tugas matematika kelompok 7
TRANSCRIPT
Tugas Matematika
Integral Hal 49- 59
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 : [email protected]
http://www.polman-babel.ac.id
Kelompok 7 :
- Fery Ardiansyah
- Sarman
- Rakam Tiano
- Mirza ramadhan
Dua aturan integrasi berguna
Latihan 7.7
Cari integral tak tentu yang paling umum..
1. 3π₯4 β 5π₯3 β 21π₯2 + 36π₯ β 10 ππ₯
2. 3π₯2 β 4πππ 2π₯ ππ₯
3. 8
π‘5+
5
π‘ ππ‘
4. 1
25 β π2+
1
100 + π2 ππ
5. π5π₯ β π4π₯
π2π₯ππ₯
6. π₯7 + π₯4
π₯5 ππ₯
7. π₯7 + π₯4
π₯5 ππ₯
8. π₯2 + 4 2ππ₯ = π₯4
9. 7
π‘3 ππ‘
10. 20 + π₯
π₯ππ₯
Penyelesaian :
1. 3π₯4 β 5π₯3 β 21π₯2 + 36π₯ β 10 ππ₯ = 3π₯4 ππ₯ β 5π₯3 ππ₯ β 21π₯2 ππ₯ +
36π₯ ππ₯ β 10 ππ₯ = 3 π₯4 ππ₯ β 5 π₯3 ππ₯ β 21 π₯2 ππ₯ + 36 π₯ ππ₯ β
10 ππ₯ = 3 π₯5
5 β 5
π₯4
4 β 21
π₯3
3 + 36
π₯2
2 β 10π₯ + π =
3
5π₯5 β
5
4π₯4 β 7π₯3 +
18π₯2 β 10π₯ + π
2. 3π₯2 β 4πππ 2π₯ ππ₯ = 3π₯2 ππ₯ β 4 πππ 2π₯ ππ₯ = 3 π₯2 ππ₯ β 4 πππ 2π₯ ππ₯ =
3 π₯3
3 β 4
1
2π ππ2π₯ + π = π₯3 β 2 sin 2π₯ + π
3. 8
π‘5 +5
π‘ ππ‘ =
8
π‘5 ππ₯ + 5
π‘ππ₯ = 8 π‘β5 ππ₯ + 5
1
π‘ππ₯ = 8
π‘β4
β4+ 5 ππ π‘ + π =
β2π‘β4 + 5 ππ π‘ + π
4. 1
25βπ2+
1
100 +π2 ππ = 1
25βπ2ππ₯ +
1
100+π2 ππ₯ = 1
52+π2ππ₯ +
1
102 +π2 ππ₯ =
π ππβ1 π
5 +
1
10π‘ππβ1 π
10+ π
5. π5π₯βπ4π₯
π2π₯ ππ₯ = π3π₯ β π2π₯ ππ₯ = π3π₯ππ₯ β π2π₯ππ₯ =1
3π3π₯ β
1
2π2π₯ + π
6. π₯7+π₯4
π₯5 ππ₯ = π₯7
π₯5 ππ₯ + π₯4
π₯5 ππ₯ =
7. 1
π6+π₯2 ππ₯ = π6 + π₯2 ππ₯ = ππ π6 + π₯2 + π
8. π₯2 + 4 2ππ₯ = π₯4 + 16 + 2.π₯2. 4 ππ₯ = π₯4 + 8π₯2 + 16 ππ₯ =1
4+1π₯4+1 +
8
2+1π₯2+1 + 16π₯ + π =
1
5π₯5 +
8
3π₯3 + π
9. 7
π‘3 ππ‘ = 7π‘β
13 ππ‘ =
7
β1
3+1π‘β
1
3+1 + π =
72
3 π‘
23 + π =
21
2π‘
23 + π
10. 20+π₯
π₯ππ₯ = 20 + π₯ π₯β
12 ππ₯ = 20π₯β
12 + π₯
12 ππ₯ =
20
β12
+1π₯β
12
+1 +1
12
+1π₯
12
+1 + π =
2012
π₯12 +
13
2 π₯
32 + π = 40π₯
12 +
2
3π₯
32 + π
Integrasi dasar teknik Integrasi dengan substitusi
Latihan 8.1
Gunakan integrasi dengan substitusi untuk menemukan integral tak tentu yang paling umum.
1. 3 π₯3 β 5 4π₯2 ππ₯
2. ππ₯4π₯3 ππ₯
3. π‘
π‘2 + 7 ππ‘
4. π₯5 β 3π₯ 14 5π₯4 β 3 ππ₯
5. π₯3 β 2π₯
π₯4 β 4π₯2 + 5 4 ππ₯
6. π₯3 β 2π₯
π₯4 β 4π₯2 + 5ππ₯
7. cos 3π₯2 + 1 ππ₯
8. 3πππ 2 π₯(π ππ π₯)
π₯ ππ₯
9. π2π₯
1 + π4π₯ ππ₯
10. 6π‘2ππ‘3β2 ππ‘
PENYELESAIAN 1. 3 π₯3 β 5 4π₯2 ππ₯
u = x3 β 5 du = 3x2 dx
= π’4 ππ’
=1
5π’5 + π
=(π₯3 β 5)5
5+ π
2. ππ₯4π₯3 ππ₯
π’ = π₯4
= ππ₯4 1
4. 4π₯3 ππ₯
=1
4 ππ₯
34π₯3 ππ₯
=1
4 ππ’ ππ’
=1
4ππ’ + π
=1
4ππ₯
4+ π
3. π‘
π‘2 + 7 ππ‘
π’ = π‘2 + 7 ππ’ = 2π‘ ππ₯
π‘
π‘2 + 7ππ‘
1
2
2π‘
π‘2 + 7ππ‘
1
2
2π‘
π‘2 + 7ππ‘
1
2 ππ’
π’
1
2πΌπ π’ + π
1
2πΌπ π‘2 + 7 + π
4. π₯5 β 3π₯ 14 5π₯4 β 3 ππ₯
π’ = π₯5 β 3π₯ ππ’ = 5π₯4 β 3 ππ₯
= π’14 ππ’
= 4π’54 + π
= 4 π₯5 β 3π₯ 54 + π
5. π₯3 β 2π₯
π₯4 β 4π₯2 + 5 4 ππ₯
π’ = π₯4 β 4π₯2 + 5 ππ’ = 4π₯3 β 8π₯ ππ₯
= 1
4.4 π₯3 β 2π₯
π’4 ππ₯
= 1
4 ππ’
π’4
=1
4πΌπ π’ + π
=1
4πΌπ π₯4 β 4π₯2 + 5 + π
6. π₯3 β 2π₯
π₯4 β 4π₯2 + 5ππ₯
π’ = π₯4 β 4π₯2 + 5 ππ’ = 4π₯3 β 8π₯ ππ₯
= 4 π₯3 β 2π₯
= 1
4.
4(π₯3 β 2π₯)
π₯4 β 4π₯2 + 5 ππ₯
=1
4 ππ’
π’
=1
4πΌπ π’ + π
=1
4πΌπ π₯4 β 4π₯2 + 5 + π
9. π2π₯
1 + π4π₯ ππ₯
= π2π₯
1 + π2π₯(2) ππ₯
π’ = 1 + π2π₯ ππ’ = 2. π2π₯ ππ₯
= 1
2.
2. π2π₯
1 + π2π₯(2)
=1
2 ππ’
π’
=1
2πΌπ π’ ππ₯
=1
2πΌπ 1 + π4π₯ + π
10. 6π‘2ππ‘3β2 ππ‘
π’ = π‘3 β 2 ππ’ = 3π‘2 ππ‘
= 6π‘2ππ‘3β2 ππ‘
= 2 3π‘2 ππ‘3β2 ππ‘
= 1
3. 3 2 . 3π‘2 . ππ‘
3β2 ππ‘
=1
3 6 ππ’. ππ’
=1
3ππ’ . 6 ππ’
=1
3ππ‘
3β2. 6 + π
= 2ππ‘3β2 + π
Integrasi dengan bagian
Latihan 8.2
Gunakan integrasi dengan bagian untuk menemukan integral tak tentu yang paling umum.
1. 2π₯.sin2x dx
2. π₯3lnx dx
3. π‘ππ‘dt
4. π₯ cos x dx
5. πππ‘β1 π₯ ππ₯
6. π₯2 ππ₯ππ₯
7. π€( π€ β 3)2 ππ€
8. π₯3 ππ 4π₯ ππ₯
9. π‘ (π‘ + 5)β4 ππ‘
10. π₯ π₯ + 2 .ππ₯
PENYELESAIAN
1. 2π₯ sin 2π₯ ππ₯
Misalnya :
u = 2x du = x
dv = sin 2x dx v= sin 2π₯ππ₯ = - 1
2 cos2x
π’.ππ£ = π’π£ β π’. ππ’
2π₯ sin 2π₯ ππ₯ = (2x) (- 1
2 cos 2x ) - (β
1
2 cos 2x ) . 2x
= - 2
2 cos 2x +
1
2 cos 2x dx
= - x cos 2x + 1
2 .
1
2 sin 2x
= - x cos 2x + 1
2 . sin 2x + c
2. π₯3 ππ π₯ ππ₯
Misalnya :
U= inx du = 1
π₯ dx
dv= π₯3dx v = π₯3 ππ₯ = π₯4
4
π’.ππ£ = π’π£ β π’. ππ’
π₯3 ππ π₯ ππ₯ = (in x) (π₯4
4) -
π₯4
4 .
1
π₯ dx
= π₯4πππ₯
4 -
1
4 . π₯4
4
= π₯4πππ₯
4 -
π₯4
16 + c
3. π‘ππ‘ ππ‘
Misalnya :
U = t du = dt
dv = ππ‘ dt v = ππ‘dt = ππ‘
π’.ππ£ = π’. π£ β π’. ππ’
π‘ππ‘ ππ‘ = (t) (ππ‘) - ππ‘dt
= π‘ππ‘ - ππ‘dt
= π‘ππ‘ - ππ‘ + c
4. π₯ cos π₯ ππ₯
Misalnya :
U= x du = dx
dv = cos x dx v = cos π₯ ππ₯ = sin x
π’.ππ£ = π’. π£ β π’. ππ’
π₯ cos π₯ ππ₯ = ( x ) ( sin x ) - sin π₯ ππ₯
= sin x + cosx dx
= sin x + cosx + c
5. πππ‘β1( x ) dx
Misalnya :
U = sinπ₯β1
Du= cosπ₯β1
Subtitusi du = sinπ₯β1 du = cosπ₯β1
πππ π₯ β1
π πππ₯ β1 dx = ππ’
π’
Salve integral
= in (u) + c
Subsitusi kembali
U=sinπ₯β1
= in (sinπ₯β1) + π
6. π₯2 ππ₯ππ₯
Misalnya :
U = π₯2 du = 2x
dv = ππ₯dx v = ππ₯dx = ππ₯
π’.ππ£ = u.v - π’.du
π₯2 ππ₯ππ₯ = π₯2ππ₯- π₯2
. 2π₯
=π₯π2π₯ - 2π₯. ππ₯
=π₯π2π₯ - x+c
7. π€(π€ β 3)2ππ€
Misalnya :
U= w du= dw
dv = (π€ β 3)2ππ€ π£ = 2π€ β 6 = π€ β 3
π’.ππ£ = u.v - π’.du
π€(π€ β 3)2ππ€ = π€. π€ β 3 β π€. ππ€
= π€2 β 3π€ β1
2π€ + π
8. π₯3 ππ 4π₯ ππ₯
Misalnya :
U= in4x du= 1
4π₯ππ₯
dv= π₯3ππ₯ v = π₯3dx = 1
4π₯4
π’.ππ£ = u.v - π£.du
π₯3 ππ 4π₯ ππ₯ = in4x. 1
4π₯4- in4x .
1
4π₯ππ₯
= 1
4π₯4ππ4π₯ β
1
5π₯5 βΆ
1
2 16π₯2 + π
= 1
4π₯4ππ4π₯ -
2π₯5
80π₯2 + c
9. π‘(π‘ + 5)β4 ππ‘
Misalnya :
U= t du= dt
dv =(π‘ + 5)β4 π£ = β4π‘β3 β 20β3 = 2π‘β2 + 10β2
π’.ππ£ = u.v - π£.du
π‘(π‘ + 5)β4 ππ‘ =( t. 2π‘β2 + 10β2 ) - 2π‘β2 + 10β2 .ππ‘
= 20π‘β4 + (2π‘ + 10 + ππ‘
10. π₯ π₯ + 2 .dx
Misalnya :
U = x du = dx
Dv= π₯ + 2 dx v= (π₯ + 2)1
2 =2π₯11
2 +0.6711
2
π’.ππ£ = u.v - π£.du
π₯ π₯ + 2 .dx = x . 2π₯11
2 +0.6711
2 - 2π₯11
2 + 0.6711
2 . dx
= x.2,67π₯3
2 - (2π₯3
2 + 0,673
2) dx
= 2,67π₯23
2 - 2,67π₯6
2 + c
Integrasi dengan menggunakan tabel rumus
terpisahkan
Latihan 8.3
Gunakan tabel rumus integral dalam Lampiran C untuk menemukan integral tak tentu yang paling umum.
1. cot π₯ ππ₯
2. 1
π₯+2 (2π₯+5)ππ₯
3. πππ₯ 2ππ₯
4. π₯ cosπ₯ ππ₯
5. π₯
π₯+2 2 ππ₯
6. 3π₯ππ₯ ππ₯
7. 10 π€ + 3 ππ€
8. π‘(π‘ + 5)β1 ππ‘
9. π₯ π₯ + 2 ππ₯
10. 1
sinπ’ cosπ’ ππ’
PENYELESAIAN
1. cot π₯ ππ₯ ( Formula nomor 7) Penyelesaian :
πππ‘ π₯ ππ₯ = πππ π₯
π πππ₯ ππ₯
Misalkan : π’ = sin π₯ ππ’ = cosπ₯ ππ₯ Subsitusi ππ’ = cos π₯,π = sin π₯
cosπ₯
sin π₯ ππ₯ =
ππ’
π’
π πππ£π πππ‘πππππ ln π’ + πΆ subsitusi kembali π = sin π₯ ππ sin π₯ + π
2. 1
π₯+2 (2π₯+5)ππ₯
=1
π₯ + 2 (2π₯ + 5)=
π΄
π₯ + 2+
π΄
2π₯ + 5
π΄ =1
π₯ + 2 (2.2 + 5)=
1
9
π΅ =1
5 + 2 (2π₯ + 5)=
1
7
Sehingga :
1
π₯ + 2 2π₯ + 5 ππ₯ =
1
π₯ + 2 2π₯ + 5
=
19
π₯ + 2 ππ₯ +
19
2π₯ + 5 ππ₯
=1
9ππ π₯ + 2 +
1
7ln 2π₯ + 5 + c
3. πππ₯ 2ππ₯ = πππ₯ πππ₯ ππ₯ Missal :
U = ln x ππ’ = (1
π₯ )2
Dv = dx dv = ππ₯ v = x
(πππ₯)2 ππ₯ = π’π£ β π£ππ’ (x ln )
= (πππ₯)2. x - π₯ 1
π₯2 ππ₯
= π₯. (πππ₯)2 - 1
π₯ ππ₯
= π₯. (πππ₯)2 x - π₯β1 ππ₯
= π₯. (πππ₯)2 - 1
0π₯0 + π
= π₯. (πππ₯)2 - ~ + π
= ln x ( x ln x-x ) β (π₯ ln π₯ β π₯) . 1
π₯
=x (ln x)2 - x ln x -
4. π₯ cosπ₯ ππ₯
Penyelesaian : π = π β ππ’ = ππ₯
ππ£ = πππ π₯ β π£ = π πππ₯
π’ππ£ = π’π£ β π£ππ’
π₯πππ π₯ππ₯ = π₯π πππ₯ β π πππ₯ ππ₯
π₯πππ π₯ππ₯ = π₯π πππ₯ + πππ π₯ + π
5. π₯
π₯+2 2 ππ₯
Penyelesaian :
π₯
π₯+2 2 =π΄
π₯+2 +π΅
π₯+2 =π΄ π₯+2 +π΅
π₯+2 2
π΄ = 2
π΄ + π΅ = 0 = β2
Sehingga :
π₯
π₯ + 2 2 ππ₯ =
ππ₯
π₯ + 2 β
ππ₯
π₯ + 2 2
πππ πππ π’ = π₯ + 2 β ππ’ = ππ₯
ππ₯
π₯ + 2 β
ππ₯
π₯ + 2 2 = ππ’
π’ β
ππ’
π’2= 2ππ +
2
π’+ π
2ππ π₯ + 2 +2
π₯+2 + π
6. 3π₯ππ₯ ππ₯
U = 3x dv = ππ₯ ππ₯ ππ’
ππ₯= 3 v = ππ₯ ππ₯ = ππ₯
du = 3 dx
π’ππ£ = u.v β π£ ππ’
= (3x) . (ππ₯) β ππ₯ . 3 ππ₯ = 3x ππ₯ β 3ππ₯
7. 10 π€ + 3 dw ( Formula nomor 2)
10 π€ + 3 dw = (10 π€ + 3)1
2 dw
=1
12 + 1
(10 π€ + 3)12
+1 + π
=2
3 (10 π€ + 3)
3
2 + π
8. π‘(π‘ + 5)β1 ππ‘
= π‘
π‘+5 dt = π‘ (π‘ + 5)β1 ππ‘
Missal:
U = t + 5 U= t+5
ππ’
ππ‘ = 1 t = (u-5)
ππ’ = ππ‘ t=uβu=t+5 =5
t = 2 β u=t+5 = 7
= π‘
π‘+5 dt = π‘ (π‘ + 5)β1 ππ‘ = π’ β 5 π’β1 ππ’ = π’0 β 5π’β1 ππ’
(π’0 β 5π’)β¦β¦β¦β¦ . = π’ β π’
β5π’β1 +1 du
β5(π’1 β1
5 π₯ ) ππ₯
-5 (ln π’ - 15
0+1
π₯0+1)
-5 ( ln π‘ + 5 - 1
5 x)
-5 ln π‘ + 5 + x
9. π₯ π₯ + 2 ππ₯ πππ ππ π’ = π₯ + 2 β π₯ = π’ β 2 ππ’ = ππ₯
Sehingga integral diatas dapat menjadi :
= πππ‘ π’ β 2 π ππ’
= πππ‘ π’ β 2 π1
2 ππ’
= πππ‘ π5
2 β π1
2 ππ’
=2
7 π
27 β
2
3π
32 + πΆ
= πππ‘ (π₯ + 2)52 β
2
3 (π₯ + 2)
3 2 + πΆ