![Dua aturan integrasi berguna
Latihan 7.7
Cari integral tak tentu yang paling umum..
1. β«(3π₯4
β 5π₯3
β 21π₯2
+ 36π₯ β 10) ππ₯
2. β«[3π₯2
β 4πππ (2π₯)] ππ₯
3. β«(
8
π‘5
+
5
π‘
) ππ‘
4. β«(
1
β25 β π2
+
1
100 + π2
) ππ
5. β«
π5π₯
β π4π₯
π2π₯
ππ₯
6. β«(
π₯7
+ π₯4
π₯5
) ππ₯
7. β«(
π₯7
+ π₯4
π₯5
) ππ₯
8. β«( π₯2
+ 4)2
ππ₯ = β« π₯4
9. β«(
7
β π‘
3
) ππ‘
10. β«
20 + π₯
β π₯
ππ₯](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Tugas matematika buku kalkulus
- 1. Tugas Matematika Integral Hal 49- 59 Disusun Oleh : Nama : 1. Ricky Adi Pratama 2. Devi Yunita 3.Gustiana Kelas : 1EA 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
- 2. Dua aturan integrasi berguna Latihan 7.7 Cari integral tak tentu yang paling umum.. 1. β«(3π₯4 β 5π₯3 β 21π₯2 + 36π₯ β 10) ππ₯ 2. β«[3π₯2 β 4πππ (2π₯)] ππ₯ 3. β«( 8 π‘5 + 5 π‘ ) ππ‘ 4. β«( 1 β25 β π2 + 1 100 + π2 ) ππ 5. β« π5π₯ β π4π₯ π2π₯ ππ₯ 6. β«( π₯7 + π₯4 π₯5 ) ππ₯ 7. β«( π₯7 + π₯4 π₯5 ) ππ₯ 8. β«( π₯2 + 4)2 ππ₯ = β« π₯4 9. β«( 7 β π‘ 3 ) ππ‘ 10. β« 20 + π₯ β π₯ ππ₯
- 3. Penyelesaian : 1. β«(3π₯4 β 5π₯3 β 21π₯2 + 36π₯ β 10) ππ₯ = β« 3π₯4 ππ₯ β β« 5π₯3 ππ₯ β β« 21π₯2 ππ₯ + β« 36π₯ ππ₯ β β« 10 ππ₯ = 3β« π₯4 ππ₯ β 5β« π₯3 ππ₯ β 21β« π₯2 ππ₯ + 36β« π₯ ππ₯ β 10β« ππ₯ = 3( π₯5 5 ) β 5 ( π₯4 4 ) β 21( π₯3 3 ) + 36( π₯2 2 ) β 10π₯ + π = 3 5 π₯5 β 5 4 π₯4 β 7π₯3 + 18π₯2 β 10π₯ + π 2. β«[3π₯2 β 4πππ (2π₯)] ππ₯ = β« 3π₯2 ππ₯ β β« 4 πππ (2π₯) ππ₯ = 3 β« π₯2 ππ₯ β 4 β« πππ (2π₯) ππ₯ = 3 ( π₯3 3 ) β 4( 1 2 π ππ2π₯) + π = π₯3 β 2 sin 2π₯ + π 3. β« ( 8 π‘5 + 5 π‘ ) ππ‘ = β« 8 π‘5 ππ₯ + β« 5 π‘ ππ₯ = 8 β« π‘β5 ππ₯ + 5 β« 1 π‘ ππ₯ = 8 π‘β4 β4 + 5 ππ| π‘| + π = β2π‘β4 + 5 ππ| π‘| + π 4. β« ( 1 β25βπ2 + 1 100 +π2 ) ππ = β« 1 β25 βπ2 ππ₯ + β« 1 100+π2 ππ₯ = β« 1 β52 +π2 ππ₯ + β« 1 102 +π2 ππ₯ = π ππβ1 ( π 5 )+ 1 10 π‘ππβ1 π 10 + π 5. β« π5π₯ βπ4π₯ π2π₯ ππ₯ = β«( π3π₯ β π2π₯) ππ₯ = β« π3π₯ ππ₯ β β« π2π₯ ππ₯ = 1 3 π3π₯ β 1 2 π2π₯ + π 6. β« ( π₯7 +π₯4 π₯5 ) ππ₯ = β« π₯7 π₯5 ππ₯ + β« π₯4 π₯5 ππ₯ = 7. β« 1 ( π6 +π₯2) ππ₯ = β«( π6 + π₯2) ππ₯ = ππ| π6 + π₯2| + π 8. β«( π₯2 + 4)2 ππ₯ = β« π₯4 + 16 + 2. π₯2 .4 ππ₯ = β« π₯4 + 8π₯2 + 16 ππ₯ = 1 4+1 π₯4+1 + 8 2+1 π₯2+1 + 16π₯ + π = 1 5 π₯5 + 8 3 π₯3 + π 9. β« ( 7 β π‘3 ) ππ‘ = β« 7π‘β 1 3 ππ‘ = 7 β 1 3 +1 π‘ β 1 3 +1 + π = 7 2 3β π‘ 2 3 + π = 21 2 π‘ 2 3 + π 10. β« 20+π₯ β π₯ ππ₯ = β«(20+ π₯) π₯β 1 2 ππ₯ = β« (20π₯β 1 2 + π₯ 1 2) ππ₯ = 20 β 1 2 +1 π₯β 1 2 +1 + 1 1 2 +1 π₯ 1 2 +1 + π = 20 1 2 π₯ 1 2 + 1 3 2β π₯ 3 2 + π = 40π₯ 1 2 + 2 3 π₯ 3 2 + π
- 4. Integrasi dasar teknik Integrasi dengan substitusi Latihan 8.1 Gunakan integrasi dengan substitusi untuk menemukan integral tak tentu yang paling umum. 1. β«3( π₯3 β 5)4 π₯2 ππ₯ 2. β« π π₯4 π₯3 ππ₯ 3. β« π‘ π‘2 + 7 ππ‘ 4. β«( π₯5 β 3π₯) 1 4 (5π₯4 β 3) ππ₯ 5. β« π₯3 β 2π₯ ( π₯4 β 4π₯2 + 5)4 ππ₯ 6. β« π₯3 β 2π₯ π₯4 β 4π₯2 + 5 ππ₯ 7. β«cos(3π₯2 + 1 ) ππ₯ 8. 3πππ 2 β π₯(π ππβ π₯) β π₯ ππ₯ 9. β« π2π₯ 1 + π4π₯ ππ₯ 10. β«6π‘2 π π‘3 β2 ππ‘
- 5. PENYELESAIAN 1. β«3( π₯3 β 5)4 π₯2 ππ₯ u = x3 β 5 du = 3x2 dx = β« π’4 ππ’ = 1 5 π’5 + π = (π₯3 β 5)5 5 + π 2. β« π π₯4 π₯3 ππ₯ π’ = π₯4 = β« π π₯4 1 4 .4π₯3 ππ₯ = 1 4 β« π π₯3 4π₯3 ππ₯ = 1 4 β« π π’ ππ’ = 1 4 π π’ + π = 1 4 π π₯4 + π 3. β« π‘ π‘2 + 7 ππ‘ π’ = π‘2 + 7 ππ’ = 2π‘ ππ₯ β« π‘ π‘2 + 7 ππ‘
- 6. β« 1 2 2π‘ π‘2 + 7 ππ‘ 1 2 β« 2π‘ π‘2 + 7 ππ‘ 1 2 β« ππ’ π’ 1 2 πΌπ| π’| + π 1 2 πΌπ( π‘2 + 7) + π 4. β«( π₯5 β 3π₯) 1 4 (5π₯4 β 3) ππ₯ π’ = ( π₯5 β 3π₯) ππ’ = 5π₯4 β 3 ππ₯ = β« π’ 1 4 ππ’ = 4π’ 5 4 + π = 4( π₯5 β 3π₯) 5 4 + π 5. β« 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₯ = β« 1 4 . 4( π₯3 β 2π₯) π’4 ππ₯ = 1 4 β« ππ’ π’4 = 1 4 πΌπ| π’| + π = 1 4 πΌπ( π₯4 β 4π₯2 + 5) + π
- 7. 6. β« π₯3 β 2π₯ π₯4 β 4π₯2 + 5 ππ₯ π’ = π₯4 β 4π₯2 + 5 ππ’ = 4π₯3 β 8π₯ ππ₯ = 4( π₯3 β 2π₯) = β« 1 4 . 4(π₯3 β 2π₯) π₯4 β 4π₯2 + 5 ππ₯ = 1 4 β« ππ’ π’ = 1 4 πΌπ| π’| + π = 1 4 πΌπ( π₯4 β 4π₯2 + 5) + π 9. β« π2π₯ 1 + π4π₯ ππ₯ = β« π2π₯ 1 + π2π₯(2) ππ₯ π’ = 1 + π2π₯ ππ’ = 2. π2π₯ ππ₯ = β« 1 2 . 2. π2π₯ 1 + π2π₯(2) = 1 2 β« ππ’ π’ = 1 2 πΌπ| π’| ππ₯ = 1 2 πΌπ 1 + π4π₯ + π
- 8. 10. β«6π‘2 π π‘3β2 ππ‘ π’ = π‘3 β 2 ππ’ = 3π‘2 ππ‘ = β« 6π‘2 π π‘3β2 ππ‘ = β« 2(3π‘2) π π‘3β2 ππ‘ = β« 1 3 . 3(2).(3π‘2). π π‘3β2 ππ‘ = 1 3 β«6 ππ’. π π’ = 1 3 π π’.6 ππ’ = 1 3 π π‘3β2.6 + π = 2π π‘3β2 + π
- 9. Integrasi dengan bagian Latihan 8.2 Gunakan integrasi dengan bagian untuk menemukan integral tak tentu yang paling umum. 1. β« 2π₯.sin2x dx 2. β« π₯3 lnx dx 3. β« π‘π π‘ dt 4. β« π₯ cos x dx 5. β« πππ‘β1 ( π₯) ππ₯ 6. β« π₯2 π π₯ ππ₯ 7. β« π€( π€ β 3)2 ππ€ 8. β« π₯3 ππ (4π₯) ππ₯ 9. β« π‘ (π‘ + 5)β4 ππ‘ 10. β« π₯β π₯ + 2 . ππ₯
- 10. PENYELESAIAN 1. β« 2π₯ sin2π₯ ππ₯ Misalnya : u = 2x du = x dv = sin 2x dx v= β« sin 2π₯ππ₯ = - 1 2 cos2x β« π’. ππ£ = π’π£ ββ« π’. ππ’ β« 2π₯ sin2π₯ ππ₯ = (2x) (- 1 2 cos 2x ) - β«(β 1 2 cos 2x ) . 2x = - 2 2 cos 2x + 1 2 β« cos 2x dx = - x cos 2x + 1 2 . 1 2 sin 2x = - x cos 2x + 1 2 . sin 2x + c 2. β« π₯3 ππ π₯ ππ₯ Misalnya : U= inx du = 1 π₯ dx dv= π₯3 dx v = β« π₯3 ππ₯ = π₯4 4 β« π’. ππ£ = π’π£ ββ« π’. ππ’ β« π₯3 ππ π₯ ππ₯ = (in x) ( π₯4 4 ) - β« π₯4 4 . 1 π₯ dx = π₯4 πππ₯ 4 - 1 4 . π₯4 4 = π₯4 πππ₯ 4 - π₯4 16 + c 3. β« π‘π π‘ ππ‘ Misalnya : U = t du = dt dv = π π‘ dt v = β« π π‘ dt = π π‘ β« π’. ππ£ = π’. π£ ββ« π’. ππ’
- 11. β« π‘π π‘ ππ‘ = (t) (π π‘ ) - β« π π‘ dt = π‘π π‘ - β« π π‘ dt = π‘π π‘ - π π‘ + c 4. β« π₯ cos π₯ ππ₯ Misalnya : U= x du = dx dv = cos x dx v = β« cos π₯ ππ₯ = sin x β« π’. ππ£ = π’. π£ ββ« π’. ππ’ β« π₯ cos π₯ ππ₯ = ( x ) ( sin x ) - β« sin π₯ ππ₯ = sin x + cosx dx = sin x + cosx + c 5. β« πππ‘β1 ( x ) dx Misalnya : U = sinπ₯β1 Du= cosπ₯β1 Subtitusi du = sinπ₯β1 du = cosπ₯β1 β« πππ π₯β1 π πππ₯β1 dx = β« ππ’ π’ Salve integral = in (u) + c Subsitusi kembali U=sinπ₯β1 = in (sinπ₯β1 ) + π 6. β« π₯2 π π₯ ππ₯ Misalnya : U = π₯2 du = 2x dv = π π₯ dx v = β« π π₯ dx = π π₯ β« π’. ππ£ = u.v - β« π’.du β« π₯2 π π₯ ππ₯ = π₯2 π π₯ -β« π₯ 2 . 2π₯ =π₯π2π₯ -β« 2π₯. ππ₯ =π₯π2π₯ - x+c 7. β« π€(π€ β 3)2 ππ€ Misalnya : U= w du= dw
- 12. dv = (π€ β 3)2 ππ€ π£ = β«(2π€ β 6 ) = π€ β 3 β« π’. ππ£ = u.v - β« π’.du β« π€(π€ β 3)2 ππ€ = π€. ( π€ β 3) β β« π€. ππ€ = ( π€2 β 3π€) β 1 2 π€ + π 8. β« π₯3 ππ (4π₯ ) ππ₯ Misalnya : U= in4x du= 1 4π₯ ππ₯ dv= π₯3 ππ₯ v = β« π₯3 dx = 1 4 π₯4 β« π’. ππ£ = u.v - β« π£.du β« π₯3 ππ (4π₯ ) ππ₯ = in4x. 1 4 π₯4 -β« in4x . 1 4π₯ ππ₯ = 1 4 π₯4 ππ4π₯ β 1 5 π₯5 βΆ 1 2 16π₯2 + π = 1 4 π₯4 ππ4π₯ - 2π₯5 80π₯2 + c 9. β« π‘(π‘ + 5)β4 ππ‘ Misalnya : U= t du= dt dv =(π‘ + 5)β4 π£ = β« β4π‘β3 β 20β3 = 2π‘β2 + 10β2 β« π’. ππ£ = u.v - β« π£.du β« π‘(π‘ + 5)β4 ππ‘ =( t. 2π‘β2 + 10β2 ) - β« 2π‘β2 + 10β2 . ππ‘ = 20π‘β4 + (2π‘ + 10 + ππ‘ 10. β« π₯β π₯ + 2 .dx Misalnya : U = x du = dx Dv=β π₯ + 2 dx v= β«(π₯ + 2) 1 2 =2π₯ 1 1 2 +0.67 1 1 2 β« π’. ππ£ = u.v - β« π£.du β« π₯β π₯ + 2 .dx = x . 2π₯ 1 1 2 +0.67 1 1 2 - β« 2π₯ 1 1 2 + 0.67 1 1 2 . dx = x.2,67π₯ 3 2 - (2π₯ 3 2 + 0,67 3 2) dx = 2,67π₯ 2 3 2 - 2,67π₯ 6 2 + c
- 13. 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. β« cot π₯ ππ₯ 2. β« 1 ( π₯+2) (2π₯+5) ππ₯ 3. β« ( πππ₯)2 ππ₯ 4. β« π₯ cos π₯ ππ₯ 5. β« π₯ ( π₯+2)2 ππ₯ 6. β« 3π₯π π₯ ππ₯ 7. β« β10 π€ + 3 ππ€ 8. β« π‘(π‘ + 5)β1 ππ‘ 9. β« π₯ β π₯ + 2 ππ₯ 10. β« 1 sin π’ cos π’ ππ’
- 14. PENYELESAIAN 1. β« cot π₯ ππ₯ ( Formula nomor 7) Penyelesaian : β« πππ‘ π₯ ππ₯ = β« πππ π₯ π πππ₯ ππ₯ Misalkan : π’ = sin π₯ ππ’ = cos π₯ ππ₯ Subsitusi ππ’ = cos π₯, π = sin π₯ β« cos π₯ sin π₯ ππ₯ = β« ππ’ π’ π πππ£π πππ‘πππππ ln| π’| + πΆ subsitusi kembali π = sin π₯ ππ|sin π₯| + π 2. β« 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 : β« 1 ( π₯ + 2) (2π₯ + 5) ππ₯ = β« 1 ( π₯ + 2) (2π₯ + 5)
- 15. = β« 1 9 ( π₯ + 2) ππ₯ + β« 1 9 (2π₯ + 5) ππ₯ = 1 9 ππ| π₯ + 2| + 1 7 ln|2π₯ + 5| + c 3. β« ( πππ₯)2 ππ₯ = β«( πππ₯) ( πππ₯) ππ₯ Missal : U = ln x β ππ’ = ( 1 π₯ )2 Dv = dx dv =β« ππ₯ v = x β«(πππ₯)2 ππ₯ = π’π£ β β« π£ππ’ (x ln ) = (πππ₯)2 . x - β« π₯ 1 π₯2 ππ₯ = π₯. (πππ₯)2 - β« 1 π₯ ππ₯ = π₯. (πππ₯)2 x - β« π₯β1 ππ₯ = π₯. (πππ₯)2 - 1 0 π₯0 + π = π₯. (πππ₯)2 - ~ + π = ln x ( x ln x-x ) β β«(π₯ ln π₯ β π₯) . 1 π₯ =x (ln x)2 - x ln x - 4. β« π₯ cos π₯ ππ₯ Penyelesaian : π = π β ππ’ = ππ₯ ππ£ = πππ π₯ β π£ = π πππ₯ β« π’ππ£ = π’π£ β β« π£ππ’
- 16. β« π₯πππ π₯ππ₯ = π₯π πππ₯ β β« π πππ₯ ππ₯ β« π₯πππ π₯ππ₯ = π₯π πππ₯ + πππ π₯ + π 5. β« 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 : β« π₯ ( π₯ + 2)2 ππ₯ = β« ππ₯ ( π₯ + 2) β ππ₯ ( π₯ + 2)2 πππ πππ π’ = π₯ + 2 β ππ’ = ππ₯ β« ππ₯ ( π₯ + 2) β β« ππ₯ ( π₯ + 2)2 = β« ππ’ π’ β β« ππ’ π’2 = 2ππ + 2 π’ + π 2ππ( π₯ + 2) + 2 ( π₯+2) + π 6. β« 3π₯π π₯ ππ₯ U = 3x dv = π π₯ ππ₯ ππ’ ππ₯ = 3 v = β« π π₯ ππ₯ = π π₯ du = 3 dx β« π’ππ£ = u.v ββ« π£ ππ’
- 17. = (3x) . (π π₯ )β β« π π₯ . 3 ππ₯ = 3x π π₯ β 3π π₯ 7. β« β10 π€ + 3 dw ( Formula nomor 2) β« β10 π€+ 3 dw = β«(10 π€ + 3) 1 2β dw = 1 1 2 + 1 (10 π€ + 3) 1 2 +1 + π = 2 3 (10 π€ + 3) 3 2 + π 8. β« π‘(π‘ + 5)β1 ππ‘ =β« π‘ π‘+5 dt = β« 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 =β« π‘ π‘+5 dt = β« π‘ (π‘ + 5)β1 ππ‘ = β«( π’ β 5) π’β1 ππ’ = β« π’0 β 5π’β1 ππ’ (π’0 β 5π’) β¦β¦ β¦ β¦. = π’ β π’ β«β5π’β1 +1 du β«β5(π’1 β 1 5 π₯ ) ππ₯ -5 (ln | π’| - 1 5 0+1 π₯0+1 ) -5 ( ln | π‘ + 5| - 1 5 x)
- 18. -5 ln | π‘ + 5| + x 9. β« π₯ β π₯ + 2 ππ₯ πππ ππ π’ = π₯ + 2 β π₯ = π’ β 2 ππ’ = ππ₯ Sehingga integral diatas dapat menjadi : = πππ‘ ( π’ β 2)β π ππ’ = πππ‘ ( π’ β 2) π 1 2 ππ’ = πππ‘ (π 5 2) β π 1 2 ππ’ = 2 7 π 2 7 β 2 3 π 3 2 + πΆ = πππ‘ (π₯ + 2) 5 2 β 2 3 (π₯ + 2) 3 2 + πΆ