Tóm tắt các dạng toán và bài tập Nguyên hàm – Tích phân – Nguyễn Thanh Sơn Toán 12

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Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
Nguyªn hµm - tÝch ph©n vµ c¸c øng dông
a.tÝnh tÝch ph©n b»ng ®Þnh nghÜa
Ph¬ng ph¸p:
1. §Ó x¸c ®Þnh nguyªn hµm cña hµm sè f(x), Chóng ta cÇn chØ ra ®îc hµm sè F(x)
sao cho:
F’(x) = f(x).
¸p dông b¶ng c¸c nguyªn hµm c¬ b¶n, c¸c hµm sè s¬ cÊp .
Neáu gaëp daïng caên thöùc ñöa veà daïng soá muõ phaân theo coâng thöùc:
,( 0)
n
mn
m
xxm=≠
Neáu gaëp daïng
()
n
Px
x
thöïc hieän pheùp chia theo coâng thöùc:
1
,( ); ,( )
mm
mn
nnnm
xx
x
mn mn
xxx
=>= <.
Coâng thöùc ñoåi bieán soá (loaïi 2):
Tích phaân daïng:
()
().'()
f
gx g xdx
Ñaët g(x) = u => g’(x)dx = du
(())'() ()
f
gx g xdx fudu=
∫∫
.
2. Mét sè d¹ng c¬ b¶n:
1.
Sö dông c«ng thøc c¬ b¶n:
1. Daïng : ñaët u = ax + b du = adx dx=
()(1,0)ax b dx a
α
α
+≠
1
du
a
()
()
1
!
1
()
1(1)
ax b
u
ax b dx u du C C
aa a
α
α
αα
αα
+
+
+
+= = += +
++
∫∫
2. Daïng : ñaët
()
1
,( 0, 1)
nn
ax b x dx a
α
α
+≠
11
11
1
1
..
1(
()
(1) (1)
n
u=ax
nn
n
nn
bduanxdxxdx du
an
uaxb
ax b x dx u du C C
an na na
αα
αα
αα
−−
++
+⇒ = =
+
+==+=
++
∫∫
)
+
3.
Daïng
:
). cos sin ( 1)
axdx
α
α
( Ñaët
1
1
cos sin ) cos sin cos
(1)
u x du xdx x xdx u du x C
αα α
α
+
=⇒= = = +
+
∫∫
). cos ( 1)
sin x
bxdx
α
α
(Ñaët
1
1
sin cos sin
1
du=cos xdx sin x
ux xdxudu x
ααα
α
+
=⇒ = = +
+
∫∫
C
4.
Daïng
:
1
ln ( 0)
dx
ax b C a
ax b a
=++
+
Neáu gaëp :
()Px
ax b
+
vôùi baäc : laøm baøi toaùn chia.
() 1Px
GV: NguyÔn Thanh S¬n
1
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
5.
Daïng
:
2
cos ( )
dx
x
abtgx+
Ñaët
22
111
;l
cos cos ( )
2
dx
co s
bdx dx du
u a btgx du du a btgx C
xxb xabtgxbub
=+ = = = = + +
+
∫∫
n
2.
Coâng thöùc
:
()
'( )
ln
u
ux u
a
auxdx adu C
a
==+
∫∫
3.
Coâng thöùc ñoåi bieán soá (loaïi 1):
Tích phaân daïng:
( )
().'()
f
gx g xdx
Ñaët g(x) = u => g’(x)dx = du
(())'() ()
f
gx g xdx fudu
=
∫∫
4.
Coâng thöùc
:
2
2
2
1
). ln .( 0)
2
). ln
du u a
aCa
ua aua
du
buukC
uk
α
=+
−+
=+++
+
5.
Coâng thöùc :
2
22
ln
22
xx k k
x
kdx x x k C
+
+= + +++
3. Mét sè d¹ng thêng gÆp:
1. Tích phaân daïng:
22
22
1).
(mx+n)dx dx (mx+n)dx
2). 3). 4).
dx
ax bx c ax bx c
ax bx c ax bx c
++ ++
+ ++
∫∫
+
Tuyø vaøo moãi daïng aùp duïng caùc coâng thöùc tính tích phaân chæ trong baûng sau:
Töû soá baäc nhaát Töû soá haèng soá
Maãu soá khoâng caên
ln
du
uC
u
= +
22
1
ln
2
= +
−+
du u a
C
ua aua
Maãu soá coù caên
2
du
uC
u
= +
2
2
ln= +++
+
du
uukC
uk
Söû duïng haèng ñaúng thöùc:
222
22
2
()()
22
22
aa
xaxx
bb
ax bx a x
aa
+=+
⎛⎞
+= +
⎜⎟
⎝⎠
GV: NguyÔn Thanh S¬n
2
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
4. TÝch ph©n cña c¸c ph©n thøc h÷u tØ:
32
ax b A B C
cx dx ex x x m x n
+
=+ +
++
Giaûi daïng naøy ta coù hai caùch:
Caùch 1: Ñoàng nhaát hai veá: Cho taát caû caùc heä soá chöùa x cuøng baäc baèng nhau.
Caùch 2: Gaùn cho x nhöõng giaù trò baát kyø. Thöôøng thì ta choïn giaù trò ñoù laø
nghieäm cuûa maãu soá
5. TÝch ph©n cña c¸c hµm sè lîng gi¸c:
1.
Daïng
:
cos , , 1). sin , cos
n n
11
sin cosaxdx= sinaxdx=- , 2). co s
aa
n
x
dx xdx ax C ax C xdx++
∫∫
Phöông phaùp:
n = chaün : haï baëc
2
2
1cos2
cos
2
1cos2
sin
2
1
sin cos sin 2
2
x
x
x
x
xx
+
=
=
=
n leõ:
Vieát:
21 2 2
cos cos cos (1 sin ) cos
pp p
x
dx x xdx x dx
+
==
Ñaët
sin cosuxdux=⇒=dx
2.
Daïng
:
sin cos
mn
uud
u
u
a. m,n cung chaün: haï baäc.
b. m,n leû (moät trong hai soá leû hay caû hai cuøng leû).
Neáu m leû
: Ta vieát: thay
1
sin sin sin
mm
uu
=
1
22 2
2
sin 1 cos (1 cos ) sin
m
va sin
m
uuu u
=− = u
Neáu m, n leû
: laøm nhö treân cho soá muõ naøo beù
3.
Daïng
: hay
n
tg xdx
cot
n
gxdx
Chuù yù:
22
2
() (1 ) (1 )
cos
2
dx
co s
dx
d tgx tg x dx tg x dx tgx C
x
x
==+ =+ =+
∫∫
Töông töï:
22
2
(cot ) (1 ) (1 )
sin
2
dx
sin
dx
d gx cotg x dx cotg x dx cotgx C
x
x
= −=+ =+ =+
∫∫
Ngoaïi tröø:
sin
ln cos
cos
(u=cosx)
xdx
tgxdx x C
x
==+
∫∫
Ñeå tính:
n
tg xdx
Phöông phaùp
:
Laøm löôïng
2
(1)tg x + xuaát hieän baèng caùch vieát:
GV: NguyÔn Thanh S¬n
3
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
2222 242 12
* ( 1) ( 1) ... ... ( 1) ( 1) ( 1) 1
nn n n
tg x tg x tg x tg tg x tg x
−−
=+++++++
n
21 23 2 25 2 2 2 1
* ( 1) ( 1) ... ... ( 1) ( 1) ( 1)
nn n n n
tg x tg x tg x tg tg x tgx tg x tgx
−−
= +− ++++ ++
4.
Daïng
: hay
2
(1)tg x dx+
2
cos
n
dx
x
Ta vieát:
2212
(1)(1)(1)
n
tg x dx tg x tg x dx
+= + +
Ñaët u = tgx
22
(1) (1)
2n
(tg x+1) dx
n
du tg x dx u du
=+ =+
1
Chuù yù:
22
2
1
1(1
cos
2n
dx
,
co s
n
tg x tg x dx
)
x
x
=+ = +
∫∫
5.
Daïng
:
cos
m
n
cotg x
, or
sin x
m
n
tg x
dx dx
x
∫∫
Phöông
phaùp
:
Neáu n chaün
: Thay
2
2
222
1
(1 ) ; (1 ) (1 ) ( 1)
cos cos
m
tg
nnn
mm
nn
xdx
tg x tg x tgx dx tg x tgx tgx dx
xx
=+ = + = + +
∫∫
Ñaët:
2
2
2
(1 )
m
2
n
tg x
du=(1+tg x)dx
cos x
n
m
u tgx dx u u du
=⇒ = +
∫∫
Neáu m leû vaø n leû :
1
1
.
cos cos cos
m
n
tgx tg x tgx
x
xx
=
Ñaët
1
cos
tgx
du=
cosx
udx
x
=⇒
Thay:
11
21
22
221
111
1(1)..(1)
cos cos cos cos
n
tgmx
cos x
mm
n
n
tgx
tgx dx dx u u du
xxxx
−−
=− = =
∫∫
6.
Daïng
:
sin cos ;
sinmxsinnxdx ; cosmxcosnxdx
mx nxdx
∫∫
Aùp duïng caùc coâng thöùc bieán ñoåi:
[]
[]
[]
sin( ) sin( )
cos( ) s( )
cos( ) cos( )
1
sinmxcosnx=
2
1
sinmxsinnx=
2
1
cosmxcosnx=
2
mnx mnx
mnxcomnx
mnx mnx
•++
•−
•−+
+
+
GV: NguyÔn Thanh S¬n
4
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
I. TÝnh c¸c tÝch ph©n bÊt ®Þnh.
Bµi 1: Dïng c¸c c«ng thøc c¬ b¶n tÝnh c¸c tÝch ph©n sau:
1/
2
1
(3x 2x )dx
x
+−
2/
2
x3
dx
x
3/
4
3
2( x )dx
x
4/
3
4
1
(3 x 4 x )dx
x
−+
5/
x
x
3
2
e
e(2 )dx
3x
6/
x2x3x
2.3 4 dx
7/
cos (1 t )
x
gx dx+
8/
2
2
(4sinx )dx
cos x
9/
2
x
2cos dx
2
10/
22
dx
cos xsin x
Bµi 2: TÝnh c¸c tÝch ph©n sau ®©y:
1/ 2/
10
x(x 1) dx
2
12
()
x1(x1)
++
dx
3/
2
xx 9dx+
4/
22
4
8x
dx
(x 1)+
5/
3. x
e
dx
x
6/
xx
dx
2
ln
7/ 8/
sin7x.cos3x.dx
4
cos xdx
9/
3
sin x
dx
cos x
10/
22
cos2x
dx
sin x.cos x
II: TÝnh c¸c tÝch ph©n x¸c ®Þnh sau:
Ph¬ng ph¸p
:
() () () ()
b
a
b
a
f
xdx Fx Fb Fa==
.
1. C¸c ph¬ng ph¸p tÝnh tÝch ph©n.
¸p dông b¶ng c¸c nguyªn hµm c¬ b¶n, c¸c hµm sè s¬ cÊp .
TÝnh tÝch ph©n b»ng ph¬ng ph¸p ph©n tÝch.
TÝnh tÝch ph©n b»ng ph¬ng ph¸p ®æi biÕn d¹ng I.
TÝnh tÝch ph©n b»ng ph¬ng ph¸p ®æi biÕn d¹ng II.
TÝnh tÝch ph©n b»ng ph¬ng ph¸p ®æi biÕn d¹ng III.
TÝnh tÝch ph©n b»ng ph¬ng ph¸p tÝch ph©n tõng phÇn.
TÝnh tÝch ph©n b»ng ph¬ng ph¸p sö dông nguyªn hµm phô.
Mét sè thñ thuËt ®æi biÕn kh¸c, tÝch ph©n chøa biÓu thøc gi¸ trÞ tuyÖt ®èi...
2. Chøng minh bÊt ®¼ng thøc tÝch ph©n
GV: NguyÔn Thanh S¬n
5
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
§Ó chøng minh bÊt ®¼ng thøc tÝch ph©n , ta thêng sö dông chñ yÕu 4 tÝnh chÊt
sau: víi c¸c hµm sè f(x), g(x) liªn tôc trªn [a;b] ta cã:
1. NÕu
[
]
() 0, ;
f
xxa≥∀ b
th×
() 0
b
a
fxdx
2. NÕu
[
]
() (), ;
f
xgxxab≥∀
th×
() ()
bb
aa
f
xdx gxdx
∫∫
DÊu ®¼ng thøc chi x¶y ra khi f(x) = g(x),
[
]
;
x
ab∀∈
3.
NÕu
[
]
() , ;mfx Mxab≤≤
th×
() () ()
b
a
mb a f xdx M b a ≤≤
4.
() () .
bb
aa
f
xdx f x dx
∫∫
Bµi 1: TÝnh c¸c tÝch ph©n x¸c ®Þnh sau:
1/ 2/
2
234
0
(3x 2x 4x )dx−+
1
32
1
(x 3x)dx
−+
3/
4
x
4
0
(3x e )dx
4/
2
2
3
1
x2x
dx
x
5/
0
2
1
xx5
dx
x3
−−
6/
5
2
dx
x1 x2
+−
7/
1
2x
x
0
e4
dx
e2
+
8/
3
2
0
4sin x
dx
1cosx
π
+
9/
3
0
sin x.c os3xdx
π
10/
2
4
2
6
2tg x 5
dx
sin x
π
π
+
11/
2
0
cos2x
dx
sinx cosx
π
12/
4
2
0
sin ( x)dx
4
π
π
Bµi 2: TÝnh c¸c tÝch ph©n cã chøa trÞ tuyÖt ®èi sau:
1/
2
2
x1dx
2/
4
2
1
x6x9d−+
x
3/
4
2
1
x3x2d
−+
x
4/
1
x
1
e1d
x
GV: NguyÔn Thanh S¬n
6
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
5/
3
3
(3 x)dx
+
6/
0
2
2
xx1dx
+
7/
0
cosx dx
π
8/
3
4
4
cos2x 1dx
π
π
+
9/
0
cosx sinxdx
π
10/
3
x
0
24d
x
Bµi 3: Chøng minh c¸c B§T sau:
1/
3
0
3x1dx≤+
6
2/
1
2
0
45
1
22
x
dx
+
≤≤
3/
2
2
0
dx
12
x1
≤≤
+
4/
2
2
4
5
3sinxdx
24
π
π
π π
≤+
5/
3
4
2
4
dx
432sinx
π
π
ππ
≤≤
2
6/
2
2
0
3
tg x 3dx
42
π
π π
+≤
7/
2
2
sin x
2
0
edxe
2
π
π
π
≤≤
8/
22
x1 2x
11
edx edx
+
9/
22
32
00
sin xdx sin xdx
ππ
∫∫
10/
22
00
sin2xdx 2 sinxdx
ππ
∫∫
B: Ph¬ng ph¸p ®æi biÕn:
Ph¬ng ph¸p:
1. Daïng
:
11
(, )
nm
R
xxdx
Ñaët
1
mn mn-1
x=t dx=mnt dt
mn
tx=⇒
2. Daïng:
11
(),()
nm
R
ax b ax b dx
⎡⎤
++
⎢⎥
⎣⎦
Ñaët
1
mn mn-1
mn
mn
t=(ax+b) ax+b=t dx= t dt
a
⇒⇒
3. Daïng
:
dx
R(lnx)
x
ñaët
ln
dx
du =
x
ux=⇒
()
dx
R(lnx)
x
R
udu⇒=
∫∫
GV: NguyÔn Thanh S¬n
7
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
4. Daïng: ñaët
x
R(e )dx
()
du
Ru
u
⇒⇒ =
∫∫
xx x
du
u=e du=e dx dx= R(e )dx
u
5. Daïng :
2
(, )
R
xax bxcdx++
Ñöa tam thöùc
2
ax bx c+ +
veà daïng: hay.
222
u+m,u-m
2
22
m-u
Ñoåi tích phaân thaønh 1 trong caùc daïng sau:
.
22
22
22
1). R(u, m -u )du
2). R(u, m +u )du.
3). R(u, m -u )du.
Neáu döôùi daáu tích phaân coù chöùa
22
m-u
ñaët
22
u=msint m -u =mcost
22
m+u
ñaët
22
m
u=mtgt m +u =
cost
22
u-m
ñaët
22
m
u= u -m =mtgt
cost
6. Daïng
:
2
()
dx
mx n ax bx c++
+
Gaëp tích phaân naøy ñaët:
1
t=
mx+n
Bµi 1: TÝnh c¸c tÝch ph©n sau b»ng ph¬ng ph¸p ®æi biÕn lo¹i I
1/
1
2
0
2x
dx
1x+
2/
4
2
0
x x 9dx+
3/
10
2
dx
5x 1
4/
1
0
x1 xdx
5/
5
0
x. x 4dx+
6/
7
3
0
x
dx
x1+
7/
5
32
0
x. x 4dx+
8/
2
2
3
3
0
3x
dx
1x+
9/
2
x
1
dx
1e
10/
4
x
1
dx
x.e
11/
tgx 2
4
2
0
e
dx
cos x
π
+
12/
e
1
13lnx
dx
x
+
GV: NguyÔn Thanh S¬n
8
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
13/
e
2
1
1lnx
dx
x
+
14/
6
0
1 4sin x.cosxdx
π
+
15/
4
2
6
1
cotgx(1 )dx
sin x
π
π
+
16/
2
2
0
cos x.sin2xdx
π
17/
/6
22
0
sin2x
dx
2sin x cos x
π
+
18/
/2
3
2
0
cosx.sin x
dx
1sinx
π
+
19/
8
2
3
1
dx
xx 1+
20/
/3
3
0
cosx.sin x.dx
π
Bµi 2 : TÝnh tÝch ph©n b»ng ph¬ng ph¸p ®æi biÕn lo¹i II:
1/
0
2
1
1
x
dx
2/
3
2
23
0
1
dx
(1 x )
3/
2
22
1
x4xdx
4/
1
2
5
dx
x4x
7
+ +
5/
2
2
0
4
dx
x +
6/
4/ 3
2
3
2
x4
dx
x
7/
1
2
2
dx
xx 1
8/
6
2
23
dx
xx 9
9/
6
2
1
dx
xx1
++
10/
3
2
2
1
93x
dx
x
+
11/
1/2
1
1x
dx
1x
+
12/
2
2
x2
dx
x1
+
13/
1
22
0
dx
(x 1)(x 2)++
14/
3
2
0
dx
x3
+
Bµi 3 : TÝnh tÝch ph©n c¸c hµm sè höu tØ:
1/
2
1
dx
x(2x 1)+
2/
2
2
1
dx
x6x9
+
3/
2
1
6x 7
dx
x
+
4/
1
42
0
x
dx
xx1++
GV: NguyÔn Thanh S¬n
9
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
5/
4
2
3
x1
dx
x3x2
+
−+
6/
1
2
0
xdx
(x 1)+
7/
6
22
0
sin2xdx
2sin x cos x
π
+
8/
3
2
6
cosx
dx
sin x 5sin x 6
π
π
−+
9/
2
0
dx
(x 1)(x 2)++
10/
3
2
2
1
93x
dx
x
+
11/
1/2
2
0
dx
4x 4x 3−−
12/
4
32
4
2
(x x x 1)dx
x1
+−+
13/
2
0
dx
(x 1)(x 2)++
14/
2001
2 2001
xdx
(x 1)+
15/
1/2
42
0
dx
x2x−+
1
16/
1
3
0
3dx
1x+
c: Ph¬ng ph¸p tÝch ph©n tõng phÇn:
Coâng thöùc:
.. .
bb
b
a
aa
udv uv vdu=−
∫∫
Coâng thöùc cho pheùp thay moät tích phaân
udv
phöùc taïp baèng 1
tích phaân
ñôn giaûn hôn.
vdu
Coâng thöùc duøng khi haøm soá döôùi daáu tích phaân coù daïng:
Daïng tích soá:
Haøm soá logaric.
Haøm soá löôïng giaùc.
* Daïng
vôùi f(x) laø haøm
n
xf(x) ,ln,sin,cos.
x
exxx
Khi tính choïn:
Haøm soá phöùc taïp ñaët baèng u.
Haøm soá cos tích phaân ñöôïc cho trong baûng tích phaân thöôøng
duøng laøm
dv
Bµi 1: Dïng ph¬ng ph¸p tÝch ph©n tõng phÇn h·y tÝnh:
GV: NguyÔn Thanh S¬n
10
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
1/ 2/
0
xsinxdx
π
1
22x
0
(x 1) e dx+
3/
4
2
6
x sin2xdx
π
π
4/
e
2
1
(xlnx) dx
5/
4
2
0
x(2 cos x 1)dx
π
6/
3
2
4
xdx
sin x
π
π
7/
e
2
1/ e
ln x
dx
(x 1)+
8/
4
x
1
edx
9/
2
4
0
xcos xdx
π
10/
3
2
0
ln(x x 1)dx++
11/
12/
1
22x
0
(x 1) .e dx+
2
2
0
(x 1).sinx.dx
π
+
13/
2
2
1
ln(1 x)
dx
x
+
14/
4
0
x.sin x.cosx.dx
π
Bµi 2: TÝnh c¸c tÝch ph©n sau:
1/
e
2
1
ln x
dx
x
2/
2
e
1
x ln xdx
3/
2
e
1
ln x
dx
x
⎛⎞
⎜⎟
⎝⎠
4/
e
2
1
ln xdx
5/
6/
e
2
1
(xlnx) dx
2
x
0
e(x sinx)dx
π
+
7/
8/
x2
0
esin(x)dx
π
π
x
0
x
esin dx
2
π
9/
x
(1 sin x)e
dx
1cosx
+
+
10/
22
2
2
3
1x
dx
x
+
D: øng dông h×nh häc cña tÝch ph©n
GV: NguyÔn Thanh S¬n
11
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
Bµi 1: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®êng (P): y = x
2
- 2x + 2 ;tiÕp tuyÕn (d)
cña nã t¹i ®iÓm M(3;5) vµ Oy.
Bµi 2: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®êng (P): y = x
2
+ 2x vµ ®êng th¼ng (d):
y = x + 2.
Bµi 3: Cho hµm sè y =
2
3x 5x 5
x1
+
(C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) ; tiÖm
cËn cña nã vµ x = 2 ; x= 3.
Bµi 4: Cho hµm sè y =
()( )
2
x1x2+−
(C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) vµ
®êng th¼ng : x - y + 1 = 0.
Bµi 5: Cho hµm sè y =
4
2
x3
x
22
−−
(C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) vµ trôc
hoµnh.
Bµi 6: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®êng (P): y
2
= 4x vµ ®êng th¼ng d : 4x
- 3y - 4 = 0 .
Bµi 7: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®êng (P): y
2
+ x - 5 = 0 vµ ®êng th¼ng d
: x + y - 3 = 0 .
Bµi 8: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®êng y = 0 ; y = tgx ; y = cotgx
.
(0 x )≤≤π
Bµi 9: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®êng (C): x
2
+ y
2
= 8 vµ ®êng (P): y
2
=
2x .
Bµi 10: TÝnh thÓ tÝch h×nh trßn xoay do h×nh ph¼ng giíi h¹n bëi c¸c ®êng : y =
4
x
vµ y = -x + 5 quay quanh Ox.
Bµi 11: Cho hµm sè y =
2
x3x
x2
3+ +
+
(C) . Gäi (H) lµ phÇn h×nh ph¼ng giíi h¹n bëi (C)
trôc Ox vµ hai ®êng th¼ng x = -1 , x = 0. TÝnh thÓ tÝch khèi trßn xoay t¹o thµnh khi (H)
quay mét vßng xung quanh Ox.
Bµi 12: Cho hµm sè y =
2
xx
x1
1+ +
+
(C) . Gäi (H) lµ phÇn h×nh ph¼ng giíi h¹n bëi (C) trôc
Ox vµ hai ®êng th¼ng x = 0, x = 1. TÝnh thÓ tÝch khèi trßn xoay t¹o thµnh khi (H) quay
mét vßng xung quanh Ox.
Bµi 13: TÝnh thÓ tÝch vËt thÓ trßn xoay ®îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi :
y =
x
, y = 2 - x vµ y = 0 khi ta quay quanh (D) quanh Oy.
Bµi 14: TÝnh thÓ tÝch vËt thÓ trßn xoay ®îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi :
GV: NguyÔn Thanh S¬n
12
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
y = , x = 1 vµ y = 0 (
x
xe
0x1
) khi ta quay quanh (D) quanh Ox.
Bµi 15: TÝnh thÓ tÝch vËt thÓ trßn xoay ®îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi : y =
sinx , y = cosx , x =
2
π
(0 x )
2
π
khi ta quay quanh (D) quanh Ox.
Bµi 16:
TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®êng sau:
1/
vµ x = -1; x = 2.
2
y0;yx 2x==
2/
2
yx4x3=−+
yx3= +
3/
2
x
y4
4
=−
2
x
y
42
=
4/
ln x
y;y0;x1
2x
===
xe
=
.
5/
2
yxx 1;Ox=+
x1=
.
E. D¹ng thêng gÆp trong c¸c k× thi §H-C§
Bµi 1: TÝnh c¸c tÝch ph©n sau:
1/
1
3
2
0
1
x
dx
x +
2/
ln3
3
0
(1)
x
x
edx
e +
3/
0
2
3
1
(
x
1)
x
ex
++
dx
4/
2
6
35
0
1 cos .sin .cos .
x
xxd
π
x
5/
23
2
5
4
dx
xx+
6/
1
32
0
1
x
xdx
7/
2
4
0
12sin
12sin2
x
dx
x
π
+
8/
ln5
2
ln 2
1
x
x
edx
e
9/
ln5
ln 2
(1).
1
xx
x
ee
dx
e
+
10/ −+
2
22
0
(3x 1) x 3x 4 dx
Bµi 2
: Cho hµm sè: f(x) =
3
.
(1)
x
a
bx e
x
+
+
T×m a, b biÕt f’(0)=-22 vµ
1
0
() 5fxdx=
Bµi 3: TÝnh c¸c tÝch ph©n sau:
GV: NguyÔn Thanh S¬n
13
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
1/
2
2
0
x
xdx
2/
2
1
3
0
.
x
x
edx
3/
2
1
1
ln .
e
x
x
dx
x
+
4/
3
1
(cos )
1
x
dx
xx
+
+−
5/
1
2
0
(1) 1
x
dx
xx++
6/
2
0
sin .sin 2 .sin3 .
x
xxd
π
x
7/
2
44
0
cos2 (sin cos )
x
xx
π
+
dx
8/
2
5
0
cos .
x
dx
π
9/
+
+
3
53
2
0
x2x
dx
x1
10/
1
23
0
(1 x ) dx
Bµi 3: TÝnh c¸c tÝch ph©n sau:
1/
2
33
0
(cos sin)
x
xdx
π
2/
3
7
84
2
12
x
dx
xx+−
3/
22
1
ln
e
x
xdx
4/
3
1
ln
e
x
dx
x
5/
2
0
4cos 3sin 1
4sin 3cos 5
xx
dx
xx
π
−+
++
6/
9
3
1
1
x
xdx
7/
2
3
0
1
32
x
dx
x
+
+
8/
1
2
0
(2)
x
x
xe dx
+
9/
π
+
4
6
0
1tgx
dx
cos2x
10/
+++
3
1
x3
dx
3x 1 x 3
Bµi 4: TÝnh c¸c tÝch ph©n sau:
1/
2
0
22
xdx
x
x++
2/
2
1
21
dx
x
x +
3/
1
2
0
ln(1 )
1
x
dx
x
+
+
4/
2
0
sin
sin cos
x
dx
x
x
π
+
5
0
.sin
x
xdx
π
6/
2
23
0
sin .cos .
x
xdx
π
GV: NguyÔn Thanh S¬n
14
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
7/
1
13ln.ln
e
x
x
dx
x
+
8/
3
32
0
1
x
xdx+
9/
−+
+
2
4
2
0
xx1
dx
x4
10/
+−
3
7
84
2
x
dx
1x 2x
Bµi 5: TÝnh c¸c tÝch ph©n sau:
1/
3
53
2
0
2
1
xx
dx
x
+
+
2/
3
3
0
1
ln .
x
x
dx
x
+
3/
1
2
0
(1)
x
x
edx+
4/
3
2
4
cos 1 cos
tgx
dx
x
x
π
π
+
5/
2
2
1
1
2
x
dx
x
⎛⎞
⎜⎟
+
⎝⎠
6
2
0
sin
1cos
xx
dx
x
π
+
7/
1
0
1
x
dx
e+
8/
4
2
0
.
x
tg xdx
π
9/
π
+
2
44
0
cos2x(sin x cos x)dx 10/
π
⎛⎞
+
⎜⎟
⎝⎠
4
0
x
1tgxtg sinxdx
2
Bµi 6: TÝnh c¸c tÝch ph©n sau:
1/
5
3
(2 2)
x
xd
+−−
x
2/
2
2
2
0
.
(2)
x
xe
dx
x +
3/
4
1
2
54
dx
x
++
4/
1
22
0
(4 2 1).
x
x
xed−−
x
5/
2
22
0
4
x
xdx
6/
1
2
0
25
dx
xx2+ +
7/
2
0
sin 2
cos 1
x
dx
x
π
+
8/
1
2
0
(1)
x
dx
x +
9/
π
+
4
sin x
0
(tgx e cosx)dx 10/
π
+
2
22
0
sinx
dx
x
sin x 2cosx.cos
2
Bµi 7: TÝnh c¸c tÝch ph©n sau:
GV: NguyÔn Thanh S¬n
15
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n
LuyÖn Thi §¹i Häc vµ Cao §¼ng
1/
2004
2
2004 2004
0
sin
sin cos
x
dx
x
x
π
+
2/
3
2
0
4sin
1cos
x
dx
x
π
+
3/
2
0
sin 2 .cos
1cos
x
x
dx
x
π
+
4/
2
0
sin 2 sin
13cos
x
x
dx
x
π
+
+
5/
2
sin
0
(cos)cos.
x
exx
π
+ dx
6/
3
2
6
cos
sin 5sin 6
x
dx
xx
π
π
−+
7/
2
2
1
x
dx
xx+−
8/
2
0
co x
dx
s
7cos2
x
π
+
9/
(
++
)
0
2x
3
1
e x 1 dx
x 10/
π
2
3
2
0
xsin x
dx
sin2xcos x
Bµi 8: TÝnh c¸c tÝch ph©n sau.
1/
1
2004
1
sin .
x
xdx
2/
2
0
.sin .cos .
x
xx
dx
π
3/
2
3
0
.cos .
x
xdx
π
4/
4
2
44
0
cos x
cos sin
x
x
π
+
5/
3
2
0
sin
cos
x
x
dx
x
π
+
6/
1
2
0
.
x
tg xdx
7/ CM:
0
2
0
2
sin sin
x
x
dx dx
xx
π
π
>
8/ CM:
44
0
2
sin cos
dx
xx
π
π π
+
<<
9/
π
e 10/
2
3x
0
sin5xdx
π
xc x
2
4
0
os dx
Chóc c¸c em lµm bµi tèt !
GV: NguyÔn Thanh S¬n
16
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Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng
Nguyªn hµm - tÝch ph©n vµ c¸c øng dông
a.tÝnh tÝch ph©n b»ng ®Þnh nghÜa Ph−¬ng ph¸p:

1. §Ó x¸c ®Þnh nguyªn hµm cña hµm sè f(x), Chóng ta cÇn chØ ra ®−îc hµm sè F(x)
sao cho: F’(x) = f(x).
• ¸p dông b¶ng c¸c nguyªn hµm c¬ b¶n, c¸c hµm sè s¬ cÊp . n
• Neáu gaëp daïng caên thöùc ñöa veà daïng soá muõ phaân theo coâng thöùc: m n m
x = x , (m ≠ 0) • P x
Neáu gaëp daïng ( ) thöïc hieän pheùp chia theo coâng thöùc: n x m m xx m n 1 = x , (m > n); =
, (m < n) . n n nm x x x
Coâng thöùc ñoåi bieán soá (loaïi 2):
Tích phaân daïng: f (g(x)).g '(x)dx
Ñaët g(x) = u => g’(x)dx = du
f (g(x))g '(x)dx = f (u)du ∫ ∫ . 2.
Mét sè d¹ng c¬ b¶n:
1. Sö dông c«ng thøc c¬ b¶n: 1. Daïng : α
(ax + b) dx(α ≠ 1, a ≠ 0) ñaët u = ax + b ⇒ du = adx dx= ∫ ⇒ 1 du a α α + + + α α u (ax b) 1 ! 1
(ax + b) dx = u du = ∫ ∫ ( + = + α + ) C C a a 1 (α +1)a 2. α Daïng : ∫( n ax b) n 1 x − +
dx, (a ≠ 0,α ≠ 1) ñaët n nn− 1 1 1
u=ax + b du = . a . n x dx x dx = du an α 1 + n α 1 + α − α u ax + b n n 1 ( 1
(ax + b) x dx = u du = + C = C ∫ ∫ ) + an na(α +1) na(α +1) 3. Daïng: α
a). cos sin xdx(α ≠ −1) ∫ ( Ñaët α α −1 α 1
u = cos x du = − sin xdx) ⇒ cos x sin xdx = − u du = cos + x + C ∫ ∫ (α +1) α b). sin c
x os xdx(α ≠ 1 − ) ∫ (Ñaët α α 1 α 1 u sin x du=cos xdx sin x cos xdx u du sin + = ⇒ ⇒ = = x + ∫ ∫ α C +1 4. dx Daïng: 1
= ln ax + b + C(a ≠ 0) ∫ ax + b a
Neáu gaëp : P(x) vôùi baäc P(x) ≥ 1 : laøm baøi toaùn chia. ax + b
GV: NguyÔn Thanh S¬n 1
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 5. dx Daïng: ∫ Ñaët 2
cos x(a + btgx) bdx dx 1 dx 1 du 1
u = a + btgx du = ⇒ = du; =
= l a + btgx + C ∫ ∫ n 2 2 2 cos x cos x b
co s x(a + btgx) b u b 2. Coâng thöùc: u a u ( x) a u '(x) u dx = a du = + C ∫ ∫ ln a
3. Coâng thöùc ñoåi bieán soá (loaïi 1):
Tích phaân daïng: f (g(x)).g '(x)dx
Ñaët g(x) = u => g’(x)dx = du
f (g(x))g '(x)dx = f (u)du ∫ ∫ 4. Coâng thöùc : du 1 u a a). = ln + C.(a ≠ 0) ∫ α 2 u a 2a u + a du 2 b).
= ln u + u + k + C ∫ 2u +k 5. Coâng thöùc : 2 x x + k k 2 2 x + k dx =
+ ln x + x + k + C 2 2
3. Mét sè d¹ng th−êng gÆp: 1. dx (mx+n)dx dx (mx+n)dx Tích phaân daïng: 1). 2). 3). 4). ∫ ∫ ∫ ∫ 2 2 2 2
ax + bx + c
ax + bx + c
ax + bx + c
ax + bx + c
Tuyø vaøo moãi daïng aùp duïng caùc coâng thöùc tính tích phaân chæ trong baûng sau: Töû soá baäc nhaát Töû soá haèng soá Maãu soá khoâng caên du = 1 − ln u + C ∫ = ln + ∫ du u a C u 2 2 u a 2a u + a Maãu soá coù caên
du = 2 u + C ∫ 2 = ln + + + ∫ du u u k C u 2 u + k a a 2 2 2
x + ax = (x + ) − ( ) 2 2
Söû duïng haèng ñaúng thöùc: 2 2 ⎡ b b ⎤ ⎛ ⎞ ⎛ ⎞ 2
ax + bx = a x + − ⎜ ⎟ ⎜ ⎟ ⎥ ⎢⎝ 2a ⎠ ⎝ 2a ⎠ ⎣ ⎥⎦
GV: NguyÔn Thanh S¬n 2
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng
4. TÝch ph©n cña c¸c ph©n thøc h÷u tØ: ax + b A B C = + + 3 2
cx + dx + ex x x m x n
Giaûi daïng naøy ta coù hai caùch: −
Caùch 1: Ñoàng nhaát hai veá: Cho taát caû caùc heä soá chöùa x cuøng baäc baèng nhau. −
Caùch 2: Gaùn cho x nhöõng giaù trò baát kyø. Thöôøng thì ta choïn giaù trò ñoù laø nghieäm cuûa maãu soá
5. TÝch ph©n cña c¸c hµm sè l−îng gi¸c: 1. Daïng: n n 1 1 n
cos xdx, sin xdx, 1
). cosaxdx= sin ax + C, sinaxdx=- cos ax + C, 2). co s ∫ ∫ ∫ aa xdx ∫ Phöông phaùp: ⎧ 1+ cos 2x 2 cos x = ⎪ 2 ⎪ ⎪ − 1 cos 2x n = chaün : haï baëc 2 sin ⎨ x = 2 ⎪ ⎪ 1 sin x cos x = sin 2x ⎪⎩ 2 n leõ: Vieát: 2 p 1 + 2 p 2 cos = cos cos = (1− sin )p xdx x xdx x cos dx
Ñaët u = sin x du = cos xdx
2. Daïng: sinm cosn u ud u
a. m,n cung chaün: haï baäc.
b. m,n leû (moät trong hai soá leû hay caû hai cuøng leû).
Neáu m leû: Ta vieát: m m 1 sin u sin − = u sin u thay m 1 − 2 2 m 2 2
sin u = 1− cos u va sin u = (1− cos u) sin u
Neáu m, n leû: laøm nhö treân cho soá muõ naøo beù 3. Daïng: n tg xdx hay ∫ cot n g xdx ∫ Chuù yù: dx dx 2 2 d (tgx) =
= (1+ tg x)dx
= (1+ tg x)dx = tgx + C ∫ ∫ 2 2 cos x co s x Töông töï: dx dx 2 2 d (cot gx) = −
= −(1+ cotg x)dx
= (1+ cotg x)dx = −cotgx + C ∫ ∫ 2 2 sin x sin x xdx Ngoaïi tröø: sin tgxdx =
= ln cos x + C (u=cosx) ∫ ∫ cos x Ñeå tính: n tg xdx Phöông phaùp: Laøm löôïng 2
(tg x +1) xuaát hieän baèng caùch vieát:
GV: NguyÔn Thanh S¬n 3
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 2 2 −2 2 2n−4 2 n 1 − 2 * n n tg x = tg
x(tg x +1) − tg
(tg x +1) + ... + ... + (−1)
(tg x +1) + ( 1)n − 1 2 1 − 2 −3 2 2n−5 2 n−2 2 n 1 * n n tg x tg
x(tg x 1) tg (tg x 1) ... ... ( 1)
tgx(tg x 1) ( 1) − = + − + + + + − + + − tgx 4. dx Daïng: 2
(tg x +1)dx hay ∫ ∫ 2 cos n x Ta vieát: 2 2 n 1 − 2
(tg x +1)dx = (tg x +1) (tg x +1)dx ∫ ∫ Ñaët u = tgx 2 2 n 2 ( 1) (tg x+1) dx ( 1)n du tg x dx u − = + ⇒ = + 1 du ∫ ∫ Chuù yù: 1 dx 2 2 = 1+ , = (1 n tg x + tg x) dx ∫ ∫ 2 2n cos x co s x m m 5. tg x cotg x Daïng: dx, or dx ∫ ∫ n n cos x sin x
Phöông phaùp: Neáu n chaün : Thay n m tg n n −2 1 xdx 2 2 m 2 m 2
= (1+ tg x) ;
= tg x(1+ tgx) dx = tg x(1+ tgx) (tgx +1)dx ∫ ∫ ∫ cosn x cosn x m n Ñaët: tg x −2 2 m 2 2 u = tgx du=(1+tg x)dx
dx = u (1+ u ) ∫ n cos x dum 1 − tgx tgx tg x tgx Neáu m leû vaø n leû : = . Ñaët 1 u = du= dx n 1 cos x cos − x cos x cos x cosx Thay: m 1 − m 1 − 1 tgmx 1 1 tgx 2 n 1 2 2 tgx = −1 dx = ( −1) . . dx = (u −1) u du ∫ ∫ ∫ 2 n 2 n 1 cos x cos x cos x cos − x cos x
6. Daïng: sin mx cos nx ; dx
sinmxsinnxdx ; cosmxcosnxdx ∫ ∫ ∫
Aùp duïng caùc coâng thöùc bieán ñoåi: 1
sinmxcosnx= [sin(m + n)x + sin(m n)x] 2 1
sinmxsinnx= [cos(m n)x cos(m + n)x] 21
cosmxcosnx= [cos(m n)x + cos(m + n)x] 2
GV: NguyÔn Thanh S¬n 4
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng I.
TÝnh c¸c tÝch ph©n bÊt ®Þnh.
Bµi 1: Dïng c¸c c«ng thøc c¬ b¶n tÝnh c¸c tÝch ph©n sau:
1 x − 3 1/ 2 (3x + 2x − )dx ∫ 2/ dx ∫ x 2 x 3 1 3/ 2( x − )dx ∫ 4/ 3 4 (3 x − 4 x + )dx ∫ 4 x x −x e 5/ x e (2 − )dx ∫ 6/ x 2 x 3x 2 .3 4 dx ∫ 3 2 3 x 2 7/
cos x(1+ t gx)dx ∫ 8/ (4sin x − )dx ∫ 2 cos x x dx 9/ 2 2 cos dx ∫ 10/ ∫ 2 2 2 cos x sin x
Bµi 2: TÝnh c¸c tÝch ph©n sau ®©y: 1 2 1/ 10 x(x −1) dx 2/ ∫ ( − ) ∫ dx 2 x + 1 (x +1) 8x 3/ 2 x x + 9dx ∫ 4/ dx ∫ 2 2 4 (x + 1) 3. x e dx 5/ dx ∫ 6/ ∫ x x 2 ln x 7/ sin 7x.cos 3x.dx 8/ ∫ 4 cos xdx ∫ sin x cos 2x 9/ dx ∫ 10/ dx ∫ 3 cos x 2 2 sin x.cos x
II: TÝnh c¸c tÝch ph©n x¸c ®Þnh sau:
Ph−¬ng ph¸p: b a
f (x)dx = F (x) = F (b) − F (a) ∫ . b a 1. C¸c
ph−¬ng ph¸p tÝnh tÝch ph©n.
• ¸p dông b¶ng c¸c nguyªn hµm c¬ b¶n, c¸c hµm sè s¬ cÊp .
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ph©n tÝch.
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ®æi biÕn d¹ng I.
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ®æi biÕn d¹ng II.
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ®æi biÕn d¹ng III.
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p tÝch ph©n tõng phÇn.
• TÝnh tÝch ph©n b»ng ph−¬ng ph¸p sö dông nguyªn hµm phô.
• Mét sè thñ thuËt ®æi biÕn kh¸c, tÝch ph©n chøa biÓu thøc gi¸ trÞ tuyÖt ®èi... 2.
Chøng minh bÊt ®¼ng thøc tÝch ph©n
GV: NguyÔn Thanh S¬n 5
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng
§Ó chøng minh bÊt ®¼ng thøc tÝch ph©n , ta th−êng sö dông chñ yÕu 4 tÝnh chÊt
sau: víi c¸c hµm sè f(x), g(x) liªn tôc trªn [a;b] ta cã:
b 1.
NÕu f (x) ≥ 0, x ∀ ∈[ ;
a b] th× f (x)dx ≥ 0 ∫a b b 2.
NÕu f (x) ≥ g(x), x ∀ ∈[ ;
a b] th× f (x)dx g(x)dx ∫ ∫ a a
DÊu ®¼ng thøc chi x¶y ra khi f(x) = g(x), x ∀ ∈[ ; a b] 3.
NÕu m f (x) ≤ M , x ∀ ∈[ ;
a b] th× b
m(b a) ≤
f (x)dx M (b a) ∫ − a b b 4.
f (x)dx f (x) d . x ∫ ∫ a a
Bµi 1: TÝnh c¸c tÝch ph©n x¸c ®Þnh sau: 2 1 1/ 2 3 4 (3x − 2x + 4x )dx 2/ 3 2 (−x + 3x) dx ∫ ∫ 0 1 − 4 x 2 2 x − 2x 3/ 4 (3x − e )dx ∫ 4/ dx ∫ 3 x 0 1 0 2 x − x − 5 5 dx 5/ dx ∫ 6/ ∫ x − 3 x −1 + x − 2 1 − 2 π 1 2 x e − 4 2 3 4sin x 7/ dx ∫ 8/ dx ∫ x e + 2 1+ cos x 0 0 π π 3 4 2 2tg x + 5 9/ sin x.cos 3xdx ∫ 10/ dx ∫ 2 π sin x 0 6 π π 2 cos 2x 4 π 11/ dx ∫ 12/ 2 sin ( − x)dx ∫ sin x − cos x 4 0 0
Bµi 2: TÝnh c¸c tÝch ph©n cã chøa trÞ tuyÖt ®èi sau: 2 4 1/ x − 1 dx ∫ 2/ 2
x 6x + 9dx −2 1 4 1 3/ 2
x 3x + 2 dx 4/ x e − 1 d ∫ x −1 −1
GV: NguyÔn Thanh S¬n 6
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 3 0 5/ (3 + x )dx ∫ 6/ 2 x x + 1 dx ∫ −3 −2 3π π 4 7/ cos x dx ∫ 8/ cos 2x + 1dx0 π 4 π 3 9/ cos x sin xdx ∫ 10/ x 2 4 dx 0 0
Bµi 3: Chøng minh c¸c B§T sau: 3 1 2 x + 4 5 1/ 3 ≤ x + 1dx ≤ ∫ 6 2/ 1 ≤ dx ≤ ∫ 2 2 0 0 π 2 dx 2 π 5π 3/ 1 ≤ ≤ 2 ∫ 4/ 2 ≤ 3 + sin xdx ≤ ∫ 2 x + 1 2 π 4 0 4 3π π 4 π dx π 2 3π π 2 5/ ≤ ≤ ∫ 6/ ≤ tg x + 3dx ≤ ∫ 2 4 − π 3 2sin x 2 4 2 0 4 π 2 π π 2 2 2 sin x + 7/ 2 ≤ e dx ≤ e ∫ 8/ x 1 2x e dx ≤ e dx ∫ ∫ 2 0 1 1 π π π π 2 2 2 2 3 2 9/ sin xdx ≤ sin xdx ∫ ∫ 10/ sin 2xdx ≤ 2 sin xdx ∫ ∫ 0 0 0 0
B: Ph−¬ng ph¸p ®æi biÕn:
Ph−¬ng ph¸p: 1 1 1
1. Daïng: ( n, m R x x )dx ∫ Ñaët mn mn-1 mn t = x
x=t dx=mnt dt 1 1 2. ⎡ ⎤
Daïng: ∫ ⎢( + )n ,( + )m R ax b ax bdx ⎣ ⎦ 1 Ñaët mn mn mn mn-1
t=(ax+b) ax+b=t dx= t dt a 3. dx dx Daïng : dx R(lnx)
ñaët u = x du = R(lnx) = x ln x ( ) ∫ x R u du
GV: NguyÔn Thanh S¬n 7
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 4. Daïng: x R(e )dx ñaët ∫ x x du x du
u=e du=e dx dx=
R(e )dx = R(u) ∫ ∫ u u 5. Daïng : 2
R(x, ax + bx + c )dx ∫ Ñöa tam thöùc 2 2 2 2 2 2
ax + bx + c veà daïng: u +m ,u -m2 hay. m -u
Ñoåi tích phaân thaønh 1 trong caùc daïng sau: 2 2 1). R(u, m -u ) . du2 2 2). R(u, m +u )du.2 2 3). R(u, m -u )du.
Neáu döôùi daáu tích phaân coù chöùa 2 2 m -u ñaët 2 2 u=msint m -u =mcost 2 2m +u ñaët 2 2 m u=mtgt m +u = cost 2 2 m u -m ñaët 2 2 u= u -m =mtgt cost dx
6. Daïng : ∫
Gaëp tích phaân naøy ñaët: 1 t= 2
(mx + n) ax + bx + c mx+n
Bµi 1: TÝnh c¸c tÝch ph©n sau b»ng ph−¬ng ph¸p ®æi biÕn lo¹i I 1 2x 4 1/ dx ∫ 2/ 2 x x + 9dx ∫ 2 1 + x 0 0 10 dx 1 3/ ∫ 4/ x 1 − xdx ∫ 5x −1 2 0 5 7 x 5/ x. x + 4dx ∫ 6/ dx ∫ 3 x +1 0 0 5 2 2 3x 7/ 3 2 x . x + 4dx ∫ 8/ dx ∫ 3 3 + 0 0 1 x 2 dx 4 dx 9/ ∫ 10/ ∫ x 1 e− − x x.e 1 1 π 4 tgx+2 e e 1 + 3ln x 11/ dx ∫ 12/ dx ∫ 2 cos x x 0 1
GV: NguyÔn Thanh S¬n 8
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng π e 2 1 + ln x 6 13/ dx ∫ 14/ 1+ 4sin x.cos xdx ∫ x 1 0 π π 4 1 2 15/ cot gx(1 + )dx ∫ 16/ 2 co s x.sin 2xdx ∫ 2 π sin x 0 6 π / 6 π sin 2x / 2 3 cos x.sin x 17/ dx ∫ 18/ dx ∫ 2 2 2sin x + cos x 2 1 + sin x 0 0 8 π 1 / 3 19/ dx ∫ 20/ 3 cos x.sin x.dx ∫ 2 + 3 x x 1 0
Bµi 2 : TÝnh tÝch ph©n b»ng ph−¬ng ph¸p ®æi biÕn lo¹i II: 3 0 2 1 1/ 2 1− x dx ∫ 2/ dx ∫ 2 3 − 1 − 0 (1 x ) 2 1 dx 3/ 2 2 x 4 − x dx ∫ 4/ ∫ 2 x + 4x + 1 5 − 7 2 4 / 3 2 dx x − 4 5/ ∫ 6/ dx ∫ 2 3 x 0 x + 4 2 −1 dx 6 dx 7/ ∫ 8/ ∫ 2 2 − − − 2 x x 1 2 3 x x 9 6 dx 3 2 9 + 3x 9/ ∫ 10/ dx ∫ 2 + + 2 x 1 − x x 1 1 1/ 2 1+ x 2 x + 2 11/ dx ∫ 12/ dx ∫ 1 − x 2 x −1 1 − 1 dx 3 dx 13/ ∫ 14/ ∫ 2 2 (x + 1)(x + 2) 2 x + 3 0 0
Bµi 3 : TÝnh tÝch ph©n c¸c hµm sè höu tØ: 2 dx 2 dx 1/ ∫ 2/ ∫ x(2x + 1) 2 x − 6x + 9 1 1 2 6x + 7 1 x 3/ dx ∫ 4/ dx ∫ x 4 2 x + x + 1 1 0
GV: NguyÔn Thanh S¬n 9
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 4 x + 1 1 xdx 5/ dx ∫ 6/ ∫ 2 x − 3x + 2 2 (x + 1) 3 0 π π 6 sin 2xdx 3 cos x 7/ ∫ 8/ dx ∫ 2 2 2sin x + cos x 2 − + π sin x 5sin x 6 0 6 2 dx 3 2 9 + 3x 9/ ∫ 10/ dx ∫ (x + 1)(x + 2) 2 x 0 1 1/ 2 dx 4 3 2 (x + x − x + 1)dx 11/ ∫ 12/ ∫ 2 4x − 4x − 3 4 x − 1 0 2 2 dx 2001 x dx 13/ ∫ 14/ ∫ (x + 1)(x + 2) 2 2001 (x + 1) 0 1/ 2 dx 1 3dx 15/ ∫ 16/ ∫ 4 2 x − 2x + 1 3 1 + x 0 0
c: Ph−¬ng ph¸p tÝch ph©n tõng phÇn:
b b Coâng thöùc: b
u.dv = u.v − . v du ∫ ∫ a a a
• Coâng thöùc cho pheùp thay moät tích phaân udv ∫ phöùc taïp baèng 1
tích phaân vdu ñôn giaûn hôn. ∫
• Coâng thöùc duøng khi haøm soá döôùi daáu tích phaân coù daïng: − Daïng tích soá: − Haøm soá logaric.
− Haøm soá löôïng giaùc. * Daïng n
x f(x) vôùi f(x) laø haøm x
e , ln x, sin x, cos . x • Khi tính choïn:
− Haøm soá phöùc taïp ñaët baèng u.
− Haøm soá cos tích phaân ñöôïc cho trong baûng tích phaân thöôøng duøng laøm dv
Bµi 1: Dïng ph−¬ng ph¸p tÝch ph©n tõng phÇn h·y tÝnh:
GV: NguyÔn Thanh S¬n 10
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng π 1 1/ x sin xdx 2/ 2 2x (x + 1) e dx ∫ ∫ 0 0 π 4 e 3/ 2 x sin 2xdx ∫ 4/ 2 (x ln x) dx ∫ π 1 6 π π 4 3 xdx 5/ 2 x(2cos x −1)dx ∫ 6/ ∫ 2 π sin x 0 4 e ln x 4 7/ dx ∫ 8/ x e dx ∫ 2 (x +1) 1/ e 1 2 π 4 3 9/ x cos xdx ∫ 10/ 2 ln(x + x +1)dx ∫ 0 0 π 1 2 11/ 2 2 x (x + 1) .e dx 12/ ∫ 2 (x +1).sin x.dx ∫ 0 0 π 2 ln(1+ x) 4 13/ dx ∫ 14/ x.sin x.cos x.dx ∫ 2 x 1 0
Bµi 2: TÝnh c¸c tÝch ph©n sau:
2 e e ln x 1/ dx ∫ 2/ x ln xdx ∫ 2 x 1 1 e 2 ⎛ e ln x ⎞ 3/ dx ∫⎜ ⎟ 4/ 2 ln xdx ∫ ⎝ x ⎠ 1 1 π e 2 x 5/ 2 (x ln x) dx 6/ ∫ e (x + sin x)dx ∫ 1 0 π π x 7/ x 2 e sin ( x π )dx 8/ ∫ x e sin dx ∫ 2 0 0 x (1 + sin x)e 2 2 2 1 + x 9/ dx ∫ 10/ dx ∫ 1 + cos x 2 x 3
D: øng dông h×nh häc cña tÝch ph©n
GV: NguyÔn Thanh S¬n 11
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng
Bµi 1: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (P): y = x2 - 2x + 2 ;tiÕp tuyÕn (d)
cña nã t¹i ®iÓm M(3;5) vµ Oy.
Bµi 2: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (P): y = x2 + 2x vµ ®−êng th¼ng (d): y = x + 2. 2
3x 5x + 5
Bµi 3: Cho hµm sè y = x
(C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) ; tiÖm 1
cËn cña nã vµ x = 2 ; x= 3.
Bµi 4: Cho hµm sè y = ( + )( − )2 x 1 x
2 (C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) vµ
®−êng th¼ng : x - y + 1 = 0. 4 x 3
Bµi 5: Cho hµm sè y = 2
x − (C) . TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi (C) vµ trôc 2 2 hoµnh.
Bµi 6: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (P): y2 = 4x vµ ®−êng th¼ng d : 4x - 3y - 4 = 0 .
Bµi 7: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (P): y2 + x - 5 = 0 vµ ®−êng th¼ng d : x + y - 3 = 0 .
Bµi 8: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng y = 0 ; y = tgx ; y = cotgx
(0 x ≤ π) .
Bµi 9: TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng (C): x2 + y2 = 8 vµ ®−êng (P): y2 = 2x .
Bµi 10: TÝnh thÓ tÝch h×nh trßn xoay do h×nh ph¼ng giíi h¹n bëi c¸c ®−êng : y =
4 vµ y = -x + 5 quay quanh Ox. x 2
x + 3x + 3
Bµi 11: Cho hµm sè y = x +
(C) . Gäi (H) lµ phÇn h×nh ph¼ng giíi h¹n bëi (C) 2
trôc Ox vµ hai ®−êng th¼ng x = -1 , x = 0. TÝnh thÓ tÝch khèi trßn xoay t¹o thµnh khi (H)
quay mét vßng xung quanh Ox. 2 x + x + 1
Bµi 12: Cho hµm sè y = x +
(C) . Gäi (H) lµ phÇn h×nh ph¼ng giíi h¹n bëi (C) trôc 1
Ox vµ hai ®−êng th¼ng x = 0, x = 1. TÝnh thÓ tÝch khèi trßn xoay t¹o thµnh khi (H) quay mét vßng xung quanh Ox.
Bµi 13: TÝnh thÓ tÝch vËt thÓ trßn xoay ®−îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi :
y = x , y = 2 - x vµ y = 0 khi ta quay quanh (D) quanh Oy.
Bµi 14: TÝnh thÓ tÝch vËt thÓ trßn xoay ®−îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi :
GV: NguyÔn Thanh S¬n 12
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng y = x
xe , x = 1 vµ y = 0 ( 0 x 1 ) khi ta quay quanh (D) quanh Ox.
Bµi 15: TÝnh thÓ tÝch vËt thÓ trßn xoay ®−îc t¹o thµnh do h×nh ph¼ng (D) giíi h¹n bëi : y = π π sinx , y = cosx , x =
(0 x
) khi ta quay quanh (D) quanh Ox. 2 2 Bµi 16:
TÝnh diÖn tÝch h×nh ph¼ng giíi h¹n bëi c¸c ®−êng sau: 1/ 2
y = 0; y = x − 2x vµ x = -1; x = 2. 2/ 2
y = x − 4x + 3 vµ y = x + 3 2 x 2 x 3/ y = 4 − vµ y = 4 4 2 ln x 4/ y = ; y = 0; x = 1 vµ x = e . 2 x 5/ 2 y = x x +1;Ox vµ x = 1.
E. D¹ng th−êng gÆp trong c¸c k× thi §H-C§
Bµi 1: TÝnh c¸c tÝch ph©n sau: 1 3 x dx ln 3 x e dx 1/ ∫ 2/ ∫ 2 x +1 x 3 0 0 (e +1) π 0 2 3/ 2 3 x( x e + x +1) ∫ dx 4/ 6 3 5 1− cos x.sin . x cos . x dx 1 − 0 2 3 dx 1 5/ ∫ 6/ 3 2 x 1− x dx ∫ 2 5 x x + 4 0 π 4 2 1− 2 sin x ln 5 2 x e dx 7/ dx ∫ 8/ ∫ 1+ 2 sin 2x x 0 ln 2 e −1 ln 5 2 ( x e +1). x e 9/ dx ∫ 10/ 2 (3x − 2 1) x + 3x − ∫ 4 dx x − ln 2 e 1 0 a
Bµi 2: Cho hµm sè: f(x) = + . x bx e 3 (x +1) 1
T×m a, b biÕt f’(0)=-22 vµ f (x)dx = 5 ∫ 0
Bµi 3: TÝnh c¸c tÝch ph©n sau:
GV: NguyÔn Thanh S¬n 13
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng 2 1 1/ 2 x x dx ∫ 2/ 2 3. x x e dx ∫ 0 0 e 2 x +1 1 3/ ln . x dx ∫ 4/ 3 (cos x + )dxx x +1 − x 1 π 1 2 x 2 5/ dx ∫ 6/ sin . x sin 2 . x sin 3 . x dx (x +1) x +1 0 0 π π 2 2 7/ 4 4
cos 2x(sin x + cos x) ∫ dx 8/ 5 cos . x dx ∫ 0 0 3 5 x + 3 2x 1 9/ ∫ dx 10/ 2 3 (1 − x ) dx ∫ 2 0 x + 1 0
Bµi 3: TÝnh c¸c tÝch ph©n sau: π 2 3 7 x 1/ 3 3
( cos x − sin x )dx ∫ 2/ dx ∫ 8 4 1+ x − 2x 0 2 e e ln x 3/ 2 2 x ln xdx ∫ 4/ dx ∫ 3 x 1 1 π
2 4 cos x − 3sin x +1 9 5/ dx ∫ 6/ 3 x 1− xdx
4 sin x + 3cos x + 5 0 1 2 x +1 1 7/ dx ∫ 8/ 2 ( + 2 ) −x x x e dx ∫ 3 3x + 2 0 0 π 6 1 + 4 tg x 3 x − 3 9/ ∫ dx 10/ ∫ dx cos2x 3 x 1 x 3 1 + + + 0 −
Bµi 4: TÝnh c¸c tÝch ph©n sau: 2 xdx 2 dx 1/ ∫ 2/ ∫ 2 + x + 2 − x x 2x +1 0 1 π 1 ln(1+ x) 2 sin x 3/ dx ∫ 4/ dx ∫ 2 1+ x sin x + cos x 0 0 π π 2 5 .s x in xdx ∫ 6/ 2 3 sin . x cos . x dx ∫ 0 0
GV: NguyÔn Thanh S¬n 14
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng e 1+ 3ln x.ln x 3 7/ dx ∫ 8/ 3 2 x 1+ x dxx 1 0 2 4 x − x + 1 3 7 x 9/ ∫ dx 10/ ∫ dx 2 x + 4 1 + 8 x − 4 2x 0 2
Bµi 5: TÝnh c¸c tÝch ph©n sau: 3 5 3 x + 2x 3 3 x +1 1/ dx ∫ 2/ ln . x dx ∫ 2 + x 0 x 1 0 π 1 3 tgx 3/ 2 ( +1) x x e dx ∫ 4/ dx ∫ 2 + 0 π cos x 1 cos x 4 2 2 ⎛ π x −1 ⎞ x sin x 5/ dx ∫⎜ ⎟ 6 dx ∫ ⎝ x + 2 ⎠ 2 1+ cos x 1 − 0 π 1 dx 4 7/ ∫ 8/ 2 . x tg xd x ∫ 1 x + e 0 0 π π 2 4 ⎛ x ⎞ 9/ 4 cos2x(sin x + ∫ 4 cos x)dx 10/ 1 + ∫⎜ tgxtg ⎟sinxdx 2 0 ⎝ ⎠ 0
Bµi 6: TÝnh c¸c tÝch ph©n sau: 5 2 2 . x x e 1/
( x + 2 − x − 2 )dx 2/ dx ∫ 2 − (x + 2) 3 0 4 2 1 3/ dx ∫ 4/ 2 2 (4 − 2 −1). x x x e dx + + − x 5 4 1 0 2 1 dx 5/ 2 2 x 4 − x dx ∫ 6/ ∫ 2 2x + 5x + 2 0 0 π 2 sin 2x 1 x 7/ dx ∫ 8/ dx ∫ cos x +1 2 (x +1) 0 0 π π 4 2 sin x 9/ (tgx + ∫ sin x e cos x)dx 10/ ∫ dx x 0 2 sin x + 2 0 2 cos x.cos 2
Bµi 7: TÝnh c¸c tÝch ph©n sau:
GV: NguyÔn Thanh S¬n 15
Chuyªn ®Ò: Nguyªn hµm-TÝch ph©n LuyÖn Thi §¹i Häc vµ Cao §¼ng π π 2 2004 sin x 2 3 4 sin x 1/ dx ∫ 2/ dx ∫ 2004 2004 sin x + cos x 1+ cos x 0 0 π π 2 sin 2 . x cos x 2 sin 2x + sin x 3/ dx ∫ 4/ dx ∫ 1+ cos x 1+ 3cos x 0 0 π π 2 3 cos x 5/ sin ( x e + cos x) cos . x dx ∫ 6/ dx ∫ 2 − + π sin x 5sin x 6 0 6 π 2xdx 2 cos x 7/ ∫ 8/ dx ∫ 2 x + x −1 7 + cos 2x 0 π 0 3 2 xsin x 9/ ( 2x e + 3 x + 1)dx ∫x 10/ ∫ dx 2 − sin 2x cos x 1 0
Bµi 8: TÝnh c¸c tÝch ph©n sau. 1 π 1/ 2004 x sin . x dx ∫ 2/ 2 .s x in . x cos . xdx 1 − 0 π 2π 2 4 cos x 3/ 3 .c x os . x dx ∫ 4/ ∫ 4 4 cos x + sin x 0 0 π 3 x + sin x 1 5/ dx ∫ 6/ 2 . x tg xdx ∫ 2 cos x 0 0 π 2 0 π sin x sin x dx 7/ CM: dx > dx ∫ 8/ CM: ∫ π < < 2π ∫ x 4 4 π x sin x + cos x 0 0 2 π π2 2 4 9/ ∫e3x sin5xdx 10/ ∫ x cos xdx 0 0
Chóc c¸c em lµm bµi tèt !
GV: NguyÔn Thanh S¬n 16
Document Outline

  • Nguyªn hµm - tÝch ph©n vµ c¸c øng dông
  • Bµi 1: TÝnh c¸c tÝch ph©n sau b»ng ph­¬ng ph¸p ®æi biÕn lo¹