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GIO TRNH MN HC CHNG TRNH KHNG CHUYNSTT 1 2 3 4 5 MN HC GHI CH

TN MN HC M S THI LNG CHNGTR NH IU KIN TIN QUYT M T MN HC

VI TCH PHN A1S tn ch: 05 ( 01 tn ch ng vi 15 tit) L thuyt: 75 tit Thc hnh: 0 tit Tng cng: 75 tit Ton ph thong Vi tch phn A1 c thit k trong nhm kin thc c bn. Cung cp kin thc i cng v tp hp, quan h v logic suy lun. Trang b cho sinh vin su kt qu c bn v Gii tch ton hc thc s cn thit cho vic tip cn cc mn chuyn ngnh: Hm s; Gii hn; lin tc; Php tnh vi, tch phn ca hm mt bin; Kho st s hi t , phn k ca chui s dng; tnh tng ca chui hm hi t. Sinh vin tip cn nhng kin thc trn thng qua vic kt hp bi ging trn lp, t hc v tm hiu thm trong cc ti liu. Trang b kin thc ton hc bc u gip sinh vin lm quen vi mt vi ng dng ton hc trong tin hc v cuc sng. - Hin din trn lp: 10 % im ( Danh sch cc bui tho lun v bi tp nhm). Vng ba bui khng c cng im ny. - Kim tra KQHT: 20 % im ( 2 bi kim tra gia v cui mn hc: C ba thang im: 2.0 ( hai chn); 1.0 ( mt trn); 0,0: (khng chn). - Kim tra ht mn: 70% im ( Bi thi ht mn) * Lu : Danh sch cc bui tho lun v hai bi kim tratrang 1

IM T

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c hy khi danh sch im thi ht mn c cng b.

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CU TRC MN HC

KQHT 1: Xc nh cc kin thc c bn v gii hn dy s v dy hm mt bin s KQHT 2: Kho st hm s, tnh gn ng gi tr ca mt biu thc bng ng dng vi phn. KQHT 3 : Tnh tch phn i bin, tng phn v ng dng tnh din tch hnh phng, di cung phng v th tch vt th trn xoay. KQHT 4: Kho st mt s bi ton v s hi t hay phn k bng vn dng l thuyt tch phn suy rng loi I, loi II KQHT 5: Kho st s hi t, phn k ca chui s dng. KQHT 6: Tnh tng ca chui hm hi t.* Thc hnh: Lm bi tp trn lp+ Hot ng theo nhm+ Tho lun

KQHT 1 S tn ti vn TON PH THNG

KQHT 2 Phn tch vn VI TCH PHN A1

KQHT 3 Tng hp vn TON PHC V CHUYN NGNH KQHT 6 ng dng trong cuc sng

KQHT 4 Thc trin vn

KQHT 5 ng dng trong Ton

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KT QU V CC BC HC TPKt qu hc tp/ hnh thc nh gi 1. Xc nh cc kin thc c bn v gii hn dy s v dy hm mt bin s. nh gi: Bi tp dng l thut Dng k hiu logc. + t : Trnh by c chnh xc t nht mt trong ba nh ngha v gii c mt v d. * Gii hn dy s; * Gii hn hm s; * Hm mt bin s lin tc ti mt im. Cc bc hc tp 1.1 Hy dng k hiu logic ton hc trnh by: + nh ngha gii hn dy. + nh ngha gii hn hm. 1.2 Trnh by t nht hai v d mang tch cht l thuyt. 1.3 Th no l hm s s cp lin tc ti im, trong khong, on? Kho st tnh lin tc mt s hm v d mang tch cht l thuyt. 1.4. Trnh by cc khi nim v cc? 2. Kho st hm s v tnh gn ng gi tr ca hm mt bin s bng vi phn. nh gi : Dng k thut +Lp s ch T. + t: Hon thnh c hai trong nm yu cu: * Vit ng 9 cng thc o hm c bn mang tnh tng qut. * Vit chnh xc biu thc vi phn ton phn hm mt bin. 2.1 o hm, vi phn hm mt bin l g? Ging v khc nhau ra sao? Phng tin, ti liu, ni hc v cch nh gi cho tng bc hc + Bng en + Kin thc c bn v gii hn Ph thng Trung hc. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho: + Nguyn nh Tr; T Ngc t Ton cao cp T2. + L vn Ht- i hc kinh t-Ton cao cp P2. + Hc trong phng. + Tr li cu hi v bi tp ngn. + Bng en + Kin thc c bn v gii hn Ph thng Trung hc. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho + Nguyn nh Tr; T Ngc t Ton cao cp T2. + L vn Ht- i hc kinh t-Ton cao cp P2. + Hc trong phng. + Tr li cu hi v bi tp ngn. + Bng, phn + Kin thc Ph thng Trung hc. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho + Nguyn nh Tr; T Ngc t Ton cao cp T2. + L vn Ht- i hc kinh t-Ton cao cp P2. + Hc trong phng. + Tr li cu hi v bi tp ngn. + Bng, phn. + Kin thc Ph thng Trung hc + Hc trong phng. + Tr li cu hi v bi tp ngn.

2.2 Cng thc c bn.

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* Vit chnh xc cng th khai trin TayloreMaclaurence. * Gii chnh xc c kim tra li bng my tnh cm tay mt v d tnh gn ng gi tr mt biu thc bng vi phn cp mt. * Gii hon chnh mt v d: kho st hm s. 3. Tnh tch phn i bin, tng phn v ng dng tnh din tch hnh phng, di cung phng v th tch vt th trn xoay nh gi: Cu hi ngn t: * Tr li c: Ti sao tch phn i bin v tch phn tng phn l hai tch phn thng dng? * Gii ng t nht mt v d ng dng cng thc tnh: din tch hnh phng, di cung phng v th tch vt th trn xoay.

2.3 ng dng. + Kho st hm s. + Thit lp phng trnh tip tuyn. + Tnh gn ng gi tr mt biu thc. * Bng vi phn cp 1. * Bng khai trin TayloreMaclaurence

+ Bng, phn. + Kin thc Ph thng Trung hc * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho + Nguyn nh Tr; T Ngc t Ton cao cp T2. + L vn Ht- i hc kinh t-Ton cao cp P2. + Hc trong phng. + Tr li cu hi v bi tp ngn. + Giy A4, A0, vit lng, bng keo * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho + L Phng Qun -Vi tch phn A1 i hc Cn th. + Nhm cc tc gi- Bi tp Gii tch- i hc Cn th. + Hc trong phng + Tr li cu hi v bi tp ngn. + Giy A4, A0, vit lng, bng keo. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho + Nhm cc tc gi - Bi tp Gii tch- i hc Cn th. + Nguyn Vit ng -Trn Ngc Hi- Ton cao cp C1 i hc m bn cng TP H Ch Minh. + L vn Ht- Ton cao cp P2 i hc kinh t. + Hc trong phng + Tr li cu hi v bi tp ngn.

3.1 nh ngha tch phn hm mt bin? Nu li cc cng thc tnh: + Din tch hnh phng. + di cung phng. + Th tch vt th trn xoay. 3.2 C bao nhiu pp tnh tch phn hm mt bin? Pp no l hiu hiu qu nht? Ti sao? 3.3 Bi tp ng dng.

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4. Kho st mt s bi ton v s hi t hay phn k bng vn dng l thuyt tch phn suy rng loi I, loi II nh gi : Cu hi ngn Bi tp thc hnh dng vit t: Tr li ng hai trong bn vn sau: * Trong hai loi tch phn suy rng loi I v loi II tch phn no d kho st? Ti sao? * Trng hp no s dng cng thc gn ng kho st s ht t hay phn k ca tch phn suy rng? * Vit chnh xc t nht hai tiu chun xt s hi t hay phn k ca tch phn? * Xt ng t nht mt v d s hi t hay phn k ca tch phn suy rng?

4.1 Trnh by nh ngha tch phn suy rng lai I; loi II ? Cc loi tch phn ny ging v khc tch phn chng trnh ph thng nhng im no? 4.2 Nu cc tiu chun xt s hi t v phn k ca tch phn suy rng? Ti sao tch phn

+ Bng, phn. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho + Vi tch phn A1 L Phng Qun-i hc Cn th + Hc trong phng + Tr li cu hi v bi tp ngn.

+ Giy A0, vit lng, bng keo. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho: + Nhm cc tc gi - Bi tp Gii b tch- i hc Cn th. f ( x)d ( x) + Nguyn Vit ng -Trn Ngc a+ Hi- Ton cao cp B v C i hc cn trn b v cn m bn cng TP H Ch Minh. di l a + ? + Hc trong phng + Tr li cu hi v bi tp ngn.

4.3 p dng xt s t v phn k ca s v d tch phn suy rng.

+ Giy A0, vit lng, bng keo. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho: + Nhm cc tc gi - Bi tp Gii tch- i hc Cn th. + Nguyn nh Tr-Ton Cao cp T 2. + Hc trong phng. + Tr li cu hi v bi tp ngn. + Bng, phn, Giy A0, vit lng, bng keo. + Kin thc v chui s, chui hm. * Ti liu chnh: Vi tch phn A1 + Hc trong phng. + Tr li cu hi ngn.

5. Kho st s hi t, phn k ca chui s dng. nh gi : Cu hi ngn Bi tp gii theo nhm. t: Gii thch ng

5.1 Th no l mt mt chui s? Chui hm? Chui an du? Chui ly tha?

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ba yu cu: * S hi t hay phn k ca chui s dng * Chui hm hi t, phn k, hi tuyt i hay bn hi t? * Xt ng t nht hai v d v s hi t hay phn k ca chui?

5.2 Trnh by cc tiu + Giy A0, vit lng, bng keo. chun v s hi, * Ti liu chnh: Vi tch phn A1 phn k ca chui? * Cc ti liu tham kho: + L Phng Qun - Vi tch phn A1 - i hc Cn th. + Hc trong phng. + Tr li cu hi v bi tp ngn. 5.3 ng dng: Xt s hi t, phn k ca mt vi chui s dng? + Giy A0, vit lng, bng keo. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho: + Nhm cc tc gi - Bi tp Gii tch- i hc Cn th. + Nguyn nh Tr-Ton Cao cp T 2. + Ton cao cp B v C Nguyn Vit ng -Trn Ngc Hi- i hc m bn cng TP H Ch Minh. + Giy A0, vit lng, bng keo. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho: + L Phng Qun-Vi tch phn A1 i hc Cn th. + Trn Ngc Lin -Vi tch phn A1 i hc Cn th. + L vn Ht- Ton cao cp P2 i hc kinh t. + Hc trong phng. + Tr li cu hi v bi tp ngn.

6. Tnh tng ca mt chui hm hi t. nh gi : Cu hi ngn Bi tp thc hnh gii theo nhm: *Yu cu: Gii ng bi ton: Tm min hi t v bn knh hi t t suy tng ca chng t: * Lp lun chnh xc * ng thi gian theo qui nh ca GV ra . * C s hp tc cc thnh vin trong nhm.

6.1 Th no l chui ly tha? Ngi ta thng kho st chui hm c tm hay chui hm khng c tm?

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* ng kt qu. 6.2 Trnh by cc bc gii bi ton tm min hi t?

6.3 Trnh by cc bc gii bi ton tnh tng ca mt chui hm hi t?

+ Giy A0, vit lng, bng keo. * Ti liu chnh: Vi tch phn A1 * Cc ti liu tham kho: + Bi tp Gii tch- Nhm cc tc gi - i hc Cn th. + Ton Cao cp T2-Nguyn nh Tr. + Ton cao cp C1 Nguyn Vit ng -Trn Ngc Hi- i hc m bn cng TP H Ch Minh. + Ton cao cp P2- L Vn Hti hc kinh t TP HCM. + Hc trong phng. + Tr li cu hi v bi tp ngn.

K HOCH NH GI MN HCKt qu hc tp 1. 2. 3. 4. 5. 6. Thi lng ging dy 11,0 11,0 9,0 11,5 9,0 8,5 Hnh thc nh gi Mc yu cu t c Gii tp Gii tp Gii tp Gii tp Gii tp Gii tp c bi c bi c bi c bi c bi c biVit Thao tc Bi tp v nh Thc tp thc t ti T hc

X X X X X X

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HNH THC THI GIAN NI DUNG NH GI

Thi ( t lun) . 90 pht. Trng tm:

- Cc bi ton tnh gii hn; Xt tnh lin tc; gin on ca dy s v dy hm mt bin s. - Cc bi ton v kho st hm s; tip tuyn; vi phn ton phn v ng dng vi phn tnh gn ng. - Cc khai trin TayLor v Maclaurance. - Cc bi ton v tch phn c bit: tch phn dng phng php i bin s v tch phn tng phn. - Cc bi tp v din tch hnh phng; di cung phng v th tch vt th trn xoay. - Xt s hi t v phn k ca tch phn suy rng loi I v loi II. - S hi t hay phn k ca chui s dng - Chui hm hi t, phn k, hi tuyt i hay bn hi t - Tm min hi t v bn knh hi t t suy tng ca chng.

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NI DUNG CHI TIT MN HC KQHT 1: Xc nh cc kin thc c bn v gii hn dy s v dy hm mt bin sLim f ( x) = ? hu hnx x0

im n 1: Xt cc BT gii hn dng

Limx

f ( x) = ? v hn g ( x)

Lim U ( x) V ( x ) = ?x ?

Lim f ( x) = f ( x 0 ) nh nghax x0

* im n 2: Xt cc BT lin tc Tm tham s hm s lin tc, gin on ti im.

BC HC 1: Trnh by cc kin thc b sung v cc trng sBi hng dn CC TRNG S I. TP CC S: Tp s t nhin: N = { ; 2;...} 1 Tp s nguyn: Z = {0; 1; 2;...} Tp s hu t: Q = x sao cho x = p ; p, q Z , q 0 q

Mt s hu t bao gi cng vit c di dng mt s thp phn hu hn hay s thp phn v hn tun hon. V d 1:1 3 = 0,25 ; = 0,75. 4 4 7 7 = 1,1666... ta c th vit = 1,1(6) 6 6 15 15 = 1,363636... hay = 1, (36) 11 11

Ngc li, cho mt s thp phn hu hn hay v hn tun hon th n s biu din mt s hu t no . S thp phn hu hn a0,a1a2an s biu th s hu ta a a p = a 0 + 1 + 22 + L + nn q 10 10 10

S thp phn v hn tun hon a0,a1,a2an (b1b2bm) s biu th s hu ta b a a 10 m n b1 b2 p ( + 2 + L + mm ) = a 0 + 1 + 22 + L + nn + m 10 10 q 10 10 1 10 10 10

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Nhn xt: Mt s thp phn hu hn cng c th c xem l s thp phn v hn tun hon, chng hn: hon. nh ngha 1 Mt s biu din c di dng mt s thp phn v hn khng tun hon c gi l s v t. Tp cc s v t k hiu l: I V d 22 = 1,414213562... ; Tp s thc R = Q I = 3,141592653... 1 = 0,25000... hay 4 1 = 0,25(0) 4

Nh vy c s ng nht gia tp s hu t v tp cc s thp phn v hn tun

ng thng thc ( trc s ): Trn ng thng ly im O lm gc v chn vect n v OE = e . s x l s thc khi v ch khi tn ti duy nht mt im M thuc ng thng sao cho OE = xe . Khi im M c gi l im biu din hnh hc ca s thc x trn ng thng v ng thng c gi l ng thng thc hay trc s. 0 1 x O E M Hnh 1.1 II S PHC S phc l s c dng: z = a + ib. Trong a, b R, i l n v o vi i2 = - 1. Ta k hiu: a = Rez gi l phn thc; b = Imz gi l phn o. C l tp hp tt c cc s phc. S phc z = a + ib c th biu din hnh hc l mt im M(a; b) trn mt phng Oxy. S phc z = a ib oc gi l s phc lin hp ca s phc z = a + ib, hai s phc lin hp i xng nhau qua Ox.

y

Php ton: Cho 2 s phc z1 = a1 + ib1; z2 = a2 + ib2, khi ta c:z1 .z 2 = (a1 a 2 b1b2 ) + i (a1b2 + a 2 b1 ) z1 a1 a 2 + b1b2 b a a1b2 ; = +i 1 2 2 2 2 2 z2 a 2 + b2 a 2 + b2 Re z1 = Re z 2 z1 = z 2 Im z1 = Im z 2 z1 z 2 = (a1 + a 2 ) + i (b1 + b2 )

b z = a + ib r

M(a; b)

O -b

a

xz = a ib

(z 2 0)H 1.2

Ch : Ta thc hin cc php ton theo quy tc chung thun tin hn. V d 3: (1 3i) + (- 2 + 7i) = - 1 + 4i ( 1 i)(2 + i) = 2 + i 2i i2 = 3 i1 4i 4i = = 4 + i (4 + i )(4 i ) 17

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Dng lng gic ca s phc Ta biu din s phc z = a + ib bi vect OM , gi r = OM = a 2 + b 2 l moun ca s phuc z, k hiu: z . Gc = Ox, OM c xc nh sai khc nhau 2k ; k Z gi l argumen, K hiu: Argz. Ta c tg =b . a T ngha hnh hc, ta c a = r cos ; b = r sin z = r (cos + i sin ) .

(

)

V d 4: Biu din s phc z = 1 + i di dng lng gic. Gii Ta c: r = 12 + 12 = 2 , tg = 1 = z = 2 cos + i sin .4 4 4z = r (cos + i sin ); z1 = r1 (cos 1 + i sin 1 ); z 2 = r2 (cos 2 + i sin 2 ) .

Cho

cc

s

phc

z1 .z 2 = r1 .z 2 [cos(1 + 2 ) + i sin (1 + 2 )] z1 .z 2 = z1 z 2 ; Arg (z1 .z 2 ) = Argz1 + Argz 2 + 2k z1 r1 = [cos(1 2 ) + i sin (1 2 )] z 2 r2 z = u un = z

z n = r n [cos n + i sin n ] z n = z ; Arg (z n ) = nArgz + 2kn n

z1 z z1 = ; Arg 1 = Argz1 Argz 2 + 2k z z2 z2 2

Biu din u di dng u = (cos + i sin ) . Ta c: u n = z n (cos n + i sin n ) = r (cos + i sin ) = n r n = r + k 2 ; k = 0; n 1 n = + k 2 = n + k 2 + k 2 u = n r cos + i sin ; k = 0; n 1 n n

V d 5: Tnh 1/ A = (1 + i )20 . 2/ u = 4 1 + i Gii 1/ Ta c: A = 2 cos 2/z2 = 4 + k 2 2 cos 4 + i sin 4 4

4

+ i sin + k 2 4

10 10 A = 2 (cos 5 + i sin 5 ) = 2 . 4

8 + k 8 + k 8 = 2 cos + i sin ; k = 0; 3 16 16 u = 4 1 + i c 4 gi tr: 9 9 + i sin u1 = 8 2 cos 16 16

u 0 = 8 2 cos + i sin 16 16 17 17 + i sin u 2 = 8 2 cos 16 16 Vi tch phn A1

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25 25 + i sin u 3 = 8 2 cos 16 16

III. KHONG - LN CN. nh ngha 2: Khong l tp hp cc s thc ( hay cc im ) nm gia hai s thc ( hay hai im ) no . Phn loi khong: Khong hu hn: Khong ng: [a, b] = {x R \ a x b} Khong m: (a, b ) = {x R \ a < x < b} Khong na ng, na m: (a, b] = {x R \ a < x b}

[a, b ) = {x R \ a x < b}

Khong v hn:

( , a ) = {x R \ x < a}; ( , a ] = {x R \ x a} (b, + ) = {x R \ x > b}; [b, + ) = {x R \ x b}

nh ngha 3 Gi s a l mt s thc, khong m (a - , a + ) (vi > 0) c gi l ln cn bn knh ca a. ( ) a a + a - Hnh 1.3 Cu hi cng c: Hy dng ging Vence biu din cc trng s m bn hc?

BC HC 2: Trnh by cc nh ngha v gii hn dy s. Bi hng dn HM S - GII HN LIN TC * HM S I. HM S. 1. nh ngha 1: Cho X R , mt hm s f xc nh trn X l mt quy tc sao cho ng vi mi gi tr ca bin x thuc X c duy nht mt gi tr thc ca bin y . K hiu y = f(x) x c gi l bin c lp, y c gi l bin ph thuc. X c gi l min xc nh ca hm s, k hiu l Df . Tp Y = {y R \ y = f ( x), x D f } c gi l min gi tr ca hm s, k hiu Rf V d 1: Khi nui mt con b, quan st qu trnh tng trng ca b ta c mi lin h gia thi gian nui t (ngy) v trng lng m (kg) ca con b l mt hm s m = m(t). 2. nh ngha 2: th ca hm s y = f(x) l tp hp cc im M( x, f(x)) trong h to Descartes. G = {M ( x, f ( x), x D} 3. Cc tnh chtVi tch phn A1 trang 13

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a. Hm s n iu nh ngha 3: Hm s y = f(x) c gi l tng ( hay tng nghim ngt ) trn tp E Df , nu vi mi x1, x2 E , x1 < x2 th f(x1) f(x2) ( hay f(x1) < f(x2). Hm s y = f(x) c gi l gim ( hay gim nghim ngt ) trn tp E Df , nu vi mi x1, x2 E , x1 < x2 th f(x1) f(x2) ( hay f(x1) > f(x2). Hm s y = f(x) c gi l hm s n iu ( hay n iu nghim ngt) trn E Df nu n tng hoc gim ( hay tng nghim ngt hoc gim nghim ngt ) trn E. Nu ta s dng thut ng trn m khng nhc n tp E th coi nh E = Df . V d 2: Hm s y = f(x) = x2 gim nghim ngt trn (- , 0] v tng nghim ngt trn[0, + ). Tht vy, gi s x1, x2 [0, + ) v x1 < x2 . Khi ta c f(x1) f(x2) = x12 x22 = ( x1 x2 )( x1 + x2 ) < 0 f(x1) < f(x2) Vy hm s y = x2 tng nghim ngt trn [0, + ) . Chng minh tng t ta c hm s y = x2 gim nghim ngt trn (- , 0] . b. Hm s chn v hm s l. nh ngha 4: Tp X c gi l tp i xng qua gc to O nu vi bt k x X th x X. Ngi ta thng gi tt l tp i xng. nh ngha 5: Cho hm s y = f(x) xc nh trn tp i xng X, khi ta c: Hm s y = f(x) l hm s chn nu vi mi x thuc X th f(-x) = f(x). Hm s y = f(x) l hm s l nu vi mi x thuc X th f(-x) = - f(x). V d 3: 1. Hm s f(x) = x2 l hm s chn trn R. 2. Hm s g(x) = x3 l hm s l trn R. Tht vy, vi mi x R , ta c: f(-x) = (- x)2 = x2 = f(x) g(-x) = (- x)3 = - x3 = - f(x) Ch : th ca hm s chn i xng qua trc tung, th ca hm s l i xng qua gc to . c. Hm s b chn. nh ngha 6: Hm s y = f(x) c gi l b chn di trn tp X Df nu tn ti s a R sao cho f(x) a x X. Hm s y = f(x) c gi l b chn trn trn tp X Df nu tn ti s b R sao cho f(x) b x X. Hm s y = f(x) c gi l b chn trn tp X Df nu n va b chn trn va b chn di, tc l tn ti hai s a, b R sao cho a f(x) b x X. Ch : th ca hm s b chn s nm gia hai ng thng y = a v y = b. V d 4: Hm s f(x) =4 b chn trn tp X= [1, + ). x 4 4 Tht vy, vi mi x X ta lun c: f(x) = > 0 v f(x) = < 4 x x 4 Vy hm s f(x) = b chn trn tp X= [1, + ). xtrang 14

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d. Hm s tun hon. nh ngha 7: Hm s y = f(x) c gi l hm s tun hon nu tn ti s t 0 sao cho vi mi x Df ta lun c x t Df v f(x + t) = f(x). S dng T nh nht (nu c) trong cc s t ni trn c gi l chu k ca hm s tun hon. V d 5: 1. Cc hm s y = sinx v y = cosx tun hon vi chu k T = 2 . 2. Cc hm s y = tgx v y = cotgx tun hon vi chu k T = . 2 . 3. Cc hm s y = sin(ax + b) v y = cos(ax + b) tun hon vi chu k T =a

Tht vy, xt hm s f(x) = sin(ax + b). Gi tn ti s t 0 sao cho f( x + t) = f(x) x R sin[a(x + t) + b] = sin(ax + b) x R sin[a(x + t) + b] - sin(ax + b) = 0 x R 2cos(ax + sin at =0 2 at at + b)sin = 0 x R 2 2

at = k , k Z\{0} 2 2k t= , k Z\{0} a

S T dng nh nht ng vi k = 1 ( hoc k = -1), do ta c T =

2 l chu k a

ca hm s f(x) = sin(ax + b). Cc hm s cn li chng minh tng t. ( coi nh bi tp) e. Hm s hp v hm s ngc. nh ngha 8: Cho hai hm s f(x) v g(x) tho Rf Dg , khi hm s hp ca f(x) v g(x) l hm s h(x) c xc nh h(x) = g[f(x)] vi mi x Df . K hiu h = g o f . V d 6: Cho hai hm s f(x) = x2 v g(x) = 2x . Hy xc nh hm s g o f v f o g. Gii g o f = g[f(x)] = g(x2) = 2 f o g = f[g(x)] = f(2x) = (2x)2 = 22x nh ngha 9: Cho hm s y = f(x) tho: vi mi x1, x2 Df v x1 x2 ta lun c f(x1) f(x2). Khi hm s ngc ca hm s f, k hiu f 1 c xc inh bi: x = f 1(y)3

x2

vi y = f(x). V du 7: Ham so y = x3 co ham ngc la y = x . Ch : 1. Nu g l hm ngc ca hm f th Dg = Rf v Rg = Df . 2. th ca hai hm s ngc nhau i xng qua ng thng y = x. 3. ieu kien e ham y = f(x) co ham ngc la ham f phai n ieu trong mien xac nh cua noVi tch phn A1 trang 15

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f. Hm s s cp. nh ngha 10: Cc hm s s cp c bn l cc hm s : Hm s lu tha: y = x ( R). Hm s m: y = ax ( 0 < a 1 ) Hm s logarithm: y = logax ( 0 < a 1 ) Cc hm s lng gic: y = sinx , y = cosx , y = tgx , y = cotgx Cc hm lng gic ngc: y = arcsinx , y = arccosx , y = arctgx , y = arccotgx i. y = arcsinx:y = sinx l hm tng nghim ngt trn [ ; ] nn n c hm ngc: x2 2

= arcsiny. Hm ngc ca y = sinx ( th ca hm y = sinx (ii. y = arccosx:

x ) qua ng thng y = x. 2 2

x ) l y = arcsinx, th ca n i xng vi 2 2

y = cosx l hm gim nghim ngt trn [0; ] nn n c hm ngc x = arccosy. Hm ngc ca hm y = cosx (0 x ) l y = arccosx, th ca n i xng vi th ca hm s y = cosx (0 x ) qua ng thng y = x. iii. y = arctgx: y = tgx l hm tng nghim ngt trn ( ; ) nn n c hm ngc:2 2

x = arctgy. Hm ngc ca hm y = tgx ( th ca hm y = tgx (iv.

x ) qua ng thng y = x. 2 2

< x < ) l y = arctgx, th ca n i xng vi 2 2

y = arccotgx: y = cotgx gim nghim ngt trn (0,) nn n c hm ngc x = arccotgy. Hm ngc ca hm y = cotgx (0 < x < ) l y = arccotgx, th ca n i xng vi th ca y = cotgx (0 < x < ) qua ng thng y = x . nh ngha 11: Hm s s cp l nhng hm s c to thnh bi mt s hu hn cc php ton i s thng thng ( cng, tr, nhn, chia vi mu khc khng) v php ly hm hp t nhng hm s s cp c bn v cc hng s.

y = cos 4 x + sin( x +V d 8:

4

)+3

y = 2x + x 4 + 2 y = 5 x 2 lg 3 x + 1

II. GII HN CA DY S + Cc nh ngha nh ngha 1 Cho hm s f xc nh trn tp N = {1, 2, 3., n}, khi cc gi tr ca hm f ng vi n = 1, 2, 3, . lp thnh mt dy s: f(1), f(2), f(3),., f(n) . Nu ta t xn = f(n) (n = 1, 2, 3.) th dy s ni trn c vit thnh: x1, x2, x3, ., xn. hay vit gn {xn}. Mi s x1, x2, x3, . c gi l s hng ca dy s {xn}, xn gi l s hng tng qut. V d 1:Vi tch phn A1 trang 16

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a. {xn}, vi xn = a n: a, a, a. b. {xn}, vi xn = (-1)n : -1, 1, -1, 1, , (-1)n nh ngha 2: S a c gi l gii hn ca dy s {xn} nu > 0 cho trc (b ty ), tn ti s t nhin N sao cho: n > N th xn a < .

K hiu: lim x n = a hay xn a khi n . n nh ngha 3: - Nu dy {xn} c gii hn l mt s hu hn a th ta ni dy s {xn} hi t hay hi t v a. - Nu dy {xn} khng hi t th ta ni dy s{xn} phn k. V d 2: Chng minh rng lim x n = lim n Gii.

n =1 n +1 n

Vi mi

> 0, ta xt x n 1 = n 1 = 1 < n > 1 1 n +1 n +1 n =1 n +1 n1

Vy > 0 (b ty ), N = [ -1]sao cho n > N Vy lim xn = lim n

n 1 < n +1

nh ngha 4: Dy s {xn} c gi l dy s dn ti khi n nu M > 0, ln ty , Nsao cho n > N th x n > M . V d 3: Chng minh rng lim x n = lim 5 n = n n

K hiu: lim x n = hay xn khi n . n

Gii:

Xt x

= 5n = 5n > M n > log M 5 M > 0 , ln ty : N = [log M ] : n > N 5n > M 5nn

Vy: lim 5 n = + Cc tnh cht. 1. Nu dy s {xn} c gii hn th gii hn l duy nht. 2. Nu dy s {xn} c lim x n = a v a > p (hay a < q) th tn ti s dng Nn

sao cho n > N x n > p (hay xn < q). 3. Nu dy {xn } c gii hn th n b chn, tc l tn ti s M > 0 sao cho x n M, n . 4. Gi s {xn}, {yn} l nhng dy s c gii hn th: - Nu xn = yn th lim x = lim yn

n

n

n

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- Nu xn yn th lim x lim yn

n

n

n

n

5. Cho ba dy s {xn}, {yn}, {zn} tho xn yn zn n. Khi , nu lim x n = lim z n = a th lim y n = a .n n

6. Gi s {xn}, {yn} l cc dy s hi t, khi ta c : Dy s {xn yn} cng hi t v lim (x n y n ) = lim x n lim y n . n n n Dy s {xn . yn} cng hi t v lim x .y = lim x n . lim y n . Dy s {xn . yn} cng hi t v lim x .y = lim x n . lim y n . n n n n Dy s {k xn} cng hi t v lim kxn n n

n n

n

n

lim n = k n x n .

x lim x n x Dy s n cng hi t v lim n = n ( lim y n 0 ) n y lim n n y n n yn

Cu hi cng c: Dng k hiu logc Ton hc trnh by nh ngha gii hn dy s?

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BC HC 3: Trnh by cc vn v gii hn hm.Bi hng dn:

GII HN HM

I. GII HN HM CA HM S a. Cc nh ngha. Trong phn ny ta lun gi s f(x) l hm s c xc nh trong ln cn im x0, khng nht thit phi xc nh ti x0. nh ngha 1: Ta ni hm s f(x) c gii hn l L nu vi mi dy s {xn} trong ln cn ca x0 tho: x n x0 n v lim x n = x th lim f(x n ) = L . K hiu: lim f(x) = L hay f(x) L khi x x0.xx0

n

0

n

nh ngha 2: S L c gi l gii hn ca hm s f(x) khi x x0 nu vi mi > 0 cho trc ( b ty ) tn ti s dng sao cho vi mi x tho 0 < x x

f(x) L < .

0

< ta c

nh ngha 3: S L c gi l gii hn phi ( tri ) ca hm s f(x) khi x x0 nu vi mi > 0 cho trc ( b ty ) tn ti s dng sao cho vi mi x tho x < x < x + ( x < x < x ) ta c f(x) L < .

0 0 0 K hiu: lim f(x) = L ( lim f(x) = L ).xx + 0xx 0

0

nh ngha 4: S L c gi l gii hn ca hm s f(x) khi x nu vi mi > 0 (b ty ) tn ti s M > 0 (ln ty ) sao cho vi mi x tho x > M ta c

f(x) L < .K hiu: lim f(x) = L hay f(x) L khi x .x

V d 1: 1. Chng minh: lim sin x = 0 .

x2 9 2. Chng minh: lim = 6. x 3 x 33. Chng minh: lim Gii:x x

x0

1

= 0.

1. V x 0 ta c th ch rt: x 0, b ty : 2 = > 0 : 0 < x 0 = x < sinx 0 = sinx x < Vy lim sin x = 0x0

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2. Khi x 3 x 3 0 ta c:

x2 9 x 3

6 = (x + 3) 6 = x 3 <

> 0, > : 0 < x 3 < Vy: lim

x2 9 x 3

6 < .

x2 9 =6 x 3 x3 1 1 1 1 3. Xt: 0 = = , x x x 1 1 M = > 0 : x > M 0 < . x 1 Vy lim = 0 x x

vi mi

> 0 (b ty )

b. Cc tnh cht. Da vo gii hn ca dy s, nh ngha gii hn ca hm s, ta suy ra cc tnh cht sau: 1. Nu f(x) c gii hn th gii hn l duy nht. e) Nu hm s f(x) c gii hn l L khi xx0 v L > a (hay L < a ) th trong mt ln cn no ca x0 (khng k x0) ta c f(x) > a (hay f(x) 0 sao cho f(x) cng cp vi [g(x)]r th ta ni rng f(x) l VCB (VCL) cp r i vi g(x). V d 2: Khi x 0 th 1 cos x v x2 l hai VCB cng cp vi nhau. V lim 1 cos x = lim x 0 x 0 x22. sin 2 x2 x x sin 2 = lim( 2 )2. 1 = 1 . x 0 x 2 2 2

Quy tc ngt b VCB cp cao: Gi s f(x), g(x) l hai VCB khi x x0 , ng

thi f(x), g(x) u l tng ca nhiu VCB th gii hn ca t s s gia hai VCB c cp thp nht t s v mu s.Vi tch phn A1 trang 23

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x + sin 2 x + tg3 x x 1 V du 3: lim = lim = 3 7 x0 x 0 3x + 4 x + 5x 3x 3 nh ngha 3 Gi s f(x), g(x) l hai VCB khi x x0 . Ta bo chng l cc VCB tng ng khi f ( x) = 1 . K hiu: f(x) g(x). x x0 . nu lim x x0 g ( x ) V d 4: Khi x 0 th sin x x ; ex 1 x; ln (1 + x) x. Ch : Nu trong qu trnh no : 1(x) 2 ( x) cn 1(x) 2 ( x) th trong qu trnh ( x) ( x) = lim 2 . y: lim 1 1 ( x) 2 ( x) V d 5: sin 5x 5x 5 1) lim = lim = x 0 sin 3x x0 3x 3 ln(1 + 2x) 2x 2 2) lim 3 x = lim = x0 x0 e 1 3x 3 b. Cc tnh cht 1) Tng ca hai VCB l mt VCB (khi x x0 ) . 2) Tch ca mt VCB vi mt i lng b chn l mt VCB (khi x x0). lim 3) x x f ( x ) = L (hu hn)khi v ch khi f(x)L = (x) l VCB khi x x0.0

HM S LIN TC

I. Cc nh Ngha nh ngha 1: Cho hm s f(x) xc nh ti x0 v trong ln cn x0, khi hm f(x) c gi l lin tc ti x0 nu lim f (x ) = f (x 0 ) .x x 0

nh ngha 2: Cho hm s f(x) xc nh ti x0 v trong ln cn x0, khi hm f(x) c gi l lin tc ti x0 nu lim f = 0.x 0

Vi x = x x0 gi l s gia ca i s x. f = f(x) f(x0) = f(x0 + x) f(x0), gi l s gia ca hm f(x) ng vi x ti x0. nh ngha 3: Hm f(x) c gi l lin tc tri ( phi )ti im x0 nu: Hm f(x) xc nh ti im x0 v trong ln cn tri (phi ) im x0. lim f ( x) = f ( x0 ) ( lim+ f ( x) = f ( x0 ) ).x x0

x x0

nh ngha 4 - Hm f(x) c gi l lin tc trong khong (a; b) nu f(x) lin tc ti mi x thuc khong (a; b). - Hm f(x) c gi l lin tc trn [a; b] nu f(x) lin tc trong khong (a; b) v lin tc phi ti x = a v lin tc tri ti x = b. nh ngha 5: Hm s f(x) c gi l gin on ti x0 nu n khng lin tc ti x0 v x0 c gi l im gin on ca hm f(x). Ngi ta chia cc im gin on ca f(x) lm hai loi:Vi tch phn A1 trang 24

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+ Nu x0 l im gin on ca hm s v gii hn tri, phi ca hm s f(x) khi x dn ti x0 u l hu hn th x0 gi l im gin on loi mt ca hm s f(x), cn = lim f ( x) lim f ( x) c gi l bc nhy ca f(x) ti x0.+ x x0 x x0

c bit: Nu lim f ( x) = lim f ( x) c gi l im gin on b c.+ x x0 x x0

+ Cc im gin on khng phi l im gin on loi mt th gi l im gin on loi hai. V d 1: x 2 khi x 1 ti im x = 1. Xt s lin tc tri, phi ca hm s f ( x) = 3 x + 1 khi x < 1

Gii * lim f ( x) = lim x = 1 = f (1) f (x ) lin tc phi ti x = 1 . + +2 x 1 x 1

* lim f ( x ) = lim 3 x + 1 = 4 f (1) f (x ) khng lin tc tri ti x = 1. x 1 x 1

Ch : iu kin cn v cho hm f(x) lin tc ti x0 l hm f(x) phi lin tc tri v lin tc phi ti x0 . II. Tnh lin tc ca hm s s cp - Mi hm s s cp f(x) nu xc nh x0 v trong ln cn ti x0 th f(x) lin tc ti x0. - Mi hm s cp f(x) lin tc ti mi im trong min xc nh ca n. V d 1: 1) f(x) = xn ( x N ) lin tc ti x. 2) f (x ) =

1 lin tc ti x 1. x 1

3) f (x ) = x 2 1 lin tc ti mi x 1 x 1 x 1 . III. Cc php tnh v hm lin tc ti cng mt im. 1) Nu f1(x), f2(x) l nhng hm s lin tc ti im x0 th tng, hiu (f1(x) f2(x)); tch (f1(x) . f2(x)); thng

f1 (x ) ( f2(x) 0) cng l nhng hm s lin tc ti f2 (x )

im x0. 2) Nu u = u(x) l hm s lin tc ti x = x0, cn hm f(u) lin tc ti u = u0 th hm f[u(x)] cng l lin tc ti x0. ngha hnh hc ca khi nim lin tc: Nu hm s y = f(x) lin tc trn [a; b] th th ca n l mt ng cong lin khng b ngt qung ni hai im A(a, f(a)); B(b, f(b)). Nhng tnh cht quan trng ca hm f(x) lin tc trn [a, b]: i. Nu hm f(x) lin tc trn [a, b] th n b chn trn [a, b]. ii. Nu hm f(x) lin tc trn [a; b] th n gi tr nh nht v gi tr ln nht. Cu hi cng c:Vi tch phn A1 trang 25

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1. Hy nu nh ngha hm s lin tc ti mt im, trong khong, trn on? 2. Hy cho bit tnh cht quan trng ca hm s lin tc trn mt on?

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KQHT 2 : Kho st hm s v tnh gn ng gi tr ca hm mt bin s bng ng dng vi phn, bng khai trin Taylore Maclaurence.o hm: y- Vi phn: dy Mi lin h y v dy. Tnh yca y = u ( x) v ( x ) Tnh vi phn ton phn Gii BT ng dng: + Tnh gn ng. + Kho st hm s. + Tm GTLN+ GTNN. + PT tip tuyn

im n:Xt cc vn v o hm

BC HC 1: Trnh by php tnh o hm hm mt bin Bi hng dn:

O HMI. Cc nh ngha nh ngha 1: Gi s y = f(x) l hm s xc nh ti im x0 v trong ln cn ca im x0. Nu gii hn lim c gi l o hm ca hm s f(x) ti im x0. K hiu: f(x0) . Ch : Ta c th k hiu o hm ca hm s di cc dng sau: y ; f(x0) ; y 'dy df ( x) ; ; f (x). dx dxf ( x 0 + x) f ( x0 ) y = lim tn ti hu hn th gii hn x 0 x x 0 x

Gi tr o hm ca hm s ti im x0 c biu din nh sau:x = x0

;

dy dx

;x = x0

df ( x) dx

.x = x0

nh ngha 2: Gi s hm s y = f(x) xc nh ti x0 v ti x > x0 ( hay x < f ( x0 + x) f ( x0 ) lim+ = f +' ( x0 ) x0 ). Nu gii hn ( hay x 0 x f ( x0 + x) f ( x0 ) lim = f ' ( x0 ) ) tn ti hu hn th gii hn c gi l o hm x 0 x phi ( hay o hm tri ) ca hm f(x) ti im x0. nh ngha 3: * Hm s f(x) c o hm trn khong (a , b) nu n c o hm ti mi im thuc khong . Hm s f(x) c o hm trn on [a,b] nu n c o hm trn khong (a , b) v c o hm phi ti a, c o hm tri ti b. V d 1: Dng nh ngha, tnh o hm ca hm s y = f(x) = ax + b.Vi tch phn A1 trang 27

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Ta c f ' ( x) = lim

x 0

t bit: Nu f(x) = C th f(x) = 0. II. Cc nh l. nh l 1: iu kin cn v hm s y = f(x) c o hm ti x l hm s f(x) c o hm tri v o hm phi bng nhau. nh l 2: Gi s hm s f(x) xc nh ti x0 v trong ln cn ca n. Khi nu hm f(x) c o hm ti x0 th n lin tc ti x0. Ch : Nu hm s f(x) lin tc ti x th cha th suy ra n c o hm ti x. V d 2: Hm s f(x) = x lin tc ti x = 0 nhng khng c o hm ti y. III. ngha ca o hm 1. ngha hnh hc. Cho hm s y = f(x) c th (C), trn (C) ly hai im M0(x0, y0), M(x, y).V tr gii hn nu c ca cc tuyn M0M khi M M0 dc theo th (C) c gi l tip tuyn ca (C) ti im M0. Vi x = x x 0; y = y y 0 ta c t s

[a( x + x) + b] (ax + b ) = lim ax = a . f ( x + x) f ( x) = lim x 0 x 0 x x x

Gii:

ca tip tuyn. Theo nh ngha ca o hm th f(x0) l h s gc ca tip tuyn vi th hm s ti im M0(x0, y0).

y l h s gc x y l h s gc ca cc tuyn M0M. Khi M M0 th x 0 v gii hn nu c ca x

y M

(C) M0 O Hnh 2.1 2. ngha vt l Xt mt cht im M chuyn ng trn trc Ox sao cho ti thi im t th S(t) l khong cch i s OM . Sau khong thi gian t tc l ti thi im t + t chtVi tch phn A1 trang 28

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im v tr M vi khong cch i s OM = S(t + t), khi qung ng i ca cht im trong khong thi gian t l S( t + t ) S(t). Do vn tc trung bnh ca cht im trong khong thi gian t l t sS ' (t ) = lim S (t + t ) S (t ) l vn tc tc thi ca cht im ti thi im t. t 0 t S (t + t ) S (t ) . By gi gi tr t

IV. Qui tc tnh o hm. nh l 1: Gi s f(x), g(x) l cc hm s c o hm ti x, khi cc hm tng, hiu, tch, thng ca chng cng c o hm ti x v:

[ f ( x) g ( x)]' = f ' ( x) g ' ( x) [ f ( x).g ( x)]' = f ' ( x).g ( x) + f ( x).g ' ( x) f ( x) f ' ( x).g ( x) f ( x).g ' ( x) = g ( x) g 2 ( x) '

( g ( x) 0)

nh l 2: Nu hm s u = u(x) c o hm ti x0, hm f(u) xc nh trong khong cha im u0 = u(x0) v hm f(u) c o hm ti im u0 th hm hp h(x) = f[u(x)] c o hm ti im x0 v h(x0) = h(u0).u(x0). nh l 3: Gi s hm y = f(x) c hm ngc l f 1(x). Nu hm f(x) c o hm ti x0 vf ' ( x0 ) 0 th f 1(x) c o hm ti y0 = f(x0) v f

( ) (y ) =1 '0

1 . f ( x0 )'

V. BNG O HM CA CC HM S S CP.f(x) x ; un a x ; au e x ; eu log a x ; lnx; logauf ' ( x)

x 1 ; nuun-1 a x ln a ; uaulna. e x lne = ex ; ueulne = ueu1 ( 1 a>0) x ln a

; sinx; sinu cosx; cosu tgx; tgu cotgx; cotgu arcsinx; arcsinu arccosx; arccosu

1 u' ( 1 u>0); ;(x>0) u log u x

cosx; ucosu - sinx; -ucosu1 u' ; cos 2 x cos 2 x u' 1 ; 2 sin 2 x sin x u' 1 1 u2 u' ; 1 x2 1 u2trang 291 x2 1

;

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arctgx; arctgu arccotgx; arccotgu

1 u' ; 1+ x2 1+ u2 1 u' ; 1+ u2 1+ x2

VI. o hm cp cao. nh ngha : Gi s hm s y = f(x) c o hm y = f(x) trong khong (a, b), ta gi f(x) l o hm cp 1ca hm f(x). Bn thn f(x) cng l hm s nn n c th c o hm, nu hm f(x) c o hm ti x thuc khong (a, b) th ta gi o hm ca hm f(x) l o hm cp 2 ca hm f(x) v k hiu y ' ' = f ' ' ( x ) =d 2 ( y) d 2 f . = dx 2 dx 2

Tng qut: o hm cp n ca hm f(x) l o hm ca o hm cp (n 1) ca n. K hiu: y ( n ) = f ( n ) ( x) =d n ( y) d n f . = dx n dx n

Cu hi cng c - Hy trnh by nh ngha o hm, cc nh l, ngha hnh hc ca o hm bng s trc quan?

BC HC : 2. Xc nh th no l vi phn? Mi quan h vi phn o hm v cc nh l c bn ca vi phn. 3. Trnh by mt s ng dng ca php tnh vi phn.Vi tch phn A1 trang 30

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4. Vit khai trin Taylore Maclaurence. Bi hng dn:

VI PHNI. NH NGHA VI PHN. nh ngha : Cho hm s f(x) xc nh ti x0 v trong ln cn ca n. Cho x mt s gia x tu , nu ti x0 s gia ca hm s y = f(x0 + x) f(x0) vit c di dng: y = A x + (x) trong A l i lng khng ph thc vo x v (x) l v cng b bc cao hn x ( ngha l (x) 0 khi x 0 ) th ta ni hm s f(x) kh vi ti im x0 v i lng A x c gi l vi phn ca hm s ti im x0. K hiu: dy = A x . Nhn xt: T nh ngha ta suy ra y = dy + (x) hay y dy = (x) . Vy nu f(x) kh vi th s gia ca hm s sai khc vi phn mt lng v cng b khng ng k. Do ta c: y dy khi x 0 . Vi phn cp hai ca hm f(x) l vi phn ca vi phn cp mt, k hiu: d2f(x). Vi phn cp n ca hm f(x) l vi phn ca vi phn cp n - 1 ca hm f(x), k hiu: dnf(x). Ta c: dnf(x) = f(n)(x).dxn. II. MI LIN H GIA VI PHN V O HM. nh l 1: iu kin cn v hm s y = f(x) kh vi ti im x0 l f(x) c o hm hu hn ti im x0. Ch : Vi phn ca hm f(x) thng c vit di dng df = f ( x0 ) x * QUI TC TNH VI PHN. nh l 2: 1. Gi s f(x), g(x) l cc hm s kh vi, khi ta c: d(f g) = df dg d(fg) = gdf + fdg'

f gdf fdg d = g g2

( g 0)

2. Gi s y =f(u) v u = u(x) l nhng hm s kh vi, khi ta c df[u(x)] = f [u(x)] = f (u).u (x).dx = f (u).du * CNG THC TNH XP X. Theo nhn xt sau nh ngha: Nu f(x) kh vi ti im x0 v f ' ( x0 ) 0 th y f ' ( x0 )x hay f ( x0 + x) f ( x0 ) + f ' ( x0 )x V d: Tnh gn ng Ta c3

3

28

Gii:1 1 28 = 3 271 + = 3 3 1 + 27 27

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33 x2 dng cng thc tnh gn ng ta c: f ( x0 + x) f ( x0 ) + f ' ( x0 )x 1 1 f (1 + ) f (1) + f ' (1). 27 27 1 1 1 3 1+ 1+ . 27 3 27 1 1 1 Vy 3 28 31 + . = 3 + 3,04 27 3 27

Xt hm s f(x) =

x f ' ( x) =

1

, Chn x0 = 1 v x =

1 . Khi p 27

III. CC NH L C BN CA PHP TNH VI PHN. nh ngha : Hm s f(x) t cc i ( hay cc tiu) ti im x0 (a, b) Df nu tn ti mt ln cn ca im x0 sao cho vi mi x thuc ln cn ta c:f ( x) f ( x0 ) ( hay f ( x) f ( x 0 ) )

im x0 gi l im cc i ( hay cc tiu) ca hm s, im cc i hay cc tiu gi chung l im cc tr. Gi tr hm s ti im cc i ( hay cc tiu) gi l gi tr cc i ( hay cc tiu) v gi chung l gi tr cc tr. nh l 1: (Fermat) Nu hm s f(x) xc nh trong khong (a, b), t cc i hay cc tiu ti im x0 (a, b) v tn ti f ' ( x0 ) th f ' ( x0 ) = 0. nh l 2: (Rolle) Nu hm s f(x) lin tc trn on [a, b] v kh vi trn khong (a, b) v f(a) = f(b) th tn ti t nht mt im c (a, b) sao cho f(c) = 0. nh l 3: (Lagrange) Nu hm s f(x) lin tc trn on [a, b] v kh vi trong khong (a, b) th tn ti t nht mt im c (a, b) sao cho f ' (c) =f (b) f (a ) . ba

nh l 4: (Cauchy) Nu cc hm s f(x), g(x) lin tc trn on [a, b], kh vi trn khong (a, b) vg ' ( x) 0 x (a, b) th tn ti t nht mt im c (a, b) sao cho f ' (c) f (b) f (a ) . = g ' (c) g (b) g (a)

nh l 5: (Taylor) Nu hm s f(x) kh vi n cp (n +1) trong ln cn ca im x0 th x , x x0 tn ti s c nm trong khong gia x v x0 sao chof ( x) = f ( x0 ) + f ' ( x0 ) f '' ( x 0 ) f ( n ) ( x0 ) ( x x 0 ) n + Rn ( x ) ( x x0 ) 2 + K + ( x x0 ) + n! 2! 1!f ( n +1) (c) ( x x 0 ) n +1 ( vi c nm gia x v x0 ). (n + 1) !n

Trong sai s Rn(x) gi l phn d Lagrange xc nh bi :Rn ( x) =

Khi cng thc trn c vit li f ( x) = k =0

f

(k )

( x0 ) ( x x0 ) k + Rn ( x). k!

Cng thc ny gi l cng thc Taylor. a thc Pn ( x) = k =0 n

f

(k )

( x0 ) ( x x0 ) k gi l a thc Taylor. k!trang 32

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Khi x0 = 0 th cng thc Taylor c dng f ( x) = k =0

f

(k )

(0) k x + Rn ( x ) k!

( By gi phn d l: Rn ( x) =

f (c) n +1 x ), gi l cng thc Maclaurin. (n + 1) !

( n +1)

* Mt s cng thc khai trin Maclaurin. 1. f(x) = ax. ln ln 2 a 2 ln n a n ax = 1+ x + x +K+ x + Rn ( x ) 1! 2! n! Vi Rn ( x) = 2. f(x) = ex.a c ln n +1 a n +1 x ( c nm gia 0 v x ). (n + 1)!

x2 xn e = 1+ x + +K+ + Rn ( x) 2! n!x

Vi Rn ( x) = 3. f(x) = sinx.sin x = x

x n +1 c e (n + 1)!

( c nm gia 0 v x ).

x3 x5 x 2 n 1 + K + (1) n 1 + R2 n 1 ( x) 3! 5! (2n 1) ! sin[c + (2n + 1) ] 2 x 2 n +1 (2n + 1)!

Vi R2 n 1 ( x) =

( c nm gia 0 v x ).

4. f(x) = cosx. x2 x4 x 2n cos x = 1 + K + (1) n + R2 n ( x ) 2! 4! 2n ! cos[c + (n + 1) ] 2 n + 2 x ( c nm gia 0 v x ). Vi R2 n ( x) =(2n + 2)!

5. f(x) = ln(x + 1). n x2 x3 x4 ( n 1) x ln( x + 1) = x + + K + (1) + Rn ( x ) 2 3 4 n

Vi Rn ( x) =

x n +1 (n + 1)(1 + c) n +1

( c nm gia 0 v x ).

6. f(x) = (1 + x) .(1 + x) = 1 + x +

( 1)2!(n + 1) !

x2 +K+

( 1)( 2)K ( n + 1)n!

x n + Rn ( x )

Vi Rn ( x) = V d:

( 1) K ( n)

(1 + c) ( n +1) x n +1

( c nm gia 0 v x ).

1) Dng khai trin Macluarin ca hm ex tnh gn ng gi tr s e vi sai s nh hn10 3. 2) Dng khai trin Macluarin ca hm sinx tnh gn ng gi tr sin10 vi sai s nh hn 10 5.trang 33

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Gii: 1) p dng cng thc khai trin Macluarin ca hm f(x) = ex vi x = 1, ta c: 1 1 e 1+1+ +K+ 2! n! Vi sai s = Rn ( x) = M =ec ; c (0,1) (n + 1) !

3 ec c (0,1) (n + 1) ! (n + 1) !

1 1 1 1 1 + + + + 2! 3! 4! 5! 6! 2) p dng cng thc khai trin Macluarin ca hm f(x) = sinx vi x= 10, ta c: (10 ) 3 (10 ) 5 (10 ) 2 n 1 sin 10 10 + K + (1) n 1 3! 5! (2n 1)! sin[c + (2n + 1) ] 2 x 2 n +1 ; c (0, ) Vi sai s = R2 n 1 ( x) =

< 10 3 th ta ch cn ly n = 6, khi e 1 + 1 +

(2n + 1) !

180

2 n +1 sin[c + (2n + 1) ] 2 x 2 n +1 x c (0, ) M = (2n + 1) ! (2n + 1) ! 180 2 n +1

180 th = R2 n 1 ( ) Vi x = 180 (2n + 1) ! 180

< 10

5

th ta ch cn ly n = 1, khi sin 1

0

( )3 180 3!(180)3

nh l 6: ( Qui tc LHospital th nht ) Gi s :

1. f(x) v g(x) l cc hm s kh vi trong ln cn ca im x0 lim lim 2. x x f ( x) = x x g ( x) = 0 .0 0

3. g ( x) 0 trong ln cn ca x0.'

f ' ( x) = A ( hu hn hay v hn ) ' 0 g ( x) f ( x) lim = A. Khi x x 0 g ( x) nh l 7: ( Qui tc LHospital th hai ) Gi s : f(x) v g(x) l cc hm s kh vi trong ln cn ca im x0 . lim lim 2. x x f ( x) = x x g ( x) = . lim 4. x x0 0

3. g ( x) 0 trong ln cn ca x0.'

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lim 4. x x

f ' ( x) = A ( hu hn hay v hn ) ' 0 g ( x) f ( x) lim = A. Khi x x 0 g ( x)

V d: Tnhx3 1) lim ; x 0 x sin xxn 2) lim x x + e Gii:2

x 3 2 2 x 3x 3x 2 = lim = lim = 6 lim = 6 1) lim x 0 x sin x x 0 1 cos x x 0 x 0 x x sin 2 2 sin 2 2 2 n n 1 n2 x nx n (n 1) x n! = lim = K = lim x = 0 2) lim x = lim x x x + e x + e x + x + e e

Cu hi cng c: 1. Hy dng s ch T phn bit mi quan h gia o hm v vi phn. 2. Hy vit biu thc vi phn ton phn v cng thc tnh xp x. 3. Vit khai trin Taylore- Maclaurence.

BC HC 3: Trnh by mt s ng dng ca php tnh vi phn. Bi hng dn:

MT S NG DNG CA PHP TNH VI PHN1. Kho st tnh n iu ca hm s. nh l 1: Gi s hm s f(x) kh vi trn (a, b), iu kin cn v f(x) tng ( hay gim ) trn khong (a, b) l f ' ( x) 0 ( hay f ' ( x) 0 ) vi mi x (a, b). * Cc tr ca hm s. nh l 2: ( iu kin cn ) Nu hm s f(x) t cc tr ti x0 v kh vi ti x0 th f ' ( x) = 0 . nh ngha: im x0 Df c gi l im ti hn ca hm s f(x) nu f(x) khng kh vi ti x0 hoc f ' ( x) = 0 . im ti hn loi f ' ( x) = 0 cn gi l im dng ca hm s. nh l 3: ( iu kin th nht ca cc tr )Vi tch phn A1 trang 35

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Gi s hm s y = f(x) lin tc trong ln cn ca im x0, c o hm trong ln cn ( c th tr im x0 ). Nu x0 l im ti hn ca hm s v f ' ( x) i du t dng sang m ( t m sang dng ) khi i qua x0 th x0 l im cc i ( cc tiu ). nh l 4: ( iu kin th hai ca cc tr ) Gi s hm s y = f(x) c o hm lin tc n cp hai trong ln cn ca im x0 v f ' ( x) = 0 . Khi nu f '' ( x0 ) < 0 ( f '' ( x0 ) > 0 ) th x0 l im cc i ( cc tiu ). V d: Tm cc tr ca hm s f ( x) = x 1 x 2 Gii: Min xc nh Df = [-1,1] 1 2x 2 2 ' =0 x= f ( x) = 2 2 1 x Bng xt du f xf ' ( x)

1 -

2 2

2 2

1 -

+

0 CT

+ 0 C

f(x)

Vy hm s t cc tiu ti x = 1 , 2

2 2 2 , t cc i ti x = ; fCT = f( )= 2 2 2

fC = f(

2 1 )= . 2 2

2. Bi ton tm gi tr nh nht, ln nht ca hm s f(x) lin tc trn on [a, b]. tm gi tr nh nht, ln nht ca hm s f(x) lin tc trn on [a, b] ta thc hin cc bc sau: 1. Tm cc im ti hn ca hm s f(x) trong khong (a, b). 2. Tnh gi tr ca hm s ti cc im trn v tnh f(a), f(b). 3. Gi tr ln nht, nh nht trong cc gi tr trn l gi tr ln nht, nh nht ca hm s f(x) lin tc trn on [a, b].

V d 1: Tm gi tr ln nht, nh nht ca hm s f(x) = x3 3x + 4 trn [-3, 2]. Gii: Ta c f ' ( x) = 3 x 2 3 = 0 x = 1Vi tch phn A1 trang 36

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f(1) = 2 ; f(-1) = 6 ; f(-3) = -14 ; f(2) = 6 Gi tr ln nht ca hm s l 6 t ti x = -1; x = 2 ( fmax = 6) v gi tr nh nht ca hm s l -14 t ti x = -3 ( fmin = -14). V d 2: Ngi ta mun thit k mt ci lon hnh tr ng c din tch ton phn l S. Hy xc nh kch thc ca lon sao cho th tch ca n ln nht. Gii: Gi x, y (x, y > 0) ln lt l bn knh y v chiu cao ca lon. Ta c: S 2 x 2 Din tch ton phn ca lon l: S = S2 y + Sxq = 2 x 2 + 2 x y y = 2 x Th tch ca lon l: V = x 2 y = x 2 S 2 x 2 2 x S = x x3 2

Bi ton tr thnh tm x sao cho V(x) = Ta c V ' ( x) =

S x x 3 t gi tr ln nht. 2

S S 3 x 2 = 0 x = . 2 6 S 6

Bng bin thin: x 0 V(x)

0 + 0

S 6

+

C

Vy V t gi tr ln nht khi x =

S y=2 6

S 6

V d 3: Ngi ta mun thit k mt ci thng hnh ch nht (vi hai y l hnh vung) vi th tch cn t c l V. Hi kch thuc cnh y v chiu cao bng bao nhiu th tit kim nguyn liu nht. Gii: Gi x, y (x, y > 0) ln lt l kch thuc cnh y v chiu cao ca thng. Ta c: Th tch ca thng l: V = x2y y =V x2 4V x

Din tch ton phn ca thng l: S = S2 y + Sxq = 2x2 + 4xy = 2x2 + Bi ton tr thnh tm x sao cho S(x) = 2x2 + Ta c S ' ( x) = 4 x Bng bin thin: x 4V x

t gi tr nh nht.

4V =0 x=3V x2

0

3

V

+

S ' ( x)Vi tch phn A1

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0

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S(x) CT Vy V t gi tr nh nht khi x = 3 V y = 3 V V d 4: Gi s AB l mt on thng trn b bin v L l mt o nh ngoi khi (AL vung gc vi AB), ngi ta mun mc mt ng dy cp t L n B. Hy xc nh v tr ca im C trn on AB sao cho tng gi tin cp ( tnh trn n v ngn ng ) l nh nht ? Bit rng: Phn cp di nc gi 500 ngn ng/km, phn cp trn b gi 300 ngn ng/km, AL = 5 km, AB = 10 km. Gii Gi AC = x km ( 0 x 10 ) CB = 10 - x V AL vung gc AB nn LC = x 2 + 5 2 Tng tin cp: 500 x 2 + 5 2 + 300(10 - x) Xt hm s t(x) = 500 x 2 + 5 2 + 300(10 - x) t ' ( x) = Cho t(x) = 0 x = 500 xx 2 + 52 300

15 15 ; Ta c: t = 5000; t (0) = 5500; t (10) = 2500 5 . 4 4 15 , tc l ta cn chn im C Vy: t(x) t gi tr nh nht ( t(x)min = 5000) khi x = 4

cch A l 3,75 km . 3. Kho st v v th hm s. kho st v v th hm s ta thc hin cc bc sau: 1. Tm min xc nh ca hm s, tnh o hm cp 1 t suy ra tnh n iu, cc tr ca hm s. 2. Tnh o hm cp 2 kho st tnh li lm, im un ca th. th hm s y = f(x) gi l lm ( hay li ) nu f ' ' ( x) > 0 (hay f ' ' ( x) < 0) . im (x0, f(x0)) gi l im un ca th hm s y = f(x) nu: A i. th ca hm s y = f(x) c mt tip tuyn ti x0. ii. Tnh li, lm ca hm s tri ngc nhau hai pha ca x0. 3. Tm cc ng tim cn ca th hm s thng qua cc gii hn c bit. Nu lim f ( x) = th x = a l ng tim cn ng.xa

B

Nu lim [ f ( x) (a x + b)] = 0 th y = ax + b l ng tim cn ngang (a = 0)x

hoc ng tim cn xin ( a 0) ca hm s. 4. Tm cc im t bit: cc im cc tr, im un, im giao ca th vi cc trc to . 5. Lp bng bin thin 6. V th hm s. Cu hi cng c: 1. Kho st v v th hm s c bao nhiu bc? 2. Hy cho bit bi ton tm gi tr ln nht v gi tr nh nht ca hm s lin tc trn on [ a; b] gm nhng bc no?Vi tch phn A1 trang 38

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PHN HNG DN THC HNH KQHT 2 : Kho st hm s v tnh gn ng gi tr ca hm mt bin s bng ng dng vi phn vi phn.Trang thit b + vt liu cung cp cho hc vin: 1. Giy A4, A3, A0. 2. Vit lng Cc bc thc hnh: Ch 1: Kho st v v th hm s 1. Dng s trc quan tm tt cc bc kho st v v th hm s. 2. Kho st v v th ca mt s hm s cp c bn. 3. So snh vi cc hm tng kho st vi chng trnh ph thng.

Ghi chp / Bo co kt qu:

Kt lun / Tho lun:

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TIU CHUN CHO BI THC HNH (CHECKLIST)Tiu ch C Khng

1. Min xc nh. 2. o hm cp mt xt tnh tng gim v cc tr ca hm s 3. Tim cn hoc tnh li lm. 4. Bng bin thin 5. im t bit 6. V th 7. Kh hay d so vi cc bi ton kho st ca ph thng. 8. Tch cc tham gia tho lun nhmNhn xt: ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Ch 2: Bi ton tm gi tr nh nht, ln nht ca hm s f(x)lin tc trn on [a, b]. 1. Dng s trc quan tm tt cc bc tm cc tr hm s lin tc trn on [a, b]. 2. Tm gi tr ln nht v gi trnh nht mt s hm s cp c bn.


Kt lun / Tho lun:

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1. Min xc nh. 2. o hm cp mt xt tnh tng gim. 3. Cho o hm cp mt trit tiu. 4. Bng bin thin 5. Xt cc gi tr hai u on. 6. So snh cc gi tr cc i, cc tiu v cc gi tr hai u on. 7. Tch cc tham gia tho lun nhmNhn xt: ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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Ch 3: ng dng vi phn tnh gn ng gi tr ca mt biu thc1. Dng s trc quan tm tt cc bc tnh gn ng gi tr ca mt biu thc. 2. Tnh gn ng gi tr ca mt s biu thc.* Ghi chp / Bo co kt qu:

Kt lun / Tho lun:

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1. Chn hm. 2. Chn x0 = ? suy ra x = ? c nh hay khng? 3. Tnh o hm cp 1. 4. Tnh cc gi tr hm v gi tr o hm ti x0. 5. Cng thc A = f ( x0 ) + f ' ( x0 ) x. 6. Tch cc tham gia tho lun nhmNhn xt: --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

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KQHT 3: Tnh tch phn i bin, tng phn, din tch hnh phng, di cung phng v th tch vt th trn xoay.o hm: y- Tch phn Mi lin h y v Tnh im n: Xt cc vn v tch phn

f ( x)dxbng:

f ( x)dx.

f ( x)dx

+ Cng thc c bn; +Bng PP i bin v tng phn Gii BT ng dng tnh: + Din tch hnh phng; + di cung phng; + Th tch vt th trn xoay.

BC HC 1: TCH PHN BT NH Bi hng dn:CHNG III

PHP TNH TCH PHN CA HM MT BIN 1 TCH PHN BT NH

I. NGUYN HM V TCH PHN BT NH nh ngha 1: Hm F(x) c gi l nguyn hm ca hm f(x) trn khong (a, b) nu F ' ( x) = f ( x) x (a, b) . V d 1:

Hm F ( x) =

x3 l nguyn hm ca hm f(x) = x2 vi mi x v F ' ( x) = f ( x) x . 3

nh l 1: Nu hm F(x) l nguyn hm ca hm f(x)trn khong (a, b) th (F(x) + C) cng l nguyn hm ca hm f(x). Ngc li, mi nguyn hm ca hm f(x) trn khong (a, b) u c th biu din di dng (F(x) + C). nh ngha 2: Tp hp tt c cc nguyn hm ca hm f(x) trn khong (a, b) c gi l tch phn bt nh ca hm f(x). K hiu: f ( x)dx .

Theo nh l 1 nu hm f(x) c nguyn hm l F(x) thx3 V d 2: x dx = + C 32

f ( x)dx = F ( x) + C .

nh l 2: Cho f(x) v g(x) l cc hm s c nguyn hm trn khong (a,b), khi :Vi tch phn A1 trang 45

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1. 3.

2. d f ( x)dx = f ( x)dx .

[ f ( x)dx] = f ( x) .'

f ( x)dx = f ( x) + C hay df ( x) = f ( x) + C . 4. f ( x)dx = f ( x)dx ( 0) . 5. [ f ( x) + g ( x)]dx = f ( x)dx + g ( x)dx .'

V d 3: x +

2 1 2 dx = x x + 4 x + C . dx = x dx + 2 3 x x

Bng cc tch phn:

adx = ax + C x dx =

; 0dx = C ;

dx x +1 1 + C ( 1) ; dx = ln x + C ; = ln x + 1 + C. +1 x x +1 ax eu ; e x dx = e x + C ; eu dx = ' + C. a x dx = +C ln a u dx 1 xa sin xdx = cos x + C ; x 2 a 2 = 2a ln x + a + C. dx 2 cos xdx = sin x + C ; x 2 + k = ln x + x + k + C. dx dx 1 a+x sin 2 x = cot gx + C ; a 2 x 2 = 2a ln a x + C. dx 1 2 a2 = tgx + C ; x 2 a 2 dx = x a 2 ln x x 2 a 2 + C. cos 2 x 2 2 dx x 2 a2 x 2 2 2 1 + x 2 = arctgx + C ; a x dx = a a x + 2 arcsin 2 + C;(a > 0). dx x dx dx x 1 x 2 = arcsin x + C ; sin x = ln tg 2 + C; cosx = ln tg ( 2 + 4 ) + C;

II. CC PHNG PHP TNH TCH PHN 1. Phng php i bin s nh l 1: Nu f ( x)dx = F ( x) + C th f [ (t )] ' (t )dt = F [ (t )] + C vi (t ) l hm

s c o hm lin tc. Dng 1: Gi s F(x) l mt nguyn hm ca hm f(x), nu hm s hp f[u(x)] vi u(x) l hm kh vi th f [u ( x)]u ' ( x)dx = f (u )du = F (u ) + C = F [u ( x)] + C .V du 1:sin 4 x +C . 4 3 2 2 2 2. sin xdx = sin x. sin xdx = (cos x 1)( sin x)dx = (cos x 1)d (cos x)

1. sin 3 x. cos xdx = sin 3 x.(sin x ) dx = sin 3 xd (sin x) ='

Vi tch phn A1

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cos 3 x = cos xd (cos x) d (cos x) = cos x + C . 32

dx 1 1 1 1 dx dx = 2 x a x + a dx = 2a x a x + a 2a a 1 d ( x a) d ( x + a) 1 1 xa = x a x + a = 2a [ln x a ln x + a + C ] = 2a ln x + a + C 2a x d dx dx 1 1 x x a 1 = = 2 = arctg + C1 = arctg + C 4. 2 2 2 a a a x +a x a x a a 2 + 1 +1 a a

3.

x

2

5/

dx a2 x2

=

1 a

dx x 1 a2

=

x d a x 1 a2

= arcsin

x +C a

Dng 2: Cho f ( x)dx , gi s x = x(t) kh vi v c hm ngc.

f ( x)dx = f [x(t )]x (t )dt = F (t ) + C = F [t ( x)] + C .'

' Nu f [x(t )].x (t ) c nguyn hm l hm F(t) th

V d 2: Tnh I =

a 2 x 2 dx , , ta c: 2 2

Gii

t x = asint vi t I =

a 2 x 2 dx =

a 2 a 2 sin 2 t ( a cos t ) dt =

a

2

cos t 1 sin 2 t dt = a 2 cos 2 tdt

a2 a 2 sin 2t a2 a2 (1 + cos 2t )dt = t + sin t cos t + C = t + +C = 2 2 2 2 2a2 x 1 arcsin + x a 2 x 2 + C 2 2 2 2. Phng php tch phn tng phn: nh l 2: =' Cho cc hm u(x), v(x) kh vi v u ( x).v( x) c nguyn hm. Khi

u ( x).v ' ( x) cng c nguyn hm v u ( x ).v ' ( x )dx = u ( x).v( x ) u ' ( x).v( x )dx .Ch : V du = u ' ( x )dx v dv = v ' ( x)dx nn cng thc trn thng c vit di dng udv = uv vdu V d 3: Tnh I = x 3 ln xdx

Vi tch phn A1

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Ta c I =

1 1 4 1 4 4 4 3 ln xd ( x ) = 4 x ln x x d (ln x) = 4 x ln x x dx 4 x4 1 = x 4 ln x + C 4 4

[

Gii

] [

]

III. TCH PHN CA CC HM S N GIN 1. Tch phn ca hm s hu t Dng 1:

(ax + b)1. I 1 =

dx

n

trong a, b l cc hng s v n = 1, 2, 3.

V d 1: Tnh cc tch phn sau:dx (ax + b) dx 2. I 2 = (ax + b) n

(n 1)

1. I 1 = 2.I2 =

dx 1 d (ax + b) 1 1 = = (ln ax + b + C1 ) = ln ax + b + C (ax + b) a (ax + b) a a

Gii

1 (ax + b)1 n dx 1 1 (ax + b)1 n = (ax + b) n d (ax + b) = + C1 = +C a 1 n (ax + b) n a a 1 n

Dng 2:

x

2

dx trong a, b l cc hng s v n = 1, 2, 3. + ax + b

V d 2: Tnh cc tch phn sau:dx x + x +1 dx 2. I 2 = 2 x + 4x + 4 dx 3. I 3 = 2 3x 2 x 1

1. I 1 =

2

Gii

1.I1 = dx1 3 x+ + 2 4 2

=

1 d x + 2 2 1 3 x + + 2 2 2

1 2 2 + C = 2 arctg 2 x + 1 + C = arctg 3 3 3 3 2 dx dx ( x + 2)1 2 1 2. I 2 = 2 = = +C = +C 2 1 2 x+2 x + 4x + 4 ( x + 2)

x+

Vi tch phn A1

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3.I3 = 1 3 dx = 2 1 2 x x 3 3

1 2 x 1 dx 13 1 3( x 1) 3 3 = = ln + C1 = ln + C1 2 2 1 2 3 34 4 3x + 1 1 2 x + x 3 3 3 3 1 x 1 ln 3 1 x 1 = ln + + C1 = ln +C 4 3x + 1 4 4 3x + 1

Cch khc:

Ta c

1 1 A B = = + 3x 2 x 1 ( x 1)(3x + 1) x 1 3x + 12

1 A = 4 3A + B = 0 1 = A(3x + 1) + B( x 1) = (3 A + B) x + A B A B = 1 B = 3 4 3 1 4 1 3 1 dx dx I3 = 4 + x 1 3x + 1 dx = 4( x 1) 4(3 x + 1) dx = 4 x 1 3 3x + 1 1 1 x 1 = (ln x 1 ln 3x + 1 ) + C = ln +C 4 4 3x + 1

Dng 3:

x

2

Ax + B dx trong A, B, a, b l cc hng s, n = 1, 2, + ax + b

v a2 4b < 0.

V d 3: Tnh I = Ta cI =

x 1 dx x + x +12

Giidx 1 2x + 1 3 1 2x + 1 x 2 + x + 1 dx = 2 x 2 + x + 1 dx 3 x 2 + x + 1 2 2 dx 1 d ( x + x + 1) 3 = 2 2 2 2 x + x +1 x + x +1 1 3 dx = ln x 2 + x + 1 +C 2 2 2 2 1 3 x + + 2 2 1 2x + 1 = ln x 2 + x + 1 3arctg +C 2 3

Dng 4:Ax 2 + Bx + C (ax 2 + bx + c)(dx + e)dx trong A, B, C, a, b, c, d, e l cc hng s, n = 1, 2, e 3 v b24ac< 0, x = khng l nghim ca phng trnh d

Ax2+Bx+C = 0.

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V d 4: Tnh cc tch phn sau:

1. I 1 =

dx x( x 2 + 1)

x2 x +1 dx 2. I 2 = ( x + 1)( x 2 + x + 1)

1. Ta c

1 x( x + 1)2

=

A Bx + C + 2 1 = A( x 2 + 1) + x( Bx + C ) = ( A + B) x 2 + Cx + A x x +1A + B = 0 A = 1 C = 0 B = 1 A = 1 C = 0

Gii

Vydx x 1 = 2 dx 2 x( x + 1) x x +1 dx xdx 1 d ( x 2 + 1) = 2 = ln x 2 +C 2 x x +1 x +1 1 = ln x ln x 2 + 1 + C 2 2 x x +1 A Bx + C = + 2 2 ( x + 1)( x + x + 1) x + 1 x + x + 1 I1 =

2. Ta c

x 2 x + 1 = A( x 2 + x + 1) + ( x + 1)( Bx + C ) x 2 x + 1 = ( A + B) x 2 + ( A + B + C ) x + A + C A + B = 1 A = 3 A + B + C = 1 B = 2 A + C = 1 C = 2 x2 x +1 dx 2x + 2 2x + 1 + 1 3 dx = dx 2 2 Vy I 2 = dx = 3 2 x +1 x + x +1 ( x + 1)( x + x + 1) x + 1 x + x + 1 (2 x + 1)dx dx = 3 ln x + 1 2 2 +C x + x +1 x + x +1 d ( x 2 + x + 1) dx = 3 ln x + 1 2 +C 2 2 x + x +1 1 3 x + + 2 2 2 2x + 1 = 3 ln x + 1 ln( x 2 + x + 1) arctg +C 3 3

Dng 5:dx trong m l hng s v n = 1, 2,. + m2 )n dx V d 5: Tnh I = 2 (x + m2 )2

(x

2

Vi tch phn A1

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Giidx 1 Ta c I = 2 = 2 2 2 m (x + m ) x + m x2 x2 1 x2 + m2 2 dx = 2 dx 2 dx (x 2 + m2 )2 m (x + m2 )2 (x + m2 )2 2 1 dx x dx = 2 2 x + m2 (x2 + m2 )2 m 2 2

Tnh

x 2 dx 1 xd ( x 2 + m 2 ) = 2 (x 2 + m2 )2 2 (x + m2 )2 t u = x du = dx d (x 2 + m2 ) 1 dv = 2 v= 2 2 2 (x + m ) x + m2 x 2 dx x 1 dx ( x 2 + m 2 ) 2 = 2( x 2 + m 2 ) + 2 x 2 + m 2

VyI= dx x 2 dx 2 2 x + m2 (x + m2 )2 dx x dx 1 1 = 2 2 2 + 2 2 2 2 x +m m x +m x + m2 1 m2 1 2m 2 1 = 2m 2 = 2 x 2 x x dx + 2 2 x + m2 +m x 1 x + arctg + C 2 m m +m

2. Tch phn ca hm s lng gic Dng 1: R(sin x, cos x)dx trong R(sinx, cosx) l hm hu t theo sinx, cosx.

Ta s hu t ho tch phn bng cch t t = tg , khi

x 2

2t 1 t2 2dt ; cos x = ; dx = 1+ t2 1+ t2 1+ t2 V R (sin x, cos x ) dx s tr thnh tch phn hm hu t. sin x =V d 1: Tnh I = dx 3 + 5 cos x

Gii

2t x ; t t = tg sin x = 1+ t2 2I = = 3+52 dt 1+ t 2 1 t 2 1+ t 2

1 t2 2dt cos x = ; dx = 1+ t2 1+ t2

=

dt 1 t+2 = ln +C 2 4t 4 t2

x 1 tg 2 + 2 ln x +C 4 tg 2 2

Dng 2:Vi tch phn A1 trang 51

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R(sin x, cos x)dx , ta xt cc trng hp sau.Trng hp 1: Nu R(sinx, cosx) = R(- sinx, - cosx) th ta t t = tgx. V d 1: Tnh cc tch phn sau: dx 1/ I 1 = 2 sin x 3 cos 2 x dx 2/ I 2 = 2 sin x cos 2 x + sin 2 x Gii t2 dt 1 ; sin 2 x = ; dx = 1/ t t = tgx cos 2 x = 2 2 1+ t 1+ t 1+ t2 dt dt 1+ t 2 = 2 I1 = t 2 t 3 3 1+1t 2 1+ t 2= =

dt t2 1

( 3)

2

=

1 2 3

ln

t 3 +C t+ 3

2 3

ln

tgx 3 + C. tgx + 3

2/ Ta c I 2 = t t = tgx cos 2 x =

dx sin 2 x cos 2 x + 2 sin x cos x

1 t2 dt t 1 ; sin 2 x = ; dx = ; sin x = ; cos x = 2 2 2 1+ t 1+ t 1+ t 1+ t2 1+ t2dt 1+ t 22

I1 = =

t 1+t 2

2 1+1t 2 + 1+tt2

=

dt dt = t 2 + 2t 1 (t + 1) 2 +C

( 2)

2

=

1 2 2

ln

(t + 1) 2 +C (t + 1) + 2

1 2 2

ln

tgx + 1 2 tgx + 1 + 2

Trng hp 2: Nu R(sinx, cosx) = -R(- sinx, cosx) th ta t t = cosx. sin 3 x dx V d 2: Tnh I = 1 + cos 2 x Gii t t = cosx dt = - sinxdxI = sin 2 x 1 t2 t2 +1 2 2 ( sin xdx) = dt = 2 dt = 1 2 dt = t 2arctgt + C 2 2 1 + cos x 1+ t t +1 t + 1

= cos x 2arctg (cos x ) + C

Trng hp 3: Nu R(sinx, cosx) = -R(sinx, - cosx) th ta t t = sinx. dx V d 3: Tnh I = sin 2 x cos xVi tch phn A1 trang 52

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Gii

t t = sinx dt = cosxdxI = cos dx dt dt dt t 2 + (1 t 2 ) + 2 dt = = 2 = 2 2 2 2 2 2 sin x cos x 1 t t t (1 t ) t (1 t )

1 1 sin x 1 dt dt 1 t 1 1 +C + C = ln + 2 = ln 2 sin x + 1 sin x 2 t +1 t t 2 1 t 3. Tch phn ca hm s v t = Dng 1:

R ( x,

n

ax + b )dx , ta t t = n ax + +b .

V d 1: Tnh I =

1+

dx3

x +1Gii

x = t 1 t t = 3 x + 1 dx = 3t 2 dt 3

I =

t2 3t 2 dt 1 = 3 t 1 + dt = 3 t + ln t + 1 + C 2 1+ t t + 1 3 ( x + 1) 2 = 3 3 x + 1 + ln 3 x + 1 + 1 + C 2

Ch : Nu

R ( x,

n1

ax + b , 2 ax + b ,...) dx , ta tdx

n

t = n ax + b

vi n = BCNN(n1, n2,).V d 2: Tnh I = 6

x (3 x + 1)

Gii

x = t t t = 6 x dx = 6t 5 dt 5 6t dt t 2 dt 1 I = 3 2 = 6 2 = 6 1 2 dt = 6(t + arctgt ) + C t (t + 1) t +1 t + 1

= 6 6 x arctg 6 x + C Dng 2:

(

)

ax 2 + bx + c ax 2 + bx + c V d 2: Tnh cc tch phn sau:

dx

;

Ax + B

dx

1. I 1 = 2. I 2 = 3. I 3 =

dx x + 2x + 5 dx2

1 x x2 5x + 3 x 2 + 4 x + 10

dx

Vi tch phn A1

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1. I 1 = 2. I 2 = 3. I 3 = 5 2

dx

(x + 1)2 + 4dx5 4

=

d ( x + 1)d (x +22

Gii

(x + 1)2 + 2 2

= ln x + 1 + x 2 + 2 x + 5 + C= arcsin x+5 2 1 2

(x +

1 2 2

)

==

( )5 2

1 2

)

(x +

1 2 2

)

+ C = arcsin

2x + 1 5

+C

(2 x + 4) 7 x + 4 x + 102

5 d ( x + 4 x + 10) d ( x + 2) x 2 + 4 x + 10 7 ( x + 2) 2 + ( 6 ) 2 2

= 5 x 2 + 4 x + 10 7 ln ( x + 2) + x 2 + 4 x + 10 + C

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Dng 2:

R ( x,

ax 2 + bx + c )dx . ax 2 + bx + c = t + a x ).

Ta dng php th Euler (i ) Nu a > 0, t ax 2 + bx + c = t a x ( hoc

(ii ) Nu c > 0, t ax 2 + bx + c = xt + c ( hoc ax 2 + bx + c = xt c ). (iii ) Nu phng trnh ax2 + bx + c = 0 c hai nghim thc x1, x2 th ax2 + bx + c = a(x x1)(x x2), khi ta t ax 2 + bx + c = t ( x x1 ) V d 2: Tnh I = dx x + x2 x +1

Gii

t

t 1 x = 2t 1 x2 x +1 = t x 2 dx = 2 t t + 1 (2t 1) 2 2

I =

dx x + x2 x +1

=

1 1 2(t 2 t + 1) 3 3t dt = 2 + dt = 2 2 2 t (2t 1) t (2t 1) t

3 8

[4(2t 1)] 3 2(2t 1) 2

dt

dt 3 d (2t 1) 2 3 d (2t 1) 3 3 1 = 2 + = 2 ln t ln 2t 1 +C 2 2 4 (2t 1) 2 4 (2t 1) t 8 (2t 1) 3 3 1 = 2 ln x + x 2 x + 1 ln 2 x + 2 x 2 x + 1 1 +C 2 4 2x + 2 x 2 x + 1 1

Ch : tnh R( x, ax 2 + bx + c )dx ta c th dng php i bin s lng gic.2 b b 2 4ac Ta c ax + bx + c = a x + 2a 4a 2 2

t u = x +

b ; 2a

m2 =

b 2 4ac b 2 4ac ; n2 = . 4a 2 4a 2

Khi tch phn trn c a v cc dng: (1) R( x, u 2 m 2 )dt ( khi b2 4ac 0 v a > 0 ). (2) R( x, m 2 u 2 )dt ( khi b2 4ac 0 v a < 0 ). (3) R( x, u 2 + n 2 )dt ( khi b2 4ac < 0 v a > 0 ). i vi cc tch phn ny ta c th dng php i bin s bng cc t: (1) u =m ; (2) u = msint; (3) u = ntgt sin t

Ch : Bng phng php tch phn tng phn ta tnh c:

x 2 a 2 dx =

x 2 a2 x a2 ln x + x 2 a 2 + C 2 2trang 55

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Cu hi cng c: 1. Theo bn hiu th no l tch phn ? 2. Tch phn bt nh l g ? Bn hiu th no v hng s C trong kt qu ca tch phn bt nh ? 3. Mc ch i bin s l g ? Lm sao bn bit t bin mi l ng hay sai ? 4. Bn hiu ngha tch phn tng phn l nh th no ? V cho bit vn chnh trong tch phn tng phn l g ? 5. Theo bn c bao nhiu phng php tnh tch phn? V phng php no thng hay ng dng nhiu nht ? 6. Theo bn c cn quan tm n tch phn cc dng khc hay khng ? Ti sao ?

x a2 x 2 2 a x dx = a x + arcsin + C 2 2 a2 2

Vi tch phn A1

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PHN HNG DN THC HNHKQHT 3 : Tnh tch phn i bin, tng phn, din tch hnh phng, di cung phng v th tch vt th trn xoay. BC HC 1: Tch phn bt nh

Trang thit b + vt liu cung cp cho hc vin: 1. Giy A4, A3. 2. Vit long

Cc bc thc hnh: 1. Bn hy lit k 10 cng thc tch phn m bn cho rng l c bn ? 2. Lit k cc phng php tnh tch phn bt nh ? 3. Mc ch i bin s l g ? Lm sao bn bit t bin mi l ng hay sai ? 4. Bn hy cho 02 bi tp v tch phn bt nh c ng dng hai phng php gii ?


Kt lun / Tho lun:

Vi tch phn A1

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TIU CHUN CHO BI THC HNH (CHECKLIST)Tiu ch C Khn g

1. C 10 cng thc tch phn c bn 2. C hai v d mi v d c cch gii ng dng 01 phng php 3. C s dng hai phng php tnh tch phn 4. C lm cho hm s di du tch phn n gin hn theo bin mi khng ? 5. Vi phn theo bin mi c xut hin trong hm s di du tch phn c khng ? 6. i bin trong tch phn bt nh c tr li bin c khi v kt qu 7. Tch phn bt nh c tnh theo vi phn hay o hm 8. Tch phn i bin, tch phn tng phn u tnh c l nh tch phn c bn 9. Bn c phn no hiu r rng hn v cch tnh tch phn so vi chng trnh ph thngNhn xt: -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Vi tch phn A1

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BC HC 2: Tch phn xc nh Bi hng dn: 2. TCH PHN XC NH

I. NH NGHA TCH PHN XC NH1. Bi ton din tch hnh thang cong Cho hm s y = f(x) lin tc, n iu v khng m trn on [a, b]. Xt hnh thang ABCD c gii hn bi cc ng thng x = a, x = b, trc ox v ng cong y = f(x). Ta chia on [a, b] mt cch tu thnh n on nh bi cc im chiaa = x0 < x1 < x2 < K < xk < xk +1 < K < xn = b

Trn mi on nh c chia [xi-1, xi ] ta dng mt hnh ch nht vi chiu rng l xi = xi xi 1 v chiu cao l f ( i ) ( vi i ( xi 1 , xi ) ). Tng din tch ca n hnh ch nht trn l: S n = f ( i ).xi ( chnh l din tch hnh bc thang nh hnh v H 3.1).i =1 n

Nhn xt: Din tch ca hnh bc thang gn bng din tch ca hnh thang cong ABCD khi n cng ln v cc on c chia cng nh. Do din tch S ca hnh

thang ABCD cho l: S = lim S n = n y D C

max xi 0

lim

f ( )xi =1 i

n

i

A O a

xi-1 H 3.1

B xi b

x

2. nh ngha tch phn xc nh Cho f(x) l hm s xc nh trn on [a, b], chia on [a, b] mt cch tu thnh n on nh bi cc im chia a = x 0 < x1 < x 2 < K < x < x k +1 < K < x n = b .

t d = max{xi } ( vi xi = xi xi 1 ), i = 1n.

Trn mi on [xi-1, xi] ly im i ( i = 1n )tu , lp tng I n = f ( i )xii =1

n

v gi l tng tch phn ca hm f(x) trn [a, b]. Tng im chia ln v hn ( n ) sao cho d 0 , nu trong qu trnh In I

( hu hn ) m khng ph thuc vo cch chia on [a, b] v cch ly im i th I c gi l tch phn xc nh ca hm f(x) trn [a, b]. K hiu: I = f ( x)dx = lim f ( i )xia d 0 i =1 b n

Khi ta ni hm f(x) kh tch trn [a, b].Nhn xt:Vi tch phn A1 trang 59

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1.

f ( x)dxa

b

nu c th ch ph thuc vo hm f(x) v hai cn a, b khng ph thuc vo

bin s, tc l 2.b b

f ( x)dx = f (t )dt .a a

b

b

Khi nh ngha tch phn xc nh ta coi a < b. Nu a > b tha

a

f ( x)dx = f ( x)dx

v khi a = b thb

f ( x)dx = f ( x)dx = 0 .a a

b

a

3. Theo nh ngha tch phn xc nh th din tch hnh thang cong m ta xt l:

S = f ( x)dx .a

4. T nh ngha trn ngi ta chng minh c cc nh l sau: nh l 1: Mi hm s f(x) lin tc trn [a, b]u kh tch trn on . nh l 2: Nu trn on [a, b], hm s f(x) b chn v ch c mt s im gin on th n kh tch trn on . nh l 3: Nu hm s f(x) n iu v b chn trn on [a, b] th n kh tch trn on . nh ly 4: ( Cc tnh cht ca hm kh tch ) 1. Nu hm s f(x) kh tch trn on [a, b] th cc hm f (x) v k.f(x) cng kh tch trn on [a, b]. 2. Nu hai hm s f(x) v g(x) kh tch trn on [a, b] th tng, hiu v tch ca chng cng kh tch trn on [a, b]. 3. Nu hm s f(x) kh tch trn on [a, b] th n kh tch trn mi on [ , ] [a, b] . Ngc li, nu ta chia on [a, b] thnh cc on nh v f(x) kh tch trn tng on nh th f(x) kh tch trn on [a, b].2. TNH CHT CA TCH PHN XC NH Gi s f(x) v g(x) l cc hm kh tch trn on [a, b], khi :

1. 2. 3.

[ f ( x) g ( x)]dx = f ( x)dx g ( x)dx .a a a

b

b

b

kf ( x)dx = k f ( x)dx .a a

b

b

f ( x)dx = f ( x)dx + f ( x)dx .a a c

b

c

b

4. Nu f ( x) g ( x) x [a, b] th

f ( x)dx g ( x)dx .a a

b

b

5.

f ( x )dx a a

b

b

f ( x ) dx .trang 60

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QT7.1/PTCT1-BM-7

6. Nu m f ( x) M

x [a, b] th m(b a) f ( x)dx M (b a) .a

7. ( nh l gi tr trung bnh ca hm s ) Nu hm s f(x) kh tch trn on [a, b] v m f ( x) M

x [a, b] th tn ti s

[m, M ] sao cho f ( x)dx = (b a) .a

b

c bit: Nu hm s f(x) lin tc trn on [a, b] th tn ti s

c [a, b ] sao cho f ( x)dx = f (c)(b a ) .b a

Gi tr f (c) = K hiu: f .

1 f ( x)dx c gi l gi tr trung bnh ca hm s f(x). ba a

b

3. CNG THC C BN CA TCH PHN XC NH Gi s hm s f(x) kh tch trn on [a, b], khi f(x) cng kh tch trn on

[a, x] [a, b]. Ngha l tn ti tch phn x.x

f (t )dta

x

v n l mt hm s theo bin

K hiu: F ( x) = f (t )dt . Khi hm F(x) c cc tnh cht sau:a

1/ Nu hm f(x) kh tch trn on [a, b] th F(x) lin tc trn on . 2/ Nu hm f(x) lin tc ti x th hm F(x) c o hm ti x v F ' ( x) = f ( x) . nh l : ( Cng thc Newton-Leibniz ) Nu hm s f(x) lin tc trn [a, b] v F(x) l mt nguyn hm ca n th

f ( x)dx = F ( x)a

b

b a

= F (b) F (a ) .

Nhn xt: Cng thc ny cho php tnh tch phn xc nh thng qua nguyn hm ca hm f(x) m khng cn s dng nh ngha, v nguyn tc ta c th tch c tch phn xc nh. 4. NG DNG CA TCH PHN XC NH 4.1. Tnh din tch hnh phng Nu hm s f(x) lin tc trn on [a, b] th din tch hnh phng gii hn bi th ca hm s y = f(x) v cc ng thng x = a ; x = b ; y = 0 c tnh theo cng thc:S=a b

b f ( x)dx khi f ( x) 0 a f ( x) dx = b . f ( x)dx khi f ( x) 0 a

Vi tch phn A1

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o Nu cc hm s f(x) v g(x) lin tc trn on [a, b] th din tch hnh phng gii hn bi th ca cc hm s y = f(x) ; y = g(x) v cc ng thng x = a ; x = b

c tnh theo cng thc: S = f ( x) g ( x) dx .a

b

Nu phng trnh ng cong cho di dng x = ( y ) , ( y ) lin tc trn on x=0 c tnh theo cng thc: S = ( y ) dy .a b

[a, b] th din tch hnh phng gii hn bi cc ng x = ( y ) ; y = a ; y = b v

Nu ng cong cho bi phng trnh tham s S = f ( x) dxa b

x = (t ) th cng thc y = (t )

t2

tr thnh

(t ). (t ) dt' t1

trong t1, t2 ln lt l nghim ca cc phng

trnh a = (t ) , b = (t ) v (t ) , (t ) , ' (t ) l cc hm s lin tc trn on [t1, t2]. V d : Tnh din tch hnh phng gii hn bi cc ng y = x2 ( x 0 ) v y = 2 - x. Gii Giao im ca cc ng y = x2 ( x 0 ) v y = 2 - x l nghim ca h y = x 2 (x 0) y = 2 x x = 1 y = 1

Vy din tch cn tm lS = (2 x) x dx = 2 0 0 1 1

[

x2 x3 (2 x) x dx = 2 x 2 32

]

1

=0

7 (vdt) 2

4.2. Tnh di ng cong phng Cung cho bi ng cong c phng trnh y = f(x), trong f(x) l hm s n tr v c o hm lin tc trn on [a, b]. di cung AB, vi A(a, f(a)) v B(b, f(b))

c tnh theo cng thc: l = 1 + [ f ' ( x)] dx .b

2

a

(t )

x = (t ) Cung cho bi ng cong c phng trnh y = (t )

(a t b ) , trong

v (t ) l cc hm s c o hm lin tc trn on [a, b]. di cung AB, vi A( ( a ), ( a )) v B ( (b), (b)) c tnh theo cng thc:

Vi tch phn A1

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QT7.1/PTCT1-BM-7' 2

L=

[ (t )] + [ (t )] dt (vd).b ' 2 a

V d : Tnh di cung ca ng cycloide

x = a(t sin t ) (0 t 2 ) y = a (1 sin t )

Gii

x ' (t ) = a(1 cos t ) Ta c ' y (t ) = a sin t x ' (t ) + y ' (t ) = a 2 (2 2 cos t ) = 4a 2 sin 2 Vy di cung cn tm l :2 2 t 2

[

] [2

]

2

t 2

t = 2a sin 2

t t t l = 2a sin dt = 4a sin 2 d ( 2 ) = 4a cos 2 0 = 8a (vd). 2 0 0

4.3. Tnh th tch vt th Vt th bt k: L vt th c gii hn bi mt mt cong kn vi hai mt phng x = a; x = b vung gc vi ox. Gi s S(x) l din tch thit din gia vt th v mt phng vung gc vi ox ti x ( x [a, b] ) v S(x) l hm s lin tc trn on [a, b]. Khi

th tch ca vt th c tnh theo cng thc: V = S ( x)dx .a

b

Vt th trn xoay: L vt th c to ra khi quay hnh thang cong gii hn bi ng y = f(x), x = a, x = b v y = 0 quanh trc ox. Khi th tch vt th trn xoay c tnh theo cng thc: V x = f 2 ( x)dx .a b

Ch : Vt th trn xoay c to ra khi quay hnh thang cong gii hn bi

ng y = f(x), x = a, x = b v y = 0 quanh trc oy. Khi th tch vt th trn xoay c tnh theo cng thc: V y = 2 xf ( x)dx .V d : Tnh th tch vt th trn xoay do hnh phng gii hn bi ng y = 2x - x2 v y = 0 khi: 1/ Xoay quanh trc ox. 2/ Xoay quanh trc oy. Gii Ta c ng y = 2x - x2 ct trc ox ti x = 0 v x = 2 nn ta c:a b

Vi tch phn A1

trang 63

Trng i hc Tr VinhVx = f 2 ( x)dx0 2

QT7.1/PTCT1-BM-7

= (2 x x 2 ) 2 dx

2

1/

0

= (4 x 4 x + x )dx2 3 4 0

2

.

4 x3 x5 = x4 + 3 5

16 = 15 02

2/Vy = 2 xf ( x)dx = 2 x(2 x x 2 )dx0 0 2 2

2 x 3 x 4 2 8 = = 2 (2 x x )dx = 2 3 4 0 3 0 2 2 3

.

4.4. Tnh din tch mt trn xoay Mt trn xoay l mt mt cong sinh ra do ta quay quanh trc ox mt cung ng cong phng AB c phng trnh y = f(x), x [a, b] [ vi f(x) l hm s n tr v c o hm lin tc trn on [a, b] ; A(a, f(a)), B(b, f(b))]. Din tch mt trn xoay c tnh theo cng thc:

S = 2 f ( x ) 1 + f ' ( x ) dx .a

b

[

]

2

Ch : 1/ Nu quay ng cong phng quanh trc oy th:

S = 2 x 1 + f ' ( x ) dx .2/ Nu ng cong phng cho bi phng trnh x = ( y ) , x [a, b] ( vi hm s ( y ) l hm s n tr v c o hm lin tc trn [a, b] ). Khi ta c:' Khi quay quanh trc ox: S = 2 y 1 + ( y ) dy . a' Khi quay quanh trc oy: S = 2 ( y ) 1 + ( y ) dy . a b

b

[

]

2

a

b

[

]

2

[

]

2

x = y 2 V d : Tnh din tch mt to nn khi quay ng parbol 0 y 1quanh trc ox. Gii

x ' ( y) = 2 y 2 1 + x ' ( y) = 1 + 4 y 2 Ta c y = x (do y 0)

[

]

Vi tch phn A1

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QT7.1/PTCT1-BM-7

S = 2 y 1 + 4 y 2 dy0

1

Vy din tch cn tm l:

= =

40

1

1 + 4 y 2 d (1 + 4 y 2 )3 2

.

(1 + 4 y 2 )4 .3 2

1

=0

6

(5 5 1)

Cu hi cng c: 1. Th no l tch phn xc nh ? 2. Hy trnh by cc cng thc c bn ca tch phn xc nh ? 3. Hy vit bn cng thc ng dng ca tch phn xc nh ?

Vi tch phn A1

trang 65

Trng i hc Tr Vinh

QT7.1/PTCT1-BM-7

BC HC 2: Tch phn xc nh


Cc bc thc hnh: 1. Cho hai v d: mt v d ng dng phng php i bin s, mt v d ng dng mt trong bn cng thc ng dng ca tch phn tng phn xc nh ? 2. Hy so snh kt qu ca tch phn xc nh v tch phn bt nh?


Kt lun / Tho lun:

Vi tch phn A1

trang 66

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QT7.1/PTCT1-BM-7

TIU CHUN CHO BI THC HNH (CHECKLIST)Tiu ch C Khn g

1. C hai v d: mt v d ng dng phng php i bin s, mt v d ng dng mt trong bn cng thc ng dng ca tch phn tng phn xc nh? 2. ng dng phng php i bin s trong qu trnh tnh tch phn khng? 3. C s dng mt trong bn cng thc ng dng ca tch phn xc nh khng ? 4. Kt qu tch phn xc nh c hu hn hay v hn ? 5. ng thi gian qui nh 6. Kt qu chnh xc v lp lun logc 7. S hp tc ca cc thnh vin trong nhmNhn xt: ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

---------------------------------------------------------------------------------------

Vi tch phn A1

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QT7.1/PTCT1-BM-7

BC HC 3: ng dng tch phn tnh din tch hnh phng, di cung phng v th tch vt th trn xoay?.


Cc bc thc hnh: 1. Cho mt v d ng dng mt trong bn cng thc ng dng ca tch phn tng phn xc nh ? 2. Hy cho bit s bt li khi s dng cng thc tnh di ca ng cng phng?


Kt lun / Tho lun:

Vi tch phn A1

trang 68

Trng i hc Tr Vinh

QT7.1/PTCT1-BM-7


1. C mt v d tnh di cung phng khng? 2. C s dng o hm cp mt khng ? 3. C s dng cng thc tnh di cung phng chnh xc khng ? 4. Kt qu tch phn xc nh c hu hn hay v hn ? 5. ng thi gian qui nh. 6. Kt qu chnh xc v lp lun logic. 7. S hp tc ca cc thnh vin trong nhm.Nhn xt: ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

------------------------------------------------------------------------------------------

Vi tch phn A1

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KT QU HC TP 4: Kho st mt s bi ton v hi t hay phn k bng s vn dng l thuyt tch phn suy rng loi I, loi II* nh ngha tch phn suy rng:+

a/

f ( x)dx : loi I.b a+

* im n: Xt cc vn vTch phn suy rng.

b/

f ( x)dx : loi II.

* Cc tiu chun hi t, phn k.

*Gii BT ng dng xt s hit v phn k ca tch phn: + Bng PP tnh trc tip. + Bng cc tiu chun.

Bi hng dn:

TCH PHN SUY RNGI. TCH PHN SUY RNG LOI MT nh ngha: Gi s hm f(x) xc nh trn [a, + ) v kh tch trn mi on [a, b]. Gii hn

( nu c ) ca tch phn trn [a, + ) , k hiu:+ +

f ( x)dxa

b

khi b + gi l tch phn suy rng ca hm f(x)

f ( x)dx .a b+

Vy

f ( x)dx =a

lim

f ( x)dx .a+

b

Nu lim Nu lim Tng t,+

b+

a b a

b

f ( x)dx hu hn th

f ( x)dxa

hi t v hm f(x) kh tch trn [a, + ) .+

b+ a

f ( x)dx v hn hoc khng tn ti th f ( x)dx phn .a b

f ( x)dx = +

lim

f ( x)dx ( Tnh hi t v phn k cng tng t ).b+

a

f ( x)dx =

a

f ( x)dx +

f ( x)dx .a+

Tch phn

f ( x)dx hi t khi f ( x)dxa+

v

f ( x)dx

a

hi t.

V d: 1/ Tnh I 1 =Vi tch phn A1

xe0

x2

dx .

trang 70

Trng i hc Tr Vinh+

QT7.1/PTCT1-BM-7

2/ Xt s hi t ca tch phn I 2 = Gii 1/b 1b 2 2 I1 = lim xe x dx = lim e x d ( x 2 ) b + b + 0 20 . 1 x 2 b 2 1 1 = lim e = lim 1 e b = b + b + 2 2 0 2

a

dx x

(a > 0, > 0) .

(

)

b + khi < 1 x 1 dx 1 1 1 1 . I 2 = lim = lim 1 = 1 blim (b a ) = a b+ x + b+ khi > 1 a a 1 Nu = 1 th b

2. Nu 1 th

+

I 2=+

a

b dx dx = lim = lim ln x a = lim (lnb ln a) = +. b+ x b+ a x b+

b

( )

Vy I 2 =

a

dx x

(a > 0, > 0) hi t khi

0 < 1 v phn k khi > 1 .

II TCH PHN SUY RNG LOI HAI nh ngha: Gi s f(x) l hm b chn v kh tch trn mi on [a, b ] ( > 0, b tu )

nhng khng b chn trn on [b , b] . Gii hn (nu c) ca tch phn khi 0 gi l tch phn suy rng ca hm f(x) trn on [a, b]. K hiu:bb

b

f ( x)dxa

f ( x)dxa

Vy

f ( x)dx = lim f ( x)dx . a 0 a

b

b

lim Nu 0lim Nu 0

a

f ( x)dx hu hn th

f ( x)dxa

b

hi t v hm f(x) kh tch trn [a, b] .b

b

f ( x)dx v hn hoc khng tn ti th f ( x)dx phn .a a

Tng t, nu f(x) l hm kh tch v b chn trn mi on [a + , b] nhng khng b chn trn on [a, a + ] th tng t ). f ( x)dx = f ( x)dx + f ( x)dx ( nu f(x) khng b chn ti c [a, b ] ).a a c b c b

f ( x)dx = lim f ( x)dx a 0 a+

b

b

( Tnh hi t v phn k cng

Vi tch phn A1

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QT7.1/PTCT1-BM-7

Tch phnV d:

f ( x)dxa

b

hi t khi

f ( x)dxa

c

v

f ( x)dxc

b

hi t.

1/ Tnh I 1 =

1

1

dx 1 x2

.1

2/ Xt s hi t ca tch phn I 2 = Gii 1/

dx . 1 x 0

I1 = 20

1

dx 1 x2

1

= 2 lim 01

0

dx 1 x2

= 2 lim arcsin x 0 0

(

) = 2 lim arcsin(1 ) . 0

= 2 arcsin1 = 2/1

I 2 = lim 0

0

1 d (1 x) dx = lim 1 x 0 1 x 01

= lim ln(1 x) 0 = lim( ln ) = + 0 0 Vy I 2 = 1

.

dx phn k. 1 x 0

3. IU KIN HI T CA TCH PHN SUY RNG nh l: Gi s f(x) v g(x) l cc hm kh tch trn mi on hu hn [a, b] v 0 f ( x) g ( x) x a , khi ta c:

Nu Nu

+

g ( x)dx hi t tha a

+

a

+

f ( x)dx hi t v+

a

+

f ( x)dx

g ( x)dx .a

+

f ( x)dx phn k th g ( x)dx phn k .a + +

nh l: Gi s f(x) v g(x) l cc hm khng m v kh tch trn mi on hu hn [a,f ( x) b]. Khi , nu lim = k (0 < k < + ) th cc tch phn x + g ( x)

f ( x)dxa

v

g ( x)dxa

cng

hi t hoc cng phn k. nh l:+

Nu+

a

+

f ( x) dx hi t th

f ( x)dx hi t.a +

nh ngha:

f ( x)dx c gi l hi t tuyt i nu a a

f ( x) dx hi t.

Vi tch phn A1

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Trng i hc Tr Vinh

QT7.1/PTCT1-BM-7+

k.

+

f ( x)dxa

c gi l hi t tuyt i nu

f ( x)dxa

+

hi t v

a

f ( x) dx phn

V d: Xt s hi t ca cc tch phn sau y:+

a)

1

1

(1 + x) 3 (1 + x )2

x 1: f(x) = v 0, x(0; 1] khi x +0 esin x 1 ln(1 + 3 x ) x ; esin x 1 sin x x11 3

Suy ra:

1

1 x .x1 2

2

3

1 0 (1 + x ) 3 (1 + x 2 ) 7 1 = v = > 1 6 x7 6

phi hi t.

1

3

1 3

1 ln(1 + 3 x ) x lim sin x = lim = lim = + x +0 0 x +0 e 1 x x 1 1 ln(1 + 3 x ) Khi x +0: l mt VCL ngang cp vi = . esin x 1 x ( x 0) 2 V = < 1 th tch phn suy rng phi hi t. 3 + + dx sin xdx 4) x1: sin3x 13 v 3 hi t, 3 x x x 0 x 0 + + sin x nn theo nh l 3 dx hi t tc sin xdx hi t tuyt i. x x3 0 01 3 2 3 2 3 2 3

Cu hi cng c: 1. Th no l tch phn suy rng ? 2. C my loi tch phn suy rng ?Vi tch phn A1 trang 73

Trng i hc Tr Vinh

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3. Hy cho bit iu kin hi t ca tch phn suy rng? BC HC 2: Tch phn suy rng


Cc bc thc hnh: 1. Cho hai v d: mt v d v tch phn suy rng loi I v mt v d v tch phn suy rng loi II 2. Hy s dng iu kin hi t ca tch phn suy rng xt s hi v phn k ca hai v d va cho ?


Kt lun / Tho lun:

Vi tch phn A1

trang 74

Trng i hc Tr Vinh

QT7.1/PTCT1-BM-7

Vi tch phn A1

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Trng i hc Tr Vinh

QT7.1/PTCT1-BM-7


1. C mt v d tch phn suy rng loi I v mt v d tch phn suy rng loi II khng? 2. C s dng tiu chun hi ca tch phn khng ? 3. ng thi gian qui nh 4. S hp tc cc thnh vin trong nhm 5. Kt qu chnh xc v lp lun logicNhn xt: ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

------------------------------------------------------------------------------------

Vi tch phn A1

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KQHT 5: Kho st s hi t, phn k ca chui s dng.Chui s - Chui hm: + nh ngha; + Cc php ton; + Cc tiu chun hi t. im n: Xt cc vn v Chui s v chui hm So snh chui s v chui hm Gii bi tp xt s hi t v phn k: + Chui s; + Chui hm.

BC HC 1: Chui s, chui s dng, chui an du. Bi hng dn: CHNG V 1 KHI NIM M U nh ngha 5.1

L THUYT CHUI

Cho dy s thc (un), n = 1, 2, 3... Biu thc

un =1

n

= u1 + u 2 + ... + u n + ...

(1)

c gi l chui. un c gi l s hng tng qut hay s hng th n ca chui (1). Tng Sn = u1 + u2 + ... + un c gi l tng ring ca chui (1). Nu Sn c gii hn S th chui (1) gi l chui hi t v c tng l S. Ta vit:S = un .n =1

Chui (1) khng hi t th gi l phn k. V d 5.1: Xt s hi t ca cc chui sau:

1/ 2/ Gii

qn =0 n =1

n

.1

ln1 + n . 1 q n = 1 q n khi q 1 khi q = 1

1/ Ta c: S n = 1 + q + q 2 + ... + q n 1 - Nu q < 1 lim S n =n n

1 chui hi t. 1 q - Nu q > 1 lim S n = chui phn k.

- Nu q = 1 lim S n = n

chui phn k.trang 77

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Trng i hc Tr Vinh

QT7.1/PTCT1-BM-7

- Nu q = 1 Sn khng c gii hn chui phn k. Vy

qn =0

n

hi t khi q < 1 v phn k khi q 1 .

n +1 = ln (n + 1) ln n n S n = (ln 2 ln 1) + (ln 3 ln 2 ) + ... + [ln (n + 1) ln n] = ln(n + 1)

2/ Ta c : u n = ln

lim S n = + chui phn k.n

* Cc nh l: nh l 5.1: Nu

un =1

n

hi t th lim u n = 0 .n

H qu: Nu lim u n 0 thn

un =1 n

n

phn k.

V d 5.2:

n + 1 phn k v lim n + 1 = 1 0 .n =1

n

n

nh l 5.2: Nu hai chui

u n vn =1

vn hi t th cc chuin =1 n =1 n =1

au n ;n =1

(un =1

n

+ v n ) cng

hi t v

un =1

n

= a u n , (u n + v n ) = u n + v n .n =1 n =1

nh l 5.3:

Cho hai c

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