1 TRNG I HC CNG NGH SI GN BAN KHOA HC C BN B MN TON BI GING TON CAO CP C1 (H I HC) Bin son: TS TRN NGC HI TP H CH MINH 2009 LU HNH NI B 2 Li ni u _____________________ p bi ging Ton cao cp C1 (H i hc) c bin son trn c s cng mn hc ca Trng i hc Cng Ngh Si Gn; nhm p ng yu cu nng cao cht lng ging dy trong giai on nh trng thc hin o to theo hc ch tn ch.TpbigingnychangnidungmtcgigingdyTrngihc Cng Ngh Si Gn v cc trng i hc khc. Tc gi by t lng cm n i vi cc ngnghipBanKhoahcCbn-TrngihcCngNghSiGnng vin, ng gp nhiu kin qu bu cho vic bin son. Tuy vy, thiu st vn khng th trnh khi. Tc gi rt mong nhn c nhng nhn xt gp ca qu ng nghip cho tp bi ging ny v xin chn thnh cm n.Tp. H Ch Minh, thng 09 nm 2009 Tc gi T3 MC LC CHNG 1. PHP TNH VI PHN HM MT BIN A. HM S1. HM S S CP C BN ........................................................................................... 5 2. HM S S CP.......................................................................................................... 9 B. GII HN1. NH NGHA V TNH CHT................................................................................. 10 2. HM TNG NG ............................................................................................... 12 3. V CNG B (VCB) - V CNG LN .................................................................... 16 4. DNG V NH 1.................................................................................................... 22 C. LIN TC 1. NH NGHA V TNH CHT .................................................................................. 23 2. HM S LIN TC TRN MT ON ................................................................... 25 D - O HM V VI PHN 1.KHI NIM O HM ............................................................................................. 27 2. PHNG PHP TNHO HM ........................................................................... 30 3. VI PHN ....................................................................................................................... 34 4. O HM V VI PHN CP CAO .......................................................................... 36 5. QUI TC LHOSPITAL ............................................................................................... 38 6. KHAI TRIN TAYLOR ............................................................................................... 43 7. NG DNG .................................................................................................................. 47 BI TP ........................................................................................................................... 53 CHNG 2. PHP TNH TCH PHN HM MT BIN A - TCH PHN BT NH1.KHI NIM V TCH PHN BT NH ................................................................ 59 4 2. CC PHNG PHP TNH TCH PHN ................................................................. 61 3. TCH PHN HM HU T ......................................................................................... 67 4. TCH PHN HM LNG GIC ............................................................................. 71 5. TCH PHN HM V T ............................................................................................ 73 B -TCH PHN XC NH - TCH PHN SUY RNG 1.TCH PHN XC NH ............................................................................................ 78 2. TCH PHN SUY RNG ............................................................................................ 84 3. NG DNG CA TCH PHN .................................................................................. 88 4. KHI NIM V PHNG TRNH VI PHN ........................................................... 90 BI TP ........................................................................................................................... 95 CHNG 3. PHP TNH VI PHN HM NHIU BIN 1. KHI NIM V HM NHIU BIN........................................................................ 99 2. O HM RING ..................................................................................................... 102 3. O HM RING CA HM HP ........................................................................ 104 4. O HM RING CA HM N .......................................................................... 105 5. VI PHN ..................................................................................................................... 107 6.CC TR .................................................................................................................... 109 7. CC TR C IU KIN ......................................................................................... 110 8. GI TR LN NHT- GI TR NH NHT .......................................................... 113 9. MT S BI TON KINH T ................................................................................. 115 BI TP ......................................................................................................................... 118 5 CHNG 1 PHP TNH VI PHN HM MT BIN A. HM S1. HM S S CP C BN 1.1. Hm ly tha y = x ( : Const) MinxcnhDcahmsy=x phthucvo.Trnghplsvt,tacD = [0; +) nu > 0; D = (0; +) nu < 0.1.2. Hm s m: y = ax (0 < a 1 : Const) Hm s y = ax c min xc nh D = R, min gi tr l (0; +).1.3. Hm s logarit: y = logax (0 < a 1 : Const) Hm s y = logax c min xc nh D = (0; +), min gi tr l R. Nhc li mt s cng thc: Vi 0 < a, b 1; x, x1, x2 > 0 vy, R, ta c: V d: Tnh A = log1325. Gii: 13ln25 A log 25 1, 254947126ln13= = . aa ya alog xa 1 2 a 1 a 21a a 1 a 22a aa ay log x1)x a . a c bie t, log 1 0;log a 1.x 02) a x.3) log (x x ) = log (x ) + log (x ).x4) log ( ) = log (x ) - log (x ).x1a c biet, log ( ) = - log (x).x5) log (x ) =log (x).6) l= = = =>=aaa a babae101og (x) =log (x) (0).7) log x = log b.log x;log x log x =.log b8) lnx = log x : LogaritNepe cua x. lgx = log x : Logarit tha p pha n cua x. 6 1.4. Hm s lng gic v hm ngc 1.4.1. Hm y = sinx v y =arcsinx: Vi 1 a 1, ta nh ngha: sin a;arcsina.2 2 = = Khi arcsina (1 a 1) c xc nh duy nht. Nh vy, y= arcsinx l hm s c tnh cht sau: Min xc nh: D = [1;1]. Min gi tr:[ ; ].2 2 [ ; ], a [ 1;1]; sin a arcsina .2 2 = = y = arcsinx l hm s l, ngha larcsin(x) = arcsinx. V d: arcsin(1/2) = /6; arcsin( 3 /2) = arcsin( 3 /2) = /3; arcsin(1/2) =/6; arcsin(3/4) = arcsin(3/4) 0,848062079; arcsin(4) khng tn ti. 1.4.2. Hm y = cosx v y =arccosx:7 Vi 1 a 1, ta nh ngha: cos a;arccosa0 . = = Khiarccosa(1a1)cxcnhduynht.Nhvy,y=arccosxlhmsc tnh cht sau: Min xc nh: D = [1;1]. Min gi tr:[0; ]. [0; ], a [ 1;1]; cos a arccosa . = = arccos( x) = arccosx. V d: arccos(1/2) = /3; arccos( 3 /2) = arccos( 3 /2) = /6 = 5/6; arccos( 2 /2) = arccos( 2 /2)= 3/4; arccos(3/4) = - arccos(3/4) 2,418858406; arccos( 4) khng tn ti. 1.4.3. Hm y = tgx v y =arctgx: 8 Vi a R, ta nh ngha: tg a;arc tga.2 2 = = < < Khi arctga c xc nh duy nht. Nh vy, y= arctgx l hm s c tnh cht sau: Min xc nh: D = R. Min gi tr:( ; ).2 2 ( ; ), a , tg a arctga .2 2 = = Ry = arctgx l hm s l, ngha larctg(x) = arctgx. V d: arctg1 = /4; arctg( 3 /3) = arctg( 3 /3) = /6; arctg(1)=/4; arctg(3/4) 0,643501108; arctg( 4) 1,3258. 1.4.4. Hm y = cotgx v y =arccotgx: Vi a R, ta nh ngha: cotg a;arc cotga0 . = = < < Khi arccotga c xc nh duy nht. Nh vy, y= arccotgx l hm s c tnh cht sau: Min xc nh: D = R. Min gi tr:(0; ). (0; ), a , cot g a arc cot ga . = = Rarccotg(x) = arccotgx. V d: arccotg1 = /4; arccotg( 3 /3) = arccotg( 3 /3) = /3 = 2/3; 9 arccotg( 3 ) = arccotg( 3 ) = /6 = 5/6;arccotg(3/4) = /2 arctg(3/4) 0,927295218 arccotg(4) = /2 arctg(4) /2 + arctg4 2,89661399. trong ta s dng tnh cht sau: 1.4.5. Tnh cht: 1) Vimi1 x 1, arcsinx + arccosx = /2. 2) Vi mi x, arctgx + arccotgx = /2. 2. HM S S CP Hm s s cp l hm s c xy dng t cc hm hng v cc hm s s cp c bn qua cc php ton i s: cng, tr, nhn, chia v php hp ni nh x. V d:y ln(1 2x) = +l mt hm s s cp. sin6xne u x < 0;y xcos3xne u x0.= khng l hm s s cp. 10 B. GII HN 1. NH NGHA V TNH CHT 1.1. nh ngha. 1) Cho hm s f(x) xc nh trn mt khong chax0 (c th loi tr x0). Ta ni f(x) c gii hn l L R khi x tin vx0, nu f(x) c th gn L ty khi x tin st n x0. K hiu: 00x xlimf (x) Lhayf(x) Lkhi x x= . Chnh xc hn, theo ngn ng ton hc, ta c: 00x x0 0 0limf (x) L 0, 0, x , 0 |x x | |f (x) L| 0, 0, x , x x x x |f (x) L|= > > < < < > > < < + < RR Minh ha: 2) Cho hm s f(x) xc nh trn mt khong c dng (a;x0). Ta ni f(x) c gii hn l L R khix tin v x0 bn tri, nu f(x) c th gn L ty khi x tin st n x0 v pha bn tri. K hiu: 00x xlimf (x) Lhayf(x) Lkhi x x= . Chnh xc hn, theo ngn ng ton hc, ta c: 00x xlimf (x) L 0, 0, x , 0 x x |f (x) L|= > > < < < R Minh ha: 3) Cho hm s f(x) xc nh trn mt khong c dng (x0;b). Ta ni f(x) c gii hn l L R khix tin v x0 bn phi,nu f(x) c th gn L ty khi x tin st n x0 v pha bn phi. K hiu: 00x xlimf (x) Lhayf(x) Lkhi x x++= . Chnh xc hn, theo ngn ng ton hc, ta c: 11 00x xlimf (x) L 0, 0, x , 0 x x |f (x) L|+= > > < < < R Minh ha: Nh vy, t cc nh ngha trn ta suy ra; 000x xx xx xlimf (x) L; limf (x) L limf (x) L.+= = = 4) Tng t, ta nh ngha c cc gii hn: 0 0 0x x x x x xlimf (x) ; limf (x) ; limf (x) ; ... = + = = . 1.2. nh l. Cho cc hm s f(x), g(x) khi x x0. Khi , vi a, b R, ta c: 1)Nu f(x) a, g(x) b th : f(x) + g(x) a + b; f(x) g(x) a b; f(x)g(x) ab; f(x)/g(x) a/b (nu b 0). 2)Nu f(x) a, g(x) th f(x) + g(x) . 3)Nu f(x) +, g(x) + th f(x) + g(x) +. 4)Nu f(x) a 0, g(x) th f(x)g(x) . 5)Nu f(x) , g(x) th f(x)g(x) . 6)Nu f(x) a 0, g(x) 0 th f(x)/g(x) . 7)Nu f(x) a, g(x) + th f(x)/g(x) 0. 8)Nu f(x) , g(x) b th f(x)/g(x) . 9)Nu f(x) a > 1, g(x) + th f(x)g(x) +. Nu f(x) a vi 0 < a < 1, g(x) + th f(x)g(x) 0. 10) Nu f(x) a th|f(x)| |a|. 11) f(x) 0 |f(x)| 0. 12) (Gii hn kp) Gi sf(x) h(x) g(x), x kh gn x0 vf(x) a; g(x) a. Khi h(x) a. 12 1.3. nh l. Cho f(x) l mt hm s s cp xc nh ti x0. Khi 00x xlimf (x) f (x ).=V d: 1) x21 cos2x 1 coslim 2.sinxsin2 = =

2)x 01 cos2xlim sinx+= x 0 x 0(vlim(1 cos2x) 1 cos0 2va limsinx sin0 0) + = + = = = 1.4. Cc dng v nh trong gii hn: C tt c 7 dng v nh trong gii hn, l:0 00;0 ; ; ;1 ;0 ; .0 1)Dng : Khif(x) + ( ) vg(x) + ( ) th ta ni lim (f(x) g(x)) c dng v nh . 2)Dng0 : Khif(x) 0 vg(x) th ta ni lim f(x)g(x) c dng v nh0 (Lu : f(x) 0 khng c ngha l f(x) 0). 3)Tng t cho 5 dng cn li. Ta ni cc dng trn l cc dng v dnh v khng c qui tc chung xc nh gi tr ca gii hn nu ch da vo cc gii hn thnh phn. tnh cc gii hn c dng v nh, ta cn bin i lm mt i dng v nh, gi l kh dng v nh. 2. HM TNG NG 2.1. nh ngha. Cho cc hm s f(x), g(x) xc nh v khng trit tiutrn mt khong chax0 (c th loi tr x0). Ta ni f(x) tng ng vi g(x) khi x x0, k hiuf(x) g(x) khi x x0, nu 0x xf (x)lim 1.g(x)=Nh vy,0x xf (x)f (x) g(x) lim 1g(x)(f (x), g(x) 0) = Cc tnh cht sau c tha: 1)f(x) f(x). 2)f(x) g(x) g(x) f(x). 13 3)f(x) g(x) v g(x) h(x) f(x) h(x). 2.2. nh l. 1) Nu f(x) L R, L 0, thf(x) L. 2)Nu f(x) g(x) v g(x) A th f(x) A. 3)Nu 1 12 2f (x) g (x);f (x) g (x).th 1 2 1 21 12 2f (x)f (x) g (x)g (x);f (x) g (x).f (x) g (x) 4)Nu f(x) g(x) th n nf (x) g(x) (gi s cc cn c ngha). Ch : Ta khng th vit f(x) 0 hay f(x) (ngay c khif(x) 0 hay f(x) )v iu ny v ngha! 1 12 2f (x) g (x);f (x) g (x).1 2 1 21 2 1 2f (x) f (x) g (x) g (x);f (x) f (x) g (x) g (x).+ + Chng minh: 1) Nu f(x) L R, L 0, th f (x)lim 1L=nn f(x) L ( y L c xem nh hm hng). 2) Nu f(x) g(x) v g(x) A th f (x)f (x) g(x) 1.A Ag(x)= = . 3) Gi s 1 12 2f (x) g (x);f (x) g (x). Khi 1 21 2f (x) f (x)lim lim 1.g (x) g (x)= =t 1 2 1 21 2 1 21 2 1 21 2 1 2f (x)f (x) f (x) f (x)lim lim . lim 1.1 1;g (x)g (x) g (x) g (x)f (x) / f (x) f (x) f (x)lim lim / lim 1/ 1 1.g (x) / g (x) g (x) g (x)= = == = = Suy ra 1 2 1 21 12 2f (x)f (x) g (x)g (x);f (x) g (x).f (x) g (x) 4)Gi s f(x) g(x). Khi nnnnf (x) f (x)lim lim 1 1.g(x)g(x)= = = Suy ra n nf (x) g(x) . 14 2.3.Mt s gii hn v tng ng c bn: GII HN TNG NG x 0sinxlim 1x=(x: rad) sinx x khi x0 (x: rad) 2x 01 cosx 1limx 2=(x: rad)1 cosx12 x2khi x0 (x: rad) x 0tgxlim 1x=(x: rad) tgx x khi x0 (x: rad) x 0arcsinxlim 1x=arcsinx x khi x0 x 0arctgxlim 1x=arctgx x khi x0 xx 0e 1lim 1x=ex 1 x khi x0 x 0ln(1 x)lim 1x+=ln(1+ x) x khi x0 x 0(1 x) 1limx+ = (1+x)1 x khi x0 ( 0) x xx xlime ; lime 0.+ = + = x x 0limln x ; limln x .++ = + = x x2 2limtgx ; limtgx . + = + = x xlimarctgx ; limarctgx .2 2+ = = ( )x1xx x 01lim 1 e; lim 1 x e.x + = + = Khi x:anxn + an1xn1+...+amxm amxm Khi x 0:anxn + an1xn1+...+amxm amxm (m < n; an 0; am 0) V d. Tnh cc gii hn sau: 2 21 2 2x x 0 x 18 63 8 7 4xlncos2x (x 5x 4) arcsin(x x)a) L lim ;b) L lim ;(x 3x) sinx(e e)(1 4x 3)3x 5x 4x 2c) L lim .x 5x 14x 1 + = =+ + += + + Gii. 1 2x 0lncos2xa) L lim(x 3x) sin x=+. Khix0 ta c lncos2x = ln[1 + (cos2x 1)] cos2x 1 (1/2)(2x)2= 2x2(1) 15 x2 + 3x 3x (2) sinxx (3) T (2) v (3) ta suy ra: (x2 + 3x)sinx 3x.x = 3x2(4) T (1) v (4) ta suy ra:22 2lncos2x 2x 2(x 3x) sinx 3x 3= + . Do 12L3= . 2 22xx 1(x 5x 4) arcsin(x x)b) L lim(e e)(1 4x 3) + = . t t = x 1 x = t+1 . Khix1 ta c t 0. Do 2 2 2 22x t x 1 t 0(x 5x 4) arcsin(x x) (t 3t) arcsin(t t)L lim lim .(e e)(1 4x 3) e(e 1)(1 1 4t) + += = + Khit0 ta c: t2 3t 3t, (1) arcsin(t2 + t) t2 + t t. (2) T (1) v (2) ta c: (t2 3t) arcsin(t2 + t) 3t.t 3t2.(3) Mt khc, et 1 t(4) 1211 1 4t 1 (1 4t) (4t) 2t2 + = + = (5) T (4) v (5) ta c:t 2e(e 1)(1 1 4t) et( 2t) 2et + = (6) T (3) v (6) ta suy ra: 2 2 22t(t 3t) arcsin(t t) 3t 3.2e 2ete(e 1)(1 1 4t) + +Do 23L2e= . 8 63 8 7 4x3x 5x 4x 2c) L lim .x 5x 14x 1 + += + +Khi x ta c 3x8 5x6 + 4x + 2 3x8

x8 5x7 + 14x4 + 1 x8

Suy ra 8 6 88 7 4 83x 5x 4x 2 3x3.x 5x 14x 1 x + + + +Do 3L 3. =16 3. V CNG B (VCB)-V CNG LN 3.1. V CNG B (VCB) 1)nh ngha. Ta ni f(x) l mt VCB khixx0 nu 0x xlimf (x) 0= . 2) So snh hai VCB: Cho f(x) v g(x) l VCB khi x x0. Gi s 0( )lim .( )=x xf xLg x

a) Nu0 L =th ta ni VCB f(x) c cp cao hn VCB g(x). b) Nu= Lth ta ni VCB f(x) c cp thp hn VCB g(x). c) Nu0 L < < + th ta ni hai VCB f(x) v g(x) c cng cp. 3) Bc ca VCB khi x 0: Cho f(x) l mt VCB khi x0. Ta ni VCB f(x) c cp khi chn x lm VCB chnh nu:f(x) ax khix0 trong a 0v > 0. Nhnxt:Ccnhnghatrong2)v3)tngthchnhaukhitasosnhhaiVCBkhi x 0. V d: Khix0,1 cos4x l mt VCB cp 2 v 2 211 cos 4x (4x) 8x .2 = va co cung cap thap cao hn 4) Tng (hiu) hai VCB: Cho f(x), g(x) l hai VCB khi x x0. a) Nu f(x) v g(x) khng c cng cp th f(x)ne u f(x) coca p tha p hn g(x);f(x) + g(x) g(x)ne u f(x) coca p cao hn g(x). b) Nu f(x) v g(x) c cng cp nhng khng tng ng th f(x) g(x) l VCB c cng cp viVCB f(x), hn na11 11f (x) f (x)f (x) g(x) f (x) g (x).(*)g(x) g (x) c bit, cho f(x), g(x) l hai VCB khi x0 c cp ln lt l , : f(x) ax(a 0); g(x) bx (b 0). Khi 17 ax ne u; (a b)xne u=;ab 0. Ch : Trng hp hai VCB f(x) v g(x) tng ng v f(x) f1(x), g(x) g1(x)th f(x) g(x) l VCBc cp ln hn VCB f(x) nhng (*) khng cn ng. 5) Qui tc gi li VCB cp b nht (Qui tc ngt b VCB cp cao): Gi s khi xx0, VCB f(x) c phn tch thnh tng ca nhiu VCB, trong ch c mt VCB cp thp nht l f0(x). Khi :f(x) f0(x) khix0. Ch : Trng hp c nhiu VCB cp b nhttrong phn tch ca f(x) th ta gp cc VCB li, xem nh l mt VCB v dng tnh cht 4b) trn kho st cp ca VCB , sau mi c th p dng qui tc trn. 3.2. V CNG LN (VCL) 1)nh ngha: Ta ni f(x) l mt VCL khixx0 nu 0x xlimf (x)= . 2) So snh hai VCL: Cho f(x) v g(x) l VCL khi x x0. Gi s 0( )lim .( )=x xf xLg x

a) Nu0 L =th ta ni VCL f(x) c cp thp hn VCL g(x). b) Nu= Lth ta ni VCL f(x) c cp cao hn VCL g(x). c) Nu0 L < < + th ta ni hai VCL f(x) v g(x) c cng cp. 3) Bc ca VCL khi x : Cho f(x) l mt VCL khi x . Ta ni VCL f(x) c cp khi chn x lm VCL chnh nu:f(x) ax khix trong a 0v > 0. Nhnxt:Ccnhnghatrong2)v3)tngthchnhaukhitasosnhhaiVCLkhix . V d: Khix ,2x3 9x2 + 5x + 19 VCL cp 3 v 3 2 32x 9x5x 19 2x . + + 4) Tng (hiu) hai VCL: Cho f(x), g(x) l hai VCL khi x x0. a) Nu f(x) v g(x) khng c cng cp th f(x)ne u f(x) coca p cao hn g(x);f(x) + g(x) g(x)neu f(x) coca p thap hn g(x). 18 b) Nu f(x) v g(x) c cng cp nhng khng tng ng th f(x) g(x) l VCL c cng cp viVCL f(x), hn na11 11f (x) f (x)f (x) g(x) f (x) g (x).(*)g(x) g (x) c bit, cho f(x), g(x) l hai VCL khi x c cp ln lt l , : f(x) ax(a 0); g(x) bx (b 0). Khi ax ne u>;f (x) g(x) bx ne u 1.x 2x 3 = + + Gii. Trn (; 0), y trng vi hm f(x) = 21 cos6xx. V f(x)l hm s s cp xc nh vi mi x 0, nn y lin tc trn ( ; 0). 24 Trn (0; 1), y trng vi hm g(x) = ax + b. V g(x)l hm s s cp xc nhvi mix R, nn y lin tc trn (0; 1). Trn (1; +), y trng vi hm h(x) = 2ln xx 4x 3 +. V h(x)l hm s s cp xc nhvi mi x > 0, x 1, nn y lin tc trn (1; +). Suy ra y lien tuctai x = 0;y lien tuc tren Ry lien tuc tai x = 1. (1) x 0x 0x 0y lien tuc ben traitai x = 0; y lien tuc tai x = 0y lie n tuc be n pha ita i x = 0.limy = y(0); limy = y(0).lim + 2x 022x 01 cos6x = b;xlim(ax+b) = b.1(6x)2lim= bxb 18(2)+ = x 1x 1x 1y lie n tu c be n tra itai x = 1; y lie n tu c ta i x = 1y lien tuc ben phaitai x = 1.limy = y(1); limy = y(1).lim + 2x 12x 1t 0t 0(ax b)= a+b;ln xlim= a+b.x 2x 3ln xlim= a+bx 2x 3ln(1 t)lim= a+b (t = x-1)t(t 4)tlim4t+++++ + + ++ = a+b 1a+b = (3)4 T(1), (2) v (3) ta suy ra: 25 1a+b= ;y lie n tu c tre n 4b = 18.71a=- ; 4b = 18. R 2. HM S LIN TC TRN MT ON 2.1. nh l. Cho hm s f(x) lin tc trn on [a; b]. Khi 1)f(x) t gi tr ln nht v gi tr nh nht trn [a; b], ngha l 1 2 1 2M, m , x [a; b], m f (x) M;x , x , f (x ) M; f (x ) m. = =RR 2)f(x) t mi gi tr trung gian gia gi tr ln nht M v gi tr nh nht m trn [a; b], ngha l 0 0m k M, x [a; b], f (x ) k = . Minh ha: 2.2. H qu. 1) Cho hm s f(x) lin tc trn on [a; b]. Gi sf(a)f(b) < 0, ngha l f(a) v f(b) tri du. Khi phng trnh f(x) = 0 c nghim trn khong (a; b), ngha l tn ti x0 (a; b) sao cho f(x0) = 0. 2) Gi s hm s f(x) lin tc trn khong (a; b) v phng trnh f(x) = 0 v nghim trn khong ny. Khi f(x) khng i du trn khong (a; b). V d 1. Chng minh mi phng trnh i s bc l: anxn + an-1xn-1+...+ a0 = 0(an 0)(1) vi n nguyn dng l, lun lun c nghim thc. Gii. t f(x) = anxn + an-1xn-1+...+ a0 = 0. 26 Khi x , f(x) anxn (do n l), nn tn ti a < 0 kh b sao cho f(a) < 0.Khi x +, f(x) anxn +, nn tn ti b > 0 kh ln sao cho f(b) > 0. V hm s f(x) lin tc trn [a; b] v f(a)f(b) < 0 nn theo h qu trn, phng trnh f(x) = 0 c nhim trn [a; b] vdo phng trnh (1) c nghim thc. V d 2. Gii bt phng trnh: 2 2(x 6x 5)(1 lnx)(x x 4x 20) 0(1) + + 0.t 2 2f(x) (x 6x 5)(1 lnx)(x x 4x 20) = + + .Tacf(x)lintctrn (0; +). Hn na, 2 222f (x) 0 (x 6x 5)(1 lnx)(x x 4x 20) 0x 6x 5 0 x 1 x 5 1 lnx 0 x ex 5x x 4x 20 0 = + + = + = = = = == + = Ta lp bng xt du: x01e5 +f(x)-0+0-0 - Lu rng, do tnh lin tc, f(x) khng i du trn mi khong ca bng xt du. Do ta ch cn th mt gi tr ca mi khong vo f(x) bit du ca f(x) trn cc khong ny. T bng xt du trn ta suy ra: f(x) 0 < x < 1 haye < x < 5 hayx > 5. Do , bt phng trnh (1) c tp nghim l: S = (0;1) (e; 5) (5; + ). 27 D- O HM V VI PHN 1. KHI NIM O HM 1.1. nh ngha. 1) Cho hm f(x) xc nh trn mt khong cha x0. Khi cho x0 mt s gia xkh b th s gia tng ng ca f(x)l y = f(x0 + x) f(x0). Lp t s 0 0f (x x) f (x ) yx x+ = . Nu t s ny cgii hn l A R khi x 0 th ta ni f(x) c o hm ti x0 v Al o hm ca f(x) ti x0, k hiu f '(x0) = A.Nh vy, 0 00x 0 x 0f (x x) f (x ) yf '(x ) lim limx x + = = 2) Tng t, ta nh ngha: f(x) c o hm bn tri ti x0, k hiu 0f (x ), nu tn ti gii hn hu hn: 0 00x 0f (x x) f (x )f (x ) lim .x+ = f(x) c o hm bn phi ti x0, k hiu 0f (x )+, nu tn ti gii hn hu hn: 0 00x 0f (x x) f (x )f (x ) lim .x+++ = 3) f(x) c o hm trn (a,b) nu f(x) c o hm ti mi x0 (a,b). 4) f(x) c o hm trn [a,b] nu f(x) c o hm trn (a,b) v c o hm bn phi ti a, o hm bn tri ti b. 1.2. ngha hnh hc ca o hm: o hm f '(x0) chnh l h s gc ca tip tuyn vi ng cong (C): y = f(x) ti im M0(x0,y0)(C). Do phng trnh ca tip tuyn vi ng cong (C): y = f(x) ti im M0(x0,y0)(C) l: 0 0 0y - y= f '(x )(x-x ) 1.3. ngha kinh t ca o hm: 1) nh ngha: Bin t ca i lngy = f(x) theo i lng x ti x0, k hiu Mxy(x0), l bin i ca i lng y khi i lng x tng ln 1 n v ti x0. 2) Biu thc ton hc ca bin t: Gi s ti x = x0 ta cho x mt s gia l x n v. Khi bin i ca i lng y = f(x) l y = f(x0 + x) f(x0). Do khi x tng 1 n v th bin i trung bnh ca i lngy = f(x) l 28 0 0f (x x) f (x ) yx x+ = . bit chnh xc bin i ca i lng y = f(x) khi x tng 1 n vti trng thi(x0,y0) ta phichuynquagiihnkhix0.Theonhnghatrn,binichnhlbint Mxy(x0) ca y = f(x) theo x ti x0 nn0 0x 0 0x 0 x 0f (x x) f (x ) yM y(x ) lim lim y'(x )x x + = = = . Nh vy, bin t ca i lng y = f(x) theo i lng x ti x0 chnh l o hm y(x0) = f(x0) ca y = f(x) ti x0: x 0 0M y(x ) y'(x ) =Tng qut, bin t ca i lng y = f(x) theo i lng x chnh l o hmy = f(x) ca y = f(x): xM y y' =Ch : Trong thc t, bin t Mxy(x0) ca y = f(x) theo x ti x0 xp x bng bin i ca y khi i lng x tng ln 1 n v t trng thi x = x0. V d: Xt m hnh sn xut mt loi sn phm. Khi hm tng chi ph C = C(Q) l hm theo tng sn phm Q.Chi ph bin t l: MC(Q) = C(Q). Chng hn, vi hm tng chi ph: C = Q3 + 2Q2+10 ta c chi ph bin t: MC(Q) = C(Q)= 3Q2+4Q. Ti Q = 100, ta c MC(100) = 30400. Nh vy, khiangsnxutvitngsnlngQ0=100,nutngtngsnlng1nvthnhQ1= 101, thtng chi ph s tng thm 30400 (Thc t l chi ph tng thm C(Q1) - C(Q0)= 30703). 3) bin i tuyt i v bin i tng i:Xt i lng x. Ti x = x0, cho x mt s gia x th x nhn gi tr mi l x0+x. Ta ni x l bin i tuyt i ca x ti x0 v t s 0xx l bin i tng ica x ti x0. bin i tng i thng c tnh bng %. 4) H s co gin: H s co gin ca i lng y = f(x) theoi lngx tix0, k hiu yx(x0), l bin i tng i ca y khi x tng tng i ln 1%. 5) Biu thc ton hc ca h s co gin: Gi s ti x0 ta cho x mt s gia l x n v. Khi : - bin i tuyt i ca x ti x0 l x. - bin i tng ica x ti x0 l29 0x.100%x. - bin i tuyt i ca y ti x0 l: y = f(x0 + x) f(x0). - bin i tng i ca y ti x0(y0 = f(x0)) l: 0 00 0f (x x) f (x ) y.100% .100%y y+ =Do , ti x = x0, khi x tng tng i 1% th bin i tng i trung bnh ca i lng y = f(x) l 0 000y.100%y x y. %xx y.100x=. bit chnh xc bin i tng i ca i lng y = f(x) khi x tng 1 n vti trng thi(x0,y0) ta phi chuyn qua gii hn khi x 0, Theo nh ngha trn, bin i tng i chnh l h s co gin yx(x0) ca i lng y = f(x) theo i lngx tix0 nn0 0yx 0 0x 00 0x x y(x ) lim . y'(x )x y y = =. Nh vy, h s co gin yx(x0) ca i lng y = f(x) theo i lngx tix0 nh bi: 0yx 0 00x(x ) y'(x )y = Tng qut, h s co gin yx(x) ca i lng y = f(x) theo i lngx nh bi: yxx(x) y'(x)y = V d. Xt m hnh sn xut mt loi sn phm. Khi hm cu QD = Q(P) l hm gim theo n gi P. H s co gin DQ Pthng c vit tt l D. Ta c: DPQ'(P) 0Q = Do 31 x2 2y1 1 1 1y' .x' cos y1 sin y 1 x= = = = Vy 21(arcsinx) '1 x= (1 < x < 1). Tng t, ta c 2x 11)' x (arccos = (1 < x < 1) 2) y = arctgx (x R, /2 < y < /2) l hm ngc ca hm x = tgy. Vi mi /2 < y < /2, ta c 2 2y 21x' 1 tg y 1 x .cos y= = + = + Do x 2y1 1y .x 1 x = =+ 2x 11)' arctgx ( Vay+= . Tng t, ta c 21(arc cot gx) .1 x = + 2.4. Bng o hm: O HM HM S f(x)O HM HM S f(u) vi u = u(x) 1(C) = 0 (C: Const)(C)(C: Const) 2(x) = x 1 (: Const)(u) = uu 1(: Const) 1( x)2 x = u( u)2 u =21 1x x = 21 uu u = 3(ex) = ex(eu) = ueu 32 4(ax) = axlna(0 < a: Const)(au) = uaulna(0 < a: Const) 51(ln|x|)x =u(ln|u|)u =6 a1(log |x|)x lna =(0 < a 1: Const) au(log |u|)u lna =(0 < a 1: Const) 7(sinx) = cosx(sinu) = ucosu 8(cosx) = sinx(cosu) = usinu 9 221(tgx) 1 tg xcos x = = +22u(tgu) u (1 tg u)cos u = = +10 221(cot gx) (1 cot g x)sin x = = +22u(cot gu) u (1 cot g u)sin u = +11 21(arcsinx)1 x = 2u(arcsinu)1 u = 12 21(arccos x)1 x = 2u(arccos u)1 u = 13 21(arc tgx)1 x =+ 2u(arc tgu)1 u =+ 14 21(arc cotgx)1 x =+ 2u(arc cotgu)1 u =+ 2.5. o hm ca hm s dng y = uvvi u = u(x); v = v(x) tnh o hm ca hm s trn ta tin hnh nh sau: Ly logarit c 2 v ca y = uv, ta c: lny = vlnu(1) Ly o hm 2 v ca (1), ta c: y' u'v' lnu v .y u= +Do vu' u'y' (v' lnu v )y (v' lnu v )u .u u= + = +V d:Tnh o hm ca hm y = xsinx. Gii. Ly logarit c 2 v ca y = xsinx, ta c lny = sinxlnx (1) 33 Ly o hm 2 v ca (1), ta c y' sin xcos x ln xy x= + . Do sinxsinxy' x (cosxlnx )x= +. 2.6. o hm ca hm n Xt phng trnhF(x,y) = 0(1) Gi s y = y(x) (x D) l hm s tha F(x,y(x)) = 0 vi mi x D. Ta ni y l hm n c xc nh bi phng trnh (1). Tacthtmohmy'cahmnyxcnhbiphngtrnh(1),theoxvy,m khng cn xc nh biu thc tngminh ca hm s y = y(x), bng cch ly o hm hai v ca (1) theo bin x, trong y l mt hm theo bin x. Ch rng khi ly o hm nh vy ta phi s dng nh l v o hm hm hp. V d 1: Tm o hm y' = y'(x) ca hm n y = y(x) xc nh bi phng trnh tgy = xy. Gii. Ly o hm hai v ca phng trnh tgy = xy ta c (1 + tg2y)y =y + xy Suy ra(1 + x + tg2y)y =y. T 2yy'1 x tg y= +.

V d 2: Tm o hm y' = y'(0) ca hm n y = y(x) xc nh bi phng trnh x3 xy xey + y 1 = 0.(2) Gii. Ly o hm hai v ca phng trnh x3 xy xey + y 1 = 0. ta c 3x2 y xy ey xeyy + y = 0.(3) Thx=0vo(2)tacy=1.Thx=0,y=1vo(3)tac1e+y=0.Suyray(0) =1 + e. 2.6. o hm ca hm s cho bi phng trnh tham s Gi s hm s y ph thuc bin s x khng trc tip m thng qua mt bin s trung gian t: x (t);y (t)= = v hm s x = (t) c hm ngc t = 1(x), hn na cc hm, v 1 u c o hm. Khi hm sy =[1 (x)] c o hm theo x. Tht vy, ta c yt = yx.xt. Suy ra 34 txtyyx = V d 1. Tm o hm y= y(x) ca hm s y = y(x) cho bi phng trnh tham s: 2x ln(1 t );y 2t 2arctgt. = += Gii. Ta c 2tx 2t tt222y (2t 2arctgt)1 ty t.2t x (ln(1 t ))1 t + = = = = ++ V d 2. Tm o hm y= y(2) ca hm s y = y(x) cho bi phng trnh tham s: t2x 2e ;y t t . == + Gii. Ta c 2tx t tt tty (t t ) 1 2ty .x (2e ) 2e + + = = = Ti x = 2 ta c 2et = 2 nn t = 0. Suy ra y'(2) = 1/2. 3. VI PHNCho hm s f(x) c o hm ti x0. t0 00f (x x) f (x )( x) f (x ).x+ = Khi ,( x) 0 khi x 0 , v0 0 0 0f (x x) f (x ) f (x ) x x ( x)f (x ) x o( x) + = + = + , trong o( x) x ( x) = . Ch rng o( x)( x) 0 khi x 0x= nno( x) l mt VCB cp cao hn VCB x khix 0. Ta ni f(x) kh vi ti x0 v vi phn ca f(x) ti x0 lf '(x0)x theo nh ngha sau: 3.1. nh ngha. Cho hm s f(x) xc nh trn mt khong cha x0. Ta ni f(x) kh vi ti x0 nu tn ti mt hng s A v mt hm so( x) l VCB cp cao hn VCB x khix 0 sao cho vi mi x kh b ta c 35 0 0f (x x) f (x ) A x o( x) + = + . Khi i lng Ax c gi l vi phn ca f(x) ti im x0, k hiu l df(x0). Nh vy, df(x0) = Ax. L lun trn cho thy nu f(x) c o hm ti x0 th f(x) kh vi ti x0 v df(x0) = f'(x0)x. Tng qut hn, ta c kt qu sau: 3.2. nh l. Hm s f(x) kh vi ti x0 khi v ch f(x) c o hm ti x0. Khi vi phn ca f(x) ti x0 l df(x0) = f (x0)x.3.3. Biu thc ca vi phn: T kt qu trn, ta c vi phn ca f(x) nh bi: df(x) = f (x)x. Nhn xt rng vi g(x) = x th g(x) = 1, do dg(x) = 1.x = x, ngha l dx = x. Do ta c biu thc ca vi phn ca f(x) nh sau: df(x) = f (x)dx Ch . Do cng thc trn, ta c: df (x)f (x) .dx = 3.4. ngha ca vi phn v cng thc tnh gn ng: Cho hm s f(x) kh vi ti x0. Khi vi mi x kh b ta c 0 0 0f (x x) f (x ) f (x ) x o( x) + = + . Vo( x) l VCB cp cao hn VCB x khix 0 nn khi x kh b ta c 0 0 0f (x x) f (x ) f (x ) x. + Ni cch khc, khix kh b, s gia f(x0) ca f(x) ti x0 gn bng vi phn df(x0) ca f(x) ti x0 v ta c cng thc tnh gn ng: 0 0 0f (x x) f (x ) df (x ) + + Vd.Chohmsy=arctgx.Tmccviphndyvdy(1).pdng:Tnhgnng arctg(1,02). Gii. 1) Vi phn dy = y'dx = 2x 11+dx. 2) Vi phn dy(1) = y'(1)dx = 21 1dx dx1 1 2=+. 3) Ta tnh gn ng arctg(1,02) nh sau: tx0 = 1; x = 0,02. Ap dng cng thc tnh gn ng cho hm s y = arctgx, ta c 36 y(1,02) y(1) + dy(1). Do arctg(1,02) arctg1 + (1/2).0,02= /4 + 0,01. 3.5. nh l. Cho cc hm s u = u(x) v v = v(x) c cc vi phn l du v dv. Ta c 1d(u + v) = du + dv 2d(ku) = kdu; (k: Const) 3d(uv) = udv + vdu 4 2u vdu udvd (v 0)v v = 4. O HM V VI PHN CP CAO 4.1. o hm cp cao 1) nh ngha. Gi shm s f(x) c o hm f (x). Ta cn gi f (x) l o hm cp mt ca f(x).Nu hm s f (x) li c o hm th o hm c gi l o hm cp hai ca f(x),k hiu l f (x) hay f(2)(x). Nh vy, f (x) = [f (x)]. Tng qut, o hm ca o hm cp (n1) ca f(x) c gi l o hm cp n ca f(x),k hiu l f(n)(x). Nh vy, (n) (n 1)f (x) [f (x)] =2) nh l. Gi s cc hm u = u(x) v v = v(x) c cc o hm cp n lu(n) = u(n)(x); v(n) = v(n)(x). Ta c (n) (n) (n)(n) (n)n(n) k (k) (n k)nk=0a) (u + v) = u + v ;b) (ku)= ku ; (k: Const)c) (uv)=C u v trong u(0) = u v v(0) = v. V d. Tm o hm cp n ca cc hm s sau: a)y = xn;b) y = sinx; c) y cosx;d)1yx a=+ (a: const)e) y = x2sinx. Gii. a) Viy = xn, ta c y = nxn1, y = n(n 1)xn2, y = n(n 1)(n 2)xn3, ...,y(n) = n(n 1)(n 2)...3.2.1 = n!,37 y(k) = 0; k > n. b) Viy = sinx, ta c y' = cosx = sin(x + 2), y'' = cos(x + 2) = sin(x + 2 + 2) = sin(x + 22). Tng qut, y(n) = sin(x + n2).c) Tng t, viy = cosx, ta c: y(n) = cos(x + n2). d) Vi 1yx a=+, ta c 22 31 2y ; y ( 1)(x a) (x a) = = + + Ta chng minh (n) nn 1n!y ( 1)(x a)+= +.(1) Vin = 1, (1) ng. Gi s (1) ng vi n = k, ngha l(k) kk 1k!y ( 1) .(x a)+= + Vi n = k + 1, ta c k(k 1) k k k 1k 1 2(k 1) k 2k! k!(k 1)(x a) (k 1) !y ( 1) ( 1) ( 1) .(x a) (x a) (x a)+ ++ + + + + += = = + + + Vy (1) cng ng vi n = k + 1. Ta kt lun(n) nn 1n!y ( 1)(x a)+= + vi mi n 1. e) t u = x2, v = sinx y = uv. Theo cc kt qu trn ta c u = 2x, u = 2, u(k) = 0, k 3; v(m) = sin(x + m2).Suy ra38 2 2y 2xsinx x sin(x ) x sin(x ) 2xcos(x ).2 2 2 = + + = + +Vin 2 ta cn(n) (n) k (k) (n k) (n) 1 (n 1) 2 (n 2)n n nk 022 2y (uv) C u v uv C u v C u v x sin(x n ) 2nxsin(x (n 1) ) n(n 1) sin(x (n 2) ).2 2 2 (x n n) sin(x n ) 2nxcos(x n ).2 2 = = = = + + = + + + + + = + + + Kt lun(n) 2 2y (x n n) sin(x n ) 2nxcos(x n ), n 1.2 2 = + + + 4.2. Vi phn cp cao Gi shm s f(x) c vi phn df(x)= f'(x)dx. Ta cn gi df(x) l vi phn cp mt ca f(x).Nu hm s f (x) kh vi th df(x) = f (x)dx c vi phn v vi phn c gi l vi phn cp hai ca f(x),k hiu d2f(x). Ta c d2f(x) = d(df(x)) = d[f (x)dx] = f (x)dx.dx= f (x)dx2. Vy d2f(x) = f(2)(x)dx2.Tng qut, vi phn ca vi phn cp (n 1) ca f(x) c gi l vi phn cp n ca f(x), k hiu dnf(x). Ta c n (n) nd f(x) = f (x)dxV d. Viy = sinx, ta c dny = y(n) dxn = sin(x + n2)dxn. 5. QUI TC LHOSPITAL 5.1. nh l (Qui tc LHospital).Xtgiihn x Af (x)limg(x)cdngvnh 00hoc (ngha l: f(x)0, g(x)0 hoc f(x), g(x)). Gi s tn ti gii hn x Af (x)lim L.g (x)= Khi x Af (x)lim Lg(x)=. 5.2. Ch . 1) Nu sau khi sdng Qui tc LHospital m gii hn vn cn dng v nh00 hoc th ta c th sdng tip qui tc ny. Lu : Nn kt hp viqui tc thay th hm tng ng vic tnh o hm c d dng hn. 39 2) Qui tc LHospital ch c p dng trc tip cho gii hn thuc hai dng v nh00 v . i vi cc dng v nh khc, mun p dng ta cn a v mt trong hai dng v nh trn m ta c th tm tt trong bng sau: BNG P QUI TC LHOSPITAL TM GII HNDNG V NHGII HN BIN IQUI TC LHOSPITAL 0/0 x Af (x)L limg(x)= x Af (x)L limg (x)= / x Af (x)L limg(x)= x Af (x)L limg (x)= 0. x AL limf (x)g(x)= x Af (x)L lim1g(x)= hay x Ag(x)L lim1f (x)= x Af (x)L lim1g(x)= hay x Ag (x)L lim1f (x)= x AL lim[f (x) g(x)]= x A1 1g(x) f (x)L lim1f (x)g(x)=x A1 1g(x) f (x)L lim1f (x)g(x) = 1 g(x)x AL limf (x)= x AKlim[f (x) 1]g(x)L e= x A[f (x) 1]K lim1g(x)= hay x Ag (x)K lim1f (x) 1= 1 00 0 g(x)x AL limf (x)= x AKlim g(x) lnf (x)L e= ( )x Alnf (x)K lim1g(x)= 40 V d. Tnh cc gii hn sau: x x1x 0e e 2xL lim .x sin x = 2x 1xln|cos( )|2L lim .xtg( )2= 232x 01L lim( cot g x).x= 10 x4xL limx e .= 1x xx5x 03 4L lim2 += ( )3ln|sin(2 x)|6x 2L lim x 2+= ln(1 2x)7x 0L lim(cot gx)++= Gii. x x1x 0e e 2x1) L lim .x sinx =TathyL1cdngvnh 00.ApdngQuitc LHospital, ta c: x x x x x x1x 0 x 0 x 0x xx 0L' Hosp L' Hospe e 2x e e 2 e eL lim lim limx sinx 1 cos x sinxL' Hospe elim 2.cos x + = += = 2x 1xln|cos( )|22) L lim .xtg( )2= Ta thy L2 c dng v nh. Ap dng Qui tc LHospital, ta c: 2x 1 x 1x 12xsin( )2 2x xln|cos( )| cos( )x x2 2L lim lim limsin( )cos( ) 0.x 2 2tg( )2 2xcos ( )2 = = = = 23 2x 013) L lim( cot g x).x= Ta thy L3 c dng v nh . Ta bin i 41 2 222 2 2 2 2 2 21 1 1 t g x x (t gx x)(t gx x)cot g x .x x t g x x tg x x tg x + = = = Khi x 0, ta ctgx + x ~ 2x vx2tg2x ~ x4. Do 22 2 2 4 31 (t gx x)(t gx x) 2x(t gx x) 2(t gx x)cot g xx x tg x x x+ = = Suy ra 3 3x 0t gx xL 2limx= .TathybygigiihnL3cdngvnh 00.ApdngQuitc LHospital, ta c 2 23 3 2 2x 0 x 0 x 0t gx x (1 t g x) 1 2 t g x 2L 2lim 2lim lim .x 3x 3 x 3 + = = = = 10 x4x4) L limx e .=Ta thy L4 c dng v nh0.. Ta bin i 101010 x 104 4 x xx x x10x xL limx e lim lim (K ) .ee = = = = trong 4 xx10xK lime= . Ta thy K4 c dng v nh. Ap dng Qui tc LHospital, ta c 4 x xx x10 10x 1K lim lim 0.1e e10 = = = Suy ra L4 = (K4)10 = 0. 1x xx5x 03 45) L lim2 += . Ta thy L5 c dng v nh1. Ta c x xx 011 3 4x xlim lnxx 25x 03 4L lim e2 + += = Xt gii hn ( )x xx x5x 0 x 0ln 3 4 ln21 3 4K lim ln lim .x 2 x + += = Ta thy K5 c dng v nh00. Ap dng Qui tc LHospital, ta c: ( )x xx xx x5x 0 x 0 x 03 ln3 4 ln4ln 3 4 ln2ln3 ln4 13 4K lim lim . lim ln12 ln 12.x 1 2 2 ++ ++= = = = =42 Suy ra 5K ln 125L e e 12 2 3 = = = =. ( )3ln|sin(2 x)|6x 26) L lim x 2 .+= Ta thy L5 c dng v nh00. Ta c ( )( )x 233lim ln x 2ln|sin(2 x)|ln|sin(2 x)|6x 2L lim x 2 e++= = Xt gii hn ( )( )6x 2 x 2ln x 23K lim ln x 2 3limln|sin(2 x)| ln|sin(2 x)|+ + = = . Ta thy K6 c dng v nh. Ap dng Qui tc LHospital, ta c ( )6x 2 x 2 x 21ln x 2sin(2 x) 1x 2K 3lim 3lim 3lim 3.cos(2 x) ln|sin(2 x)| 2 x cos(2 x)sin(2 x)+ + + = = = = Suy ra 6K 36L e e . = =ln(1 2x)7x 07) L lim(cot gx) .++= Ta thy L7 c dng v nh0. Ta c x 0lim (1 2x) ln(cot gx)ln(1 2x)7x 0L lim(cot gx) e++++= = Xt gii hn 7x 0K limln(1 2x) ln(cot gx)+= +. Ta thy K7 c dng v nh0.. Khi x 0+, ta cln(1+2x) ~ 2x, do ln(1+2x)ln(cotgx) ~ 2x ln(cotgx). Suy ra7x 0 x 0ln(cot gx)K lim2xl n(cot gx) 2lim1x+ + = =. Ta thy by gi gii hn K7 c dng v nh. Ap dng Qui tc LHospital, ta c 43 22 27x 0 x 0 x 0 x 0 x 021( )sin xln(cot gx) x x x cot gxK 2lim 2lim 2lim 2lim 2lim 0.1 1 sinxcos x xcos x cos xx x+ + + + + = = = = = = Suy ra 7K 07L e e 1. = = = 6. KHAI TRIN TAYLOR 6.1. nh l (Taylor). Cho hm s f(x) c o hm n cp n + 1 lin tc trn on [a,b]. Khi vi mi x0 [a,b], ta c (n) (n 1)2 n n 1 0 0 00 0 0 0 0f '(x ) f ''(x ) f (x ) f (c)f(x) f(x ) (x x ) (x x ) ... (x x ) (x x ) (1)1! 2! n! (n 1)!++= + + + + + + vi mi x [a,b], trong c nm gia x0 v x. Ta gi (1) l khai trin Taylor n cp n ca f(x) ti x0vi phn d di dng Lagrange: (n 1)n 1n 0f (c)R (x) (x x )(n 1)!++= +. ChrngRn(x)lmtVCBcpcaohnVCB(xx0)nkhixx0nntacthvitRn(x) = o((x x0)n). Nh vy, (1) cn c vit di dng: (n)2 n n 0 0 00 0 0 0 0f '(x ) f ''(x ) f (x )f(x) f(x ) (x x ) (x x ) ... (x x ) o((x x ) ) (1')1! 2! n!= + + + + + Ta gi (1) l khai trin Taylor n cp n ca f(x) ti x0vi phn d di dng Peano. 6.2.KhaitrinMacLaurin.KhaitrinTaylorcaf(x)tix0 =0cgilkhaitrin MacLaurin ca f(x). Nh vy, khai trin MacLaurin n cp n ca f(x) nh bi: (n) (n 1)2 n n 1f '(0) f ''(0) f (0) f (c)f(x) f(0) x x ... x x (2)1! 2! n! (n 1)!++= + + + + ++ hay (n)2 n nf '(0) f ''(0) f (0)f(x) f(0) x x ... x o(x ) (2')1! 2! n!= + + + + + trong c nm gia 0 v x; o(xn) l mt VCB cp cao hn VCB xn khi x 0. 6.3. Khai trin MacLaurin ca mt s hm s cp: 44 2 n n 1x c2 nx nx x x xe 1 ... e1! 2! n! (n 1)!x x xe 1 ... o(x )1! 2! n!+= + + + + ++= + + + + +3 5 2k 1 2k 3k3 5 2k 1k 2k 2x x x xsinx x ... ( 1) sin[c (2k 3) ]3! 5! (2k 1)! (2k 3)! 2x x xsinx x ... ( 1) o(x )3! 5! (2k 1)!+ +++= + + + + ++ += + + ++2 4 6 2k 2k 2k2 4 6 2kk 2k 1x x x x xcosx 1 ... ( 1) cos[c (k 1) ]2! 4! 6! (2k)! (2k 2)!x x x xcosx 1 ... ( 1) o(x )2! 4! 6! (2k)!++= + + + + + + += + + + +n 12 nn 22 n n1 x1 x x ... x1 x (1 c) 11 x x ... x o(x )(x 1)1 x++= + + + + + = + + + + + < n 12 n n n 1n 22 n n n1 x1 x x ... ( 1) x ( 1)1 x (1 c) 11 x x ... ( 1) x o(x ) (x 1)1 x+++= + + + + += + + + > + 2 3 n n 1n 12 3 nnx x x xln(1 x) x ...2 3 n (n 1)(1 c) x x xln(1 x) x ... o(x )(x 1)2 3 n++ = + = + < 45 2 3 n n 1n 1 nn 12 3 nn 1 nx x x xln(1 x) x ... ( 1) ( 1)2 3 n (n 1)(1 c)

x x xln(1 x) x ... ( 1) o(x )(x 1)2 3 n+++ = + + + + + ++ = + + + + > 3 5 2k 1k 2k 2x x xarctgx x ... ( 1) o(x )3 3 2k 1++= + + ++ 3 5 61 2tgx x x x o(x ) ( x ) 3 15 2 2 = + + + < < 6.4. ng dng 1) Tnh xp x. Ta thng dngkhai trin MacLaurin tnh xp x gi tr ca hm f(x) sau khi chn n ln phn d Rn(x) c tr tuyt i khng vt qu sai s cho php. V d. Tnh cos25o chnh xc n 0,00001. Gii. Xt khai trin MacLaurin ca cosx: 2 4 6 2k 2k 2kx x x x xcos x 1 ... ( 1) cos[c (k 1) ]2! 4! 6! (2k)! (2k 2)!+= + + + + + + + Phn d ca khai trin l: 2k 2nxR (x) cos[c (k 1) ](2k 2)!+= + + + Vi x = 25o=536, ta c:2k 22k 2 2k 2nx |x| 1 5|R (x)|cos[c (k 1) ](2k 2)! (2k 2)! (2k 2)! 36++ + = + + = + + + Chn k = 2, ta c: 2k 2n1 5|R (x)| 0, 000016! 36+ < . Vy ta c th tnh cos25o chnh xc n 0,00001 nh cng thc: 2 4x xcos x 12! 4! + 46 ngha l2 4o5 55 36 36cos25 cos 1 0, 9063236 2! 4! = + . 2) Tnh gii hn dng v nh:V d. Tnh cc gii hn sau: 2 41x 02 2cos x x 2xL lim .x(x tgx) += x 3 22x 06e x 3x 6x 6L lim .x sinx+ = 3 432 3x 03x 3arctgx x xL lim .6ln(1 x) 6x 3x 2x += + + + Gii. 1) 2 41x 02 2cos x x 2xL lim .x(x tgx) += Khix 0, ta c 2 42 4 5 2 4 4 5 43 3 3 45 4x x 23 232 2cosx x 2x 2 2(1 o(x )) x 2x = x o(x ) x ,2! 4! 12 12x x x xx tgx x (x o(x )) o(x ) x(x tgx) .3 3 3 3 + = + + + + = + + = + nn 42 4423x2 2cos x x 2x 2312.x x(x tgx) 43 + Vy123L .4= x 3 22x 06e x 3x 6x 62) L lim .x sinx+ =Khix 0, ta c 2 3x 3 2 3 3 2 3 3 33 3 34 4x x6e x 3x 6x 6 6(1 x o(x )) x 3x 6x 6 2x o(x ) 2x ,2! 3!x x xx sinx x (x o(x )) o(x ) .3! 6 6+ = + + + + + = + = + = + nn 47 x 3 2 336e x 3x 6x 6 2x12.x sinx x6+ Vy L2 = 12. 3 43 2 3x 03x 3arctgx x x3) L lim .6ln(1 x) 6x 3x 2x += + + +Khix 0, ta c 33 4 3 4 3 4 4 4 4 42 3 42 3 4 2 3 4 4 4x3x x x 3arctgx x x 3x x x 3(x o(x ))x o(x ) x ,3x x x 3 36ln(1 x) 6x 3x 2x 6( x o(x )) 6x 3x 2x x o(x ) x .2 3 4 2 2 + + = + + = + + + + = + + + + = +

Suy ra 3 4 3 4 42 343x x x 3arctgx x x x 236ln(1 x) 6x 3x 2x 3x2 + + + + +. Vy32L3= . 7. NG DNG7.1. Tnh n iu - Cc tr - Tnh li lm - im un - GTLN - GTNN Sinh vin t n7.2. Bi ton lp k hoch sn xuat t li nhun ti a Biton:Gismtxnghipsnxutcquynmtloisnphm.Bithmcul QD=D(P) (P l n gi) v hm tng chi ph l C=C(Q) (Q l sn lng). Hy xc nh mc sn lng Q x nghip t li nhun ti a. Phngphpgii:VimcsnlngQ,bnhtsnphm,xnghipcnbntheo n gi P sao cho QD = Q. Do D(P) = Q P = D1(Q). Khi : -Doanh thu ca x nghip l:R(Q) = P.Q= D1(Q).Q -Li nhun ca x nghip l:(Q) = R(Q) C(Q) = Q.D1(Q) C(Q) Ta cn xc nh gi tr Q > 0 (Q) t cc i. Thng thng ta ch cn tm Q = Q0 > 0 sao cho '(Q0) = 0 v ''(Q0)< 0, hn na, ph hp vi thc t, ti Q = Q0 ta phi c li nhun, n gi v tng chi ph u dng.48 Vd:Mtxnghipsnxutcquynmtloisnphm.BithmcuQD = 1656 -P2(Plngi)vhmtngchiphlC=Q377.Q2+1000Q+40000(Qlsn lng). Hy xc nh mc sn lng Q x nghip t li nhun ti a. Gii.VimcsnlngQ,bnhtsnphm,xnghipcnbntheongiPsao cho:QD = Q 1656 P2= Q P =1312 2Q. Khi : -Doanh thu ca x nghip l:R(Q) = P.Q = (1312 2Q)Q. -Li nhun ca x nghip l:(Q) = R(Q) C(Q) = (1312 2Q)Q (Q3 77Q2 + 1000Q + 40000) = Q3 + 75Q2 + 312Q 40000.Cn xc nh gi tr Q > 0 (Q) t cc i. Ta c: '(Q) =3Q2 + 150Q +312. Suy ra: '(Q) = 0 3Q2 + 150Q + 312 = 0 Q = 2 (loi) hay Q = 52. Ta cng c: ''(Q) = 6Q + 150nn ''(52) < 0. Suy ra (Q) t cc i ti Q = 52. Khi ta c cc s liu sau u ph hp: -Li nhun l = 38416 > 0. -n gi l P = 1208 > 0. -Tng chi ph l C = 24400 > 0. Kt lun: t li nhun cao nht, x nghip cn sn xut vi mc sn lng Q = 52. Khi linhun tng ng l = 38416. 7.3.Bi ton thu doanh thu Biton:Gismtxnghipsnxutcquynmtloisnphm.Bithmcul QD=D(P) (P l n gi) v hm tng chi ph l C=C(Q) (Q l sn lng). Hy xc nh mc thu t trn mt n v sn phm c th thu c nhiu thu nht t x nghip. Phng php gii: Vi mc thu t trn mt n v sn phm, x nghip s nh mc sn lngQphthucvotsaochotlinhuntia.VimcsnlngQ,bnhtsn phm, x nghip cn bn theo n gi P sao cho QD = Q. Do D(P) = Q P = D1(Q). Khi : 49 -Doanh thu ca x nghip l:R(Q) = P.Q= D1(Q).Q -Tin thu x nghip phi np l: T(t)= Qt. -Li nhun ca x nghip l:(Q) = R(Q) C(Q) Qt = D1(Q).Q C(Q) Qt. Nhnitrn,tacnxcnhQsaocho(Q)tcci.KhiQ=Q(t)(Qph thuc vo t) v tin thu m x nghip phi np lT = Q(t)t. thu c nhiu thu nht t x nghip ta cn xc nh gi tr t > 0 T = Q(t)t t cc i. Ch rng ph hp vi thc t, ti gi tr t tm c ta phi c sn lng, n gi, li nhun v tng chi ph u dng.V d. Mt x nghip sn xut c quyn mt loi sn phm. Bit hm cu l QD= 2000 P (P l n gi) v hm tng chi ph l C = Q2 + 1000 Q + 50 (Q l sn lng). Hy xc nh mc thu t trn mt n v sn phm c th thu c nhiu thu nht t x nghip. Gii.Vi mcsnlngQ,bnhtsnphm,xnghipcnbntheongiPsao cho:QD = Q 2000 P = Q P = 2000 Q. Khi : -Doanh thu ca x nghip l:R(Q) = P.Q= (2000 Q)Q. -Tin thu x nghip phi np l: T(t) = Qt. -Li nhun ca x nghip l:(Q)= R(Q) C(Q) Qt=(2000 Q)Q (Q2 + 1000 Q + 50) Qt = 2Q2 + (1000 t) Q 50. Mc sn lng c nh ra sao cho (Q) t cc i. Ta c: '(Q) = 4Q + 1000 t. Suy ra: '(Q) = 0 4Q + 1000 t = 0 Q =1000 t4. V ''(Q) = 4 < 0 nn (Q) t cc i tiQ =1000 t4. Khi tin thu m x nghip phi np l: T(t) = Qt = 21000t t4. Ta cn xc nh gi tr t > 0 T(t) t cc i. Ta c 50 T'(t) =1000 2t4. Suy raT'(t) = 0 1000 2t4 = 0 t =500. V T''(t)= 1/2< 0 nn T(t) t cc i ti t= 500. Khi ta c cc s liu sau u ph hp: -Sn lng lQ = 125 > 0. Tin thu thu c l T = 62500. -n gi l P = 1875 > 0. -Li nhun l = 31200 > 0. -Tng chi ph l C = 140675 > 0. Kt lun: thu c nhiu thu nht t x nghip, cn nh mc thu trn mt n v sn phm l t = 500. Khi tin thu thu c l T = 62500. 7.4. Bi ton thu nhp khu Bi ton: Cho bit hm cung v hm cu ca mt loi sn phm trong th trng ni a ln lt l QS = S(P) v QD = D(P) (P l n gi). Bit rng gi bn ca loi sn phm trn thtrngquctcngvichiphnhpkhu(nhngchatnhthunhpkhu)lP1 0 v t + P1 < P0. Do c c quyn, cng ty s nhp sn phm trn bn vi n gi P tha t + P1 < P < P0 vi s lng l QD QS = D(P)S(P). Khi li nhun m cng ty thu c l:(P) = (P P1 t)[D(P) S(P)]. Tt nhin cng ty s chn n gi li nhun t cao nht. Do ta cn xc nh P sao cho (P) t cc i. Khi P = P(t) (P ph thuc vo t) v tin thu m cng ty phi np l: T(t) =t[D(P(t)) S(P(t))]. thu cnhiu thu nhtt cng ty ta cnxcnhgitr t > 0T(t)t cci.Mc thu t phi tha t + P1 < P0 v ph hp vi thc t, ta phi c cc i lng tng ng nhn gi, lng cung, lng cu u dng.V d. Cho bit hm cung v hm cu ca mt loi sn phm trong th trng ni a ln lt l QS = P 200 v QD = 4200 P (P l n gi). Bit rng gi bn ca loi sn phm trnthtrngquctcngvichiphnhpkhu(nhngchatnhthu)lP1 =1600.Mt cng ty c c quyn nhp loi sn phm trn. Hy xc nh mc thu nhp khu t trn mt n v sn phm thu c t cng ty nhiu thu nht. Gii. Trc ht ta tm n gi ti im cn bng trong th trng ni a. Ta c: QS = QD P 200 = 4200 P P = 2200. Vy n gi ti im cn bng trong th trng ni a l P0 = 2200. 51 Gi t lmc thu nhp khu trn mt n v sn phm. iu kin: t > 0; 1600 + t < 2200(*). Khi : n gi P tha1600 + t < P < 2200 (**) v ta c - Lng hng m cng ty nhp v l: QD QS = (4200 P) (P 200) = 4400 2P. - Li nhun m cng ty thu c l:(P)= (P P1 t)[ QDQS] = (P 1600 t)( 4400 2P) = 2P2 + 2(3800 + t)P 4400(1600 + t). n gi P c nh ra sao cho (P) t cc i. Ta c: '(P) = 4P + 2(3800 + t). Suy ra: '(P) = 0 4P + 2(3800 + t) = 0 tP =1900 +.2 V ''(P) = 4 < 0 nn (P) t cc i ti tP =1900 + 2. Khi tin thu m cng ty phi np l: T(t) = t[ QD QS] = t (4400 2P) = t(600 t). Ta cn xc nh t T(t) t cc i. Ta c: T'(t) =600 2t. Suy ra T'(t) = 0 600 2t = 0 t =300. V T''(t)= 2< 0 nn T(t) t cc i ti t= 300 vi T(t) = 90000. Kim tra ta thy iu kin (*); (**) c tha v cc s liu sau u ph hp: -n gi l P = 2050 > 0. -Lng cung QS = 1850 > 0. -Lng cu l QD = 2150 > 0. Kt lun: thu c nhiu nht thu nhp khut cng ty, cn nh mc thu trn mt n v sn phm l t = 300. Khi tin thu thu c l T = 90000. 7.5. Bi ton thu xut khu Bi ton. Cho bit hm cung v hm cu ca mt loi sn phm trong th trng ni a ln lt l QS = S(P) v QD = D(P) (P l n gi). Bit rng gi bn ca loi sn phm trn th trng quc t tr i chi ph xut khu (nhng cha tr thu xut khu) l P1 > P0, trong P0 l n gi ti im cn bng ca th trng ni a. Mt cng ty c c quyn xut khu loi sn phm trn. Hy xc nh mc thu xut khu t trn mt n v sn phm thu c t cng ty nhiu thu nht (Gi s khi lng xut khu ca cng ty khng nh hng n gi bn trn th trng quc t). 52 Phngphpgii:Gitlmcthuxutkhutrnmtnvsnphm.Mcthut phi tha iu kin t > 0 v P1 t > P0. Do c c quyn, cng ty s thu mua sn phm trn vi n gi P tha P0 < P < P1 t vi s lng lQS QD = S(P) D(P). Khi li nhun m cng ty thu c l:(P) = (P1 P t)[ S(P) D(P)]. Tt nhin cng ty s chn n gi mua li nhun t cao nht. Do ta cn xc nh P sao cho (P) t cc i. Khi P = P(t) (P ph thuc vo t) v tin thu m cng ty phi np l: T(t) =t[S(P(t)) D(P(t))]. thu c nhiu thu nht t cng ty ta cn xc nh gi tr t > 0 T(t) t cc i. Mc thu t phi tha P1 t > P0 v ph hp vi thc t, ta phi c cc i lng tng ng nhn gi mua, lng cung, lng cu u dng.V d. Cho bit hm cung v hm cu ca mt loi sn phm trong th trng ni a ln lt l QS = P 200 v QD = 4200 P (P l n gi). Bit rng gi bn ca loi sn phm trnthtrngqucttrchiphxutkhu(nhngchatrthu)lP1 =3200.Mtcngty c c quyn xut khu loi sn phm trn. Hy xc nh mc thu xut khu t trn mt n v sn phm thu c t cng ty nhiu thu nht. Gii. Trc ht ta tm n gi ti im cn bng trong th trng ni a. Ta c QS = QD P 200 = 4200 P P = 2200. Vy n gi ti im cn bng trong th trng ni a l P0 = 2200. Gi t lmc thu xut khu trn mt n v sn phm. iu kin: t > 0; 3200 t > 2200(*). Khi : Cng ty s thu mua vi n gi P tho: 2200 < P < 3200 t (**) - Lng hng m cng ty xut khu l: QS - QD = (P 200) (4200 P) = 2P 4400. - Li nhun m cng ty thu c l:(P)= (P1 P t)(QS QD) = (3200 P t)(2P 4400) = 2P2 + 2(5400 t)P 4400(3200 t). n gi P c nh ra sao cho (P) t cc i. Ta c '(P) = 4P + 2(5400 t). Suy ra: '(P) = 0 4P + 2(5400 t) = 0 tP 27002= . V ''(P) = 4 < 0 nn (P) t cc i ti tP27002= . Khi tin thu m cng ty phi np l T(t) = t(QS QD)= t (2P 4400) = t(1000 t). Ta cn xc nh t T(t) t cc i. Ta c T'(t) =1000 2t. 53 Suy ra T'(t) = 0 1000 2t = 0 t =500. V T''(t)= 2< 0 nn T(t) t cc i ti t= 500 vi T(t) = 250000. Kim tra ta thy iu kin (*) c tha v cc s liu sau u ph hp: -n gi l P = 2450 > 0 v tho (**). -Lng cung QS = 2250 > 0. -Lng cu l QD = 1750 > 0. Kt lun: thu c nhiu nht thu xut khu t cng ty, cn nh mc thu trn mt n v sn phm l t = 500. Khi tin thu thu c l T = 250000. BI TP 1.Tnh cc gii hn sau: 2 x 0(1 cos x)a) limln(cos4x) . x 01 3sinx 1 tgx 2b) limsin2x + + + 2 32 x 01 cosx ln(1 tg 2x) 2arcsin xc) lim1 cos4x sin x + + + + 3 2 23 2x 0arcsin(x tg 3x) 2arcsin xd) lim1 cos 2x sin x + + + 2 2x 2 43 x 0(x 2x 4)(1 cos2x) (e 1) xe) limln(cos4x) x + + + ++. 22 2 2 x 0(x 3x 4) ln(c os x) cos2x 1f ) lim(x 2x 2)(sin2x x ) + + + + + + x 22 x 0(cos2x e )(x 1 c os x)g) limx(cos3x cos x) ln(1 e cos x) + + 23 2 (x 1)2x 2 4 2 x 1x x(x 1)(x 2x sin ) sin e2 2h) lim(e e )(x 1) ln x + + + 23 2 (x 1)3 2x 1(x 1) ln(x 2) x 4x 5x 1 ei ) lim(1 cos x)(x 1) x 2x 2 1++ + + + + + ++ + + + + 3 x 2 3 22x 4 x 2tg(x 3x 2) 7 x 3e x 3x 4j) limx(e e ) sin x cos x 1 + + + + + + 54 2.Tnh cc gii hn sau: 2 2 2 2 xa) lim( x x x x x x x x)++ + + . 2 2 2 2 xb) lim( x x x x x x x x)+ + + 3 2 3 2 3 3 xc) lim( 3x 3x x 1 3x x 1)+ + + + 3 3 3 2 2 3 xd) limx( 2x x 2x 1 1 x 2x )+ + + + 3 3 3 4 2 3 xe) limx( x x x 1 2x 1 1 x x )+ + + + + 3.Tnh cc gii hn sau: 22 xx 3x 2a) limx 5x 1 + + . cot gx x 0b) lim(sinx cos x)+ 32 cot g x x 0c ) lim(cos2x x )++ 32 cot g x x 0d ) lim(cos2x x )+ 32 cot g x x 0e) lim(cos2x x )+ 4.nh cc tham s a, b cc hm s sau lin tc ti cc im c ch ra: 2x3e cos 2xneux 0;a) yx 4xaneu x = 0. =+ ti x = 0. 22ln(cos3x) neux < 0;xb) y ax b ne u 0 x1;1arctg( ) ne u x > 1.x 2x 3= + + ti x = 0 v x = 1. 5.nh cc tham s a, b cc hm s sau lin tc trn R: 31arctg ne ux 2;a) y (x 2)a ne u x = 2.= 55 222sin x ne ux < 1;x 3x 2b) y ax bx 1 neu 1 x2;ln(x 4x 5)ne u x > 2.2 2 x += + + + + 6. Tm o hm y = y(x) ca cc hm s sau: ln2xxsin3x1a) y=(xcos2x) b) y =x+x 7. Tm o hm y = y(x) ca cc hm n y = y(x) nh bi: x3 2 ya) y= x+arctgy.b) y = 1 + ye .c) x ln y x e 0. Toxa c nh y (0).d) ycosx + sinx + lny = 0. T oxac nh y ( ).2 + = 8. Tm cc o hm y = y(x0) v y = y(x0) ca cc hm s y = y(x) c cho di dng tham s sau: 020t02x ln(1 t)a)tai x ln2y2t 2arctgtxarctgtb) ta i xt3y2x2ec)tai x2yt t = +== = ===== + 9. Chng minh rng hm s1xsin khi x 0y x0khi x = 0 = lin tc ti x = 0 nhng khng c o hm bn tri ln o hm bn phi ti im ny. 10. Chng minh rng hm s21x sin khi x 0yx0 khi x=0 = c o hm trn R. 11. Cho y = 5x . Tm dy v dy(32). Tnh gn ng531 . 56 12. Cho y = arc tg x. Tm dy v dy(1). Tnh gn ngarc tg 1, 05. 13.Tnh cc gii hn sau: x 0x ac sin xa) limx tgx. x 02tgx tg2xb) limx(1 cos3x) 35 x 02(tgx sinx) xc) limx x 0ln|sin2x|d) limln|sin3x| x 01 1e) lim ( )ln(1 x) x+. 1 x2 x 0ln(1 x) 1f ) lim ( )x x++ n x xg) limx e tg(1 x) x 1h) lim (ln(x 1))+ 2/ lnsinx x 0i ) lim(sin3x)+ ln(x 2)2 x 2x 2x 3j) limx 1+ + + 32 2x x 0ln (1 x) sin xk) lim 1 e+ x 2 x 0x arctgxl) lim 2e x 2x 2 14. Tm o hm cp n ca cc hm s sau: 23 xx42a) y = xsinxb ) y = x cosxxc) y =x e d) y = ex 1e) y =x lnx f) y = x 2x 3++ 15. Tm khai trin MacLaurin ca cc hm s sau: 565661a) y = en so hang x .1 sin xb) y = cos(sin2x) e n soha ng x .c) y = arctg(sin3x) en so hang x .d) y = ln(cos2x) en so ha ng x .e) y = arctg(1 cos x) e n sohang x . 16.Tm khai trinTaylor ti x0 ca cc hm s sau n s hng (x x0)5: 57 20 03 x0 0x40 02a) y = xsinx; x = b) y = x cosx; x =6 3xc) y=x e; x =1d) y =; x =1ex 1e) y=x lnx; x =1. f) y =; x =2.x 2x 3 ++ 17. Tnh gn ng chnh xc n 10-6: a) cos41ob) ln1,5. 18. Xc nh cp ca cc v cng b sau y khi chn x lm v cng b chnh: 2 4 23 3 5a) 2 2cos x x2x .b) 2x 2ln(1 x) x .c) x 3tgx x . d) 30x 15arctg2x 40x 96x . + + + + 19.Tmcckhongtnggimvcctrcacchmsysauy,ngthitmgitrln nht v nh nhtca y trn tp D tng ng: a)y = ) 2 1 ( x x ; D = [1/4 , 1]; [1/4 , 1); (1/4 , 1); (1/4 , 1]; [1/4 , + ) . b)y =| | ln 6 2 /2x x xe

D = [1 , 4]; (1 , 4]; [1 , 4); (1 , 4); [1 ,+ ); ( , 1). c)y =x xe x523

D = [4/3 , 2]; (4/3 , 2); [4/3 , 2); (4/3 , 2); ( , 4/3); [2 , + ); R. d)y =4 / 1 x x +

D = [1 , 4]; [1, 4); (1 , 4]; (1 , 4); (1, + ); [4 , + ). e)y =2 31 52+ x xx

D = [2, 0]; (2, 0); [2, 0); (2 , 0]; (2, + ); ( ,0] f) 42x 1yx 1+=+

D = [1, 1];[2, 0); (2, 0]; (2, 0);R. g) 24x 1yx 1+=+

D = [1, 1];[0, 2); ( 0, 2]; (0, 2);R. 20. Tmcc khong li lm v im un ca th ca cc hm ssau y: a)y =| | ln22xx+;b) y =xxe/ 1 ;c) y = (x+2)e1/x. 58 21. Mt x nghip sn xut c quyn mt loi sn phm. Bit hm cu QD = 300 P (P l n gi) v hm tng chi ph l C = Q3 19Q2 + 333Q + 10 (Q l sn lng). Hy xc nh mc sn lng Q x nghip t li nhun ti a.22. Mt x nghip sn xut c quyn mt loi sn phm. Bit hm cu l QD= 2640 P (P l ngi)vhmtngchiphlC=Q2+1000Q+100(Qlsnlng).Hyxcnhmc thu t trn mt n v sn phm c th thu c nhiu thu nht t x nghip. 23. Cho bit hm cung v hm cu ca mt loi sn phm trong th trng ni a ln lt l QS =P200vQD=1800P(Plngi).Bitrnggibncaloisnphmtrnth trng quc t cng vi chi ph nhp khu (nhng cha tnh thu) l P1 = 500. Mt cng ty c cquynnhploisnphmtrn.Hyxcnhmcthunhpkhuttrnmtnvsn phm thu c t cng ty nhiu thu nht. 24. Cho bit hm cung v hm cu ca mt loi sn phm trong th trng ni a ln lt l QS = P 20 v QD = 400 P (P l n gi). Bit rng gi bn ca loi sn phm trn th trng quct trchiphxutkhu(nhngcha trthu)lP1 =310.Mt cng tycc quyn xut khu loi sn phm trn. Hy xc nh mc thu xut khu t trn mt n v sn phm thu c t cng ty nhiu thu nht. 59 CHNG 2 PHP TNH TCH PHN HM MT BIN A-TCH PHN BT NH 1. KHI NIM V TCH PHN BT NH 1.1. nh ngha nguyn hm. Hm s F(x) l mt nguyn hm ca f(x) trn (a,b) nu F(x)= f(x),x(a,b). V d. 1) 4x4 l mt nguyn hm ca x3trn R. 2) cosx l mt nguyn hm ca sinx trn R. Khininnguynhmcaf(x)mkhngchrkhong(a,b)thtahiulnguyn hm ca f(x) trn cc khong xc nh ca f(x). 1.2. nh l. ChoF(x) l mt nguyn hm ca f(x) trn (a,b). Khi 1)Vi mi hng s C, F(x) + C cng l nguyn hm ca f(x) trn (a, b). 2) Ngc li, mi nguyn hm ca f(x) trn (a,b) u c dng F(x) + C. 1.3. nh ngha tch phn bt nh Tphpttcccnguynhmcaf(x)cgiltchphnbtnhcahmf(x),k hiu lf (x)dx. Nu bit F(x) l mt nguyn hm ca f(x) th: f (x)dx F(x) C. = + V d.43xx dx C;4= +sin xdx cos x C. = + 1.4. Tnh cht 1) Nu f(x) c nguyn hm th ( )f (x)dx f (x). = 2)2) f (x)dx f (x) C. = + 3) Vi k l hng s, ta c kf (x)dx k f (x)dx C. = + 4) [f (x) g(x)]dx f (x)dx g(x)dx. + = + 60 1.5. Bng cc tnh phn c bn1xx dx C(-1 : Const)1+= + + dx2 x Cx= + 2dx 1Cx x= + dxln | x | Cx= +

x xe dx = e + C xxaa dx= +C lna(0 < a 1: Const) sinxdx =cosx + C cosxdx = sinx+ C 22dx(1 + tg x)dx cos x = tgx +C= 22dx(1+ cotg x)dxsin x cot gx C== + tgxdx ln|cos x| C = + cot gxdx ln|sin x| C = +

2 2dx xarcsin+ Caa x= (0< a: Const) 22dxn x x h Cx h= + + ++/ (h: Const) 2 2dx 1 xarctg Ca x a a(0 a: Const)= ++ 2 2dx 1 x aln C2a x aa x+= +

(0 a: Const) 2 2d x 1 x al n C2 a x a x a= ++ (0 a:Const) 2 2 2 2 21 1 xa x dx x a x a arcsin C (0 < a: Const)2 2 a = + + 2 2 21 1x h dx x x h h.ln|x x h | C(h: Const)2 2+ = + + + + + Ch . Nuf (x)dx F(x) C = + th vi a 0 vb l cc hng s, ta c( )1f ax b dx F(ax b) Ca+ = + +. V d. 3x 4 3x 41e dx e C.3 = + 61 2. CC PHNG PHP TNH TCH PHN 2.1. Phng php phn tch Mun tnh tch phn bt nh ca mt hm s f(x) ta dng cc tnh cht ca tch phn v phn tch f(x) a tch phn cn tnh v cc dng tch phn c bn. V d. Tnh cc tch phn sau: 1/ 2x 1 21)dx x dx xdx=x x 2 x C.3x+= + + +

4 42 22 2 2 23x x 16 16 16 dx2)dx dx (x 4 )dx x dx 4 dx 16x 4 x 4 x 4 x 4x x=4x 8arctg C.3 2 += = + = ++ + + + + + 1 1 13)sin5x.sin3xdx (cos2x cos8x)dx cos2xdx cos8xdx2 2 21 1 sin2x sin8x C.4 16= = = + 21 cos4x 1 1 14)sin 2x dx dx (1 cos4x)dxx sin4x C.2 2 2 8= = = + 2 2 2 4 3 54 45)(1 2x ) dx (1 4x 4x )dx x x x C3 5+ = + + = + + + 10 11 111 1 16)(1 2x) dx (1 2x) C (1 2x) C.2 11 22+ = + + = + + 2.2. Phng php i bin s1.i bin s dng 1: Gi s tch phn c dng:[ ]I f u(x) u'(x)dx = , trong u(x) v u'(x) lin tc. t t = u(x) dt = u'(x)dx. Ta c [ ]I f u(x) u '(x)dx f (t)dt = = (1) Tnh tch phn sau cng trong (1) theo t, sau thay t = u(x) suy ra I. 2.i bin s dng 2: Xt tch phnI f (x)dx = . t x = (t), trong (t) c o hm '(t) lin tc v x =(t) c hm ngc t = 1(x). Khi dx ='(t)dt v I f (x)dx f[ (x)] (t)dt = = (2) Tnh tch phn sau cng trong (2) theo t, sau thay t = 1(x) suy ra I. V d. Tnh cc tch phn sau: 2 3 41) I x (3 2x ) dx. = + 62 3 2 2dta tt 3 2xdt 6x dxx dx .6= + = = Suy ra 5 3 541 t (3 2x )I t dt C C6 30 30+= = + = +. 22x 12) I dx.x x 3+=+ t2t x x 3 dt (2x 1)dx. = + = + Suy ra I = 2dtln t C ln x x 3 C.t= + = + + 2xdx3) I .x x 3=+ Ta c22 2J1 2x 1 1 dx 1 1I dx ln|x x 3| J.2 2 2 2 x x 3 x x 3+= = + + + _ Xt 2dxJx x 3=+ . Ta c 2 21 13x x 3 (x ) .2 4+ = + Suy ra 2 2221 1 13d(x ) xdx dx 12 2 2J n C1 13 x x 313 1 131 13 (x )2. x(x )2 42 2 22 21 2x 1 13 ln C.13 2x 1 13+ + = = = = ++ + + ++ + = ++ + / Vy 21 1 2x 1 13I ln|x x 3| ln C.22 13 2x 1 13+ = + ++ + 4xdx4) I .1 x=+ 2dtatt = x dt 2xdx xdx2 = = . Suy ra 63 I= 2 2 421 dt 1 1I ln t 1 t C ln x 1 x C.2 2 21 t= = + + + = + + ++ 2ln x 15) I dx.x ln x+= t t = lnx xdxdt = . Suy ra I = 2 2 2t 1 1 t ln xI dt (t )dt ln t C ln ln x C.t t 2 2+= = + = + + = + + 3x 56) I dx.4x 1+=+

2t 1 1att =4x 1 x dx tdt4 2+ = = . Suy ra 232 3t 13 51 3 17 t 17t 1 174I tdt ( t )dt (4x 1) 4x 1 C.t 2 8 8 8 8 8 8+= = + = + = + + + + 2 27) I a x dx(0 < a: Const). = t x = asint,t2 2 t = arc sinax. Khi 2 2a x a|cos t| a cos t; dx a cos tdt. = = =Suy ra ( )2 2 2 2 2 21 1 1 1 1I a cos tdt a 1 cos2t dt a (t sin2t) C a t a sin2t C.2 2 2 2 4= = + = + + = + + Mt khc, 2 2 2 2 21 1 1 1a sin2t a sint cos t a sinta cos t x a x .4 2 2 2= = = Vy 2 2 21 x 1I a arcsin x a x C.2 a 2= + + 2.3. Phng php tch phn tng phn Cho cc hm s u = u(x) v v= v(x) co1ca1c o hm u = u(x) v v = v(x) lin tc. Khi (uv) = uv + uv nn uv= (uv) vu. Suy ra uv dx (uv) dx u vdx uv u vdx. = = Ta chng minh cng thc tch phn tng phn: 64 uv dx uv vu dx = Ta cn vit cng thc trn di dng: udv uv vdu = Ch . 1) tnh g(x)h(x)dx bng phng php tch phn tng phn c 2 cch t:du g (x)dxu g(x)v h(x)dx(thng chon C = 0) dv h(x)dx = = = = hocdu h (x)dxu h(x)v g(x)dx(thng chon C = 0) dv g(x)dx = = = = Ta thng chn cch t no tnh cvdu. 2)ivimtsbiton,saukhipdngtchphntngphn,tacmhthcc dng f (x)dx F(x) f (x)dx, (1 : Const). = + Khi ( )1f x dx F(x) C.1= + 3)Cctchphnsauydctnhbngphngphptchphntngphnviccht tng ng ( y p(x) l a thc theo x c a l hng s): LOICCH T axp(x) sinaxdx, p(x) cosaxdx, p(x)e dx,... u = p(x); dv = sinaxdx (cosaxdx, eaxdx,) p(x) lnaxdx, p(x)arctgaxdx, p(x) arcsinaxdx,... u = lnax (arctgax,arcsinax,); dv = p(x)dx V d. Tnh cc tch phn sau: 1) I x cos xdx = . t u x du dxdv cos xdx v sin x = = = = Suy raI xsin x sinxdx xsinx cos x C. = = + + 2xdx2) I .sin x= t 2u xdu dxdxv cot gx dvsin x= = = = Suy ra 65 cos x d(sinx)I xcot gx cot gxdx xcot gx dx xcot g x xcot g x ln sinx C.sinx sinx= + = + = + = + + x3) I e sin xdx = . t ====x xe vdx x cos dudx e dvx sin u 1x xISuy raI e sinx e cos xdx. = _ Tnh I1 : t = ===x xe vdx x sin dudx e dvx cos u Suy ra x x x1I e cos x e sin xdx e cos x I. = + = +Vy x xI e sin x e cos x I. = T x1I e (sinx cos x) C.2= + 4) I x ln xdx( 1 : Const)= . t1dxduu ln xxdv x dx xv1 +== == + . Ta c 1 1 1 1 12x x dx x x x xI ln x ln x dxln x C1 1 x 1 1 1 ( 1)+ + + + += = = + + + + + + + 2 3x5) I x e dx. = t ====3evxdx 2 dudx e dvx ux 3x 32. Suy ra 12 3x3xIx e 2I xe dx.3 3= _ Tnh I1: t====3evdx due dvx ux 3x 3 .Ta c 3x 3x3x 3x1xe 1 xe 1I e dx e C3 3 3 9= = + Vy 66 2 3 x 3 x3 x 3 x 2x e 2 xe 1 1I ( e ) C e (9x 6x 2) C3 3 3 9 27= + = + +

6) I x arc tg x dx. = t22dxduu arc tgx1 xdv x 1 xv2== + = +=.Ta c 2 2 221 x 1 x dx 1 x 1I arctgx arctgx x C.2 2 2 2 1 x+ + += = ++ 2 27) I a x dx(0 < a: Const) =

t2 22 2xdxduu a xa xdv dxv x = = = =.Ta c 2 2 2 22 2 2 22 2 2 22 2 2 2 2 2 2 22 2x (a x ) aI x a x dx = x a x dxa x a xdx x x a x a x dx ax a x a arcsin I.aa x = = + = + Suy ra 2 2 21 1 xI x a x a arcsin C.2 2 a= + +28)I x h dx (h: Const) = + = t22xdxduu x hx hdv dxv x = = + += =.Ta c 2 22 22 22 2 2 22x (x h) hI x x h dx = x x h dxx h x hdx x x h x hdx hx x h h.ln|x x h| I.x h+ = + + + += + + + = + + + + + Suy ra 2 21 1I x x h h.ln | x x h |+ C.2 2= + + + +67 3. TCH PHN HM HU T 3.1. Tch phn ca cc phn thc n gin Xt cc tch phn c dng sau: ( )k kAI dxx a=, ( )m m2Mx NJ dxx px q+=+ +, trong A, M, N, a, p, q R ; k , mnguyn dng vp2 4q < 0. 1) ( )k kAI dxx a=c tnh nh sau: 1AI dx Aln x a C.x a= = + ( ) ( ) ( )2 k k 1A AI dx C (k > 1).x a k 1 x a= = + 2)Tnh tch phn ( )1 m2Mx NJ dxx px q+=+ +: Ta c 222p pxpx qx q .2 4 + + = + + V p2 4p < 0 nn 2pq 04 > . t 2pa q4= . Thc hin i bin pt x dt dx.2= + =Ta cx2 + px + q = t2 + a2 vMx + N = Mt + 2MpN . Do J1= + +=+ ++dta t2MpN Mtdxp px xN Mx2 2 2= + ++2 2 2 2a tdt2MpNa ttdt 22M = ( )+ +++Catarctg2MpNa1a ta t d2M2 22 2=( ) Catarctg2MpNa1a t ln2M2 2+ + += ( )22 2M 2N Mp 2x pln x px q arctg C22 4q p 4q q ++ + + + 3)Tnh tch phn ( )m m2Mx NJ dxx px q+=+ + (m > 1): Bin i ging nh J1 ta c 68 ( )( )( ) ( )m mmm m m m 2 22 2 2 2 2K LMpMt N2Mx N M 2tdt Mp dtdtJ dx N .2 2 t ax px q t a t a + += = = + + + + + + _ _Ta tnh Km bng cch i bintdt 2 du a t u2 2= + = .

m 2 2 m m m 1 2 2 m 12tdt du 1 1K 1 C C.(t a ) u (m 1)u (m 1)(t a ) = = = + = ++ + Ta tnh m 2 2 mdtL(t a )=+bng cng thc truy hi nh sau: 4)Tnh tch phn m 2 2 mdtL(t a )=+ (m nguyn dng) t 2 2 m 2 2 m 11 2mtu du dt(t a ) (t a )dv dt v t+ = = + + = = . Ta c ( ) ( )2m m m 12 2 2 2Lt tL 2m dtt a t a+= ++ +_, 2 2 22 2m m 1 2 2 m 1 2 2 m 2 2 m 1(t a ) a dt dtL dt aL a L .(t a ) (t a ) (t a )+ + ++ = = = + + + Do 2m m m 1 2 2 mtL 2mL 2ma L .(t a )+= + + Suy ra m 1 m 2 2 2 m 21 t 2m 1 1L L2m 2ma (t a ) a+= ++ y l cng thc truy hi tnh Lm, trong 1 2 2dt 1 tL arctg C.a a t a= = ++ 3.2. Tch phn cc hm hu t Hm hu t l mt hm s c dng: nn 1 0mm 1 0x a ... x a ax b ... x b b) x ( Q) x ( P) x ( f+ + ++ + += =(1) 69 vi ai, bi R v an, bm0 v P(x), Q(x) khng c nghim chung. Ta thy nu bc ca P(x) ln hn hoc bng bc ca Q(x) (m n) th bng cch chia t cho mu ta c th biu din (1) di dng: 21P (x)f (x) P (x)Q(x)= + , trong P1(x), P2(x) l cc a thc theo x vi bc ca P2(x) b hn bc ca Q(x). V P1(x) l a thc nn tch phn P1(x) tnh c d dng. V vy ta gi thit rng f(x) c dng (1) vi bc ca tbhnbccamu(m 0). 77 Gii. t 21 dtx dx .t t= = Ta c222dtdt 1 t x 1tI arc sin C arc sin C1 1 2 2 x 2 2 (1 t)1t t t+ += = = + = + + 78 B -TCH PHN XC NH - TCH PHN SUY RNG 1. TCH PHN XC NH 1.1. nh ngha. Xt hm s f(x) lin tc trn on [a,b]. Chia on [a,b] thnh cc on nh bi cc im x0, x1,..., xn nh sau: b x x ... x x x an 1 n 2 1 0= < < < < < = Trn mi on nh[ ]i 1 ix , x ly mt im ity i 1 i ix x (i 1, n) =v t i i i 1x x x (i 1, n). = =Lp tng nn 1 1 0 1 2 1 n n n 1 i ii 1I f ( )(x x ) f ( )(x x ) ... f ( )(x x ) f ( ) x .== + + + = Xt gii hn: i inn i in ni 1max x 0 max x 0lim I lim f ( ) x .+ += = Nu gii hn trn tn ti, hu hn v bng I R th ta ni f(x) kh tch trn [a,b] v I c gi l tch phn xc nh ca) x ( f trn on[a,b], k hiu l baI f (x)dx = . Ta gi: A l cn di; b l cn trn; f(x) l hm s ly tch phn; f(x)dx l biu thc di du tch phn. Ch . 1) Tch phn xc nh khng ph thuc vo k hiu bin s di du tch phn, ngha l b b ba a af (x)dx f (t)dt f (u)du ... = = = 2)Ta cng t0 dx ) x ( faa=;79 b aa bf (x)dx f (x)dx. = 1.2. Cc tnh cht ca tch phn xc nh 1)) const k ( dx ) x ( f . k dx ) x ( f . kbaba= = 2)[ ]+ba) x ( g ) x ( f dx= dx ) x ( g dx ) x ( fbaba + . 3) Vi a, b, c bt k ta c dx ) x ( f dx ) x ( f dx ) x ( fbccaba + = . (Gi s cc tch phn trn u tn ti). 4) Nu[ ]f (x) g(x) , x a, b , thdx ) x ( g dx ) x ( fbaba .c bit, nu f(x) 0, x [a,b], th baf (x)dx 0 . 4) Nu [ ]m f (x) M, x a, b thba1m f (x)dx M.b a Ta gi ba1f (x)dxb a l gi tr trung bnh ca f(x) trn [a,b]. 1.3. nh l (Tch phn xc nh vi cn trn bin thin). Gi s hm s f(x) lin tc trn [a,b]. Khi dt ) t ( f ) x ( Fxa= l mt nguyn hm ca f(x) trn [a,b], ngha l F(x) = f(x), x [a,b]. Ch . T kt qu trn ta suy ra vi (x) l hm kh vi, ta c (x)af (t)dt f ( (x)) (x). = V d. Tnh gii hn sau: 80 2x2 3t010x 0(t 2t)(lncos t)(e 1)dtL limx = Gii. Ta thy L c dng v nh 0/0. p dng Qui tc lHospital ta c ( )222xx2 3t2 3t0010x 0 x 0102 2 24 2 2 3x 29 9x 0 x 0(t 2t)(lncos t)(e 1)dt(t 2t)(lncos t)(e 1)dtL lim limxx12x x 3x 2x2 (x 2x )(lncos(x ))(e 1)(x ) 3lim lim .510x 10x = = = = = 1.4. nh l (Cng thc Newton Leibniz). Nu hm s f(x) lin tc trn [a,b] v F(x) l mt nguyn hm ca f(x) trn [a,b] th bbaaf (x)dx F(x) F(b) F(a) = = V d. 111 21dx1) arctgx arctg1 arctg( 1) .4 4 2 1 x = = = + =+e e 22 3 e11 1ln x 1 12)dx ln xd( nx) ln x .x 3 3= = = /. 44 420 001 cos2x 1 1 1 1 23) sin xdx dx x sin2x ( ) .2 2 2 2 4 2 8 = = = = 44 44233 3 3dx 1 1 x 2 44) ( )dx ln x 2 ln x 1 ln ln .x 2 x 1 x 1 3 x 3x 2 = = = = +

1.5. Phng php i bin s Dng 1: Xt tch phn baI f (x)dx = vi f(x) lin tc trn [a,b]. t t = (x) tha 1) (x) c o hm lin tc trn [a,b]. 2) f(x)dx tr thnh g(t)dt trong g(t) l mt hm lin tc trn on c hai u mt l (a) v (b). 81 Khi (b) ba (a)f (x)dx g(t)dt.= Dng 2: Xt tch phndx ) x ( fba vi f(x) lin tc trn [a,b]. t x = (t) tha 1) (t) c o hm lin tc trn [,]. 2) a = () v b = ( ). 3) Khi t bin thin trn [,] th x bin thin trn [a,b].Khi [ ]baf (x)dx f (t) (t)dt = . V d. Tnh cc tch phn sau: 220a) 4 x dx ;30b) x 1 xdx +; 220cos xc) dx1 sin x+; 21 x21ed) dxx. Gii.220a)I 4 x dx. = tx 2sin t( 2 t 2) = . Ta c dx = 2costdt; 24 x 2cos t. =i cn x02 t0/2 Suy ra 22 220 0 01 cos2t sin2tI 4 cos tdt 4 dt 2 t .2 2 += = = + = 30b)I x 1 xdx. = + t2t 1 x x t 1 dx 2tdt. = + = =i cn x03 t12 Suy ra 225 32 211t t 116I 2(t 1)t dt 2( ) .5 3 15= = = 82 220cos xc)I dx.1 sin x=+a t t sin x dt cos xdx = = . i cn x0/2 t01 11020dtSuy ra I arctg .41 t= = =+ 2 1 x21ed)I dx.x= t 21 1t dt dx.xx= = i cn x12 t11/2 Suy ra 1/ 21/ 2t t11I ( e )dt e e e = = = . 1.6. Phng php tch phn tng phn Gisu=u(x)vv=v(x)lnhnghmscohmlintc trong[a,b].Khit cng thc tch phn tng phn trong tch phn bt nh ta suy ra cng thc tch phn tng phn trong tch phn xc nh nh sau: b bbaa audv (uv) vdu = V d. Tnh cc tch phn sau: e1a) lnxdx;2ob) xcos xdx; / 2x0c) e cos xdx; 10d) arc tg xdx. Gii.e1a)I lnxdx = . t 83 dxu lnx duxdv dxv x = = = = Suy ra ee e1 11dxI xlnx x e x e e 1 1.x= = = + =. 2ob) I xcos xdx.= tu x du dxdv cos xdx v sin x = = = = Suy ra 22 20 00I =xsin x sin xdx cos x 0. = = / 2x0c) I e cos xdx.= tx xu e du e dxdv cos xdx v sin x = = = = Suy ra 1/ 2 / 2/ 2x x / 2 x00 0II e sinx e sinxdx e e sinxdx. = = _ Xt I1. tx xu e du e dxdv sin xdx v cos x = = = = Suy ra 2x 2 x1 00I e cos x e cos xdx 1 I.= + = +. 2Vay I e (1 I).= +Do 2e 1I .2=84 10d) I arc tg xdx. = t2dxu arctgx du1 xdv dxv x= = + == Suy ra ( )21 111 202 200 0d 1 xx 1 1 1I = xarc tgx dx = ln 1 x ln2.4 2 4 2 4 21 x 1 x+ = + = + + V d. Chng minh rng nu hm s f(x) lin tc trn [a,a] th aaa00ne u f(x) laha m solef (x)dx2 f (x)dxneu f(x) la ham so chan= Gii.Ta c a 0 aa a 0If (x)dx f (x)dx f (x)dx = + _ Xt I. t t = x dt = dx. Ta c 0 0 aa a 0I f (x)dx f ( t)dt f ( t)dt.= = = Suy ra a a a a a aa 0 0 0 0 0a0f (x)dx f ( t)dt f (x)dx f ( x)dx f (x)dx [f ( x) f (x)]dx0ne u f(x) laha m sole 2 f (x)dxneu f(x) la ham so chan= + = + = += 2. TCH PHN SUY RNG 2.1. Tch phn suy rng vi cn v hn (loi I) 2.1.1. nh ngha. Gi s hm s f(x) xc nh trn [a; +) v kh tch trn mi on hu hn [a,b]. Ta nh ngha 85 tta af (x)dx lim f (x)dx (1)++= v gi l tch phn suy rng ca hm s f(x) trn [a; +). Tch phn suy rng c gi l hi t (tng ng, phn k) khi gii hn trong v phi ca (1) tn ti v hu hn (tng ng, khng c gii hn hoc c gii hn v cng).Tng t nh ngha tch phn suy rng ca hm s f(x)trn [ ; a): a auuf (x)dx lim f (x)dx= v trn ( ; +): a a tu ta u af (x)dx f (x)dx f (x)dx lim f (x)dx lim f (x)dx+ + + = + = + (*) (a c chn ty , tch phn suy rng s khng ph thuc vo cch chn a). Trong (*), nu c hai gii hn u tn ti hu hn th tch phn suy rngf (x)dx+ mi hi t. Ngc li, nu c t nht mt trong hai gii hn khng tn ti (hoc bng v cng) thf (x)dx+ phn k. Ta thy rng tch phn suy rng l gii hn ca tch phn xc nh khi cho cn tch phn dn ti v cc. V vy tnh tch phn suy rng ta c th dng cng thc Newton-Leibniz nh sau: af (x)dx F(x) F( ) F(a),a++= = + trong F(x) l mt nguyn hm ca f(x) v xF( ) limF(x).++=Tng t, ta c aaf (x)dx F(x) F(a) F( )= = v f (x)dx F(x) F( ) F( )++= = + vi( )xF( ) limF x++= ;( )xF( ) limF x= . V d 1. 0 2x0dxa) arc tgx limarc tgx arctg0 .2 1 x+++= = =+

002xdxb) arctgx arctg0 limarctgx .2 1 x = = =+ 86 02 2 20dx dx dxc) .2 2 1 x 1 x 1 x+ + = + = + = + + + . V d 2. Chng minh rng 1dxIx+ = hi t vi > 1 v phn k vi 1. Gii. 1) Vi 1 ta c ( ) ( )1 1x11dx 1 1 1I lim .1 x 1 x 1 x++ + = = = + Nu < 1 th ( )1x1lim I1 x += = + nn I phn k. Nu > 1 th ( )1x1 1lim 0 I1 1 x += = nn I hi t. 2) Vi = 1 ta c 1 x1dxI lnx lim ln x ln1x+++= = = = + nn I phn k. 2.2. Tch phn ca hm khng b chn (loi II) 2.2.1. nh ngha. 1) Nu hm s f(x) lin tc trn [a, b) v khng b chn ti b, ngha l( ) =x f limb x (khi x = b cn c gi l im bt thng ca f(x)), th ta t b tt ba af (x)dx lim f (x)dx= . 2) Nu hm s f(x) lin tc trn (a, b] v khng b chn ti a, ngha l ( ) =+x f lima x (ngha l x = a l im bt thng ca f(x)), th ta t b bu aa uf (x)dx lim f (x)dx+= . 3) Nu hm s f(x) khng b chn ti im c (a,b) v lin tc ti mi x [a,b]\{c} th ta tb c b t bt c u ca a c a uf (x) dx f (x) dx f (x) dx lim f (x)dx lim f (x)dx (*) + = + = +

87 4) Nu cc gii hn trn tn ti v hu hn th ta ni cc tnh phn suy rngtng ng hi tu, ngc li ta ni chng phn k. Ch rng trong (*), tch phn suy rng baf (x) dxch hi t khi c hai gii hn tng ng u tn ti hu hn. Ch . Nu F(x) l mt nguyn hm ca f(x) th ta cng c cng thc tng t nh cng thc Newton-Leibniz nh sau: Vi F(x) l mt nguyn hm ca f(x), ta c a) Nu hm s f(x) lin tc trn [a,b) v c im bt thng l x = b th bbaaf (x)dx F(x) F(b ) F(a)= = , trong ( )x bF(b ) limF x= . b)Nu hm s f(x) lin tc trn (a, b] v c im bt thng l x = a th bbaaf (x)dx F(x) F(b) F(a )++= = , trong x aF(a ) limF(x).++=V d 1. Xt =211 xxdxJc im bt thng x = 1. Ta c ( )( )( )3x 1 1xdx dx 2dx x 1dx x 1 2 x 1 C.3x 1 x 1x 1 += = + = + + Do ( ) ( )23 3t 112 2 2 8J x 1 2 x 1 2 lim t 1 2 t 1 .3 3 3 3++ = + = + + = Vy J hi t v J = 8/3. V d 2.Tch phn suy rng ( )=2021 xdxIc im bt thng l x = 1. Ta c 2 11 22 20 1I Idx dxI(x 1) (x 1)= + _ _ ( )1112x 10 0dx 1 1I lim 1 .x 1 x 1x 1 = = = = + Vy I1 phn k nn I cng phn k (ta khng cn kho st I2). 88 V d 3. Chng minh rng badxJ(x a) = (a < b) hi t khi < 1 v phn k khi 1. Gii. 1) Vi 1 ta c ( ) ( ) ( )bb1 1 1x aaadx 1 1 1J lim .(x a) 1 (x a) 1 (b a) 1 (x a)++ = = = + Nu < 1 th ( ) ( )1 1x a1 1lim 0 J1 (x a) 1 (b a)+ = = nn J hi t. Nu > 1 th ( )1x a1lim J1 (x a)+ = + = + nn J phn k. 2) Vi = 1 ta c bba x aadxJ ln|x a| ln(b a) lim ln|x a|x a+ + = = = = + nn J phn k. 3. NG DNG CA TCH PHN 3.1. Tnh din tch hnh phng Dintchhnhphnggiihnbiccngthngx=a,x=b,y=f1(x),y=f2(x),vi f1(x), f2(x) l cc hm s lin tc trong [a,b] c tnh theo cng thc: ( ) ( )b1 2aS f x f x dx = (1) tnhtchphntrong(1)tacngiiphngtrnhhanhgiaoimf1(x)=f2(x) tm tt c cc nghim x1< x2