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Kyriakos
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Core Mathematics 4
Complete notes for C4

2014-2015

4

6

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Contents
1
...
Differentiation

9 – 16

Implicit differentiation

9

Differentiation of parametric functions

10 – 11

problems with rate of change

12 – 16

3
...
Binomial Expansion

21 – 24

5
...
Vectors

2

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3

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1 Coordinate geometry
Parametric equations
Example:
Suppose we have thrown a ball horizontally from a specific height 10 to examine its motion
...
To fully describe the path of
the ball, we need two equations, one for the vertical component and one for the horizontal
component
...

Suppose now that 𝑥 is the length of the ball and 𝑦 the height of the ball and also that the
velocity of the ball is constant 5𝑚/𝑠
...


𝑡
𝑥
𝑦

0
10
50

1
15
45

2
20
30

3
25
5

The red dots represent the points we found
and the curve the motion of the ball
...
1
Draw the curve given by the parametric equations 𝑥 = 1 + 𝑡 2 ,

𝑡
𝑥
𝑦

-3
10
-6

-2
5
-4

-1
2
-2

0
1
0

1
2
2

2
5
4

𝑦 = 2𝑡,

𝑡 𝜖ℝ

3
1
6

Example 1
...
3
Find the Cartesian equation of the line with parametric equations 𝑥 = 1 + 𝑡 2 and 𝑦 = 2𝑡
𝑦

𝑦 = 2𝑡 ⇒ 𝑡 = 2

𝑦 2

𝑥 = 1 + 𝑡 2 ⇒ 𝑥 = 1 + (2 ) ⇒ 𝑥 = 1 +

𝑦2
4

𝑦 2 = 4𝑥 − 1

⇒

Example 1
...

1

𝑡

-1

−2

𝑥

0

−

𝑦

-3

-2

1
4

1
4
3
−
16
−

−

3
2

0
0
-1

1
4
5
16
−

1
2

1
2

1

3
4

2

0

1

5

𝑦 = 2𝑡 − 1, 1 ≤ 𝑡 ≤ 1
...
5
Find the Cartesian equation of the line with parametric equations 𝑥 = 5 cos 𝜃 and 𝑦 = 2 sin 𝜃
𝑥 = 5 cos 𝜃 ⇒ cos 𝜃 =
𝑦 = 2 sin 𝜃 ⇒ sin 𝜃 =

𝑥 2
5

𝑥
5

𝑦 2
2

sin2 𝜃 + cos2 𝜃 = 1 ⇒ ( ) + ( ) = 1

𝑦
2

⇒

𝑡

𝑥
5

0

0

2

-5

0

0

-2

5

=1

𝑦

0
𝜋
2
𝜋
3𝜋
2
2𝜋

𝑥2
𝑦2
+
25
4

0

Example 1
...

Then sketch the graph in Cartesian form
...
7
The curve with parametric equations 𝑥 = 5 cos 𝑡, 𝑦 = 2sin 𝑡, 0 ≤ 𝑡 ≤ 2𝜋 meets the 𝑥 axis at
the points 𝐴 and 𝐵
...

Since the graph meets the 𝑥 axis → 𝑦 = 0
...
8
A curve has parametric equations = 3t 2 −4 , 𝑦 = 2𝑡
...
The
line and the curve intersect at 𝐴 and 𝐵
...

Substituting 𝑦 = 3𝑥 ⇒ 2𝑡 = 3(3t 2 −4) ⇒ 9𝑡 2 − 12 = 2𝑡 ⇒ 9𝑡 2 − 2𝑡 − 12 = 0
𝑡=
𝑡1 =

1+√109
9

𝑡2 =

𝑡1 ⟶ 𝑥 = 0
...
70

2 ± √4 − 4 ∙ (9) ∙ (−12) 2 ± √436 2 ± 2√109
=
=
18
18
18

1−√109
9

𝑦 = 2
...
85,2
...
10

𝐵(−0
...
10)

Example 1
...
Find the
value of 𝑎
𝑥 = 2 ⇒ 𝑒 𝑡 = 2 ⇒ ln 𝑒 𝑡 = ln 2 ⇒ 𝑡 = ln 2
𝑎

𝑦 = 5 ⇒ 𝑦 = 𝑒 𝑎𝑡 = 𝑒 𝑎 ln 2 = 𝑒 ln 2 = 2 𝑎 = 5

ln 5

⇒ ln 2 𝑎 = ln 5 ⇒ 𝑎 = ln 2 ≈ 2
...
1
1
...
Find the
coordinates of these points
...

2
...
𝐴 and 𝐵 are points of the circle with coordinates
(2 cos 𝜃 , 2 sin 𝜃) and (2,0) respectively
...

3
...
Find in Cartesian form the parametric equations of the curve and identify the graph
...
Plot the graphs with parametric equation
a)
b)

𝑥 = 2 cos 𝑡
𝑥 = 2 cos 𝑡

𝑦 = 2 sin 𝑡
𝑦 = 3 sin 𝑡

c)
d)

𝑥 = 𝑡 + 1 𝑦 = 2𝑡 2 − 3
𝑥 = 𝑡 2 + 1 𝑦 = 2𝑡 4 − 3

0 ≤ 𝑡 ≤ 2𝜋
𝜋
0≤ 𝑡≤2
0≤ 𝑡≤1
0≤ 𝑡≤1

6
...
Find the coordinates of the points 𝐴, 𝐵 and 𝐶
...
The graph with parametric equations 𝑥 = 𝑡 5 + 3𝑡 + 4
4 intersect at the point 𝑃
...

8
...
Find the coordinates of the points 𝐴, 𝐵 and 𝐶
...
The graph with parametric equations 𝑥 = 𝑡 3 − 2𝑡 𝑦 = 5𝑡 and the line 𝑦 = 𝑥 intersect
at the points 𝑂, 𝐴 and 𝐵 where 𝑂 is the origin
...
The curve with parametric equations 𝑥 = 𝑎 cos 2𝜃 , 𝑦 = 6 sin 𝜃 0 ≤ 𝜃 ≤ 2𝜋 passes
through the point (4,3)
...


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2 Differentiation
Reminder of the differentiation formulas from 𝑪 𝟑
1
...


𝑑
(√𝑓(𝑥))
𝑑𝑥

3
...


𝑑
[𝑎 𝑓(𝑥) ]
𝑑𝑥

= 𝑎 𝑓(𝑥) ∙ 𝑓 ′ (𝑥) ∙ ln 𝑎

5
...


𝑑
(𝑢
𝑑𝑥

7
...


𝑑
(sin
𝑑𝑥

9
...


𝑑
(tan f(x))
𝑑𝑥

= sec 2 f(x) ∙ f ′ (x)

11
...


𝑑
(sec f(x))
𝑑𝑥

= sec f(x) ∙ tan f(x) ∙ f ′ (x)

13
...
However functions can also be implicitly defined
...
g
...
For example 𝑥 2 + sin 𝑦 = 𝑥 3 𝑦 7 is differentiated:

2𝑥 + (cos 𝑦)

𝑑𝑦
𝑑𝑦
𝑑𝑦
𝑑𝑦
= 3𝑥 2 𝑦 7 + 7𝑥 3 𝑦 6
⇒ (cos 𝑦)
− 7𝑥 3 𝑦 6
= 3𝑥 2 𝑦 7 − 2𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑦
3𝑥 2 𝑦 7 − 2𝑥
⇒
=
𝑑𝑥 cos 𝑦 − 7𝑥 3 𝑦 6
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Differentiation of Parametric equations
It is often necessary to find the rate of change of a function defined parametrically or the
𝑑𝑦
gradient of a curve at any point etc, that is, we want to calculate
...

Chain rule:
𝑑𝑦
𝑑𝑦 𝑑𝑡
=
∙
𝑑𝑥
𝑑𝑡 𝑑𝑥
Therefore we can calculate

𝑑𝑦
𝑑𝑥

by
𝑑𝑦
𝑑𝑦 𝑑𝑡
𝑑𝑦 𝑑𝑥
=
∙
=
÷
𝑑𝑥
𝑑𝑡 𝑑𝑥
𝑑𝑡
𝑑𝑡

Example 2
...


Example 2
...
3

Determine the coordinates of the stationary points of the function 𝑥 = 𝑡𝑒 2𝑡 𝑦 = 𝑡 2 𝑒 −𝑡 , 𝑡 > 0
...
4
Determine the equation of the tangent to the curve 𝑥 = 𝑡 ln 𝑡 𝑦 = 𝑡 2 at the point 𝑡 = 1
𝑑𝑦
𝑑𝑥
𝑚 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = 2

at 𝑡 = 1,

𝑡=1

𝑑𝑦
2𝑡
2(1)
2
= 𝑑𝑡 =
=
= =2
𝑑𝑥 ln 𝑡 + 1 ln(1) + 1 1
𝑑𝑡

𝑦 = 1 and 𝑥 = 0

𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) ⇒ 𝑦 − 1 = 2(𝑥 − 0) ⇒ 𝑦 = 2𝑥 + 1

Exercise 2
...
Differentiate the following implicit functions
a)

𝑥 2 + 𝑥𝑦 = 3

b)

𝑦 3 + sin 𝑦 = 3𝑥

c)

ln(sin 𝑦 3 )
3𝑥
𝑦
𝑥

=4

d) sin 𝑥 + cos 𝑥 2 = 3

e) √ 𝑥𝑦 = 𝑥 2 + 3

f)

g) 𝑒 3𝑦 + 𝑒 3𝑥 = 4

h) ln[sin(√ 𝑦)] = 4

i)

3𝑦+3𝑥 =7

k) ln22 𝑥 2 + √ 𝑦 = 1

l)

𝑘 2𝑦 = 3

o)

𝑦= 𝑎𝑥

j)

csc 3𝑥 + sec 𝑦 = 𝑦

m) 𝑦 = 𝑥 𝑥

n)

𝑦 = sinx 𝑥

= 3 cos 𝑥

𝑥

2
...
Find the stationary point of the curve 𝑥(𝑡) = 3𝑒 3𝑡 + 2, 𝑦(𝑡) = 𝑡 2
4
...
Prove that the derivative of 𝑦 = 𝑎3𝑥 is 3𝑎3𝑥 ln 𝑎
6
...

Example 2
...
The tank is being filled with
water at a constant rate 45 liters per second
...
6
A point 𝑀(𝑥, 𝑦) is moving on the line 2𝑥 + 3𝑦 = 5
...
5

2

8

= − 3 (2𝑡 2 + 1)′ = − 3 (4𝑡) = − 3 𝑡

2

1
...
5

8

𝑦 ′ (𝑡) = − 3 𝑡 = − 3 (3) = −8

-1
...


-1

-0
...
5

-0
...
5

-2

12

1

1
...
5

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Kyriakos

Example 2
...
05 𝑚3 /
𝑚𝑖𝑛
...
The height if the tank
is 12 𝑚 and the diameter at the top is 4 𝑚
...
2 𝑚/𝑚𝑖𝑛
when the height of the water is 2
...

𝑘(𝑡)

𝑉𝑐𝑜𝑛𝑒 =
2 𝑚

𝜋𝑟 2 ℎ
3

From similar triangles we know that

12 𝑚

𝑉𝑐𝑜𝑛𝑒
𝑑ℎ
𝑑𝑡

𝑟
ℎ

=

2
12

1
6

⇒ 𝑟= ℎ

1 2
𝜋 (6 ℎ) ℎ
𝜋𝑟 2 ℎ
𝜋ℎ3
=
=
=
3
3
108

= 0
...
5
𝑑𝑉
𝑑𝑉 𝑑𝑡
𝑑𝑉
𝑑𝑉 𝑑ℎ
=
∙
⇒
=
∙
𝑑ℎ
𝑑𝑡 𝑑ℎ
𝑑𝑡
𝑑ℎ 𝑑𝑡

0
...
2) =
𝜋 𝑚/𝑚𝑖𝑛
𝑑𝑡
𝑑ℎ 𝑑𝑡 144
144
That means that the water is increasing at this rate
Now let 𝑘(𝑡) be the rate at which water is pumped in
...
05
𝑑𝑡
5
5
𝜋 = 𝑘(𝑡) − 0
...
05 ⇒ 𝑘(𝑡) = 0
...
8
Two cars 𝐴 and 𝐵 are moving across two perpendicular roads 𝑂𝐴 and 𝑂𝐵 and they approach
to the crossing of the road with speed 50 𝑘𝑚/ℎ and 100 𝑘𝑚/ℎ respectively
...
(Hint: 𝑉 = 𝑆 ∙ 𝑡)

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𝑦

𝑂𝐴 = 𝑥(𝑡)
𝐵

𝑂𝐵 = 𝑦(𝑡)

using Pythagoras theorem
𝑦(𝑡)2 + 𝑥(𝑡)2 = 𝑆(𝑡)2

𝐴𝐵 = 𝑆(𝑡)
𝑆(𝑡)

we also have that at time 𝑡0 ,

𝑑𝑥
𝑑𝑡

= 50 𝑘𝑚/ℎ

𝑑𝑦
𝑑𝑡

𝑦(𝑡)

= 100 𝑘𝑚/ℎ

(Velocity is the change of displacement divided
𝑂

𝑥(𝑡)

𝑥(𝑡0 ) = 0
...
6 𝑘𝑚

𝑦(𝑡)2 + 𝑥(𝑡)2 = 𝑆(𝑡)2

𝑥

by the change of time )
𝑦(𝑡0 )2 + 𝑥(𝑡0 )2 = 𝑆(𝑡0 )2 ⇒ (0
...
6)2 = 𝑆(𝑡0 )2
⇒ 𝑆(𝑡0 ) = 1

Differentiating the relation 𝑦(𝑡)2 + 𝑥(𝑡)2 = 𝑆(𝑡)2 ⇒ 2𝑦
2𝑦

𝑑𝑦
𝑑𝑡

+ 2𝑥

𝑑𝑥
𝑑𝑡

= 2𝑆

𝑑𝑆
𝑑𝑡

𝑑𝑦
𝑑𝑥
𝑑𝑆
𝑑𝑆
𝑑𝑆
+ 2𝑥
= 2𝑆
⇒ 2(0
...
8)(50) = 2 (1)
⇒
= 100 𝑘𝑚/ℎ
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡

Useful formulas for volume and surface areas
...
2
1
...

2
...

3
...

At some point 𝑀 the rate of change of 𝑥(𝑡) of 𝑀 is 3 times the rate of change of 𝑦(𝑡)
...

4
...
Find the rate of change of its
surface area when the radius is 𝑅 = 17 𝑐𝑚
10

5
...
If the radius
of its base is 80𝑐𝑚, calculate the rate of increase of the volume of the water in the
tank
6
...
A car is
passing exactly below the balloon when the balloon is at a height of 39𝑚 and is
moving horizontally with a constant velocity 30 𝑚/𝑠𝑒𝑐
...

7
...
The balloon has a hole on it were the water is
leaking
...
Calculate the rate of change of the radius of the sphere and the rate of
change of its surface area
...
(past exam question, June 2013)
In an experiment testing solid rocket fuel, some fuel is burned and the waste
products are collected
...

Let 𝑥 be the mass of waste products in kg, at time 𝑡 minutes after the start of the
experiment
...

The differential equation connecting 𝑥 and 𝑡 may be written in the form

where 𝑀 is a constant
...


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Kyriakos
Given that initially the mass of waste products is zero,
b) Solve the differential equation, expressing 𝑥 in terms of 𝑘, 𝑀 and 𝑡
...


9
...
At time 𝑡 seconds
2
liquid is leaking from the container at a rate of 15 V cm3 s1, where V cm3 is the
volume of liquid in the container at that time
...


Given that 𝑉 = 1000 when 𝑡 = 0,
b) find the solution of the differential equation, in the form 𝑉 = 𝑓(𝑡)
...


(June 2002 past paper)
A Pancho car has value £V at time t years
...

a) By forming and solving an appropriate differential equation, show that V = Aekt,
where A and k are positive constants
...

b) Find, to the nearest £100, an estimate for the value of the Pancho when it is 10
years old
...

c) Find the approximate age of the Pancho when it becomes ‘scrap’
...
(January 2002 past exam question)
A radioactive isotope decays in such a way that the rate of change of the number N
of radioactive atoms present after t days, is proportional to N
...

b) Show that the general solution may be written as N = Aekt, where A and k are
positive constants
...

c) Find the value of k
...


16

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3 Partial fractions
As we already know, if we want to add multiple fractions we must create the same
denominator and eventually express all the fractions as a single fraction
...
g
...
e multiple fractions
...


Example 3
...
we will find them by comparing terms
𝑥+5
𝐴
𝐵
𝐴(𝑥 − 3)
𝐵(𝑥 − 2)
=
+
=
+
(𝑥 − 2)(𝑥 − 3)
𝑥 − 2 𝑥 − 3 (𝑥 − 2)(𝑥 − 3) (𝑥 − 3)(𝑥 − 2)

⇒ 𝑥 + 5 = 𝐴(𝑥 − 3) + 𝐵(𝑥 − 2)
When 𝑥 = 3 → 8 = 0 + 𝐵(1) ⇒ 𝐵 = 8
𝑥 = 2 → 7 = 𝐴(−1) + 0 ⇒ 𝐴 = −7
⇒

𝑥+5
8
7
=
−
(𝑥 − 2)(𝑥 − 3)
𝑥−3 𝑥−2

Example 3
...

Now since (𝑥 − 3)2 has an exponent of 2 we need two fractions for this
...
Therefore,
𝑥−3
𝐴
𝐵
𝐶
=
+
+
2 (𝑥 − 5)
2
(𝑥 − 3)
(𝑥 − 3)
(𝑥 − 3) (𝑥 − 5)
𝑥−3
(𝑥−3)2 (𝑥−5)

𝐴(𝑥−5)
𝐵(𝑥−3)(𝑥−5)
+ (𝑥−3)2
(𝑥−5)
(𝑥−5)

= (𝑥−3)2

𝐶(𝑥−3)2
(𝑥−5)

+ (𝑥−3)2

17

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Kyriakos

𝑥−3
(𝑥−3)2 (𝑥−5)

=

𝐴(𝑥−5)+𝐵(𝑥−3)(𝑥−5)+𝐶(𝑥−3)2
(𝑥−3)2 (𝑥−5)

𝑥 − 3 = 𝐴(𝑥 − 5) + 𝐵(𝑥 − 3)(𝑥 − 5) + 𝐶(𝑥 − 3)2
When 𝑥 = 3 → 0 = −2𝐴 ⇒ 𝐴 = 0
1

𝑥 = 5 → 2 = 4𝐶 ⇒ 𝐶 = 2
1

𝑥 = 0 → −3 = −5𝐴 − 3𝐵 + 9𝐶 ⇒ −3 = −3𝐵 + 9 (2) ⇒ −3𝐵 = −

15
2

5

⇒ 𝐵=2

5
1
𝑥−3
0
5
1
2
2
=
+
+
=
+
2 (𝑥 − 5)
2
(𝑥 − 3)
(𝑥 − 3)
(𝑥 − 3) (𝑥 − 5) 2(𝑥 − 3) 2(𝑥 − 5)

Example 3
...


Here, we have in the denominator a quadratic expression that can’t be factorized any further
...
4 (improper fractions)
An improper fraction is one where the degree of the numerator is greater than or equal to
the denominators
...

Since the fraction is improper we are going to use the long division
...
1
1
...
Express

3𝑥+5
(𝑥−2)(𝑥+3)

3
...
Express

𝑥 3 +4𝑥 2 −1
𝑥 3 (𝑥−1)2

5
...
Express

3𝑥+5
(𝑥 2 +4)(5𝑥−2)

7
...
Express

𝑥 2 −𝑥
(𝑥−2)2 (𝑥+2)

9
...
Express

3𝑥+2
(𝑥 2 +1)(𝑥+3)

11
...
Express

𝑥 5 −2𝑥 4 +𝑥 3 +𝑥+5
𝑥 3 −2𝑥 2 +𝑥−2

13
...
Given that

(4𝑥 2 +1)
(2𝑥+1)(2𝑥−1)

into partial fractions
into partial fractions and find its derivative
into partial fractions

into partial fractions
into partial fractions

3𝑥 2 +7𝑥+12

into partial fractions
...

2

2

b) Hence show that the derivative is (2𝑥+1)2 − (2𝑥−1)2

20

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Kyriakos

4 The Binomial Expansion
(1 + 𝑥) 𝑛 = 1 + 𝑛𝑥 +

𝑛(𝑛 − 1) 2
𝑛(𝑛 − 1)(𝑛 − 2) 3
𝑥 +
𝑥 + ⋯+
2!
3!

𝑛

𝐶𝑟 𝑥 𝑟

We have met binomial expansion before in C2
...
We notice that
one of the two terms must be 1
...

If 𝒏 is negative or fractional the expansion is infinite but it might be convergent or
divergent
An infinite series is convergent if |𝑥| < 1
...
We are going to see more in the examples
...
1
Expand (1 + 𝑥)5
(1 + 𝑥)5 = 1 + 5𝑥 +

5∙4 2 5∙4∙3 3 5∙4∙3∙2
𝑥 +
𝑥 +
+ 𝑥5
2!
3!
4!

Since 𝑛 is positive integer we can see that the series is finite
...
2
1

Expand the function (1+5𝑥)4 up to and including 𝑥 4 and find the range for it to be convergent
...
Now we need to find
the range for the expansion to be convergent
...
Hence the expansion converge when < 𝑥 <
5
5
5
5
Example 4
...
4
Find the expansion of √1 − 5𝑥 up to and including the term 𝑥 3
...
05 to find an approximation for √3
...
05) ≈ 1 − 2 𝑥 −

25
8

𝑥2 −

125
16

5

𝑥 3 = 1 − 2 (0
...
05)2
8

−

125
(0
...
75 ≈
⇒√
≈
⇒√
≈
⇒
⇒ √3 ≈
√3 ≈
1
1024
100 1024
100
1024 10
1024
512
2
√3 ≈

887
≈ 1
...
5
Expand

𝑥+2
√1−𝑥

up to and including 𝑥 3
...
6
Method 1 (partial Fractions first)
3

Expand (𝑥+1)(1−3𝑥) up to and including 𝑥 3
3
𝐴
𝐵
𝐴(1 − 3𝑥) + 𝐵(𝑥 + 1)
=
+
=
(1 + 𝑥)(1 − 3𝑥)
(𝑥 + 1)(1 − 3𝑥)
𝑥 + 1 1 − 3𝑥
3 = 𝐴(1 − 3𝑥) + 𝐵(𝑥 + 1)
3

For 𝑥 = −1 → 3 = 4𝐴 ⇒ 𝐴 = 4
1

4

9

𝑥=3 →3=3 𝐵 ⇒ 𝐵=4
3
(𝑥+1)(𝑥−2)
3
(1 −
4

3

1

9

1

3

9

= 4 ( 𝑥+1) + 4 (1−3𝑥 ) = 4 (𝑥 + 1)−1 + 4 (1 − 3𝑥)−1 =

𝑥+

−1(−2)
2!

𝑥2 +

(−1)(−2)(−3)
3!

9

𝑥 3 + ⋯ ) + 4 (1 + 3𝑥 +

22

−1(−2)
(−3𝑥)2
2!

+

(−1)(−2)(−3)
3!

(−3𝑥)3 … )

𝐶4
3
4

−

Kyriakos
3
4

𝑥+

3
4

𝑥2 −

3
4

9
4

𝑥3 + +

27
4

𝑥+

81
4

𝑥2 +

243
4

𝑥3

= 3 + 6𝑥 + 21𝑥 2 + 60𝑥 3 + ⋯

Method 2 (Directly)
3
(𝑥+1)(1−3𝑥)

= 3[(𝑥 + 1)(1 − 3𝑥)]−1 = 3 (𝑥 + 1)−1 (1 − 3𝑥)−1

= 3 (1 − 𝑥 +

−1(−2)
2!

(−1)(−2)(−3)

𝑥2 +

3!

𝑥 3 + ⋯ ) (1 + 3𝑥 +

−1(−2)
(−3𝑥)2
2!

+

(−1)(−2)(−3)
3!

(−3𝑥)3 … )

3(1 + 3𝑥 + 9𝑥 2 + 27𝑥 3 − 𝑥 − 3𝑥 2 − 9𝑥 3 + 𝑥 2 + 3𝑥 3 − 𝑥 3 ) = 3(1 + 2𝑥 + 7𝑥 2 + 20𝑥 3 )
3 + 6𝑥 + 21𝑥 2 + 60𝑥 3 + ⋯
Since there is more than one expansion, for the overall fraction to be valid we have to find
the validity for both
...
7
Find the first four terms in the binomial expansion of (5 + 3𝑥)1/3
...
1

24

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Kyriakos

5 Integration
Formulas
α) ∫ 𝑎 𝑑𝑥 = 𝑎𝑥 + 𝑐

Properties

β) ∫ 𝑓(𝑥) 𝑛 𝑓 ′ (𝑥) 𝑑𝑥 =

(𝑎𝑥+𝑏) 𝑛+1
𝑎(𝑛+1)

γ) ∫(𝑎𝑥 + 𝑏) 𝑛 𝑑𝑥 =
δ) ∫
ε) ∫

𝑓′ (𝑥)
𝑓(𝑥)
𝑓′ (𝑥)
√𝑓(𝑥)

𝑓 𝑛+1 (𝑥)
+
(𝑛+1)

𝑐

+ 𝑐

1
𝑎𝑥+𝑏

∫ 𝑎 ∙ 𝑓(𝑥)𝑑𝑥 = 𝑎 ∫ 𝑓(𝑥)𝑑𝑥

𝑑𝑥 = ln|𝑓(𝑥)| + 𝑐
𝑑𝑥 = 2√𝑓(𝑥) + 𝑐

στ) ∫ 𝑒 𝑓(𝑥) 𝑓 ′ (𝑥) = 𝑒 𝑓(𝑥) + 𝑐
ζ)∫

∫ 𝑓(𝑥) ± 𝑔(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥

1
𝑎

𝑑𝑥 = ln|𝑎𝑥 + 𝑏| + 𝑐

η) ∫ 𝑒 𝑎𝑥+𝑏 𝑑𝑥 =

𝑒 𝑎𝑥+𝑏
𝑎

Useful trigonometric formulas
sin2 𝑥 + cos 2 𝑥 = 1
1 + tan2 𝑥 = sec 2 𝑥

+ 𝑐

𝑎𝑥

θ) ∫ 𝑎 𝑥 𝑑𝑥 = ln 𝑎 + 𝑐

1 + cot 2 𝑥 = csc 2 𝑥

ι) ∫ sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐

sin2 𝑥 =

1−cos 2𝑥
2

cos2 𝑥 =

1+cos 2𝑥
2

1

ια)∫ sin 𝑎𝑥 𝑑𝑥 = − 𝑎 cos 𝑥 + 𝑐
ιβ)∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐
1

ιγ)∫ cos 𝑎𝑥 𝑑𝑥 = 𝑎 sin 𝑥 + 𝑐
ιδ)∫ sec 2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
1

ιε)∫ sec 2 𝑎𝑥 𝑑𝑥 = 𝑎 tan 𝑥 + 𝑐
ιστ)∫ csc 2 𝑥 𝑑𝑥 = − cot 𝑥 + 𝑐
1

ιζ)∫ csc 2 𝑎𝑥 𝑑𝑥 = − cot 𝑥 + 𝑐
𝑎
ιη)∫ sec 𝑥 tan 𝑥 𝑑𝑥 = sec 𝑥 + 𝑐
1

ιθ)∫ sec 𝑥 𝑎 tan 𝑎𝑥 𝑑𝑥 = 𝑎 sec 𝑥 + 𝑐
κ)∫ csc 𝑥 cot 𝑥 𝑑𝑥 = − csc 𝑥 + 𝑐
1

κα)∫ csc 𝑎𝑥 cot 𝑎𝑥 𝑑𝑥 = − 𝑎 csc 𝑥 + 𝑐

25

𝐶4

Kyriakos

Example 5
...
1

1
...

Example 5
...
2
cos 𝑥

∫ √1+sin

𝑥

𝑑𝑥

Substitute 𝑢 = √1 + sin 𝑥

𝑢 = √1 + sin 𝑥 ⇒ 𝑢2 = 1 + sin 𝑥 ⇒ 2𝑢 𝑑𝑢 = cos 𝑥 𝑑𝑥
cos 𝑥

∫ √1+sin

𝑥

𝑑𝑥 = ∫

2𝑢
𝑢

𝑑𝑢 = ∫ 2 𝑑𝑢 = 2𝑢 + 𝑐

⇒ 2√1 + sin 𝑥 + 𝑐

Example 5
...
4
∫ sin4 𝑥 cos 𝑥 𝑑𝑥

using the substitution sin 𝑥 = 𝑢

𝑢 = sin 𝑥 ⇒ 𝑑𝑢 = cos 𝑥 𝑑𝑥
∫ sin4 𝑥 cos 𝑥 𝑑𝑥 = ∫ 𝑢4 𝑑𝑢 =

𝑢5
sin5 𝑥
+ 𝑐=
+ 𝐶
5
5

Example 5
...
2
1
...
In the square bracket is the
substitution you need to use
...


a) ∫ 2𝜒(𝜒 2 + 4)5 𝑑𝑥

[𝜒 2 + 4]

b) ∫ sin5 𝑥 cos 𝑥 𝑑𝑥

[cos 𝑥]

d) ∫ 𝑒 𝑥 (𝑒 𝑥 + 4)3 𝑑𝑥

[√𝑥 2 + 4]

f)

[cos 𝑥]

h) ∫ tan3 𝑥 ∙ sec 2 𝑥 𝑑𝑥

[ln 𝑥]

j)

∫ 𝑒 cos 𝑥 ∙ sin 𝑥

[ln 𝑥]

l)

∫ (𝑥 2 +2)3 𝑑𝑥

[tan 𝑥]

n) ∫ sin 𝜃 (1 + cos 𝜃 )3 𝑑𝜃

o) ∫ 1+𝑥4 𝑑𝑥

[𝑥 2 ]

p) ∫ 1+𝑒 2𝑥 𝑑𝑥

q) ∫ cos(3𝑥 + 4) 𝑑𝑥

[3𝑥 + 4]

r) ∫(𝑥 + 4)5 𝑑𝑥

s) ∫
𝑑𝑥
1−2𝑥

[1 − 2𝑥]

t)

u) ∫ 2𝑥√1 + 𝑥 2

[1 + 𝑥 2 ]

v) ∫ 2𝑥 𝑒 𝑥

w) ∫
𝑑𝑥
√2𝑥 2 +1

[2𝑥 2 + 1]

x) ∫ 4
𝑑𝑥
√𝑥 +16

y) ∫ 𝑥 sin(2𝑥 2 ) 𝑑𝑥

[2𝑥 2 ]

z) ∫(2𝑥 + 2)𝑒 𝑥

[𝑥 3 ]

bb) ∫ √7𝑥 − 3 𝑑𝑥

sin 𝑥

c) ∫ cos7

𝑥

𝑑𝑥

1

e) ∫ 𝑥√𝑥 2 + 4 𝑑𝑥
sin 𝑥

g) ∫ 1+cos2
i)

∫

k) ∫

ln6 𝑥
𝑥

𝑑𝑥

𝑥

𝑑𝑥

𝑑𝑥
𝑥√1−ln2

𝑥

m) sec 2 𝑥 tan3 𝑥 𝑑𝑥
2𝑥

1

4𝑥

𝑥2

aa) ∫
𝑑𝑥
√1−𝑥 6

∫

𝑒𝑥
𝑥2

𝑑𝑥

𝑑𝑥

𝑥3

𝑒𝑥

∫

sin √ 𝑥
√𝑥

𝑑𝑥
2 −5

𝑑𝑥

𝑥3

29

2 +2𝑥+3

𝑑𝑥

𝐶4

Kyriakos

Integration by partial fractions
Example 5
...

Example 5
...
3
Integrate using partial fractions
𝑥 2 +2𝑥−1

i)

∫ (𝑥+1)(𝑥 2 +1) 𝑑𝑥

ii)

∫ (𝑥−2)(𝑥+1)2 𝑑𝑥

iii)

∫ (𝑥−2)(𝑥+1) 𝑑𝑥

iv)

∫

v)

∫ (𝑥+1)(𝑥−2)2 𝑑𝑥

vi)

∫

vii)

∫ (𝑥−1)(𝑥−2) 𝑑𝑥

viii)

∫ (1+𝑥)(1+𝑥 2 ) 𝑑𝑥

ix)

∫

x)

∫

xi)

∫

xii)

∫

xiii)

∫

xiv)

∫ (𝑥+1)(𝑥 3 +1) 𝑑𝑥

xv)

∫

xvi)

∫

2𝑥 2 +1

2𝑥−3

3𝑥−4

𝑑𝑥

𝑥 2 −5𝑥+6

𝑥 2 +4𝑥−15

𝑥 5 −𝑥+3
𝑥 2 −1

𝑑𝑥

𝑥3

2𝑥

𝑥+4
𝑥 2 +𝑥−2
2𝑥−4
𝑥 2 +𝑥−6

𝑑𝑥
𝑑𝑥

3𝑥 2 +3𝑥+1
𝑥 3 +2𝑥 2 +𝑥

𝑑𝑥

𝑥 3 −4𝑥 2 −3𝑥+3
𝑥 2 −3𝑥
𝑥3

𝑑𝑥

𝑑𝑥

𝑥 2 +1
1

6𝑥+13
𝑥 2 +5𝑥+6
𝑥3
𝑥 2 +4

𝑑𝑥

𝑑𝑥

31

𝐶4

Kyriakos

Integration by parts

∫ 𝒖 𝒅𝒗 = 𝒖 ∙ 𝒗 − ∫ 𝒗 𝒅𝒖
The method occurs from the differentiation of the multiplication of two functions (𝑢 ∙ 𝑣)
Reminder: (𝑢 ∙ 𝑣)′ = 𝑢′ 𝑣 + 𝑢𝑣′ therefore 𝑢𝑣 ′ = (𝑢 ∙ 𝑣)′ − 𝑢′ 𝑣
Now integrating both sides we get ∫ 𝑢𝑣 ′ 𝑑𝑥 = ∫(𝑢 ∙ 𝑣)′ − 𝑢′ 𝑣 𝑑𝑥 ⇒ ∫ 𝑢 𝑑𝑣 = 𝑢 ∙ 𝑣 − ∫ 𝑣 𝑑𝑢
Using the method:
Step 1: Decide which of the two is 𝑢 and which one is 𝑣
Step 2: Put 𝑣 into the differential (𝑑𝑥)
...
To remove something from the differential we differentiate it
...


Example 5
...
9
∫ 𝑥 2 cos 𝑥 𝑑𝑥
Step 1: 𝑢 = 𝑥 2

Step 2:

𝑑𝑣 = cos 𝑥

∫ 𝑥 2 cos 𝑥 𝑑𝑥 = ∫ 𝑥 2 𝑑(sin 𝑥)

Step 3:
𝑥 2 sin 𝑥 − ∫ sin 𝑥 𝑑(𝑥 2 ) = 𝑥 2 sin 𝑥 − ∫ 2𝑥 sin 𝑥 𝑑𝑥
Since I have integration again I will have to use the method again
...


32

𝐶4

Kyriakos

𝑥 2 sin 𝑥 − ∫ 2𝑥 sin 𝑥 𝑑𝑥 = 𝑥 2 sin 𝑥 − 2 ∫ 𝑥 𝑑(− cos 𝑥 ) = 𝑥 2 sin 𝑥 − 2(−𝑥 cos 𝑥 − ∫ −cos 𝑥 𝑑𝑥)
⇒ 𝑥 2 sin 𝑥 + 2𝑥 cos 𝑥 − 2 ∫ cos 𝑥 𝑑𝑥 = 𝑥 2 sin 𝑥 + 2𝑥 cos 𝑥 − 2 sin 𝑥 + 𝐶
= sin 𝑥 (𝑥 2 − 2) + 2𝑥 cos 𝑥 + 𝐶
Example 5
...
4
1
...

i)

∫(𝑥 + 2)𝑒 𝑥 𝑑𝑥

ii)

∫ ln 𝑥 𝑑𝑥

iii)

∫ 𝑥 𝑒 −𝑥 𝑑𝑥

iv)

∫ 𝑒 −𝑥 sin 2𝑥 𝑑𝑥

v)

∫ sec 3 𝑥 𝑑𝑥

vi)

∫ 𝑥 cos 3𝑥 𝑑𝑥

vii)

∫(ln 𝑥 )2 𝑑𝑥

viii)

∫ 𝑥 2 𝑒 3𝑥 𝑑𝑥

ix)

∫ 𝑥 𝑒 6𝑥 𝑑𝑥

x)

∫ 𝑥 3 ln 𝑥 𝑑𝑥

xi)

∫ arctan 𝑥 𝑑𝑥

xii)

∫

xiii)

∫ √ 𝑥 ln 𝑥 𝑑𝑥

xiv)

∫ sin 𝑥 cos 𝑥 𝑑𝑥

xv)

∫ 𝑒 𝑥 cos 𝑥 𝑑𝑥

xvi)

∫ 𝑥 √ 𝑥 + 1 𝑑𝑥

33

ln 𝑥
𝑥2

𝑑𝑥

𝐶4

Kyriakos

Integration using trigonometric identities
Example 5
...
12
∫ tan2 𝑥 𝑑𝑥
∫ tan2 𝑥 𝑑𝑥 = ∫ sec 2 𝑥 − 1 𝑑𝑥 = ∫ sec 2 𝑥 𝑑𝑥 − ∫ 𝑑𝑥 = tan 𝑥 − 𝑥 + 𝐶

Exercise 5
...
Integrate the Following
a) ∫ cos2 𝑥 𝑑𝑥
c) ∫ sin2

b) ∫ tan2 3𝑥 𝑑𝑥

𝑑𝑥
𝑥 cos2 𝑥

d) ∫ sin3 𝑥 𝑑𝑥

e) ∫ cos3 𝑥 𝑑𝑥

f)

g) ∫ sec 4 𝑥 𝑑𝑥

h) ∫ cos2 3𝑥 𝑑𝑥

∫ cot 2 𝑥 𝑑𝑥

∫ tan3 𝑥 𝑑𝑥

tan 𝑥
𝑥

𝑑𝑥

j)

∫ sec3

k) ∫ sin2 𝑥 cos3 𝑥 𝑑𝑥

l)

∫ cot 6 𝑥 + cot 8 𝑥 𝑑𝑥

m) ∫ √1 + sin 2𝑥 𝑑𝑥

n) ∫ √10 + 10 cos 10𝑥 𝑑𝑥

o) ∫ cot 3 𝑥 𝑑𝑥

p) ∫
𝑑𝑥
√4−𝑥 2

i)

q) ∫ √9 − 𝑥 2 𝑑𝑥

𝑥2

𝑑𝑥

r) ∫ 2
√𝑥 −16

[sub: 𝑥 = 3 sin 𝜃]

34

[sub: 𝑥 = 2 sin 𝜃]
[sub: 𝑥 = 4 sec 𝜃]

𝐶4

Kyriakos

Trapezium rule
𝑦

𝑏

ℎ

𝐴𝑟𝑒𝑎 = ∫𝑎 𝑓(𝑥)𝑑𝑥 ≈ 2 [𝑦0 + 𝑦 𝑛 + 2(𝑦1 + 𝑦2 + 𝑦3 + ⋯ + 𝑦 𝑛−1 )]
Where ℎ =
𝑎

𝑏

𝑥

𝑏−𝑎
𝑛

and 𝑛 is the number of strips and hence

𝑛 + 1 vertical lines
...
13
Given that
𝜋

∫ sin 𝑥 𝑑𝑥
𝜋/2

a) Find the exact value of the area under the graph when

𝜋
2

< 𝜒< 𝜋

b) Use the trapezium rule with 3 strips to find an approximation of that area
c) Calculate the percentage error

a)

𝜋

𝜋

𝛢 = ∫ 𝜋/2 sin 𝑥 𝑑𝑥 = [− cos 𝑥] 𝜋𝜋 = (− cos 𝜋 + cos 2 ) = 1
2

b) ℎ =

𝜋−
3

𝜋
2

=

𝜋
6

𝜒 = 𝜋/2

𝜒 = 2𝜋/3

𝜒 = 5𝜋/6

𝜒= 𝜋

𝑦0 = 1

𝑦1 = √3/2

𝑦2 = 1/2

𝑦3 = 0

𝜋
ℎ
6 [1 + 0 + 2 (√3 + 1)] = 𝜋 [√3 + 2] ≈ 0
...
𝑝
𝐴 ≈ [𝑦0 + 𝑦3 + 2(𝑦1 + 𝑦2 )] =
2
2
2
2
12
c) percentage error

=

=

|𝑒𝑥𝑎𝑐𝑡 𝑣𝑎𝑙𝑢𝑒−𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛|
𝑒𝑥𝑎𝑐𝑡 𝑣𝑎𝑙𝑢𝑒

× 100%

1 − 0
...
023 ∙ 100% = 2
...
6
1
...

2
...

𝑥
𝑦

0
1

0
...
216

1
𝑝

1
...
413

2
𝑞

a) Find the value of 𝑝 and the value of 𝑞
...
(past paper P2 June 2001)
A measure of the effective voltage, 𝑀 volts, in an electrical circuit is given by
1

2
dt
𝑀 =  V

0

2

where 𝑉 volts is the voltage at time 𝑡 seconds
...

𝑡
𝑉
𝑉2

0
−48

1/4
207

1/2
37

3/4
−161

1
−29

a) Use the trapezium rule with five values of V 2 to estimate the value of M
...
Find the exact value of

𝐶4

Kyriakos

Area under curve


If the area we want is above the 𝑥 axis
𝑏

𝐴𝑟𝑒𝑎 = ∫ 𝑓(𝑥)𝑑𝑥 = [𝑔(𝑥)] 𝑏 = 𝑔(𝑏) − 𝑔(𝑎)
𝑎
𝑎



If the area is below the 𝑥 axis
𝑏

𝐴𝑟𝑒𝑎 = − ∫ 𝑓(𝑥)𝑑𝑥 = −[𝑔(𝑥)] 𝑏 = − [𝑔(𝑏) − 𝑔(𝑎)]
𝑎
𝑎

Example 5
...


Example 5
...

2
2
2
2
4 0 2
2

0

0

0

Example 5
...
1 ≤ 𝑥 ≤ 1
1

𝐴 = − ∫ ln 𝑥 𝑑𝑥 = −[𝑥 ln 𝑥 − 𝑥]1
0
...
1
(exercise 5
...
1 ∙ ln 0
...
1)]
𝐴 = 1 + 0
...
33 𝑠
...
d
...
17
Find the area stated below
6

𝐴1 = ∫ (𝑥 − 4)2 + 3 𝑑𝑥
2
10

𝐴2 = ∫ (𝑥 − 8)2 + 3
6
𝐴 = 𝐴1 + 𝐴2
𝑨𝟏

6

𝑨𝟐

6

(𝑥 − 4)3
8
8
16
52
𝐴1 = ∫(𝑥 − 4) + 3 𝑑𝑥 = [
+ 3𝑥] = + 18 − (− + 6) =
+ 12 =
𝑠
...
𝑢
3
3
3
3
3
6
6

𝐴 = 𝐴1 + 𝐴2 =

52 52 104
+
=
𝑠
...
18
Calculate the volume of revolution of the graph generated by the curve 𝑦 = 𝑥 2 the 𝑂𝑥 axis
and the line 𝑥 = 1 when is rotaded 2𝜋 by the 𝑥 axis
...


Example 5
...
From 0𝑏

2

𝑉 = 𝜋 ∫𝑎 𝑓 2 (𝑥) − 𝑔2 (𝑥) 𝑑𝑥 = 𝜋 ∫ 8𝑥 − 𝑥 4 𝑑𝑥
0
𝑉 = 𝜋[

8𝑥 2
2

−

2
𝑥5
]
5 0

= 𝜋 [16 −

32
]
5

=

48
5

𝜋 𝑐
...
20
Calculate the area given by the curve with parametric equation 𝑦(𝑡) = tan 𝑡 , 𝑥(𝑡) = sin 𝑡
0 ≤ 𝑡 ≤ 2𝜋 the line 𝑥 = 1 and the positive 𝑦 axis
At (0,0):
𝑥(𝑡) = sin 𝑡 = 0 ⇒ 𝑡 = 0
𝑦(𝑡) = tan 𝑡 = 0 ⇒ 𝑡 = 0
At 𝑥 = 1:
𝑥(𝑡) = sin 𝑡 = 1 ⇒ 𝑡 =

𝜋
2

𝑡1

𝜋/2

𝜋/2

𝜋/2

𝑑𝑥
sin 𝑡
𝐴 = ∫ 𝑦(𝑡) ∙ ( ) 𝑑𝑡 = ∫ tan(𝑡) ∙ (cos 𝑡) 𝑑𝑡 = ∫
∙ (cos 𝑡) 𝑑𝑡 = ∫ sin 𝑡 𝑑𝑡
𝑑𝑡
cos 𝑡
𝑡0

0

𝜋/2

𝐴 = −[cos 𝑡]0

0

0

𝜋
= − [(cos ( )) − (cos(0))] = 0 + 1 = 1 𝑠
...
21
Calculate the area given by the curve with parametric equation 𝑦(𝑡) = sin 3𝑡 , 𝑥(𝑡) = cos 2𝑡
0 ≤ 𝑡 ≤ 2𝜋 the line 𝑥 = 1 and the positive 𝑦 axis
...

2
4
5
5
2 2
10
10

𝑏

𝑉 = 𝜋 ∫ 𝑦 2 (𝑡) ∙ (
𝑎

𝑑𝑥
) 𝑑𝑡
𝑑𝑡

Example 5
...

From 0 ≤ 𝑥 ≤ 1
...
𝑢

𝐶4

Kyriakos

Exercise 5
...
Find the area of the region bounded by the curves 𝑦 = 𝑒 𝑥 , 𝑦 = 𝑥 from 𝑥 = 0 to 𝑥 = 1
...
Find the area enclosed by the curves 𝑦 = 𝑥 2 and 𝑦 = 2𝑥 − 𝑥 2
3
...
Find the area enclosed by the curves 𝑦 = tan2 𝑥, 𝑦 = √ 𝑥
5
...
Find the area enclosed by the curves 𝑦 = cos 𝜋𝑥,

𝑦 = 4𝑥 2 − 1

7
...
Suppose that 0 < 𝑐 < 2
...
Find the volume of the solid obtained by rotating about the 𝑥 axis the area under the
curve 𝑦 = √ 𝑥 from 0 ≤ 𝑥 ≤ 1
10
...

11
...
Area volume lvkns;lns; differentia l

42

𝐶4

Kyriakos

First Order Differential equations
Method of separable variable
Suppose we have the differential equation 𝑔(𝑦)𝑦′ = 𝑓(𝑥)
...
23
Solve the differential equation 𝑦 ′ =

3𝑥 2 −1
3+2𝑦

𝑑𝑦 3𝑥 2 − 1
=
⇒ (3 + 2𝑦)𝑑𝑦 = (3𝑥 2 − 1)𝑑𝑥 ⇒ ∫(3 + 2𝑦)𝑑𝑦 = ∫(3𝑥 2 − 1)𝑑𝑥
𝑑𝑥
3 + 2𝑦
⇒ 3𝑦 + 𝑦 2 = 𝑥 3 − 𝑥 + 𝑐
Example 5
...
25
Consider a mouse population that reproduces at a rate proportional to the current
population, with a rate equal to 0
...

When owls are present, they eat the mice
...

a) Write a differential equation describing the mouse population in the presence of owls
...
)
b) Hence solve the differential equation

a)

𝑑𝑝
𝑑𝑡

= 0
...
5(𝑝 − 900) ⇒ ∫
𝑑𝑝 = ∫ 0
...
5𝑡+𝑐 ⇒ 𝑝 = 𝑒 0
...
8
1
...


iii
...


𝑑𝑦
𝑑𝑥

=

𝑥
𝑦

6𝑥 2
2𝑦+cos 𝑦

ii
...


𝑦 ′ = 𝑥𝑒 −𝑦
𝑒 𝑦 sin2 𝜃
𝑦 sec 𝜃

vii
...


𝑑𝑝
𝑑𝑡

= 𝑡2 𝑝 − 𝑝 + 𝑡2 − 1

xi
...


𝑦′ = 𝑥2 𝑦

vi
...


(𝑦 2 + 𝑥𝑦 2 )𝑦′ = 1

x
...


𝑑𝑦
𝑑𝑥

= 𝑥𝑦 + 2𝑥 + 𝑦 + 2

xiv
...


𝑑𝑢
𝑑𝑡

=

2𝑡+sec2 𝑡
2𝑢

= 6𝑦 2 𝑥

𝑥√𝑥 2 +1
𝑦𝑒 𝑦

3𝑦 ′ + [sin(𝑥) + 3𝑒 𝑎𝑥 ]𝑦 = 0

2
...


𝑑𝑦
𝑑𝑥

=

ln 𝑥
𝑥𝑦

iii
...


𝑑𝑦
𝑑𝑥

v
...


𝑦′ =

3

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Title: C4 full notes for Getting an A*
Description: This note are made with everything in C4 syllabus. It has lots of examples explanations and a lot of exercises for practice all chapters

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