Tugas mtk 10
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tugas kisi-kisi test akhir sem.2 mtk2
![1 . (10
6
5
+ 73
) = (10 6モ5 +
7
3)
=
1
11
11 +
6
4
モ4 +
3
10
10
3 +c
=
1
11
11 +
3
2
モ4 +
3
10
10
3 +c
2. [cos(9 11) + 2(6 8)]
=
1
9
sin(9 11) +
1
6
tan(6 8) +
3. Denganmenggunakancara subsitusi
6+2
= (6 + 2)
1
2
Misalkan :
= 6 + 2
= 2
=
1
2
(6 + 2)
1
2
=
1
2 .
1
2
=
2
.
1
2
=
1
2
.
1
2
=
1
2
1
2
+1
1
2
+1
+
=
1
2
1
2
1
2 +
=(6 + 2)
1
2 +](https://image.slidesharecdn.com/tugasmtk10-150712214046-lva1-app6891/85/Tugas-mtk-10-2-320.jpg)
![B=
5
12
A=
7
12
Sehingga:
( モ7)(+5)
=
( モ7)
+
( +5)
=
7
12
( モ7)
+
5
12
( +5)
=
7
12
x-7 +
5
12
x+5 + C
8. ( 45
1 + 3 +
1
3
) = ( 45
1 + 3 + モ3 )
=
1
5
[5 +
3
2
2
1
2
モ2]5
1
= (
1
5
55 +
3
2
52
1
2
52 ) (
1
5
15 +
3
2
12
1
2
12 )
=(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 = 2 1
Y = 3x + 9
Dit = Luas daerah
Jawab:
2 1 = 3 + 9
2 1 3 9 = 0
2 3 10 = 0
(x-5) (x+2) = 0](https://image.slidesharecdn.com/tugasmtk10-150712214046-lva1-app6891/85/Tugas-mtk-10-5-320.jpg)
![X= 5 v x=-2
L= (3 + 9 ) (25
2 1)
= 3
5
2 2 + 10
=
3
2
[2
1
3
3 + 10] 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)
= (
225250+300
6
) (
18+860
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 = ( 2 22)
= (2 (
2
0
1
3
)2 ) dy
= 2
2
0
1
9
2 dy
=
2
0
8
9
2 dy
=[(
8
9
2+1
2+1)] 2
0
= (
8
27
3) 2
0
= (
8
27
23-)-(
8
27
. 03) =
64
27
=2
10
27](https://image.slidesharecdn.com/tugasmtk10-150712214046-lva1-app6891/85/Tugas-mtk-10-6-320.jpg)
Tugas mtk 10
- 1. TUGAS MATEMATIKA Kisi-Kisi Test Akhir Sem.2 MTK2 Di susun oleh : Nama : Larasati 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/
- 2. 1 . (10 6 5 + 73 ) = (10 6モ5 + 7 3) = 1 11 11 + 6 4 モ4 + 3 10 10 3 +c = 1 11 11 + 3 2 モ4 + 3 10 10 3 +c 2. [cos(9 11) + 2(6 8)] = 1 9 sin(9 11) + 1 6 tan(6 8) + 3. Denganmenggunakancara subsitusi 6+2 = (6 + 2) 1 2 Misalkan : = 6 + 2 = 2 = 1 2 (6 + 2) 1 2 = 1 2 . 1 2 = 2 . 1 2 = 1 2 . 1 2 = 1 2 1 2 +1 1 2 +1 + = 1 2 1 2 1 2 + =(6 + 2) 1 2 +
- 3. 4. Denganmenggunkancara subsitusi (2 + 5)cos(22 + 10 + 8 ) Misalkan U =22 + 10 + 8 = 4 + 10 Dx= 1 4+10 (2 + 5)cos(22 + 10 + 8 ) =(2 + 5)cos 1 4+10 du (2+5) 2 (2+5) cos 1 2 cos u du = 1 2 sinu du = 1 2 sin(2 2+ 10x +8 ) + c 5. Integral parsial 2. sin(10 + 3) dx Misalkan: u= 2x du =2dx dv =sin(10x +3 ) v=sin(10 + 3) = 1 10 cos(10 + 3) = = =2.sin(10 + 3) =2 ( 1 10 cos(10 + 3)) 1 10 cos(10 + 3). 2 = 1 5 .cos(10 + 3) + 2 100 sin(10 + 3) +
- 4. = 1 5 .cos(10 + 3) + 1 50 sin(10 + 3) + 6. Denganmenggunakantable 2 7 Turunan U Integral dv +2 -2x +2 -0 7 1 7 7 1 49 7 1 363 7 = 2 ( 1 7 7) -2x . 1 49 7 + 2 ( 1 369 7)+ = 2 1 7 7 -2x . 1 49 2 363 7 7 + 2 + = 1 7 2 7 -2x . 1 49 2 363 7 7 + 2 + 7.Integral fungsi rasional 22モ35 22モ35 = ( モ7)(+5) = (モ7) + (+5) = ( +5)+(モ7) ( モ7)( +5) 基+5+巨モ7) ( モ7)( +5) A+B = 1 x5 5A+5B = 5 5A +B =0 x1 5A-7B = 0 12B=5
- 5. B= 5 12 A= 7 12 Sehingga: ( モ7)(+5) = ( モ7) + ( +5) = 7 12 ( モ7) + 5 12 ( +5) = 7 12 x-7 + 5 12 x+5 + C 8. ( 45 1 + 3 + 1 3 ) = ( 45 1 + 3 + モ3 ) = 1 5 [5 + 3 2 2 1 2 モ2]5 1 = ( 1 5 55 + 3 2 52 1 2 52 ) ( 1 5 15 + 3 2 12 1 2 12 ) =(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 = 2 1 Y = 3x + 9 Dit = Luas daerah Jawab: 2 1 = 3 + 9 2 1 3 9 = 0 2 3 10 = 0 (x-5) (x+2) = 0
- 6. X= 5 v x=-2 L= (3 + 9 ) (25 2 1) = 3 5 2 2 + 10 = 3 2 [2 1 3 3 + 10] 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) = ( 225250+300 6 ) ( 18+860 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 = ( 2 22) = (2 ( 2 0 1 3 )2 ) dy = 2 2 0 1 9 2 dy = 2 0 8 9 2 dy =[( 8 9 2+1 2+1)] 2 0 = ( 8 27 3) 2 0 = ( 8 27 23-)-( 8 27 . 03) = 64 27 =2 10 27