
+ 4𝑒 4𝑥 𝑦 = 𝑒 −3𝑥 𝑒 4𝑥
𝑑𝑥
𝑑𝑦
∫ 𝑒 4𝑥
+ 4𝑒 4𝑥 𝑦 𝑑𝑥 = ∫ 𝑒 𝑥 𝑑𝑥
𝑑𝑥
𝑦𝑒 4𝑥 = 𝑒 𝑥 + 𝑐
𝑦 = 𝑒 −4𝑥 (𝑒 𝑥 + 𝑐)
When 𝑦 = 3, 𝑥 = 0
3 = 𝑒 4(0) (𝑒 0 + 𝑐)
2=𝑐
𝑦 = 𝑒 −4𝑥 (𝑒 𝑥 + 2)
Linear Differential Equations with
Constant Coefficients
Homogenous Differential Equations
𝑎
𝑑𝑦
d2 𝑦
𝑑𝑦
+ 𝑏𝑦 = 0 and 𝑎 2 + 𝑏
+ 𝑐𝑦 = 0
𝑑𝑥
𝑑𝑥
𝑑𝑥
are first and second order homogenous equations
where 𝑎, 𝑏 and 𝑐 are constants. The solution of
these equations is called the complementary
function (C.F).
LESSON 7
Solve the differential equation
5
𝑑𝑦
− 2𝑦 = 0
𝑑𝑥
SOLUTION
𝑑𝑦
= 2𝑦
𝑑𝑥
𝑑𝑦 2
= 𝑦
𝑑𝑥 5
1
2
𝑑𝑦 = 𝑑𝑥
𝑦
5
5
1
2
∫ 𝑑𝑦 = ∫ 𝑑𝑥
𝑦
5
ln 𝑦 =
2
𝑥+𝑐
5
2
𝑒 ln 𝑦 = 𝑒 5𝑥+𝑐
2
𝑦 = 𝑒 𝑐 𝑒 5𝑥
2
𝑦 = 𝐴𝑒 5𝑥
In general, the solution (complementary function)
of a first order differential equation is of the form
𝑎
𝑑𝑦
+ 𝑏𝑦 = 0
𝑑𝑥
is
𝑏
𝑦 = 𝐴𝑒 −𝑎𝑥
Auxiliary Quadratic Equation
Given the equation
𝑑2𝑦
𝑑𝑦
𝑎 2+𝑏
+ 𝑐𝑦 = 0
𝑑𝑥
𝑑𝑥
If the auxiliary quadratic equation has a
repeated root , 𝛼,
𝑦 = 𝑒 𝛼𝑥 (𝐴𝑥 + 𝐵)
The complementary function is determined by the
roots of the nature of the roots of the quadratic
is the general solution of the differential
equation.
auxiliary equation which is
𝑎𝑢2 + 𝑏𝑢 + 𝑐 = 0
LESSON 8
Solve the equation
𝑑2𝑦
𝑑𝑦
−5
+ 4𝑦 = 0
𝑑𝑥 2
𝑑𝑥
SOLUTION
𝑑2 𝑦
𝑑𝑦
−5
+ 4𝑦 = 0
2
𝑑𝑥
𝑑𝑥
Auxiliary equation
2
𝑢 − 5𝑢 + 4 = 0
𝑦 = 𝐴𝑒 + 𝐵𝑒
𝑑2𝑦
𝑑𝑦
−2
+ 5𝑦 = 0
𝑑𝑥 2
𝑑𝑥
SOLUTION
𝑑2 𝑦
𝑑𝑦
+2
+ 5𝑦 = 0
2
𝑑𝑥
𝑑𝑥
Auxiliary equation
𝑢2 + 2𝑢 + 5 = 0
𝑦 = 𝑒 𝑥 (𝐴 cos 2𝑥 + 𝐵 sin 2𝑥)
4𝑥
If the auxiliary quadratic equation has
real and distinct roots, 𝛼 and 𝛽 then
If the auxiliary quadratic equation has
complex roots of the form 𝛼 ± 𝛽𝑖,
𝑦 = 𝑒 𝛼𝑥 (𝐴 cos 𝛽𝑥 + 𝐵 sin 𝛽𝑥)
𝑦 = 𝐴𝑒 𝛼𝑥 + 𝐵𝑒 𝛽𝑥
is the general solution of the differential
equation.
LESSON 9
Solve the equation
2
4
𝑑 𝑦
𝑑𝑦
−4
+𝑦=0
2
𝑑𝑥
𝑑𝑥
SOLUTION
is the general solution of the differential
equation.
Non – Homogeneous Differential Equations
Non – homogeneous first – order and second –
order differential equation are of the form
𝑎
𝑑2 𝑦
𝑑𝑦
4 2−4
+𝑦=0
𝑑𝑥
𝑑𝑥
Auxiliary equation
4𝑢2 − 4𝑢 + 1 = 0
𝑢=
1
(twice)
2
1
Solve the equation
𝑢 = 1 ± 2𝑖
𝑢 = 1, 4
𝑥
LESSON 10
𝑦 = 𝑒 2𝑥 (𝐴 + 𝐵𝑥)
𝑎
𝑑𝑦
+ 𝑏𝑦 = 𝑓(𝑥)
𝑑𝑥
𝑑2𝑦
𝑑𝑦
+𝑏
+ 𝑐𝑦 = 𝑓(𝑥),
2
𝑑𝑥
𝑑𝑥
𝑓(𝑥) ≠ 0
The particular integral is any solution of these
types of differential equations.
𝑓(𝑥) is a Polynomial
SOLUTION
LESSON 11
𝑑2 𝑦
− 4𝑦 = 3𝑥 2 − 2𝑥 + 1
𝑑𝑥 2
Solve the equation
𝑑2𝑦
𝑑𝑦
−5
+ 6𝑦 = 𝑥
𝑑𝑥 2
𝑑𝑥
SOLUTION
It seems sensible to think that the
solution is of the form 𝑦 = 𝐶𝑥 + 𝐷
2
𝑦 = 𝐶𝑥 + 𝐷 →
𝑑𝑦
𝑑 𝑦
=𝐶 →
=0
𝑑𝑥
𝑑𝑥 2
Substituting into original equation
Auxiliary equation
𝑢2 − 4 = 0
𝑢 = ±2
C.F:𝑦 = 𝐴𝑒 −2𝑥 + 𝐵𝑒 2𝑥
Particular Integral:
0 − 5𝐶 + 6(𝐶𝑥 + 𝐷) = 𝑥
6𝐶𝑥 − 5𝐶 + 6𝐷 = 𝑥
𝑑𝑦
𝑑2 𝑦
= 2𝐶𝑥 + 𝐷 →
= 2𝐶
𝑑𝑥
𝑑𝑥 2
Substituting into original equation
6𝐶 = 1
𝐶=
𝑦 = 𝐶𝑥 2 + 𝐷𝑥 + 𝐸 →
2𝐶 − 4(𝐶𝑥 2 + 𝐷𝑥 + 𝐸) = 3𝑥 2 − 2𝑥 + 1
1
6
−4𝐶𝑥 2 − 4𝐷𝑥 + 2𝐶 − 4𝐸 = 3𝑥 2 − 2𝑥 + 1
−5𝐶 + 6𝐷 = 0
−4𝐶 = 3
5
𝐷=
36
3
4
1
→𝐷=
2
→𝐶=−
−4𝐷 = −2
1
5
6
36
So 𝑦 = 𝑥 +
is a solution of the given equation,
but it cannot be the complete solution since it
does not contain any arbitrary constant. However,
it must be part of the complete solution, and is
called a particular integral (P.I). The remainder of
2𝐶 − 4𝐸 = 1
→𝐸=−
3
1
5
4
2
8
P.I is 𝑦 = − 𝑥 2 + 𝑥 −
5
8
General solution:
3
1
5
𝑦 = 𝐴𝑒 −2𝑥 + 𝐵𝑒 2𝑥 − 𝑥 2 + 𝑥 −
4
2
8
the solution can be found by considering the
simpler differential equation
𝑑2𝑦
𝑑𝑦
−5
+ 6𝑦 = 0
2
𝑑𝑥
𝑑𝑥
𝑓(𝑥) is a Trigonometric Function
whose solution is 𝑦 = 𝐴𝑒 2𝑥 + 𝐵𝑒 3𝑥
LESSON 13
Thus the complete solution is
differential equation
1
5
𝑦 = 𝐴𝑒 2𝑥 + 𝐵𝑒 3𝑥 + 𝑥 +
6
36
which is the combination of the complementary
function and the particular integral.
LESSON 12
Solve the differential equation
𝑑2𝑦
− 4𝑦 = 3𝑥 2 − 2𝑥 + 1
𝑑𝑥 2
4
Find the complete solution of the
𝑑2𝑦
𝑑𝑦
−5
+ 𝑦 = cos 𝑥 − sin 𝑥
2
𝑑𝑥
𝑑𝑥
SOLUTION
4
𝑑2 𝑦
𝑑𝑦
−5
+ 𝑦 = cos 𝑥 − sin 𝑥
2
𝑑𝑥
𝑑𝑥
Auxiliary equation
Trial Solution
4𝑢2 − 5𝑢 + 1 = 0
𝑓(𝑥) = 𝑚 sin 𝑞𝑥 + 𝑛 cos 𝑞𝑥
(4𝑢 − 1)(𝑢 − 1) = 0
𝑢=
1
,1
4
C.F is 𝑦 = 𝐴𝑒 −𝑥 + 𝐵𝑒 −3𝑥
1
C.F is 𝑦 = 𝐴𝑒 4𝑥 + 𝐵𝑒 𝑥
Trial Solution
𝑓(𝑥) = 𝑚 sin 𝑞𝑥 + 𝑛 cos 𝑞𝑥
𝑦 = 𝐶 cos 𝑥 + 𝐷 sin 𝑥
𝑑𝑦
= −𝐶 sin 𝑥 + 𝐷 cos 𝑥
𝑑𝑥
𝑦 = 𝐶 sin 2𝑥 + 𝐷 cos 2𝑥
𝑑𝑦
= 2𝐶 cos 2𝑥 − 2𝐷 sin 2𝑥
𝑑𝑥
𝑑2 𝑦
= −𝐶 cos 𝑥 − 𝐷 sin 𝑥
𝑑𝑥 2
Substituting into original equation
𝑑2 𝑦
= −4𝐶 sin 2𝑥 − 4𝐷 cos 2𝑥
𝑑𝑥 2
4(−𝐶 cos 𝑥 − 𝐷 sin 𝑥) − 5(−𝐶 sin 𝑥 + 𝐷 cos 𝑥)
Substituting into original equation
+𝐶 cos 𝑥 + 𝐷 sin 𝑥 = cos 𝑥 − sin 𝑥
−4𝐶 sin 2𝑥 − 4𝐷 cos 2𝑥 + 4(2𝐶 cos 2𝑥 − 2𝐷 sin 2𝑥)
+3(𝐶 sin 2𝑥 + 𝐷 cos 2𝑥) = 65 sin 2𝑥
(−4𝐶 − 5𝐷 + 𝐶) cos 𝑥 + (−4𝐷 + 5𝐶 + 𝐷) sin 𝑥
= cos 𝑥 − sin 𝑥
(−4𝐶 − 8𝐷 + 3𝐶) sin 2𝑥 + (−4𝐷 + 8𝐶 + 3𝐷) cos 2𝑥
−3𝐶 − 5𝐷 = 1
= 65 sin 2𝑥
−3𝐷 + 5𝐶 = −1
(−𝐶 − 8𝐷) sin 2𝑥 + (8𝐶 − 𝐷) cos 2𝑥 = 65 sin 2𝑥
4
17
1
𝐷=−
17
−𝐶 − 8𝐷 = 65
} → 𝐶 = −1 & 𝐷 = −8
8𝐶 − 𝐷 = 0
𝐶=−
Particular Integral is 𝑦 = − sin 2𝑥 − 8 cos 2𝑥
General solution is
Particular Integral is
𝑦=−
4
1
cos 𝑥 −
sin 𝑥
17
17
When 𝑥 = 0, 𝑦 = 3,
The complete solution is
𝑦=−
𝑦 = 𝐴𝑒 −𝑥 + 𝐵𝑒 −3𝑥 − sin 2𝑥 − 8 cos 2𝑥
1
4
1
cos 𝑥 −
sin 𝑥 + 𝐴𝑒 4𝑥 + 𝐵𝑒 𝑥
17
17
𝑑𝑦
𝑑𝑥
=7
𝑦 = 𝐴𝑒 −𝑥 + 𝐵𝑒 −3𝑥 − sin 2𝑥 − 8 cos 2𝑥
𝑑𝑦
= −𝐴𝑒 −𝑥 − 3𝐵𝑒 −3𝑥 − 2 cos 2𝑥 + 16 sin 2𝑥
𝑑𝑥
LESSON 14
(i) Solve the D.E
𝑑2 𝑦
𝑑𝑥 2
+4
𝑑𝑦
𝑑𝑥
3 = 𝐴𝑒 0 + 𝐵𝑒 0 − sin(0) − 8 cos(0)
+ 3𝑦 = 65 sin 2𝑥
(ii) Hence, find the particular solution for which
𝑑𝑦
𝑦 = 3,
= 7 when 𝑥 = 0
𝑑𝑥
3=𝐴+𝐵−8
𝐴 + 𝐵 = 11
7 = −𝐴𝑒 0 − 3𝐵𝑒 0 − 2 cos(0) + 16 sin(0)
SOLUTION
7 = −𝐴 − 3𝐵 − 2
Auxiliary equation
𝐴 + 3𝐵 = −9
𝑢2 + 4𝑢 + 3 = 0
𝑢 = −1, −3
𝐴 + 𝐵 = 11
} → 𝐵 = −10 & 𝐴 = 21
𝐴 + 3𝐵 = −9
Particular Solution is
𝑦 = 21𝑒 −𝑥 − 10𝑒 −3𝑥 − sin 2𝑥 − 8 cos 2𝑥
𝑓(𝑥) is an Exponential Function
LESSON 15
5 = 𝑒 0 (𝐴 cos 0 + 𝐵 sin 0) + 2𝑒 0
3=𝐴
𝑑𝑦
= 11; 𝑥 = 0
𝑑𝑥
𝑑𝑦
= −𝑒 −𝑥 (𝐴 cos 3𝑥 + 𝐵 sin 3𝑥)
𝑑𝑥
Solve the differential equation
𝑑2𝑦
𝑑𝑦
+2
+ 10𝑦 = 26𝑒 𝑥
2
𝑑𝑥
𝑑𝑥
given that 𝑦 = 5 and
𝑑𝑦
𝑑𝑥
= 11 when 𝑥 = 0. Give
+ 𝑒 −𝑥 (−3𝐴 sin 3𝑥 + 3𝐵 cos 3𝑥)
+ 2𝑒 𝑥
11 = −𝑒 0 (𝐴 cos 0 + 𝐵 sin 0) + 𝑒 0 (−3𝐴 sin 0 + 3𝐵 cos 0) + 2𝑒 0
11 = −1(3) + 1(3𝐵) + 2
your answer in the form 𝑦 = 𝑓(𝑥).
12 = 3𝐵
SOLUTION
𝐵=4
𝑦 = 𝑒 −𝑥 (3 cos 3𝑥 + 4 sin 3𝑥) + 2𝑒 𝑥
Auxiliary equation:
𝑢2 + 2𝑢 + 10 = 0
𝑢=
−2 ± √22 − 4(1)(10)
2(1)
LESSON 16
Solve the differential equation
𝑑2𝑦
𝑑𝑦
−2
− 3𝑦 = 2𝑒 −𝑥
𝑑𝑥 2
𝑑𝑥
𝑢 = −1 ± 3𝑖
Complementary function:
given that 𝑦 → 0 as 𝑥 → ∞ and that
𝑦 = 𝑒 −𝑥 (𝐴 cos 3𝑥 + 𝐵 sin 3𝑥)
𝑥 = 0.
Particular Integral:
SOLUTION
𝑑𝑦
𝑑𝑥
= −3 when
Auxiliary equation:
Trial Solution
Both the trial 𝑓(𝑥) and the
original 𝑓(𝑥) have the same
form. Coefficients differ.
𝑢2 − 2𝑢 − 3 = 0
(𝑢 − 3)(𝑢 + 1) = 0
𝑢 = −1, 3
C.F: 𝑦 = 𝐴𝑒 −𝑥 + 𝐵𝑒 3𝑥
𝑦 = 𝐶𝑒 𝑥
𝑑𝑦
= 𝐶𝑒 𝑥
𝑑𝑥
𝑑2 𝑦
= 𝐶𝑒 𝑥
𝑑𝑥 2
In this case, the particular integral would be of the
form 𝑦 = 𝐴𝑒 −𝑥 but since this is already included in
the complementary function we have to use
𝑦 = 𝐴𝑥𝑒 −𝑥 .
𝐶𝑒 𝑥 + 2𝐶𝑒 𝑥 + 10𝐶𝑒 𝑥 = 26𝑒 𝑥
13𝐶𝑒 𝑥 = 26𝑒 𝑥
𝐶=2
𝑦 = 2𝑒 𝑥
Trial Solution
Both the trial 𝑓(𝑥) and the
original 𝑓(𝑥) have the same
form. Coefficients differ.
General solution:
𝑦 = 𝑒 −𝑥 (𝐴 cos 3𝑥 + 𝐵 sin 3𝑥) + 2𝑒 𝑥
𝑦 = 𝐶𝑥𝑒 −𝑥
Particular solution:
𝑑𝑦
= 𝐶𝑒 −𝑥 − 𝐶𝑥𝑒 −𝑥
𝑑𝑥
𝑦 = 5; 𝑥 = 0
𝑑2 𝑦
= −𝐶𝑒 −𝑥 − 𝐶𝑒 −𝑥 + 𝐶𝑥𝑒 −𝑥
𝑑𝑥 2
= −2𝐶𝑒 −𝑥 + 𝐶𝑥𝑒 −𝑥
Differential Equations Requiring a
Substitution
LESSON 17
(−2𝐶𝑒 −𝑥 + 𝐶𝑥𝑒 −𝑥 ) − 2(𝐶𝑒 −𝑥 − 𝐶𝑥𝑒 −𝑥 ) − 3𝐶𝑥𝑒 −𝑥 = 2𝑒 −𝑥
1
−4𝐶𝑒 −𝑥 = 2𝑒 −𝑥
(i) Show that by using the substitution 𝑦 = , the
−4𝐶 = 2
differential equation
𝑑𝑦
+ 2𝑦 = 𝑥𝑦 2
𝑑𝑥
may be written in the form
𝑑𝑧
− 2𝑧 = −𝑥
𝑑𝑥
(ii) Find the general solution of
𝑑𝑧
− 2𝑧 = −𝑥
𝑑𝑥
and hence find the general solution of
𝑑𝑦
+ 2𝑦 = 𝑥𝑦 2
𝑑𝑥
𝐶=−
𝑧
1
2
Particular Integral:
1
𝑦 = − 𝑥𝑒 −𝑥
2
General Solution:
1
𝑦 = 𝐴𝑒 −𝑥 + 𝐵𝑒 3𝑥 − 𝑥𝑒 −𝑥
2
𝑦 → 0 as 𝑥 → ∞ and that
𝑑𝑦
𝑑𝑥
= −3 when 𝑥 = 0
𝑥 → ∞: 𝑒 −𝑥 → 0
SOLUTION
Therefore we have 𝑦 = 𝐵𝑒 3𝑥
(i) 𝑦 =
𝑦 → 0: 𝐵 = 0
𝑑𝑦
= −3, 𝑥 = 0
𝑑𝑥
1
𝑧
𝑑𝑦 𝑑𝑦 𝑑𝑧
=
×
𝑑𝑥 𝑑𝑧 𝑑𝑥
=−
From general solution (𝐵 = 0):
𝑑𝑦
1
1
= −𝐴𝑒 −𝑥 + 𝑒 −𝑥 + 𝑥𝑒 −𝑥
𝑑𝑥
2
2
1
1
−3 = −𝐴𝑒 0 − 𝑒 0 + (0)𝑒 0
2
2
5
𝐴=
2
Particular solution:
𝑦=
5 −𝑥 1 −𝑥
𝑒 − 𝑥𝑒
2
2
1 𝑑𝑧
×
𝑧 2 𝑑𝑥
𝑑𝑦
+ 2𝑦 = 𝑥𝑦 2
𝑑𝑥
(−
1 𝑑𝑧
1
1 2
)
(
)
+
2
(
)
=
𝑥
(
)
𝑧 2 𝑑𝑥
𝑧
𝑧
𝑑𝑧
− 2𝑧 = −𝑥
𝑑𝑥
(ii) Auxiliary equation:
𝑢−2=0
𝑢=2
Complementary function: 𝑧 = 𝐴𝑒 2𝑥
Particular Integral
Let 𝑧 = 𝐵𝑥 + 𝐶
𝑑𝑧
=𝐵
𝑑𝑥
𝑑𝑢
𝑑𝑦
=𝑦+𝑥
𝑑𝑥
𝑑𝑥
∴ 𝐵 − 2(𝐵𝑥 + 𝐶) = −𝑥
𝐵 − 2𝐶 − 2𝐵𝑥 = −𝑥
𝑑 2 𝑢 𝑑𝑦 𝑑𝑦
𝑑2𝑦
=
+
+
𝑥
𝑑𝑥 2 𝑑𝑥 𝑑𝑥
𝑑𝑥 2
−2𝐵 = −1
𝐵=
1
2
=2
𝑑𝑦
𝑑2𝑦
+𝑥 2
𝑑𝑥
𝑑𝑥
1
− 2𝐶 = 0
2
𝑥
𝑑2𝑦
𝑑𝑦
+ 2(3𝑥 + 1)
+ 3𝑦(3𝑥 + 2) = 18𝑥
2
𝑑𝑥
𝑑𝑥
1
=𝐶
4
𝑥
𝑑2𝑦
𝑑𝑦
𝑑𝑦
+ 6𝑥
+2
+ 9𝑥𝑦 + 6𝑦 = 18𝑥
2
𝑑𝑥
𝑑𝑥
𝑑𝑥
1
1
2
4
Particular Integral: 𝑧 = 𝑥 +
(𝑥
1
1
2
4
Complete solution: 𝑧 = 𝐴𝑒 2𝑥 + 𝑥 +
𝑦=
1
𝑧
→
𝑧=
1
𝑦
1
1
1
= 𝐴𝑒 2𝑥 + 𝑥 +
𝑦
2
4
1 4𝐴𝑒 2𝑥 + 2𝑥 + 1
=
𝑦
4
𝑦=
4
4𝐴𝑒 2𝑥 + 2𝑥 + 1
𝑑2𝑦
𝑑𝑦
𝑑𝑦
+ 2 ) + 6 (𝑦 + 𝑥 ) + 9𝑥𝑦 = 18𝑥
2
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑2𝑢
𝑑𝑢
+6
+ 9𝑢 = 18𝑥
2
𝑑𝑥
𝑑𝑥
(b) Auxiliary equation
𝑚2 + 6𝑚 + 9 = 0
𝑚 = −3, −3
Complementary function: 𝑢 = 𝑒 −3𝑥 (𝐴𝑥 + 𝐵)
Particular Integral
Let 𝑢 = 𝐶𝑥 + 𝐷
LESSON 18
It is given that 𝑥 ≠ 0, 𝑦 satisfies
the differential equation
𝑑2𝑦
𝑑𝑦
𝑥 2 + 2(3𝑥 + 1)
+ 3𝑦(3𝑥 + 2) = 18𝑥
𝑑𝑥
𝑑𝑥
(a) Show that the substitution 𝑢 = 𝑥𝑦 transforms
this differential equation into
𝑑2𝑢
𝑑𝑢
+6
+ 9𝑢 = 18𝑥
2
𝑑𝑥
𝑑𝑥
(b) Hence find the general solution of the
differential equation
𝑥
𝑑2𝑦
𝑑𝑦
+ 2(3𝑥 + 1)
+ 3𝑦(3𝑥 + 2) = 18𝑥
𝑑𝑥 2
𝑑𝑥
giving your answer in the form 𝑦 = 𝑓(𝑥).
SOLUTION
(a) 𝑢 = 𝑥𝑦
𝑑𝑢
=𝐶
𝑑𝑥
𝑑2𝑢
=0
𝑑𝑥 2
0 + 6𝐶 + 9(𝐶𝑥 + 𝐷) = 18𝑥
6𝐶 + 9𝐷 + 9𝐶𝑥 = 18𝑥
9𝐶 = 18
𝐶=2
6𝐶 + 9𝐷 = 0
9𝐷 = −12
𝐷=−
4
3
𝑢 = 2𝑥 −
4
3
General solution: 𝑢 = 𝑒 −3𝑥 (𝐴𝑥 + 𝐵) + 2𝑥 −
𝑥𝑦 = 𝑒 −3𝑥 (𝐴𝑥 + 𝐵) + 2𝑥 −
1
1 𝑑 2 𝑦 𝑑𝑡
𝑑2 𝑦
− 𝑑𝑦
2
2
=
(𝑡
+
2𝑡
)
𝑑𝑥 2
𝑑𝑡
𝑑𝑡 2 𝑑𝑥
4
3
4
3
1
= (𝑡 −2
𝐵
4
𝑦 = 𝑒 −3𝑥 (𝐴 + ) + 2 −
𝑥
3𝑥
=2
(b) 𝑥
LESSON 19
1
(ii)
𝑑𝑦
𝑑𝑥
1
= 2𝑡 2
𝑑2𝑦
𝑑𝑥 2
= 4𝑡
𝑑𝑦
4𝑡
𝑑𝑡
𝑑2𝑦
𝑑𝑡 2
+2
𝑑𝑦
𝑑𝑥 2
1
(b) Hence show that the substitution 𝑥 = 𝑡 2
transforms the differential equation
𝑥
𝑑2𝑦
𝑑𝑦
− (8𝑥 2 + 1)
+ 12𝑥 3 𝑦 = 12𝑥 5
2
𝑑𝑥
𝑑𝑥
− (8𝑥 2 + 1)
𝑑𝑦
𝑑𝑥
+ 12𝑥 3 𝑦 = 12𝑥 5
1
1
1 𝑑𝑦
1 2
1 4
𝑑2 𝑦
𝑑𝑦
+2
− (8𝑡 2 + 𝑡 −2 ) (2𝑡 2 ) + 12 (𝑡 2 ) 𝑦 = 12 (𝑡 2 )
2
𝑑𝑡
𝑑𝑡
𝑑𝑡
4𝑡
𝑑𝑡
𝑑𝑦
𝑑2𝑦
+ 4𝑡 2
𝑑𝑡
𝑑𝑡
𝑑2 𝑦
1 𝑑𝑦
− (8𝑥 + )
+ 12𝑥 2 𝑦 = 12𝑥 4
2
𝑑𝑥
𝑥 𝑑𝑥
(a) Given that 𝑥 = 𝑡 2 , 𝑥 > 0, 𝑡 > 0 and 𝑦 is a
function of 𝑥, show that:
(i)
𝑑2𝑦
1 𝑑2𝑦
1
𝑑𝑦
+ 2𝑡 2 2 ) (2𝑡 2 )
𝑑𝑡
𝑑𝑡
4𝑡
𝑑2𝑦
𝑑𝑦
+ (2 − 16𝑡 − 2)
+ 12𝑡𝑦 = 12𝑡 2
𝑑𝑡 2
𝑑𝑡
𝑑2𝑦
𝑑𝑦
− 16𝑡
+ 12𝑡𝑦 = 12𝑡 2
2
𝑑𝑡
𝑑𝑡
𝑑2𝑦
𝑑𝑦
−4
+ 3𝑦 = 3𝑡
𝑑𝑡 2
𝑑𝑡
(c) Auxiliary equation
into
𝑑2𝑦
𝑑𝑦
−4
+ 3𝑦 = 3𝑡
2
𝑑𝑡
𝑑𝑡
(c) Hence find the general solution of the
differential equation
𝑥
𝑑2𝑦
𝑑𝑦
− (8𝑥 2 + 1)
+ 12𝑥 3 𝑦 = 12𝑥 5
𝑑𝑥 2
𝑑𝑥
giving your answer in the form 𝑦 = 𝑓(𝑥).
𝑢2 − 4𝑢 + 3 = 0
𝑢 = 1, 3
Complementary function: 𝑦 = 𝐴𝑒 𝑡 + 𝐵𝑒 3𝑡
𝑦 = 𝐶𝑡 + 𝐷
𝑑𝑦
=𝐶
𝑑𝑡
𝑑2𝑦
=0
𝑑𝑡 2
SOLUTION
1
(a) (i) 𝑥 = 𝑡 2
𝑑𝑦 𝑑𝑦 𝑑𝑡
=
×
𝑑𝑥 𝑑𝑡 𝑑𝑥
0 − 4𝐶 + 3(𝐶𝑡 + 𝐷) = 3𝑡
−4𝐶 + 3𝐶𝑡 + 3𝐷 = 3𝑡
Now,
𝑑𝑥 1 −1
= 𝑡 2
𝑑𝑡 2
→
1
𝑑𝑡
= 2𝑡 2
𝑑𝑥
1 𝑑𝑦
𝑑𝑦
= 2𝑡 2
𝑑𝑥
𝑑𝑡
(ii)
𝑑
𝐶=1
−4𝐶 + 3𝐷 = 0
𝑑𝑦
𝑑
𝑑𝑥 𝑑𝑥
𝑑𝑡
( )=
3𝐶 = 3
1 𝑑𝑦
(2𝑡 2
𝑑𝑡
)
𝑑𝑡
𝑑𝑥
−4 + 3𝐷 = 0
3𝐷 = 4
𝐷=
4
3
radius is increasing at a rate of 0.5
metres per second.
𝑦=𝑡+
4
3
SOLUTION
𝑡
General Solution: 𝑦 = 𝐴𝑒 + 𝐵𝑒
3𝑡
+𝑡+
4
𝑦 = 𝐴𝑒
+ 𝐵𝑒
4
+𝑥 +
3
2
Mathematical Modelling
LESSON 20
(a) A pond is initially empty and is then filled
gradually with water. After 𝑡 minutes, the
depth of the water, 𝑥 metres, satisfies the
differential equation
𝑑𝑥
√4 + 5𝑥
=
𝑑𝑡 5(1 + 𝑡)2
Solve this differential equation to find 𝑥 in
terms of 𝑡.
(b) Another pond is gradually filling with water,
after 𝑡 minutes, the surface of the water forms
a circle of radius 𝑟 metres. The rate of change
of the radius is inversely proportional to the
area of the surface of the water.
(i)
Write down a differential equation, in
the variables 𝑟 and 𝑡 and a constant of
proportionality, which represents
how the radius of the surface of the
water is changing with time.
(You are not required to solve your
differential equation.)
(ii)
𝑑𝑡
=
√4+5𝑥
5(1+𝑡)2
1
√4 + 5𝑥
Since 𝑥 = 𝑡 , 𝑡 = 𝑥 2
3𝑥 2
𝑑𝑥
3
1
2
𝑥2
(a)
When the radius of the pond is 1
metre, the radius is increasing at a
𝑑𝑥 =
1
𝑑𝑡
5(1 + 𝑡)2
1
1
∫(4 + 5𝑥)−2 𝑑𝑥 = ∫(1 + 𝑡)−2 𝑑𝑡
5
1
(4 + 5𝑥)2 1
= [−(1 + 𝑡)−1 ] + 𝑐
1
5
5( )
2
2
1 1
√4 + 5𝑥 = − (
)+𝑐
5
5 1+𝑡
When 𝑥 = 0, 𝑡 = 0
2
1
1
√4 + 5(0) = − (
)+𝑐
5
5 1+0
1=𝑐
2
1 1
√4 + 5𝑥 = 1 − (
)
5
5 1+𝑡
5 1 1
√4 + 5𝑥 = − (
)
2 2 1+𝑡
2
5
1
4 + 5𝑥 = [ −
]
2 2(1 + 𝑡)
2
5
1
5𝑥 = [ −
] −4
2 2(1 + 𝑡)
2
1 5
1
4
𝑥= [ −
] −
5 2 2(1 + 𝑡)
5
(b) (i)
𝑑𝑟
𝑑𝑡
=
𝑘
𝜋𝑟 2
(ii) 𝑟 = 1,
𝑑𝑟
𝑑𝑡
= 4.5
9
𝑘
=
2 𝜋(1)2
9 𝑘
=
2 𝜋
9𝜋
𝑘=
2
rate of 4.5 metres per second. Find
the radius of the pond when the
𝑑𝑟
9
=
𝑑𝑡 2𝑟 2
When
𝑑𝑟
𝑑𝑡
=
1
2
1
9
=
2 2𝑟 2
𝑟2 = 9
𝑟=3
LESSON 2
The number of bacteria in a
liquid culture is observed to grow at a rate
proportional to the number of cells present. At the
beginning of the experiment there are 10,000 cells
and after three hours there are 500,000. How
many will there be after one day of growth if this
unlimited growth continues? What is the doubling
time of the bacteria?
SOLUTION
Let 𝑦(𝑡) represent the number of bacteria present
at time 𝑡. Then the rate of change is
𝑑𝑦
= 𝑘𝑦
𝑑𝑡
where 𝑘 is the constant of proportionality
1
𝑑𝑦 = 𝑘 𝑑𝑡
𝑦
1
∫ 𝑑𝑦 = ∫ 𝑘 𝑑𝑡
𝑦
ln 𝑦 = 𝑘𝑡 + 𝑐
𝑒 ln 𝑦 = 𝑒 𝑘𝑡 + 𝑐
𝑦 = 𝑒 𝑘𝑡+𝑐
𝑦 = 𝑒 𝑐 𝑒 𝑘𝑡
𝑦 = 𝐴𝑒 𝑘𝑡 where 𝐴 = 𝑒 𝑐
When 𝑡 = 0, 𝑦 = 𝑦0 (initial population)
Also, 𝑦 = 𝐴𝑒 𝑘(0) = 𝐴
Therefore, 𝐴 = 𝑦0
𝑦 = 𝑦0 𝑒 𝑘𝑡
Using the information from the problem we now
determine 𝑘
500 000 = 10 000𝑒 3𝑘
50 = 𝑒 3𝑘
ln 50 = 3𝑘
1
𝑘 = ln 50 ≅ 1.304
3
𝑦 = 10 000𝑒 1.304𝑡
Doubling time refers to the amount of time for the
bacteria to double in number from its original
number.
2𝑦0 = 𝑦0 𝑒 1.304𝑡
2 = 𝑒 1.304𝑡
ln 2 = 1.304𝑡
ln 2
𝑡=
≅ 0.532 hours
1.304