Tugas Kisi-Kisi MTK Tes 2 (Monica R.)1. MATEMATIKA 2
KISI-KISI TES 2
Disusun Oleh :
Nama : Monica Roselina
NPM : 003 14 18
Prodi : Teknik Elektronika
Kelas : 1 EA
Semester : 2 (Genap)
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Kawasan Industri Air Kantung Sungailiat, Bangka 33211
Telp. (0717) 93586, Fax. (0717) 93585
Email :polman@polman-babel.ac.id
Website :www.polman-babel.ac.id
TAHUN AJARAN 2014/2015
2. 1. Hitunglah β« (π₯12
β
12
π₯5 + βπ₯103
) ππ₯
β«(π₯12
β
12
π₯5
+ β π₯103
) ππ₯
= β« π₯12
β 12π₯β5
+ π₯
10
3 ππ₯
=
1
13
π₯13
β
12
β4
π₯β4
+
1
13
3
π₯
13
3 + πΆ
=
1
13
π₯13
+ 3π₯β4
+
3
13
π₯
13
3 + πΆ
=
1
13
π₯13
+
3
π₯4
+
3
13
β π₯133
+ πΆ
2. Hitunglah β«[cos(7π₯ β 12) + π ππ2(9π₯ β 15)] ππ₯
β«[cos(7π₯ β 12) + π ππ2(9π₯ β 15)] ππ₯
=
1
7
sin(7π₯ β 12)+
1
9
tan(9π₯ β 15) + πΆ
3. Dengan menggunakan cara substitusi hitunglah β«
π₯2
β3+π₯3 ππ₯
β«
π₯2
β3 + π₯3
ππ₯
= β« π₯2
.(3 + π₯3)β
1
2 ππ₯
π’ = 3 + π₯3
β
ππ’
ππ₯
= 3π₯2
β ππ₯ =
ππ’
3π₯2
β« π₯2
. (3 + π₯3)β
1
2 ππ₯ = β« π₯2
. π’
β
1
2 .
ππ’
3π₯2
=
1
3
β« π’
β
1
2 ππ’ =
1
3
.
1
1
2
π’
1
2 + πΆ
=
2
3
β π’ + πΆ =
2
3
β3 + π₯3 + πΆ
4. Dengan menggunakan cara substitusi hitunglah β«(2π₯ + 2)cos(5π₯2
+ 10π₯ + 8) ππ₯
β«(2π₯ + 2)cos(5π₯2
+ 10π₯ + 8) ππ₯
π’ = 5π₯2
+ 10π₯ + 8 β
ππ’
ππ₯
= 10π₯ + 10 β ππ₯ =
ππ’
10π₯ + 10
β«(2π₯ + 2)cos(5π₯2
+ 10π₯ + 8) ππ₯ = β«(2π₯ + 2).cos π’ .
ππ’
10π₯ + 10
= β«(2π₯ + 2).cos π’ .
ππ’
5(2π₯ + 2)
=
1
5
β«cos π’ ππ’
=
1
5
sin π’ + πΆ =
1
5
sin(5π₯2
+ 10π₯ + 8) + πΆ
3. 5. Hitunglah integral parsil dari β« 2π₯. sin(12π₯ + 4) ππ₯
β«2π₯. sin(12π₯ + 4) ππ₯
π’ = 2π₯ β
ππ’
ππ₯
= 2 β ππ’ = 2ππ₯
ππ£ = sin(12π₯ + 4) ππ₯ β π£ = β«sin(12π₯ + 4) ππ₯ = β
1
12
cos(12π₯ + 4)
β« π’. ππ£ = π’. π£ β β« π£ ππ’
β«2π₯. sin(12π₯ + 4) ππ₯ = 2π₯. β
1
12
cos(12π₯ + 4) β β« β
1
12
cos(12π₯ + 4). 2ππ₯
= β
1
6
π₯ cos(12π₯ + 4) + 2 [
1
12
12
sin(12π₯ + 4)] + πΆ
= β
1
6
π₯ cos(12π₯ + 4) +
1
72
sin(12π₯ + 4) + πΆ
6. Dengan menggunakan bantuan table hitunglah integral dari β« π₯3
πβ5π₯
ππ₯
+
π₯3
πβ5π₯
-
3π₯2
β
1
5
πβ5π₯
+
6π₯ 1
25
πβ5π₯
-
6
β
1
125
πβ5π₯
+ 0
1
625
πβ5π₯
= β
1
5
π₯3
πβ5π₯
β
3
25
π₯2
πβ5π₯
β
6
125
π₯πβ5π₯
β
6
625
πβ5π₯
+ πΆ
turunan integral
4. 7. Hitung integral fungsi rasional dari β«
3π₯
π₯2β2π₯β15
ππ₯
3π₯
π₯2 β 2π₯ β 15
=
3π₯
(π₯ β 5)(π₯ + 3)
=
π΄
( π₯ β 5)
+
π΅
( π₯ + 3)
π₯ β 5 = 0 β π₯ = 5 β π΄ =
3.5
(5 + 3)
=
15
8
π₯ + 3 = 0 β π₯ = β3 β π΅ =
3. β3
(β3 β 5)
=
9
8
β«
3π₯
π₯2 β 2π₯ β 15
ππ₯ = β«
15
8
( π₯ β 5)
ππ₯ + β«
9
8
( π₯ + 3)
ππ₯
=
15
8
ln| π₯ β 5| +
9
8
ln| π₯ + 3| + πΆ
8. Hitunglah integral tentu dari β« (π₯4
+ 5π₯ +
1
π₯3)
4
1
ππ₯
β«(π₯4
+ 5π₯ +
1
π₯3
)
4
1
ππ₯ = β«(π₯4
+ 5π₯ + π₯β3
)
4
1
ππ₯
=
1
5
π₯5
+
5
2
π₯2
β
1
2
π₯β2
=
1
5
π₯5
+
5
2
π₯2
β
1
2π₯2
= (
1
5
. 45
+
5
2
. 42
β
1
2.42
) β (
1
5
. 15
+
5
2
. 12
β
1
2.12
)
= (
1024
5
+ 40 β
1
32
) β (
1
5
+
5
2
β
1
2
)
=
1024
5
β
1
5
β
1
32
β
4
2
+ 40 =
1023
5
β
1
32
+ 38
=
32736 β 5 + 6080
160
=
38811
160
9. Tentukan luas daerah yang dibatasi oleh kurva π¦ = π₯2
+ 4dan garis π¦ = βπ₯ + 16
π¦1 = π¦2 β π₯2
+ 4 = βπ₯ + 16
π₯2
+ π₯ β 12 = 0
( π₯ + 4)( π₯ β 3) = 0
π₯ = β4 ππ‘ππ’ π₯ = 3
πΏ = β«(βπ₯ + 16) β ( π₯2
+ 4)
3
β4
ππ₯
= β«(βπ₯2
β π₯ + 12)
3
β4
ππ₯ = β
1
3
π₯3
β
1
2
π₯2
+ 12π₯
= (β
1
3
. 33
β
1
2
. 32
+ 12.3) β (β
1
3
. β43
β
1
2
. β42
+ 12. β4)
= (β9 β
9
2
+ 36) β (
64
3
β 8 β 48)
= 27 β
9
2
β
64
3
+ 56 = β
64
3
β
9
2
+ 83
5. =
β128 β 27 + 498
6
=
343
6
π ππ‘π’ππ ππ’ππ
10. Tentukanlah volume benda yang terbentuk dengan memutar mengelilingi sumbu -y
dari daerah yang dibatasi oleh π¦ = 3π₯, π¦ = π₯, π¦ = 0 dan garis π¦ = 3
π¦ = 3π₯ β π₯ =
1
3
π¦
π¦ = π₯ β π₯ = π¦
π = π β«( π₯1
2
β π₯2
2)
3
0
ππ¦
= π β«(π¦2
β (
1
3
π¦)
2
)
3
0
ππ¦ = π β« (π¦2
β
1
9
π¦2
)
3
0
ππ¦
= π β«
8
9
π¦2
3
0
ππ¦ = π [
8
9
3
π¦3
]
= π [
8
27
π¦3
] = π [
8
27
. 33
β
8
27
. 03
]
= π[8 β 0] = 8π π ππ‘π’ππ π£πππ’ππ