Tich phan (Ly thuyet va bai tap)

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<p>Tran S TungTch phan Trang 1 Nhac lai Gii han ao ham Vi phan 1.Cac gii han ac biet: a)=x 0sinxlim 1x He qua: =x 0xlim 1sinx</p> <p>=u(x) 0sinu(x)lim 1u(x) =u(x) 0u(x)lim 1sinu(x) b) xx1lim 1 e, x Rx + = He qua:1xx 0lim(1 x) e.+ =x 0ln(1 x)lim 1x+=xx 0e 1lim 1x-=2.Bang ao ham cac ham so s cap c ban va cac he qua:(c) = 0 (c la hang so) 1(x )' xa a-= a1(u )' u u'a a-= a21 1'x x = - 21 u''u u = - ( )1x '2 x=( )u'u '2 u=x x(e )' e =u u(e )' u' .e =x x(a )' a . lna =u u(a )' a . lna . u' =1(ln x )'x=u'(ln u )'u=a1(log x ')x. lna=au'(log u )'u. lna=(sinx) = cosx(sinu) = u.cosu 221(tgx)' 1 tg xcos x= = +22u'(tgu)' (1 tg u).u'cos u= = +221(cot gx)' (1 cot g x)sin x-= = - +22u'(cot gu)' (1 cot g u).u'sin u-= = - +3.Vi phan:Cho ham so y = f(x) xac nh tren khoang (a ; b) va co ao ham taix (a; b) . Cho so gia Dx tai x sao chox x (a; b) + D . Ta goi tch y.Dx (hoac f(x).Dx) la vi phan cua ham so y = f(x) tai x, ky hieu la dy (hoac df(x)). dy = y.Dx(hoac df(x) = f(x).Dx Ap dung nh ngha tren vao ham so y = x, thdx = (x)Dx = 1.Dx = Dx V vay ta co:dy = ydx (hoac df(x) = f(x)dx) Tch phanTran S TungTrang 2 NGUYEN HAM VA TCH PHAN 1.nh ngha:Ham so F(x) c goi la nguyen ham cua ham so f(x) tren khoang (a ; b) neu moi x thuoc (a ; b), ta co: F(x) = f(x). Neu thay cho khoang (a ; b) la oan [a ; b] th phai co them: F'(a ) f(x) va F'(b ) f(b)+ -= = 2.nh ly:Neu F(x) la mot nguyen ham cua ham so f(x) tren khoang (a ; b) th : a/Vi moi hang so C, F(x) + C cung la mot nguyen ham cua ham so f(x) tren khoang o. b/Ngc lai, moi nguyen ham cua ham so f(x) tren khoang (a ; b) eu co the viet di dang: F(x) + C vi C la mot hang so. Ngi ta ky hieu ho tat ca cac nguyen ham cua ham so f(x) laf(x)dx. Do o viet:f(x)dx F(x) C = + Bo e: Neu F(x) = 0 tren khoang (a ; b) th F(x) khong oi tren khoang o. 3.Cac tnh chat cua nguyen ham: ( )f(x)dx ' f(x) = af(x)dx a f(x)dx (a 0) = [ ] f(x) g(x) dx f(x)dx g(x)dx + = + [ ] [ ] f(t)dt F(t) C f u(x) u'(x)dx F u(x) C F(u) C (u u(x)) = + = + = + = 4.S ton tai nguyen ham:nh ly: Moi ham so f(x) lien tuc tren oan [a ; b] eu co nguyen ham tren oan o. 8o| 1. NGUYEN HAM Tran S TungTch phan Trang 3 BANG CAC NGUYEN HAM Nguyenhamcuaca chamsoscap thng gap Nguyenhamcuacachamsohp (di ay u = u(x)) dx x C = +du u C = + 1xx dx C ( 1)1a+a= + a -a + 1uu du C ( 1)1a+a= + a -a + dxln x C (x 0)x= + duln u C (u u(x) 0)u= + = x xe dx e C = + u ue du e C = + xxaa dx C (0 a 1)lna= + &lt; uuaa du C (0 a 1)lna= + &lt; cosxdx sinx C = +cos udu sin u C = + sinxdx cosx C = - +sin udu cos u C = - + 22dx(1 tg x)dx tgx Ccos x = + = + 22du(1 tg u)du tgu Ccos u = + = + 22dx(1 cot g x)dx cot gx Csin x = + = - + 22du(1 cot g u)du cot gu Csin u = + = - + dxx C (x 0)2 x= + &gt; duu C (u 0)2 u = + &gt; 1cos(ax b)dx sin(ax b) C (a 0)a+ = + + 1sin(ax b)dx cos(ax b) C (a 0)a+ = - + + dx 1ln ax b Cax b a= + ++ ax b ax b1e dx e C (a 0)a+ += + dx 2ax b C (a 0)a ax b= + + + Tch phanTran S TungTrang 4 Von de 1:Ik0 0I0h 080IE0 hkN 8k08 0I0h 08hk Bai toan 1: CMR F(x) la mot nguyen ham cua ham so f(x) tren (a ; b) PHNG PHAP CHUNG Ta thc hien theo cac bc sau:+Bc 1: Xac nh F(x) tren (a ; b) +Bc 2: Chng to rangF'(x) f(x) vi x (a; b) = " Chu y: Neu thay (a ; b) bang [a ; b]th phai thc hien chi tiet hn, nh sau:+Bc 1: Xac nh F(x) tren (a ; b) Xac nh F(a+) Xac nh F(b) +Bc 2: Chng to rang F'(x) f(x), x (a ; b)F'(a ) f(a)F'(b ) f(b)+-= " == V du 1: CMR ham so:2F(x) ln(x x a) = + +vi a &gt; 0la mot nguyen ham cua ham so 21f(x)x a=+ tren R. Giai: Ta co: 2222 22x1(x x a)'2 x aF'(x) [ln(x x a)]'x x a x x a++ ++= + + = =+ + + + 22 2 2x a x 1f(x)x a(x x a) x a+ += = =+ + + + Vay F(x) vi a &gt; 0 la mot nguyen ham cua ham so f(x) tren R. V du 2: CMR ham so:x2e khi x 0F(x)x x 1 khi x 0 = + + la mot nguyen ham cua ham so: 2x khi x 1f(x)2 khi x 1 = &gt; tren R. Giai: e tnh ao ham cua ham so F(x) ta i xet hai trng hp:a/Vix 1 , ta co:2x khi x 1F'(x)2 khi x 1&lt; = &gt; b/Vi x = 1, ta co: e ham so F(x) co ao ham tai iem x = 1, trc het F(x) phai lien tuc tai x = 1, do o :x 1 x 1limF(x) limF(x) f(1) a b 1 b 1 a (1)- + = = + = = -ao ham ben trai cua ham so y = F(x) tai iem x = 1. 2x 1 x 1f(x) F(1) x 1F'(1) = lim lim 2.x 1 x 1- - -= =- - ao ham ben phai cua ham so y = f(x) tai iem x0 = 0. x 1 x 1 x 1F(x) F(1) ax b 1 ax 1 a 1F'(1 ) lim lim lim a.x 1 x 1 x 1+ + ++ - + - + - -= = = =- - - Ham so y = F(x) co ao ham tai iem x = 1F'(1 ) F'(1 ) a 2.- + = = (2) Thay (2) vao (1), ta c b = 1. Vay ham so y = F(x) co ao ham tai iem x = 1, neu va ch neu a = 2, b = 1. Khi o: F(1) = 2 = f(1) Tom lai vi a = 2, b = 1 th F(x) la mot nguyen ham cua ham so f(x). V du 4: Xac nh a , b , c e ham so:-= + +2 2xF(x) (ax bx c)ela mot nguyen ham cua 2 2xF(x) (2x 8x 7)e-= - - +tren R. Giai: Ta co: 2x 2 2xF'(x) (2ax b)e 2(ax bx c)e- -= + - + +2 2x2ax 2(a b)x b 2c e- = - + - + - Do o F(x) la mot nguyen ham cua f(x) tren R F'(x) f(x), x R = "- + - + - = - + - " 2 22ax 2(a b)x b 2c 2x 8x 7, x R a 1 a 1a b 4 b 3b 2c 7 c 2= = - = = - - = - = Vay -= - +2 2xF(x) (x 3x 2)e . Tran S TungTch phan Trang 7 BAI TAP Bai 1.Tnh ao ham cua ham so xF(x) ln tg2 4p = + T o suy ra nguyen ham cua ham so 1f(x)cos x= . Bai 2.Chng to rang ham so 2ln(x 1), x 0F(x)x0 , x 0 + = = la mot nguyen ham cua ham so 22 22 ln(x 1), x 0f(x)x 1 x1 , x 0 +- =+ = Bai 3.Xac nh a, b, c sao cho ham so 2 xF(x) (ax bx c).e-= + +la mot nguyen ham cua ham so 2 xf(x) (2x 5x 2)e-= - +tren R. S:a = 2 ; b = 1 ; c = 1. Bai 4. a/Tnh nguyen ham 3 22x 3x 3x 7F(x) cua f(x) va F(0) 8.(x 1)+ + -= =+ b/Tm nguyen ham F(x) cua 2xf(x) sin va F .2 2 4p p = = S:a/2x 8F(x) x ;2 x 1= + ++b/ 1F(x) (x sinx 1)2= - +Bai 5. a/Xac nh cac hang so a, b, c sao cho ham so: </p> <p>2F(x) (ax bx c) 2x 3 = + + -la mot nguyen ham cua ham so: 220x 30x 7 3f(x) tre n khoang ;2 2x 3- + = + - b/Tm nguyen ham G(x) cua f(x) vi G(2) = 0. S:a/ a 4; b 2; c 1; = = - = b/ 2G(x) (4x 2x 10) 2x 3 22. = - + - - Tch phanTran S TungTrang 8 Von de 2:Ik0 0I0h 080IE0 hkN 8k08 IE0 80 0008 8k080k0 080IE0 hkN 00 8k0 V du 1: CMR , neuf(x)dx F(x) C = + th 1f(ax b)dx F(ax b) C v i a 0.a+ = + + Giai: Ta luon co: 1f(ax b)dx f(ax b)d(ax b) vi a 0.a+ = + + Ap dung tnh chat 4, ta c:1 1f(ax b)dx (ax b)d(ax b) F(ax b) C (pcm)a a+ = + + + + . Ghi chu: Cong thc tren c ap dung cho cac ham so hp:f(t)dt F(t) C f(u)du F(u) C, vi u u(x) = + = + = V du 2: Tnh cac tch phan bat nh sau:a/ 3(2x 3) dx +b/4cos x.sinxdxc/xx2edxe 1 +d/2(2lnx 1)dxx+ Giai: a/Ta co: 4 43 31 1 (2x 3) (2x 3)(2x 3) dx (2x 3) d(2x 3) . C C.2 2 4 8+ ++ = + + = + = + b/Ta co: 54 4cos xcos x.sinxdx cos xd(cos x) C5= - = - + c/Ta co: x xxx x2e d(e 1)dx 2 2ln(e 1) Ce 1 e 1+= = + ++ + d/Ta co: 22 3(2lnx 1) 1 1dx (2lnx 1) d(2lnx 1) (2lnx 1) C.x 2 2+= + + = + + V du 3: Tnh cac tch phan bat nh sau:a/ 2x2sin dx2b/2cot g xdxc/ tgxdxd/3tgxdxcos x Giai: a/Ta co: 2x2sin dx (1 cosx)dx x sinx C2= - = - + b/Ta co: 221cot g xdx 1 dx cot gx x Csin x = - = - - + c/Ta co: sinx d(cosx)tgxdx dx ln cosx Ccosx cosx= = - = - + Tran S TungTch phan Trang 9 d/Ta co: 33 4 4 3tgx sinx d(cosx) 1 1dx dx cos x C C.cos x cos x cos x 3 3cos x-= =- = - + = - + V du 4: Tnh cac tch phan bat nh sau:a/ 2xdx1 x +b/ 21dxx 3x 2 - + Giai: a/Ta co: 222 2x 1 d(1 x ) 1dx ln(1 x ) C1 x 2 1 x 2+= = + ++ + b/Ta co: 21 1 1 1dx dx dxx 3x 2 (x 1)(x 2) x 2 x 1 = = - - + - - - - x 2ln x 2 ln x 1 C ln C.x 1-= - - - + = +- BAI TAP Bai 6.Tm nguyen ham cua cac ham so:a/ 2xf(x) cos ;2= b/ 3f(x) sin x.S: a/1(x sinx) C ;2+ + b/ 31cos x cos x C.3- + +Bai 7.Tnh cac tch phan bat nh : a/ x xe (2 e )dx;-- b/ xxedx ;2c/ 2x x xx2 .3 .5dx10. d/ 2 5xxe 1dx;e-+e/ xxedxe 2 + S: a/ x2e x C; - + b/ xxeC;(1 ln2)2+-c/ x6Cln6 +d/ 2 6x x1e e C;6- -- - + e/ xln(e 2) C + + . Bai 8.Tnh cac tch phan bat nh : a/ 4 4x x 2 dx-+ +;b/ 3 5x xdx ;c/ 2x x 1dx +; d/ 2001(1 2x) dx; - e/ 3 4lnxdxx- S: a/ 3x 1C;3 x- + b/ 5 75x C;7+ c/ 2 21(x 1) x 1 C3+ + +; d/ 20021 (1 2x). C;2 2002-- + e/ 1(3 4lnx) 3 4lnx C.6+ + +Tch phanTran S TungTrang 10 Von de 3:Ik0 0I0h 080IE0 hkN 8k08 Ph0008 PhkP Phk0 0h Phngphapphantchthcchatlaviecs dungcacongnhatthcebienoibieu thc di dau tch phan thanh tong cac bieu thc ma nguyen ham cua moi bieu thc o co the nhan c t bang nguyen ham hoac ch bang cac phep bien oi n gian a biet. Chuyquantrong:iemmauchotlaphepphantchlacothe rutraytngchorieng mnh t mot vai minh hoa sau: Vi 3 2 6 3f(x) (x 2) th vie t lai f(x) x 4x 4. = - = - +Vi 2x 4x 5 2f(x) th viet lai f(x) x 3x 1 x 1- += = - +- -. Vi 21 1 1f(x) th vie t la i f(x)x 5x 6 x 3 x 2= = -- + - - Vi 1 1f(x) th vie t la i f(x) ( 3 2x 2x 1)2 2x 1 3 2x= = - - ++ + - Vi x x 2 x x xf(x) (2 3 ) th vie t la i f(x) 4 2.6 9 . = - = - +Vi 3f(x) 8cos x.sinx th vie t la i f(x) 2(cos3x 3cosx).sinx = = +2cos3x.sinx 6cosx.sinx sin4x sin2x 3sin2x sin4x 2sin2x. = + = - + = + 2 2tg x (1 tg x) 1 = + - 2 2cot g x (1 cot g x) 1 = + - n 2n2 2x (1 x ) 1 1x1 x 1 x+ += ++ +. o ch la mot vai minh hoa mang tnh ien hnh. V du 1: Tnh tch phan bat nh: 2002I x(1 x) dx. = - Giai: S dung ong nhat thc :x = 1 (1 x)ta c:2002 2002 2002 2003x(1 x) [1 (1 x)](1 x) (1 x) (1 x) . - = - - - = - - -Khi o: 2002 2003 2002 20032003 2004I (1 x) dx (1 x) dx (1 x) d(1 x) (1 x) d(1 x)(1 x) (1 x)C.2003 2004= - - - = - - - + - -- -= - + + Tong qua t: Tnh tch phan bat nh:I x(ax b) dx, vi a 0a= + S dung ong nhat thc: 1 1x .ax [(ax b) b]a a= = + -Tran S TungTch phan Trang 11 Ta c: 11 1x(ax b) [(ax b) b)(ax b) [ (ax b) d(ax b) (ax b) d(ax d)]a aa a a+ a+ = + - + = + + - + + Ta xet ba trng hp : Vi a = 2, ta c: 1 221I [ (ax b) d(ax b) (ax b) d(ax b)]a- -= + + - + + </p> <p>21 1[ln ax b ] C.a ax b= + + ++ Vi a = 1, ta c: 12 21 1I [ d(ax b) (ax b) d(ax b)] [ax b ln ax b ] C.a a-= + - + + = + - + + ViR \ { 2; 1}, a - -ta c:2 121 (ax b) (ax b)I [ ] C.a 2 1a+ a++ += + +a + a+ V du 2: Tnh tch phan bat nh: 2dxIx 4x 3=- + Giai: Ta co:21 1 1 (x 1) (x 3) 1 1 1. .x 4x 3 (x 3)(x 1) 2 (x 3)(x 1) 2 x 3 x 1- - - = = = - - + - - - - - - Khi o:- - = - = - = - - - + - - - - 1 dx dx 1 d(x 3) d(x 1) 1I . [ ' .(ln x 3 ln x 1) C2 x 3 x 1 2 x 3 x 1 2 -= +-1 x 3ln C.2 x 1 V du 3: Tnh tch phan bat nh: dxIx 2 x 3=+ + - Giai: Kh tnh vo t mau so bang cach truc can thc, ta c: 1 12 23 31 1I ( x 2 x 3)dx [ (x 2) d(x 2) (x 3) d(x 3)]5 52[ (x 2) (x 3) ] C.15= + + - = + + + - -= + + - + V du 4: Tnh tch phan bat nh: 2dxI .sinx. cos x= Giai: S dung ong nhat thc:2 2sin x cos x 1, + =Tch phanTran S TungTrang 12 Ta c:2 22 2 2 2211 sin x cos x sinx 1 sinx 12. .x xsinx. cos x sinx.sin x cos x sinx cos xcos tg2 2+= = + = +Suy ra: 2 22x 1d tgsinx d(cosx) 1 x2 2I dx dx ln tg C.x x xcos x cos x cosx 2cos tg tg2 2 2 = + = - + = + + V du 5: Tnh tch phan bat nh: 4dxI .cos x= Giai: S dung ket qua:2dxd(tgx)cos x =ta c: 2 2 32 21 dx 1I . (1 tg x)d(tgx) d(tgx) tg xd(tgx) tgx tg x C.cos x cos x 3= = + = + = + + BAI TAP Bai 9.Tm ho nguyen ham cua cac ham so: a/ 2 3f(x) (1 2x ) ; = - b/ 3 x 232 x x e 3xf(x)x- -= ; c/ 2(2 x)f(x) ;x+= d/ 1f(x)3x 4 3x 2=+ - + S: a/ 3 5 712 8x 2x x x C5 7- + - +;b/ x4e ln x C;3x x- - + +c/3 3 2 2 624 36 x x x x x C;7 5+ + + d/ 3 31(3x 4) (3x 2) C.9 - + + + Bai 10. Tm ho nguyen ham cua cac ham so: a/ 21f(x) ;x 6x 5=- +b/ 24x 6x 1f(x) ;2x 1+ +=+ c/ 3 24x 4x 1f(x) ;2x 1+ -=+d/ 324x 9x 1f(x) ;9 4x- + +=- S: a/ 1 x 5ln C;4 x 1-+- b/ 21x 2x ln 2x 1 C;2+ - + +c/ 3 22 1 1 1x x x ln 2x 1 C3 2 2 4+ - - + + ;d/ 2x 1 2x 3ln C.2 12 2x 3-- ++ Bai 11. Tm ho nguyen ham cua cac ham so: Tran S TungTch phan Trang 13 a/ 2(sinx cos x) ; +b/cos 2x . cos 2x ;3 4p p - + c/ 3cos x;d/ 4cos x; e/ 4 4sin x cos x; + f/ 6 6sin 2x cos 2x. +S: a/ 1x cos2x C2- +;b/ 1 7 1sin 5x sin x C10 12 2 12p p + + - + c/ 3 1sinx si n3x C;4 12+ + d/ 3 1 1x si n2x si n4x C;8 4 31+ + +e/ 3 sin4xx C;4 16+ + f/ 5 3x sin8x C.8 64+ +Tch phanTran S TungTrang 14 Von de 4:Ik0 0I0h 080IE0 hkN 8k08 Ph0008 PhkP 00I 8IE0 80 Phng phap oi bien so c s dung kha pho bien trong viec tnh cac tch phan bat nh. Phng phap oi bien so e xac nh nguyen ham co hai dang da tren nh ly sau: nh ly: a/Neu f(x)dx F(x) C va u (x) = + = j la ham so co ao ham th f(u)du F(u) C = +. b/Neu ham so f(x) lien tuc th khi at x = j(t) trong o j(t) cung vi ao ham cua no (j(t) la nhng ham so lien tuc, ta se c: f(x)dx f[ (t)]. '(t)dt. = j j T o ta trnh bay hai bai toan ve phng phap oi bien nh sau: Bai toan 1: S dung phng phap oi bien so dang 1 tch tch phan bat nh I f(x)dx. = PHNG PHAP CHUNG Ta thc hien theo cac bc:+Bc 1: Chon x = j(t), trong o j(t) la ham so ma ta chon cho thch hp. +Bc 2: Lay vi phan dx = j(t)dt +Bc 3: Bieu th f(x)dx theo t va dt. Gia s rang f(x)dx = g(t)dt +Bc 4: Khi oI g(t)dt. = Lu y: Cac dau hieu dan ti viec la chon an phu kieu tren thong thng la: Dau hieuCach chon 2 2a x -x a sint vi t2 2x x cost v i 0 tp p= - = p 2 2x a -ax v i t ; \ {0}sint 2 2ax v i t [0; ] \ { }cost 2 p p = - p = p 2 2a x +x a tgt vi t2 2x a cot gt vi 0 tp p= - &lt; , ta xet hai trng hp : Vi x &gt; 1 at: 1x ; 0 tsin2t 4p= &lt; &lt; Suy ra:22cos2tdtdxsin 2t= 2 2 2 23 3 32x dx 2dt 2(cos t sin t) dtsin 2t 8sin t cos tx 1+= - = -- 2 22 2 21 1 1 1(cot gt. tgt. )dt4 sin t cos t sint cost1 1 1 2 1(cot gt. tdt. )4 sin t cos t tgt cos t1 d(tgt)[ cot gt.d(cot gt) tgt.d(tgt) 2 ].4 tgt=- + += - + += - - + + Khi o:1 d(tgt)I [ cot gt.d(cot gt) tgt.d(tgt) 2 ]4 tgt= - - + + 2 2 2 22 21 1 1 1 1( cot g t tg t 2ln tgt ) C (cot g t tg t) ln tgt C4 2 2 8 21 1x x 1 ln x x 1 C.2 2= - - + + + = - - += - - - - + Vi x &lt; 1 e ngh ban oc t lam Chu y:Trong v du tren s d ta co: 2 2 2 2cot g t tg t 4x x 1 va tgt x x 1 - = - = - -la bi: 4 4 22 22 2 2 2 2cos t sin t 4cos2t 4 1 sin 2t 4 1cot g t tg t 1cos t.sin t sin 2t sin 2t sin2t sin 2t- -- = = = = -tgt = -= = = -2 22sint 2sin t 1 cos2t 1 cos 2tcos t 2sint. cost sin2t sin2tsin 2t= - -21 11sin2tsin 2t Tch phanTran S TungTrang 16 V du 3: Tnh tch phan bat nh: 2 3dxI(1 x )=+ Giai: at:x tgt; t2 2p p= - &lt; &lt; . Suy ra:32 22 3dt dx cos tdtdx &amp; cos tdt.cos t cos t(1 x )= = =+ Khi o: 2xI costdt sint C C1 x= = + = ++ Chu y:1.Trong v du tren s d ta co:2 21 xcos t va sint1 x 1 x= =+ + la bi:22cos t costt cost 0x2 2 sint tgt. cost1 x=p p - &lt; &lt; &gt; = =+ 2.Phng phap tren c ap dung e giai bai toan tong quat: 2 2 2k 1dxI , v i k Z.(a x )+= + Bai toan 2: S dung phng phap oi bien so dang 2 tch tch phanI f(x)dx. = PHNG PHAP CHUNG Ta thc hien theo cac bc:+Bc 1: Chon t = y(x), trong o y(x) la ham so ma ta chon cho thch hp +Bc 2: Xac nh vi phan= y dt '(x)dx.+Bc 3: Bieu th f(x)dx theo t va dt.Gia s rang f(x)dx = g(t)dt +Bc 4: Khi oI g(t)dt. = Dau hieuCach chon Ham so mau cot la mau so Ham sof(x, (x) j t (x) = jHam a.sinx b. cosxf(x)c.sinx d. cosx e+=+ + x xt tg (v i cos 0)2 2= Ham 1f(x)(x a)(x b)=+ + Vi x + a &gt; 0 &amp; x + b &gt; 0, at:t x a x b = + + + Vi x + a &lt; 0 &amp; x + b &lt; 0, at:t x a x b = - + - - Tran S TungTch phan Trang 17 V du 4: Tnh tch phan bat nh: 3 2 8I x (2 3x ) dx. = - Giai: at: 2t 2 3x = - .Suy ra: dt 6xdx = 3 2 8 2 2 8 8 9 82 t 2 t 1 1x (2 3x ) dx x (2 3x ) xdx .t . dt (t 2t )dt.3 3 6 18- - - = - = = - = - Khi o: 9 8 10 9 10 91 1 1 2 1 1I (t 2t )dt t t C t t C18 18 10 9 180 81 = - = - + = - + V du 5: Tnh tch phan bat nh: 2x dxI1 x=- Giai: at: 2t 1 x x 1...</p>