Thiết kế bài giảng giải tích 12 (tập 2) phần 1

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TRAN V I N H

I hiet ke bai giang

GIAI TICH ] 2
TAP HAI

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^ * ' r >'» 3 c

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N H A X U A T BAN H A N 6 I



TRAN VINH

THIET KE BAI GIANG

GIAI TICH

TAP HAI

NHA XUAT BAN HA NOI



Chi/dNq III

NGUYEN HAM - TICH PHAN VA UNC DUNG
Phan 1
NHJtXG

VAX D E CUA CHMfONG

I. NOI DUNG
Noi dung chinh cua chucung 3 :
Nguyen ham : Dinh nghia ; tinh chat; cac nguyen ham ccf ban ; cac phucmg phap
tinh nguyen ham.
Tich phan : Dinh nghia ; cac tinh chat cua tich phan ; cac phuang phap tinh
tich phan.
" Lftig dung cua tich phSn : Bai toan dien tich, bai toan thi tich.
n . MUC TIEU
1. Kien thiirc
Nam dugc toan bo kien thiic co ban trong chuong da neu tren, cu the :
Nam viing dinh nghia nguyen ham, cac nguyen ham co ban, cac tinh chat ciia
nguyen ham.
• Dinh nghia tich phan, cac tinh chat ciia tich phan, ung dung ciia tich phan, moi
quan he giiia tich phan va nguyen ham.
M6t s6' ling dung tich phan trong hinh hoc : Tinh dugc dien tich hinh phang,
the tich vat the trong khong gian.
2. KT nang

van dung cac nguyen ham co ban de tinh cac nguyen ham.


Van dung thanh thao cong thiic Niuton - Laibonit de tinh tich phan. Moi
quan he giiia dao ham va nguyen ham.
Van dung tich phan de tinh dien tich hinh phang va the tich ciia vat the.
3. Thai do
Tu giac. tich cue, dgc lap va chii dgng phat hien ciing nhu ITnh hoi kien" thiic
trong qua trinh hoat dgng.
Cam nhan dugc su cSn thiet cua dao ham trong viec khao sat ham so.
Cam nhan dugc thuc te cua toan hgc, nhat la doi vdi dao ham.


P H a n 2.
CAC B A I SOA]!!^

§1. N g u y e n ham
(tiet 1, 2, 3, 4, 5)
I. MUC TIEU
1. Kien thurc
HS nam duac :
Nh6 lai each tinh dao ham cua ham sd.
• Dinh nghia nguyen ham.
• Cac tinh chat ciia nguyen ham.
Mot so' nguyen ham co ban.
Cac phuong phap tinh nguyen ham : Phuong phap doi bien sd va phuong
phap nguyen ham tiing phan.
2. KT nang
HS tinh thanh thao cac nguyen ham co ban.
Tinh dugc nguyen ham dua vao phuong phap doi bien sd va phuong phap

nguyen ham tiing phan.
3. Thai do
Tu giac, tich cue trong hgc tap.
Biet phan biet ro cac khai niem co ban va van dung trong tiing trudng hgp cu the.
" Tu duy cac va'n de cua toan hgc mot each Idgic va he thdng.
n . C H U A N B I C U A G V VA H S
1. Chuan bj ciia GV
Chuan bi cac cau hoi ggi mo.
Chuan bi pha'n mau, va mdt sd dd diing khac.


2. Chuan bj cua HS
Can dn lai mot sd kien thiic da hgc ve dao ham.
ra.

P H A N PHOI THCJI L U O N G
Bai nay chia lam 5 tiet:
Tiet 1 : Tic dau den hit miic 2 phdn I.
Tiet 2 : Tiep theo den het phdn I.
Tiet 3 : Tiep theo den het muc I phdn II.
Tiet 4 : Tiep theo den het phdn II.
Tiet 5 : Bdi tap

IV TIEN TRINH DAY

HOC

A. DAT VAN OE
Cau hoi 1
Xet tinh diing - sai cua cac cau sau day :

a) Ham sd y = In(cosx) cd dao ham y' = -tanx.
b) Ham sd y = In(cosx) cd dao ham y' = -cotx.
Cau hoi 2
C h o h a m s d y = 3''""
a) Hay tinh dao ham cua ham sd da cho.
b) Chiing minh rang ham sd y = x3''"'' cd dao ham la y' = 3''""
GV:
Ham y = xS^'"" ggi la nguyen ham ciia ham sd y' = 3^'""

B. BAi Mdl
I

NGUYEN H A M VA TINH CHAT


HOATDONC1
1. Nguyen ham
• Thuc hien f \ 1 trong 5'
Hoat dong cua HS

Hoat dgng cua GV

Ggi y tra loi cau hoi 1

Cau hoi 1
Tim mot ham sd F(x) ma
F(x) = 3x2

GV ggi mot vai HS tra Idi. Bai toan
nay cd nhieu dap sd.

Tong quat : F(x) = x^ + C trong do
C la hang sd bat ki.

Cau hoi 2

Ggi y tra Idi cau hoi 2

Tim mot ham sd F(x) ma
FYY^

—

Lam tuong tu cau a.
F(x) =

r vx; —
cos X

In X

- ^

cos X

• GV neu dinh nghia :
Cho hdm sof(x) xdc dinh tren K
Ham soF(x) duac ggi Id nguyen hdm cda hdm sof(x) tren K neu
F '(x) - f(x) vai mgi x e K
• GV neu va thuc hien vf du 1, GV cd the lay mdt vai vi du khac.
HI. Tim nguyen ham ciia ham sd y = x.

H2. Tim nguyen ham cua ham sd y = x
H3. Tim nguyen ham cua ham sd y = x
H4. Tim nguyen ham ciia ham sd y = x"

4
• Thuc Men f\2

trong 5'.


Hoat dong ciia GV

Hoat dong ciia HS

Cau hoi 1

Ggi y tra loi cau hoi 1
GV ggi mot vai HS tra Idi. Bai toan
nay cd nhieu dap sd.

Tim mot ham sd F(x) ma
F(x) = 2x.

Tong quat : F(x) = x^ + C trong dd
C la hang sd bat ki.
Cau hoi 2

Ggi y tra loi cau hoi 2

Tim mot ham sd F(x) ma

Lam tuong tu cau a.

V{x)=-.

F(x) = hix + C.
X

H5. Tim nguyen ham ciia ham sd y = sin x.
H6. Tim nguyen ham cua ham sd y = cosx.
H7. Tim nguyen ham ciia ham sd y

1

2Vx
N/2

H8. Tim nguyen ham ciia ham sd y = x
• GV neu dinh li 1:
Neu F(x) Id mot nguyen hdm cua hdm sof(x) tren K thi vai moi hang so
C, hdm soG(x) = F(x) + C cUng Id mot nguyen hdm cda f(x) tren K
H9. Biet ham sd cd mdt nguyen ham la y = sin x. Hay tim nguyen ham cua ham sd dd.
HIO. Biet ham sd cd mdt nguyen ham la y = cosx. Hay tim nguyen ham cua ham sd dd.
1
H l l . Biet ham sd cd mdt nguyen ham la y = ^^ '^ . Hay tim nguyen ham cua ham sd dd.
H12. Biet ham sd cd mdt nguyen ham la y = ^

. Hay tim nguyen ham ciia ham sd dd.

• Thuc hien Sgr 3 trong 5'.

Hoat dgng ciia GV
Cau hdi 1

Hoat dgng ciia HS
Ggi y tra loi cay hoi 1


Tinh dao ham ciia ham sd :

{G{x)y = [Fix) + C]'

y = G(x).

= F'(x) + C' =

fix),xeK.

Ggi y tra loi cau hoi 2

Cau hoi 2

GV tu ket luan.

Hay ket luan.
GV neu dinh li 2:

Neil F(x) Id mot nguyen hdm cua hdm sof(x) tren K thi moi nguyen
hdm cua f(x) tren K deu co dgng F(x) + C, vai C Id mot hang so.
De chiing minh dinh li, GV neu va'n 66 de HS chiing minh.
GV neu ki hieu :

^f(x)dx - Fix) + C.
• GV neu chii y trong SGK.
• De thuc hien vi du 2, GV cd the neu cac vi du khac hoac cho HS tu neu vi du va
dat cac cau hdi sau :
H13.Tinh J3xdx.
H14.Tinh Jkdx.
H15.Tinh f-dx.
Hld.Tinh

f-^dx
•'2Vx
HOAT DONG 2

2. Tinh chat ciia nguyen ham
• GV neu tinh chat 1:
( lf{x)dx)'

= fix) ; jf'ix)dx

HI7. Hay chiing minh cac tinh chat tren.

- fix) + C.


H18. Tinh ftanxdx.
• GV neu va cho HS thuc hien vi du 3 hoac cd the lay nhiing vi du khac.
• GV neu tinh chat 2 :
\kfix)dx = k

\fix)dx

De chiing minh tinh chat nay, GV cSn dua ra cac cau hdi sau :
HI9. Tinh dao ham hai ve.
H20. Chiing minh dao ham hai ve bang nhau.
• GV neu tinh chat 3 :
j[fix) ± gix)]dx = \fix)6x

±

jgix)dx.

• Thuc hien "pt 4 trong 5'
Hoat dgng cua GV
Cau hoi 1
Tinh dao ham cua ham
so d mdi ve.

Cau hoi 2
Hay lam tuong tu ddi
vdi trudng hgp dau trir.

Hoat dgng ciia HS
Ggi y tra loi cau hoi 1
[\fix)Ax+

\gix)6x\

= [\fix)6x)

+[\gix)dx]

H22. Tinh [(cos x + tan x)dx .
H23. Tinh [(cosx - vx)dx .
H24.Tinh |(x^+x + l)dx.

10

+ gix).

Ggi y tra loi cau hoi 2
[lfix)dx

- jgix)dx]

= fix) - gix).

• GV neu va thuc hien vi du 4. GV cd the thay bdi vi du khac.
H21. Tinh J (cos x + sin x)dx .

^fix)


nOATiyDNG3
3. Sii ton tai nguyen ham
• GV n6u dinh li 3:
Moi hdm sof(x) lien tuc tren K deu co nguyen hdm tren K
• Thuc hidn vi du 5:
2

H25. Chiing minh ham sd y = x ^ cd nguyen ham. Tinh nguyen ham cua ham sd dd:
H26. Chiing minh ham sd y = —z— cd nguyen ham. Tinh nguyen ham ciia ham sd dd.
sin X
• GV cho HS tinh nguyfen ham va dien vao bang sau :
••"';:-v-.'^- fix)

'"\--:.:,jr

:

fix) + C

0
ax«-^
1
x
e^
a * l n a ia>

0,a^l)

cosx
—siruc
1
cos x
1
sin x

11

• Thuc hien vi du 6 trong 5'
Cau a.
Hoat dgng ciia GV

Hoat dgng ciia HS
Ggi y tra loi cau hoi 1

Cau hdi 1
Tinh nguyen ham ciia ham so:
y = ly}
Cau hoi 2

Ggi y tra loi cau hdi 2

Tinh nguyen ham ciia ham sd:
1
Cau hoi 3

[2x2dx = - x ^
J
3

1

^

'

[-^L=dx- [x 3dx-3x3

Ggi y tra loi cau hoi 3

Tinh nguyen ham ciia ham so HS tu tinh.
da cho.
cau b. HS tu tinh tuong tu.
II. PHUONG PHAP TINH NGUYEN HAM
HOAT DONG 4
1. Phuong phap doi bien sd
• Thuc hien ^ . 6 trong 5'
Hoat dgng cua GV
Cau hdi 1
Dat u = X - 1, tinh du
Cau hoi 2
Tinh |(x-l)'°dx.

12

Hoat dgng ciia HS
Ggi y tra loi cau hoi 1
Ta cd du = u'dx = dx.
Ggi y tra loi cau hoi 2


[(x-iyOdx^[u'Odu = - u i ' + C

= l(x-i)"+C
U

caub.

Hoat dgng ciia GV

Hoat dgng cua HS

Cau hdi 1

Ggi y tra loi cau hdi 1

Dat x = e' tinh dt.

Ta cd t = Inx => dt = —dx
X

Cau hoi 2
Tinh [

Ggi y tra loi cau hdi 2

dx.

•' X

[^^dx=[tdt=^2^C=^ln2x + C
J
2
2
J X
• GV neu dinh li 1:
Neil \fiu)du
tuc thi

= Fiu) + C vdu = u(x) la hdm so co dao hdm lien

\fiuix))u

'ix) dx = Fiuix))

+ C.

H27. Hay chiing minh dinh ii tren.
• GV neu he qua:
f
\fiax+
J

1
b)dx =—Fiax+
a

b) + C (a ^ 0).

H28. Hay chiing minh he qua tren.
• GV cho HS thuc hien vi du 7. GV cd the thay bdi vi du tuong tu.
• Ddi vdi chii y trong SGK, GV neu va nhSii manh dieu nay :
Mgi bien sau khi thay ddi trong qua trinh tinh toan, song ket qua cud'i ciing phai la
bien ban dau.
• Thuc hien vi du 8 trong 5'

13

Hoat dgng ciia GV

Hoat dgng ciia HS

Cau hdi 1

Ggi y tra Idi cau hdi 1

Nen dat bien nao bdi bien u.

Datu = x - 1.

Cau hdi 2

Ggi y tra Idi cau hdi 2
Tinh nguyen ham ciia ham sd
HS tu tinh.
da cho.

H29.Tinh f(3x + l)dx
H30.Tinh f(x + l)dx
H31.Tinh jtanxdx.
HOATD0NG5
2. Phifong phap nguyen ham tiimg phSn
• Thuc hien -^

7trong 5'

Hoat dgng ciia GV

Cau hdi 1
Tinh

\ix cos xYdx.

Hoat dgng cua HS
Ggi y tra loi cau hdi 1
Ta cd
Ux cos x)'dx = x cos x + Ci

Cau hdi 2

Ggi y tra Idi cau hdi 2

Tinh fcosxdx.

[cosxdx = sinjc + C2,
Ggi y tra loi cau hdi 3

Cau hdi 3

\x sin xdx = -x cos x + sin x + C

Tinh [xsinxdx.
• GV neu dinh li 2 :

Neu hai hdm sou = u(x) vdv = v(x) co dao hdm lien tuc tren K thi
\uix)v\x)dx
14

= uix)vix) -

^u\x)vix)dx.


H32. Hay chiing minh dinh li tren.
• GV neu chii y : [udi; - uv -

\vdu.

• Thuc hien vi du 9 trong 7' Day la vi du quan trgng, GV nen hudng dan cu the
cau a.
Hoat dgng ciia GV
Cau hdi 1
Dat u va dv hgp If.
Cau hdi 2
Van dung dinh If, hay tfnh
nguyen ham ciia ham sd tren.

Hoat dgng ciia HS
Ggi y tra loi cau hdi 1
Dat u = X, dv = e^'dx
Ggi y tra loi cau hdi 2
Jxe^'dx = xe'' - fe''dx = xe" - e " + C.

caub.
Hoat dgng ciia GV
Cau hdi 1
Dat u va dv hgp If.
Cau hdi 2

v a n dung dinh If hay tfnh
nguyen ham ciia ham so tren.

Hoat dgng cua HS
Ggi y tra loi cau hdi 1
Dat u = X, dv = cosxdx
Ggi y tra loi cau hdi 2
[x cos X d X = X sin X + cos x + C.

Cau c.
Hoat dgng cua GV
Cau hdi 1
Dat u va dv hgp If.
Cau hdi 2
Van dung dinh If hay tfnh
nguyen ham cua ham sd tren.

Hoat dgng ciia HS
Ggi y tra loi cau hdi 1
Dat u = Inx, dv = dx
Ggi y tra loi cau hdi 2
[in X d X = X In X - j dx = x In
-x x+-C.

15


• Thuc hien GL 8 trong 5'
GV cho HS tu dien vao bang. Ket qua nhu sau:

jPix)e''dx

[P(x)cosxdx

JP(x) In xdx

u

P(x)

P{x)

Inx

dv

e^dx

cosxdx

Prxjdx

HOAT DONG 6

TOM T ^ B^l H9C
1. Cho ham so fix) xac dinh tren K
Ham sd F(x) dugc ggi la nguyen ham ciia ham sd fix) tren K neu F 'ix) - fix)
vdi mgi x e K
2. Neu Fix) la mdt nguyen ham cua ham sd fix) tren K thi vdi mdi hang so C,
ham sd G(x) = Fix) + C ciing la mdt nguyen ham ciia fix) tren K

h) Neu Fix) la mdt nguyen ham cua ham so fix) tren K thi mgi nguyen ham cua
fix) tren K deu cd dang Fix) + C, vdi C la mdt hang sd.
3. ( jfix)dx)'
\kfix)dx

- fix)

= k\fix)dx\;

;

lf'ix)dx

- fix) + C.

I \[fix)±gix)\dx

= \fix)dx±

\gix)dx.

5. Mgi ham sd fix) lien tuc tren K deu cd nguyen ham tren K
6.

16

1"

fi^

J0dx = C

la^dx =
+ C (a > 0, a ^ 1)
J
In a

\dx = X + C

Icosxdx = sinx+ C


fr"Hr

J

a +1

r " + l -I-r' rr/3t

H

jsinxdx = -cosx + C

— dx = Inlxl + C
Jx
' '

—dx = tanx + C
•'cos X

fe''dx = e* + C

—dx = - c o t x + C
•"sin X

7. Ne'u \fiu)du = Fiu) + C va M = u(x) la ham sd cd dao ham lien tuc thi
jfiuix))u'ix)dx

= Fiuix)) + C.

8. Neu hai ham sd u = uix)va.v - vix) cd dao ham lien tuc tren AT thi
[u(x)v'(x)dx = u(x)v(x)- [u'(x)v(x)dx;

iudv = uv -

wdu

HOAT DONG 7

M9T SO C^a HOI TR^C NGHIEM ON T6P B^l 1
Cdu L Cho ham sd y = x Hay dien diing sai vao cac cau sau

(b) Ham sd chi cd mdt nguyen ham.

D
D

(c) Ham so chi cd nguyen ham la — x
4

D

(d) Ham sd cd vd sd nguyen ham dang — x + C.
4
Trd Idi.

D

(a) Ham so ludn cd nguyen ham.

a

b

c

d

D

S

S

D

17

Cdu 2. Cho ham sd y = Vx Hay dien diing sai vao cac cau

sau

(a) Ham sd ludn cd nguyen ham.

Q

(b) Ham sd chi cd mdt nguyen ham.

[J

(c) Ham sd chi cd nguyen ham la — x ^

D

(d) Ham sd cd vd sd nguyen ham dang — x ^ + C.

D

Trd Idi.
a

b

c

d

D

s

S

D

Cdu 3. Cho ham sd y = x + cosx. Hay dien diing sai vao cac cau sau

D
D

(a) Ham sd ludn cd nguyen ham.
(b) Ham sd chi cd mdt nguyen ham.
1

7

(c) H a m sd chi cd nguyen h a m la — x + sin x.

D

(d) H a m sd cd vd sd nguyen h a m dang .—x^ + sin x + C.

D

Trd Idi.
a

b

c

d

D

S

S

D

Cdu 4. Ham sd nao sau day cd nguyen ham la 2x
(a)y=x'+2 ;

(b)y=2x;

(c) y = 2 ;

(d)y= VST.

Trd led. (c).

18


Cdu 5. Ham sd nao sau day cd nguyen ham la Vx
(a)y= y = ^ ^ ;
2Vx

(b) y = - x 2 ;
3

(c)y = x2;

(d)y=-.
X

Trd Idi. (a).
Cdu 6. Ham sd nao sau day cd nguyen ham la - cos 2x
(a) y = sin2x ;

(b) y = —sin2x;

(c) y = -sin2x;
(d)y=cos2x.
Trd Idi. (b).
Cdu 7. Ham sd nao sau day cd nguyen ham la cos 2x
(a) y = sin2x ;

(b) y = — s i n 2 x ;

(c)y = -sin2x;
Trd Idi. (b).

(d)y=sin2x.

Cdu 8. Ham so nao sau day cd nguyen ham la sin2x
(a) y = sin2x ;

(b) y = —cos2x;
2
(d)y=sin2x.

(c) y = -sin2x;
Trd Idi. (b).

Cdu 9. Ham so nao sau day cd nguyen ham la e"
(a)y = e ' ' ;

(b)y=^e2'';

(c)y = lnx;

(d)y=e'"''

Trd Idi. (a).
Cdu 10. Ham so nao sau day cd nguyen ham la In x
(a)y = l n x ;

(b)y=-;

(c)y = - l n x ;

(d)y=e'"''

X

Trd Idi. (b).

19


HOAT DONG 8

naCTNG D^N Bfil T6P SBai 1. Hudng ddn. Dua vao dinh nghia nguyen ham.
a) e ^ va -e

la nguyen ham ciia nhau.

HS tu tinh dao ham cua mdi ham sd tren de chiing minh.
b) Lam tuong tu cau a.
2

sin X la mdt nguyen ham ciia sin2x.
c) Lam tuong tu cau a.
4^

e^ la mdt nguyen ham cua

1
X)

Bai 2. Hudng ddn. Su dung cac tinh chat ciia nguyen ham.
cau a. Chia tii cho miu, sau dd sii dung tinh chat ciia nguyen ham ciia ham sd y = x"
5
7
2

Ddp sd. — x3 +—x^ +7;X^ + C .
5
7
2
. 2'' + In 2 - 1
cau b. Ddp so
+ C.

e''an2-l)
cau c. Su dung cong thiic lugng giac va nguyen ham ciia ham sd lugng giac.
Ddp sd. -2cot2x + C.
cau d. Sii dung cdng thiic lugng giac va nguyen ham cua ham sd lugng giac.
Ddp sd.

1

1

cos8x + cos2x + C.

cau e. Sir dung cong thiic lugng giac va nguyen ham cua ham sd lugng giac.
Ddp sd. tanx - x + C .
cau g. Dat 3 - 2x = u.
Ddpsd. - - e ^ ' ^ ^ + C .
2
20


1
1/ 1

•
(1 + x)(l - 2x) ~ 3 1 + X

cauh.

1 +X

Ddp sd. — In

l-2x

+

1 - 2x

+ C.

Bai 3. Hudng ddn. Su dung cac tinh chat ciia nguyen ham.
cau a. Dat u = 1- x

.- (1-x) 10

Ddp sd.

10

- + C.

cau b. Dat u = 1 + x'
5

Ddpsd - ( l + x ^ ) 2 + c

cau c. Dat t = cos X
1
Ddp sd. — cos x + C
4
cau d. Dat t = cos x
Ddp sd.

-1

-+C

\ + e'
Bai 4. Hudng ddn. Sii dung cac tinh chat ciia nguyen ham.
Cau a. Ap dung tinh nguyen ham tirng phan : u = I n ( l + x), dt; = xdx.
Ddp sd \ix^

- l ) l n ( l + x) - j x ^ + I + C..
2

X

Cau b. u = X + 2x - 1, du = e dx
Ddp sd e"" (x^ - 1) + C.
cau c.u=x,dv

- sin(2x + l)dx.

X
1
Ddp sd. -— cos(2x + 1) + — sin(2x + 1) + C
cau d. Dat u = 1 - X, dy = cosxdx.
Ddp sd. (1 - x) sin x - cos x + C.

21


§2. Tich phan
(tiet 6, 7, 8, 9, lO)
I. MUC TIEU
1. Kien thiic
HS nam dugc ;
Khai niem tich phan la gi?
- Dien tich hinh thang cong ?
- Dinh nghia tich phan.
• Cac tfnh chat cua tich phan.
• Cac phuong phap tinh tich phan
- Phuong phap ddi bien sd.
- Phuong phap nguyen ham tiing phSn.
Mdi quan he giiia tich phan va nguyen ham.
2. KT nang
van dung thanh thao cac tfnh chat ciia tich phan.
Tfnh dugc tfch phan bang phuong phap ddi bien sd, thanh thao trong viec
ddi bien sd.
Tinh dugc tfch phan nhd phuong phap tfch phan tijrng phin.
3. Thai do
Tu giac, tfch cue trong hgc tap.
Biet phan biet rd cac khai niem co ban va van dung trong tiing trudng hgp cu th^.

Tu duy cac va'n de ciia toan hgc mdt each Idgic va he thdng.
H. CHUAN BI CUA GV VA HS
1. Chuan bj cua GV
Chudn bi cac cau hdi ggi md.
Chudn bi cac hinh tir 45 den hinh 50.
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Chudn bi pha'n mau va mdt sd dd dung khac.
2. Chuan bi ciia HS
can dn lai mot sd kien thiic da hgc ve dao ham va dn tap bai 1.
HI. PHAN PHOI

TH6I

LUONG

Bai nay chia lam 5 tiet:
Tiet 1 : Tit dau den hit mud

phdn I.

Tiet 2 : Tiep theo din het muc 2phdn I.
Tiet 3 : Phdn II.
Tii't 4 : Muc 1 phdn III.
Tii't 3 :Muc 2 phdn III.
IV. TIEN TRINH DAY - HOC
A. OAT VAN OE
Cau hoi 1
Neu dinh nghia va tinh chdt ciia nguyen ham.

Neu nhiing van de co ban cua phuong phap ddi bie'n sd va phuong phap
tinh nguyen ham timg phdn.
Cau hdi 2
Tinh nguyen ham ciia cac ham sd sau day:
a)y = x^-3x + 3 ;

b)y = x.e''

Cau hdi 3
Tim nguyen ham ciia cac ham sd sau day:
a) y = sinx + xcosx.

b) y = xlnx

B. BAI Mdl
I - KHAI NIEM TICH PHAN
HOAT DONG 1
1. Dien tich hinh thang cong
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