A. Evaluasi Pengertian atau Ingatan 1. C. x6+c
6x5dx = 6 6 x6+c = x6+c 2. B. n a xn+c , dengan n≠0
c x n a dx axn1 n11 1 1 = n a xn+c , dengan n≠0 3. B. c x n a n 1 ) 1 (
dx ax dx x a n n c x n a n 1 ) 1 ( c x n a n 1 ) 1 ( 4. A. 4 1 z4-z3+ 2 1 z2-3z+c
(z33z2z3)dx = 4 1 z4-z3+ 2 1 z2-3z+c 5. B. x3c 1 1 4 5
x dx x3 c 1 3 1 1 3 4 3 5 3 5 x3c 1 1 4 5 6. A. x c x 2 3 3 2 2
c x x dx x x 2 3 2 1 3 2 1 2 ) 2 ( 2 1 c x x 2 3 3 2 2 7. C. 0
x2x4 dxx x3x5c 3 1 ) 5 1 ( A=1 B=1 → A-B=0 8. A. 0
dx
x dx x x x ) 1 2 ( 2 2 2 3 c x x 2 A=1 B= -1 → (A+B)2=0 9. C. 0
x x dx x x xc 3 1 3 2 3 1 3 1 6 5 5 3 2 4 A= 3 1 B= 3 2 D= 3 1 → A+B+D=0 10. A. c t t 3 3 1 1
dt t t dt t t 4 2 4 2 1 1 1 c t t 3 3 1 1B. Evaluasi Pemahaman dan Penguasaan Materi 1. a. x x x x3 c 8 3 8 3 8 3 5 8 3 8 3 b. x x x x 2x 5 3 2 2 3 5 3 5 3 2 c x x 2 5 3 53 c. x11x10 x10 x10c 10 1 10 1 d. x4 x5 x5 x5c 5 3 5 3 3 3 e. x x x c x 2 2 1 f.23xx32x3x2x4 c x x x x x x2 2 4 2 2 4 4 1 2 3 4 1 2 3 g. 32xx43x2x2 x5 c x x x x x x2 5 2 5 5 1 3 5 1 3 h.4x33x22x54x43x32x25x c x x x x x x x x43 25 4325
BAB I
INTEGRAL
Latihan Kompetensi
Siswa 1
2. a.
x dxx c x x dx 12 2 3 2 c x 2 b.
tt2 dttt2 t3c 3 1 3 ) 2 3 ( c.
dx
xx xdx x x x 5 4 3 5 2 2 5 2 5 c x x x 4 3 2 4 1 4 5 c x x x 4 3 12 3 1 4 5 d.
(4x33x26x1)dx c x x x x 4 3 2 2 e. c x x dx x x
2 12 32 3 1 2 f.
12x36x21
dx c x x x 4 3 2 3 g.
dx x x dx x x3 23 5 3 10 3 5 c x x 4 2 5 4 5 h.
c x x x dx x x x 1 5 1 1 2 2 5 2 4 3. a.
x dx
x54dx x94c 9 4 4 5 b.
c x dx x dx x 3 1 3 4 3 1 3 4 c. dx x x x c x x
5 3 5 3 2 2 1 2 3 d.
dx x x x 2 2 1 c x x x x 4 4 1 2 e.
dx
y
dx y y y 2 2 2 2 2 1 1 2 c x y x 2 1 3 f.
dx x x x c x x 2 3 2 2 1 2 g.
dx x x x 5 5 3 1 c x x x x 112 11 10 2 2 h.
x3 x2dx x6 x3c 6 1 3 4. a.
x212dx
x42x21
dx c x x x 5 3 3 2 5 1 b.
(2x)3dx
812x6x2x3
dx c x x x x 2 3 4 4 1 2 6 8 c.
1x3 2dx
12x3x6
dx c x x x 4 7 7 1 2 1 d.
(4x2)2x2
dx
16x2 8x4 x6dx c x x x 3 5 7 7 1 5 8 3 16 e.
x3 x2
dx x6x3c 6 1 ) 3 ( f.
(1x3)2x
dx
x2x4x7
dx c x x x 2 5 8 8 1 5 2 2 1 5. a.
(1x3)2x3
dx
x32x6x9
dx c x x x 4 7 10 10 1 7 2 4 1 b.
x xdx x x x2 xc 5 2 3 2 ) 1 ( c.
dx x x 3 1
x32 3x12 3x12 x32dx c x x x x 2 1 2 1 2 3 2 5 2 6 2 5 2 d.
dx
x x
dx x x 2 ) 1 ( 12 2 c x x x 12 32 3 2 2 2 e.
x2x1
2dx
x4 2x3 3x2 2x 1dx c x x x x x 5 4 3 2 2 1 5 1 f.
dx x x x 1)( 2) (
x32 x 2x12dx c x x x 52 32 12 4 3 2 5 2A. Evaluasi Pemahaman dan Penguasaan Materi 1. a.
dxx x c x x x 4 2 3 4 6 2 3 3 4 b.
c x x dx x x x 1 4 1 4 3 6 c.
27x73x545x3 2x
dx c x x x x x 2 3 2 4 45 2 1 8 27 8 6 4 d.
x2
x35x23x 3
dx c x x x x 6 5 4 3 3 3 4 3 6 1 e.
dz
z z z dz z z 2 1 2 3 2 7 2 12 2 c z z z 2 1 2 5 2 9 2 5 4 9 2 f.
x3 x
2dx c x x x 7 2 2 1 9 4 7 1 92 g.
t 2t 2dt c t t t 3 2 2 5 5 2 4 3 1 h.
dr r r r c r r r 2 3 2 5 2 7 3 2 5 4 7 2 12 i.
dx x x 4 5 3 2 4 c x x 4 1 3 5 16 5 3 j.
dx x x x 1 2 c x x 2 1 2 5 2 5 4 2. a. x dx x
1 2 3 32 3 c x x x 52 5 4 2 3 2 b.
dx x x34 23 3 2 c x x 8 14 9 13 c.
dx x x 3 3 2 6 4 c x x 53 23 9 5 12 d.
3 x(x dx1)3
x x3 x43 x13 dx 7 3 10 3 3 c x x x x 3 43 7 3 10 3 13 4 3 7 9 10 9 13 3 e.
x(1 dxx)3
x12 3x32 3x52 x72 dx c x x x x 2 9 2 7 2 5 2 3 9 2 7 6 5 6 3 2 f.
dx x x x 2 3 4 7 5 3 2 c x x x 7 5 14 3 21 2 3 g.
dx x x 4)2 3 (
9x32 24x12 16x12dx c x x x 2 1 2 3 2 5 32 16 5 18 h.
dx x x x 4)3 3 ( 2 3 64 144 108 27 3 2 x x x x
x x x x 2dx 3 2 1 2 1 2 3 64 144 108 27 c x x x x 2 1 2 1 2 3 2 5 128 288 72 5 54 i.
(x dx1)4
x4 4x3 6x2 4x2 1dx c x x x x x 5 4 3 3 3 4 2 5 1 j.
(9t11)5dt
(9 11) (9 11) 9 1 5 t d t c t (9 11) 54 1 3. a.
(x1)(x3)dx 5 3
x x dx 5 9 5 12 5 3 2 c x x x 5 9 5 6 5 1 3 2Latihan Kompetensi
Siswa 2
b.
dx x x 3 2 1
x8 3x3 3 x3 dx c x x x x 9 4 2 2 1 3 4 3 9 1 c.
dt t t 4 2 2 1
1 2t 2 t 4 dt c t t t 1 3 3 1 2 d.
x xdx x x x x 1 1 2
x xdx x x 2 2 2 1
x4 x3 1 x2 dx c x x x x 5 4 1 4 1 5 1 4. a.
x57 x55 x3dx
x295dx x345 c 34 5 b.
x410x53 xdx
x215dx x265 c 26 5 c.
x35 7 x53 x2 xdx
x196dx x256 c 25 6 d.
x67 5 x2 xdx 2 2
2635x1263219dx
2635x42179dx c x 635 42170 491 70 2 5. a.
dx x x x410 53 3
dx x15 68 3 c x 15 53 53 45 b.
dx x x x 4 102 233 2 dx x
39 113 30 1 10 11 3 2 c x 83 3 2 3 30 83 30 1 10 11 c.
dx x x 5 3 2 4 2 dx x6 1 3 1 2
c x 6 7 3 1 7 2 6 d.
2x37 25 2 3xdx dx x70 211 70 1 35 41 3 2
c x 79 281 70 1 35 41 70 281 3 2B. Evaluasi Kemampuan Analisis
1.
dx x x2 ( 1)2 1 1 1 = ... = ... 2.
x2x1
5dx
dx x x x x x x x x x 1 5 15 30 46 56 45 30 10 2 3 4 6 6 7 8 10 c x x x x x x x x x x 2 3 4 5 6 7 8 9 11 2 5 3 15 4 30 5 46 6 56 7 45 8 30 9 10 11 c x x x x x x x x x x 2 3 4 5 6 7 8 9 11 2 5 5 2 15 5 46 3 28 7 45 4 15 9 10 11 3. y x2 3 x2 3... dengan x≠0 ... 3 3 3 4 2 2 4 x x x y y x y4 4 3 4 3 4 3x y y y 3 4 33x y
ydx 33x43dx c x 73 7 3 334. y x2 x2 x2... = ...
= ...
A. Evaluasi pengertian atau Ingatan 1. D. 10
(34x3x2)dx3x2x2x3c f(-3)= -9-18+27+c=10 c=10 2. C. 2x3-x2+x-10 f(x)=
(6x22x1)dx= 2x3-x2+x+c f(2)=16-4+2+c=4 14+c=4 c= -10 f(x)= 2x3-x2+x-10 3. D. 5
xdx x x c x f( ) 3 2 19 4 4 2 ) 4 ( c f c=3 3 2 ) (x x x f f(1)=5 4. A. x4-x2-5 y1=f1(x)=
(12x2 dx2) = 4x3-2x+c1 y=f(x)=
(4x32xc1)dx = x4-x2+c1x+c2 f(0)=c2= -5 f(2)= 16-4+2c1-5=7 c1=0 f(x)= x4-x2-5 5. C. x2-2x y1=f1(x)=
2dx=2x+c 1 f1(x)= 2+c1=0 c1= -2 f1(x)=2x-2 f(x)=
2x2dx = x2-2x+c2 f(0)=02-2∙0+c 2 c2=0 f(x)= x2-2x 6. B. -x 1 +3 f(x)=
c x dx x 1 1 2 f(1)= -1+c=2 c=3 f(x)= -x 1 +3 7. B. x3+2x2+x-2 f1(x)=
6x dx4 =3x2+4x+c 1 f(x)=
(3x24xc1)dx = x3+2x2+c 1x+c2 f(1)=1+2+c1+c2= 2 c1+c2= -1 ... (a) f(-2)= -8+8-2c1+c2= -4 -2c1+c2= -4 ... (b) a dan b dieliminasi, didapat: c1=1 dan c2= -2 f(x)= x3+2x2+x-2 8. B. 2 1 15 f1(x)=
f11(x)dx
(2x3)dx =x2-3x+c1 5=0-0+c1→ c1=5 f1(x)=x2-3x+5 f(x)=
f1(x)dx
(x23x5)dx = 3 2 2 5 2 3 3 1 c x x x 5=0-0+0+c2→ c2=5 f(x)= 5 5 2 3 3 1 3 2 x x x f(3)= (3) 5(3) 5 2 3 ) 3 ( 3 1 3 2 = 2 1 15Latihan Kompetensi
Siswa 3
9. C. 11 y=
(4x dx1) =2x2+x+c 0=8+2+c1→ c1= -10 y=2x2+x-10 x=3 → y=18+3-10=11 10. D. 3 1 5 5 x y
x dx x x c y 2 3 3 1 ) 1 ( 0=0-0+c → c=0 y= 3 1 x3-x m=y1(2)=22-1=3 y(2)= 3 2 2 3 8 ) 2 ( 3 3 2 x y 3 1 5 3 x yB. Evaluasi Pemahaman dan Penguasaan Materi 1. a. F(x)=
(6x dx4) = 3x2-4x+c 3=0-0+c → c=3 Jadi F(x)= 3x2-4x+3 b. F(x)=
(3x26x2)dx = x3-3x2+2x+c 7=0-0+0+c → c=7 Jadi F(x)= x3-3x2+2x+7 c. F(x)=
c x dx x 1 1 2 2 1 3 2 1 3 cc Jadi F(x)= 2 1 3 1 x d. F(x)=
x2 dx 4 1 c x x 3 3 1 4 1 c 3 ) 1 ( 3 1 ) 1 ( 4 1 2 12 11 1 12 1 2 cc Jadi F(x) 12 11 1 3 1 4 1 3 x x e. F(x)=
(x1)(x2)dx =
(x23x2)dx = x x 2xc 2 3 3 1 3 2 c 6 2 27 9 2 3 c=0 Jadi F(x)= x x 2x 2 3 3 1 3 2 f. F(x)=
c x x dx x x 2 2 1 2 2 2 9=2+1+c → c=6 Jadi F(x)= 2 6 2 1 2 x x g. F x
xdx x xc 3 2 ) ( c 4 4 3 2 0 3 16 c Jadi, F x x xc 3 2 ) ( h.
dx x x x F 2 3 2 3 ) ( c x x 2 4 3 4 4 1 2 2 4 3 1 c Jadi, 4 1 2 2 4 3 ) ( 4 x x x f i.
dx x x x x F( ) 4 3 33 c x x x 5 2 2 2 3 2 3 5 1 c 2 3 2 3 5 1 1 5 4 1 c Jadi, 5 4 1 2 3 2 3 5 1 ) ( 5 2 2 x x x x F j. F(x)
(x2)3dx
(x3 6x2 12x 8)dx c x x x x 2 6 8 4 1 4 3 2 c 4 16 24 16 1 5 c Jadi, 2 6 8 5 4 1 ) (x x4x3x2x F2. a. g1 x
x dx x2 xc 2 3 ) 1 3 ( ) ( b. g x
xdx x xc 3 2 ) ( 1 c.
dx x x dx x x x g 3 3 4 1 1 1 ) ( c x 2 1 2 1 2 d. g1(x)
(2x3)dx c x x 2 3 e. g x
x43 c ) ( 1 c x x 23 7 3 f. g1 x
x2x dx x3 x2c 2 1 3 1 ) ( ) ( 3. a. y
2xdxx2c 2 1 32cc 2 2 x y b. y
x dx x2xc 2 1 ) 1 ( 2 1 3 1 2 1 3 cc 2 1 3 2 1 2 x x y c. y
x dx x2xc 2 1 ) 1 ( 2 1 1 1 2 1 3 cc 2 1 1 2 1 2 x x y d. y
x dx x2xc 2 3 ) 1 3 ( 2 1 2 1 2 3 3 cc 2 1 2 2 3 2 x x y e. y
x2dx x3xc 3 1 1 3 2 3 1 3 1 3 cc 3 2 3 3 1 3 x x y f.
c x dx x y 12 1 1 1 0cc 1 1 x y g. y
xdx x xc 3 2 c 4 4 3 2 5 3 1 c 3 1 3 2 x x y h. y
3x22x1
dx c x x x 3 2 1 1 1 1 0cc 1 2 3 x x x y i. y
4x36x21
dx c x x x 4 3 2 6 2 16 16 8 cc 6 2 3 4 x x x j.
c x dx x y 2 1 2 1 2 3 4 1 4 3 2 cc 3 21 x y 4. a. F(x)
(4x1)dx c x x 2 2 1 3 18 20 cc 1 2 ) ( 2 x x x F 9 1 2 8 ) 2 ( F b. F(x)
6x22x1
dx c x x x 2 3 2 5 0 0 0 5 cc 5 2 ) ( 3 2 x x x x F 19 5 2 4 16 ) 2 ( F c.
c x x dx x x F( ) 1 12 1 3 1 1 3cc 3 1 ) ( x x x F 2 1 4 3 2 1 2 ) 2 ( Fd. xdx
dx x dF 6 ) ( 1 2 3x c 0 0 0 c1c1 2 3 ) ( x dx x dF
2 3 2 3 ) (x xdx x c F 8 8 0c2c2 16 ) 2 ( 8 ) (x x3 F F e.
2 2 1 ) ( c x dx dx x dF 3 0 2 3 cc1 3 2 ) ( x dx x dF
x dx x F( ) (2 3) 2 2 3x c x 7 3 1 5 c2c2 7 3 ) (x x2x F 17 7 6 4 ) 2 ( F f.
c x dx x x F 2 1 2 1 ) ( 2 3 1 1 6 1 6 7 c c 3 1 1 2 1 ) ( x x F 12 1 1 3 1 1 4 1 ) 2 ( F g. F(x)
(x1)(4x20)dx
(4x2 16x 20)dx c x x x 8 20 3 4 3 2 3 2 36 20 8 3 4 10 cc 3 2 36 20 8 3 4 ) (x x3x2 x F 3 2 36 40 32 3 32 ) 2 ( F 3 1 25 h.
x dx dx x dF ) 1 6 ( 2 ) ( 2
(12x2 2)dx 1 3 2 4x xc 5 0 0 5 c1c1 5 2 4 ) ( 3 x x dx x dF
x x dx x F( ) 4 3 2 5 2 2 4 5x c x x 3 5 1 1 2 c2c2 3 5 ) (x x4x2x F 19 3 10 4 16 ) 2 ( fSoal Aplikasi Bidang Geometri 5. y
(2x3)dx x23xc 2 9 9 2 cc Persamaan kurva: 2 3 2 x x y 6. y
x2x2dx
x42x3x2
dx c x x x 5 4 3 3 1 2 1 5 1 0 0 0 0 0 cc Persamaan kurva: 3 4 5 3 1 2 1 5 1 x x x y 7. F(t)
(52kt)dt c kt t 2 5 0 0 0 5 0 k2cc 0 1 1 5 1 k 4 k 2 4 5 ) (t t t F 8. F(x)
3 xdx2x xc 50 9 9 2 4 cc 50 2 ) (x x x F 9. a. y
3x28x5
dx c x x x 3 4 2 5 c 27 36 15 4 2 c 2 5 4 2 3 x x x yb. Akan dibuktikan (2,0) memenuhi persamaan kuva tersebut. 2 2 5 2 4 2 03 2 2 10 16 8 0 0 0 (terbukti)
Jadi, (2,0) melalui persamaan kurva
2 5 4 2 3 x x x y 10 . y
abx dxaxbx2c 2 ) ( 0 0 0 0 cc 2 2x b ax y 1 ) 0 ( b a 1 a 2 2 ) 1 ( 2 ) 1 ( 1 ) 1 ( 2 ) 1 ( 3a b b 4 2 1 3bb 2 2 x x y 11 .
1 2 3 6xdx x c dx dy 0 12 12 1 1 c c 2 3x dx dy
2 3 2 3x dx x c y 4 8 42 2 c c 4 3 x y 12 . 1 2 3 14 5 ) (t t t F c t t t t F()5 7 23 Sumbu simetri 3 7 6 14 t c 27 343 9 49 7 3 7 5 25 7 326 7 5 ) ( 27 326 23 F t t t t c 13. f(x)
(4x3)dx c x x 2 2 3 0 ) ( 1 x f 0 3 4x 4 3 x c 4 3 3 4 3 2 8 1 2 c 8 18 8 9 8 1 4 5 8 10 c 4 5 3 2 ) (x x2x fSoal Aplikasi Bidang Mekanika 14.v(t)
adt (102t)dt 1 2 10tt c 1 2 0 0 10 0 0 ) 0 ( c v 0 1 c 2 10 ) (t t t v
t t dt t s( ) (10 2) 2 3 2 3 1 5t t c 0 ) 0 ( s 2 3 2 0 3 1 0 5 0 c 0 2 c 3 2 3 1 5 ) (t t t s Benda akan berhenti ketika s(t)0
0 3 1 5t2 t3 0 15 1 1 52 t t t=0 t=15
Benda itu akan berhenti ketika t15 15. v(t)
(182t)dt 1 2 18tt c 20 ) 0 ( v 1 0 0 20 c 20 1 c 20 18 ) (t tt2 v 20 36 108 ) 6 ( v s m 92
18 20 ) (t t2 t s 2 2 3 20 9 3 1 c t t t 81 ) 3 ( s 2 60 81 9 81 c 51 2 c 51 20 9 3 1 ) (t t3t2 t s 16 . a. s
t dt 3 1 ) 3 6 (
36 6 30 3 3 31 2 t t b.
4 2 2 2 3t tdt s
42 80 12 68 2 3 t t c.
9 4 5 dt t s 3 1 35 63 5 3 2 9 4 t t t 3 2 27 d.
5 0 ) 2 2 ( t dt s
15 2 50 2 t t e.
4 0 2 3t 2dt t s 4 0 2 3 2 2 3 3 1 t t t 3 1 5 8 24 3 64 17 . Total biaya =
(1,0640,005)xdx c x x 2 0025 , 0 064 , 1 biaya awal = 16,3 Total biaya =1,06400,002502c 3 , 16 c Total biaya =1,064x0,0025x216,3 Rata-rata biaya x x 16,3 0025 , 0 064 , 1 18. Total biaya
f(x)
2 60x 6x2 dx c x x x 2 3 2 30 2 Rata-rata biaya x c x x 2 2 30 219. Fungsi pendapatan total
R(x)
8 6x 2x2dx c x x x 2 3 3 2 3 8 Fungsi Demand: x c x x 2 3 2 3 820. Fungsi pendapatan total
c x x dx x x
32 2 3 2ln Fungsi Demand x c x x x 32 2lnC. Evaluasi Kemampuan Analisis 1. a. yax2bxc c b a 1,0) 0 ( ... (1) c b a 0 9 3 ) 0 , 3 ( ... (2) 0 4 8a b c b a 12 ) 12 , 1 ( ... (3)
(1) dieliminasi dengan (3), diperoleh: 6 b Subsitusi b6 ke 8a b4 0 24 8a 3 a 9 6 3 12 cc Jadi, persamaan kurva adalah:
9 6 3 2 x x y
b. Sketsa grafik kurva.
8 10 12 2 2 4 4 6 6 -2 -4 x
2. 21 x a dx dy 3 1 dx dy x 2 1 1 3a2a 1 2 2 x dx dy
dx x y 22 1 c x x 2 3 1 y x c 1 1 2 3 4 c 4 2 x x y 3. a(x p)(x q) dx dy apq x q p a ax dx dy ) ( 2
ax a p q x apq y 2 ( ) c apqx x q p a x a 3 2 2 ) ( 3 a x c apq q p a a 2 ( ) 2 1 3 8 0 a x c apq q p a a 2 2 ) ( 3 1 pq q p 2 2 2 ... 4. x ax b dx dy 2 3 3 3 0 abab 5 3 3 27 32 abab 8 4a 2 a 1 b
x x dx y 3 2 2 1 c x x x 3 2 5. ... 6. v
32dt32tc 50 0 32 50 ) 0 ( c v 50 c 50 4 32 ) 4 ( v 178 kaki/detik
t dt s 32 50 k t t 16 2 50 1000 0 50 0 16 1000 ) 0 ( 2 k s 1000 k 1000 50 16 ) ( 2 t t t s 1000 200 256 ) 4 ( s 1456 mA. Evaluasi Pemahaman dan Penguasaan Materi 1. a.
6u23cosu
du c u u 2 3 3sin b.
tan21
d
sec2d tan c c.
d d tan sec cos sin sec c sec d.
(2sin3cos)d c 2cos 3sin e.
d 2 cos 1 cos
coscosecd c ec cos f.
(cot2x dx1)
cosec2xdx c x cotLatihan Kompetensi
Siswa 4
g.
xdx x 2 sin 1 sin
tanx secxdx c x sec h.
x2sinx x
dx c x x x x 3 2 cos 3 1 3 2. a.
6cosx4x2
dx c x x 3 3 4 sin 6 b.
8cosx6sinx
dx c x x 8sin 6cos c.
dx x x 2 cos sin c x sec d.
dx x x 2 sin cos c ecx cose.
sin4xcos4x2sin2xcos2xdx
sin2x cos2x2dx
12dx x c f.
1 2cos d
2sin 2cos c g.
1cos2d
2cos 2sin c h.
secx sec2x dx1
secx tanxdx c x sec 3. a. F(x)
32sinxdx c x x 3 2cos c s c 3 0 2 0 0 1 1 c b. F(x)
32cosxdx c x x 3 2sin c 3 0 2 sin0 5 5 c 5 sin 2 3 ) (x x x F c. F(x)
sec2x3dx c x x tan 3 c 4 3 4 tan 4 3 2 1 c 1 3 tan ) (x xx F 4. a. F(x)
cosxsinx c x x sin cos c 2 cos 2 sin 5 4 c 4 4 cos 4 sin 4 F 2 4 b. F(x)
sinxcosxdx c x x cos sin c 3 sin 3 cos 3 2 1 2 1 c 3 2 1 2 1 3 2 1 2 1 1 c 1 4 sin 4 cos 4 F 1 2 c.
dx xc x x x F sec cos sin ) ( 2 c 3 sec 7 5 c 5 4 sec 4 F 2 5 5. a.
4cos2x2sec2x
dx
x 2xdx sec 2 2 cos 2 1 2 1 4 c x x x 2 sin2 2tanb.
cosxcosecxtanxsecx
dxc x ecx cos sec c.
sec2xcosec0x
dx c ecx x tan cosd.
4sinx2cosxcosec2xsecxtanx
dx xdx xx
x 2sin cot sec cos 4 e.
tan2x dx5
tan2 1 6dx
sec2x 6dx c x x tan 6 f.
cot2x dx7
cot2x 1 6dx
cosec2x 6dx c x x cot 6B. Evaluasi Kemampuan Analisis 1.a.
dx x x x x 2 2 sin 1 sin 2 sec tan=
tan2xsec2x2tanxsecx2tanxsecxdx
tan2x 1 sec2x 1dx
2sec2x 1dx c x x tan b.
dx x x x x 2 2 sec sin 2 sec tan
tan2x sec2x 2tanxsecx 2sinxcos2xdx
tan2x 1 sec2x 1 2tanxsecx
2cos2xd(cosx) c x x x x 3 cos 3 2 sec tan 2 c.
dx x x ecx x 2 2 sin cos 2 cos cot
cot2x cosec2x 2cotxcosecx ecxdx x cos cot 2
cot2x 1 cosec2x 1dx c x x cot2 2. a.
dx x x x x 2 3 3 cos sin cos sin ... b.
dx x x x x 2 3 3 cos sin sin cos ... c.
dx x x x x 4 2 3 3 cos cos cos sin ... d.
dx x x x x 4 2 3 3 sin sin sin cos ...A. Evaluasi Pengertian atau Ingatan 1. B.
b ba a
3 2 b a b a x x dx x
32
b ba a
3 2 2. C.
ab ab(ab) ab
3 2 a b b a a b b a x x dx x
32 =
ab ab(ab) ab
3 2 3. C. 3 1
2 1 2 2 1 2 1 ) 2 ( 2 1 dx x x dx x x 2 1 3 2 6 1 2 1 x x 6 1 2 1 3 4 2 3 1 4. 104
9 1 9 1 4 6 xdxx x
104 4 108 Latihan Kompetensi
Siswa 5
5. D. 2 1
2 6 2 6 sin cos x xdx
2 1 2 1 1 6. C. 5
12 3 ) 3 2 ( 1 2 1
n n x x dx x 12 ) 3 1 ( 3 2 n n 0 10 3 2 n n 0 ) 2 )( 5 (n n 5 n 7. C. -1
6 3 ) 2 3 ( 22 2
a a x x dx x 6 ) 3 ( 4 6 aa2 0 4 3 2 a a 0 ) 1 )( 4 (a a 1 a 8. E. 4
2 0 ) ( 9 tdt
2 1 1 0 ) ( 9 ) ( 9 t dt t
1 2 1 0 ) ( 9 ) ( 9 t dt t dt 4 ) 2 ( 2 9. D.32
26 3 1 3 1 2
t t x dx x 26 1 t t 2 2 3 3 t
3 23 3 t 10 . C. 388
7 3 3 10x 2dx x 7 3 2 4 5 2 4 1 x x x =388 11. B. -13,5
4 1 2 15 12 3x x dx 5 , 13 12 2 15 4 1 2 3 x x x 12. A. 2 xdx x 3cos sin 0
0 sin 3 cosx x 2 ) 0 1 ( 0 1 13. A. 4
2 0 cos ) tan 5 ( xdx x
2 0 sin cos 5 xdx x
2 0 cos sin 5 x x 4 ) 1 0 ( 0 5 14. A. -6
0 sin ) 3 (cotx xdx
0 sin 3 cosx xdx
0 cos 3 sinx x 6 ) 3 0 ( 3 0 15. A. -2, 0 dan 3
n dx x x 1 2 2 6 3
4 6 1 2 3 n x x x 4 6 1 1 6 2 3n n n 0 6 2 3n n n 0 ) 2 )( 3 (n n n n= -2 n=0 n=3B. Evaluasi Pemahaman dan Penguasaan Materi
1. a. 5 0 2 5 0 2 1
xdx x c ) 0 ( 2 25 c c 2 25 b. 3 1 4 3 1 3 4 1
x dx x c 20 4 1 4 81 c cc. 5 0 3 5 0 2 3 5 5
x dx x c ) 0 ( 3 625 c c 3 625 d. 3 2 3 2 2 1
c x x dx c c 2 1 3 1 6 1 e. 2 1 3 4 2 1 3 4 3
xdx x c c c 4 3 16 4 33
16 1 1,14 4 3 3 f. 3 1 2 3 1 2 2 3 2 3
x dx x x c c 2 c 2 3 6 2 27 8 g. 4 1 2 4 1 3 16 32
c x dx x 15 ) 16 ( 1 c c h.
2 1 2 3 2 3 4x x xdx
2 1 3 4 2 x x x c ) 2 1 1 ( 4 8 16 c c 30 i.
1 1 4 2 4 2x x dx 1 1 5 3 5 4 3 2 x x c 5 4 3 2 5 4 3 2 c 5 1 3 15 48 15 24 20 5 8 3 4 j. 3 2 3 3 2 2 1 1 1 1
c x x dx x x c c 18 1 3 1 2 1 1 9 2 1 18 22 18 1 6 9 18 2. a. 2
14 4 2 2 4 1
x x dx b. 6 2 1 2 ) 2 ( 2 0 2 2 0
xdx x x c.
2 0 2 2 0 2 4 4 ) 2 ( x dx x x dx 3 8 3 1 2 4 2 0 3 2 x x x d.
3 0 2 2 3 x x dx 9 3 1 3 3 0 3 2 x x x e.
2 1 3 2 1 2 1 t tdt t t dt 2 1 4 2 4 1 2 1 t t 4 1 2 4 1 2 1 4 2 f.
4 1 ) 1 ( u udu
4 1 du u u u 4 1 2 5 2 3 2 u u u u 5 2 3 2 5 64 3 16 15 186 70 5 62 3 14 15 11 7 15 116 g.
2 0 2 5 2 0 3 2x 1dx x x dx x 2 0 3 6 3 1 6 1 x x 3 40 3 8 3 32 h.
1 0 2 1 x dx x
1 0 2dx x x 6 1 3 1 2 1 1 0 3 2 x xi. 8 1 3 2 8 1 3 2 5 3
x dx x x 5 93 5 3 5 96 j.
2 1 ) 1 2 )( 5 (x x dx
2 1 2 5 9 2x x dx 2 1 2 3 5 2 9 3 2 x x x 5 2 9 3 2 10 18 3 16 2 9 3. a.
5 1 5 1 ) ( 2 ) ( 2h xdx h xdx 8 4 2 b.
5 1 3 ) (x dx h
5 1 5 1 3 ) (x dx dx h 16 ) 1 5 ( 3 4 c.
1 5 ) ( dxx h 4 ) ( 5 1
h xdx d.
5 1 ) (x kxdx h 28 ) ( 5 1
h x dx kxdx
5 1 28 2 4 22 k 2 k 4. a.
3 0 ) ( dxx f
3 2 2 0 ) ( ) (xdx f xdx f 10 5 5 b.
2 3 2 0 ) ( ) (xdx f xdx f
3 2 2 0 ) ( ) (xdx f xdx f 0 5 5 c.
2 0 2 ) ( 4f x dx
2 0 2 0 2 ) ( 4 f xdx dx 22 ) 0 2 ( 2 5 4 d.
0 2 ) ( dxx f 5 ) ( 2 0
f xdx 5. a.
2 0 ) 1 (cos dx x
2 0 sin x x 2 1 b.
4 0 2 tan 1 dx x
1 tan 4 0 x c.
0 0 cos sinxdx x
2 1 1 d.
2 0 ) sin ( dx x x 2 0 2 cos 2 1 x x ) 1 0 ( 0 8 2 1 8 2 e.
4 4 ) (cos dx x x 4 4 2 2 1 sin x x 32 2 2 1 32 2 2 1 2 2 2 f.
4 0 2 cot 1 xdx
1 cot 4 0 xg.
4 4 sin cot 2 xdx x
2 sin cos 2 4 4 x x h.
3 4 2 tan 1 xdx
3 4 tan x 3 1 ) 1 ( 3 i.
4 3 2 cot 1 xdx
4 3 cotx 1 3 3 1 3 1 1 j.
2 4 cos tan 2 xdx x
2 4 sin cos 2 xdx x
2 4 cos sin 2 x x 2 2 1 2 0 2 2 2 3 2 6. a. (2 1)
1 4 2 1
a a x x dx x 0 6 2a a 0 ) 2 )( 3 (a a a= -3 a=2 HP={-3,2} b. 2(2 3) 2 6
1 4 2 1
a a x x dx x 0 8 6 2a2 a 0 4 3 2 a a 0 ) 1 )( 4 (a a a=4 a= -1 HP={-8,3} c. (2 5) 5
1 18 2 1
a a x x dx x 0 24 5 2 a a 0 ) 3 )( 8 (a a a= -8 a=3 HP={-8,3} d. 24 2 1 ) 1 3 ( 3 1 2 3 3 2
xdx x x a 24 2 1 2 9 27 3 2 a a 48 9 54 a3a2 0 3 2 3a a 13 2 1 2 1 2 12 1 1 2 , 1 a HP= 2 13 1 , 2 13 1 7. a.
5 2 2 2 8 t t dt s 5 2 3 2 3 1 8 t t t 3 8 4 16 3 125 25 40 6 3 117 45 b.
3 1 2 8 2 dt t t s 3 1 2 8 t t 3 8 ) 8 1 ( 3 8 9 c.
t t
dt 3 0 2 2 3 0 2 3 2 2 1 3 1 t t t 2 1 10 6 2 9 9 8. a.
3 1 ) 2 )( 1 (x x dx x dx x x x
3 1 2 3 2 3 1 2 3 4 3 1 4 1 x x x 1 3 1 4 1 9 9 4 81 3 2 20 b.
dx x x x x 2 2
dx x x x 2 1 2 x x 4 1 1 c. dx x x x
3 1 2 3 2 1
3 1 2 1 1 dx x x 3 1 2 1 2 1 x x x 1 2 1 1 3 1 2 9 1 3 2 6 d.
3 2 2 2 3 4 ) 1 ( 2 2 dx x x x x x
3 2 2 2 2 1 ) 2 )( 1 ( dx x x x x 3 2 2 3 3 1 x x 4 3 8 9 9 3 1 11 e.
2 1 2 3 3 4 5 8x x x dx
2 1 2 3 4 2 5 2 x x x x 52 5 2 1 2 20 8 8 32 f.
3 3 2 2 3 2 3 2 12 5 dx x x 2 3 2 3 3 2 3 3 2 36 3 x x x
3 2 2 36 2
3 36 3 9 3 2 42 3 63
3 3 2 2
21 9. 2 1 ) 2 ( 1
n dx x 2 1 2 2 1 2 1 n x x 2 1 2 2 1 2 n n 1 4 22 n n 0 3 4 2 n n 0 ) 3 )( 1 (n n n=1 n=3 (tm) a. n2321 b. (2 1)249 n c. (2n)21 d. (2n)225 10.
a dx x 1 10 ) 5 2 (
10 5 1 2 a x x 0 4 5 2 a a 0 ) 1 )( 4 (a a a= -4 (tm) a= -1
1 0 ) 1 2 ( b dx x
1 0 2 b x x 0 2b2b 0 2 2b b 0 ) 1 )( 2 (b b b=2 b= -1 (tm) a. ( b)2(12)29 a b.(ab)2 (12)21 c.a2b2143 d.b2a2 413C. Evaluasi Kemampuan Analisis 1. a.
2 3 2 3 2 7 6x dx x 2 3 2 3 2 3 7 3 3 1 x x x 2 33 18 2 21 7 2 3 29 15 2 7 21 2 18 3 2 3 19 5 2 6 20 b.
dx c bx ax2 ax3 bx2 cx 2 3
a 3 3 b 2 2 c 2 3 c.
a a dt t t2 22
a a dt t t t t4 23 52 4 4 a a t t t t t 2 4 3 5 2 1 5 1 5 4 3 2
5 5
4 4
3 3
3 5 2 1 5 1 a a a a a a
a a
aa 2 2 2 4 a a a 8 3 10 5 2 5 3 d.
1 2 2 x x dt t 1 2 3 3 1 x x t
x x xx
8 12 12 1 3 1 3 2 3 1 3 11 4 3 8 3 2 x x x 2. f x ax3bx2cxd ) ( 0 0 ) 0 ( d f 0 0 ) 1 ( abc f c bx ax x f1( )3 22 36 36 ) 0 ( 1 c f sehingga ab36 36 a b
1 0 2 3 5 ddx cx bx ax
1 0 2 3 36 ) ( a x xdx ax 5 18 3 36 4 1 0 2 3 4 ax a x x 5 18 12 3 4 a a 1 12 a 12 a 48 36 12 b 3.
1 0 2 ax b lx mdx x
1 0 2 3 (m al)x (am bl)x bmdx lx 0 2 3 4 1 0 2 3 4 l x m alx am blx bmx = ... 4.
1 0 2 2 1 0 ) 6 6 ( 6 6 12ax adx ax b a x 0 ) 6 ( b a x bdx
1 3 3 0 2 6 2 (2 2 ) 6ax ax ax ba x 0 2 3 1 0 2 b a x bx 0 2 3 2 2 2 12aababa b 0 6 2 1 16 a b a b 12 33
1 0 2 2 2 1 0 2dx a x 2abx b dx a 1 3 1 0 2 2 3 2 2 a x a x abx b x 1 3 2 2 2a abb a 1 144 1089 12 33 3 2 2 2 a a a a a1 144 1089 396 48 144 2 2 2 2 a a a a 144 885a2 885 144 2 a 885 12 a a b 12 33 885 396 885 12 b a 885 396 885 12 b a a. 4a23ab3b2 12 , 516 885 12492 b.8a3b 885 396 3 885 96 885 1096 885 1096 atau 5. a.
x dt t t x f 1 2 3 4 ) ( x t t t 1 2 3 4 2 3 3 1 4 2 3 3 1 4 2 3 3 1 3 2 x x x 6 5 2 4 2 3 3 1 3 2 x x x 4 3 ) ( 2 1 x x x f b.
1 5 1 2 3 4 x dt t t
3
2 2
1 ) 1 5 ( 2 3 1 ) 1 5 ( 3 1 x x
5 1 1
4 x x x x x x x 15 20 2 75 5 25 3 125 3 2 2 x x x 10 2 15 3 125 3 2 10 15 125 ) ( 2 1 x x x f c.
x x dt t t 1 2 2 x x t t t 3 2 2 1 3 2
x x
x x
xx 3 3 2 2 2 1 3 2 x x 2 3 4 3 2 4 ) ( 2 1 x x f d.
3 1 2 2 dt x xt t 3 1 2 2 3 2 3 1 t xt xt ) 1 3 ( ) 1 9 ( 2 ) 1 27 ( 3 1 2 x x 3 28 4 4 2 x x 4 8 ) ( 1 x x f e.
x dx x xt 3 2 x x t x 3 2 3 2 1 3 2 9 9 2 1 3 3 4 x x x 9 2 3 3 4 ) ( 3 2 1 x x x f f.
x dt t xt 3 2 x t t x 3 2 3 3 9 9 3 2 4 x x x 9 2 3 4 ) ( 3 2 1 x x x f 6. a.
2 0 1 2 ) ( ) (x x f t dt f ) 0 ( ) 2 ( ) ( 2 f f x x f b.
x a x x dt t f1( ) 2 3 2 2 3 ) ( ) (x f a x2x f ) ( 2 3 ) (x x2 x f a f c.
x a x x dt t f 1 2 1 2 ) ( a x x f x f( ) (1) 22 ) 1 ( 2 ) (x x2 x a f f d.
x x x x f dt t f 1 2 1 7 ) ( ) ( 7 ) ( ) 1 ( ) (x f f x x2x f ) 1 ( f ... e.
1 1 1 2 ) ( dxx f 2 ) 1 ( ) 1 ( f f
x f x f dt t f 0 1 ) 0 ( ) ( ) ( 3 9x k 7.
a dt t f x x af 0 ) ( 4 ) (
a x x af dt t f 0 4 ) ( ) (
a x x af dt x f 0 4 ) ( ) ( 8.
x a dt t f ax x dt t f 0 0 2 ) ( ) ( ...A. Evaluasi Pengertian atau Ingatan 1. C.
; 0, 1 ) 1 ( 1 1 n a c b ax n a n
ax dxbn ) ( ) ( 1 b ax d b ax a n
c b ax n a n ( )1 ) 1 ( 1Penyebut tidak boleh nol, sehingga a≠0 dan n≠ -1 2. A. ( ) ; 0; 1 ) 1 ( 2 1 2 1 n a c b ax n a n
ax b dax b a dx b ax x 2 n 2 n 2 2 1 ) (
ax b
c n a n 2 1 ) 1 ( 2 1 a≠0 dan n≠ -1 3. E. 2
a x c x dx 3 3
x c x dx 3 2 3 Jadi, a=2 4. C. 6x5
56 (56 ) 6 1 6 5 xdx xd x c x x (5 6 ) 5 6 3 2 6 1 c x x c x x f (56 ) 56 9 1 6 5 ) ( 9 1 5 6 ) 6 5 ( ) (x x x f 5. A. 95x
(95 ) (95 ) 5 1 ) 5 9 ( 6 6 x d x dx x c x 7 ) 5 9 ( 35 1 x x f( )95 6. B. (73x)5c 15 1
(73 ) (73 ) 3 1 ) 3 7 ( x 4dx x4d x c x 5 ) 3 7 ( 15 1Latihan Kompetensi
Siswa 6
7. C.12cos(axb)c;a0 a
ax dxb asin( ) 1 0 ; ) cos( 1 2 ax b ca a 8. D.sin(ax ab); 0
acos(ax dxb) 0 ); sin( ax b a 9. E. x c 3 1 4 sin 4 1
x dx 3 1 4 cos = x c 3 1 4 sin 4 1 10 . D. x c 4 1 5 tan 5 1
x dx 4 1 5 sec2 c x 4 1 5 tan 5 1 11 . B. 9 7
1 0 2 3 1 x dx x
1 0 2 2 3 1 3 1 6 1 x d x
1 0 2 2 3 1 3 1 9 1 x x 9 7 9 1 9 8 12 . D.
2 2 1 3 2 1 tan d sec 45 0 2
= 1 tan (tan ) 45 0 d
= 45 0 tan 1 ) tan 1 ( 3 2 = 3 2 2 2 4 =
2 2 1 3 2 13. D. sec2c 2 1
d d 2 3 tan sec cos sin
secd(sec) c 2 sec 2 1 14. A.
t 3 t 3c 3 1 2 2
t43t2dt
t t23dt
3 ( 3) 2 1 2 2 t d t
t t c 3 3 3 1 2 2 15. D. 3 1
1 0 2 1 0 4 2 1 x dx x dx x x
1 0 2 2 ) 1 ( 1 2 1 x d x 1 0 2 2 1 ) 1 ( 3 1 x x 3 1 16. C. 8 17
1 1 3 1 4 1 dx x
1 1 3 1 4 1 1 4 1 4 x d x 1 1 4 1 4 1 x 256 81 256 635 8 17 16 34 256 544 17. E. 2
x35 x35c
9x2 x3 dx5
3 x3 5d x3 5
x x c 2 3 5 3 518 . C. 4 x31c
dx x x 1 6 3 2
1 1 2 3 3 x x d c x 4 3 1 19 . B.
8 2 8 1
4 0 4 sin cos 5 xdx x
4 0 4 ) (cos cos 5 x xd
4 0 5 cos x 1 2 8 1
8 2 8 1 20 . A.
2x214x82
x27x41c
6x21 x27x41dx
3 x2 7x 41d x2 7x 41
x x
x x c 2 2 7 41 2 7 41B. Evaluasi Pemahaman dan Penguasaan Materi
1. a.