1. TUGAS KELOMPOK MATEMATIKA
Buku Calculus Hal. 61-66
Di susun oleh :
Kelompok 8
Nama : 1. Hetty Agustina Tampubolon
2. Larasati
3. Ratna
Kelas : 1 Elektronika B
Semester : 2 (Genap)
Jurusan : Elektronika dan 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. Latihan 9.1
1. ∫ (3𝑥210
−10
+ 4𝑥 − 5) 𝑑𝑥
=
3
3
𝑥3
+
4
2
𝑥2
− 5𝑥 ] 10
−10
=𝑥3
+ 2𝑥2
− 5𝑥 ] 10
−10
=( (103 )+ 2(10) 2
− 5(10))− ((−10) 3
+ 2(−10) 2
− 5(−10))
=(1000 + 200 − 50) − (−1000 + 200 + 50)
=1150 − (−750)
=1150 + 750
=1900
2. ∫ 8 𝑑𝑥
30
−50
=8𝑥] 30
−50
=8(30)− 8(−50)
=240 + 400
=640
3. ∫
𝑥5
𝑥2
7
2
𝑑𝑥
=∫ 𝑥5
. 𝑥2
𝑑𝑥
7
2
=∫ 𝑥3
𝑑𝑥
7
2
=
1
4
𝑥 4
]7
2
=
1
4
74
−
1
4
24
=
2401
4
−
16
4
=
2385
4
=596,25
4. ∫
1
𝑡
𝑑𝑡
36
6
= ∫
36
6
𝑡−1
𝑑𝑡 = ln | 𝑡| ]36
6
=ln|36| − ln|6|
=3,58 − 1.79
=1,79
5. ∫
𝜋
0,5𝜋
sec (
5
6
θ)tan (
5
6
𝜃) 𝑑𝜃
=
6
5
sec(
5
6
𝜃) ]
𝜋
0,5𝜋
=(
6
5
sec(
5
6
. 180°))− (
6
5
sec(
5
6
. 90°))
=(
6
5
sec(150°))− (
6
5
sec(75°))

3. =
6
5
(
1
cos 150°
−
1
cos 75°
)
=
6
5
(−1,15 − 3,86) =
6
5
(−5,01)
=−6,012
6. ∫ √3
1
𝑑𝑥
√4−𝑥2 = ∫ √3
1
1
√4−𝑥2 𝑑𝑥 = ∫ √3
1
(4 − 𝑥2)−
1
2 𝑑𝑥
=
1
−2𝑥
.
1
−
1
2
+ 1
. (4 − 𝑥2)−
1
2
+1
= −
1
2𝑥
. 2. (4 − 𝑥2)
1
2
= −
1
𝑥
√4 − 𝑥2
= (−
1
√3
. √4 − (√3)
2
) − (−
1
1
. √4 − (1)2)
= (−
1
√3
. √1) − (−1. √3)
= −
1
√3
+ √3 =
−1 + 3
√3
=
2
√3
.
√3
√3
=
2
3
√3
7. ∫ (3𝑥4
− 5𝑥 3
− 21𝑥 2
+ 36𝑥 − 10) 𝑑𝑥
2
1
=
3
5
𝑥5
−
5
4
𝑥4
−
21
3
𝑥3
+
36
2
𝑥2
− 10𝑥] 2
1
=(
3
5
25
−
5
4
24
− 7.23
+ 18. 22
− 10.2 ) − (
3
5
15
−
5
4
14
− 7. 13
+ 18. 12
− 10.1)
=(
3
5
. 32 −
5
4
. 16 − 7.8 + 18.4 − 20) − (
3
5
−
5
4
− 7 + 18 − 10)
=(19,2 − 20 − 56 + 72 − 20) − (0.6 − 1,25 − 7 + 18 − 10)
=−4,8 − 0,35
=−5,15
8. ∫
5
3
( 𝑥3
ln 𝑥) 𝑑𝑥 =
Misal :
𝑢 = ln 𝑥 => 𝑑𝑢 =
1
𝑥
𝑑𝑥
𝑑𝑣 = 𝑥3
𝑑𝑥 => 𝑣 = ∫ 𝑥3
𝑑𝑥
𝑣 =
1
4
𝑥4

4. ∫ 𝑢. 𝑑𝑣 = 𝑢. 𝑣 − ∫ 𝑣 𝑑𝑢 = ln 𝑥 .
1
4
𝑥4
− ∫
1
4
𝑥4
.
1
𝑥
𝑑𝑥
=
1
4
𝑥4
ln 𝑥 −
1
4
∫ 𝑥3
𝑑𝑥
=
1
4
𝑥4
ln 𝑥 −
1
4
(
1
4
𝑥4
) + C
=
1
4
𝑥4
ln 𝑥 −
1
16
𝑥4
+ C
Jadi,
∫
5
3
( 𝑥3
ln 𝑥) 𝑑𝑥 =
1
4
𝑥4
ln 𝑥 −
1
16
𝑥4
] 5
3
= (
1
4
. 54
ln 5 −
1
16
. 54
) − (
1
4
. 34
ln 3 −
1
16
. 34
)
= (
201
4
−
625
16
) − (
89
4
−
81
16
)
=
112
4
−
544
16
= 28 − 34 = −6
9. ∫ √3
1
𝑐𝑜𝑡−1( 𝑥) 𝑑𝑥 = ∫ √3
1
𝑡𝑎𝑛( 𝑥) 𝑑𝑥 = −ln | cos 𝑥| ] √3
1
= − ln |cos√3| − (− ln |cos1 |)
= − ln1 − (− ln 1) = 0 + 0
=0
10. ∫
1
1+𝑒 𝑥
5
2
𝑑𝑥
= ∫ (1 + 𝑒 𝑥)−1
5
2
𝑑𝑥
𝑀𝑖𝑠𝑎𝑙,
𝑢 = 1 + 𝑒 𝑥
𝑑𝑢
𝑑𝑥
= 𝑒 𝑥
𝑑𝑥 =
𝑑𝑢
𝑒 𝑥
Jadi,
∫ (1 + 𝑒 𝑥)−1
5
2
𝑑𝑥 = ∫ 𝑢−1
5
2
.
𝑑𝑢
𝑒 𝑥
=
1
𝑒 𝑥
∫ 𝑢−1
5
2
𝑑𝑢
=
1
𝑒 𝑥
. ln 𝑢 =
1
𝑒 𝑥
. ln(1 + 𝑒 𝑥) =
ln(1 + 𝑒 𝑥)
𝑒 𝑥
= [
ln(1 + 𝑒5)
𝑒5
] − [
ln(1 + 𝑒2)
𝑒2
]
= 0,0337 − 0,2878 = −0,2541

5. Latihan 9.2
Diberikan ∫ 𝑓( 𝑥) 𝑑𝑥 = 12 𝑑𝑎𝑛 ∫ 𝑓( 𝑥) 𝑑𝑥 = 15
2
0
0
−2
1. ∫ 𝑓( 𝑥) 𝑑𝑥 = 0
2
2
𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑠𝑖𝑓𝑎𝑡 1, ∫ 𝑓( 𝑥) 𝑑𝑥 = 0
2
2
2. ∫ 𝑓( 𝑥) 𝑑𝑥
−2
0
𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑠𝑖𝑓𝑎𝑡 2,∫ 𝑓( 𝑥) 𝑑𝑥 = − ∫ 𝑓( 𝑥) 𝑑𝑥 = −12
0
−2
−2
0
3. ∫ 𝑓( 𝑥) 𝑑𝑥
1
1
𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑠𝑖𝑓𝑎𝑡 1,∫ 𝑓( 𝑥) 𝑑𝑥 = 0
1
1
4. ∫ 𝑓( 𝑥) 𝑑𝑥
2
−2
𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑠𝑖𝑓𝑎𝑡 3,∫ 𝑓( 𝑥) 𝑑𝑥 = ∫ 𝑓( 𝑥) 𝑑𝑥 + ∫ 𝑓( 𝑥) 𝑑𝑥 = 12 + 15 = 27
2
0
0
−2
2
−2
5. ∫ 5𝑓( 𝑥) 𝑑𝑥
0
−2
𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑠𝑖𝑓𝑎𝑡 4,∫ 5𝑓( 𝑥) 𝑑𝑥 = 5 ∫ 𝑓( 𝑥) 𝑑𝑥 = 5(12) = 60
0
−2
0
−2
6. ∫ 10𝑓( 𝑥) 𝑑𝑥
−2
2
𝐷𝑖𝑘𝑒𝑡𝑎ℎ𝑢𝑖 𝑑𝑎𝑟𝑖 𝑠𝑜𝑎𝑙 𝑛𝑜 4 𝑏𝑎ℎ𝑤𝑎 ∫ 𝑓 = 27, 𝑚𝑎𝑘𝑎 𝑚𝑒𝑛𝑔𝑔𝑢𝑛𝑎𝑘𝑎𝑛 𝑠𝑖𝑓𝑎𝑡 2 & 4,
2
−2
∫ 𝑓( 𝑥) 𝑑𝑥 = −10∫ 𝑓( 𝑥) 𝑑𝑥 = −10(27) = −270
2
−2
−2
2
Diberikan ∫ 𝑓( 𝑥) 𝑑𝑥 = −8 𝑑𝑎𝑛 ∫ 𝑔( 𝑥) 𝑑𝑥 = 22
5
1
5
1
7. ∫ [ 𝑓( 𝑥) + 𝑔( 𝑥)] 𝑑𝑥
5
1
Mengggunakan sifat 5,
∫ [ 𝑓( 𝑥) + 𝑔( 𝑥)] 𝑑𝑥 = ∫ 𝑓( 𝑥) 𝑑𝑥 + ∫ 𝑔( 𝑥) 𝑑𝑥 = −8 + 22 = 14
5
1
5
1
5
1
8. ∫ [ 𝑓( 𝑥) − 𝑔( 𝑥)]
5
1
𝑑𝑥
Menggunakan sifat 5,
∫ [𝑓( 𝑥) − 𝑔( 𝑥) 𝑑𝑥 =
5
1
∫ 𝑓( 𝑥) 𝑑𝑥 −
5
1
∫ 𝑔( 𝑥) 𝑑𝑥 = −8 − 22 = −30
5
1

6. 9. ∫
5
1
1
2
𝑓( 𝑥) 𝑑𝑥
Menggunakan sifat 4,
∫
5
1
1
2
𝑓( 𝑥) 𝑑𝑥 =
1
2
∫ 𝑓( 𝑥) 𝑑𝑥 =
5
1
1
2
(−8) = −4
10. ∫ 2𝑔(𝑥)
5
1
𝑑𝑥 + ∫ 3𝑓(𝑥)
5
1
𝑑𝑥
Menggunakan sifat 4,
∫ 2𝑔( 𝑥) 𝑑𝑥 +
5
1
∫ 3𝑓( 𝑥) 𝑑𝑥 = 2
5
1
∫ 𝑔( 𝑥) 𝑑𝑥 + 3
5
1
∫ 𝑓( 𝑥)
5
1
= 2(22)+ 3(−8)
= 44 − 24 = 20

7. Latihan 9.3
1.
𝑑
𝑑𝑥
[∫ ( 𝑡2
+ 3)−5𝑥
0
𝑑𝑡]
= ( 𝑥2
+ 3)−5
=
1
( 𝑥2 + 3)5
2.
𝑑
𝑑𝑥
[∫ √3𝑡 + 5
𝑥
1
𝑑𝑡]
= √3𝑥 + 5
3.
𝑑
𝑑𝑥
[∫ 𝑡 sin 𝑡
𝑥4
𝜋
]
= 𝑥4
sin( 𝑥4).
𝑑
𝑑𝑥
( 𝑥4)
= 𝑥4
sin( 𝑥4). 4𝑥3
= 4𝑥7
sin( 𝑥4)
4.
𝑑
𝑑𝑥
[∫ √𝑡235𝑥2
−5
𝑑𝑡]
= √(5𝑥2)23
.
𝑑
𝑑𝑥
(5𝑥2)
= √25𝑥43
.10𝑥
= 10𝑥 √25𝑥43
5.
𝑑
𝑑𝑥
[∫ ( 𝑡2
− 2𝑡 + 1)
𝑥+2
−10
𝑑𝑡
= ( 𝑥 + 2)2
− 2( 𝑥 + 2) + 1
= 𝑥2
+ 4𝑥 + 4 − 2𝑥 − 4 + 1
= 𝑥2
+ 2𝑥 + 1
6. 𝐹( 𝑥) = ∫ sin(3𝑡)𝑥
0
𝑑𝑡
𝐹′( 𝑥) =
𝑑
𝑑𝑥
[∫ sin(3𝑡)
𝑥
0
𝑑𝑡]
𝐹′( 𝑥) = sin 3𝑥
7. 𝐹( 𝑥) = ∫
1
𝑡+1
4𝑥
5
𝑑𝑡
𝐹′( 𝑥) =
𝑑
𝑑𝑥
[∫
1
𝑡 + 1
4𝑥
5
𝑑𝑡]
𝐹′( 𝑥) =
1
4𝑥 + 1
8. 𝐹( 𝑥) = ∫ 6𝑡2sin 𝑥
0
𝑑𝑡
𝐹′( 𝑥) =
𝑑
𝑑𝑥
[∫ 6𝑡2
sin 𝑥
0
𝑑𝑡]
𝐹′( 𝑥) = 6(sin 𝑥)2

8. 𝐹′( 𝑥) = 6. 𝑠𝑖𝑛2
𝑥
9. 𝐹( 𝑥) = ∫ 2𝑡4√ 𝑥
−3
𝑑𝑡
𝐹′( 𝑥) =
𝑑
𝑑𝑥
[∫ 2𝑡4
√ 𝑥
−3
𝑑𝑡]
𝐹′( 𝑥) = 2(√ 𝑥)
4
𝐹′( 𝑥) = 2𝑥2
10. 𝐹( 𝑥) = ∫ 3𝑡 − 7
2𝑥+1
−8
𝑑𝑡
𝐹′( 𝑥) =
𝑑
𝑑𝑥
[∫ (3𝑡 − 7)
2𝑥+1
−8
𝑑𝑡]
𝐹′( 𝑥) = 3(2𝑥 + 1) − 7
𝐹′( 𝑥) = 6𝑥 + 3 − 7
𝐹′( 𝑥) = 6𝑥 − 4

9. Latihan 9.4
1. 𝑓( 𝑥) = 2𝑥 + 6 , dan interval [-1,1]
∫
𝑏
𝑎
𝑓( 𝑥) 𝑑𝑥 = 𝑓 (𝑐)(𝑏 − 𝑎)
∫
1
−1
(2𝑥 + 6) 𝑑𝑥 = (2𝑐 + 6)(1 − (−1))
(
2
2
𝑥2
+ +6𝑥) | 1
−1
= (2c+6) (1+1)
(𝑥2
− 6𝑥 ) |
1
−1
= (2𝑐 + 6).2
((12
+ 6.1) − ((−1)2
+ 6 (−1))) = 4𝑐 + 12
7 − (−5) = 4𝑐 + 12
12 = 4𝑐 + 12
12 − 12 = 4𝑐
0 = 4𝑐
𝑐 =
0
4
= 0
2. 𝑓( 𝑥) = 2 − 5√ 𝑥 , dan interval [0,4]
∫
𝑏
𝑎
𝑓( 𝑥) 𝑑𝑥 = 𝑓 (𝑐)(𝑏 − 𝑎)
∫
4
0
(2 − 5√ 𝑥)𝑑𝑥 = (2 − 5 √ 𝑐)(4 − 0)
∫
4
0
(2 − 5𝑥
1
2 )𝑑𝑥 = (2 − 5 √ 𝑐)(4)
2𝑥 −
5
3
2⁄
𝑥
3
2|4
0
= 8 − 20√ 𝑐
(2.4 −
10
3
4
3
2 ) − (0) = 8 − 20√ 𝑐
8 −
80
3
= 8 − 20√ 𝑐
20√ 𝑐 = 8 − 8 +
80
3
=
80
3
60√ 𝑐 = 80
√ 𝑐 =
80
60
=
4
3
(√ 𝑐)2
= (
4
3
)
2
𝑐 =
16
9
= 1,778
3. 𝑓( 𝑥) =
4
𝑥3 , dan interval [1,4]
∫
𝑏
𝑎
𝑓( 𝑥) 𝑑𝑥 = 𝑓 (𝑐)(𝑏 − 𝑎)
∫
4
1
(
4
𝑥3) 𝑑𝑥 =
4
𝑐3
(4 − 1)
∫
4
1
(4𝑥−3) 𝑑𝑥 =
4
𝑐3
(3)
(
4
−2
𝑥−2
) | 4
1
=
12
𝑐3

10. (−2 𝑥−2) |
4
1
=
12
𝑐3
(−2 .4−2) − (−2 1−2) =
12
𝑐3
−2
16
− (−2) =
12
𝑐3
−0,125 + 2 =
12
𝑐3
1,875 =
12
𝑐3
1,875 𝑐3
= 12
𝑐3
=
12
1,875
𝑐3
= 6,4
𝑐 = √6,43
= 1,86
4. 𝑓( 𝑥) = sin 𝑥 , dan interval [0, 𝜋]
∫
𝑏
𝑎
𝑓( 𝑥) 𝑑𝑥 = 𝑓 (𝑐)(𝑏 − 𝑎)
∫
𝜋
0
(sin 𝑥) 𝑑𝑥 = sin 𝑐 ( 𝜋 − 0)
(− cos 𝑥) | 𝜋
0
= 𝜋 sin 𝑐
(− cos 𝜋) − (− cos0) = 𝜋 sin 𝑐
(− cos 180°)− (−cos0) = 180°sin 𝑐
1 + 1 = 180sin 𝑐
2 = 180. sin 𝑐
𝑠𝑖𝑛 𝑐 =
2
180
= 0,01
𝑐 = 𝑎𝑟𝑐 sin 0,01
= 0,57
5. 𝑓( 𝑥) =
1
𝑥
, dan interval [1,3]
∫
𝑏
𝑎
𝑓( 𝑥) 𝑑𝑥 = 𝑓 (𝑐)(𝑏− 𝑎)
∫
3
1
(
1
𝑥
) 𝑑𝑥 =
1
𝑐
(3 − 1)
∫
3
1
(
1
𝑥
) 𝑑𝑥 =
1
𝑐
(2)
ln| 𝑥| ] 3
1
=
2
𝑐
ln|3| − ln |1| =
2
𝑐
1,1 − 0 =
2
𝑐
1,1 𝑐 = 2
𝑐 =
2
1,1
= 1,82

11. 6. 𝑓( 𝑥) = 𝑥2
, dan interval [-2,2]
1
𝑏−𝑎
∫
𝑏
𝑎
𝑓( 𝑥) 𝑑𝑥
=
1
2−(−2)
∫
2
−2
(𝑥2
) 𝑑𝑥
=
1
4
(
1
3
𝑥3
) | 2
−2
=
1
4
((
1
3
23
) − (
1
3
(−2)3
))
=
1
4
(
8
3
− (−
8
3
))
=
1
4
.
16
3
=
4
3
7. 𝑓( 𝑥) =
1
𝑥
, dan interval [1,3]
1
𝑏 − 𝑎
∫ 𝑓( 𝑥)
𝑏
𝑎
𝑑𝑥
=
1
3−1
∫
1
𝑥
3
1
𝑑𝑥 =
1
2
[ln 𝑥] | 3
1
=
1
2
[ln3 − ln 1] =
1
2
(1,0986) = 0,5493
8. 𝑓( 𝑥) = cos 𝑥 , dan interval [
−𝜋
2
,
𝜋
2
]
1
𝑏 − 𝑎
∫ 𝑓( 𝑥)
𝑏
𝑎
𝑑𝑥
=
1
𝜋
2
−(
−𝜋
2
)
∫ cos 𝑥
𝜋
2
−𝜋
2
𝑑𝑥 =
1
𝜋
[sin 𝑥] |
𝜋
2
−𝜋
2
=
1
𝜋
[sin(90°) − sin(−90°)] =
1
𝜋
(1 − (−1)) =
2
𝜋
9. 𝑓( 𝑥) =
9
2
√ 𝑥 , dan interval [1,4]
1
𝑏 − 𝑎
∫ 𝑓( 𝑥)
𝑏
𝑎
𝑑𝑥
=
1
4 − 1
∫
9
2
4
1
√ 𝑥𝑑𝑥 =
1
3
∫
9
2
𝑥
1
2
4
1
𝑑𝑥
=
1
3
[
9
2⁄
3
2⁄
𝑥
3
2]|
4
1
=
1
3
[3. 𝑥√ 𝑥]|
4
1
=
1
3
[3.4. √4 − 3.1. √1]
=
1
3
(24 − 3) =
1
3
(21) = 7

12. 10. 𝑓( 𝑥) = 𝑒 𝑥
, dan interval [0,1]
1
𝑏 − 𝑎
∫ 𝑓( 𝑥)
𝑏
𝑎
𝑑𝑥
=
1
1 − 0
∫ 𝑒 𝑥
1
0
𝑑𝑥 = 1( 𝑒 𝑥)|
1
0
= 𝑒1
− 𝑒0
= 2,718 − 1 = 1,718
Selesai