![3
INTEGRAL TENTU
Definisi : β« π( π₯) ππ₯ = [πΉ( π₯)] π
ππ
π
β = πΉ( π) β πΉ(π)
Contoh soal :
Tentukan integral tentu berikut ini :
1. β« ( π₯ + 2) ππ₯ = [1
2
π₯2
+ 2π₯]1
44
1
= (
1
2
(42
))+ 2(4)β (
1
2
(12
))+ 2(1))
= (8 + 8) β (
1
2
+ 2)
= 16 β 2
1
2
= 13
1
2
π ππ‘π’ππ ππ’ππ
Penyelesaian dengan cara manual yaitu :
π( π₯) = π₯ + 2
π¦ = π₯ + 2
x 1 2 3 4
y 3 4 5 6](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-3-320.jpg)
![4
πΏ =
( π + ππ‘ππ ) π‘
2
=
(6 + 3)3
2
=
27
2
= 13
1
2
π ππ‘π’ππ ππ’ππ
2. β« π₯2
ππ₯ = [
1
3
π₯3
]1
33
1
= (
1
3
(32
))β (
1
3
(12
))
= (
1
3
.27 β (
1
3
.1)
= 9 β
1
3
=
27β1
3
=
26
3
= 8
2
3
π ππ‘π’ππ ππ’ππ
Penyelesaian dengan cara manual yaitu :
π( π₯) = π₯2
π¦ = π₯2
x 1 2 3 0
y 1 4 9 0
1
4
9
1 2
1
3
Tidak ada bentuknya](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-4-320.jpg)
![6
-2
-1
2
7
-3
-2
-1
0
1
2
3
4
5
6
7
8
0 1 2 3 4
Contoh soal :
1. ππππ‘π’πππ ππ’ππ πππππβ π¦πππ πππππ‘ππ π πππβ ππ’ππ£π π( π₯) = π₯2
β 2 ππππ π₯ =
0 πππ π₯ = 3
Penyelesaian dengan cara cepat :
πΏ = β« (π₯2
β π₯)ππ₯ = [
1
3
π₯3
β 2π₯]0
3
3
0
= (
1
3
.33
β 2 .3) β 0
= (
27
3
β 6) = 9 β 6 = 3 π ππ‘π’ππ ππ’ππ
Penyelesaian dengan cara manual :
π¦ = π₯2
β π₯
π¦ = π₯2
β 2
0 = π₯2
β 2
2 = π₯2
π₯ = β2
πΏ1 = ββ« ( π₯2
β 2) ππ₯
β2
0
= β [
1
3
π₯3
β 2π₯]
β2
0
= β (
1
3
(β2)3
β 2(β2))β 0
= β (
2β2
3
β 2β2)
= β (
2β2 β 6β2
3
)
= β (β
4β2
3
) =
4β2
3
π ππ‘π’ππ ππ’ππ
x 0 1 2 3
y 0 -1 2 7](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-6-320.jpg)
![7
πΏ2 = β β« ( π₯2
β 2)
3
β2
ππ₯ = [
1
3
π₯3
β 2π₯]
3
β2
= (
1
3
(33
)β 2.3) β (
1
3
(β2
3
) β 2(β2))
= (9 β 6) β (
2β2
3
β 2β2)
= 3 β
4β2
3
π ππ‘π’ππ ππ’ππ
πΏ = πΏ1 + πΏ2
=
4β2
3
+ 3 β
4β2
3
= 3 π ππ‘π’ππ ππ’ππ
2. ππππ‘π’πππ ππ’ππ πππππβ π¦πππ πππππ‘ππ π πππβ β« (2π₯3
β 3) ππππ π₯ = 1 πππ π₯ = 3
Penyelesaian dengan cara cepat :
πΏ = β« (2π₯3
β 3)ππ₯ =
3
1
[
2
3
π₯3
β 2π₯]
3
β2
= (
2
3
(33
) β 3(3)) β (
2
3
(13
) β 3(1))
= (
2
3
(27) β 9) β (
2
3
(1)β 3)
= (18 β 9) β (
2
3
β 3)
= (18 β 9) β (
2β9
3
)
= 9 β (β
7
3
)
= 9
7
3
π ππ‘π’ππ ππ’ππ
Penyelesaian dengan cara manual :
π¦ = 2π₯2
β 3
x 0 1 2 3
y -3 -1 5 15](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-7-320.jpg)
![8
π¦ = 2π₯2
β 3
0 = 2π₯2
β 3
3 = 2π₯2
π₯ = β
3
2
πΏ1 = β« (2π₯2
β 3) ππ₯
3
β
3
2
= [
2
3
π₯3
β 3π₯]
3
β
3
2
= (
2
3
(33
) β 3(3)) β (
2
3
(β3
2
3
) β 3 (β
3
2
))
= (
2
3
(27)β 9) β ((
2
3
.
3
2
β
3
2
)β 3β
3
2
)
= (18β 9) β (1β
3
2
β 3β
3
2
)
= 9 β (β2β
3
2
)
= 9 + 2β
3
2
π ππ‘π’ππ ππ’ππ
πΏ2 = β« (2π₯2
β 3) ππ₯
β
3
2
1
= [
2
3
π₯3
β 3π₯]
β
3
2
1
= (
2
3
(β3
2
3
) β 3 (β
3
2
)) β (
2
3
(13
) β 3(1))
= (
2
3
.
3
2
β
3
2
β 3β
3
2
) β (
2
3
(1)β 3)
15
5
-1
-3
1 2 3](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-8-320.jpg)
![10
πΏ = β« (π₯ β π₯2
)ππ₯ =
1
0
[
1
2
π₯2
β
1
3
π₯3
]
1
0
= (
1
2
(12)β (
1
3
(13
)) β 0
=
1
2
β
1
3
=
3β2
6
=
1
6
π ππ‘π’ππ ππ’ππ](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-10-320.jpg)
![12
Contoh soal :
Hitunglah volume benda putar suatu daerah yang dibatasi oleh kurva y = x β 1 dengan sumbu
x pada x = 3
Penyelesaian:
π¦ = 2π₯2
β 3
π = π β« (π₯ β 1)2
ππ₯ = π β« (π₯2
β 2π₯ + 1)ππ₯
3
1
3
1
= π β« [
1
3
π₯3
β π₯2
+ π₯]
3
1
3
1
= π(1
3
(33
)β32
+3) β (1
3
(13
)β12
+1)
= π(9β9+3) β (
1
3
β1+1)
= π (3 β
1
3
)
= 2
2
3
π π ππ‘π’ππ π£πππ’ππ
x 0 1 2 3 4
y -1 0 1 2 3](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-12-320.jpg)
![13
Soal-soal latihan :
1. Kurva π¦ = π₯2
, π¦ = 0 dengan π¦ = 3 putar dengan sumbu π¦
Jawaban:
π = π β« (β π¦)2
ππ¦ = π β« π¦ ππ¦
3
0
3
0
= π [
1
2
π¦2
]
0
3
= π [(
1
2
(32
)) β 0]
= π [
1
2
(9)]
= π (
9
2
)
=
9
2
π = 4
1
2
π π ππ‘π’ππ π£πππ’ππ
2. π¦ = 2π₯ + 2 dengan sumbu π₯ pada π₯ = 0 dan π₯ = 2
Jawaban:
y 0 1 2 3 4
x 0 1 β2 β3 2
y 0 1 2 3
x 2 4 6 8](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-13-320.jpg)
![14
π = π β« (2π₯ + 2)2
ππ₯ = π β« (4π₯2
+ 8π₯ + 4)ππ₯
2
0
2
0
= π [
4
3
π₯3
+ 4π₯2
+ 4π₯]
0
2
= π [(
4
3
(23
)+4(22
)4(2)) β 0]
= π [(
4
3
(8)+4(4)+8) β 0]
= π (
32
3
+16+8)
= π (
32+48+24
3
)
= π(104
3
) = 104
3
π = 32
3
π π ππ‘π’ππ π£πππ’ππ
3. π¦ = 3π₯ β 2, π¦ = 0 dan π¦ = 2
Jawaban:
π¦ = 3π₯ β 2
π¦ + 2 = 3π₯
π₯ =
π¦+2
3
y 0 1 2 3
x
2
3
1
4
3
5
3](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-14-320.jpg)
![15
π = π β« (
π¦+2
3
)
2
2
0
ππ¦ = π β« (
π¦2
+4π¦+4
3
)
2
0
ππ¦
= π β« (
1
9
π¦2+
4
9
π¦+
4
9
)
2
0
ππ¦
= π [
1
9
3
π¦3
+
4
9
3
π¦2
+
4
9
π¦]
0
2
= π [
1
9
.
1
3
π¦3
+
4
9
.
1
2
π¦2
+
4
9
π¦]
0
2
= π [
1
27
π¦3
+
4
18
π¦2
+
4
9
π¦]
0
2
= π [(
1
27
(23
)+
4
18
(22
) +
4
9
(2))β 0]
= π [
1
27
(8)+
4
18
(4) +
4
9
(2)]
= π [
8
27
+
16
18
+
8
9
]
= π [
16+48+48
54
]
= π [
112
54
]
=
112
54
π
= 2
4
54
π π ππ‘π’ππ π£πππ’ππ](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-15-320.jpg)
![16
4. π¦ = π₯2
+ 1, π¦ = 1 dan π¦ = 2
Jawaban:
π¦ = π₯2
+ 1
π¦ β 1 = π₯2
π₯ = βπ¦ β 1
π = π β« (β π¦β1)
2
2
1
ππ¦ = π β« ( π¦β1)
2
1
ππ¦
= π [
1
2
π¦2
β π¦]
1
2
= π [(
1
2
(22
) β 2) β (
1
2
(12
)β 1)]
= π [(
1
2
(4) β 2) β (
1
2
(1)β 1)]
= π [(2β 2) β (β
1
2
)]
= π [0 β (β
1
2
)]
=
1
2
π π ππ‘π’ππ π£πππ’ππ
y 1 2 3
x 0 1 β2](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-16-320.jpg)
![18
π₯( π₯ β 2) = 0
π₯ = 0 β π₯ β 2 = 0
π₯ = 0 β¨ π₯ = 2
π = π β« ((3π₯ β π₯2
)2
β π₯2) ππ₯
2
0
= π β« (9π₯2
β 6π₯4
+ π₯3
+ π₯2) ππ₯
2
0
= π β« (8π₯2
β 6π₯3
+ π₯4)
2
0
= π [
8
3
π₯3
β
6
4
π₯4
+
1
5
π₯5
]
2
0
= π [
8
3
(2)3
β
6
4
(2)4
+
1
5
(2)5
] β 0
= π [
8
3
(8)β
6
4
(16) +
1
5
(32)]
= π [
64
3
β
96
4
+
32
5
]
= π [
1280 β 1440 + 384
60
]
= π [
224
60
]
= π
56
15
= 3
11
15
π π ππ‘π’ππ π£πππ’ππ](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-18-320.jpg)
![22
Pertemuan ke-6
Senin, 28 Maret 2016
INTEGRAL LIPAT 2
Integral lipat dua ini, bisa dirumuskan sebagai berikut:
1. β¬ π( π₯, π¦) ππ΄ = β« π΄( π¦)
π2
π2
ππ¦
π
π
2. β¬ π(π₯, π¦)
π
π
ππ΄ = β« [β« π( π₯, π¦) ππ₯
π2
π2
] ππ¦
π2
π1
3. β¬ π( π₯, π¦) ππ΄ = β« [β« π( π₯, π¦) ππ¦
π2
π2
] ππ₯
π2
π1
π
π
Contoh soal :
1. Hitung β¬ (2π₯ + 3π¦) ππ΄,
π
π
R daerah π₯ππ¦ yang hanya memuat titik-titik ( π₯, π¦) dengan
batas 1 β€ π₯ β€ 2 dan 0 β€ π¦ β€ 3
Jawaban:
Diketahui bahwa π1 = 1, π1 = 2 dan π2 = 0, π2 = 3
β¬(2π₯ + 3π¦) ππ΄ = β« β«(2π₯ + 3π¦) ππ₯ ππ¦
2
1
3
0
π
π
= β« [β«(2π₯ + 3π¦
2
1
)ππ₯]
3
0
ππ¦
= β«[ π₯2
+ 3π₯π¦]
2
1
3
0
ππ¦](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-22-320.jpg)
![23
= β«[((2)2
+ 3(2) π¦)β ((1)2
+ 3(1) π¦)] ππ¦
3
0
= β«(4 + 6π¦ β 1 β 3π¦) ππ¦
3
0
= β«(3π¦ + 3) ππ¦
3
0
= [
3
2
π¦2
+ 3π¦]
3
0
= [
3
2
(3)2
+ 3(3)] β 0
=
3
2
(9)+ 9 β 0
=
27
2
+
18
2
=
45
2
= 22
1
2
β¬(2π₯ + 3π¦) ππ΄ = β« β«(2π₯ + 3π¦) ππ¦ ππ₯
3
0
2
1
π
π
= β« [β«(2π₯ + 3π¦) ππ¦
3
0
] ππ₯
2
1
= β« [2π₯π¦ +
3
2
π¦2
]
3
0
ππ₯
2
1
= β« [2π₯(3)+
3
2
(3)2
] β 0 ππ₯
2
1
= β« 6π₯ +
27
2
ππ₯
2
1
= [3π₯2
+
27
2
π₯]
2
1
= [3(2)2
+
27
2
(2)] β [3(1)2
+
27
2
(1)]
= [3(4) +
54
2
] β [3 +
27
2
]](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-23-320.jpg)
![24
= [12 + 27] β [
6
2
+
27
2
]
= 39 β [
33
2
]
=
78
2
β
33
2
=
45
2
= 22
1
2
Latihan soal:
1. Hitungβ« β« ( π₯2
+ π¦) ππ₯ ππ¦
4
0
8
0
Jawaban:
β¬( π₯2
+ π¦) ππ΄ = β«β«( π₯2
+ π¦) ππ₯ ππ¦
4
0
8
0
π
π
= β« [β« ( π₯2
+ π¦) ππ₯
4
0
] ππ¦
8
0
= β«[
1
3
π₯3
+ π₯π¦]
4
0
ππ¦
8
0
= β«[
1
3
(4)3
+ (4) π¦] β [
1
3
(0)3
β (0) π¦] ππ¦
8
0
= β«[
1
3
(64) + 4π¦] β 0 ππ¦
8
0
= β«[
64
3
+ 4π¦] ππ¦
8
0
= [
64
3
π¦ + 2π¦2
]
8
0
= [
64
3
(8) + 2(8)2
β (0)]
=
512
3
+ 128
=
512 + 384
3
=
896
3
= 298
2
3](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-24-320.jpg)
![25
2. β« β« π₯
π₯
0
π¦3
ππ¦ ππ₯
2
1
Jawaban:
β¬( π₯ + π¦3) ππ΄ = β« β«( π₯ + π¦3) ππ¦ ππ₯
π₯
0
2
1
π
π
= β«[β«( π₯ + π¦3) ππ¦
π₯
0
] ππ₯
2
1
= β«[ π₯π¦ + π¦3]
π₯
0
ππ₯
2
1
= β«[ π₯( π₯) + π₯3] β 0 ππ₯
2
1
= β«( π₯2
+ π₯3) ππ₯
2
1
= [22
+ 23] β [12
+ 13]
= (4 + 8) β (1 + 1)
= 12 β 2
= 10
3. β« β« ππ₯ ππ¦
π¦2
0
1
0
Jawaban:
β« β« ππ₯ ππ¦
π¦2
0
1
0
= β«[β« (1) ππ₯
π¦2
0
] ππ¦
1
0
= β«[ π₯]
π¦2
0
1
0
ππ¦
= β« π¦2
ππ¦
1
0
= [
1
3
π¦3
]
1
0
= [
1
3
(1)3
β
1
3
(0)3
]
=
1
3](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-25-320.jpg)
![26
4. β¬ ( π₯π¦ + π¦2) ππ΄, π
= ( π₯, π¦):0 β€ π₯ β€ 2 ,1 β€ π¦ β€ 3
π
π
Jawaban:
Diketahui bahwa π1 = 0, π1 = 2 dan π2 = 1, π2 = 3
β¬( π₯π¦ + π¦2) ππ΄ = β«β«( π₯π¦ + π¦2) ππ₯ ππ¦
2
0
3
1
π
π
= β« [β«( π₯π¦ + π¦2) ππ₯
2
0
] ππ¦
3
1
= β«[ π₯2
π¦ + π¦2]
2
0
ππ¦
3
1
= β«[22
π¦ + 2π¦2] β [02
π¦ + 0π¦2] ππ¦
3
1
= β«[4π¦ + 2π¦2] β [ π¦ + π¦2] ππ¦
3
1
= β«(3π¦2
β 3π¦) ππ¦
3
1
= [
3
3
π¦3
β
3
2
π¦2
]
3
1
= [π¦3
β
3
2
π¦2
]
3
1
= [33
β
3
2
(32
)] β [13
β
3
2
(12
)]
= [27 β
3
2
(9)]β [1 β
3
2
(1)]
= [27 β
27
2
] β [1 β
3
2
]
= [
54β27
2
] β [
2β3
2
]
= [
27
2
] β [
β1
2
]
=
27
2
+
1
2
=
28
2
= 14](https://image.slidesharecdn.com/rangkumankalkulusindah-170403051828/85/Rangkuman-Materi-Kalkulus-26-320.jpg)
Rangkuman Materi Kalkulus
- 1. 1 Pertemuan ke-1 Senin, 22 Februari 2016 INTEGRAL TAK TENTU Definisi : β« π( π₯) ππ₯ = π( π₯)+ πΆ Kebalikan diferensial atau turunan : π( π₯) = π₯ π β πβ²( π₯) = π . π₯ πβ1 Kaidah-kaidah : a. β« ππ₯ = π₯ + πΆ b. β« πππ₯ = ππ₯ + πΆ c. β« π₯ π ππ₯ = 1 π+1 π₯ π+1 + πΆ β π β β1 d. β« ππ₯ π ππ₯ = π π+1 π₯ π+1 + πΆ β π β β1 Contoh-contoh : 1. β« 3π₯2 ππ₯ = π₯3 + πΆ 2. β« 3ππ₯ = 3π₯ + πΆ 3. β« π₯4 ππ₯ = 1 4+1 π₯4+1 + πΆ = 1 5 π₯5 + πΆ 4. β« 4π₯5 ππ₯ = 4 5+1 π₯5+1 + πΆ = 4 6 π₯6 + πΆ = 2 3 π₯5 + πΆ 5. β« (6π₯2 + 7π₯ β 2) ππ₯ = 6 3 π₯3 + 7 2 π₯2 β 2π₯ + πΆ = 2π₯3 + 3 1 2 π₯2 β 2π₯ + πΆ 6. β« (2π₯ + 4)2 ππ₯ = β« (4π₯2 + 16π₯ + 16)2 ππ₯ = 4 3 π₯3 + 16 2 π₯2 + 16π₯ + πΆ = 1 1 3 π₯3 + 8π₯2 + 16π₯ + πΆ 7. β« ( π₯ β 5)( π₯ + 6) ππ₯ = β« ( π₯2 + π₯ β 30) ππ₯ = 1 3 π₯3 + 1 2 π₯2 β 30π₯ + πΆ
- 2. 2 8. β« 2 3 π₯2 β 4π₯ 1 2 β 1 3 π₯ ππ₯ = β« 2 3 π₯2 β β« 4π₯ 1 2 β β« 1 3 π₯ ππ₯ = β« 2 3 2 + 1 π₯2+1 β 4 1 2 +1 π₯ 1 2 +1 β 1 3 1 + 1 π₯1+1 + πΆ = β« 2 3 3 π₯3 β 4 3 2 π₯ 2 3 β 1 3 2 π₯2 + πΆ = 2 3 . 1 3 π₯3 β 4 1 . 2 3 π₯ 2 3 β 1 3 . 1 2 π₯2 + πΆ = 2 9 π₯3 β 8 3 π₯ 2 3 β 1 6 π₯2 + πΆ 9. β« (β3π₯ + 4)2 ππ₯ = β« (9π₯2 β 24π₯ + 16)ππ₯ = 9 3 π₯3 β 24 2 π₯2 +16π₯ + πΆ = 3π₯3 β 12π₯2 + 16π₯ + πΆ 10. β« (2π₯ + 1)( π₯ β 2) ππ₯ = β« (2π₯2 β 4π₯ + π₯ β 2)ππ₯ = β« (2π₯2 β 3π₯ β 2)ππ₯ = 2 3 π₯3 β 3 2 π₯2 β 2π₯ + πΆ = 2 3 π₯3 β 1 1 2 π₯2 β 2π₯ + πΆ Sifat-sifat : a. β« πππ₯ = π β« ππ₯ β β« ππ(π₯) = π β« π(π₯)ππ₯ b. β« (π( π₯) Β± π( π₯))ππ₯ = β« π( π₯) ππ₯ Β± β« π( π₯) ππ₯ Contoh soal : 1. β« 2π₯2 ππ₯ = 2β« π₯2 ππ₯ = 2. 1 3 π₯3 + πΆ = 2 3 π₯2 + πΆ 2. β« (2π₯2 β 3π₯) ππ₯ = β« 2π₯2 ππ₯ β β« 3π₯ππ₯ = 2β« π₯2 ππ₯ β 3β« π₯ππ₯ = 2. 1 3 π₯3 + πΆ1 β 3. 1 2 π₯2 + πΆ2 = 2 3 π₯3 + πΆ1 β 3 2 π₯2 + πΆ2 = 2 3 π₯3 β 1 1 2 π₯2 + πΆ1 + πΆ2 = 2 3 π₯3 β 1 1 2 π₯2 + πΆ
- 3. 3 INTEGRAL TENTU Definisi : β« π( π₯) ππ₯ = [πΉ( π₯)] π ππ π β = πΉ( π) β πΉ(π) Contoh soal : Tentukan integral tentu berikut ini : 1. β« ( π₯ + 2) ππ₯ = [1 2 π₯2 + 2π₯]1 44 1 = ( 1 2 (42 ))+ 2(4)β ( 1 2 (12 ))+ 2(1)) = (8 + 8) β ( 1 2 + 2) = 16 β 2 1 2 = 13 1 2 π ππ‘π’ππ ππ’ππ Penyelesaian dengan cara manual yaitu : π( π₯) = π₯ + 2 π¦ = π₯ + 2 x 1 2 3 4 y 3 4 5 6
- 4. 4 πΏ = ( π + ππ‘ππ ) π‘ 2 = (6 + 3)3 2 = 27 2 = 13 1 2 π ππ‘π’ππ ππ’ππ 2. β« π₯2 ππ₯ = [ 1 3 π₯3 ]1 33 1 = ( 1 3 (32 ))β ( 1 3 (12 )) = ( 1 3 .27 β ( 1 3 .1) = 9 β 1 3 = 27β1 3 = 26 3 = 8 2 3 π ππ‘π’ππ ππ’ππ Penyelesaian dengan cara manual yaitu : π( π₯) = π₯2 π¦ = π₯2 x 1 2 3 0 y 1 4 9 0 1 4 9 1 2 1 3 Tidak ada bentuknya
- 5. 5 Pertemuan ke-2 Senin, 29 Februari 2016 LUAS DAERAH Kuadran I Kuadran II Kuadran III Kuadran IV
- 6. 6 -2 -1 2 7 -3 -2 -1 0 1 2 3 4 5 6 7 8 0 1 2 3 4 Contoh soal : 1. ππππ‘π’πππ ππ’ππ πππππβ π¦πππ πππππ‘ππ π πππβ ππ’ππ£π π( π₯) = π₯2 β 2 ππππ π₯ = 0 πππ π₯ = 3 Penyelesaian dengan cara cepat : πΏ = β« (π₯2 β π₯)ππ₯ = [ 1 3 π₯3 β 2π₯]0 3 3 0 = ( 1 3 .33 β 2 .3) β 0 = ( 27 3 β 6) = 9 β 6 = 3 π ππ‘π’ππ ππ’ππ Penyelesaian dengan cara manual : π¦ = π₯2 β π₯ π¦ = π₯2 β 2 0 = π₯2 β 2 2 = π₯2 π₯ = β2 πΏ1 = ββ« ( π₯2 β 2) ππ₯ β2 0 = β [ 1 3 π₯3 β 2π₯] β2 0 = β ( 1 3 (β2)3 β 2(β2))β 0 = β ( 2β2 3 β 2β2) = β ( 2β2 β 6β2 3 ) = β (β 4β2 3 ) = 4β2 3 π ππ‘π’ππ ππ’ππ x 0 1 2 3 y 0 -1 2 7
- 7. 7 πΏ2 = β β« ( π₯2 β 2) 3 β2 ππ₯ = [ 1 3 π₯3 β 2π₯] 3 β2 = ( 1 3 (33 )β 2.3) β ( 1 3 (β2 3 ) β 2(β2)) = (9 β 6) β ( 2β2 3 β 2β2) = 3 β 4β2 3 π ππ‘π’ππ ππ’ππ πΏ = πΏ1 + πΏ2 = 4β2 3 + 3 β 4β2 3 = 3 π ππ‘π’ππ ππ’ππ 2. ππππ‘π’πππ ππ’ππ πππππβ π¦πππ πππππ‘ππ π πππβ β« (2π₯3 β 3) ππππ π₯ = 1 πππ π₯ = 3 Penyelesaian dengan cara cepat : πΏ = β« (2π₯3 β 3)ππ₯ = 3 1 [ 2 3 π₯3 β 2π₯] 3 β2 = ( 2 3 (33 ) β 3(3)) β ( 2 3 (13 ) β 3(1)) = ( 2 3 (27) β 9) β ( 2 3 (1)β 3) = (18 β 9) β ( 2 3 β 3) = (18 β 9) β ( 2β9 3 ) = 9 β (β 7 3 ) = 9 7 3 π ππ‘π’ππ ππ’ππ Penyelesaian dengan cara manual : π¦ = 2π₯2 β 3 x 0 1 2 3 y -3 -1 5 15
- 8. 8 π¦ = 2π₯2 β 3 0 = 2π₯2 β 3 3 = 2π₯2 π₯ = β 3 2 πΏ1 = β« (2π₯2 β 3) ππ₯ 3 β 3 2 = [ 2 3 π₯3 β 3π₯] 3 β 3 2 = ( 2 3 (33 ) β 3(3)) β ( 2 3 (β3 2 3 ) β 3 (β 3 2 )) = ( 2 3 (27)β 9) β (( 2 3 . 3 2 β 3 2 )β 3β 3 2 ) = (18β 9) β (1β 3 2 β 3β 3 2 ) = 9 β (β2β 3 2 ) = 9 + 2β 3 2 π ππ‘π’ππ ππ’ππ πΏ2 = β« (2π₯2 β 3) ππ₯ β 3 2 1 = [ 2 3 π₯3 β 3π₯] β 3 2 1 = ( 2 3 (β3 2 3 ) β 3 (β 3 2 )) β ( 2 3 (13 ) β 3(1)) = ( 2 3 . 3 2 β 3 2 β 3β 3 2 ) β ( 2 3 (1)β 3) 15 5 -1 -3 1 2 3
- 9. 9 = (1β 3 2 β 3β 3 2 ) β ( 2β9 3 ) = β2β 3 2 β (β 7 3 ) = β2β 3 2 + 7 3 π ππ‘π’ππ ππ’ππ πΏ = πΏ1 + πΏ2 = 9+2β 3 2 + (β2β 3 2 ) + 7 3 = 9 7 3 π ππ‘π’ππ ππ’ππ Contoh soal : πΆπππ ππ’ππ πππππβ π¦πππ πππππ‘ππ π ππ’ππ£π π¦ = π₯2 πππ π¦ = π₯ π¦ = π₯2 π₯ = π₯2 π¦ = π₯ π₯ β π₯2 = 0 π₯( π₯ β 1) = 0 π₯ = 0 , π₯ = 1 π¦ = π₯2 π¦ = π₯ x 0 1 2 1 y 0 1 2 1 x 0 1 2 1 y 0 1 2 1
- 10. 10 πΏ = β« (π₯ β π₯2 )ππ₯ = 1 0 [ 1 2 π₯2 β 1 3 π₯3 ] 1 0 = ( 1 2 (12)β ( 1 3 (13 )) β 0 = 1 2 β 1 3 = 3β2 6 = 1 6 π ππ‘π’ππ ππ’ππ
- 11. 11 Pertemuan ke-3 Senin, 7 Maret 2016 VOLUME BENDA PUTAR π·πππ’π‘ππ ππππππ π π’πππ’ π₯, ππππ: π = π β« π¦2 ππ₯ π 0 ππ‘ππ’ π = π β« π2 (π₯)ππ₯ π 0 π·πππ’π‘ππ ππππππ π π’πππ’ π¦, ππππ: π = π β« π₯2 ππ₯ π π ππ‘ππ’ π = π β« π2 (π¦)ππ₯ π π
- 12. 12 Contoh soal : Hitunglah volume benda putar suatu daerah yang dibatasi oleh kurva y = x β 1 dengan sumbu x pada x = 3 Penyelesaian: π¦ = 2π₯2 β 3 π = π β« (π₯ β 1)2 ππ₯ = π β« (π₯2 β 2π₯ + 1)ππ₯ 3 1 3 1 = π β« [ 1 3 π₯3 β π₯2 + π₯] 3 1 3 1 = π(1 3 (33 )β32 +3) β (1 3 (13 )β12 +1) = π(9β9+3) β ( 1 3 β1+1) = π (3 β 1 3 ) = 2 2 3 π π ππ‘π’ππ π£πππ’ππ x 0 1 2 3 4 y -1 0 1 2 3
- 13. 13 Soal-soal latihan : 1. Kurva π¦ = π₯2 , π¦ = 0 dengan π¦ = 3 putar dengan sumbu π¦ Jawaban: π = π β« (β π¦)2 ππ¦ = π β« π¦ ππ¦ 3 0 3 0 = π [ 1 2 π¦2 ] 0 3 = π [( 1 2 (32 )) β 0] = π [ 1 2 (9)] = π ( 9 2 ) = 9 2 π = 4 1 2 π π ππ‘π’ππ π£πππ’ππ 2. π¦ = 2π₯ + 2 dengan sumbu π₯ pada π₯ = 0 dan π₯ = 2 Jawaban: y 0 1 2 3 4 x 0 1 β2 β3 2 y 0 1 2 3 x 2 4 6 8
- 14. 14 π = π β« (2π₯ + 2)2 ππ₯ = π β« (4π₯2 + 8π₯ + 4)ππ₯ 2 0 2 0 = π [ 4 3 π₯3 + 4π₯2 + 4π₯] 0 2 = π [( 4 3 (23 )+4(22 )4(2)) β 0] = π [( 4 3 (8)+4(4)+8) β 0] = π ( 32 3 +16+8) = π ( 32+48+24 3 ) = π(104 3 ) = 104 3 π = 32 3 π π ππ‘π’ππ π£πππ’ππ 3. π¦ = 3π₯ β 2, π¦ = 0 dan π¦ = 2 Jawaban: π¦ = 3π₯ β 2 π¦ + 2 = 3π₯ π₯ = π¦+2 3 y 0 1 2 3 x 2 3 1 4 3 5 3
- 15. 15 π = π β« ( π¦+2 3 ) 2 2 0 ππ¦ = π β« ( π¦2 +4π¦+4 3 ) 2 0 ππ¦ = π β« ( 1 9 π¦2+ 4 9 π¦+ 4 9 ) 2 0 ππ¦ = π [ 1 9 3 π¦3 + 4 9 3 π¦2 + 4 9 π¦] 0 2 = π [ 1 9 . 1 3 π¦3 + 4 9 . 1 2 π¦2 + 4 9 π¦] 0 2 = π [ 1 27 π¦3 + 4 18 π¦2 + 4 9 π¦] 0 2 = π [( 1 27 (23 )+ 4 18 (22 ) + 4 9 (2))β 0] = π [ 1 27 (8)+ 4 18 (4) + 4 9 (2)] = π [ 8 27 + 16 18 + 8 9 ] = π [ 16+48+48 54 ] = π [ 112 54 ] = 112 54 π = 2 4 54 π π ππ‘π’ππ π£πππ’ππ
- 16. 16 4. π¦ = π₯2 + 1, π¦ = 1 dan π¦ = 2 Jawaban: π¦ = π₯2 + 1 π¦ β 1 = π₯2 π₯ = βπ¦ β 1 π = π β« (β π¦β1) 2 2 1 ππ¦ = π β« ( π¦β1) 2 1 ππ¦ = π [ 1 2 π¦2 β π¦] 1 2 = π [( 1 2 (22 ) β 2) β ( 1 2 (12 )β 1)] = π [( 1 2 (4) β 2) β ( 1 2 (1)β 1)] = π [(2β 2) β (β 1 2 )] = π [0 β (β 1 2 )] = 1 2 π π ππ‘π’ππ π£πππ’ππ y 1 2 3 x 0 1 β2
- 17. 17 Pertemuan ke-4 Senin, 14 Maret 2016 VOLUME BENDA PUTAR DI ANTARA DUA KURVA π·πππ’π‘ππ ππππππ π π’πππ’ π₯, ππππ: π = π β« π2( π₯) β π2( π₯) ππ₯ ππ‘ππ’ π = π β«( π¦1 2 β π¦2 2) ππ₯ π·πππ’π‘ππ ππππππ π π’πππ’ π¦, ππππ: π = π β« π2( π¦) β π2( π¦) ππ¦ ππ‘ππ’ π = π β«( π₯1 2 β π₯2 2) ππ¦ Contoh Soal: 1. Hitung luas daerah yang dibatasi oleh kurva π¦ = π₯ , dan π¦ = 3π₯ β π₯2 jika bidang diputar pada sumbu x Jawaban: π¦ = π₯ π¦ = 3π₯ β π₯2 π₯ = 3π₯ β π₯2 π₯2 + π₯ β 3π₯ = 0 π₯2 β 2π₯ = 0 π¦ = π₯ x 0 1 2 y 0 1 2 π¦ = 3π₯ β π₯2 x 0 1 2 y 0 2 2
- 18. 18 π₯( π₯ β 2) = 0 π₯ = 0 β π₯ β 2 = 0 π₯ = 0 β¨ π₯ = 2 π = π β« ((3π₯ β π₯2 )2 β π₯2) ππ₯ 2 0 = π β« (9π₯2 β 6π₯4 + π₯3 + π₯2) ππ₯ 2 0 = π β« (8π₯2 β 6π₯3 + π₯4) 2 0 = π [ 8 3 π₯3 β 6 4 π₯4 + 1 5 π₯5 ] 2 0 = π [ 8 3 (2)3 β 6 4 (2)4 + 1 5 (2)5 ] β 0 = π [ 8 3 (8)β 6 4 (16) + 1 5 (32)] = π [ 64 3 β 96 4 + 32 5 ] = π [ 1280 β 1440 + 384 60 ] = π [ 224 60 ] = π 56 15 = 3 11 15 π π ππ‘π’ππ π£πππ’ππ
- 19. 19 Pertemuan ke-5 Senin, 21 Maret 2016 INTEGRAL SUBSTITUSI Substitusi = menggantikan 1. Misal: Tentukan β« ( π₯3 + 6π₯)5(6π₯2 + 2) ππ₯ Jawaban: Misal π’ = π₯3 + 6π₯ ππ’ ππ₯ = 3π₯2 + 6 ππ’ = 3π₯2 + 6 ππ₯ 2ππ’ = 2(3π₯2 + 6) ππ₯ 2ππ’ = 6π₯2 + 12 ππ₯ β΄ β« ( π₯3 + 6π₯)5(6π₯2 + 12) ππ₯ β« π’5 . 2ππ’ = 2β« π’5 ππ’ = 2. 1 6 π’6 + πΆ = 2 1 6 ( π₯3 + 6π₯)6 + πΆ = 1 3 ( π₯3 + 6π₯)6 + πΆ 2. Hitung β« ( π₯2 + 4)10 π₯ ππ₯ Jawaban: Misal π’ = π₯2 + 4 ππ’ ππ₯ = 2π₯ ππ’ = 2π₯ ππ₯ π₯ ππ₯ = ππ’ 2 π₯ ππ₯ = 1 2 ππ’ NOTASI SIGMA (β)sigma = jumlah 1. 12 + 22 + 33 + β¦β¦ + 1002 = β π2100 π=1 2. π1 + π2 + π3 + β¦β¦ π π = β πππ π=1 β΄ β« ( π₯2 + 4)10 π₯ ππ₯ = β« π’10 . 1 2 ππ’ = 1 2 β« π’10 ππ’ = 1 2 . 1 11 π’11 + πΆ = 1 22 ( π₯2 + 4)11 + πΆ
- 20. 20 Sifat-sifat 1. β πΆππ = πΆ β πππ π=1 π π=1 2. β ( ππ Β± ππ) = β πππ π=1 Β± β πππ π=1 π π=1 3. β πΆπ = π1 + π2 + π3 + β¦β¦ + π = π. πΆπ π=1 β πΆ = π. πΆ π π=1 Soal-soal latihan. Tentukan : 1. β πΎ πΎ2+1 = β―4 πΎ=1 Jawaban: ( 1 12 + 1 ) + ( 2 22 + 1 ) + ( 3 32 + 1 ) + ( 4 42 + 1 ) = 1 2 + 2 5 + 3 10 + 4 17 = 85 + 68 + 51 + 40 170 = 244 170 = 1 74 170 2. β (β4)100 π=1 = β― Jawaban: 100(-4)=-400 3. Jika β ππ = 60100 π=1 dan β ππ = 11100 π=1 . Hitunglah β (2ππ β 3ππ + 4)100 π=1 Jawaban: β 2 ππ β β3 ππ + β 4 100 π=1 100 π=1 100 π=1 = 2β ππ β 3β ππ + β 4 100 π=1 100 π=1 100 π=1 = 2(60)β 3(11)+ 100(4) = 120 β 33 + 400 = 487 JUMLAH KHUSUS 1. β π = 1 + 2 + 3 + β― + π = π( π+1) 2 π π=1 2. β π2 = 12 + 22 + 32 + β― + π2 = π( π+1)(2π+1) 6 π π=1
- 21. 21 3. β π3 = 13 + 22 + 33 + β― + π3 = ( π( π+1) 2 ) 2 π π=1 4. β π4 = 14π π=1 + 24 + 34 + β― + π4 = π( π+1)(6π3 +9π2 +πβ1) 30 Hitunglah β 2π( π β 5)10 π=1 jawaban : β 2π( π β 5)10 π=1 = β 2π2 β 10π10 π=1 = β 2π2 β β 10π10 π=1 10 π=1 = β π210 π=1 β β 10π10 π=1 = 2 (10(10+1)(2(10)+1) 6 = 2 (110 .21) 6 = 2 2310 6 = 2(385)
- 22. 22 Pertemuan ke-6 Senin, 28 Maret 2016 INTEGRAL LIPAT 2 Integral lipat dua ini, bisa dirumuskan sebagai berikut: 1. β¬ π( π₯, π¦) ππ΄ = β« π΄( π¦) π2 π2 ππ¦ π π 2. β¬ π(π₯, π¦) π π ππ΄ = β« [β« π( π₯, π¦) ππ₯ π2 π2 ] ππ¦ π2 π1 3. β¬ π( π₯, π¦) ππ΄ = β« [β« π( π₯, π¦) ππ¦ π2 π2 ] ππ₯ π2 π1 π π Contoh soal : 1. Hitung β¬ (2π₯ + 3π¦) ππ΄, π π R daerah π₯ππ¦ yang hanya memuat titik-titik ( π₯, π¦) dengan batas 1 β€ π₯ β€ 2 dan 0 β€ π¦ β€ 3 Jawaban: Diketahui bahwa π1 = 1, π1 = 2 dan π2 = 0, π2 = 3 β¬(2π₯ + 3π¦) ππ΄ = β« β«(2π₯ + 3π¦) ππ₯ ππ¦ 2 1 3 0 π π = β« [β«(2π₯ + 3π¦ 2 1 )ππ₯] 3 0 ππ¦ = β«[ π₯2 + 3π₯π¦] 2 1 3 0 ππ¦
- 23. 23 = β«[((2)2 + 3(2) π¦)β ((1)2 + 3(1) π¦)] ππ¦ 3 0 = β«(4 + 6π¦ β 1 β 3π¦) ππ¦ 3 0 = β«(3π¦ + 3) ππ¦ 3 0 = [ 3 2 π¦2 + 3π¦] 3 0 = [ 3 2 (3)2 + 3(3)] β 0 = 3 2 (9)+ 9 β 0 = 27 2 + 18 2 = 45 2 = 22 1 2 β¬(2π₯ + 3π¦) ππ΄ = β« β«(2π₯ + 3π¦) ππ¦ ππ₯ 3 0 2 1 π π = β« [β«(2π₯ + 3π¦) ππ¦ 3 0 ] ππ₯ 2 1 = β« [2π₯π¦ + 3 2 π¦2 ] 3 0 ππ₯ 2 1 = β« [2π₯(3)+ 3 2 (3)2 ] β 0 ππ₯ 2 1 = β« 6π₯ + 27 2 ππ₯ 2 1 = [3π₯2 + 27 2 π₯] 2 1 = [3(2)2 + 27 2 (2)] β [3(1)2 + 27 2 (1)] = [3(4) + 54 2 ] β [3 + 27 2 ]
- 24. 24 = [12 + 27] β [ 6 2 + 27 2 ] = 39 β [ 33 2 ] = 78 2 β 33 2 = 45 2 = 22 1 2 Latihan soal: 1. Hitungβ« β« ( π₯2 + π¦) ππ₯ ππ¦ 4 0 8 0 Jawaban: β¬( π₯2 + π¦) ππ΄ = β«β«( π₯2 + π¦) ππ₯ ππ¦ 4 0 8 0 π π = β« [β« ( π₯2 + π¦) ππ₯ 4 0 ] ππ¦ 8 0 = β«[ 1 3 π₯3 + π₯π¦] 4 0 ππ¦ 8 0 = β«[ 1 3 (4)3 + (4) π¦] β [ 1 3 (0)3 β (0) π¦] ππ¦ 8 0 = β«[ 1 3 (64) + 4π¦] β 0 ππ¦ 8 0 = β«[ 64 3 + 4π¦] ππ¦ 8 0 = [ 64 3 π¦ + 2π¦2 ] 8 0 = [ 64 3 (8) + 2(8)2 β (0)] = 512 3 + 128 = 512 + 384 3 = 896 3 = 298 2 3
- 25. 25 2. β« β« π₯ π₯ 0 π¦3 ππ¦ ππ₯ 2 1 Jawaban: β¬( π₯ + π¦3) ππ΄ = β« β«( π₯ + π¦3) ππ¦ ππ₯ π₯ 0 2 1 π π = β«[β«( π₯ + π¦3) ππ¦ π₯ 0 ] ππ₯ 2 1 = β«[ π₯π¦ + π¦3] π₯ 0 ππ₯ 2 1 = β«[ π₯( π₯) + π₯3] β 0 ππ₯ 2 1 = β«( π₯2 + π₯3) ππ₯ 2 1 = [22 + 23] β [12 + 13] = (4 + 8) β (1 + 1) = 12 β 2 = 10 3. β« β« ππ₯ ππ¦ π¦2 0 1 0 Jawaban: β« β« ππ₯ ππ¦ π¦2 0 1 0 = β«[β« (1) ππ₯ π¦2 0 ] ππ¦ 1 0 = β«[ π₯] π¦2 0 1 0 ππ¦ = β« π¦2 ππ¦ 1 0 = [ 1 3 π¦3 ] 1 0 = [ 1 3 (1)3 β 1 3 (0)3 ] = 1 3
- 26. 26 4. β¬ ( π₯π¦ + π¦2) ππ΄, π = ( π₯, π¦):0 β€ π₯ β€ 2 ,1 β€ π¦ β€ 3 π π Jawaban: Diketahui bahwa π1 = 0, π1 = 2 dan π2 = 1, π2 = 3 β¬( π₯π¦ + π¦2) ππ΄ = β«β«( π₯π¦ + π¦2) ππ₯ ππ¦ 2 0 3 1 π π = β« [β«( π₯π¦ + π¦2) ππ₯ 2 0 ] ππ¦ 3 1 = β«[ π₯2 π¦ + π¦2] 2 0 ππ¦ 3 1 = β«[22 π¦ + 2π¦2] β [02 π¦ + 0π¦2] ππ¦ 3 1 = β«[4π¦ + 2π¦2] β [ π¦ + π¦2] ππ¦ 3 1 = β«(3π¦2 β 3π¦) ππ¦ 3 1 = [ 3 3 π¦3 β 3 2 π¦2 ] 3 1 = [π¦3 β 3 2 π¦2 ] 3 1 = [33 β 3 2 (32 )] β [13 β 3 2 (12 )] = [27 β 3 2 (9)]β [1 β 3 2 (1)] = [27 β 27 2 ] β [1 β 3 2 ] = [ 54β27 2 ] β [ 2β3 2 ] = [ 27 2 ] β [ β1 2 ] = 27 2 + 1 2 = 28 2 = 14