# KISI-KISI

15 Jul 2015
1 sur 6

### KISI-KISI

• 1. 1 . (𝑥10 − 6 𝑥5 + 𝑥73 ) 𝑑𝑥 = (𝑥10 − 6𝑥−5 + 𝑥 7 3) 𝑑𝑥 = 1 11 𝑥11 + 6 4 𝑥−4 + 3 10 𝑥 10 3 +c = 1 11 𝑥11 + 3 2 𝑥−4 + 3 10 𝑥 10 3 +c 2. cos 9𝑥 − 11 + 𝑠𝑒𝑐2 6𝑥 − 8 𝑑𝑥 = 1 9 sin 9𝑥 − 11 + 1 6 tan 6𝑥 − 8 + 𝑐 3. Dengan menggunakan cara subsitusi 𝑥 6+𝑥2 𝑑𝑥 = 𝑥(6 + 𝑥2 ) 1 2 𝑑𝑥 Misalkan : 𝑢 = 6 + 𝑥2 𝑑𝑢 𝑑𝑥 = 2𝑥 𝑑𝑢 = 1 2𝑥 𝑑𝑢 𝑥(6 + 𝑥2 ) 1 2 𝑑𝑥 = 𝑥 𝑈 −1 2 . 1 2𝑥 𝑑𝑢 = 𝑥 2𝑥 . 𝑈 −1 2 𝑑𝑢 = 1 2 . 𝑈 −1 2 𝑑𝑢 = 1 2 −1 2 +1 𝑈 −1 2 +1 + 𝐶 = 1 2 1 2 𝑈 1 2 + 𝐶
• 2. =(6𝑥 + 𝑥2 ) 1 2 + 𝐶 4. Dengan menggunkan cara subsitusi 2𝑥 + 5 cos⁡(2𝑥2 + 10𝑥 + 8 ) 𝑑𝑥 Misalkan U =2𝑥2 + 10𝑥 + 8 𝑑𝑢 𝑑𝑥 = 4𝑥 + 10 Dx= 1 4𝑥+10 𝑑𝑢 2𝑥 + 5 cos⁡(2𝑥2 + 10𝑥 + 8 ) 𝑑𝑥 = 2𝑥 + 5 cos 𝑢 1 4𝑥+10 du 2𝑥 + 5 2 2𝑥 + 5 cos 𝑢 𝑑𝑢 1 2 cos u du = 1 2 sin u du = 1 2 sin (2𝑥 2 + 10x +8 ) + c 5.Integral parsial 2𝑥. sin⁡(10𝑥 + 3) dx Misalkan : u= 2x du =2dx dv =sin (10x +3 ) v= sin 10𝑥 + 3 𝑑𝑥 = − 1 10 cos 10𝑥 + 3 = 𝑈𝑑𝑣 = 𝑢𝑣 − 𝑣 𝑑𝑢 = 2𝑥. sin 10𝑥 + 3 𝑑𝑥 =2𝑥 − 1 10 cos 10𝑥 + 3 − − 1 10 cos 10𝑥 + 3 . 2 𝑑𝑥
• 3. =− 1 5 𝑥. cos 10𝑥 + 3 𝑑𝑥 + 2 100 sin 10𝑥 + 3 + 𝐶 =− 1 5 𝑥. cos 10𝑥 + 3 + 1 50 sin 10𝑥 + 3 + 𝐶 6. Dengan menggunakan table 𝑥2 𝑒−7𝑥 𝑑𝑥 Turunan U Integral dv +𝑥2 -2x +2 -0 𝑒−7𝑥 − 1 7 𝑒−7𝑥 1 49 𝑒−7𝑥 − 1 363 𝑒−7𝑥 𝑢𝑑𝑣 = 𝑥2 ( − 1 7 𝑒−7𝑥 ) -2x . 1 49 𝑒−7𝑥 + 2 − 1 369 𝑒−7𝑥 + 𝑐 = −𝑥2 1 7 𝑒−7𝑥 -2x . 1 49 − 2 363 𝑒−7𝑥 𝑒−7𝑥 + 2 + 𝑐 = − 1 7 𝑥2 𝑒−7𝑥 -2x . 1 49 − 2 363 𝑒−7𝑥 𝑒−7𝑥 + 2 + 𝑐 7.Integral fungsi rasional 𝑥 𝑥2 − 2𝑥 − 35 𝑑𝑥 𝑥 𝑥2−2𝑥−35 = 𝑥 𝑥−7 (𝑥+5) = 𝐴 (𝑥−7) + 𝐵 (𝑥+5) = 𝐴 𝑥 + 5 + 𝐵(𝑥 − 7) 𝑥 − 7 𝑥 + 5 𝐴𝑥 + 5𝐴 + 𝐵𝑥 − 7𝐵) 𝑥 − 7 𝑥 + 5 A+B = 1 x5 5A+5B = 5
• 4. 5A +B =0 x1 5A-7B = 0 12B=5 B= 5 12 A= 7 12 Sehingga : 𝑥 𝑥 − 7 (𝑥 + 5) 𝑑𝑥 = 𝐴 𝑥 − 7 𝑑𝑥 + 𝐵 𝑥 + 5 𝑑𝑥 = 7 12 𝑥−7 𝑑𝑥 + 5 12 𝑥+5 𝑑𝑥 = 7 12 𝑙𝑛 x-7 + 5 12 𝑙𝑛 x+5 + C 8. ( 𝑥45 1 + 3𝑥 + 1 𝑥3 ) 𝑑𝑥 = ( 𝑥45 1 + 3𝑥 + 𝑥−3 ) 𝑑𝑥 = 1 5 [𝑥5 + 3 2 𝑥2 − 1 2 𝑥−2 ]5 1 = ( 1 5 55 + 3 2 52 − 1 2 𝑥5−2 ) –( 1 5 15 + 3 2 12 − 1 2 1−2 ) =(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 = 𝑥2 − 1 Y = 3x + 9 Dit = Luas daerah Jawab : 𝑥2 − 1 = 3𝑥 + 9 𝑥2 − 1 − 3𝑥 − 9 = 0 𝑥2 − 3𝑥 − 10 = 0
• 5. (x-5) (x+2) = 0 X= 5 v x=-2 L= 3𝑥 + 9 – (𝑥25 −2 − 1) 𝑑𝑥 = 3𝑥 − 5 −2 𝑥2 + 10 𝑑𝑥 = 3 2 [𝑥2 − 1 3 𝑥3 + 10𝑥] 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 = 225−250+300 6 − 18+8−60 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 sumbu y Jawab = V = 𝜋 𝑥2 − 𝑥22 𝑑𝑦 𝑑 𝑒 = 𝜋 (𝑦2 − ( 2 0 1 3 𝑦)2 ) dy = 𝜋 𝑦2 − 2 0 1 9 𝑦2 dy = 𝜋 − 2 0 8 9 𝑦2 dy
• 6. =𝜋[( 8 9 2+1 𝑦2+1 )] 2 0 = 𝜋( 8 27 𝑦3 ) 2 0 = 𝜋( 8 27 23 -)-( 8 27 . 03 ) = 𝜋 64 27 =2 10 27 𝜋