BÀI TẬP VỀ NHÀ (Giới hạn, tích phân và ứng dụng) Tính các tích phân sau:
Bài 1
3 2
0
4sin
1 cos
x
x
π
=
+
∫
Bài 2 :
1
3
xdx I
x
=
+
∫
Bài 3 :
I = ∫ x x + dx
Bài 4 :
2
4
sinx cos
1 sin 2
x
x
π
=
+
∫
Bài 5 :
( )
ln 3
3 0
1
x x
e dx I
e
=
+
∫
Bài 6 :
2 0
s inx
1 3cos
dx I
x
π
=
+
∫
Bài 7 :
1
01 x
dx I
e
=
+
∫
Trang 2Bài 8 :
0 3
−
Bài 9 :
2
ln 5
1
x x
e dx I
e
=
−
∫
Bài 10 :
1
I = ∫ − c x c xdx
Bài 11 :
0
2
x dx I
=
∫
Bài 12 :
ln 2 0
1
x
I = ∫ e − dx
Bài 13:
0
sin
π
=
+
∫
Bài 14:
0
1
I = ∫ x − x dx
Bài 15:
Trang 3
2 sinx 0
.sin 2
π
= ∫
Bài 16:
2
1 ln
e
I = ∫ x xdx Bài 17:
1
0
−
∫ 101 99
7x 1
2x + 1
Bài 18:
2
0
(x 1)sin 2x
π
+
∫
2
2 1
ln(x 1) x
+
∫
Bài 20:
2
2 0
dx
4 x +
∫
Bài 21: Tính các giới hạn sau đây:
Trang 4
( ) ( ) ( )
x 0 m n
x 1 100 50
x 1
20 2
10
x 0
3
x 0
3 2
x 0
3
x 0
1 x 1 2x 1 3x 1
*Bµi1: lim
x
x 1
*Bµi 2 : lim
x 1
x 2x 1
*Bµi 3 : lim
x 2x 1
x x 2
*Bµi 4 : lim
x 12x 16
x 9 x 16 7
*Bµi 5 : lim
x
2 1 x 8 x
*Bµi 6 : lim
x 2x 1 1 3x
*Bµi 7 : lim
x
1 4x 1 6x 1
*Bµi 8 : lim
→
→
→
→
→
→
→
→
−
−
− −
3 2
x 0
3 4
x 7
3 2
x 0
8x 1 10x 1 x
2x 1 x 1
*Bµi 9 : lim
sin x
x 2 x 20
*Bµi10 : lim
x 9 2
1 4x 1 6x
*Bµi11: lim
x
→
→
→
+ −
Trang 5( ) 3
2
x 2
x 0
2
x 0
2
x 0
x sin x sin x
2
x 0
x
x
x
4 x Bµi12 : lim
x cos 4 sin sin sinx Bµi13 : lim
x
1 cos x cos2x Bµi14 : lim
x
1 cos x cos 2x cos 2010x Bµi15 : lim
x
ln sin x cos x Bµi16 : lim
x
Bµi17 : lim
x
x 3 Bµi18 : lim
x 1 Bµi19 : lim
→
→
→
→
→∞
−
→
→+∞
→+∞
− π
−
−
+
− +
3 x
2 2
x 0
3
x 0
3 2
x
tan x sin x Bµi 20 : lim
x
Bµi 21: lim
x
1 tan x 1 sin x Bµi 22 : lim
x
Bµi 23 : lim
sin(x 1)
→∞
→
→
→∞
−
+ −
−
……….Hết………
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Trang 6HDG CÁC BTVN
Bài 1
3 2
0
4sin
1 cos
x
x
π
=
+
∫
HDG:
2
2 0
4sin 4sin (1 cos )
4sin 2sin 2 cos 2 4 cos 2 2
0
π
−
+
Bài 2 :
1
3
xdx I
x
=
+
∫
HDG
2
0
1 1
1
0
x
−
+ −
∫
Bài 3 :
HDG
3
2 2 1
2 2 1 2
tdt
x t
−
Bài 4 :
Trang 72
4
sinx cos
1 sin 2
x
x
π
=
+
∫
HDG
2
2 1
tdt
Bài 5 :
( )
ln 3
3 0
1
x x
e dx I
e
=
+
∫
HDG
2
2 3 2
2
2 1
2
x
tdt
e tdt
I
Bài 6 :
2 0
s inx
1 3cos
dx I
x
π
=
+
∫
HDG
4 1
: 1 3cos 3sin
3sin ln
ln 4
dt
x t
t
−
Bài 7 :
Trang 81
01 x
dx I
e
=
+
∫
HDG
1
0
2
1 ln(1 ) ln 2 ln
1
x x
x
e
e e
e
+
Bài 8 :
0 3
−
HDG
3
7 4
1 3 0
0
∫
Bài 9 :
2
ln 5
1
x x
e dx I
e
=
−
∫
HDG
2
3
2 2 1
2
2 20
1
x
tdt
e t
∫
Bài 10 :
Trang 92 6 3 5
1
I = ∫ − c x c xdx
HDG
1 6 6
1
0
∫
Bài 11 :
0
2
x dx I
=
∫
HDG
2
2
3
Bài 12 :
ln 2 0
1
x
I = ∫ e − dx
HDG
2
2
1
2 1
x
t
π
+
−
Bài 13:
Trang 102
0
sin
π
= +
∫
HDG
2
:
2
π
−
Bài 14:
1 5 ( 3)6
0
1
I = ∫ x − x dx
HDG
2
3
1
dt
x
t t
−
Bài 15:
2 sinx
0
.sin 2
π
= ∫
HDG
Trang 11
2 sinx 0
2
0
sin
ó : 2 sin cos
2 2 2 2 2 2 2
0
x
π
π
π
π
=
∫
∫
Bài 16:
2
1
ln
e
I = ∫ x xdx
HDG
2
1
:
1
3
e
dx u
v
=
=
Bài 17:
( )
1
0
−
∫ 101 99
7x 1
2x + 1
HDG
2
100
100
1
−
+
Bài 18:
Trang 122
0
(x 1)sin 2x
π
+
∫
HDG
0
du dx
2
2
π
= +
Bài 19:
2
2 1
ln(x 1) x
+
∫
HDG
2
1 2
dx
2
+
=
∫
Bài 20:
2
2 0
dx
4 x +
∫
HDG
2
2
π
Bài II:
Trang 13( ) ( ) ( )
x 0
x 0
x 0
m n
x 1
x 1
1 x 1 2x 1 3x 1
*Bµi1: lim
x
1 x 1 2x 1 3x 1 x 1 2x 1 x 1 2x 1 x 1 x 1
lim
x
1 x 1 2x 1 3x 1 1 x 1 2x 1 x
lim
x 3x 1 x 1 2x 2x 1 x x 3 1 x 1 2x 2 1 x 1
x 1
*Bµi 2 : lim
x 1
x 1 x x
lim
→
→
→
→
→
=
=
−
−
100 50
x 1
x 1 x x x 1 m
lim
n
x 1 x x x 1 x x x 1
x 2x 1
*Bµi 3 : lim
x 2x 1
x 1 2(x 1) x 1 x x x 1 2 98 49
48 24
x 1 2(x 1) x 1 x x x 1 2
→
20 2
10
x 0
*Bµi 4 : lim
*Bµi 5 : lim
x
→
→
− −
− +
+ + + − + − + + −
x 0
x
→
+ +
Trang 14
( ) ( ) ( ) ( )
3
x 0
3 2
x 0
2
*Bµi 6 : lim
x
12 12
*Bµi 7 : lim
x
→
→
+ − −
−
+ − +
2
2
2
x 0
3
x 0
2x 1 (x 1)
1 4x 1 6x 1 8x 1 10x 1
*Bµi 8 : lim
x
1 4x 1 6x 1 8x
lim
→
→
→
+ + +
− +
+ + + + + +
=
x 0
3
x
lim
x
1 4x 1
x
→
→
+
+ −
x 0 n
2
1 2nx 1
x
→
→
= = + +
+ −
Trang 15( ) ( ) ( ) ( )
x 0
3 4
x 7
4 4
4
t 2
2x 1 x 1
*Bµi 9 : lim
sin x 2x 1 1 x 1 1 2x 1 1 x 1 1
x 2 x 20
*Bµi10 : lim
x 9 2
t
§ Æt t x 9 x t 9 I lim
→
→
→
+ −
( ) ( )
( ) ( )
3 4
3
2 4
3
2
3
3
x 0
7 t 11
t 2
t 4 t 2
t 16
t 7 3
t 2 t 11 3 t 11 9
t 4 t 2 16 32 176 lim
3 27 27
t 11 3 t 11 9
1 4x 1 6x
*Bµi11: lim
→
→
−
−
2
2 3
2
2
x
1 4x (1 2x) 1 6x (1 2x) 1 4x (1 2x)
x 1 4x (1 2x)
x 1 6x (1 2x) 1 6x (1 2x)
4x (3 2x) lim
x 1 6x (1 2x) 1 6x (1 2x)
→
−
1 12 7
2 3 2
÷
Trang 16• Bài 3:
2
x 2
x 0
x 0
4 x
1 Bµi1: lim
x cos 4
t(t 4) t(t 4)
§ Æt : t x 2 x t 2 I lim lim
t
cos
4
4 2 t(t 4) t(t 4) (t 4) 16
t 4
4 sin sin sinx
2 Bµi 2 : lim
x sin sin sinx
lim
sin si
→
→
→
−
π π
π π
π
−
=
2
x 0
2
2
sin sinx sinx
nx sin x x
1 cos x cos2x
3 Bµi 3 : lim
x
cosx 1 cos2x
1 cos x cos x cos x cos 2x 1 cosx
x
2 sin cos x 1 cos 2x 1 2 cos x.sin x
2
2
x x 1 cos 2x 1 cos 2x x
4
2
→
=
−
−
−
−
÷
2
1
= + =
2
x 0
2
x 0
2010
2
2
2
1 cosx cos2x cos2010x
4 Bµi 4 : lim
x
1 cosx cosx cos x cos 2x cos x cos 2x cos2010x
lim
x cosx 1 cos2x
1 cosx
nx
2 sin
n 2
→
→
−
−
=
−
−
−
(1 22 32 20102)
Trang 17
x
2
2010(2010 1)(2.2010 1)
12
ln sin x cos x
5 Bµi 5 : lim
x
ln sin x cos x ln sin x cos x sin 2x
ln sin x cos x ln 1 t sin 2x
Mµ : lim lim 1 Víi t sin 2x vµ lim 1
I 1.1 1
→∞
=
+
−
cosx cos3x
2
x 0 cosx cos3x
x 0
cosx cos3x
x 0
6 Bµi 6 : lim
x
lim
lim
−
→
−
→
−
→
−
−
x 0
2
x 0
x x
x x
1 cos2x
x
x 3
7 Bµi 7 : lim
x 1
−
→
→
→+∞
→+∞
− = − =
− = ⇒ = + = +
+
= + ÷ = ⇒ = − → +∞ ⇒ → +∞
⇒
2
= + ÷ = + ÷ + ÷ =
Trang 18
( )
x
x
2
3 3 2
2
3
2 x
3 3
2
2
3x
3
1 1
x x
→+∞
→+∞
→+∞
− +
− +
3
x 0
2 2
2 2
x 0
2
1 2
tan x sin x
9 Bµi 9 : lim
x
1
2
10 Bµi10 : lim
x
→
→
= −
−
−
−
÷
−
2 2
x 0
x
2
2 2 x
4.
2
→
Trang 19( ) ( )
3
x 0
2 2
x 0
3 2
x 1
x 1
11 Bµi11: lim
x
x sin
.2
4.
2
12 Bµi12 : lim
sin(x 1)
lim
→
→
→
→
−
÷
+ −
−
−
− +
x 1
sin(x 1)
x 1 sin(x 1)
x 1
→
−
−
− = ⇒ =
−
……….Hết………
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