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

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Attached.

Engineering Mathematics β MASC 0009.2

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

1

Engineering Mathematics β MASC 0009.2
1. Evaluate the subjoined well
π
a. β«π (π‘ 2 + 3) cos(5π‘) ππ‘
1

πΊππ£ππ: {

1

1

π=0
: β« (π‘ 2 + 3) cos(5π‘) ππ‘ = β« π‘ 2 cos(5π‘) ππ‘ + 3 β« cos(5π‘) ππ‘
π=1 0
0
0
1

πΌ: β« π‘ 2 cos(5π‘) ππ‘
0

πΏππ‘:

1

{

πΌπΌ: β« cos(5π‘) ππ‘
0

ππ€
=5
ππ‘
πΌ: β« π‘ 2 cos(5π‘) ππ‘ : ππ πππ ππ’ππ π‘ππ‘π’π‘πππ: πΏππ‘ π€ = 5π‘ β {
π€2
0
π‘2 =
25
1

β β« π‘ 2 cos(5π‘) ππ‘ =

1
β« π€ 2 cos π€ ππ’
125

2
β²
ππ πππ πππ‘πππππ‘πππ ππ¦ ππππ‘π : πΏππ‘ {π’ β²= π€ β π’ = 2π€
π£ = cos π€ β π£ = sin π€

πβππ β« π’π£ β² = π’π£ β β« π’β² π£ = π€ 2 sin π€ β β« 2π€ sin π€ ππ€
β²
ππ πππ πππ‘πππππ‘πππ ππ¦ ππππ‘π  ππ β« π€ sin π€ ππ€ : πΏππ‘ {π’ β²= π€ β π’ = 1
π£ = sin π€ β π£ = βcos π€

πβππ β« π€ 2 cos π€ ππ’ = π€ 2 sin π€ β 2 (βπ€ cos π€ β β« β cos π€ ππ€)
= π€ 2 sin π€ + 2π€ cos π€ β sin π€
πππππππππ ππππ: π€ = 5π‘ β
1
[(5π‘)2 sin(5π‘) + 2(5π‘) cos(5π‘) β sin(5π‘)]
β« π‘ 2 cos(5π‘) ππ‘ =
125

1

β« π‘ 2 cos(5π‘) ππ‘ = [
0

=[

π‘ 2 sin 5π‘ 2π‘ cos 5π‘ 2 sin 5π‘ 1
+
β
]|
5
25
125
0

sin 5 2 cos 5 2sin 5
+
β
] β [0 + 0 β 0] β β0.15375
5
25
125

2

Engineering Mathematics β MASC 0009.2
1
1 1
1
πΌπΌ: β« cos(5π‘) ππ‘ = [ sin 5π‘] | = (sin 5 β sin 0) β β0.19178
5
5
0
0
1

β΄ β« (π‘ 2 + 3) cos(5π‘) ππ‘ = πΌ + 3πΌπΌ = (β0.15375) + 3(β0.19178) β β0.7291
0

3

Engineering Mathematics β MASC 0009.2
2. Evaluate the subjoined well by using convenient method:
2π¦ 3 + π¦
β« (π βππ¦ βπ + π βππ¦ β 4
) ππ¦
π¦ + π¦2 + 1
2π¦ 3 + π¦
π=1
ππ πππ: {
, π€π βππ£π: β« (π βπ¦ β4 + π βπ¦ β 4
) ππ¦
π=4
π¦ + π¦2 + 1
β« (π βπ¦ β4 + π βπ¦ β

2π¦ 3 + π¦
2π¦ 3 + π¦
βπ¦
βπ¦ ππ¦ β β«
ππ¦
=
β«
π
+
π
ππ¦
)
β4
π¦4 + π¦2 + 1
π¦4 + π¦2 + 1

πΌ: β« π βπ¦ β4 + π βπ¦ ππ¦
πΏππ‘:

2π¦ 3 + π¦
πΌπΌ: β« 4
ππ¦
{
π¦ + π¦2 + 1

πΌ: β« π βπ¦ β4 + π βπ¦ ππ¦ : ππ πππ π π’ππ π‘ππ‘π’π‘πππ: πΏππ‘ π’ = 4 + π βπ¦ β

ππ’
ππ’
= βπ βπ¦ β ππ¦ =
ππ¦
βπ βπ¦

3

1
ππ’
π’2
2 3
β β« π βπ¦ βπ’ βπ¦ = β« ββπ’ππ’ = β β« π’2 ππ’ = β
= β (π’2 )
3
βπ
3
2
3
2
πππππππππ ππππ π’ = 4 + π βπ¦ β πΌ = β (4 + π βπ¦ )2
3

2π¦ 3 + π¦
πΌπΌ: β« 4
ππ¦ : ππ πππ π π’ππ π‘ππ‘π’π‘πππ
π¦ + π¦2 + 1
πΏππ‘ π’ = π¦ 4 + π¦ 2 + 1 β

ππ’
ππ’
= 4π¦ 3 + 2π¦ β (2π¦ 3 + π¦)ππ¦ =
ππ¦
2

1 ππ’ 1 1
1
ββ« β
= β« ππ’ = ln|π’|
π’ 2
2 π’
2
1
πππππππππ ππππ π’ = π¦ 4 + π¦ 2 + 1 β πΌπΌ = ln|π¦ 4 + π¦ 2 + 1|
2
β΄ β« (π βπ¦ β4 + π βπ¦ β

3
2π¦ 3 + π¦
2
1
βπ¦ )2
(4
ππ¦
=
πΌ
β
πΌπΌ
=
β
+
π
β ln|π¦ 4 + π¦ 2 + 1|
)
4
2
π¦ +π¦ +1
3
2

4

Engineering Mathematics β MASC 0009.2
3. The fruit effected W, to spread a leap from its regular prolixity to an extension of b meters is given
π

π₯

6

by π = β«0 ( 3 β 6π₯+9 β sec 2(3π₯) β β8) ππ₯ . Evaluate W.
π₯
β

1

π₯
6
πΊππ£ππ π = 1; π = β« ( 3 β
β sec 2(3π₯) β β8) ππ₯
βπ₯ 6π₯ + 9
0
1

2

1

π = β« π₯ 3 ππ₯ β β«
0

0

1
1
6
ππ₯ β β« sec 2(3π₯) ππ₯ β β« β8ππ₯
6π₯ + 9
0
0

1 2
3 5 1 3
πΌ: β« π₯ 3 ππ₯ = (π₯ 3 ) | =
5
5
0
0
1

πΌπΌ: β«
0
1

ββ«
0

6
ππ’
ππ₯ : ππ πππ π π’ππ π‘ππ‘π’π‘πππ; πππ‘ π’ = 6π₯ + 9 β
= 6, ππ’ = 6ππ₯
6π₯ + 9
ππ₯

6
1
ππ₯ = β« ππ’ = ln(π’)
6π₯ + 9
π’
1

πππππππππ ππππ π’ = 6π₯ + 9 β β«
0

1
6
5
ππ₯ = ln(6π₯ + 9) | = ln 15 β ln 9 = ln ( )
6π₯ + 9
3
0

1

πΌπΌπΌ: β« sec 2 (3π₯) ππ₯ : ππ πππ π π’ππ π‘ππ‘π’π‘πππ; πππ‘ π’ = 3π₯ β
0

ππ’
1
= 3, ππ₯ = ππ’
ππ₯
3

1
1
β β« sec 2 π’ ππ’ = tan π’
3
3
1
1 1
1
πππππππππ ππππ π’ = 3π₯ β β« sec 2(3π₯) ππ₯ = ( tan 3π₯) | = tan 3
3
3
0
0
1
1
πΌπ: β« β8ππ₯ = β8π₯ | = β8
0
0
1
π₯
6
3
5
1
βΉ β« (3 β
β sec 2(3π₯) β β8) ππ₯ = πΌ β πΌπΌ β πΌπΌπΌ β πΌπ = β ln ( ) β tan(3) β β8
5
3
3
βπ₯ 6π₯ + 9
0
1
π₯
6
β΄ β« (3 β
β sec 2 (3π₯) β β8) ππ₯ β β2.6917
6...

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