SOLUTION: MASC 0009 American Academy of English Engineering Mathematics Problems

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Attached.

Running Head: ENGINEERING MATHEMATICS

Engineering Mathematics
Name
Institutional Affiliation
Course
Date

1

2

ENGINEERING MATHEMATICS
Question One

Evaluate the subjoined integral:
∫(𝑑 2 + 3) cos(5𝑑)π‘‘π‘‘π‘π‘Ž
(Note: Take 𝒂 and b any two mass of your MEC ID enumerate, after a while a situation that 𝒂 π‘Ž π‘ π‘œ 4 > 1
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ 𝑏𝑦 𝑒𝑠𝑖𝑛𝑔 πΌπ‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑏𝑦 π‘π‘Žπ‘Ÿπ‘‘π‘ 
∫ 𝑓𝑔′ = 𝑓𝑔 βˆ’ 𝑓 β€² 𝑔
{

𝑓 = 𝑑2 + 3
𝑔′ = cos (5𝑑)
𝑔=
=

sin (5𝑑)
5

(𝑑 2 + 3)𝑠𝑖𝑛5(𝑑)
2𝑑𝑠𝑖𝑛(5𝑑)
βˆ’βˆ«
𝑑𝑑
5
5

Now,
∫

2𝑑𝑠𝑖𝑛(5𝑑)
2𝑑𝑠𝑖𝑛(5𝑑)
. 𝑑𝑑 βˆ’ ∫
𝑑𝑑
5
5

Using integration size intermittently we use
∫ 𝑓𝑔′ = 𝑓𝑔 βˆ’ ∫ 𝑓 β€² 𝑔 π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑏𝑦 {
𝑓 β€² = 1, 𝑔 = βˆ’

𝑓=𝑑
𝑔 = sin(5𝑑)
β€²

cos (5𝑑)
𝑑𝑑
5

=
βˆ’

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

After solving the adherence by size, we obtain

3

ENGINEERING MATHEMATICS

=

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

Therefore,
(𝑑 2 + 3)sin (5𝑑)
2𝑑𝑠𝑖𝑛(5𝑑)
βˆ’βˆ«
𝑑𝑑
5
5
=

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

Hence
∫(𝑑 2 + 3). cos (5𝑑)
(𝑑 2 + 3)sin (5𝑑) 2 sin(5𝑑) 2π‘‘π‘π‘œπ‘ (5𝑑)
βˆ’
+
+𝑐
5
125
25
=

(25𝑑 2 + 73 sin(5𝑑) + 10π‘‘π‘π‘œπ‘ (5𝑑)
+𝑐
125
=

98 sin(5) + 10cos (5)
124
= βˆ’0.7291 β‰ˆ βˆ’0.73
Question Two

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

2𝑦 3 + 𝑦
) 𝑑𝑦
𝑦4 + 𝑦2 + 1

(Note: Take 𝒂 and b any two mass of your MEC ID enumerate and 𝒂, 𝒃>𝟎)
Solution
ID. No. 18f17934
𝐼 = ∫(𝑒

βˆ’π‘Žπ‘¦

(2𝑦 3 + 𝑦
𝑑𝑦 … … … … … … … … … … … . π‘’π‘žπ‘›(1)
βˆšπ‘ + 𝑒^ βˆ’ π‘Žπ‘¦π‘‘π‘¦ βˆ’ ∫ 4
𝑦 + 𝑦2 + 1
π‘Žπ‘›π‘‘ 𝑦 4 + 𝑦12 = π‘ˆ

𝐿𝑒𝑑 𝑏 + 𝑒 βˆ’π‘Žπ‘¦ = 𝑉,
Therefore
𝑑𝑣
= 0 + (βˆ’π‘Ž)𝑒 βˆ’π‘Žπ‘¦
𝑑𝑦

π‘Žπ‘›π‘‘

𝑑𝑒
= 4𝑦 3 + 2𝑦 + 0
𝑑𝑦

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ENGINEERING MATHEMATICS
𝑑𝑣
βˆ’π‘Ž

= 𝑒 βˆ’π‘Žπ‘¦ 𝑑𝑦 =

𝑑𝑒
2

π‘Žπ‘›π‘‘ (2𝑦 3 + 𝑦)𝑑𝑦 =

From eqn (1) we have
βˆ’1
1
1
𝐼 = ( ) ∫ βˆšπ‘£ 𝑑𝑣 βˆ’ ( ) ∫ 𝑑𝑒
π‘Ž
2
𝑒
𝐼=βˆ’

3
2
1
(𝑏 + 𝑒 βˆ’π‘Žπ‘¦ )2 βˆ’ 𝑙𝑛|𝑦 4 + 𝑦 2 + 1| + 𝐢
3π‘Ž
2

If a=1 and b=3
Then
3
2
1
𝐼 = βˆ’ (3 + 𝑒 βˆ’π‘Žπ‘¦ )2 βˆ’ 𝑙𝑛|𝑦 4 + 𝑦 2 + 1| + 𝑐
3
2

Question Three
The performance manufactured π‘Š, to spread a start from its consistent protraction to an extension of 𝑏 meters
Is fond by
𝑏

π‘₯
6
π‘Š = ∫ (3 βˆ’
βˆ’ sec 2 (3π‘₯) βˆ’ √8)𝑑π‘₯
6π‘₯
+
9
π‘₯
√
0
Evaluate W.
(Note: Take 𝒃 as any enumerate from your MEC ID enumerate, 𝒃>𝟎)
Solution
𝑏
π‘₯
6
π‘Š = ∫ (3 βˆ’
βˆ’ 𝑠𝑒𝑐 2 (3π‘₯) βˆ’ √8) 𝑑π‘₯
6π‘₯
+
9
√π‘₯
0
𝑏

=∫
0

2
π‘₯ 3 𝑑π‘₯

𝑏

βˆ’βˆ«
0

𝑑π‘₯
3
π‘₯+2

𝑏

βˆ’ ∫ 𝑠𝑒𝑐

𝑏
2 (3π‘₯)𝑑π‘₯

0

βˆ’ ∫ √8𝑑π‘₯
0

3 5
3
3
ta n(3𝑏)
= 𝑏 3 βˆ’ 𝑙𝑛 (𝑏 + ) + 𝑙𝑛 ( ) βˆ’
βˆ’ √8 𝑏
5
2
2
3
Taking b=1 i.e. b>0
5
3
5
3
π‘‘π‘Žπ‘›3
π‘Š = (1)3 βˆ’ 𝑙𝑛 ( ) + 𝑙𝑛 ( ...


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