1. Tugas Matematika
Kisi-kisi test 2
Disusun Oleh :
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
TAHUN AJARAN 2014/2015
Industri Air Kantung Sungailiat 33211
Bangka Induk, Propinsi Kepulauan Bangka Belitung
Telp : +62717 93586
Fax : +6271793585 email : polman@polman-babel.ac.id
http://www.polman-babel.ac.id
Nama : Rakam Tiano
Kelas : 1 EB
Prodi : Elektronika
Semester : 2 ( Genap)
3. 1 . (𝑥10
−
6
𝑥5 + 𝑥73
) 𝑑𝑥 = (𝑥10
− 6𝑥−5
+ 𝑥
7
3) 𝑑𝑥
=
1
11
𝑥11
+
6
4
𝑥−4
+
3
10
𝑥
10
3 +c
=
1
11
𝑥11
+
3
42
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. Dengan menggunakan cara 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
3
2 + 𝑐 =
1
3
𝑢
3
2 +c
=
1
3
( 6 + 𝑥2
)
3
2 + 𝑐
4. Dengan menggunkan cara subsitusi
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
sin u du
=
1
2
sin (2𝑥 2
+ 10x +8 ) + c
4. 5. Integral parsial
2𝑥. sin(10𝑥 + 3) dx
Mislkan :
U = 2x dv =sin (10x +3 )
du =2dx v = sin 10𝑥 + 3 𝑑𝑥 = −
1
10
cos(10𝑥 3)
Type equation here.
𝑈𝑑𝑣 =Uv - 𝑉𝑑𝑢 =
2 𝑥 . sin 10𝑋 + 3 𝑑𝑥 = 2𝑥 (
1
10
cos 10𝑥 + 3 —
1
10
cos 10𝑥 + 3 . 2 𝑑𝑥
=
1
5
𝑥 cos 10𝑥 + 3 +
2
100
sin (10x +3) + c
=
1
5
𝑥 cos 10𝑥 + 3 +
1
50
sin (10x +3) + c
6. Dengan menggunakan table
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
+ 𝑐
= −
1
7
𝑥2
𝑒−7𝑥
-
2
49
𝑥 𝑒 −7𝑥
−
2
363
𝑒−7𝑥
+ 𝑐
7. Integral fungsi rasional
𝑥
𝑥2 − 2𝑥 − 35
𝑑𝑥
5. 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
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
6. 𝑥2
− 3𝑥 − 10 = 0
(x-5) (x+2) = 0
X= 5 v x=-2
= 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 :
y = 3x y= x
y= 0 y= 2
Dit : Volume benda mengelilingi sumbu y
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
