Kalkulus-2_Unpam

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h e l O R .BennyWahyuadi ,S.T. ,M M.. ,MBA. I N D : 4N 0 41 01 06 20

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S A T L U K A F TEKNIK A P S A T I S R E V I N U MULANG I R A U N A J 02 11

R A T N A G N E P A T A K Puj idan syukur senantiasapenu ils panjatkan kehadira tAllah SWT . a i n u r a k n a d t a m h a r n a h a p m il s a t a -Nya ,sholawa tdan salamsenantiasa a y n a g r a u l e k n a d W A S h a ll u l u s a R s a t a h a r u c r e t sehingga ditengah -s a t i n i t u r n a d n a k u b i s e k h a g n e t penu ils serta dengan segala k e k urangannya , dapa tdisusun modu lsederhana yang diharapkan . l a r g e t n I s u l u k l a K i r a j a l e p m e m m a l a d a w s i s a h a m u t n a b m e m t a p a d Modu l in i dimaksudkan untuk memberikan beka l kepada a w s i s a h a m Jurusan Teknik Informatika Fakultas Teknik Universitas g n a l u m a P yang sedang mengikut iperku ilahan Kalkulus .Kekurangan s il u n e p ” n a t u t n u t ‘ i d a j n e m i n i l u d o m a y n a n r u p m e s m u l e b n a d n a u h a t e g n e p k a y n a b a m i r e n e m a w s i s a h a m a y n s u r a h e s g n a y a g g n i h e s g n a t n e t Kalkulus Integra l dar i modu l in i belum dapa t terwujud u r u l e s hnya . Terselesaikannya penu ilsan modu l in i tentu tidak terlepas dar i n a k e r n a u t n a b -rekan seprofes id iUnivers tias Pa um lang ,lebih-lebih a r e g e s k u t n u s il u n e p i s a v i t o m i d a j n e m g n a y a w s i s a h a m n a k i a s e l e y n e m modu lsederhanaini .Terima kasih teruntuk rist i ku L ,ia a g u j n a d ibuku yang juga telah memberikan dorongan dan inspiras i . i n i l u d o m n a t a u b m e p a m a l e s g n a j n a p Semoga bahan ajar yang telah dituangkan dalam modu lini ,akan a n a s i d n a f a li h k e k n a d n a g n a r u k e K . a w s i s a h a m i g a b a n u g r e b t a g n a s i n i s Insyaallah diperbaik idikemudianhar.i Pamulang, 3Januar i2011 Penu ils, R .BennyWahyuadi

I S I R A T F A D n a m a l a H l u p m a S n a m a l a H .. .. .. .... .. .... .. .... .. .. .... .. .... .. .. .... .. .... .. .... 1 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . r a t n a g n e P a t a K . 2 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. i s I r a t f a D .... .. .... .. .... .. 3 I b a B PENDAHULUAN 1 . 1 Turunan . ... .. .... .. .. .... .. .... .. .. .... .. .... .. .. .... .. .... .. .... .. 4 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . n a n u r u t it n A 2 . 1 . .. .... .. .... .. .... .. .. .... 9 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. u t n e t r e T l a r g e t n I 3 . 1 ... .. .. .... .. .. 17 I I b a B TEKNIKI NTEGRAL 2 .1TeknikSubstitus i. .. .... .. .... .. .. .... .. .... .. .. .... .. .... .. .... .. 28 2.2I ntegra lFungs iTrigonometri .. .. ... .. .... .. .... .. .. .... .. .. 34 2 .3TeknikSubstitus iFungs iTrigonometri .. .. .... .. .. .... 45 2.4I ntegra lParsial .... .. .... .. .. .... .. .... .. .. .... .. .... .. .. .... .. .... 5 7 2.5I ntegra lFungs iRasional . .... .. .... .. .... .. .. .... ... .. .. .... .. 16 2 .6Integra lFungs iRasiona lyangMemua tFungs i i r t e m o n o g i r T ... . .. .... .. .. .... .. .. .... .. .... .. .... .. .. .... .. .... .. .. .... 73 I b a B I I INTEGRALTIDAKWAJAR 3 . 1 Pengertian . .. .. .... .. .. ... .... .. .. .... .. .... .. .... .. .. .... .. .... .. .. 79 3. 2 Integra lTidakWajardengan BatasDiskontinu . .. 8 1 3 . 3 Integra lTidakWajardengan BatasTak Hingga .. 8 5 b a B VI RUMUS-RUMUSI NTEGRAL .. .... .. .... .. .. .... .. .... .. .... .. .. .... 9 1 V b a B TRANSFORMAS ILAPLACE 5 .1De ifnis iTransforms iLaplace .. .... .. .... .. .. .... .. ... .. .... 1 01 . 5 2Syara tCukupTransformas iLaplaceAda ... .. ... .... .. 61 0 5 .3MetodeTransformas iLaplace . .... .. .... .. .. .... .. .. .... .. 61 0 5 .4Sfiat-s fia tTransformas iLaplace .. .... .. .. .... .. .. .... .. .. 1 08 A K A T S U P R A T F A D .. .... .. .... .. .. .... .. .... .. .. .... .. .... .. .... .. .. .. 1 22

I B A B N A N U R U T I T N A 1 . 1 Turunan t g n a t n e t n a s a h a b m e P urunan tidak dapa t dipisahkan dar i , i s g n u f g n a t n e t n a it r e g n e p baik f ungs ieksp ilsi tmaupun f ungsii mp ilsit . i s g n u f h a l a d a t i s il p s k e i s g n u F yang secara umum penuilsannya h a l a d a t i s il p m i i s g n u f n a k g n a d e s , ) x ( f = y k u t n e b m a l a d n a k a t a y n i d i s g n u f yang secara umum penu ilsannya dinyatakan dalam bentuk . 0 = ) y , x ( f . i n i h a w a b i d i s g n u f h o t n o c a p a r e b e b n a k i t a h r e P . 1 y=2- 2 3x . 2 y=3_x2 _4_x3 . 3 y= x x x . 4 x2 _y+ 2 _{– 25=0} . 5 yx 2 _x+ 2_{y – 2=0} . 6 x2 _{– 2x+y}2 _{+4y– 5=0} , t i s il p s k e i s g n u f h a l a d a 3 n a d , 2 , 1 o n i s g n u f , s a t a i d h o t n o c a d a P g n a y i s g n u f a u m e S . ti s il p m i i s g n u f h a l a d a 6 n a d , 5 , 4 h o t n o c n a k g n a d e s m a l a d a y n n a s il u n e p h a b u i d t a p a d t i s il p s k e k u t n e b m a l a d s il u t i d d g n a y i s g n u f a u m e s k a d it i p a t e t n a k a , t i s il p m i k u t n e b ituils dalam n a k i t a h r e P . t i s il p s k e k u t n e b m a l a d h a b u i d t a p a d t i s il p m i k u t n e b i s g n u f i r a d a y n t u j n a l e S . s a t a i d 5 h o t n o c -fungs i tersebut , dapa t . a y n n a n u r u t n a k u t n e t i d i s i n if e D n a g n e d n a k i s a t o n i d g n a y n i a l i s g n u f h a l a d a ) x ( f = y i s g n u f n a n u r u T d i d n a d ) x ( ’f e ifnisikanoleh

= ) x ( ’f x x f x x f x '  '  o ' ) ( ) ( m il ₀ ,asalkan ilmitnyaada. + x ( l a s i M 'x)= t ,maka 'x = – x t a n e r a K 'xo0maka t ox n i a l k u t n e b m a l a d n a k a t a y n i d t a p a d s a t a i d n a n u r u t i s i n if e d a g g n i h e S = ) x ( ’f x t x f t f x t   o ) ( ) ( m il ,asalkan ilmitnyaada. n a g n e d n a k a t a y n i d ) x ( f = y n a n u r u t k u t n u n i a l i s a t o N ,D f(x) x d y d x ,df_d(_xx) . , t i s il p m i k u t n e b m a l a d n a k a t a y n i d i u h a t e k i d g n a y i s g n u f a k i J h a d i a k n a k a n u g g n e m n a g n e d n a k u k a li d t a p a d a y n n a n u r u t a k a m g n i s a m n a k l a i s n e r e fi d n e m a r a c n a g n e d u t i a y l a i s n e r e ff i d -masing . t u b e s r e t i s g u f m a l a d l e b a i r a v Beriku tin idiberikan beberapa contoh n e n e m tukant urunanf ungs ieksp ilsi tdani mp ilsit . h o t n o C n a k u t n e T x d y d _{fungsi-fungs iberiku.t} . 1 y= x C+ Berdasarkan de ifnis id iatasdiperoleh x x f x x f x d y d x '  '  o ' ) ( ) ( m il 0 = x x x x x '  '  o 'ilm0 = x x x x x '  '  o 'ilm0 . x x x x x x  '   '  = 0 m il o 'x _{{ }} ) ( ) ( x x x x x x x  '  '  '  =

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x x x x x x _{' ' } ' o 'ilm0 = x x x xo _{' } ' 1 m il 0

= x 2 1 . 2 y= ) 1 ( 3 x  Berdasarkande ifnis id iatasdiperoleh x x f x x f x d y d x '  '  o ' ) ( ) ( m il 0 = x x x x x '   '   o ' 1 3 ) 1 ( 3 m il 0 = } ) 1 ( ) 1 ( { ) 1 ( 3 ) 1 ( 3 m il 0 x x x x x x x x '   ' '     o ' = ) 1 ( ) 1 1 ( 3 m il 0 x x x   '  o ' = ₂ ) 1 ( 3 x   i s g n u F -fungs iyang mempunya iturunan sebagaimana d jielaskan fi d g n a y i s g n u f t u b e s i d s a t a i d h o t n o c a d a p ferensiable (dapa t . ) n a k n u r u t i d x = y a k ij , a m a s g n a y a r a c n a g n e D n _{makat urunannyaditentukan oleh :} x x f x x f x d y d x '  '  o ' ) ( ) ( m il 0 = x x x x m i L n n x '  '  o ' ) ( 0 = x x x x x n n n x x n n x x n xn n n n n n x '  '   '    '   '     o ' ) ( .. . ) ( ! 3 ) 2 ( ) 1 ( ) ( ! 2 ) 1 ( m il 3 3 2 2 1 0 = x x x x n n n x x n n x x n n n n n x ' '   '    '   '    o ' ) ( .. .. ) ( ! 3 ) 2 ( ) 1 ( ) ( ! 2 ) 1 ( m il 3 3 2 2 1 0 = ( ) .... ( ) ] ! 3 ) 2 ( ) 1 ( ) ( ! 2 ) 1 ( [ m il 1 2 3 2 1 0     o ' '   '    '   n n n n x x x x n n n x x n n x n = nxn1

. 3 x2_y2_25 0 g n i s a m n a k l a i s n e r e fi d n e m n a g n e D -masingvariabel ,diperoleh: x ( d 2) _+d(y2_{) - d(25)=d(0)} 0 2 2  œ xdx ydy œx+y x d y d _{= 0} œ y x x d y d _ . 4 Tentukan x d y d _{dari x}2_y+xy2_{– 2=0} œd(x2_y)+d(xy2_{) – d(2) =d(0)} œ(x2_{dy+2xydx)+(2xydy+y}2_dx)=0 œ(2xy+y2_{)dx+(2xy +x}2_)dy=0 œ x d y d _{= -} 2 2 2 2 x y x y y x   ) x ( w = w n a d , ) x ( v = v , ) x ( u = u l a s i m , m u m u a r a c e S adalah g n a y i s g n u f masing-masing dapa tditurunkandan csebarangb liangan n a n u r u t i s i n if e d n a k a n u g g n e m n a g n e d a k a m , l a e r dapa tditentukan . t u k i r e b i a g a b e s i s g n u f n a n u r u t m u m u s u m u r a p a r e b e b . 1 x d d _(c)=0 . 2 x d d _(x)=1 . 3 x d d x ( n_)=nxn-1 . 4 x d d _(un_)=nun- 1 x d d _{( u)} . 5 x d d _(u+v)= x d d _(u)+ x d d _{( v)} . 6 x d d _{( - u v)=} x d d _(u)- x d d _{( v)}

. 7 x d d _{(u rv r w r .. .)=} x d d _{(u) r} x d d _{(v) r} x d d _{(w) r ...} . 8 x d d _(cu)=c x d d _{( u)} . 9 x d d u = ) v u ( x d d v + ) v ( x d d ) u ( . 0 1 x d d _(uvw)=uv x d d _(w)+uw x d d _(v)+vw x d d _{( u)} . 1 1 x d d ₍ v u₎₌ 2 v dx v d u x d u d v  t a fi s i t k u B -s fia t d i atas diserahkan kepada pembaca sebaga i . n a h i t a l Selanjutnya ,denganmenggunakande ifnisit urunan x d y d ₌ x x f x x f x '  '  o ' ) ( ) ( m il ₀ ,dapa tditunjukkan beberapa turunan fungs i .i n i h a w a b i d i r t e m o e g y =cosx ,maka x d y d = x x f x x f x '  '  o ' ) ( ) ( m il 0 = x x x x x '  '  o ' s o c ) ( s o c m il 0 = x x x x x x x x '  '   '   o ' 2 ) ( n i s 2 ) ( n i s 2 m il 0 = 2 n i s 2 ) 2 ( n i s 2 m il 0 x x x x x ' ' '   o ' = - sinx. : n i a l g n a y i r t e m o n o g i r t i s g n u f n a n u r u t h e l o r e p i d , g o l a n A . 1 x d d _(sinx)=cosx . 2 x d d _(cosx)=-sinx

. 3 x d d _{(tan x)=sec}2_x . 4 x d d _{(co tx)=- ccs} 2_x . 5 x d d x n a t x c e s = ) x c e s ( . 6 x d d _{(cscx)=-cscxco tx} n a n u r u t i t n A 2 . 1 n a n u r u t i t n A merupakan bailkan dar iturunan , sehingga untuk . i s g n u f n a n u r u t n a g n e d n a k t i a k i d s u r a h a y n i r a j a l e p m e m = y a k ij , n a n u r u t i s i n if e d t u r u n e M x maka x x d y d 2 1 _. h e l o r e p i d , a m a s g n a y a r a c n a g n e D . 1 Jikay= x+3maka x x d y d 2 1 _. . 2 Jikay= x - 3maka x x d y d 2 1 _. . 3 Jikay= x - 100maka x x d y d 2 1 . 4 Jikay= x + 7 1 _maka x x dy d 2 1 _{,danseterusnya .} = y k u t n u , n i a l a t a k n a g n e D x +C ,CR maka x x d y d 2 1 _. n a s il u n e p a k a m , n a n u r u t i r a d n a k il a b n a k a p u r e m n a n u r u t i t n a a n e r a K A n a g n e d n a k a n a h r e d e s i d t a p a d s a t a i d k u t n e b x ¸¸ ¹ · ¨¨ © § x 2 1 _{= x C .} = y i s g n u f a w h a b i t r a r e b i n i l a H x C ,dengan C Rmempunya i n a n u r u t x x d y d 2 1 _.

= ) x ( f i r a d n a n u r u t i t n a u a t a _¸¸ ¹ · ¨¨ © § x 2 1 _{adalahF(x)= x +C ,CR .} i s g n u F -fungs iyangdapa tditentukanantiturunannyadisebuti ntegrable . ) n a k l a r g e t n i r e t ( A k u t n e b , m u m u h i b e l g n a y l a h m a l a D x ¸¸ ¹ · ¨¨ © § x 2 1 _{= x C .} n a g n e d n a k a t a y n i d

³

_¸¸ ¹ · ¨¨ © § x 2 1 _{dx= x C .} a y n n a n u r u t i t n a n a d ) x ( f = y l a s i m , i d a J adalahF(x)+C ,maka

³

f(x)dx =F(x)+C ,C Real . Bentuk

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f(x) dx = F(x) + C ,f(x) disebu tintegran dan F(x) + C . n a n u r u t i t n a t u b e s i d . 1 a m e r o e T a k i J rsebarangbliangan rasiona lkecual i-1, maka:

³

   C r x x d xr r 1 1 . a k ij a y n t a b i k A r = -1maka

_³

_xr_dx

_³

_x1_dx =

_³

dx x 1 _{=l n x C} i t k u B k u t n e b r e b g n a y l i s a h u t a u s n a k g n a b m e g n e m k u t n U

³

f(x)dx =F(x)+C ,C Rea.l a w h a b n a k k u j n u n e m p u k u c a t i K ) ( ] ) ( [F x C f x D_x  s a t a i d s u s a k m a l a D r r r x _rx C _r n x x D _«¬ª  _»¼º  » ¼ º « ¬ ª _   ) 1 ( 1 1 1 1

2 a m e r o e T i s g n u f ) x ( g n a d ) x ( f l a s i M -fungs iyang integrable dan C sebarang : a k a m a t n a t s n o k . 1

_³

Cf(x)dx =C

_³

f )(x dx, . 2

³

[f(x)g(x)]dx

³

f(x)dx

³

g(x)dx, . 3

_³

[f(x)g(x)]dx

_³

f(x)dx

_³

g(x)dx, i t k u B n a g n e d p u k u c , s a t a i d a m e r o e t n a k i t k u b m e m k u t n U h e l o r e p m e m a t i k a w h a b i t a m a n a d n a n a k s a u r n a k l a i s n e r e f e d n e m .i r i k s a u r i r a d n a r g e t n i . 1 Dx {C

³

f )(x dx }=CDx {

³

f )(x dx} =Cf(x) . 2 Dx{

³

f(x)dx

³

g(x)dx }=Dx

³

f(x)dxDx

³

g(x)dx =f (x)+g(x) . 3 Dx{

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f(x)dx

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g(x)dx }=Dx

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f(x)dxDx

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g(x)dx =f (x)- g(x) h o t n o C . s a t a i d l a r g e t n i t a fi s n a k r a s a d r e b t u k i r e b l a r g e t n i n a k u t n e T . 1

_³



x2 x



dx b a w a J

³



x2x



dx₌

_³

_x2_dx

_³

_xdx = 1₃x3C1₂1x2 C2 = x3 x2 C 2 1 3 1

. 2 dx x x2 ₁ 2

³

¨¨_©§  ¸¸_¹· b a w a J dx x x2 ₁ 2

³

¨¨_©§  _¹·¸¸ =

³

x 2_xx 1dx 2 4 =

³



³



³

dx x x d x x x d x x4 2 2 1 =

_³

_x7/2_dx2

_³

_x3/2_dx

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_x1/2_dx . 3 dx x x x

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3 2 ) 1 ( b a w a J x d x x x

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3 2 ) 1 ( ₌

³

3  2 _{2 1)} ( x x x x _{d x} = dx x x x d x x x d x x

³

³

³

₃ 3 2 ₃ 2  ₃ =

³

x8/3dx₂

³

x5/3dx

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x2/3dx = x11/3 x8/3 x5/3C 5 3 4 3 1 1 3 3 a m e r o e T

³

sinxdx=- cosx+C ,C Real

³

cos xdx=sinx+C ,c Real i t k u B n a k k u j n u n e m n a g n e d p u k u c s a t a i d a m e r o e t n a k i t k u b m e m k u t n U a w h a b Dx(cosx) sinx dan Dx(sinx) cosx. 4 a m e r o e T g n a y l a n o i s a R n a g n a li b n n a d e l b a i s n e r e ff i d g n a y i s g n u f ) x ( f n a k i a d n A n a k u b -1 ,maka:

>

@

>

@

³

   , 1 ) ( ) (' ) ( 1 C r x f x d x f x f r r _{C Real .} h o t n o C . 1

_³

3x 4x2 11dx b a w a J Karena _D (4_x2_11) x =6xdx ,sehinggaberdasarkan t eoremad iatas

³

3x 4x2 11dx ₌

_³

_{4 11} 2 1 _x2 _d(6x) = x  C 2 / 3 ) 1 1 4 ( 2 1 2 3/2 = ₍₄ 2 ₁₁₎3/2 3 1 _{x  +C.} . 2

³

 dy y y 5 2 3 2 Jawab KarenaDx(2y25) =4ydy ,maka

_³

 dy y y 5 2 3 2 =

³

  ydy y 5) 3 2 ( 2 1/2 =

_³

_{y }  _4ydy 4 3 ) 5 2 ( 2 1/2 =

³

_{(2y 5)} _.4ydy 4 3 2 1/2 = y  C 2 / 1 ) 5 2 ( . 4 3 2 1/2 = 2y 5C 2 3 2 . 3

³

3sin(6x2)dx Jawab Misal U=6x+2 Ÿ dU=6dxatau3dx= 2 U d _,sehingga

³

3sin(6x2)dx =

³

2 n i s U dU

= (cosU ) C 2 1 =  cos(6x2)C 2 1 . 4

_³

1cosx nsi xdx Jawab Misa lA= 1cosx Ÿ A2 1_cosx 2AdA=(-sinx)dx ,sehingga:

³

1cosx nsi xdx =

³

A(.2A)dA = - 2

_³

A2dA =  A 3 C 3 2 =  ₍₁_cosA ₎3 C 3 2 . u t n e t k a t l a r g e t n i r a s a d s u m u r a p a r e b e B . 1

_³

dx=x+C ,C Real . 2

_³

f(x)dx=F(x)+C ,C Real . 1 xr _dx= 1 1  r x r+1_{+C ,C Real ,r z- 1} . 2

_³

(u+v)dx=

_³

udx+

_³

vdx . 3

³

audu= a

³

udu . 4 x 1 _{dx=l n |x| +C=} e_{log│x│+C ,C  Real} . 5

_³

au_du= a a n l +C ,C Real . 6

_³

eu _du=eu_{+C ,C Real} . 7

_³

tanxdx=l n|secx|+C ,C Real . 8

³

secxdx=l n|secx +t anx|+C ,C Real

³

³

. 9

³

co txdx=l n|sinx|+C ,C Real . 0 1

_³

cssxdx=l n|cscx– co tx|+C ,C Real . 1 1

³

sec2_{xdx=t anx +C ,C Real} . 2 1

³

csc2_{xdx=- co tx+C ,C Real} . 3 1

³

secxt anxdx=secx +C ,C Real . 4 1

_³

cscxco txdx=-cscx+C ,C Real . 5 1

³

cosm_xdx= n n nx x n _{sin 1} s o c 1 _  

³

cos m-2_xdx . 6 1

³

sinm_xdx= n n n x x n _{cos 1} n i s 1 _   

³

sinm-2_xdx . 7 1

_³

udv=uv-

_³

vdu . 8 1

³

_{2 2} a x x d  = 2a 1 _ln a x a x   C , C +  Real . 9 1

_³

_{2 2} x a x d  = 2a 1 _ln a x a x   C , C +  Real . 0 2

³

_{2 2} x a x d  =arcsin a x _{+ C} . 1 2

_³

_{2 2} a x x d  = a 1 _{arct gn} a x _{+ C} . 2 2

³

_{2 2} a x x x d  = a 1 _arcsec a x _+C . 3 2

³

_{x }2 _a2 _dx= 2 1_{u x }2 _a2 ₊ 2 1_a2_{Ln(u+ x }2 _u2 _)+C . 4 2

³

_{x }2 _a2 _dx= 2 1_{u x }2 _a2 _- 2 1_a2_{Ln(u+ x }2 _u2 _)+C . 5 2

_³

_{x }2 _a2 _dx= 2 1_{u x }2 _a2 ₊ 2 1_a2_{Ln(u+ x }2 _u2 _)+C . 0 3

³

 2 2 _a x x d h n i s c r a = a x C +

. 1 3

³

 2 2 _a x x d _=arccosh a x_+C . 2 3

_³

_um_eau_du= m au_

_³

_um_eau a m e u a 1 1 _{d u} l a o S -soa l .t u k i r e b l a r g e t n i n a k u t n e T 1

_³

(x 23)3 2dx 2

_³

_(Sx3 ₁₎4_3Sx2dx 3

_³

₍₅x2₁₎₍₅x3 ₃x₈₎6dx 4

_³

(5x21) 5x3 3x8dx 5 x x dx x x _{2 1)} 1 2 ( 2   

³

6 dx x x x x 5 2 3 ) 5 2 ( 2 2 / 3 3    

³

7

_³

  x d x x x 1 1 4 2 4 8

_³

dx x 3 s o c 1 2 9

_³

_sin3_[(x2 ₁₎4_]cos[(x2 ₁₎4₍x2₁₎3xdx 0 1 Andaikan u=sin{(x2_1)4_} 1 1 Tentukan

_³

_sin2xdx 2 1

_³

6sin[3(x2)]dx 3 1

_³

¸ ¹ · ¨ © §x _dx 6 n i s 3 4 1

_³

(x2cos2xxsin2x)dx

3 . 1 INTEGRALTERTENTU : i s i n i f e D f l a s i M (x)suatufungs iyangdide ifnisikanpada[a,b] ,selanjutnyaf( x) n a k l a r g e t n i r e t n a k a t a k i d (integrable)pada [a,b] a k ij

¦

o ' n i i i Pilm0 ₁f(x ) x aad . Selanjutnya b

_³

a f )(x dx f ) n n a m e i R l a r g e t n I ( u t n e T l a r g e t n I t u b e s i d ( x) n a k i s i n if e d i d n a d , b e k a i r a d

³

b a f )(x dx =

¦

o ' n i i i Pilm0 ₁f(x ) x .

³

b af )(x dx a v r u k a r a t n a i d p u k a c r e t g n a y h a r e a d s a u l n a k a t a y n e m a k ij , ] b , a [ g n a l e s m a l a d x u b m u s n a d ) x ( f = y b

_³

a f )(x dxbertandanegati f . x u b m u s h a w a b i d a d a r e b g n a y h a r e a d s a u l n a k a t a y n e m a k a m : i s i n i f e D . 1 a

_³

a f )(x dx 0= . 2 b

_³

a f )(x dx = - a

_³

b f )(x dx b > a ,

s u l u k l a K r a s a D a m e r o e T a m e r o e T dasarKalkulusmemberikan kemudahanuntukmenghitung : t u b e s r e t a m e r o e t t u k i r e b , u t n e T l a r g e t n I f l a s i M ( x) kontinupada[a,b]danF( x) sebarangantit urunanf (x) , a k a m b

_³

a f )(x dx ) b ( F = – F(a) ) b ( F s il u t i d a y n t u j n a l e S – F(a)= [F ](x) b_a : h o t n o C . 1 Per ilhatkanbahwa ijkar Qdanrz -1 ,maka 1 1 1 1     

³

x dx b_rr a_rr b a r : b a w a J = ) x ( F a n e r a K 1 1   r xr x = ) x ( f i r a d n a n u r u t i t n a u t a u s r _,makamenurut s u l u k l a K r a s a d a m e r o e t 1 1 ) ( ) ( 1 1      

³

x dx F b F a b_rr a_rr b a r : t a fi s r e b u t i a y , r a e n il r o t a r e p o i a g a b e s u t n e t l a r g e t n I g n a d ) x ( f l a s i M ( x) terintegralkanpada[a,b]danksuatukonstanta , : a k a m . 1 b

_³

akf(x)dx k

³

b a

f )

(

x

d

x

. 2

f

x

g

x

d

x

b a

³

[

(

)

(

)

]



=

³

b a

f )

(

x

d

x

+ b

_³

ag )(x dx

: h o t n o C g n u t i H 2(4x 6x )dx 1 2

³

  : b a w a J x d x x d x x d x x

_³

_³

³

     2 1 2 2 1 2 1 2_{) 4 6} 6 4 ( 4= 2 1 3 2 1 2 3 6 2 _{ »}»_¼ º « « ¬ ª  » » ¼ º « « ¬ ª_{x x} = 4 ¸ ¹ · ¨ © §   ¸ ¹ · ¨ © §  3 1 3 8 6 2 1 2 4 =  21 t a f i S -Sifa tIntegra lTentu . 1 Sfia tPenambahan Selang : a m e r o e T f a k i J ( x) terintegralkan padasuatuselangyangmemua ttigat i itka , a k a m , c n a d b x d x f c a

³

( ) = b f xdx a

³

( ) + c f x dx b

³

( ) . c n a d b , a n a t u r u n u p a n a m i a g a b : h o t n o C . 1

_³

x dx

_³

x dx2

_³

x dx 1 2 1 0 2 2 0 2 _2.

_³

_{x dx}

_³

_{x dx}2

_³

_{x dx} 3 2 3 0 2 2 0 2 . 3

_³

x dx

_³

x dx

_³

x dx    2 1 2 1 0 2 2 0 2 . 2 Sfia tSimetri f a k i J (x) f ungs igenap ,yaitusuatuf ungs iyangmemenuh is fia t ( f -x)=f (x) ,maka: x d x f a a

³

 ( ) 2 = a

_³

f x dx 0 ( ) nd a

f a k i J (x) fungs iganj li ,yatiusuatuf ungs iyangmemenuh isfia t ( f -x)=- f(x), maka x d x f a a

³

 ( ) . 0 = : h o t n o C . 1

_³

_³

¸ ¹ · ¨ © § ¸ ¹ · ¨ © §  S S Scos 4 20cos 4 dx x x d x _{4 2} 4 1 . 4 s o c 8 0

³

¸¹ · ¨ © § S x d x . 2 dx x x

³

  5 5 2 5 4 0= , m u m u h i b e l a r a c e S s fiat-s fiati ntegralt ertentu adalah: a d a p u n i t n o k ) x ( g n a d ) x ( f a k i J interva l[a,b]dank Rea ldanf (x) ,g(x) : a k a m , t u b e s r e t l a v r e t n i a d a p n a k l a r g e t n i r e t . 1

_³

_ab

_³

b a x d x f k x d x f k ( ) ( ) . 2 f x g x dx f x dx bg xdx a b a b a

³

³

³

[ ( ) ( )] ( )  ( ) . 3 [f(x) g(x)]dx f(x)dx bg(x)dx, a b a b a

³

³

³

  . 4

_³

a ( ) 0 a x d x f . 5

_³



_³

a b b a x d x f x d x f( ) ( ) ,ijkab<a . 6

³

b a x d x f )(

_³



_³

b c c a x d x f x d x f( ) ( ) ,c( ba, ) . 7

_³

( ) 0,  a a x f ijkaf (-x)=-f(x) . 8

³

 a a x d x f )( =2

³

a f x dx 0 ) ( ,ijkaf (-x)=f (x)

. 9 JikaF(u)=

³

b a x d x f )( ,maka F(u) f(u) u d d . 0 1

_³

b a x d x f )( =(b- )a f(xo) untukpailngsediki tx=xo antaraadanb . . 1 1

³

d

³

b a b a x d x g x d x f( ) ( ) ijkadanhanya ijkaf (x) d g(x) untukse itap x . ] b , a [ . 2 1 D x f(t)dt f(x) a x » ¼ º « ¬ ª

³

h o t n o C l a r g e t n i li s a h n a k u t n e T . 1

_³

2 xdx 0 ) 2 ( b a w a J

_³

2 xdx 0 ) 2 ( = 2 0 2 2 2 _» ¼ º « ¬ ª  x x = _» ¼ º « ¬ ª   » ¼ º « ¬ ª  2 0 0 . 2 2 2 2 . 2 2 2 =(4+2)– (0+0)= 6 . 2

³

2  0 3 2_{(x 1)dx} x b a w a J n l a s i M ya u=(x3_1₎ du=3x2_{d x} du x2dx 3 : a g g n i h e s , 9 = u a k a m 2 = x k u t n u n a d 1 = u a k a m 0 = x k u t n U

³

2  0 3 2_{(x 1)dx} x =

³

9 1 3 u d u

= 9 1 2 6 »¼ º « ¬ ªu = _«¬ª  _»¼º 6 1 6 1 9 = 6 0 9 . 3

³

4  1 ) 1 ( u udu b a w a J = p l a s i M u œp2 _u= œ2pdp=du 1 = p a k a m 1 = u k u t n U : a g g n i h e s , 2 = p a k a m 4 = u k u t n U . 4

_³

4  1 ) 1 ( u udu =

³

2  1 2_{) .2} 1 ( p p pdp =

_³

2  1 3 2 _{2 )} 2 ( p p dp = 2 1 4 3 4 2 3 2 »¼ º «¬ ª _{p  p} = ª_«¬ 3 4º_»¼ª_«¬ 3 ₍₁₎4º_»¼ 4 2 ) 1 ( 3 2 ) 2 ( 4 2 ) 2 ( 3 2 = _«¬ª  _»¼º_«¬ª  _»¼º 4 2 3 2 8 3 6 1 = 4 0 3 3 4 1  = 4 1 3  . 5

³

 8 4 x2 15 x d x b a w a J

= A l a s i M _x2_15 _œA2 _x2_15 œ2AdA=2xdx 1 = A a k a m 4 = x k u t n U a g g n i h e s , 7 = A a k a m 8 = x k u t n U

³

 8 4 x2 15 x d x ₌

³

7 1 A A d A =

³

7 1 A d = [A] 7 1 = – 1 7 = 6 . 6

³

 0 1 625 x2 x d ₌ 10 6 5 5 n l 5 . 2 1   x x = 5 6 5 6 n l 0 1 1 5 0 1 5 0 1 n l 0 1 1      = ln11 0 1 1 3 n l 0 1 1 _ . 7 Tentukan

³

b a x d x f )( = ) x ( f n a g n e d ° ¯ ° ® ­ ! d  d d 2 , 2 1 , 2 1 0 , 2 x k u t n u x x k u t n u x k u t n u x a k i a s e l e s i d t a p a d s a t a i d l a o S n denganmenggunakans fia t

³

b a x d x f )(

_³



_³

b c c a x d x f x d x f( ) ( ) ,c( ba, ) sehingga:

³

b a x d x f )(

³



³

2 

³

1 5 2 1 0 2 2xdx dx xdx

=

> @

2 1₀

> @

₁2 5₂ 2 2 _«¬ª _»¼º  x x x = -(1 0)+(4-2)+

>

5/21

@

1 = 2 9 . 8

³

 3 3 x xd Menuru tdeifnisif ungs ihargamutlak ,bentukd iatasdapa t n a g n e d n a k a t a y n i d

³

 3 3 x dx=

³

3 0 x dx +

³

  0 3 x d x. = 0 3 2 3 0 2 2 2 »_¼ º « ¬ ª   » ¼ º « ¬ ªx x = (8/3– 0 – ) ( – 0 8/3) = 3 6 1 i s g n u f n a l a r g e t n i g n e p l i s a h n a k u t n e t , s a t a i d h o t n o c n a k r a s a d r e B -:i n i t u k i r e b i s g n u f . 1

³

8  1 3 1 xdx . 2

_³

x_(1 x₎2dx =

_³

x

>

12 xx

@

dx =

³

(x2x xx2)dx _{,dengans fiati ntegra ldiperoleh} =

³

xdx -

³

2x x dx+

³

x2dx = 2 2 3 3 5 1 2 3 1 ) 5 2 ( 2 2 1_{x C  x C  x C} = 2 3 1 2 3 5 2 3 1 ) 5 2 ( 2 2 1_{x  x  x C C C}

= x2 x₂5  x3C 3 1 ) 5 2 ( 2 2 1 h a m u r i d n a h i t a L . 2 dz z z

³

( 21)2 . 3

_³

 ds s s s 3 2 ) 1 ( . 4

_³

₍x ₂x₎3dx . 3

³

  1 1 2 2 _{4 x dx} x 2 = 1 2 2 0 4 x x 

³

xd l a s i M _{4 x} 2 _u= 4-x2_=u2 _ataux2 _{= - u4} 2 -2xdx=2uduataudx= du x u  . 4

_³

  2 2 2 4 x dx . 5

³

 3 0 1 x x d . 6

_³

4  2 2 6 1 _dx x x . 7

³

 7 2 8 x x1/3 x d . 8

³

S 2 0 2 n i s xdx

. 9

_³

/3 0 2_sin3 S x d x x . 0 1

_³

 2 / 0 3 cos2 S x x d . 1 1 1

_³

1  3 3 2x dx . 2 1

³

  9 41 1 x x _{d x} . 3 1 x ex dx

³

2 0 3 2 . 4 1

³

/4 6 / sin2 S S x x d . 5 1

³

    2 1 x2 2x 2 x d . 6 1

³

    2 1 2( 1) ) 1 ( _dx x x x . 7 1 

_³

   2 1 ( 2)2 ) 2 ( x x x d x . 8 1

³

2   1 2 ₁₎ ( n l x x dx . 9 1

_³

 4 / 0 2 sin S x x d . 0 2 

_³

    1 2 2 4 3 ) 1 ( _dx x x x . 1 2

_³

4



 



0 1 2x dx x . 2 2

_³

  3 1 3 2 3 1 xd x x x . 3 2

³

a







a x d x a 8 3 3 / 1 3 / 1

. 4 2

_³

/2 0 2_{3 sin3} s o c S x d x x . 5 2 /2sin 3xcos3xdx 0 2

³

S . 6 2 Hitunglah

³

b a x d x f )( , ijka: . a f(x)= ¯ ® ­ d    d d 2 1 , 2 ) 1 ( 2 1 0 , 2 x k u t n u x x k u t n u x . b f(x)= °¯ ° ® ­ d d  d d  2 1 , 1 1 0 , 1 2 x k u t n u x x k u t n u x . c f(x)= °¯ ° ® ­ d    d d    2 0 , 2 2 0 2 , 1 2 x k u t n u x x k u t n u x . d f(x)= x2 untuk -4dxd4 . e f(x)= x

> @

x ,untuk-1dxd2 .f f(x)=(x-

> @

x )2 . g f(x)=x2

> @

x _{,untuk-1d xd2}

I I B A B L A R G E T N I K I N K E T k u t n u n a k a n u g i d n a l a r g r e t n i g n e p k i n k e t m a c a m a p a r e b e B n a n u r u t i t n a n a k u t n e n e m suatu fungsi . Ha l in i bertujuan untuk g n a y i s g n u f l a r g e t n i n a i a s e l e s n a k u t n e n e m m a l a d n a k h a d u m e m h e l o i m a h a p i d h a d u m n a l a r g e t g n i g n e p k i n k e t r a g A . n a k u t n e t i d n a l a r g e t n i g n e p k i n k e t n a k i c n i r i d i n i b a b m a l a d a k a m , a c a b m e p t a r a y s n a g n e d d u s k a m i d -syara t yang ditentukan . Teknik-teknik : h a l a d a t u b e s r e t l a r g e t n i Teknik Substitusi , Integra l Fungs i , i r t e m o n o g i r T Teknik Substitus iFungs iTrigonometri ,Integra lParsial , l a r g e t n I Fungs iRasional ,dan Integra lFungs iRasiona lyang memua t .i r e t m o n o g i r T i s g n u f k i n k e t n a s a l e j n e p i n i t u k i r e B -teknikdalampengintegralan . 1 . 2 TeknikSubstitus i k i n k e t k u t n u n i a l h a li t s I substitus i adalah pemisalan . Teknik p i s u ti t s b u s ada umumnya digunakan untuk memudahkan selesaian ; u t i a y , u t n e t k a t l a r g e t n i s u m u r r a s a d s u m u r k u t n e b e k l a r g e t n i . a

³

_xn_dx= 1 1   n xn n n a k l a s a , C + z -1atau . b

>

f₍x₎

@

nf _('x₎dx

³

=

>

@

1 ) ( 1   n x f n n n a k l a s a , C + z - 1 m u m u n a m o d e p h a l a d a s a t a i d s u m u r a n e r a K . makai ntegrannya . s a t a i d s u m u r n a g n e d n a k i a u s e y n e m Jika belum sesua i atau h a b u i d n i k g n u m t a p a d e s a k a m s a t a i d k u t n e b i r a d g n a p m i y n e m n a g n e D . u l u h a d h i b e l r e t demikian setelah integran sesua i dengan n a k i s a k il p a g n e m n a g n e d n a k u k a li d t a p a d a y n l a r g e t n i u k a b k u t n e b u t n e t k a d it l a r g e t n i r a s a d s u m u r . Akhirnya selesaiannya dapa t . i s u t it s b u s e d o t e m n a g n e d n a k u k a li d

: t u k i r e b h o t n o c a p a r e b e b n a k i t a h r e P . 1

³

1x xd = u l a s i M 1x x u  œ 2 1 ) 1 ( ) (_u2 _{d x} d  œ x d u d u  œ2 e k r i h k a r e t k u t n e b i s u t i t s b u S

_³

1x dx ,diperoleh

³

u(2u)du= -2

³

u2du t a p a d i d r a s a d s u m u r n a g n e D

³

1x d x = -2

³

u2du = -2 u _» C ¼ º « ¬ ª 3 3 = - ₍₁x ₎3 C 3 2 . 2

³

₍₃x₁₂₎11dx A l a s i M =3x+12 d(A) =d(3x+12) d A =3dx d x = 3 A d a g g n i h e S

³

₍₃x₁₂₎11dx ₌

_³

3 1 1 dA A =

³

A11dA 3 1 = A ) C 2 1 ( 3 1 12 = A 12 C 6 3 1 = x C 6 3 ) 2 1 3 ( 12

. 3

_³

Cos22x _xd A l a s i M =2x d(A)=d(2x) d A =2dx d x = 2 A d x s o C 22

³

d x = 2 s o c 2_AdA

³

=

_³

Cos2A _dA 2 1 =

³

_cos2AdA 2 1 =

³

 AdA 2 2 s o c 1 2 1 =

³

dA

³

cos2AdA 4 1 4 1 = A AC 8 2 n i s 4 = x xC 8 4 n i s 4 2 = x xC 8 4 n i s 2 . 4

_³

4x2 4x _(4x+2)dx Jawab Misal A= 4x2_4x A2 _=4x2_{ x4} 2AdA=(8x+4)dx 2AdA=2(4x+2)dx AdA =(4x+2)dx Sehingga

_³

4x24x _(4x+2)dx=

_³

_A.AdA =

³

A2dA = A 3 C 3 1 = 3 ₄ 2 ₄ 3 1 _{x  x +C} . 5

³

4 3t t d t b a w a J = P l a s i M 3 t 4 P2_{=3 t+4 œ t=} 3 4 2_ P d(P2_)=d(3t+4) 2Pdp=3d tœd t= Pdp 3 2 _,sehingga

³

_3ttd_t ₄ =

³

 p p d p P ₎ 3 2 ( ) 3 4 ( 2 =

³

(2P 8)dp 9 1 2 . 6

³

 2 2 6 1 x x d x b a w a J Misa lU= _16x2 U2 _{=16- x}2 _œx2_{=16- U}2 d(U2_{) =d(16- x}2₎ 2Udu = -( 2x)dx d x = du x U 

³

 2 2 6 1 x x d x ₌

_³

¸¹ · ¨ © §  u x u u ) 6 1 ( 2 u d = du x u

³

  16 2 = -

³

u du x (16 ) 1 2 = 2 3 1 ₃ 6 1 _C x u C x u_{  }  = C x x x x x _{ }  _   3 6 1 ) 6 1 ( 6 1 6 1 2 2 2 = C x x x x _  _   3 ) 6 1 ( ) 6 1 ( 6 1 2 1/2 2 3/2 l a o S -soa l :i n i h a w a b i d n a l a r g e t n i g n e p l i s a h n a k u t n e T . 1

_³

t₍t₂₎3/2dt b a w a J ) 2 + t ( = M l a s i M 2 3 M2 _=(t+2)3 2MdM=3(t+2)2_{d t}

³

t₍t₂₎3/2dt ₌

_³

₂₎2 ( 3 2 . . t M d M t M = _

³

M dM t t₂₎2 2 ( 3 2 = 3 2 ₃ 1 ) 2 ( 3 2 _M t t  C+ = 2 9 2( 2) ) 2 ( 9 2 _  t t t _C+ = t t 2 C 5 ) 2 ( 9 2

. 2 dx x x

³

sin . 3

_³

1 2 3 t t d . 4

_³

 dx x x 2 n i s 2 s o c 1 2 . 5

_³

     t d t t t t t 1 3 1 3 n i s ) 1 6 ( 2 2 . 6

³

9 2 x x x d . 7

_³

x₍₃x₂₎3/2dx . 8

³

 dx x x 6 1 2 . 9

_³

xdx 3 n i s . 0 1

³

 x x d x 2 s o c 6 1 n i s . 1 1

³

cos(2x4)dx . 2 1

³

xsin(x21)dx . 3 1

³

x2cos(x31)dx . 4 1

_³

_x(x2_3)12/7_dx . 5 1

³

   _dx x x x 1 3 2 2 . 6 1

³

    _dx e e e e x x x x 2 2 2 2 . 7 1 dt e e t t

³

_ 6 3 4 . 8 1

³

 dx x x 4 4 2 . 9 1

³

_ 4 4 x x d x

. 0 2

³

sinx 12cosxdx 2 . 2 Integra lFungs iTrigonometri a r a c e s i r t e m o n o g i r t i s g n u f l a r g e t n i k i n k e t s a h a b m e m m u l e b e S i r t e m o n o g i r t i s g n u f r a s a d l a r g e t n i n a k i r e b i d i n i t u k i r e b , i c n i r h i b e l n a g n e d n a l a r g e t n i g n e p l i s a h n a k u t n e n e m k u t n u n a u c a i d a j n e m g n a y . i r t e m o n o g i r t i s g n u f k i n k e t Bentukdasart ersebu tadalah : . 1

³

sinx d x = - cosx+C . 2

³

cosx d x = sinx+C . 3

_³

tanxdx = l n secx C = - ln cosx C . 4

³

cot xdx = - ln cscx C = l n sinx C . 5

³

secx d x = l n secxtanxC . 6

³

cscxdx = l n cscxcotx C n a k r a s a d r e B bentuk d iatas selanjutnya diberikan beberapa kasus i s g n u f l a r g e t n i k u t n e b trigonometr iyang dibahas pada bagian ini , : h a l a d a a y n a r a t n a i d . A

³

sinmxdx, _dan

_³

_cosm_{xdx denganmbliangan gan ij lataugenappos tiip} m ( i d a j n e m h a b u i d m a k a m , li j n a g n a d p i ti s o p t a l u b m a k i J -1) + n a g n e d i s u t i t s b u s a y n t u j n a l e S . t a k e d r e t n a k p a n e g i d m u a t a , 1 n a k a n u g g n e m kesamaan i dentitas sin2_xcos2_x 1 _{atau sin}2_{x = - s1 co} x 2 _ataucos2_{x = 1- nsi} 2_x. k t a p a d i d t u b e s r e t i s u t i t s b u s n a g n e d a y n r i h k A esamaan antara t a p a d h a d u m n a g n e d a g g n i h e s , a y n i s a r g e t n i a d n a t n a g n e d n a r g e t n i . n a k i a s e l e s i d

: h o t n o C . 1

³

_sin3xdx Jawab

³

_sin3xdx ₌

_³

_sin(31)1_xdx =

_³

sin2x nsi x _xd =

³

(1cos2_x)_d(cos_x) =

_³

1_d(cos_x)

_³

cos2_d(cos_x) = -cosx+ _cos3x C 3 1 . 2

_³

_cos5xdx Jawab

_³

_cos5xdx ₌

_³

_cos(51)1_{xd x} =

_³

cos4x sco xdx =

³

(1sin2_x)2_d(sin_x) =

_³

(1_2sin2_{x }sin4_x)_d(sin_x) =

³

1_d(sin_x)2

³

sin2_xd(sin_x)

³

sin4_xd(sin_x) =sinx- 3x _sin5xC 5 1 n i s 3 2 . 3

³

sin5(2x)dx Jawab: Misa lu=2x ,du=2dxataudx= 2 u d Sehingga

³

³

2 n i s ) 2 ( n i s 5 _{x dx} 5_udu =

³

_sin5udu 2 1 =

³

sin u nsi udu 2 1 4

=

³

(1cos ) (cos ) 2 1 2_u 2_{d u} =

_³

(12cos cos ) (cos ) 2 1 2_u 4_{u d u} =  u 3u _sin5uC 0 1 1 n i s 3 1 s o c 2 1 =  x x sin 2xC 0 1 1 2 n i s 3 1 2 s o c 2 1 3 5 k u t n e B

³

_cosmxdx _,

_³

_sinm_{dx , ijka m bliangan bula tpositip genap ,} i s u t it s b u s n a k a n u g g n e m n a g n e d n a k u k a li d t a p a d a y n n a i a s e l e s t u d u s h a g n e t e s n a a m a s e k n i s 2x ₌ 2 2 s o c 1 x s o c n a d 2 2 s o c 1 2_x  x : h o t n o C . 1

³

_sin2xdx Karenapangkatnyagenap ,digunakankesamaansetengahsudut , a k a m

_³

_sin2xdx ₌

_³

 x_dx 2 2 s o c 1 =

³

dx

³

cos2xdx 2 1 2 1 = x xC 4 2 s o c 2 . 2

³

_cos4xdx Jawab

³

_cos4xdx ₌

_³

_(cos2_x)2 _{d x} =

³

¸ ¹ · ¨ © §  x 2_dx 2 2 s o c 1 =

_³

 x cos 2x)dx 4 1 2 2 s o c 4 1 ( 2 =

³

dx

³

xdx

³

cos 2xdx 4 1 2 2 s o c 4 1 2