# Tich phan ham phan thuc huu ti 2

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• 1. Bi s 5: TCH PHN ( TIT 1 ) I. Khi nim tch phn 1. Din tch hnh thang cong . Gii thiu cho hc sinh v cch tnh din tch ca mt hnh thang cong T suy ra cng thc : ( ) ( ) ( )0 0 0 0 lim x x S x S x f x x x = 2. nh ngha tch phn Cho hm f lin tc trn mt khong K v a, b l hai s bt k thuc K. Nu F l mt nguyn hm ca f trn K th hiu s : F(b)-F(a) c gi l tch phn ca f i t a n b , k hiu l : ( ) b a f x dx C ngha l : ( ) ( )( ) b a f x dx F b F a= Gi F(x) l mt nguyn hm ca f(x) v ( ) ( ) ( ) b F x F b F a a = th : ( ) ( ) ( )( ) b a b f x dx F x F b F a a = = Trong : - a : l cn trn , b l cn di - f(x) gi l hm s di du tch phn - dx : gi l vi phn ca i s -f(x)dx : Gi l biu thc di du tch phn II. Tnh cht ca tch phn Gi s cho hai hm s f v g lin tc trn K , a,b,c l ba s bt k thuc K . Khi ta c : 1. ( ) 0 a a f x dx = 2. ( ) ( ) b a a b f x dx f x dx= . ( Gi l tch cht i cn ) 3. ( ) ( ) ( ) b c b a a c f x dx f x dx f x dx= + 4. [ ]( ) ( ) ( ) ( ) b b b a a a f x g x dx f x dx g x dx = . ( Tch phn c mt tng hoc hiu hai tch phn bng tng hoc hiu hai tch phn ) . 5. ( ) . ( ) b b a a kf x dx k f x dx= . ( Hng s k trong du tch phn , c th a ra ngoi du tch phn c ) Ngoi 5 tnh cht trn , ngi ta cn chng minh c mt s tnh cht khc nh : 6 . Nu f(x) [ ]0 ;x a b th : [ ]( ) 0 ; b a f x dx x a b

2. Bi ging s 5: TCH PHN XC NH ( Ti liu ni b- Son : T2 nm 2012 ) 7. Nu : [ ]; : ( ) ( ) ( ) ( ) b b a a x a b f x g x f x dx g x dx . ( Bt ng thc trong tch phn ) 8. Nu : [ ];x a b v vi hai s M,N ta lun c : ( )M f x N . Th : ( ) ( )( ) b a M b a f x dx N b a . ( Tnh cht gi tr trung bnh ca tch phn ) III. CC PHNG PHP TNH TCH PHN A. PHNG PHP PHN TCH 1.Trong phng php ny , chng ta cn : K nng : Cn bit phn tch f(x) thnh tng , hiu , tch , thng ca nhiu hm s khc , m ta c th s dng c trc tip bng nguyn hm c bn tm nguyn hm ca chng . Kin thc : Nh trnh by trong phn " Nguyn hm " , cn phi nm trc cc kin thc v Vi phn , cc cng thc v php ton ly tha , php ton cn bc n ca mt s v biu din chng di dng ly tha vi s m hu t . 2. V d p dng V d 1: Tnh cc tch phn sau a/ ( )4 2 2 1 2 1 1 1 x x dx x + + b/ ( ) 1 2 3 0 1 x dx x + c/ ( ) ( ) 3 1 2 2 ln 1 2 1 x x x x dx x x + + + d/ 2 3 2 4 2 2 1 2 1 x x x dx x x + + + Gii a/ ( )4 2 2 22 2 2 2 2 2 2 1 1 1 2 1 1 2 1 1 2 1 1 1 1 1 x x x x x x x dx dx x x dx x x x x + + = + = + + + + + ( ) ( ) ( ) 2 2 2 2 2 2 2 2 1 1 2 21 3 1 1 1 1 1 5 2 1 12 2 x d x d x x x + + = + + = + b/ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 21 1 1 12 3 3 3 3 3 2 3 0 0 0 0 1 1 1 1 1 1 1 1 2 2 11 1 1 1 1 1 1 x xx x dx dx dx dx xx x x x x x x + + + = = + = + ++ + + + + + + ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 3 2 0 0 0 1 1 11 1 1 1 1 1 3 2 ln 1 2 ln 2 0 0 01 1 2 81 1 1 d x d x d x I x x xx x x + + + = + = + + = + + ++ + + c/ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 1 1 1 2 2 ln 1 ln 1 ln 11 1 1 22 1 1 1 1 2 x x x x x xx dx dx x dx xx x x x x x + + + + = + = + + + + + ( ) ( ) ( ) ( ) ( ) 3 3 3 2 1 1 ln 1 3 32 1 1 ln 1 1 131 x I x dx d x x x x x + = + + = + + = + ( )2 2 2 3 4 ln 1 3 ln 2= + + Nguyn nh S -T: 0985.270.218Trang 2 3. Bi ging s 5: TCH PHN XC NH ( Ti liu ni b- Son : T2 nm 2012 ) d/ ( ) ( ) ( ) 32 2 2 23 2 24 2 4 2 2 2 2 2 2 2 41 1 1 2 2 1 4 2 1 1 1 x x dxx x x dx dx dx x x x x x x + + = + + + + ( ) ( ) 4 2 22 2 2 4 2 2 2 2 2 11 1 1 1 1 1 1 2 4 2 1 1 4 1 12 1 d x x dx dx x x x xx x + = + + + + + = ( ) 22 2 2 21 1 1 1 1 1 1 ln 1 ln ln 4 2 1 2 1 1 12 2 2 x x x x x x x + + = + + + V d 2. Tnh cc tch phn sau a/ ( )22 0 2sin sin 1 1 osx x x dx c + b/ 3 2 2 0 sin 2 2sin 3cos x dx x x + c/ 1 2 1 1 2 ln 4 2 x dx x x + d/ 4 2 0 sinx+ 1+tanx os dx c x V d 3. Tnh cc tch phn sau a/ 2 3 3 ln 1 ln e e x dx x x + b/ ( ) 2 2 2 1 1 2 1 x dx x x + c/ 34 2 6 4 sin 2 sin 2 x dx x + d/ 3 0 sin3 . osxdxx c B. PHNG PHP I BIN S I. Phng php i bin s dng 1. tnh tch phn dng ny , ta cn thc hin theo cc bc sau 1/ Quy tc : Bc 1: t x=v(t) Bc 2: Tnh vi phn hai v v i cn Bc 3: Phn tch f(x)dx=f(v(t))v'(t)dt Bc 4: Tnh ( ) ( ) ( ) ( ) ( ) ( ) ( ) v bb a v a v b f x dx g t dt G t v a = = Bc 5: Kt lun : I= ( ) ( ) ( ) v b G t v a 2/ Nhn dng : ( Xem li phn nguyn hm ) * Ch : a. Cc du hiu dn ti vic la chn n ph kiu trn thng thng l : Du hiu Cch chn 2 2 a x sin 2 2 ost 0 t x a t t x a c = = Nguyn nh S -T: 0985.270.218 Trang 3 4. Bi ging s 5: TCH PHN XC NH ( Ti liu ni b- Son : T2 nm 2012 ) 2 2 x a [ ] ; sin 2 2 0;ost 2 a x t t a x t c = = 2 2 a x+ ( ) tan ; 2 2 cot 0; x a t t x a t t = = a x a x a x a x + + x=a.cos2t ( ) ( )x a b x x=a+( ) 2 sinb a t b. Quan trng nht l cc em phi nhn ra dng : - V d : Trong dng phn thc hu t : * ( )2 222 1 1 1 1 0 ax b a x+ 2a 2 dx dx du bx c a u k a < = = + + + + Vi : b x+ , , 2a 2 u k du dx a = = = . * p dng gii bi ton tng qut : ( ) ( )2 12 2 k dx k Z a x + + . * ( ) ( ) 2 2 2 1 1 2 2 3 1 dx dx x x x = + . T suy ra cch t : 1 3sinx t = 3/ Mt s v d p dng : V d 1: Tnh cc tch phn sau a/ 1 2 0 1 x dx b/ 1 2 2 0 1 1 2 dx x c/ 2 2 1 1 3 2 dx x x+ Gii a/ t x=sint vi : ; 2 2 t Suy ra : dx=costdt v : 0 sin 0 0 1 sin 1 2 x t t x t t = = = = = = Do : f(x)dx= ( )2 2 2 1 1 1 sin ostdt=cos 1 os2t 2 x dx tc tdt c dt = = + Vy : ( )1 2 0 0 1 os2t 1 1 1 1 1 ( ) sin 2 2 2 2 2 2 2 2 4 0 c dt f x dx t t + = = + = = b/ t : x = 1 sin ; 2 22 t t Nguyn nh S -T: 0985.270.218Trang 4 5. Bi ging s 5: TCH PHN XC NH ( Ti liu ni b- Son : T2 nm 2012 ) Suy ra : dx = x=0 sint=0 t=0 1 ostdt 1 1 1 x= sin2 22 2 2 c t t = = Do : 1 1 2 2 2 2 2 20 0 0 02 1 1 1 1 1 1 1 1 ostdt 2 12 2 2 2 2 2 211 2 1 sin 0 22 dx dx c dt t x tx = = = = = c/ V : ( ) 22 3 2 4 1x x x+ = . Cho nn : t : ( ) 1 1 2sin ; sin * 2 2 2 x x t t t = = Suy ra : dx= 2 costdt v : 1 1 1 sin 0 0 2 0; ost>0 2 1 1 6 2 sin 2 2 6 x t t t c x t t = = = = = = = = Do : f(x)dx= ( ) ( )2 2 2 1 1 1 2cos 3 2 4 1 sin4 1 dx dx tdt dt x x tx = = = + Vy : 2 6 1 0 ( ) 6 6 0 f x dx dt t = = = V d 2: Tnh cc tch phn sau a/ 2 2 1 12 4 5x x dx b/ 1 2 0 1 1 dx x x+ + c/ 5 2 2 1 4 7 dx x x + d/ ( ) 2 22 0 b a x dx a x + * Ch : tnh tch phn dng c cha ( )2 2 2 ,x a a x+ , ta cn s dng phng php i bin s : u(x)=g(x,t) V d 1 : Tnh tch phn sau 1 2 0 1 1 dx x + Gii : t : 2 2 1 1 2 t x x t x t + = = Khi : 2 2 0 1; 1 1 2 1 2 x t x t t dx t = = = = + = Do vy : ( ) 1 1 2 1 22 2 22 0 1 1 1 2 1 1 2 . ln ln 2 1 1 2 11 t t dt dx dt t t t tx + = = = = + + V d 2: Tnh tch phn : 1 2 2 0 1I x x dx= Gii Nguyn nh S -T: 0985.270.218 Trang 5 6. Bi ging s 5: TCH PHN XC NH ( Ti liu ni b- Son : T2 nm 2012 ) t : t=sinx , suy ra dt=cosxdx v khi x=0,t=0 ; Khi x=1 , t= 2 Do : f(x)dx= 2 2 2 2 2 2 1 1 os4t 1 sin . 1 sin ostdt=sin cos 4 2 c x x dx t tc t tdt dt = = Vy : I= ( ) 1 2 0 0 1 1 1 1 ( ) 1 os4t sin 4 2 8 8 4 8 2 16 0 f x dx c dt t t = = = = II. i bin s dng 2 1. Quy tc : ( Ta tnh tch phn bng phng php i bin s dng 2 theo cc bc sau : ) Bc 1: Kho lo chn mt hm s u(x) v t n bng t : t=u(x) . Bc 2: Tnh vi phn hai v v i cn : dt=u'(x)dx Bc 3: Ta phn tch f(x)dx = g[u(x)]u'(x)dx = g(t)dt . Bc 4: Tnh ( ) ( ) ( ) ( ) ( ) ( ) ( ) u bb a u a u b f x dx g t dt G t u a = = Kt lun : I= ( ) ( ) ( ) u b G t u a 2. Nhn dng : TCH PHN HM PHN THC HU T A. DNG : I= ( ) ( ) 0 ax+b P x dx a * Ch n cng thc : ln ax+b ax+b m m dx a = . V nu bc ca P(x) cao hn hoc bng 2 th ta chia t cho mu dn n ( ) 1 ( ) ( ) ax+b ax+b ax+b P x m dx Q x dx Q x dx m dx = + = + V d 1 : Tnh tch phn : I= 2 3 1 2 3 x dx x + Gii Ta c : 3 21 3 9 27 1 ( ) 2 3 2 4 8 8 2 3 x f x x x x x = = + + + Do : 2 23 2 3 2 1 1 21 3 9 27 1 1 3 9 27 13 27 ln 2 3 ln35 12 3 2 4 8 8 2 3 3 8 8 16 6 16 x dx x x dx x x x x x x = + = + + = + + V d 2: Tnh tch phn : I= 3 2 5 5 1 x dx x + Gii Ta c : f(x)= 2 5 4 1 1 1 x x x x = + + . Nguyn nh S -T: 0985.270.218Trang 6 7. Bi ging s 5: TCH PHN XC NH ( Ti liu ni b- Son : T2 nm 2012 ) Do : 3 32 2 5 5 35 4 1 5 1 1 4ln 1 5 1 4ln 1 1 2 45 x dx x dx x x x x x + = = + = + + + B. DNG : 2 ( ) ax P x dx bx c + + 1. Tam thc : 2 ( ) axf x bx c= + + c hai nghim phn bit Cng thc cn lu : '( ) ln ( ) ( ) u x dx u x u x = Ta c hai cch Cch 1: ( H s bt nh ) Cch 2: ( Nhy tng lu ) V d 3: Tnh tch phn : I= 1 2 0 4 11 5 6 x dx x x + + + . Gii Cch 1: ( H s bt nh ) Ta c : f(x)= ( ) ( ) 2 3 24 11 4 11 5 6 ( 2)( 3) 2 3 ( 2)( 3) A x B xx x A B x x x x x x x x + + ++ + = = + = + + + + + + + + Thay x=-2 vo hai t s : 3=A v thay x=-3 vo hai t s : -1= -B suy ra B=1 Do : f(x)= 3 1 2 3x x + + + Vy : ( ) 1 1 2 0 0 14 11 3 1 3ln 2 ln 3 2ln3 ln 2 05 6 2 3 x dx dx x x x x x x + = + = + + + = + + + + Cch 2: ( Nhy tng lu ) Ta c : f(x)= ( ) ( ) ( )2 2 2 2 2 5 1 2 5 1 2 5 1 1 2. 2. 5 6 5 6 2 3 5 6 2 3 x x x x x x x x x x x x x + + + + = + = + + + + + + + + + + + Do : I= 1 1 2 2 0 0 12 5 1 1 2 ( ) 2. 2ln 5 6 ln 2ln3 ln 2 05 6 2 3 3 x x f x dx dx x x x x x x x + + = + = + + + = + + + + + 2. Tam thc : 2 ( ) axf x bx c= + + c hai nghim kp Cng thc cn ch : ( ) '( ) ln ( ) ( ) u x dx u x u x = Thng thng ta t (x+b/2a)=t . V d 4 : Tnh tch phn sau : I= 3 3 2 0 2 1 x dx x x+ + Gii Ta c : ( ) 3 33 3 22 0 0 2 1 1 x x dx dx x x x = + + + t : t=x+1 suy ra : dx=dt ; x=t-1 v : khi x=0 th t=1 ; khi x=3 th t=4 . Do : ( ) ( ) 33 4 43 2 2 2 2 0 1 1 41 3 1 1 1 3 3 3 ln 2ln 2 12 21 tx dx dt t dt t t t t t t tx = = + = + + = + Nguyn nh S -T: 0985.270.218 Trang 7 8. Bi ging s 5: TCH PHN XC NH ( Ti liu ni b- Son : T2 nm 2012 ) V d 5: Tnh tch phn sau : I= 1 2 0 4 4 4 1 x dx x x + Gii Ta c : ( ) 22 4 4 4 4 1 2 1 x x x x x = + t : t= 2x-1 suy ra : 0 11 2 ; 1 12 x t dt dx dx dt x t = = = = = = Do : ( ) ( )1 1 1 1 22 2 2 0 0 1 1 1 4. 1 14 4 1 1 1 12 ln 2 14 4 1 22 1 t x x dx dx dt dt t x x t t t tx + = = = + = = +

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