SOLUTION: MASC 0009 American Academy of English Engineering Mathematics Problems Paper

[ad_1]

Attached.

Engineering Mathematics – MASC 0009.2

Engineering Mathematics – 1
MASC 0009.2
CW2 (Assignment)
Student Name
MEC Student ID 18F18046
University Name
Faculty Name

1

Engineering Mathematics – MASC 0009.2
1. Evaluate the subjoined well
𝑏
a. βˆ«π‘Ž (𝑑 2 + 3) cos(5𝑑) 𝑑𝑑
1

𝐺𝑖𝑣𝑒𝑛: {

1

1

π‘Ž=0
: ∫ (𝑑 2 + 3) cos(5𝑑) 𝑑𝑑 = ∫ 𝑑 2 cos(5𝑑) 𝑑𝑑 + 3 ∫ cos(5𝑑) 𝑑𝑑
𝑏=1 0
0
0
1

𝐼: ∫ 𝑑 2 cos(5𝑑) 𝑑𝑑
0

𝐿𝑒𝑑:

1

{

𝐼𝐼: ∫ cos(5𝑑) 𝑑𝑑
0

𝑑𝑀
=5
𝑑𝑑
𝐼: ∫ 𝑑 2 cos(5𝑑) 𝑑𝑑 : π‘ˆπ‘ π‘–π‘›π‘” π‘†π‘’π‘π‘ π‘‘π‘–π‘‘π‘’π‘‘π‘–π‘œπ‘›: 𝐿𝑒𝑑 𝑀 = 5𝑑 β†’ {
𝑀2
0
𝑑2 =
25
1

β†’ ∫ 𝑑 2 cos(5𝑑) 𝑑𝑑 =

1
∫ 𝑀 2 cos 𝑀 𝑑𝑒
125

2
β€²
π‘ˆπ‘ π‘–π‘›π‘” π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑏𝑦 π‘π‘Žπ‘Ÿπ‘‘π‘ : 𝐿𝑒𝑑 {𝑒 β€²= 𝑀 β†’ 𝑒 = 2𝑀
𝑣 = cos 𝑀 β†’ 𝑣 = sin 𝑀

π‘‡β„Žπ‘’π‘› ∫ 𝑒𝑣 β€² = 𝑒𝑣 βˆ’ ∫ 𝑒′ 𝑣 = 𝑀 2 sin 𝑀 βˆ’ ∫ 2𝑀 sin 𝑀 𝑑𝑀
β€²
π‘ˆπ‘ π‘–π‘›π‘” π‘–π‘›π‘‘π‘’π‘”π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› 𝑏𝑦 π‘π‘Žπ‘Ÿπ‘‘π‘  π‘œπ‘› ∫ 𝑀 sin 𝑀 𝑑𝑀 : 𝐿𝑒𝑑 {𝑒 β€²= 𝑀 β†’ 𝑒 = 1
𝑣 = sin 𝑀 β†’ 𝑣 = βˆ’cos 𝑀

π‘‡β„Žπ‘’π‘› ∫ 𝑀 2 cos 𝑀 𝑑𝑒 = 𝑀 2 sin 𝑀 βˆ’ 2 (βˆ’π‘€ cos 𝑀 βˆ’ ∫ βˆ’ cos 𝑀 𝑑𝑀)
= 𝑀 2 sin 𝑀 + 2𝑀 cos 𝑀 βˆ’ sin 𝑀
π‘…π‘’π‘π‘™π‘Žπ‘π‘–π‘›π‘” π‘π‘Žπ‘π‘˜: 𝑀 = 5𝑑 β†’
1
[(5𝑑)2 sin(5𝑑) + 2(5𝑑) cos(5𝑑) βˆ’ sin(5𝑑)]
∫ 𝑑 2 cos(5𝑑) 𝑑𝑑 =
125

1

∫ 𝑑 2 cos(5𝑑) 𝑑𝑑 = [
0

=[

𝑑 2 sin 5𝑑 2𝑑 cos 5𝑑 2 sin 5𝑑 1
+
βˆ’
]|
5
25
125
0

sin 5 2 cos 5 2sin 5
+
βˆ’
] βˆ’ [0 + 0 βˆ’ 0] β‰… βˆ’0.15375
5
25
125

2

Engineering Mathematics – MASC 0009.2
1
1 1
1
𝐼𝐼: ∫ cos(5𝑑) 𝑑𝑑 = [ sin 5𝑑] | = (sin 5 βˆ’ sin 0) β‰… βˆ’0.19178
5
5
0
0
1

∴ ∫ (𝑑 2 + 3) cos(5𝑑) 𝑑𝑑 = 𝐼 + 3𝐼𝐼 = (βˆ’0.15375) + 3(βˆ’0.19178) β‰… βˆ’0.7291
0

3

Engineering Mathematics – MASC 0009.2
2. Evaluate the subjoined well by using convenient method:
2𝑦 3 + 𝑦
∫ (𝑒 βˆ’π‘Žπ‘¦ βˆšπ‘ + 𝑒 βˆ’π‘Žπ‘¦ βˆ’ 4
) 𝑑𝑦
𝑦 + 𝑦2 + 1
2𝑦 3 + 𝑦
π‘Ž=1
π‘ˆπ‘ π‘–π‘›π‘”: {
, 𝑀𝑒 β„Žπ‘Žπ‘£π‘’: ∫ (𝑒 βˆ’π‘¦ √4 + 𝑒 βˆ’π‘¦ βˆ’ 4
) 𝑑𝑦
𝑏=4
𝑦 + 𝑦2 + 1
∫ (𝑒 βˆ’π‘¦ √4 + 𝑒 βˆ’π‘¦ βˆ’

2𝑦 3 + 𝑦
2𝑦 3 + 𝑦
βˆ’π‘¦
βˆ’π‘¦ 𝑑𝑦 βˆ’ ∫
𝑑𝑦
=
∫
𝑒
+
𝑒
𝑑𝑦
)
√4
𝑦4 + 𝑦2 + 1
𝑦4 + 𝑦2 + 1

𝐼: ∫ 𝑒 βˆ’π‘¦ √4 + 𝑒 βˆ’π‘¦ 𝑑𝑦
𝐿𝑒𝑑:

2𝑦 3 + 𝑦
𝐼𝐼: ∫ 4
𝑑𝑦
{
𝑦 + 𝑦2 + 1

𝐼: ∫ 𝑒 βˆ’π‘¦ √4 + 𝑒 βˆ’π‘¦ 𝑑𝑦 : π‘ˆπ‘ π‘–π‘›π‘” π‘ π‘’π‘π‘ π‘‘π‘–π‘‘π‘’π‘‘π‘–π‘œπ‘›: 𝐿𝑒𝑑 𝑒 = 4 + 𝑒 βˆ’π‘¦ β†’

𝑑𝑒
𝑑𝑒
= βˆ’π‘’ βˆ’π‘¦ β†’ 𝑑𝑦 =
𝑑𝑦
βˆ’π‘’ βˆ’π‘¦

3

1
𝑑𝑒
𝑒2
2 3
β†’ ∫ 𝑒 βˆ’π‘¦ βˆšπ‘’ βˆ’π‘¦ = ∫ βˆ’βˆšπ‘’π‘‘π‘’ = βˆ’ ∫ 𝑒2 𝑑𝑒 = βˆ’
= βˆ’ (𝑒2 )
3
βˆ’π‘’
3
2
3
2
π‘…π‘’π‘π‘™π‘Žπ‘π‘–π‘›π‘” π‘π‘Žπ‘π‘˜ 𝑒 = 4 + 𝑒 βˆ’π‘¦ β†’ 𝐼 = βˆ’ (4 + 𝑒 βˆ’π‘¦ )2
3

2𝑦 3 + 𝑦
𝐼𝐼: ∫ 4
𝑑𝑦 : π‘ˆπ‘ π‘–π‘›π‘” π‘ π‘’π‘π‘ π‘‘π‘–π‘‘π‘’π‘‘π‘–π‘œπ‘›
𝑦 + 𝑦2 + 1
𝐿𝑒𝑑 𝑒 = 𝑦 4 + 𝑦 2 + 1 β†’

𝑑𝑒
𝑑𝑒
= 4𝑦 3 + 2𝑦 β†’ (2𝑦 3 + 𝑦)𝑑𝑦 =
𝑑𝑦
2

1 𝑑𝑒 1 1
1
β†’βˆ« βˆ™
= ∫ 𝑑𝑒 = ln|𝑒|
𝑒 2
2 𝑒
2
1
π‘…π‘’π‘π‘™π‘Žπ‘π‘–π‘›π‘” π‘π‘Žπ‘π‘˜ 𝑒 = 𝑦 4 + 𝑦 2 + 1 β†’ 𝐼𝐼 = ln|𝑦 4 + 𝑦 2 + 1|
2
∴ ∫ (𝑒 βˆ’π‘¦ √4 + 𝑒 βˆ’π‘¦ βˆ’

3
2𝑦 3 + 𝑦
2
1
βˆ’π‘¦ )2
(4
𝑑𝑦
=
𝐼
βˆ’
𝐼𝐼
=
βˆ’
+
𝑒
βˆ’ ln|𝑦 4 + 𝑦 2 + 1|
)
4
2
𝑦 +𝑦 +1
3
2

4

Engineering Mathematics – MASC 0009.2
3. The fruit effected W, to spread a leap from its regular prolixity to an extension of b meters is given
𝑏

π‘₯

6

by π‘Š = ∫0 ( 3 βˆ’ 6π‘₯+9 βˆ’ sec 2(3π‘₯) βˆ’ √8) 𝑑π‘₯ . Evaluate W.
π‘₯
√

1

π‘₯
6
𝐺𝑖𝑣𝑒𝑛 𝑏 = 1; π‘Š = ∫ ( 3 βˆ’
βˆ’ sec 2(3π‘₯) βˆ’ √8) 𝑑π‘₯
√π‘₯ 6π‘₯ + 9
0
1

2

1

π‘Š = ∫ π‘₯ 3 𝑑π‘₯ βˆ’ ∫
0

0

1
1
6
𝑑π‘₯ βˆ’ ∫ sec 2(3π‘₯) 𝑑π‘₯ βˆ’ ∫ √8𝑑π‘₯
6π‘₯ + 9
0
0

1 2
3 5 1 3
𝐼: ∫ π‘₯ 3 𝑑π‘₯ = (π‘₯ 3 ) | =
5
5
0
0
1

𝐼𝐼: ∫
0
1

β‡’βˆ«
0

6
𝑑𝑒
𝑑π‘₯ : π‘ˆπ‘ π‘–π‘›π‘” π‘ π‘’π‘π‘ π‘‘π‘–π‘‘π‘’π‘‘π‘–π‘œπ‘›; 𝑙𝑒𝑑 𝑒 = 6π‘₯ + 9 β†’
= 6, 𝑑𝑒 = 6𝑑π‘₯
6π‘₯ + 9
𝑑π‘₯

6
1
𝑑π‘₯ = ∫ 𝑑𝑒 = ln(𝑒)
6π‘₯ + 9
𝑒
1

π‘…π‘’π‘π‘™π‘Žπ‘π‘–π‘›π‘” π‘π‘Žπ‘π‘˜ 𝑒 = 6π‘₯ + 9 β†’ ∫
0

1
6
5
𝑑π‘₯ = ln(6π‘₯ + 9) | = ln 15 βˆ’ ln 9 = ln ( )
6π‘₯ + 9
3
0

1

𝐼𝐼𝐼: ∫ sec 2 (3π‘₯) 𝑑π‘₯ : π‘ˆπ‘ π‘–π‘›π‘” π‘ π‘’π‘π‘ π‘‘π‘–π‘‘π‘’π‘‘π‘–π‘œπ‘›; 𝑙𝑒𝑑 𝑒 = 3π‘₯ β†’
0

𝑑𝑒
1
= 3, 𝑑π‘₯ = 𝑑𝑒
𝑑π‘₯
3

1
1
β†’ ∫ sec 2 𝑒 𝑑𝑒 = tan 𝑒
3
3
1
1 1
1
π‘…π‘’π‘π‘™π‘Žπ‘π‘–π‘›π‘” π‘π‘Žπ‘π‘˜ 𝑒 = 3π‘₯ β†’ ∫ sec 2(3π‘₯) 𝑑π‘₯ = ( tan 3π‘₯) | = tan 3
3
3
0
0
1
1
𝐼𝑉: ∫ √8𝑑π‘₯ = √8π‘₯ | = √8
0
0
1
π‘₯
6
3
5
1
⟹ ∫ (3 βˆ’
βˆ’ sec 2(3π‘₯) βˆ’ √8) 𝑑π‘₯ = 𝐼 βˆ’ 𝐼𝐼 βˆ’ 𝐼𝐼𝐼 βˆ’ 𝐼𝑉 = βˆ’ ln ( ) βˆ’ tan(3) βˆ’ √8
5
3
3
√π‘₯ 6π‘₯ + 9
0
1
π‘₯
6
∴ ∫ (3 βˆ’
βˆ’ sec 2 (3π‘₯) βˆ’ √8) 𝑑π‘₯ β‰… βˆ’2.6917
6...


[ad_2]
Source amalgamate