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Nama : Hendrik Pical TTL : Banjar Masin,26-10-1956 Pendidikan : S1 Prodi : Matematika Hobi : Menulis Alamat Web : Blokmatek.wordpress.com No.HP : 081248149394 Alamat Email : [email protected] School : SMA Kristen Kalam Kudus Jayapura Jl.Ardipura I No. 50. Telepon 0967-533467 Jayapura Papua
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MGMP MATEMATIKA
SD
SMP
SMA
SKKK JAYAPURA
Kami mohon Donasi dari saudara-saudara sekalian agar blog ini tetapEksis untuk membantu saudara-saudara sekalian agar dapat mengakses materi bahan ajar atau soal-soal dan lainnya dalam bentuk “POWERPOINT” silahkan salurkan lewat rekening Bank MANDIRI atas nama HENDRIK PICAL,A.Md,S.Sos dengan No. ac Bank1540004492181. dan konvirmasi lewat No. HP. 081248149394. Terima Kasih.
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KUDUS" KALAM KRISTEN SEKOLAH"
: VISI
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JENDELA JOHARI
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8
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masalahpemecahan
dalam Integral konsepn Menggunaka
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tentuIntegraldan
tak tentuIntegral konsep Memahami
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13
integral (aturan)sifat -sifatn menggunaka
dengan tentu integral Menentukan 3.
datar bidang didaerah luas
sebagai tertentu integraln Menjelaska 2.
sederhanaaljabar fungsi
dari tak tentuintegral Menentukan 1.
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15
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integral (aturan)
sifat-sifatn menggunakadengan tu ten
integral menentukandapat didik Peserta c.
datar bidang didaerah luas sebagaientu tert
integraln menjelaskadapat didik Peserta b.
sederhanaaljabar fungsi daritentu tak
integral menentukandapat didik Peserta a.
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tertentuIntegral b.
tak tentuIntegral a.
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Diskusi 3.
Jawab Tanya 2.
Ceramah 1.
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x y d.
x
1 y c.
x 2x y b.
2 4x xy a.
iniberikut urunan Tentukan t
:Contoh
2
2
3
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x2
1 y x y d.
x
2- y
x
1 y c.
14x y 2x y b.
43x y 24 xy a.
: Jawab
1
31
2
12
213
x
x
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F(x) F’(x)2x2x2x
---------- 2x ---
2x
2x12 x32 x
42 xC2 x
PROSES DIFERENSIAL
PROSES INTEGRAL
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C F(X) dx f(x)
: Rumusnya
diketahui (x)F' turunannya
jika F(X) semula fungsi mencari Proses
adalah tak tentuIntegral
: Konklusi / Kesimpulan
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alanPengintegr Constanta c 3.
Integran Fungsi f(x) 2.
f(x)(x)F' (bersifat)
UmumIntegral Fungsi F(x) 1.
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f(x) x(x)F' x1n
1 F(x) 3.
f(x) x(x)F' C4
1 F(x) 2.
f(x) x (x)F' 3
1 F(x) 1.
contoh-Contoh
n1n
34
23
C
x
x
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C 5x
.....-.
C ) 5...
.....(-.
......
...... dx x .4
... ....
.... dx x .3
......
... dx
...
...
3
dx .2
C...... dx 5 1.
5
5
56-
1110
C
C
C
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C 5x
1-
C )x
1(
5
1-
x5
1- dxx .4
x 11
1 dxx .3
x3
1 dx
3
1
3
dx .2
C5x dx 5 1.
5
5
5-6
1110
C
C
C
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29
11F(2)dan x(x)F' jika F(x),Tentukan 5
3
1 x
6
1 F(x)
3
1
6
2 c
c6
410 11
11 c(2)6
1 F(2)
cx6
1 F(x)
dxx F(x)
: Jawab
6
6
6
5
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logxln xdengan c,xln a dxx
a 6.
1- n dengan c,x1n
a dxax 5.
1- n dengan c,x1n
1 dxx 4.
caxdx a .3
constantaadalah a dx, f(x) a af(x)dx .2
cxdx 1.
e
1nn
1nn
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31
dx x
1 d).
dx x c).
dx x
1 b).
dx5x a).
:iniberikut tak tentuintegral-IntegralTentukan
3 2
4 3
3
4
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32
c x
c x14
5 dx 5x a).
5
144
c2x
1-
c)1
(2
1-
c2
1-
c x13-
1
dxx dx x
1 b).
2
2
2
13-
3-3
x
x
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33
cxx7
4
c7
4
c
471
cx1
43
1
dxx dx x c).
4 3
4
7
4
7
14
3
4
34 3
x
x
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34c x3
c3x
c1
32
-
1
dxx dx x
1 d).
3
3
1
13
2
3
2-
3 2
x
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35
dx f(x) a dx f(x) a .3
g(x)dx -f(x)dx dx g(x)f(x) .2
g(x)dx f(x)dx dx g(x)f(x) 1.
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36
dx5x 3.
dx )xx-(x 2.
dx )x(x 1.
iniberikut tak tentuIntegralTentukan
256
23
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37
Cx3
1x
4
1
ccx3
1x
4
1
cx3
1cx
4
1
dx x dx x dx )x(x 1.
34
2134
23
14
2323
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38
Cx3
1x
6
1x
7
1
cx3
1cx
6
1-cx
7
1
dx xdx x- dxx dx )xx-(x 2.
367
33
26
17
256256
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39
cx2
5
cx)2
15(
dxx 5 dx5x 3.
2
2
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dx 7x 4.
dx5)(x 3.
dp )p(p 2.
dx )xx(x 1.
: inidibawah IntegralTentukan
5
2
43
32
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41
cxxx
432
3232
4
1
3
1
2
1
dxx -dx xdx x dx )xx(x 1.
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42
c p 5
1 p
4
1
dpp dpp dp )p(p 2.
54
4343
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c 25x 5x x3
1
25dx dx 10x dxx
25)dx x10(x dx5)(x 3.
23
2
22
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44cx
cx
5
6
5
6
15
1
5
15
6
35
567
c x1
51
7
dx)(7x dx 7x 4.
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45
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dx x 1. 7 35 x
dx 1)x(x 2. 2
dx 2)(x 3. 3
dt 1t
1t2t 4.
2
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47c
45
7
cx
c1
738x
dxx
dxx
dx ) x.(x dx x .1
7 36
7
45
7
45
17
38
7
38
5
7
357 35
73
xx
x
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48
cx2
1x
3
2 x
4
1
dx x)x2(x dx 1)x(x 2.
234
232
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49
c8x6x2xx4
1
dx 8)12x6x(x dx 2)(x 3.
234
233
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ct t
dt 1) (2t
dt1)(t
1)1)(t(2t dt
1t
1t2t 4.
2
2
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51
xc
c
c
logln xdengan ,xln a dxx
a 6.
1- n dengan ,x1n
a dxax 5.
1- n dengan ,x1n
1 dxx .4
caxdx a .3
constantaadalah a dx, f(x) a af(x)dx .2
cxdx 1.
e
1nn
1nn
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52
dx )x(1
3.
dx )1
xx( 2.
dx x)(xx 1.
: iniberikut tak tentuIntegralTentukan
3
2
2
x
x
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53
c5
2
7
2
c5
2
7
2
c11
dx )(
dx x)(xx dx x)(xx 1.
23
2
5
2
7
12
31
2
5
2
3
2
5
22
12
12
31
2
5
xxxx
xx
xx
xx
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54
c3
2
5
2
c3
2
5
2
c1
1
dx )(x
dx )x(x dx )1
xx( 2.
2
2
3
2
5
12
11
2
3
2
1
2
3
2
1
2
1
12
11
2
3
xxxx
xx
xx
x
xx
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55
c121
dx )x2x(x
dx xx)x21(
dx )x2(1
dx )x(1
3.
13
2
13
2
16
1
16
1
13
1
13
1
3
2
6
1
3
1
3
1
3
13
2
xxx
x
x
x
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56
c5
3
7
12
2
3 3
5
6
7
3
2
xxx
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57
ydan x antarahubungan carilah 3 y dan 1, dan x
0 y 0, x Bila .24dx
yddan f(x) y Diberikan 2.
11F(2)dan x(x)F' jika F(x),Tentukan 1.
2
2
5
x
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58
3
1 x
6
1 F(x)
3
1
6
2 c
c6
410 11
11 c(2)6
1 F(2)
cx6
1 F(x)
dxx F(x)
6
6
6
5
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59 x- 4x y Jadi
1- c
0 c 4 3
c1.c4.1 3 3 y dan 1x
0 c c0 0 0 0ydan 0
cc4x y
dx )c(12x dxdx
dy y
c12x dx 24 dx dx
yd
dx
dy 24
dx
yd
3
1
1
213
22
213
12
12
2
2
2
2
x
x
xx
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60
tersebutkurva
persamaancarilah ,3dx
dyitu kurva singgung
garisgradien dan (0,4) titik melalui kurvaSebuah
2x
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61
4 xy adalah kurvapersamaan Jadi
4 c
c0 4
(0,4) melalui Kurva c, xy
dx3x y
dxdx
dy y 3x
dx
dy
3
3
3
2
2
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62
! x(t)posisi fungsiuntuk formulaTentukan
12ta(t) percepatan fungsidengan sumbu x sepanjang
bergerakdan 10 titik x pada 0) awal (kecepatan
diamkeadaan daribergerak mulai partikelSebuah
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63102t x(t)posisi fungsi formula Jadi
10 c c2.0 10
yaitu c nilaidiperoleh 10, Untuk x(0)
c2t dt 6t dt v(t) x(t) dt
dx v(t)
6t v(t) 0 c c 6.0 0
:yaitu ,c nilaidiperoleh 0,Untuk v(0)
c 6t dt 12t a(t)dtv(t)
0)dengan v(012t a(t)dt
xd
3
223
2
232
211
2
1
1
2
2
2
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64
Tan x sec xSec x5
-Cosec xCot x4
Sec xTan x3
-Cot x cosec xCosec x6
-Sin xCos x2
Cos xSin x1F’(x)F(x)No.
2
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65
c x cosec- dx x ccot x.cose 6.
c x sec dx x tan x.sec 5.
c cot x - dx x cosec 4.
c tan x dx xsec .3
c x cos- dx sin x .2
c sin x dx x cos 1.
2
2
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66
-acot(ax+b).cosec(ax+b)Cosec(ax+b)6
atan(ax+b).sec(ax+b)Sec(ax+b)5
-acosec (ax+b)Cot(ax+b)4
asec (ax+b)tan(ax+b)3
-asin(ax+b)Cos(ax+b)2
acos(ax+b)Sin(ax+b)1
F’(x)F(x)No
2
2
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67
cb)cosec(ax1
dx b)xb).cosec(acot(ax 6.
cb)axsec(a
1 dx b)b).sec(axtan(ax 5.
cb)(axcot1
dx b)(axcosec 4.
cb)(axtana
1 dx b)(axsec 3.
cb)axcos(a
1-dx b)sin(ax 2.
cb)axsin(a
1 dx b)cos(ax 1.
2
2
a
a
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68
βαCosβαCos2
1- βSin αSin 4.
βαCosβαSin2
1β Cos α Cos .3
βαSinβαSin2
1 Sinβ α Cos .2
βαSinβαSin2
1 β Cos αSin .1
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69dx3x cos 6.
dxx sin .5
dx 4x) cos4x (sin 4.
dx x)sec (tan x 3.
dx x)cos-(sin x .2
dx)4(tan 1.
:berikut tak tentuintegral-integralTentukan
2
2
2
2
2
x
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70
c 3x tan x
dx 3 dx sec
3)dxx(sex dx 3)1(tan
menjadidiubah dx)4(tan 1.
2
22
2
x
x
x
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71
c cos2x 2
1 x
ccos2x)2
1(- - x
dx2x sin -dx
dx sin2x)-(1
menjadidiubah dx x)cos-(sin x .2 2
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72
c x - x sec 2 tan x 2
dx-dx x tan x.sec2 dx x sec 2
dx )1sec.tan2sec (2
: menjadiakan disederhandx x)sec (tan x 3.
2
2
2
xxx
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73
ccos8x16
1-
c 8x) cos 8
1(-
2
1
dx8x sin 2
1
dx )8(sin2
1 dx 4x) cos4x (sin
rangkapsudut 1 kerumusdiubah dx 4x) cos4x (sin 4.
x
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74
csin2x4
1 - x
2
1
csin2x)2
1(
2
1-x
2
1
dx cos2x2
1 - dx
2
1
dx)2cos2
1
2
1( dx 2x) cos -(1
2
1
menjadidiubah dx sin .5 2
x
x
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75
c 6x sin12
1x
2
1
c 6x) sin 6
1(
2
1 x
2
1
dx 6x) (cos2
1 dx
2
1 dx 6x) cos(1
2
1
menjadi diubah dx 3xcos 6. 2
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7664 E.
10 D.
0 C.
10- B.
64- A.
.....
adalah c nilai maka 10, f(3)dan 3x-4x-3-
adalah f(x) dari pertamaturunan Diberikan 1.2
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77
14 x x-2x E.
14 x x-2x D.
10 x x-2x C.
10 x x-2x B.
10 x x-2x A.
.adalah.... f(x)
maka 4, f(2)dan 1 2x - 6x (x)' f Bila 2.
23
23
23
23
23
2
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78
5x- xE.
5-x- xD.
5-x xC.
5x- xB.
5-x- xA.
.adalah.... yaPersamaann 7.f(2)dan
5- bernilai f(x) fungsi 0, Untuk x .212dx
yd
kedua turunan mempunyai f(x) y Fungsi 3.
23
23
24
24
24
22
2
x
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79
21 E.
13 D.
11 C.
9 B.
7 A.
.adalah....
3 x paday Nilai 1. 4x adalah (2,0)tik ti
melalui yang kurvasebuah dari fungsiGradien 4.
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80
5-3x y E. 3
15-3x y D.
3
25-3x y C.
2
16 -3x y B.
6 -3x y A.
.adalah.... 2 berabsis yangik tit
di kurva pada singgung garispersamaan maka
(0,0) titik melalui kurvanya yang f(x) y fungsi
dari pertamaunan adalah tur 1 xy Apabila 5. 21
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81
dx3x.sin x sin d.
dx6x 8x.cos cos c.
dx3x sin 6x. cos b.
dx 2x 5x.cossin a.
: iniberikut tak tentuIntegral Selesaikan 1.
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82
csin4x32
1sin2x
4
1x
8
3 dx x cos b.
csin4x32
1sin2x
4
1x
8
3 dx x sin a.
: bahwaTunjukan .2
4
4
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83:oleh ditentukan tentu integral maka
f(x) dari turunan antisuatu adalah F(x)dan
bxa interval padakontinu f(x) jika Jadi
alan).Pengintegrintegrasi( dari atas batasdan
bawah batasdisebut masing-masing bdan a 2.
integrandisebut f(x) Fungsi 1.
b. x sampai a x dari
f(x), fungsi tentu Integraldisebut dx f(x) Simbolb
a
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84
F(a) - F(b) F(x) dx f(x) ba
b
a
TENTU INTEGRAL DASAR RUMUS
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85
f(u)F(u)du
d maka dx, f(x)F(u) Bila .6
bcauntuk dx, f(x)dx f(x) dx f(x) 5.
dx g(x)dx f(x)g(x) f(x) 4.
real konstantaadalah cdengan ,f(x)dx c dx f(x) c .3
dx f(x)- dx f(x) .2
0 dx f(x) 1.
u
a
b
a
c
a
b
c
b
a
b
a
b
a
b
a
b
a
a
b
a
a
b
a
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86
0
1-
24
1
2
1-
33
1
dx 2)-x(6x d. dx 3)(4x b.
dx)1(2x c. dx2x a.
inidibawah tentu integral setiap nilaiHitunglah
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87
81-3 x dx2x a. 223
12
3
1
39
5-44
32- 1232
)1(32(1)- )4(32(4)
32x dx 3)(4x b.
22
4
12
4
1
x
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88
21
21
21
4214
21
2
14
21
2
1-
3
10
10
1- 28
1)1(2)2(
dx)1(2x c.
xx
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89
2
1
22-- 0
)1(2)1(2(-1)- )0(2)0()0(2
22x dx 2)-x(6x d.
21
22132
213
0
12
213
0
1-
2
xx
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90
2
2
0
dxSin x b.
dx x Cos a.
Hitunglah
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91
1-
1 - 0
sin - πsin
sin x dx x Cos a.
2π
2
2
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92
1
10
0 Cos-- Cos
xCos - dxSin x b.
2π
0
0
2π
2
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93
36dx 16x)(x b.
4 dx x
1 a.
: iniberikut
persamaan setiap memenuhi yang p nilaiTentukan
p
2
3
p
0
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94
4 2 p
2 p
4 p2
402p2
4x2
4 dx 4 dx x
1 a.
2
p
0
p
0
2
1p
0
x
![Page 95: Nama : Hendrik Pical TTL : Banjar Masin,26-10-1956 Pendidikan : S1](https://reader035.vdocuments.pub/reader035/viewer/2022062422/56813c88550346895da62829/html5/thumbnails/95.jpg)
95
4 p
0)16p)(16(p
0 256 p32p
064p8p
36)324(p8p
362.8.2p8p
36 - 8
36dx 16x)x( b.
22
24
2441
2441
244124
41
p2
2441
p
2
3
xx
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9613 E.
26 D.
52 C.
78 B.
104 A.
..... dx x6 .19
1
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97
)a-bab(3
2 E.
)bb-a(a2
3 D.
)bb-a(a3
2 C.
)aa-b(b3
2 B.
)aa-b(b2
3 A.
dx x 2.b
a
![Page 98: Nama : Hendrik Pical TTL : Banjar Masin,26-10-1956 Pendidikan : S1](https://reader035.vdocuments.pub/reader035/viewer/2022062422/56813c88550346895da62829/html5/thumbnails/98.jpg)
98
3 E.
4 D.
5 C.
6 B.
7 A.
.adalah....n nilai maka 12 dx 3)-(2xdan 0 n Bila 3.n
1
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99 4 E.
2 D.
0 C.
2- B.
4- A.
.....dengan samadt g(t)
maka -2,dt g(t)dan 2dt g(t) Bila 4.
2
0
1
0
1
2
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1002- E.
1- D.
0 C.
1 B.
2 A.
.....dx )cos3(sin 5.π
0
xx
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101
C F(X) dx f(x)
: Rumusnya
diketahui (x)F' turunannya
jika F(X) semula fungsi mencari Proses
adalah tak tentuIntegral
: Konklusi / Kesimpulan
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102
alanPengintegr Constanta c 3.
Integran Fungsi f(x) 2.
f(x)(x)F' (bersifat)
UmumIntegral Fungsi F(x) 1.
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103
dx f(x) a dx f(x) a .3
g(x)dx -f(x)dx dx g(x)f(x) .2
g(x)dx f(x)dx dx g(x)f(x) 1.
![Page 104: Nama : Hendrik Pical TTL : Banjar Masin,26-10-1956 Pendidikan : S1](https://reader035.vdocuments.pub/reader035/viewer/2022062422/56813c88550346895da62829/html5/thumbnails/104.jpg)
104
xc
c
c
logln xdengan ,xln a dxx
a 6.
1- n dengan ,x1n
a dxax 5.
1- n dengan ,x1n
1 dxx .4
caxdx a .3
constantaadalah a dx, f(x) a af(x)dx .2
cxdx 1.
e
1nn
1nn
![Page 105: Nama : Hendrik Pical TTL : Banjar Masin,26-10-1956 Pendidikan : S1](https://reader035.vdocuments.pub/reader035/viewer/2022062422/56813c88550346895da62829/html5/thumbnails/105.jpg)
105
Tan x sec xSec x5
-Cosec xCot x4
Sec xTan x3
-Cot x cosec xCosec x6
-Sin xCos x2
Cos xSin x1F’(x)F(x)No.
2
2
![Page 106: Nama : Hendrik Pical TTL : Banjar Masin,26-10-1956 Pendidikan : S1](https://reader035.vdocuments.pub/reader035/viewer/2022062422/56813c88550346895da62829/html5/thumbnails/106.jpg)
106
c x cosec- dx x ccot x.cose 6.
c x sec dx x tan x.sec 5.
c cot x - dx x cosec 4.
c tan x dx xsec .3
c x cos- dx sin x .2
c sin x dx x cos 1.
2
2
![Page 107: Nama : Hendrik Pical TTL : Banjar Masin,26-10-1956 Pendidikan : S1](https://reader035.vdocuments.pub/reader035/viewer/2022062422/56813c88550346895da62829/html5/thumbnails/107.jpg)
107
-acot(ax+b).cosec(ax+b)Cosec(ax+b)6
atan(ax+b).sec(ax+b)Sec(ax+b)5
-acosec (ax+b)Cot(ax+b)4
asec (ax+b)tan(ax+b)3
-asin(ax+b)Cos(ax+b)2
acos(ax+b)Sin(ax+b)1
F’(x)F(x)No
2
2
![Page 108: Nama : Hendrik Pical TTL : Banjar Masin,26-10-1956 Pendidikan : S1](https://reader035.vdocuments.pub/reader035/viewer/2022062422/56813c88550346895da62829/html5/thumbnails/108.jpg)
108
cb)cosec(ax1
dx b)xb).cosec(acot(ax 6.
cb)axsec(a
1 dx b)b).sec(axtan(ax 5.
cb)(axcot1
dx b)(axcosec 4.
cb)(axtana
1 dx b)(axsec 3.
cb)axcos(a
1-dx b)sin(ax 2.
cb)axsin(a
1 dx b)cos(ax 1.
2
2
a
a
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βαCosβαCos2
1- βSin αSin 4.
βαCosβαSin2
1β Cos α Cos .3
βαSinβαSin2
1 Sinβ α Cos .2
βαSinβαSin2
1 β Cos αSin .1
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tepat.
paling yangjawaban didepan Eatau D,C,B,A,
hurufsatu salah pada (X) silang ndaBerilah ta
26) :2ah (Pengkhotb
Nya-hati
anmenyenangk yang orang kepadan kebahagiaa
dann pengetahua hikmat, memberikanAllah
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UJI KOMPETENSI 01 MGMP MATEMATIKA
SD
SMP SM
A
SKKK JAYAPURA
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32x-3x E.
c3x2x-3x D.
c2x-3x C.
c4x9x B.
c3x4x9x A.
...adalah dx 3)4x-(9x dari Hasil 1.
23
23
23
23
23
2
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113cxx
2
1 E.
cxx5
2 D.
cx5
2 C.
cx2
1x
5
2- B.
c8x A.
... dx x
x2x
4
5
5
25
3
4
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11432x-2x E.
32x-2x D.
332x-2x C.
332x-2x B.
332x-2x A.
adalah...
kurvapersamaan maka (2,-1) titik melalui kurva
dan 34x6dx
dyoleh ditentukan f(x) y
kurva pada y) titik(x,singgung garisGradien
23
23
23
23
23
2
x
x
x
x
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cx4
1xx
2
1- E.
cx4
1xx
3
1- D.
cx3
1x xC.
cx4
1xx
3
1 B.
c)xx(x3
1 A.
...adalah dx 2
1 dari Hasil
4
4
4
4
4
3
xx
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4t
22t E.
4t
22t D.
4t
22t C.
8t
22t B.
8t
23t A.
...s(t)adalah ebut jarak ters Rumus
detik). dalam(t meter 8adalah 1saat t pada
ditempuh yangjarak dan t
26t v(t)dirumuskan
yangkecepatan dengan bergerak bendaSuatu
3
33
3
2
2
22
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9 E.
8 D.
7 C.
6 B.
5 A.
... a nilai maka 1 a 24, dx 6)-(4x Diketahuia
1
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6 E.
10 D.
13 C.
16 B.
22 A.
...adalah dx )73(3x dari Nilai2
0
2 x
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119
134 E.
132 D.
130 C.
128 B.
126 A.
...F(4) maka 6 F(1)dan dx xx10 F(x) Jika
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120
4
1- E.
2
1- D.
1- C.
2- B.
4- A.
... a nilai maka 0, adan 6dx 2x)-(3 Bila2
a
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121
320 E.
368 D.
388 C.
374 B.
404 A.
... dx )210(7
3
3 xx
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122
C 10. C 5.
C 9. B 4.
C 8. C 3.
B 7. E 2.
A 6. D .1
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123