CAU III TICH PHAN UNG DUNG

- Ứng dụng của tích phân: tính diện tích hình phẳng, thể tích khối tròn xoay.. Tích phaân vaø ÖÙng duïng.[r]

(1)

Câu III (1 điểm): - Tìm giới hạn.

- Tìm ngun hàm, tính tích phân.

- Ứng dụng tích phân: tính diện tích hình phẳng, thể tích khối trịn xoay.

Tích phân Ứng dụng

Biên soạn: Thầy Nguyễn Quang Vũ Bài Tính tích phân sau.

a) A =

1

2x 9 dx 2x 3

 

 KQ:3ln531

b) A =

2

x 3x 2

dx x 3

 



 KQ:1 4ln2 2ln32 

c) A =

2

1

4x 5x 3

dx 2x 1



  



 KQ:5 ln32 4

d) A =1

0

1

x x dx x

  

 KQ:11 ln26 

Baøi Tính tích phân sau.

a) A =

2

4x dx x 3x



 

 KQ: 18ln 7ln3

b) A =

 

0

2 5x 6 x

dx

KQ:2ln2 – ln3

c) A =1 2

2 5 2 0

dx

x x



  KQ:1 ln23

d) A =

2

4 11

x dx x x



 

 KQ: 2ln3 – ln2

e) A =

0

1

(2 2) 2 3



 

 

x dx

x x KQ:

3 ln

4

f) A =

1 2

5 5 6

  

 

xx xx dx KQ:ln9

2

g) A =

2

1  7 12

x xx dx KQ:25ln2 – 16ln3 +

h) A =

0

2

2 9

3

x x x dx x x



  

 

(2)

i) A =1 2 

x x dx x

 

 KQ: ln2  32ln3

j) A =

3

1

2

2x 8x 10x 3

dx

x 4x 5

  

 

 KQ:1 22 5ln 1

k) A =

3

0

2

2x 6x 9x 9

dx

x 3x 2



  

 

 KQ: 1 33ln 19ln3

l) A =

1

2

4 1 3 2



 

x x x dx KQ:19ln2 + 13ln3 – 14ln7

m) A =

3

2

x

dx x  16

 KQ:4 – 8ln3 – 8ln5 + 8ln7

n) A =

2

x

dx 4 x

 KQ:

o) A =

1

2

x

dx x  1

 KQ:13 ln324 2

p) A =

2

5

dx x  6x 9

 KQ:52

q) A =

2

x 1

( ) dx

x 2 

 

 KQ:39 12ln24 

r) A =

0

2

2x 3

( ) dx

2x 1 

 

 KQ:19 4ln33 

s) A =2 3 3

2 1 0

x dx

x x



  KQ: -6 + 9ln2

t) A =

3

2

x 6x 9x 5

dx

x 6x 9

   

 

 KQ:

u) A =

2

1

dx x x

  

KQ: 18



v) A =

1

1 dx

x 2x 2

 KQ: 4

w) A =

 

0

2 2x 2 x

dx

KQ:



(3)

x) A = 2

3 2

x dx x

 

 KQ: 3  12 3

y) 

 

1

0

1

x dx x

I KQ:1ln2

2  z) A =1 2

0

1

x x dx x

  

 KQ:12 4

aa) A =

2 2

2

4

x x x dx x

  



 KQ: 68

bb) A =2 4 1

2 4 0

x x dx

x

 



 KQ:

1ln 2 16 17

2



  

cc) A =

1 2

3 10 2 9

x x

dx

x x

 

 KQ: ln3 ln2 12  

Bài Tính tích phân sau.

a) A =

1

2

x 3

dx (x 1)(x 3x 2)



  

 KQ:10ln2 – 5ln3 -

b) A =

3

3 3

3

x x dx x x

 

 

 KQ:3 ln52

c) A =

1

3

0

4x 1

dx

x 2x x 2



  

 KQ:1027ln2 95ln310

d) A =

2

3

3x 3x 3

dx

x 3x 2

 

 

 KQ:3 ln52

e) A =

1

3 1

dx x



 KQ:33 ln2

f) A = 3 3

1

dx x x 



KQ: 1ln 1ln3

2

 

g) A =1 3

( 3) 0

xdx x 

 KQ:

1 96

h) A =

3

(1 )

x dx x



 KQ:181

i) A =

1

4

0

1

dx x 4x 3

(4)

j) A =

1

4

0

1

dx x  3x  4

 KQ:20 ln3

k) A =

 

3

2

1

dx x x 1

 KQ:

3 12



 

l) A =

1

4

1 12

  

x xx dx KQ: 4ln 2ln3

7

 

m) A =

 

1

2 1

 

 dx

x KQ:

1

 

n) A =

2

4

1 x dx 1 x

 

 KQ:

o) A =

4

6

1 x dx 1 x

 

 KQ:3

p) A =

7

8

2

x

dx 1 x  2x

 KQ: 14ln3 ln2 96013

q) A =

 

 

2

9

1 x

dx x

KQ: 18

a) B =

1

3

0

x (x  1) dx

 KQ: 241

b) B =

1

5

0

x (1 x ) dx

 KQ:1681

c) B =

2 10

(1 )(1 2 x  x3 ) x dx

 KQ:36279705522

d) B =

2007

2

1 1 dx x x

 



 

 

 KQ: 32008 22008

2008



a) B =

3

x x  dx

 KQ: 85

b) B =

5

0

x  x dx

 KQ: 1058

(5)

c) B = 3

1

x  x dx

 KQ: 105848

d) B =

3

0

x x dx

 KQ:15 152  2

e) B =

3

x x dx

 KQ:  4687

f) B =

15

x  x dx

 KQ:27029

g) B =

2

3

(x 3) x  6x dx

 KQ:

h) B = 3 5 2 3

2

0 1

x x dx

x  



KQ:26

i) B =

2 1

2 3

0 ( 1)

x x

dx x

 



KQ: 23 28 14

j) B =

2

x

dx 2 x  2 x

 KQ:8 ln23 3

k) B =

3

2

x

dx x 1 x

 KQ: 15 151  ln2

l) B =1 2

1 0  

x

dx

x x KQ: 2 23



a) B =

x x dx x

 

 KQ: 32 10ln33 

b) B =

3

1

dx (1 x) 2x+ +3

ò KQ:ln3

2

c) B =2

1 1

1

x dx

x 

  KQ:11 4ln23 

d) B =

( 1)

x dx x

 

 KQ: 1330

e) B =

4

0

2x dx 2x



 

(6)

f) B =1

0 dx 1 3x 1

 KQ:

g) B =

 



02

3

dx x x

KQ: 4ln3 8ln 27 3  h) B =

7

2

x dx x

 

 KQ: 14110

i) B = 3

1

x dx x

 

 KQ:

46 15

j) B =

3

0 x

dx 1 x

 KQ:

a) B =

1

3

x dx x x



   

 KQ: ln3 8

b) B =

dx 2x 1  4x 1

 KQ: ln3 12 12

c) B = 10

5

dx x x 1 

 KQ: ln 1

d) B =

2

dx x x 

 KQ:

e) B =

2

1

dx x x 4

 KQ:

f) B =

1

2

1

dx (3 2x) 12x 4x

   

KQ: 12



g) B =

1

2

3

2 4  1

x dxx KQ:

3



h) B =

4

2

1

dx x 16 x

 KQ:18ln(2 3)  18ln(2 3)

i) B =

6

2

1

dx x x  9

 KQ:18

(7)

j) B =

2

3

dx x x 

 KQ:

12



k) B =

3

dx x x 

 KQ:

l) B =1 2

( 1) 1 0

x

dx

x x



  KQ:16 11 ln23 

m) B =

2 2

2

x 1

dx

x x 1

 

 KQ:1 82 3ln

n) B =

1

2

1

dx

1 x 1 x

   

KQ:1

a) B =

2

4 x dx

 KQ:2 2 ln( 1) 

b) B =1

0

1x dx

 KQ: ln( 1)2  2 

c) B =

2

9 3x dx x



 KQ

d) B =

1

2 (1 )

dx x



 KQ:

e) B =

2

x x dx



 

 KQ:

f) B =

1

2

(1 x ) dx

 KQ:316

g) B =

2 2

2

x

dx 1 x

 KQ:

1



 

h) B =

1

2

1

dx 4 x

 KQ:6

i) B =

1

2

3

1

dx x 4 x

(8)

j) B = 2

1

x 4 x dx



 KQ: 43 5 6

k) B =

2

2 2

1 x dx x



 KQ:1

4

 

l) B =

2

2 (1 )

dx x



 KQ:1

m) B =

5

2

12x 4x  8dx

 KQ:

3 12 16

 

n) B =

0

1 xx dx



 

 KQ:4 1

o) B =

1

0

3 x dx x

-+

ò KQ:

3

p- +

a) C =2

6

1 sin 2x cos 2x dx cos x sin x



 



 KQ:1

b) C =2

sin xsin 2xdx



 KQ:

2

c) C =

2

cos5xcos3xdx

  

 KQ:

d) C =4

x

sin cosxdx



 KQ:

e) C =2sin sin sin 3

0 x x xdx



 KQ:

1

f) C =4

(1 tan tan )sin 2

x

x xdx





 KQ: ln 22

g)

(9)

h) C =4 4 

cos x sin x dx





 KQ:

1

i) C =2 3

(sin xcos )x dx

 

KQ:

j) C =2cos2 (sin4 cos4 )

0 x x x dx





 KQ:0

k) C =

0

cos



 xdx KQ:3

8



l) C = sin62

0 x dx



 KQ:5

16



m) C =2cos5

0 xdx



 KQ:

8 15

n) C =





2

1

2

cos )

(cos x xdx KQ:15 48  

o) C =2

cos x.cos 4x dx 

 KQ:0

p) C =2

sin cos



 x xdx KQ:

q) C =2

sin xcos xdx

 

KQ: 32



r) C =2

cos x sin xdx 

 KQ:

8 315

s) C =2

sin cos



 x xdx KQ:

t) C =3sin2xtgxdx



 KQ:

3 ln2

8

(10)

u) C =2

sin x cos x(1 cos x) dx 



 KQ:

17 12

v) C =2  2 3

sin 2x sin x dx





 KQ:

15

w) C =

0

2

sin 2x

dx (2 sin x)

   KQ:2ln2 -

x) C =

 

2

3

cos2x dx sin x cosx



 

 KQ: 321

y) C =

 

4

3

cos2 sin cos 2

x

dx

x x



 

 KQ:29 2 1

z) C =

2

4

cos sin



 xxdx KQ:1

7

aa) C =6

0cos

tg x dx x



 KQ:  

1 10

ln

2  

bb) C =4 

1 tg x dx





 KQ:

76 105

Baøi 10 Tính tích phân sau.

a) C =2

sin cos cos

x xdx x





 KQ: ln 1

b) C =4

cos2x dx 2sin 2x





 KQ: ln34

c) C =2

4sin cosx dxx





 KQ:

Thầy Nguyễn Quang Vũ – THPT Lục Ngạn số – Baéc Giang Trang

(11)

d) C =2

0

cos x dx cos x 1 



 KQ:2

 

e) C =2

cos3 sinx x dx1





 KQ: 3ln 2

f) C =6

sin3x sin 3x dx cos3x



 

 KQ:  1 ln26 3

g) C =4

0

1 2sin sin 2xx dx



 

 KQ: ln22

h) C =

2

sin( x)

4 dx

sin( x) 4



 

  



 KQ:0

i) C =4 4

0

sin x cos x dx sin x cos x 1 



 

 KQ:

2 ln

1

  

Bài 11 Tính tích phaân sau.

a) C =

/6

2

cos

6 5sin sin



 

 xx xdx KQ:

b) C =

/

2 /

cos

11 7sin cos

x

dx

x x



  

KQ:

c) C =

/

2

cos

7 5sin cos

x dx

x x



 

 KQ:ln4

3 d) C =

/2

2

sin

3sin 4



  

 x x dxcosx KQ:

e) C =

/2

0

sin

3 4sin cos



 

 xx dx x KQ:

f) C =2

2

0

3sin 4cos 3sin 4cos



 

 xx xxdx KQ: 4ln3 63

 

g) C =

2

0

sin

sin 2cos cos

xdx

x x x





(12)

h) C =2

2

sin x.cos x dx cos x 1 



 KQ:1 ln24 2

i) C =4

4

0

sin 4 sin cos





 x x xdx KQ:ln2

j) C =2

4

sin 2x dx 1 cos x 



 KQ:4



k) C =4

6

0

sin 4 sin cos





 x x xdx KQ:2 ln43

l) C =

 

4

0

sin

sin 2 sin cos

x dx

x x x

  



 

 

  

 KQ:

4



a) C =2 4cos 3sin 1 4sin 3cos 5 0

x x dx

x x



 

   KQ:

b) C =2 7cos sin 6

4sin 3cos 5 0



 

 xx xx dx KQ:

9 ln

2



 

c)

d) C =

2

sin cos sin cos

x x dx x x



 

 

 



KQ:

e) C =

6

dx sin x.sin x

3



   

 

 

 KQ: ln23

f) C =3

4sin cos( 4)

dx x x

 

 

KQ:

Thầy Nguyễn Quang Vũ – THPT Lục Ngạn số – Bắc Giang Trang

(13)

g) C =3

6sin sin( 6)

dx x x

 

 

KQ:

h) C =3 3

4

sin cos

dx x x



 KQ:

i) C =

/3 /6sin cos

dx x x

 



KQ:

Baøi 13 Tính tích phân sau.

a) C =2 61 cos3 .sin cos5

0 x x xdx

 

 KQ:

12 91

b) C =2

sin sin 3cos

x x dx x



 

 KQ:

34 27

c) C =2

0

sin 2x sin x dx cos x 1 

 

 KQ: 2 23



d) C =3 2

4

tan cos cos

x dx

x x



 

KQ:

e) C =2

2

0

sin2x dx cos x 4sin x





 KQ:

2

f) C =4 2

6

1

dx sin x cot gx 



KQ: 2 3

g) C =2

2

cos x

dx cos x 1 



(14)

h) C =2

0

cos x

dx cos 2x 7 



 KQ:

i) C =3

0

cos cos2

xdx x



 

KQ:

j) C =2

2

cos cos

xdx x



 

KQ:

k) C =

4

sin x cosx dx sin2x



 

 KQ: ln

l) C =3

4

cos x sin x dx 3 sin 2x 



 

 KQ:

m) C =2

0

sinx cosx cos xdx

 

KQ:

a) C =2 

0

cos ) (



xdx

x KQ:

b) C =2 

x sin2x dx





 KQ:

 

c) C =4

2

x dx cos x



 KQ: 4 ln 22

d) C = 

4

01 cos2



x

xdx KQ: 1ln

8 2

 

e) C =

3

sin cos



 

x xxdx KQ:

(15)

f) C =2 

2x cos xdx





 KQ:

2

1 1

2

 

 

 

 

 

g) C =

0

sin cos

x x xdx

 

KQ:

h) C =2

x cosxdx



 KQ:

2

 

i) C =4 2

x sinx dx



 KQ:

3 1

384 32

 

 

j) C =3  

ln tgx dx sin2x



 KQ: 161 ln 32

k) C = 2

6

ln(sin ) cos xx dx

 



KQ:

l) C =

4

x sin xdx 

 KQ:

m) C =

4

x cos xdx 

 KQ:

n) C =

2 4

.cos . 0 x x dx



 KQ:

o) C =

2

4

sin

x xdx



 KQ:

2

 

p) C =

3

sin



 

 

 

 xdx KQ:

(16)

a) D =1

1 0

dx x e 

 KQ:

2 ln

1 e

 

b) D = 

 ln

0

x

e dx

KQ:ln(1 ) ln 2e e

  

c) D = 

3

1

x

e dx

KQ:ln(e2 e 1) 2   

d) D =2

1

1 ex dx



 KQ:

e) D =

x ln

x

1 e dx 1 e

 

 KQ:

f) D =

2x

x

e

dx e 1

 KQ:e2 1 ln ln(1 e2)

g) D =

2x

x

e

dx

e 1

  

 KQ:

h) D =

x

2x

(1 e ) dx 1 e

 

 KQ:

i) D =

1

2x x

0 1

dx e e

 KQ:

j) D = ln5

x x

ln3

dx e 2e

 

 KQ: ln32

k) D =

ln2 x x x

(3e 1)dx 2e e



 

 KQ:

l) D =

ln2 2x x 2x x

(e 3e )dx e 3e



 

 KQ: 12ln23 92 8ln

m)

Bài 16 Tính tích phân sau.

a) D =

ln x

e  1dx

 KQ:

b) D =ln8 1. 2 ln3

x x

e  e dx

 KQ:

(17)

c) D =

ln2 2x x

e dx e 2

 KQ: 3 83

d) D =ln5 2

1 ln2

x e dx

x e 



KQ:20

e) D =

ln2

0

x x e dx

e 

 KQ: 23

f) D =

ln3

3 0 ( 1)

x e dx

x e 

 KQ: 1

g) D =

ln x

1

dx e 1

 KQ:

h) D =

x ln

x x

0

e

dx (e 1) e  1

 KQ:

i) D =

1

4x 2x

2

2x

3e e

dx 1 e

 

 KQ:

j) D = ln5

0

x

x x

e dx e  e 

 KQ:

k) D = ln9 ln6

(3 )

x x

e e dx e e



 

 KQ:

l) D =1

0

x x x

e dx e e



 KQ:

a) D =

1 x

xe dx

 KQ:

b) D =  

2x

x e dx

 KQ:

2 3e

2



c) D =  

0 2x

x e x dx



 

 KQ: 34e2  6031

d) D =  

1

2

1

x

x e  x dx

 KQ:

2 e

(18)

e) D =

0

2

( x 1)

x e x dx



 

 KQ:

3 4e 

f) D =  

0

2 2 1)

4

( x x e xdx KQ:-1

g) D =1 2 2  0

x

x  x e dx

 KQ:4 9e

h) D =

1

2 2x

(1 x) e dx

 KQ:

2

4

e 

i) D =2 cos sin 2 0

x

e xdx



 KQ:2

j) D =2 sin2 3

sin cos

x

e x xdx

 

KQ:

e



k) D =2 sin x2

4

e sin 2x dx 



 KQ:e e

l) D =1 3 2

0

x x e dx

 KQ:12

m) D = 

2 ln

0

dx e

x x

KQ:

n) D =2 sin 

cos cos

x

e x xdx





 KQ: e



 

o) D =4 sin 

.cos

x

tgx e x dx





 KQ:

1

ln e 1

p) D =2

sin5

x

e xdx



 KQ:

3 3.e

34





a) E = 1 3ln ln

1

e x x dx

x 

 KQ:116135

(19)

b) E = e

3 ln x dx x 2ln x

 

 KQ: 103 2 113

c) E =  

e

dx x x

x

1 2ln ln

KQ: 10 11



d) E =e

ln x ln x dx x



 KQ: 3 28  

e) E = 3 ln2

ln 1 1

e x

dx

x x



 KQ:

76 15

f) E = 2 1 ln

e dx

x  x

 KQ: 6

g) E =

 

e

2

dx x ln x

 KQ: 4

a) E =  

2

1

4x ln x dx

 KQ: ln 2

b) E =2   

0

2x ln x dx 

 KQ: 24 ln3 14

c) E =  

x ln x dx

 KQ: ln44

d) E =

ln

e

x xdx

 KQ:

2 e

4



e) E =

2

3

lnx dx x

 KQ: 2ln 216

f) E = e

0

lnx dx x

 KQ: e

g) E =  

2

x ln x dx

 KQ: ln2 12

h) E =3  

0

x ln x 5 dx

 KQ: 14ln14 5ln5 92   

i) E =

ln

e

x xdx

(20)

j) E = e

x lnxdx x

  

 

 

 KQ:

3 2e 11

9 18 k) E =

3

2

3 ln

( 1)

 

x xdx KQ:ln3 18 4 ln334

l) E =3ln 2 

2 x  x dx KQ:

m) E =  

e

2

xlnx dx

 KQ: 271 5e 2  

n) E = e

Ix ln x dx KQ:

4 5e

32 

o) E = 2 2ln

1

e

x xdx

 KQ:

p) E = 

  

 

 

e

dx x x

x x

1

2 ln ln ln

KQ:

q) E =

e e

1 1

( )dx

ln x ln x 

 KQ:

a) F =

2

(x x )dx



  

 KQ:12

b) F =

3

1

x dx



 KQ:443

c) F = 

0

2 xdx

x KQ:1

d) F =x  x dx

0

2 2 3

KQ:4

e) F =

2

x  x dx

 KQ:52

f) F =

2

3

1

x 2x x dx



  

 KQ:3712

g) F =

4

6

x  x dx

 KQ:52

(21)

h) F =

1

9x  6x1dx

 KQ:56

i) F =  

0

2

3 4x 4xdx

x KQ: 32

5 15

 

j) F =

2 1/2

1

x dx x

 

 KQ:98

k) F =

3

2x  dx

 KQ:

l) F =  



2

0

2 cos

1 xdx KQ:

m) F =

0

1 sin x dx





 KQ:

n) F = sinxdx



 



KQ:

o) F =

2

1 cos xdx

 

KQ:

p) F =

2

cos x cos xdx



 



 KQ:

q) F = 3

2

cosx cosx cos xdx



 



KQ:

r) F =3 2

6

tan xcot x 2dx

 



KQ:2 ln

a) G =

1 2

1 cos ln

1

x x dx

x



  

 



 

(22)

b) G =  

1

2

ln x x dx



 

 KQ:

c) G = 2

2

cos sin

x x dx x



 

 

 KQ:

d) G =1

1

1 2x dxx



 

 KQ:

e) G =

2

sin sin3 cos5 x

x x x dx e

 

 



KQ:

f) G =

1

2 1( x 1)( 1)

dx e x

  

 KQ:

g) G =2

0

sin sin cos

x dx x x

 

KQ:

h) G =2

4

0

sin cos sin

x dx x x





 KQ:

i) G =2

0

cos cos sin

n n n

x dx x x

 

KQ:

j) G = 2

0

.sin cos

x x dx x



 

KQ:

k) G = 2

0

sin cos

x x dx x



 

KQ:

Bài 22 Tính diện tích hình phẳng tạo đồ thị hàm số sau.

a) y lnx ,y 0,x 1,x e x



    KQ:

b) ln , 0, ,

2

x

y y x e x x

    KQ:

c) 1, 0,

1

x

y y x

x

 

  

 KQ:

Thaày Nguyễn Quang Vũ – THPT Lục Ngạn số – Baéc Giang Trang

(23)

d) y x e y ;x 0; x1; x2 KQ: e) y x .ln ;2x y0; x1; x e . KQ:

f) 12 ; 12 ; ;

6

sin cos

y y x x

x x

 

    KQ:

g) y x 2,y

x

    KQ:

h) y 4 x y x2,  2 2x KQ:

i) 2,

4

y x y x  KQ:

j) y x 2 2 ,x yx24x KQ:

k) y x 22 ,x y x 2 KQ: l) y e y e x;  x; x1 KQ: m) 2, 2

2 1

x y y

x

 

 KQ:

n) ,2 2

x

y y

x

 



KQ:

o) y = ;

(1 )

1

x x

y x

 

 KQ:

p)

4

4 x2

y  ;

2

x

y KQ:

q) y 4 x x2; 3y 0

    KQ:

r) y = x2 ; y 2 x2

  KQ:

s) y2 ,x y x2  2 2x1, y2 KQ: t) y x y,  2 x y, 0 KQ: u) y x x y,   0, y0 KQ: v) y x 2 4x5, y2x4,y4x11 KQ:

w) y2 ,x y x2  2 4x 4,y8 KQ: x) y = (e + 1)x , y = (1 + ex)x KQ: y) y x y ,  2 x2 KQ:

z) y x2 4x 3 , y x 3

     KQ:

aa) y x2 5x 6 , y x 1

     KQ:

bb) y x2 3x 2 ,y x2

(24)

cc) , 3 2

y x y x   x KQ:

Bài 23 Tính thể tích khối trịn xoay hình phẳng tạo đồ thị hàm số sau xoay quanh trục Ox.

a) 2, 0, 0, 3

y x  x y x x KQ:

b) sin6 cos ,6 0, 0,

y x x y x x KQ:

c) y x 3 1, y0, x1, x1 KQ:

d) y = xlnx , y = , x = e KQ: e) yln ,x y0, x2 KQ:

f) y x y 2,  x KQ:

g) 4y = x2 ; y = x KQ:

h) y x24 ,x y x 2 KQ:

i) 2,

4

x x

y y KQ:

j) y x 2 4x6, y x2 2x6 KQ:

(25)

Định dạng
Số trang	25
Dung lượng	1,48 MB