New Progress in Senior Mathematics Module 1 Book 1 (Extended … · 2015. 5. 9. · ∴ The indefinite integral is incorrect. 6.2 (a) ∫8 , where dx =8x+C C is any constant p.220

© Hong Kong Educational Publishing Co. 142

New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 6

6

pp.214 – 227 p.214 2. xx +23 is a primitive function of .16 +x

3. 232 xx + is a primitive function of .26 2 xx +

4. xe2 is a primitive function of .2 2xe

p.216 (a) Cxdxx +=∫ 767

(b) Cxdxx

+=⎟⎠⎞

⎜⎝⎛−∫ 3

4 51

(c) Cxdxx +=∫ 321

52

15

(d) Cedte tt +=∫ 22

21

p.227 1. 13 −= xu

8)(

3

uuF

u

=

=′

2. 42 −= xu

uuF

xu

=

=′

)(

2

3. 4xu =

ueuF

xu

=

=′

)(

4 3

4. xxu += 2

6)(

12

uuF

xu

=

+=′

pp.217 – 240

6.1 (a) )ln2( Cxdxd

+ p.217

x

x2

12

=

⎟⎠⎞

⎜⎝⎛=

∴ The indefinite integral is correct.

(b) ⎥⎥⎦

⎤

⎢⎢⎣

⎡++ Cx

dxd 2

3

)2(21

2

43

)2(23

21

+=

+= •

x

x

∴ The indefinite integral is incorrect. 6.2 (a) Cxdx +=∫ 88 , where C is any constant p.220

(b) Cxdxx +=∫ 76

71 , where C is any constant

(c) Cxdxxdxx +== ∫∫ 34

31

3

43 ,

where C is any constant

(d) Cedxe xx +=∫ 66


6.3 (a) ∫ + dxxx )4( 3 p.222

212

4

21

2

2

1

4

3

4 where,24

constantsany are and where

,42

44

CCCCxxCC

CxCx

xdxdxx

+=++=

+⎟⎟⎠

⎞⎜⎜⎝

⎛++=

+= ∫ ∫

(b) ∫ +− − dxxx )236( 421

32132

3321

323

127

421

36 where,24

constantsany are and , where

,2_313

326

236

CCCCCxxx

CCC

CxCxCx

dxdxxdxx

+−=+++=

++⎟⎠⎞

⎜⎝⎛−−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+=

+−=

−

−

−∫ ∫ ∫

Indefinite Integrals

143 © Hong Kong Educational Publishing Co.


6.4 (a) ∫ + dxxx )43( p.223

Cxx

Cxx

xdxdxx

xdxx

++=

+⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛=

+=

+=

∫∫∫

23

23

2

2

2

24

33

43

43

, where C is any constant

(b) ∫ −+ dxxx )13)(2( 2

Cxxxx

C

Cxxxx

dxxdxdxxdxx

dxxxx

+−+−=

+−⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎟⎠

⎞⎜⎜⎝

⎛−⎟

⎟⎠

⎞⎜⎜⎝

⎛=

−+−=

−+−=

∫∫∫∫∫

2334

3

constantany is where

,22

634

3

263

)263(

23

4

234

23

23

(c) ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛ + dxx

xx2

Cxx

dxx

++=

+= ∫

2

)1(2


(d) ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛ +− dxx

xxx3

54 26

constantany is where,3

22

6

26

)26(

32

22

22

CCxxx

dxxxdxdxx

dxxxx

++−−=

+−=

+−=

∫∫∫∫

−

−

(e) ∫ + dxxe x )65( 22


25

65

32

22

CCxe

dxxdxe

x

x

++=

+= ∫∫

(f) dxx

xex

∫−1

Cxe

dxxex

x

+−=

−= ∫ −

ln

1


6.5 (a) )( xxedxd p.224

xx exe +=

(b) ,)( 1Cxedxexe xxx +=+∫ 1C is any constant

,Cxedxedxxe xxx +=+∫∫

,12 CxeCedxxe xxx +=++∫ 2C is any constant

∫ +−= ,Cexedxxe xxx where 21 CCC −=

6.6 Let ,13 −= xu ,3=′u .)( 8uuF = p.228

∫ − dxx 8)13(3

Cx

CCu

duu

dxuu

+−

=

+=

=

′=

∫∫

9)13(


9

9

8

8

6.7 Let .42 −= xu p.229 xu 2=′

∴ duudxxn ∫∫ =− 42 2

Cx

C

Cu

+−=

+=

23

2

23

)4(32


,32

6.8 Let .13 −= xu p.230 ,3=′u dxdxu 3=′

dxxdxx ∫∫ −− −=− )3()13(31)13( 22

Cx

CCu

duu

+−

−=

+−=

=

−

−

−∫

3)13(


31

1

1

2

6.9 (a) Let .83 −= tu p.231 ,3 2tu =′ dttdtu 23=′

dtttdttt )3(318282 3232 ⎟⎠⎞

⎜⎝⎛−=− ∫∫

Ct

C

Cu

duu

+−=

+=

×=

•

∫

23

3

23

)8(94


,32

32

312



(b) Let .2xu −= ,2xu −=′ xdxdxu 2−=′

duedxxe ux ∫∫ −=−

212

Ce

CCe

x

u

+−=

+−=

−

2


2

(c) Let .23 −= xu ,3 2xu =′ dxxdxu 23=′

∫∫ =− u

dudxx

x31

23

2

C

x

CCu

+−

=

+=

3

2ln

constantany is where,ln31

3

6.10 Method I p.232 Let .1+= xu ,1=′u dudxu =′ and .1−= ux

duuudxxx ∫∫ −=+ )1(1

Cxx

CCuu

duuu

++−+=

+−=

−= ∫

23

25

23

25

21

23

)1(32)1(

52


52

)(

Method II Let ,1+= xu then ,12 += xu ,1 2ux −= .12 =′− uu

∴ dxx

dxudu12

1+

−=′=

dxxxxdxxx ∫∫ +

+−=+

12)1(21

Cxx

CCuu

duuu

duuu

++−+=

+−=

−=

−−=

∫∫

23

25

35

24

22

)1(32)1(

52


252

)22(

)1(2

6.11 ∫ + dxex x )1(2

p.233

∫ ∫+= dxexxdx x2

For the second integral, let ,2xu = ,2xu =′ .2xdxdxu =′

∴ ∫∫ = duedxxe ux

212

Ceu +=21 , where C is any constant

∴ ∫ ∫∫ +=+ duexdxdxex ux

21)1(

2

Cex x++=

22

22

6.12 ∫ −= dxey x 12 p.236

Cxex +−= 2 , where C is any constant

Q The curve passes through (0, 6). ∴ Ce +−= 026 0 4=C ∴ The equation is .42 +−= xey x

6.13 ∫ += xdxxdxdy 52 p.237

Cxx++=

25

3

23, where C is any constant

When x = 0, .0=dxdy

∴ C++= 000 ∴ 0=C

∴ 2

53

23 xxdxdy

+=


512

25

3

1

34

23

CCxx

dxxxy

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+= ∫

When x = 0, y = 1.

1

001

1

1

=++=

CC

∴ The equation of the curve is

.16

512

34++=

xxy



6.14 (a) dtt

tx ∫ +−=

41202 p.238

dtt

t∫ +

−=4

120 2

Let .42 += tu ,2tu =′ tdtdtu 2=′

∴ ∫−

= dtu

x 12120

Ct

CCu

++−=

+−=

)4ln(60

constantany is where,||ln602

When t = 0, x = 300.

∴ C+−= 4ln60300 4ln60300 +=C ∴ 4ln60300)4ln(60 2 +++−= tx

⎟⎠⎞

⎜⎝⎛

++=

44ln60300 2t

(b) ⎟⎠⎞

⎜⎝⎛

++=

4104ln60300 2x

105= (cor. to the nearest integer) ∴ The concentration is 105 .unit/m3

6.15 (a) dtetw t∫ −= 25.06)( p.240

Ce

C

Ce

t

t

+−=

+−

=

−

−

25.0

205.0

24


,25.0

6

When t = 1, w = 1.

25.0

25.0

241

241−

−

+=

+−=

eC

Ce

∴ 25.025.0 24124 −− ++−= eew t )(241 25.025.0 tee −− −+=

(b) When t = 3,

3544.8

)(241 75.025.0

≈−+= −− eew

When t = 12,

4963.18

)(241 325.0

≈−+= −− eew

∴ The increase in weight kg )3544.84963.18( − kg 10= (cor. to the nearest kg)

pp.217 – 239 Example 6.1T p.217

(a) ⎟⎠⎞

⎜⎝⎛ +− C

xdxd 3

2

2

3

)](3[

x

xdxd

=

−−= −


(b) ⎟⎟⎠

⎞⎜⎜⎝

⎛+C

xdxd

34

23

23

21

3

2

21

34

34

x

x

Cxdxd

−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+=

−

−

∴ The indefinite integral is incorrect. Example 6.2T p.220

(a) Cxdx +−=−∫ 6)6( , where C is any constant

(b) ∫∫ = dxxdxx 21

Cx += 23


(c) Cedxe xx +=∫ 33


Example 6.3T p.222

dxxx )64( 5∫ −

2123

6

21223

1

6

21

5

64 where,432

constants are and where,326

64

64

CCCCxx

CCCxCx

dxxdxx

−=+−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−⎟

⎟⎠

⎞⎜⎜⎝

⎛+=

−= ∫∫



Example 6.4T p.222

(a) dxex

x∫ ⎟⎠⎞

⎜⎝⎛ − −374

constantany is where,

37||ln4

714

3

3

CCex

dxedxx

x

x

++=

−=

−

−∫∫

(b) dxx

xx∫ ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −3

32 54

constantany is where,5||ln4

)54( 1

CCxx

dxx

+−=

−= ∫ −

(c) xdxx∫ +− )12)(23(


22

)26(2

3

2

CCxxx

dxxx

+−−=

−−= ∫

Example 6.5T p.224

(a) )ln( 4 xxdxd

xxx

xxx

x

ln4

ln41

33

34

+=

+⎟⎠⎞

⎜⎝⎛=

(b) 1

433 ln)ln4( Cxxdxxxx∫ +=+ ,

where 1C is any constant

∫

∫

∫∫

+−=

+=++

+=+

,4

lnln4


,lnln44

lnln4

3

443

2

143

2

4

1433

Cxxxxdxx

C

CxxxdxxCx

Cxxxdxxdxx

where 213 CCC −=

∴ ,16

ln41ln

443∫ +−= Cxxxxdxx

where4

3CC =

Example 6.6T p.228

(a) Let .53 += xu ,3 2xu =′ dxxdxu 23=′

Cx

C

Cu

duudxxx

++

=

+=

=+ ∫∫

7)5(


,7

)3()5(

73

7

6263

(b) Let .2xu = ,2xu =′ .2xdxdxu =′

Ce

CCe

duedxxe

x

u

ux

+=

+=

= ∫∫

2

2


2

Example 6.7T p.229

Let .13 += xu ,3 2xu =′ dxxdxu 23=′

Cx

CCu

duu

duudxxx

++=

+=

=

=+

•

∫

∫∫

23

3

23

21

32

)1(92


31

31

311

Example 6.8T p.229

Let .21 xu += ,2=′u dxdxu 2=′

Cx

CCu

duu

duu

dxxdxx

++=

+=

=

=

⎟⎠⎞

⎜⎝⎛+=+

•

∫

∫

∫∫

34

34

31

3

33

)21(83


21

2121

)2(212121



Example 6.9T p.230

(a) Let .54 −= xu ,4 3xu =′ dxxdxu 34=′

Cx

C

Cu

duu

dxxxdxxx

+−

=

+=

=

⎟⎠⎞

⎜⎝⎛−=−

•

∫

∫∫

32)4(


,81

4141

)4(41)4()4(

84

8

7

743743

(b) Let .23 += xu ,3 2xu =′ dxxdxu 23=′

Ce

CCe

due

dxexdxex

x

u

u

xx

+=

+=

=

⎟⎠⎞

⎜⎝⎛=

+

++

∫

∫∫

3


)3(31

2

2222

3

33

(c) Let .8−= tu ,1=′u dtdtu =′

CtCCu

duu

dtt

dtx

+−=+=

=

−=

−

∫

∫∫

|8|ln4constantany is where,||ln4

14

814

84

Example 6.10T p.231

Let .43 += xu

,3=′u dxdxu 3=′ and )4(31

−= ux

Cxx

C

Cuu

duuu

duu

u

duu

udx

xx

++−+=

+⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=

−=

−=

−=

+

•

−

•

∫

∫

∫∫

21

23

21

23

21

21

)43(9

16)43(274


,2432

92

)4(92

492

)4(31

312

432

Example 6.11T p.233

∫ ++ dxxx ])1(1[ 42

∫∫∫

++=

++=

dxxxxdx

dxxxx42

42

)1(

])1([

For the second integral, let .12 += xu

,2xu =′ xdxdxu 2=′

Then ∫ ∫=+ dxudxxx 442

21)1(

∴ ∫ ++ dxxx ])1(1[ 42


,51

21

2

21

21

25

1

2

4

CC

CuCx

duuxdx

+++=

+=

•

∫∫

Cxx+

++=

10)1(

2

522

, where .21 CCC +=

Example 6.12T p.235


2

)2(

34

23

23

CCxx

duxdxx

dxxxy

++=

+=

+=

∫∫∫

Q The curve passes through (1, 2).

∴ 631

212 ++=

67

=C


.67

32

34++=

xxy

Example 6.13T p.236


)212(

2 CCxx

dxxdxdy

+−=

−= ∫

When x = 1, .0=dxdy

4260

−=+−=

CC

∴ 426 2 −−= xxdxdy


)426(

1123

2

CCxxx

dxxxy

+−−=

−−= ∫



When x = 1, y = 4.

7

4124

1

1

=+−−=

CC


.742 23 +−−= xxxy

Example 6.14T p.237

(a) dtt

h ∫ += 2)4(

1

Let 4+= tu .

,1=′u dtdtu =′

Ct

CCu

duu

duu

h

++

=

+−=

=

=

∫

∫−

41


1

2

2

When h = 6, t = 8.

1273

1216

4816

=

+−=

++

−=

C

C

C

∴ 1273

41

++

−=t

h

(b) When t = 0,

635

1273

41

=

+−=h

∴ The height is m. 6

35

Example 6.15T p.239

At

AAt

dtttC

+=

+=

=

•

∫

34

34

31

49


3)(

When t = 0, C = 40.

40

)0(4940

=

+=

A

A

∴ 4049)( 3

4

+= ttC

When t = 3,

40)3(49 3

4

+=C

500= (cor. to the nearest ten thousand)

∴ The cost is 500 thousand dollars.

6.1 pp.224– 226

p.224

1. (a) ∫ ++=+ Cxxdxx 2)12( ,


(b) ∫ ++=− −− Cexdxe xx 2)2( ,


(c) ∫ ++=+ Cexdxex xx 2342 23)49( ,


(d) ∫ +=⎟⎟⎠

⎞⎜⎜⎝

⎛ − Cx

edxx

exe xxx

2 ,


2. (a) ⎟⎟⎠

⎞⎜⎜⎝

⎛++ Cxx

dxd 2

3

3

)2(

2

2)3(31

2

2

+=+=

+=

xxxx

xx


(b) ⎥⎦⎤

⎢⎣⎡ ++ Cx

dxd 82 )1(

161

72

72

)1(

)2()1(8161

+=

+= •

xx

xx

∴ The indefinite integral is correct. 3. ∫ += Cxdx 66 , where C is any constant

4. ∫ += Cxdxx5


5. ∫ += Cedxex

x

2


6. ∫ +−=−

− Ctdxt2




7. ∫ ∫= dxxdxx 53

5 3

Cx += 58


8. ∫ ∫−

= dxxdxx

53

5 3

1

Cx += 52


9. ∫ ∫ −− = dttdtt 77 66

16

116

6 ,


CCCt

CCt

=+−=

+⎟⎠⎞

⎜⎝⎛−=

−

−

10. ∫ ∫= dxx

dxx

144

1

11

4 , ||ln4constantany is where,4 ||ln4

CCCxCCx

=+=+=

11. ∫ ∫ −− = dxedxe xx 22 33

1

2

112

3 ,23


CCCe

CCe

x

x

=+−=

⎟⎠⎞

⎜⎝⎛ +−=

−

−

12.

137

1137

34

5 ,7

15


5

CCCx

CCx

dxx

=+=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+=

= ∫

13. ∫ + dxxx )6( 34

214

521

2

4

1

5

34

6 ,23

5


,4

65

6

CCCCxx

CC

CxCx

dxxdxx

+=++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+++=

+= ∫ ∫

14. ∫ −+ − dxxx )423( 22

32113

321

321

1

3

22

23 ,42

constantsany are and , where

,4)(23

3

423

CCCCCxxx

CCC

CxCxCx

dxdxxdxx

++=+−−=

+−+−+⎟⎟⎠

⎞⎜⎜⎝

⎛+=

−+=

−

−

−∫ ∫ ∫

15. ∫ + dxx

x )52(3 5

4 3

2132

47

21

232

147

35

43

52 ,7

1578


,235

742

53

CCCCxx

CC

CxCx

dxxdxx

+=+−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+=

+=

−

−

−

∫ ∫

16. ∫ ⎟⎠⎞

⎜⎝⎛ − − dxe

xx234

21

221

22

1

2

34 ,2

3||ln4


,213) ||(ln4

314

CCCCex

CC

CeCx

dxedxx

x

x

x

−=++=

⎟⎠⎞

⎜⎝⎛ +−−+=

−=

−

−

−∫ ∫

17. ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛+ dx

xx 32

2121

23

21

221

123

21

21

32 ,634


,23322

32

CCCCxx

CC

CxCx

dxxdxx

+=++=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛++

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+=

+= ∫ ∫−

18. ∫ + dxxx )6(3

Cxx

CCxx

dxxdxx

++=

++⎟⎟⎠

⎞⎜⎜⎝

⎛=

+= ∫ ∫

23

23

2

9


183

3

183

19. ∫ +− dxee xx )2(


2

)2( 0

CCex

dxedx

dxee

x

x

x

++=

+=

+=

∫∫∫

20. ∫ ++ dxxx )1)(6(


3

67

)67(

23

2

2

CCxxx

dxxdxdxx

dxxx

+++=

++=

++=

∫ ∫∫∫



21. ∫ +− dxxx )2)(32(


2

62

)62(

23

2

2

CCxxx

dxxdxdxx

dxxx

+−+=

−+=

−+=

∫∫∫∫

22. ∫ ++ dxxx )2)(1( 2


24

22

)22(

23

4

23

23

CCxxxx

dxxdxdxxdxx

dxxxx

++++=

+++=

+++=

∫∫∫∫∫

23. ∫ ⎟⎠⎞

⎜⎝⎛ + dx

x

2

11


21

121

1

2

2

CCxxx

dxdxx

dxx

dxxx

+++−=

++=

⎟⎠⎞

⎜⎝⎛ ++=

−

∫∫∫

∫

24. ∫+ dxx

x1

constantany is where,||ln

11

CCxx

dxx

dx

dxx

++=

+=

⎟⎠⎞

⎜⎝⎛ +=

∫∫

∫

25. ∫−+ dx

xxx

2

2 43

constantany is where,4 ||ln3

43

431

1

2

2

CCxxx

dxx

dxx

dx

dxxx

+++=

++=

⎟⎠⎞

⎜⎝⎛ −+=

−

∫∫∫

∫

p.226

26. ∫+ dxxx 2)1(


12

12

2

2

CCxxx

dxx

x

dxx

xx

+++=

⎟⎠⎞

⎜⎝⎛ ++=

++=

∫

∫

27. ∫+ dtt

t2

22 )1(


12

12

3

22

2

24

CCt

tt

dtt

t

dtt

tt

+−+=

⎟⎠⎞

⎜⎝⎛ ++=

++=

∫

∫

28. ∫ + dxex 2)2(

Cxee

CCxee

dxee

xx

xx

xx

+++=

++⎟⎟⎠

⎞⎜⎜⎝

⎛+=

++= ∫

422


42

)44(

22

22

2

29. ∫− dx

ee

x

x 13 2

Cee

CCee

dxee

xx

xx

xx

++=

+−−=

−=

−

−

−∫

3

constantany is where,)(3

)3(

30. ∫−−+ dx

eeee

x

xxx

22 3

Cexe

Cdxee

xx

xx

+++=

⎟⎠⎞

⎜⎝⎛ ++=

−

−∫

422


212

22

22

31. ∫+ dxe

ex

x 2)2(


)44(

442

CCexe

dxee

dxe

ee

xx

xx

x

xx

+−+=

++=

++=

−

−∫

∫

32. ∫+ dxxe

xeex

xx

32 3

Cex

CCex

dxex

x

x

x

++=

++=

⎟⎠⎞

⎜⎝⎛ +=

•

∫

2

2

2

65||ln

32


35||ln

32

35

32



33. ∫ −− dx

xx

242


)2(2

)2)(2(

2CCxx

dxx

dxx

xx

++=

+=

−+−

=

∫∫

34. ∫ ++ dx

xx

113


)1(1

)1)(1(

25

2

2

CCxxx

dxxx

dxx

xxx

++−=

+−=

++−+

=

∫∫

35. (a) )( 3xxedxd

xx xee 33 3+=

(b) 1333 )3( Cxedxxee xxx +=+∫ ,


∫∫ +=+ 1333 3 Cxedxxedxe xxx

133

2

33

3CxedxxeCe xx

x+=++ ∫ ,


21

333

33 CCexedxxe

xxx −+−=∫

3

,93

21

333

CCC

Cexedxxexx

x

−=

+−=∫

36. (a) )ln( 5 xxdxd

xxx

xxx

x

ln5

ln51

44

45

+=

+⎟⎠⎞

⎜⎝⎛=

(b) 1544 ln)ln5( Cxxdxxxx +=+∫ ,


∫∫ +=+ 1544 lnln5 Cxxxdxxdxx

154

2

5lnln5

5CxxxdxxCx

+=++ ∫ ,


5

,255

lnln

21

554

CCC

Cxxxxdxx

−=

+−=∫

37. (a) xdxd ln

xx

xdxd

ln21

])[(ln 21

=

=

(b)


,lnln2

1

1

1

C

Cxdxxx

+=∫

1lnln1

21 Cxdx

xx+=∫

12,ln2ln1 CCCxdx

xx=+=∫

38. (a) Let 42)( xxg =′ .

By Definition 6.2, ∫ += Cdxxxg 42)( ,


( ) 44 2)(2 xxgdxxdxd

=′=∴ ∫

(b) Let xxxg ln)( =′ .

By Definition 6.2, ∫ += Cxdxxxg ln)( ,


∴

( )

xx

xgxdxxdxd

ln

)(ln

=

′=∫

39. (a) Let ).()( 2 xedxdxf x += Then xexg x += 2)( .

By Definition 6.2,

dxxfdxxedxd x ∫∫ =+ )()( 2

Cxe x ++= 2 ,


(b) Let ).ln()( 4 xxdxdxf = Then xxxg ln)( 4= .

By Definition 6.2,

,ln

)()ln(

4

4

Cxx

dxxfdxxxdxd

+=

= ∫∫

where C is any constant 40. (a) Let 3)( xxg =′ .

By Definition 6.2, ∫= dxxxg 3)( .

( )

3

3

)(

x

xgdxxdxd

=

′=∴ ∫



(b) Let )()( 3xdxdxf = . Then 3)( xxg = .

By Definition 6.2,


)(

3

3

CCx

dxxfdxxdxd

+=

=⎟⎠⎞

⎜⎝⎛

∫∫

(c) No, they are not equal.

( )∫ dxxdxd 3 is a function, but

dxxdxd

∫ ⎟⎠⎞

⎜⎝⎛ 3 is a group of functions.

6.2 pp.233 – 235

p.233

1. Let .72 −= xu ,2xu =′ .2xdxdxu =′

∫∫ =− duudxxx 882 )7(2

Cx

CCu

+−

=

+=

9)7(


92

9

2. Let .14 += xu ,4=′u .4dxdxu =′

∫∫ =+ duudxx 1010)14(4

Cx

CCu

++

=

+=

11)14(


11

11

3. Let .23 += tu ,3 2tu =′ .3 2dttdtu =′

∫∫ =+ duudttt 23 32

Ct

CCu

++=

+=

23

3

23

)2(32


4. Let .12 += eu ,xeu =′ .dxedxu x=′

∫∫ =+ duudxee xx 44)1(

Ce

CCu

x+

+=

+=

5)1(


5

5

5. Let .2xu = ,2xu =′ .2xdxdxu =′

Ce

CCe

duedxxe

x

u

ux

+=

+=

= ∫∫

2

2


2

6. Let .14 2 += xu ,8xu =′ .8xdxdxu =′

Cx

CCu

duudxxx

++

=

+=

=+ ∫∫

7)14(


)14(8

72

7

662

7. Let .42 += tu ,2tu =′ .2tdtdtu =′

∫∫ =+

duu

dtt

t32

1)4(

2

Ct

CCu

duu

++

−=

+−=

=

−

−

−∫

2)4(


22

2

3

8. Let .32 ttu += ,32 +=′ tu .)32( dttdtu +=′

∫∫ =++ duudtttt 3)32( 2

Ctt

C

Cu

++=

+=

23

2

23

)3(32


,32

9. Let .13 += xu ,3=′u .3dxdxu =′

Cx

CCu

duudxx

++

=

+=

=+

•

∫∫

21)13(


131)13(

7

7

66

10. Let .12 2 += xu ,4xu =′ .4xdxdxu =′

Cx

CCu

duuxdxx

++

=

+=

=+

•

∫∫

24)12(


141)12(

62

6

552



11. Let .52 −= tu ,2=′u .2dtdtu =′

Ct

CCu

duudtt

+−

−=

+−=

=−

−

−•

−− ∫∫

2)52(

constantany is where,2121)52(

1

1

22

12. Let .1 3xu += ,3 2xu =′ .3 2dxxdxu =′

Cx

CCu

duu

dxx

x

++

=

+=

=+ ∫∫

3|1|ln


131

1

3

3

2

p.234

13. Let .64 −= xu ,4 3xu =′ .4 3dxxdxu =′

∫∫ =− duudxxx41643

Cx

CCu

+−

=

+= •

6)6(


41

23

4

23

14. Let .843 2 ++= xxu ,46 +=′ xu .)46( dxxdxu +=′

Cxx

CCu

duu

dxxx

x

+++=

+=

=++

+∫∫

21

2

21

2

)843(2


1

843

46

15. ∫ •⎟⎟⎠

⎞⎜⎜⎝

⎛ + dxxx

x3

2

2

2 12

∫ •⎟⎠⎞

⎜⎝⎛ += dx

xx 3

2

2121

Let .21 2xu +=

,43x

u −=′ .43 dx

xdxu −=′

∫∫ −=⎟⎠⎞

⎜⎝⎛ + • duudx

xx2

3

2

2 41121

Cx

C

Cu

+⎟⎠⎞

⎜⎝⎛ +−=

+−= •

3

2

3

21121


,34

1

16. Let .52 += teu ,2 2teu =′ .2 2 dxedxu x=′

Ce

CCu

duudxee

x

xx

++

=

+=

=+

•

∫∫

10)5(


121)5(

52

5

4422

17. Let .22 teu t −= ),1(222 22 −=−=′ tt eeu .)1(2 2 dtedtu t −=′

∫∫ =−− duudtete tt 2222

21)1()2(

Cte

C

Cu

t+

−=

+= •

6)2(


,32

1

32

3

18. Let .92 2 += tu ,4tu =′ .4tdtdtu =′

Ct

CCu

duu

dtt

t

++

−=

+−

=

=+

−

−•

∫∫

12)92(


141

141

)92(

32

3

442

19. Let .ln tu =

,1t

u =′ .1 dtt

dtu =′

Ct

CCu

duudttt

+=

+=

=

•

∫∫

8)(ln


121

2)(ln

4

4

33



20. Let .baxu += ,au =′ .adxdxu =′

Ca

e

CCea

duea

dxe

bax

u

ubax

+=

+=

=

+

+

•

∫∫


1

21. Let .12 += xu ,2xu =′ .2xdxdxu =′

Ce

CCe

duexdxe

x

u

ux

+=

+=

−=

+

+ ∫∫

2


21

1

1

2

2

22. Let .3 2xu = ,6xu =′ .6xdxdxu =′

Ce

CCe

duedxxe

x

u

ux

+=

+=

= ∫∫

3


2

2

3

3

23. Let .1t

u =

,12t

u −=′ .12 dt

tdtu −=′

Ce

CCe

duedtet

t

u

ut

+−=

+−=

−= ∫∫

1

1

2


1

24. Let .32 xxu −= ,32 −=′ xu .)32( dxxdxu −=′

Ce

CCe

duedxex

xx

u

uxx

+=

+=

=−

−

− ∫∫

3

3

2

2


)32(

25. Let .33 xxu += ,33 2 +=′ xu .)1(3 2 dxxdxu +=′


,

)1(33

43 2

3

3

C

Ce

duedxexxx

xx

+=

=+

+

+ ∫∫

26. Let .12 ++= ttu ,12 +=′ tu .)12( dttdtu +=′

Ce

CCe

duedtet

tt

u

utt

+=

+=

=+

++

++ ∫∫

1

1

2

2


)12(

27. Let .xu =

,2

1x

u =′ .2

1 dxx

dxu =′

Ce

CCe

duedxx

e

x

u

ux

+=

+=

= ∫∫

2


2

28. Let .82 += xu ,2xu =′ .2xdxdxu =′

Cx

CCu

duu

dxx

x

++=

+=

=+ ∫∫

|8|ln21


121

8

2

2

29. Let .1 2xu −= ,2xu −=′ .2xdxdxu −=′

Cx

CCu

duu

dxxx

+−

−=

+−=

⎟⎠⎞

⎜⎝⎛−=

− ∫∫ •

2|1|ln3


1213

13

2

2

30. Let .1 xu −= ,1−=′u .dxdxu −=′

Cx

CCu

duu

dxx

+−−=+−=

−=− ∫∫

|1|ln2constantany is where,||ln2

121

2

31. Let .22 ttu += ,22 +=′ tu .)1(2 dttdtu +=′

Ctt

CCu

duu

dttt

t

++=

+=

=++

∫∫

|2|ln21


121

21

2

2



32. Let .12 += xeu ,2 2xeu =′ .2 2 dxedxu x=′

Ce

CCu

duu

dxe

e

x

x

x

++=

+=

=+ ∫∫

|1|ln21


121

1

2

2

2

33. Let .23 xxu += ,23 2 +=′ xu .)23( 2 dxxdxu +=′

Cxx

CCu

duu

dxxx

xdxxx

x

++

=

+=

=

++

=++

∫

∫∫

5|2|ln


151

)2(523

10523

3

3

2

3

2

34. Let .122 24 +−= xxu ),4(228 33 xxxxu −=−=′ .)4(2 3 dxxxdxu −=′

Cxx

CCu

duu

dxxx

xx

++−

=

+=

=+−

−∫∫

2|122|ln


121

1224

4

24

3

35. Let .ln xu =

,1x

u =′ .1 dxx

dxu =′

Cx

CCu

duu

dxxx

+=+=

= ∫∫

|ln|lnconstantany is where,||ln

1ln1

36. Let .2+= xu ,1=′u dxdxu =′ and .2−= ux

Cxx

Cuu

C

Cuu

duuu

duuudxxx

++−+=

+−=

+−=

−=

−=+

•••

∫

∫∫

23

25

23

25

23

25

21

23

)2(4)2(56

456


,3223

523

)2(3

)2(323

37. Let .1+= xu ,1=′u dxdxu =′ and .1−= ux

1

1

11

1 where, |1|ln |1|ln1

constant any is where, ||ln

11

11

CCCxxCxx

CCuu

duu

duu

udxx

x

+=++−=++−+=

+−=

⎟⎠⎞

⎜⎝⎛ −=

−=

+

∫

∫∫

38. Let .13 += xu

,3=′u dxdxu 3=′ and .3

1−=

ux

,)13(94)13(

274


,232

92

192

131

312

132

21

23

1

121

23

Cxx

C

Cuu

duu

u

duu

udxxx

++−+=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−=

+

∫

∫∫ ••

where 192 CC +=

39. Let .4+= tu ,1=′u dtdtu =′ and .4−= ut

Ctt

CCuu

duuu

duuudttt

++

−+

=

+−=

−=

−=+

∫∫∫

3)4(

13)4(


)4(

)4()4(

1213

1213

1112

1111

40. Let .2 tu −= ,1−=′u dtdtu −=′ and .2 ut −=

Ctt

CCuu

duuu

duu

u

duuudt

tt

+−+−

−=

++−=

−=

−−=

+−=

−+

−−

−−

−−∫

∫

∫∫

32

32

43

4

44

)2(2

)2(


)3(

3

12)2(1



41. Let .1+= xu Then xu =−1

,2

1x

u =′ dxx

dxu2

1=′

1

1

1

1

2 where, |1|ln22

|1|ln222


112

12

21

12

11

CCCxx

Cxx

CCuu

duu

duu

u

dxxx

xdxx

+=++−=

++−+=

+−=

⎟⎠⎞

⎜⎝⎛ −=

−=

+=

+

∫

∫

∫∫ •

42. Let .1ln += xu

,1x

u =′ .1 dxx

dxu =′


)1(ln22

121

21ln

2

2

CCx

u

ududxx

x

++

=

=

=+

•

∫∫

43. Let .12 += xu ,2xu =′ .2xdxdxu =′

∫ ++ dxxx )11( 2

1

23

22

1

23

22

1123

21 where,

3)1(

2

3)1(

21


21

21

)1(21

CCCxx

Cxx

CCuu

duu

+=++

+=

++

++

=

++=

+=

•

∫

44. Let .72 += xu

,2=′u dxdxu 2=′ and .2

7−=

ux

∫ −+ dxxxx )72(

,3

)72(67

10)72(


,33

247

52

41

)7(41

27

21

72

3232

5

21

2

3

123

25

221

23

2

2

Cxxx

CC

CxCuu

dxxduuu

dxxduuu

dxxdxxx

+−+−+

=

−−−−=

−−=

−−

=

−+=

••

∫∫

∫ ∫

∫∫

where )( 21 CCC +−=

45. (a) 3)52( +x

125150608

55)2()5()2()2(23

3232

231

3

+++=

+++=

xxx

xCxCx

∫ + dxx 3)52(


,12575202

)125150608(234

23

C

Cxxxx

dxxxx

++++=

+++= ∫

(b) Let .52 += xu ,2=′u .2dxdxu =′

Cx

CCu

duudxx

++

=

+=

=+

•

∫∫

8)52(


121)52(

4

4

33

(c) Yes, the difference between them is a constant. They represent the same group of functions.

6.3 pp.240 – 243

p.240

1. ∫ −= dxxy )38(


382 CCxx

dxxdx

+−=

−= ∫∫

Q The curve passes through ).1 ,2( −

11

)2(3)2(41 2

−=+−=−

CC

∴ The equation of the curve is .1134 2 −−= xxy



2. ∫ ⎟⎠⎞

⎜⎝⎛ += dx

xy 12


12

CCxx

dxdxx

++=

+= ∫∫

Q The curve passes through ).4 ,1(

∴ C++= 11ln24 3=C ∴ The equation is .3 ||ln2 ++= xxy

3. ∫ += dxey x )2(


2

CCxe

dxdxex

x

++=

+= ∫∫


∴ Ce ++= )0(22 0

1=C ∴ The equation is .12 ++= xey x

4. ∫ += dxxy 2)21(

Let .21 xu += ,2=′u .2dxdxu =′

∴ ∫= duuy 2

21

Cx

CCu

++

=

+= •

6)21(


1

2

3


∴ C++

=6

)]1(21[02

29

−=C

∴ The equation is .29

6)21( 3−

+=

xy

5. (a) ∫ += dtty )4.02(

constantany is where,4.02 CCtt ++=

When t = 0, x = 6. ∴ C++= )0(4.006 2

6=C ∴ 64.02 ++= ttx

(b) When t = 4,

6.23

6)4(4.042

=++=x

∴ The amount is 23.6 units.

6. (a) ∫ −= dteD t6.02

Ce

CCe

t

t

+−=

+−

=

−

−

6.0

6.0

310

constantany is where,6.0

2

When t = 0, .35

=D

53

1035

310

35 0

=

+=

+−=

C

Ce

∴ 53

10 6.0 +−= − teD

(b) When t = 4,

53

10 4.2 +−= −eD

7.4= (cor. to 1 d. p.) ∴ The concentration is .g/cm 7.4 3

7. ∫= dten t1.0300

constantany is where,3000 1.0 CCe t +=

When t = 0, n = 12 000.

90003000000 12 0

=+=

CCe

∴ 90003000 1.0 += ten

When t = 8, 90003000 )8(1.0 += en 677 15= (cor. to the nearest integer)

∴ The number of probiotics is 15 677. 8. ∫ −= dteP t3.015.0

constantany is where,5.0 3.0 CCe t +−= −

When t = 0, P = 6.0.

5.6

5.00.6 0

=+−=

CCe

∴ 5.65.0 3.0 +−= − teP

When t = 4, 5.65.0 2.1 +−= −eP 3.6= (cor. to 1 d. p.)

∴ The pH value is 6.3.



9. ∫ += dt

tR

2310

Let .23 += tu ,3=′u .3dtdtu =′

constantany is where,|23|ln

310

13

10

CCt

duu

R

++=

= ∫

When t = 0, R = 0.5.

2ln

3105.0

2ln3

105.0

−=

+=

C

C

∴ 2ln3

105.0 |23|ln3

10−++= tR

When t = 4,

2ln3

105.0|2)4(3|ln3

10−++=R

0.7= (cor. to 1 d. p.)

∴ The radius of the polluted area is 7.0 m. 10. ∫ −= dteN t4.048

constantany is where,120 4.0 CCe t +−= −

When t = 0, N = 300.

420

120300 0

=+−=

CCe

∴ 420120 4.0 +−= − teN

When t = 4, 420120 6.1 +−= −eN 396= (cor. to the nearest integer)

∴ There are 396 visitors after 4 hours. 11. ∫ −= dteN t2.08


When t = 0, N = 10.

50

4010 0

=+−=

CCe

∴ 5040 2.0 +−= − teN

When t = 7, 5040 4.1 +−= −eN 40= (cor. to the nearest integer)

∴ His typing speed is 40 words per minute.

12. ∫−= dttM 53

8

constantany is where,5 58

CCt +−=

When t = 0, M = 6000.

6000

)0(56000=

+−=C

C

∴ 60005 58

+−= tM

When M = 1000,

9894.741000

1000

600051000

85

58

58

≈=

=

+−=

t

t

t

∴ It will take 75 months for the price to decrease by half.

13. ∫ += dttP )400300(


800300 23

CCtt ++=

When t = 0, P = 5000.

5000

605000=

++=C

C

∴ 50003

800300 23

++= ttP

When t = 2,

6354

500023

8002300 23

=

+×+×=P

∴ The population after 2 years is 6354.

p.242

14. 12

2+= xe

dxyd


)1(

11 CCxe

dxedxdy

x

x

++=

+= ∫

When x = 0, 1−=dxdy .

2

01

1

10

−=++=−

CCe

∴ 2−+= xedxdy x


2

)2(

22

2CCxxe

dxxey

x

x

+−+=

−+= ∫

Since the point (0, 6) is on the curve,

5

006

2

20

=+−+=

CCe


.522

2+−+= xxey x



15. xdx

yd=2

2

constantany is where,2 11

2CCxxdx

dxdy

+== ∫


2

221

3

1

2

CCxCx

dxCxy

++=

⎟⎟⎠

⎞⎜⎜⎝

⎛+= ∫

Q The curve passes through (0, 3) and (2, 9).

37

32629

3003

1

1

32

2

=

++=

=++=

C

C

CC


.337

6

3++= xxy

16. 24 22

2+= xe

dxyd


)24(

112

2

CCxe

dxedxdy

x

x

++=

+= ∫

Q The slope of the tangent at (0, 1) is 4.

40=

=xdxdy

∴ 10 024 Ce ++=

21 =C

∴ 222 2 ++= xedxdy x


)222(

2222

2

CCxxe

dxxeyx

x

+++=

++= ∫

Since the curve passes through (0, 1), ∴ 2

0 001 Ce +++= 02 =C

∴ The equation of the curve is .222 xxey x ++=

17. xdx

yd−= 32

2


23

)3(

11

2CCxx

dxxdxdy

+−=

−= ∫

Since the slope of the tangent y = x + 1 is 1,

11=

=xdxdy

23

21)1(31

1

1

−=

+−=

C

C

∴ 23

23

2−−=

xxdxdy


623

23

23

12

32

2

CCxxx

dxxxy

+−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−= ∫

Since (1, 2) is on the curve,

∴ 223

61

232 C+−−=

6

132 =C


.6

1323

23

61 23 +−+−= xxxy

18. ∫ += dt

ttn

1

Let .1+= tu Then .1−= ut 1=′u and dtdtu =′

Ctt

CCuu

duuu

duu

un

++−+=

+−=

−=

−=

∫

∫−

21

23

21

23

21

21

)1(2)1(32


)(

1

When t = 0, n = 36.

3137

23236

=

+−=

C

C

∴ 31

21

23

37)1(2)1(32

++−+= ttn

When t = 5,

3137)15(2)15(

32 2

123

++−+=n

42= (cor. to the nearest integer) Q The number of leopards after 5 years is 42. 19. (a) dtel t∫ −= 3.01.2


When t = 0, l = 1.

8

71 0

=+−=

CCe

∴ 87 3.0 +−= − tel



(b) When t = 10, 87 3 +−= −el 7.7= (cor. to 1 d. p.)

∴ The length of the prawn after ten days is 7.7 cm. (c) When l = 6,

1759.43.0

72ln

72

876

3.0

3.0

≈−

=

=

+−=

−

−

t

t

e

e

t

t

∴ 5 days are needed before they can be sold.

20. (a) dtttP ∫ += 42 2

Let .42 2 += tu ,4tu =′ .4tdtdtu =′

Ct

CCu

duuP

++=

+=

=

•

∫

23

2

23

)42(61


41

41

(b) When t = 4, P = 102.

66

]4)4(2[61102 2

32

=

++=

C

C

∴ 66)42(61 2

32 ++= tP

When t = 5,

66]4)5(2[61 2

32 ++=P

132= (cor. to the nearest integer)

∴ The profit after 5 years is 132 thousand dollars. 21. (a) dtteP t∫ −−=

2

10000

Let .2tu = ,2tu =′ .2tdtdtu =′

Ce

CCe

dueP

t

u

u

+=

+−−=

−=

−

−

−• ∫

2

5000

constantany is where,)(50002110000

When t = 0, P = 6000. ∴ Ce += 050006000 1000=C

∴ 100050002

+= −teP

(b) When t = 5, 10005000

25 += −eP 1000= (cor. to the nearest dollars)

∴ The value is $1000.

22. (a) dtttt

P ∫ +++

= 92

50 2

dtttdtt ∫∫ +++

= 92

150 2

For the second integral, Let .92 += tu ,2tu =′ .2tdtdtu =′

duudttt ∫∫ =+ 21

2

219

∴ duudtt

P ∫∫ ++

= 21

21

2150

Ctt

C

Ctt

++++=

++++= •

23

2

23

2

)9(31|2|ln50


,)9(32

21|2|ln50

When t = 0, P = 10.5.

2ln505.1

)9(312ln505.10 2

3

−=

++=

C

C

∴ 2ln505.1)9(31|2|ln50 2

32 −++++= ttP

(b) The number of year from 2009 to 2016

7

20082016=

−=

2ln505.1)97(31)27ln(50 2

32 −++++=P

9.223= (cor. to 1 d. p.)

∴ The annual profit in 2016 is 2239 thousand dollars.



pp.248 – 251

p.248

1. ∫∫∫ −− +=+ dxxdxxdxxx 2323 34)34(

Cxx

C

Cxx

+−=

+−+⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

14

14

3


,)(34

4

2. ∫∫∫−

−=⎟⎟⎠

⎞⎜⎜⎝

⎛− dxxdxxdx

xx 2

121

212

Cxx

C

Cxx

+−=

+−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

21

23

21

23

234


,)2(322

3. ∫∫∫ +=+ −− dxdxedxe xx 63)63( 22


,62

3 2

C

Cxe x++−=

−

4. ∫∫ −=−+ dttdttt )1()1)(1( 2


3CCtt +−=

5. dtttdttt )2()21( 322 ∫∫ +=+

Ctt

CCtt

++=

+⎟⎟⎠

⎞⎜⎜⎝

⎛+=

23


23

43

43

6. ∫∫++

=+ dt

tttdt

tt

3

2

3

2 12)1(

Cttt

C

Cttt

dtttt

+−−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛

−+−+=

++=

−−

−−

−−−∫

22 ||ln


,2

)(2 ||ln

)2(

21

21

321

7. Let .23 −= xu ,3 2xu =′ .3 2dxxdxu =′

∫ ∫=− duudxxx 5532 )2(3

Cx

CCu

+−

=

+=

6)2(


63

6

8. Let .14 += xu ,4 3xu =′ .4 3dxxdxu =′

∫ ∫=+ duedxex ux 1 3 4

4

Ce

CCex

u

+=

+=+14


9. Let .1 2xu −= ,2xu −=′ .2xdxdxu −=′

∫ ∫−=− dxudxxx 442

21)1(

Cx

CCu

+−

−=

+−= •

10)1(


1

52

5

10. Let .1 xeu −= ,xeu −=′ .dxedxu x−=′

∫ ∫−=− dxudxee xx 1

Ce

CCu

x +−−=

+−=

23

23

)1(32


11. Let .xx eeu −+= ,xx eeu −−=′ .)2( dxedxu xx −−=′

∫ ∫=+−

−

−

duu

dxeeee

xx

xx 1

Cee

CCuxx ++=

+=− ||ln

constantany is where,||ln

12. ∫∫ +++

=+++ dx

xxxdx

xxx

1)1(

11

2

2

2

2

∫ ∫ ++= dx

xxdx

12

For the second integral, let ,12 += xu ,2xu =′ .2xdxdxu −=′

∫ ∫=+

duu

dxx

x 121

12

∴ ∫ ∫∫ +=+++ du

udxdx

xxx 1

21

11

2

2

Cxx

C

Cux

+++=

++=

|1|ln21


,||ln21

2



13. ∫∫ −+=+

= dxxdxx

xy )1(1 22

2

constantany is where,1 CCx

x +−=

Q The curve passes through (1, 2).

∴ C+−=1112

2=C

∴ 21+−=

xxy

14. ∫ += dx

xy

141

Let .14 += xu ,4=′u .4dxdxu −=′

∴ ∫= duu

y 141

Cx

CCu

++=

+=

|14|ln41


When x = 2, y = 5.

9ln

415

9ln415

−=

+=

C

C

∴ 9ln415|14|ln

41

−++= xy

59

14ln41

++

=x

15. (a) ∫ +−= dttd )41.0(

Ctt

CCtt

++−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

2

2

21.0


41.0

When t = 0, d = 0. ∴ 0=C ∴ 221.0 ttd +−=

(b) When t = 10,

199

2001)10(2)10(1.0 2

=+−=

+−=d

∴ The distance travelled is 199 m.

p.248

16. Let .2 xu −= Then .2 ux −= ,1−=′u .dxdxu −=′

∫ ∫ +−−=−+ duuudxxx )12(2)1(

Cxx

C

Cuu

duuu

duuu

+−−−=

+−=

−=

−−=

•

∫

∫

23

25

23

25

21

23

)2(2)2(52


,323

52

)3(

)3(

17. Let .1+= tu Then .1−= ut ,1=′u .dtdtu −=′

∫ − dttt 13

Ctttt

C

Cuuuu

duuuuu

duuuuu

duuu

++−+++−+=

+−⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=

−+−=

−+−=

−−=

∫

∫∫

−

23

25

27

29

23

25

27

29

21

23

25

27

123

3

)1(32)1(

56)1(

76)1(

92


,32

523

723

92

)33(

)133(

)1(

18. Let .3+= xu Then .3−= ux ,1=′u .dxdxu =′

∫ ∫ −−=+ duuudxxx 100100 )3()3(

Cxx

CCuu

duuu

++

−+

=

+−=

−= ∫

101)3(3

102)3(


3102

)3(

101102

101102

100101

19.

dxe

e

dxee

edx

e

x

x

x

x

xx

∫

∫∫

+=

+=

+•

−−

1

11

11

2

2

2

2

22

Let .12 += xeu ,2 2xeu =′ xedxu 22=′ .

∫ ∫=+ −

duu

dxe x 1

21

11

2

Ce

C

Cu

x ++=

+=

|1|ln21


,||ln21

2



20. Let .12 += xu Then .12 −= ux ,2xu =′ .2xdxdxu =′

∫ ∫+

=+

•dx

x

xx

x

x 12

1

12 2

2

2

3

121)1(

61


,232

41

)(41

121

21

223

2

21

23

21

21

Cxx

C

Cuu

duuu

duu

u

++−+=

+⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=

−=

−=

∫

∫−

•

21. Let .5ln xu =

,1x

u =′ .x

dxdxu =′

duudxx

x 5ln∫ ∫=

Cx

CCu

+=

+=

2

2

)5(ln21


22. dxx

xx∫ ⎥⎦

⎤⎢⎣

⎡

+++

)4(3)12( 2

4

∫ ∫ +++= dx

xxdxx

43)12( 2

4

For the first integral, let .12 += xu ,2=′u .2dxdxu =′

For the second integral, let .42 += xu ,2xu =′ .2xdxdxu =′

∴ dxx

xx∫ ⎥⎦

⎤⎢⎣

⎡

+++

)4(3)12( 2

4

Cxx

CCuu

duu

duu

++++

=

++=

+=

•

∫∫

|4|ln23

10)12(


521

123

21

25

5

4

23. ∫ ∫∫ −=⎥⎦⎤

⎢⎣⎡ − −− dx

xxdxxedx

xxxe xx lnln 22

For the first integral, let .2xu −= ,2xu −=′ .2xdxdxu −=′

For the second integral, let .ln xu =

,1x

u =′ .1 dxx

dxu =′

∴ dxxxxe x∫ ⎥⎦⎤

⎢⎣⎡ −− ln2

Cxe

CCue

ududue

x

u

u

+−−=

+−−=

−−=

−

∫∫

2|)|(ln

21


121

2

2

2

24. dttth ∫ ++

=1

)1ln(

Let ).1ln( += tu

,1

1+

=′t

u .1

1 dtt

dtu+

=′

∴ duuh ∫=

Ct

CCu

++

=

+=

2)]1[ln(


2

2

When t = 0, h = 2

22

)10ln(22

=

+⎟⎠⎞

⎜⎝⎛ +

=

C

C

∴ 22

)]1[ln( 2+

+=

th

25. dxe

exf x

x

∫+

= 21)(


21

)(

2

2

CCee

dxee

xx

xx

+−−=

+=

−−

−−∫

211

421

4)0(

00

=

=+−−

=

C

Cee

f

∴ 2

1121)( 2 +−−= −− xx eexf



26. dxx

exfx

∫=)(

Let .xu =

,2

1x

u =′ .2

1 dxx

dxu =′

∴ ∫= duexf u2)(

Ce

CCex

u

+=

+=

2


2

24

2

2

)4(

eC

eCe

ef

−=

=+

=

∴ 22)( eexf x −=

27. (a) 2)2(

2 +++

=+

+x

BxAx

BA

2

2+

++=

xBAAx

Q 2

22 +

++≡

+ xBAAx

xx

∴ 1=A

∴ 2

0)1(202

−==+=+

BBBA

(b) ∫ ∫ ⎟⎠⎞

⎜⎝⎛

+−=

+dx

xdx

xx

221

2

Cxx ++−= |2|ln2 ,


28. (a) )1(

)1(1 +

++=

++

xxBxxA

xB

xA

)1(

)(+++

=xx

AxBA

Q )1(

)()1(

1+++

≡+ xx

AxBAxx

∴ 1=A

10−=

=+BBA

(b) ∫ +dx

xx )1(1

constantany is where,|1|ln||ln

111

CCxx

dxxx

++−=

⎟⎠⎞

⎜⎝⎛

+−= ∫

29. (a) )12)(1(

)1()12(121 −+

++−=

−+

+ xxxBxA

xB

xA

)12)(1()2(

)12)(1(2

−+−++

=

−+++−

=

xxABxBA

xxBBxAAx

Q )12)(1(

)2()12)(1(

3−+−++

≡−+ xx

ABxBAxx

x

∴ ⎩⎨⎧

=−=+

)2....(..........0)1...(..........32

ABBA

:)2()1( −

133

==

AA

Substituting A = 1 into (2),

101

==−

BB

(b) ∫ −+ )12)(1(3

xxx

∫∫

∫

−+

+=

⎟⎠⎞

⎜⎝⎛

−+

+=

dxx

dxx

dxxx

121

11

121

11

For the second integral, Let .12 −= xu ,2=′u .2dxdxu =′

∴ ∫ −+ )12)(1(3

xxx

Cxx

C

Cux

duu

dxx

+−++=

+++=

++

= ∫ ∫

|12|ln21|1|ln


,||ln21|1|ln

121

11

30. (a) 32 −

++ x

Bx

A

)3)(2(32)(

)3)(2(23

−+−++

=

−+++−

=

xxABxBA

xxBBxAAx

Q )3)(2(32)(

)3)(2(5

−+−++

≡−+ xx

ABxBAxx

∴ ⎩⎨⎧

=−=+

)2....(..........532)1........(..........0

ABBA

:)2(2)1( −×

1555)32(22

−=−=−=−−+

AA

ABBA

Substituting 1−=A into (1),

1010

==+−=+

BBBA



(b) ∫ −+dx

xx )3)(2(5


,|3|ln|2|ln3

12

1

C

Cxx

dxxx

+−++−=

⎟⎠⎞

⎜⎝⎛

−−

+−

= ∫

31. ∫ += dt

ttT 231

Let .31 2tu += ,6tu =′ .6tdtdtu =′

∴ ∫= duu

T 161

Ct

CCu

++=

+=

|31|ln61


2

When t = 0, T = 25.

25

|ln|6125

=

+=

C

C

∴ 25|31|ln61 2 ++= tT

When t = 10,

25|)10(31|ln61 2 ++=T

0.26= (cor. to 1 d. p.)

∴ The average temperature after ten years is C.0.26 °

32 – 33 (GCE Questions) 34 – 35 (HKCEE Questions) 36 – 38 (HKASLE Questions) Extended Questions p.251

39. Let ,)( 2 cbxaxxf ++=′

where a, b and c are constants. From the graph, .1)0( =′f ∴ cba ++= )0()0(1 2

∴ 1=c ∴ 1)( 2 ++=′ bxaxxf From the graph, 0)1( =′f and .0)1( =−′f

⎩⎨⎧

=+−=++

)2....(..........01)1.....(..........01

baba

:)2()1( −

002

==

bb

Substituting b = 0 into (1),

1

010−=

=++a

a

∴ 1)( 2 +−=′ xxf


3

)1()(3

2

CCxx

dxxxf

++−=

+−= ∫

Q 1)0( =f

∴ C++= 03

013

∴ 1=C

∴ 13

)(3

++= xxxf

Open-ended Question p.251

40. Consider .)2()( 10+= xxf

.)2(10)]([ 9+= xxfdxd

∴ ∫ ++=+ 1109 )2()2(10 Cxdxx ,


∫∫

∫

++++

=

++

=+++

dxxdxxx

CC

Cxdxxx

88

12

2

108

)2()2)(1(10

where

,10

)2()11()2(

2

10

10)2( Cx

++

=

∫ ++

+++ 3

98

9)2()2)(1( Cxdxxx ,


2

10

10)2( Cx

++

=

∴ ∫ ++

−+

=++ Cxxdxxx9

)2(10

)2()2)(1(910

8

where 32 CCC −=

(or other reasonable answers)

Documents

New Progress in Senior Mathematics Module 1 Book 1 (Extended … · 2015. 5. 9. · ∴ The indefinite integral is incorrect. 6.2 (a) ∫8 , where dx =8x+C C is any constant p.220