SOLUTION: MASC 0009 2 Middle East College Engineering Mathematics Questions

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Engineering Science1

Engineering Mathematics
[By Name and I.D]

Module Name
Name of Faculty
Session
Date

2

ENGINEERING MATHEMATICS
Question One

Evaluate the subjoined integral:
∫(𝑑 2 + 3) cos(5𝑑)π‘‘π‘‘π‘π‘Ž
Answer
ID. No. 18F17988
Let a=1, and b=8
As b>a so 8>1
𝑀𝑒 π‘ β„Žπ‘Žπ‘™π‘™ π‘ π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘–π‘  π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 𝑏𝑦 π‘–π‘›π‘‘π‘’π‘Ÿπ‘”π‘Ÿπ‘Žπ‘‘π‘–π‘›π‘” 𝑏𝑦 π‘π‘Žπ‘Ÿπ‘‘π‘  𝑖. 𝑒 ∫ 𝑓𝑔′ = 𝑓𝑔 βˆ’ 𝑓 β€² 𝑔 … … . π‘’π‘žπ‘› (1)
𝑠𝑖𝑛𝑐𝑒 𝑓 = 𝑑 2 + 3, and g β€² = absorb 5 t,

to meet g we allure intergrate g β€² to finf g to be

by replacing the rates of f g f’ and g’ on eqn (1) we allure allure eqn (2) below:

(𝑑 2 + 3)𝑠𝑖𝑛5(𝑑)
2𝑑𝑠𝑖𝑛(5𝑑)
∫ 𝑓𝑔′ =
βˆ’βˆ«
𝑑𝑑
5
5
=∫

2𝑑𝑠𝑖𝑛(5𝑑)
2𝑑𝑠𝑖𝑛(5𝑑)
. 𝑑𝑑 βˆ’ ∫
𝑑𝑑
5
5

By integrating partially:
𝑓 = 𝑑; π‘Žπ‘›π‘‘ 𝑓 β€² = 1
∫ 𝑓𝑔′ = 𝑓𝑔 βˆ’ ∫ 𝑓 β€² 𝑔 π‘€β„Žπ‘’π‘Ÿπ‘’ { β€²
cos(5𝑑)
𝑔 = sin(5𝑑) ; 𝑔 = βˆ’
5
Which allure then be
βˆ’

π‘‘π‘π‘œπ‘ (5𝑑)
cos (5𝑑)
βˆ’ βˆ«βˆ’
𝑑𝑑
5
5

After substituting the equation over by magnitude, we allure get
=

sin (5𝑑) π‘‘π‘π‘œπ‘ (5𝑑) 2 sin(5𝑑) 2π‘‘π‘π‘œπ‘ (5𝑑25)
βˆ’
=
βˆ’
25
5
125
25
(𝑑 2 + 3)sin (5𝑑) 2 sin(5𝑑) 2π‘‘π‘π‘œπ‘ (5𝑑)
=
βˆ’
+
5
125
25

Also, on integrating ∫(𝑑 2 + 3). cos(5𝑑) π‘π‘Žπ‘Ÿπ‘‘π‘–π‘Žπ‘™π‘™π‘¦; we allure allure the equations below

sin (5𝑑)
5

3

ENGINEERING MATHEMATICS
(𝑑 2 + 3)sin (5𝑑) 2 sin(5𝑑) 2π‘‘π‘π‘œπ‘ (5𝑑)
(25𝑑 2 + 73 sin(5𝑑) + 10π‘‘π‘π‘œπ‘ (5𝑑)
βˆ’
+
+𝑐 =
+𝑐
5
125
25
125
=

98 sin(5) + 10cos (5)
= βˆ’0.7
124
𝐻𝑒𝑛𝑐𝑒 π‘‘β„Žπ‘’ π‘Žπ‘›π‘ π‘€π‘’π‘Ÿ 𝑖𝑠 0.7
Question Two

Evaluate the subjoined Integral by using accordant method:
∫ (𝐸 (βˆ’π‘Žπ‘¦) βˆšπ‘ + 𝑒𝑦 βˆ’π‘Žπ‘¦ βˆ’

2𝑦 3 + 𝑦
) 𝑑𝑦
𝑦4 + 𝑦2 + 1

(Note: Grasp 𝒂 and b any two bulk of your MEC ID enumerate and 𝒂, 𝒃>𝟎)
Answer
ID. No. 18F17988
Given the equation
𝑆 = ∫(𝑒

βˆ’π‘Žπ‘¦

(2𝑦 3 + 𝑦
𝑑𝑦 … … … … … … … … … … … . (π‘Ž)
βˆšπ‘ + 𝑒^ βˆ’ π‘Žπ‘¦π‘‘π‘¦ βˆ’ ∫ 4
𝑦 + 𝑦2 + 1
π΄π‘ π‘ π‘’π‘šπ‘–π‘›π‘” π‘‘β„Žπ‘Žπ‘‘
𝑀 = 𝑏 + 𝑒 βˆ’π‘Žπ‘¦
𝑧 = 𝑦 4 + 𝑦12

Then on differentiating w and z rates, we allure
𝑑𝑀
𝑑𝑦
𝑑𝑧
𝑑𝑦

= 0 + (βˆ’π‘Ž)𝑒 βˆ’π‘Žπ‘¦ … … … … . (𝑏)
= 4𝑦 3 + 2𝑦 + 0 … … … … … (𝑐)

On simplifying eqn b and c
𝑑𝑀
βˆ’π‘Ž
𝑑𝑧
2

= 𝑒 βˆ’π‘Žπ‘¦ 𝑑𝑦
= (2𝑦 3 + 𝑦)𝑑𝑦

Therefore, we allure foreclosure eqn (a), hence we have
3
βˆ’1
1
1
2
1
𝑆 = ( ) ∫ βˆšπ‘§ 𝑑𝑧 βˆ’ ( ) ∫ 𝑑𝑀 = βˆ’ (𝑏 + 𝑒 βˆ’π‘Žπ‘¦ )2 βˆ’ 𝑙𝑛|𝑦 4 + 𝑦 2 + 1| + 𝐢
π‘Ž
2
𝑀
3π‘Ž
2

4

ENGINEERING MATHEMATICS
Therefore, if our a and b rates are as follows;
𝐼𝑓 π‘Ž = 1 π‘Žπ‘›π‘‘ 𝑏 = 8
The consummate reresolution allure be as follows
3
2
1
𝑆 = βˆ’ (8 + 𝑒 βˆ’π‘¦ )2 βˆ’ 𝑙𝑛|𝑦 4 + 𝑦 2 + 1| + 𝑐
3
2

Question Three
The operation produced π‘Š, to reach a originate from its cosmical protraction to an production of 𝑏 meters
Is consecrated by
𝑏
π‘₯
6
π‘Š = ∫ (3 βˆ’
βˆ’ sec 2 (3π‘₯) βˆ’ √8)𝑑π‘₯
6π‘₯
+
9
0 √π‘₯

Evaluate W.
Answer
Since our I.D enumerate is ID. No. 18F17988, we allure grasp b=7
Therefore, consecrated the equation:
𝑏

π‘₯

6

π‘Š = ∫0 ( 3 βˆ’ 6π‘₯+9 βˆ’ 𝑠𝑒𝑐 2 (3π‘₯) βˆ’ √8) 𝑑π‘₯ π‘Žπ‘  π‘’π‘žπ‘› (π‘Ž)π‘‘β„Žπ‘’π‘›;
√π‘₯

𝑏

2

𝑏 𝑑π‘₯

= ∫0 π‘₯ 3 𝑑π‘₯ βˆ’ ∫0

3
π‘₯+
2

𝑏

𝑏

3

5

3

3

βˆ’ ∫0 𝑠𝑒𝑐 2 (3π‘₯)𝑑π‘₯ βˆ’ ∫0 √8𝑑π‘₯ = 5 𝑏 3 βˆ’ 𝑙𝑛 (𝑏 + 2) + 𝑙𝑛 (2) βˆ’

ta n(3𝑏)
3

βˆ’

√8 𝑏
Since b=7 the rate of W allure be consecrated as
5
3
3
3
π‘‘π‘Žπ‘›21
π‘Š = (7)3 βˆ’ 𝑙𝑛 (7 + ) + 𝑙𝑛 ( ) βˆ’
βˆ’ √8(7)
5
2
2
3

π‘Š = βˆ’21.64 𝐽
Question Four
Write the subjoined intricate enumerate in Polar fashion:
𝑧=

5 + √4𝑖
5 βˆ’ 6𝑖

Answer
By rationalizing the eqn consecrated over, by multiplying twain the denominator and the numerator
with (5+6i), we get
5+√4𝑖(5+6𝑖)

25βˆ’30𝑖+5√4π‘–βˆ’6√4

β‡’ (5βˆ’6𝑖)(5+6𝑖) =

25+36

5

ENGINEERING MATHEMATICS

β‡’

(25 βˆ’ 6√4) + 𝑖(30 + 5√4
25 βˆ’ 6√4
(βˆ’30 + 5(√4
=(
)+(
) 𝑖 … … … … π‘’π‘žπ‘›(π‘Ž)
61
61
61

On collecting affect stipulations in eqn (a)
𝑍=

25 βˆ’ 12 βˆ’30 + 10
13 20
+
𝑖=
βˆ’
𝑖
61
61
61 61

Hence to employ into polar fashion, we allure earliest meet the modulus i.e. r of 13/61 βˆ’ 20/61𝑖
13 2
20 2
169 + 400
√( ) + (βˆ’ ) = √
= 0.39; π‘‘β„Žπ‘’π‘  π‘œπ‘’π‘Ÿ π‘Ÿ π‘£π‘Žπ‘™π‘’π‘’ = 0.39
61
61
3721
To estimate the angle:
𝑦
20
πœƒ = tanβˆ’1 ( ) π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑦 = 20 π‘Žπ‘›π‘‘ 𝑛 = 13 = βˆ’ tanβˆ’1 ( )
𝑛
13
Hence to value the eqn in intricate fashion we allure the fashion: 𝑧 = π‘Ÿπ‘’ π‘–πœƒ = 0.39𝑒 βˆ’π‘–π‘‘π‘Žπ‘›

Question Five
Simplify the subjoined expression:
βˆšβˆ’πŸπŸ“ + πŸπ’Šπ’‚ + 𝒃(π’Šπ’‚ )𝒄 + |πŸ— + πŸ–π’Š| βˆ’ πŸ’(πŸ“ + πŸ”π’Š)βˆšβˆ’πŸ‘πŸ”
Answer
ID. No. 18F17988
Let a=988, b=8, and c=1
= βˆšβˆ’25 + 2𝑖 988 + 7(𝑖 988 )1 + |9 + 8𝑖| βˆ’ 4(5 + 6𝑖)βˆšβˆ’36
5βˆšβˆ’1 + 2(𝑖 2 )494 + 9(𝑖 2 )494 + |9 + 8𝑖| βˆ’ 6𝑖(4(5 + 6𝑖))
= 5𝑖 + 2(βˆ’1)494 + 9(βˆ’1)494 + |9 + 8𝑖| βˆ’ 24𝑖(5 + 6𝑖)
= 5𝑖 + 2 + 9 + |9 + 8𝑖| βˆ’ 120𝑖 + 144
= 5𝑖 + 11 + |9 + 8𝑖| βˆ’ 120𝑖 + 144
𝐴𝑛𝑠 = βˆ’115𝑖 + 155 + |9 + 8𝑖|
b. Express (π‘π‘œπ‘ (π‘Žπœƒ) + 𝑖𝑠𝑖𝑛(π‘Žπœƒ))𝑏 in exponential fashion (4 marks)
Answer
ID. No. 18F17988

βˆ’1 (20)
13

6

ENGINEERING MATHEMATICS
Let a =...


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