TUGAS MTK 2 KISI-KISI 2, 10 SOAL

1. TUGAS MATEMATIKA
Kisi-Kisi Test Akhir Semester 2 MTK2
Di susun oleh :
Nama : Gesti
Kelas : 1 ElektronikaB
Semester : 2 (Genap)
Jurusan : Elektronikadan Informatika
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Kawasan Industri Air Kantung Sungailiat 33211
Bangka Induk Propinsi Kepulauan Bangka Belitung
Telp : (0717) 431335 ext. 2281, 2126
Fax : (0717) 93585 email : polman@polman-babel.ac.id
http://www.polman-babel.ac.id/
1 . ∫(𝑥10 −
6
𝑥5
+ √𝑥73
) 𝑑𝑥 = ∫(𝑥10 − 6𝑥−5 + 𝑥
7
3) 𝑑𝑥

2. =
1
11
𝑥11 +
6
4
𝑥−4 +
3
10
𝑥
10
3 +c
=
1
11
𝑥11 +
3
2
𝑥−4 +
3
10
𝑥
10
3 +c
2. ∫[cos(9𝑥 − 11) + 𝑠𝑒𝑐2(6𝑥 − 8)] 𝑑𝑥
=
1
9
sin(9𝑥 − 11) +
1
6
tan(6𝑥 − 8) + 𝑐
3. Denganmenggunakancara subsitusi
∫
𝑥
√6+𝑥2
𝑑𝑥 =∫ 𝑥(6 + 𝑥2)
1
2 𝑑𝑥
Misalkan :
𝑢 = 6 + 𝑥2
𝑑𝑢
𝑑𝑥
= 2𝑥
𝑑𝑢 =
1
2𝑥
𝑑𝑢
∫ 𝑥(6 + 𝑥2)
1
2 𝑑𝑥
=∫ 𝑥 𝑈
−1
2 .
1
2𝑥
𝑑𝑢
=∫
𝑥
2𝑥
. 𝑈
−1
2 𝑑𝑢
=∫
1
2
. 𝑈
−1
2 𝑑𝑢
=
1
2
−1
2
+1
𝑈
−1
2
+1
+ 𝐶
=
1
2
1
2
𝑈
1
2 + 𝐶
=(6𝑥 + 𝑥2)
1
2 + 𝐶
4. Denganmenggunkancara subsitusi

3. ∫(2𝑥 + 5)cos(2𝑥2 + 10𝑥 + 8 ) 𝑑𝑥
Misalkan
U =2𝑥2 + 10𝑥 + 8
𝑑𝑢
𝑑𝑥
= 4𝑥 + 10
Dx=
1
4𝑥+10
𝑑𝑢
∫(2𝑥 + 5)cos(2𝑥2 + 10𝑥 + 8 ) 𝑑𝑥
=∫(2𝑥 + 5)cos 𝑢
1
4𝑥+10
du
∫
(2𝑥+5)
2 (2𝑥+5)
cos 𝑢 𝑑𝑢
∫
1
2
cos u du
=
1
2
sinu du
=
1
2
sin(2𝑥 2+ 10x +8 ) + c
5. Integral parsial
∫2𝑥. sin(10𝑥 + 3) dx
Misalkan:
u= 2x du =2dx
dv =sin(10x +3 ) v=∫sin(10𝑥 + 3) 𝑑𝑥 = −
1
10
cos(10𝑥 + 3)
=∫ 𝑈𝑑𝑣 = 𝑢𝑣 −∫ 𝑣 𝑑𝑢
=∫2𝑥.sin(10𝑥 + 3) 𝑑𝑥
=2𝑥 (−
1
10
cos(10𝑥 + 3)) − ∫−
1
10
cos(10𝑥 + 3). 2 𝑑𝑥
=−
1
5
𝑥.cos(10𝑥 + 3) 𝑑𝑥 +
2
100
sin(10𝑥 + 3) + 𝐶
=−
1
5
𝑥.cos(10𝑥 + 3) +
1
50
sin(10𝑥 + 3) + 𝐶

4. 6. Denganmenggunakantable
∫ 𝑥2 𝑒−7𝑥 𝑑𝑥
Turunan U Integral dv
+𝑥2
-2x
+2
-0
𝑒−7𝑥
−
1
7
𝑒−7𝑥
1
49
𝑒−7𝑥
−
1
363
𝑒−7𝑥
∫ 𝑢𝑑𝑣 = 𝑥2 ( −
1
7
𝑒−7𝑥) -2x .
1
49
𝑒−7𝑥 + 2 (−
1
369
𝑒−7𝑥)+ 𝑐
= −𝑥2 1
7
𝑒−7𝑥 -2x .
1
49
−
2
363
𝑒−7𝑥 𝑒−7𝑥 + 2 + 𝑐
= −
1
7
𝑥2 𝑒−7𝑥 -2x .
1
49
−
2
363
𝑒−7𝑥 𝑒−7𝑥 + 2 + 𝑐
7.Integral fungsi rasional
∫
𝑥
𝑥2−2𝑥−35
𝑑𝑥
𝑥
𝑥2−2𝑥−35
=
𝑥
( 𝑥−7)(𝑥+5)
=
𝐴
(𝑥−7)
+
𝐵
(𝑥+5)
=
𝐴( 𝑥+5)+𝐵(𝑥−7)
( 𝑥−7)( 𝑥+5)
𝐴𝑥+5𝐴+𝐵𝑥−7𝐵)
( 𝑥−7)( 𝑥+5)
A+B = 1 x5 5A+5B = 5
5A +B =0 x1 5A-7B = 0
12B=5
B=
5
12
A=
7
12

5. Sehingga:
∫
𝑥
( 𝑥−7)(𝑥+5)
𝑑𝑥 = ∫
𝐴
( 𝑥−7)
𝑑𝑥 + ∫
𝐵
( 𝑥+5)
𝑑𝑥
=∫
7
12
( 𝑥−7)
𝑑𝑥 + ∫
5
12
( 𝑥+5)
𝑑𝑥
=
7
12
𝑙𝑛 x-7 +
5
12
𝑙𝑛 x+5 + C
8.∫ ( 𝑥45
1 + 3𝑥 +
1
𝑥3
) 𝑑𝑥 =∫ ( 𝑥45
1 + 3𝑥 + 𝑥−3 ) 𝑑𝑥
=
1
5
[𝑥5 +
3
2
𝑥2 −
1
2
𝑥−2]5
1
= (
1
5
55 +
3
2
52 −
1
2
𝑥5−2 ) –(
1
5
15 +
3
2
12 −
1
2
1−2 )
=(625 +
75
2
−
1
50
) − (
1
5
+
3
2
−
1
2
)
=625- 1 +
75
2
−
1
50
−
1
5
=624 +
75
2
-
1
50
-
1
5
31200 + 1875 − 1 − 10
50
=
33064
50
= 661
14
50
9.Dik = y = 𝑥2 − 1
Y = 3x + 9
Dit = Luas daerah
Jawab:
𝑥2 − 1 = 3𝑥 + 9
𝑥2 − 1 − 3𝑥 − 9 = 0
𝑥2 − 3𝑥 − 10 = 0
(x-5) (x+2) = 0
X= 5 v x=-2

6. L=∫ (3𝑥 + 9 )– (𝑥25
−2 − 1) 𝑑𝑥
=∫ 3𝑥 −
5
−2 𝑥2 + 10 𝑑𝑥
=
3
2
[𝑥2 −
1
3
𝑥3 + 10𝑥] 5
−2
=(
3
2
52 −
1
3
53 + 10.5) − (
3
2
(−2)2 −
1
3
(−2)3 + 10. −2)
= (
75
2
−
125
3
+ 50) − (6 +
8
3
− 20)
= (
225−250+300
6
) − (
18+8−60
3
)
=
275
6
+
34
3
=
275+68
6
=
343
6
= 57
1
6
10.
Diketahu :
iy= 3x Y= x
Y= 0 y= 2
Dit : Volume benda =mengelilingi sumbuy
Jawab=
V = 𝜋 ∫ ( 𝑥2 − 𝑥22) 𝑑𝑦
𝑑
𝑒
= 𝜋∫ (𝑦2 − (
2
0
1
3
𝑦)2 ) dy
= 𝜋∫ 𝑦2 −
2
0
1
9
𝑦2 dy
= 𝜋∫ −
2
0
8
9
𝑦2 dy
=𝜋[(
8
9
2+1
𝑦2+1)] 2
0
= 𝜋(
8
27
𝑦3) 2
0
= 𝜋(
8
27
23-)-(
8
27
. 03) = 𝜋
64
27
=2
10
27
𝜋