New Progress in Senior Mathematics Module 1 …...7 New Progress in Senior Mathematics Module 1 Book...

© Hong Kong Educational Publishing Co. 166

New Progress in Senior Mathematics Module 1 Book 1 (Extended Part) Solution Guide 7

7

pp.259 – 282 7.1 When ,1=x 31)1(2 =+=y p.259 When x = 4, 91)4(2 =+=y Q 12 += xy is non-negative continuous in the

interval .41 ≤≤ x

∴ ∫ =+4

1)12( dxx Area of the trapezium

18

2)14)(93(

=

−+=

7.2 1

343

13

4 ⎥⎥⎦

⎤

⎢⎢⎣

⎡=∫

xdxx p.262

20

41

42 44

=

−=

7.3 ∫−5

1 3

2 1 dxx

x p.263

22

1

5

2

1

52

5

13

)1(211ln

)5(215ln

21||ln

2||ln

1

−−+=

⎥⎦⎤

⎢⎣⎡ +=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

−−=

⎟⎠⎞

⎜⎝⎛ −=

−

−∫

xx

xx

dxxx

1294.1= (cor. to 4 d. p.)

7.4 9

1243

5 2=∫b

dxx p.263

9

12433

1 53=

⎥⎥⎦

⎤

⎢⎢⎣

⎡•

b

x

11

1241259

12499

5

3

3

33

==

=−

=−

bb

b

b

7.5 (a) ∫∫∫ +=5

3

3

0

5

0)()()( dxxgdxxgdxxg p.266

∫ −+=3

0)2()(4 dxxg

∫ −−=3

0)2(4)( dxxg

6=

∫∫ −3

5

3

0)()( dxxgdxxg

4

)2(6

)()(3

0

5

3

=−+=

⎟⎠⎞

⎜⎝⎛−−= ∫ ∫ dxxgdxxg

(b) They are the same.

7.6 ∫−1

3|| dxx p.267

521

29

22

)(

||||

0

12

3

02

0

3

1

0

0

3

1

0

=

+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

+−=

+=

−

−

−

∫ ∫

∫ ∫

xx

xdxdxx

dxxdxx

7.7 ∫− +−2

134 )42( dxxx p.268

10111

][44

25

42

12

1

24

1

25

2

1

2

1

2

134

=

+⎥⎥⎦

⎤

⎢⎢⎣

⎡−

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

+−=

−−−

− − −∫ ∫ ∫

xxx

dxdxxdxx

7.8 ∫ +−5

0]2)(3)(4[ dxxgxf p.268

44

][2)2(3)7(4

2)(3)(4

05

5

0

5

0

5

0

=

+−−=

+−= ∫ ∫ ∫dxx

dxdxxgdxxf

Definite Integrals

167 © Hong Kong Educational Publishing Co.

Definite Integrals

7.9 Let ,12 += xu then .2xdxdu = p.271

∴ ∫ ∫=+ duudxxx 662

21)1(

Cx

Cu

++

=

+= •

14)1(

721

72

7

∴ ∫ ⎥⎥⎦

⎤

⎢⎢⎣

⎡ +=+

2

11

27262

14)1()1( xdxxx

14997 77

=

7.10 Let .12 += xu Then dxdu 2= p.273

When x = 0, u = 1. When x = 4, u = 9.

∴ ∫ ∫−

=+

4

0

9

121

21

121 duudxx

2

]2[21

192

1

=

= u

7.11 Let .ln xu = Then dxx

du 1= p.273

When x = 1, u = 0. When x = e, u = 1.

∫ ∫=e

duudxxx

1

1

03

3)(ln

41

4 0

14

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

u

7.12 (a) Let .3 xu −= Then dxdu −= and .3 ux −= p.274

When x = 0, u = 3. When x = 3, u = 0.

∴ ∫ ∫ −−=3

0

0

3)3()( duufdxxf

∫

∫−=

−=

3

0

3

0

)3(

)3(

dxxf

duuf

(b) Let .)3()( 5−= xxf Then .)3( 5xxf =−

∫ ∫−=−3

0

3

055)3( xdxdxx (from (a))

2243

6 0

36

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

x

7.13 We obtain the corresponding values of p.278 )( ixf for some values of .ix

ix 1 1.2 1.4 1.6 1.8 2

)( ixf 21 22.1 24.1 26.1 28.1 22

The sum of the areas of the trapeziums

251)28.1(

251)8.16.1(

251)6.14.1(

251)4.12.1(

251)2.11(

2222

22222

++

++

++

++

+=

34.2= (cor. to 3 sig. fig.)

∴ An approximation of the integral 34.22

12 ≈∫ dxx

Actual value

1

23

2

12

3 ⎥⎥⎦

⎤

⎢⎢⎣

⎡=

= ∫x

dxx

37

31

38

=

−=

7.14 The width of each subinterval is p.280

2.05

12=

− .

The end points of the subintervals are 1, 1.2, 1.4, 1.6, 1.8 and 2.

dxx∫2

1ln

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+++

≈

8.16.1ln

4.1ln2.1ln2

2ln1ln2.0

5751.0= (cor. to 4 d. p.)

7.15 (a) The width of each subinterval is p.282

2.05

01=

− .

The end points are 0, 0.2, 0.4, 0.6, 0.8 and 1. 1I

222

4.042.042

14042.0 −+−+⎢⎢⎣

⎡ −+−=

⎥⎥⎦

⎤−+−+ 22 8.046.04

9113.1= (cor. to 4 d. p.)



(b) )2(42

1)4(2

2 xx

xdxd

−−

=−

24 x

x

−−=

)4( 22

2x

dxd

−

23

2

2

2

22

2

2

2

)4(

4

44

4

44

)(4

x

xx

xx

xx

xxxdxdx

−

−=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−−

+−

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−−−−

−=

(c) Since ,10 ≤≤ x

0)4( 23

2 >− x

∴ 04 22

2<− x

dxd

∴ The approximation 1I is an underestimate

of I. pp.259 – 281 Example 7.1T p.259

21 xy −= is non-negative in the interval .10 ≤≤ x

∫ =−1

021 dxx area of the quarter

4π

π(1)41 2

=

×=

Example 7.2T p.262

3

55

3

43 4

4)4(∫ ⎥

⎥⎦

⎤

⎢⎢⎣

⎡−=− xxdxx

128

)3(44

3)5(44

5 44

=⎥⎥⎦

⎤

⎢⎢⎣

⎡−−

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

Example 7.3T p.262

∫ −4ln

3ln23 )( dxee x

334

333144

31

31

3131

33

3ln3ln34ln4ln3

3ln

4ln3

=

⎟⎠⎞

⎜⎝⎛ −×−⎟

⎠⎞

⎜⎝⎛ −×=

+−−=

⎥⎦⎤

⎢⎣⎡ −=

eeee

ee xx

Example 7.4T p.263

∫ =+2

12 3)3( dxax

3][ 123 =+ axx

43)1()22( 3

−==+−+

aaa

Example 7.5T p.266

(a) ∫ ∫∫ +=7

1

7

3

3

1)()()( dxxhdxxhdxxh

5)(

)(510

7

3

7

3

=

+=

∫

∫dxxh

dxxh

(b) ∫ ∫∫ ++1

3

7

1

3


0

)()()()(7

3

3

1

7

3

3

1

=

++⎟⎠⎞

⎜⎝⎛−+−= ∫∫∫∫ dxxhdxxhdxxhdxxh

Example 7.6T p.267

26188

22

)(||

0

62

4

02

6

0

0

4

6

4

=+=

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡−=

+−=

−

−− ∫∫∫xx

xdxdxxdxx

Example 7.7T p.268

35

][42

74

2

472)472(

23

2

32

2

34

3

2

3

2

3

2

33

2

3

=

+⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡=

+−=+−

−−−

−−−− ∫∫∫∫

xxx

dxxdxdxxdxxx


Definite Integrals

Example 7.8T p.268

722116

)1(2)3(7][4

)(2)(74)](2)(74[

26

6

2

6

2

6

2

6

2

−=−−=

−+−=

+−=+− ∫∫∫∫x

dxxgdxxfdxdxxgxf

Example 7.9T p.271

Let ,ln xu = .1 dxx

du =

∴ ∫ ∫= ududxxxln

Cx

Cu

+=

+=

2)(ln

22

2

5.0

2)(lnln1

1

2

1

=⎥⎥⎦

⎤

⎢⎢⎣

⎡=∫

ee xdxxx

Example 7.10T p.273

Let ,54 2 += xu .8xdxdu =

When x = 1, u = 9. When x = 2, u = 21.

∴ ududxx

x∫∫ −=

+

21

922

1 22 81

)54(

1261

91

211

81

181

9

21

=

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−−−=

⎥⎦⎤

⎢⎣⎡−=

u

Example 7.11T p.273

Let ),1ln( += xu .1

1 dxx

du+

=

When x = 1, .2ln=u When x = 3, .4ln=u

∫∫ =++ 4ln

2ln

3

1 1)1ln( ududx

xx

2

22

22

2ln

4ln2

)2(ln23

2)2(ln)2(ln4

2)2(ln)4(ln

2

=

−=

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

u

Example 7.12T p.274

(a) Let ,1+= xu ,dxdu = .1−= ux

When x = 0, u = 1. When x = 1, u = 2.

∴ udufdxxf ∫∫ −=2

1

1

0)1()(

∫ −=2

1)1( dxxf

(b) Let ,)1()( 6xxf += .)1( 6xxf =−

∫∫ =+2

161

06)1( dxxdxx (from (a))

7127

7 1

27

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

x

Example 7.13T p.277

The table shows the corresponding values of )( ixf for some values of .ix

ix 1 1.4 1.8 2.2 2.6 3

)( ixf 1 4.1 8.1 2.2 6.2 3

The sum of areas of the trapeziums

252)36.2(

252)6.22.2(

252)2.28.1(

252)8.14.1(

252)4.11(

++

++

++

++

+=

7946.2= (cor. to 4 d. p.)

∴ An approximation of the integral 7946.23

1≈= ∫ dxx

Actual value

3232

32

1

323

3

1

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

= ∫

x

dxx



Example 7.14T p.280

The width of each subinterval is .4.05

24=

−

The end points are 2, 2.4, 2.8, 3.2, 3.6 and 4.

∴ ∫ +

4

2 11 dx

ex

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

++

++

++

+++

++

=

11

11

11

11

21

11

1

4.0

6.32.3

8.22.2

42

ee

eeee

1099.0= (cor. to 4 d. p.)

Example 7.15T p.281

(a) The width of each subinterval is .1.05

5.01=

−

The end points are 0.5, 0.6, 0.7, 0.8, 0.9 and 1.

∴

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

++

+++

=

9.01ln

8.01ln

7.01ln

6.01ln

2

1ln5.0

1ln

1.01I

1543.0= (cor. to 4 d. p.)

(b) )ln(1ln xdxd

xdxd

−=⎟⎠⎞

⎜⎝⎛

22

2 11ln

1

xxdxd

x

=⎟⎠⎞

⎜⎝⎛

−=

(c) Since 15.0 ≤≤ x 125.0 2 ≤≤ x

114 2 ≥≥x

∴ 01ln2

2>⎟

⎠⎞

⎜⎝⎛

xdxd

∴ The approximation 1I is an overestimate of I.

7.1 pp.263 – 265

p.263

1. (a) ∫− =2

26dx Area of the rectangle

24

646)]2(2[

=×=

×−−=

(b) When x = 8, y = 4.

∫8

0 21 xdx = Area of the triangle

16

284

=

×=

(c) When x = 4, y = 8. When x = 8, y = 4.

∫ −8

4)12( dxx = Area of the trapezium

24

2)48)(84(

=

−+=

(d) ∫− −2

124 dxx

= Area of the semicircle

π2

)2(π21 2

=

=

2. 255

2)]([)( xFdxxf =∫

6

410)2()5(

=−=−= FF

6)()(5

2

5

2==∫∫ dxxfduuf

3. 11)]([)( ee

xFdxxf =∫

4

04)1()(

=−=

−= FeF

4. )( xxedxd

xx exe +=

∫ +3

0)( dxxee xx

3

03

3

][

e

xex

=

=

5. 1

525

1 2 ⎥⎥⎦

⎤

⎢⎢⎣

⎡=∫

xxdx

12

21

225

=

−=


Definite Integrals

6. 2323

2][)12( xxdxx −=−∫

4

)22()33( 22

=−−−=

7. 1232

12 ][3 −−

=∫ xdxx

9

)1(2 33

=−−=

8. 0

4234

0

2

23

32)32( ⎥

⎦

⎤⎢⎣

⎡+=+∫

xxdxxx

3200

2)4(3

3)4(2 23

=

+=

9. ∫3

2 31 dxx

725

)2(21

)3(21

21

2

22

2

3

2

3

2

2

3

23

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−=

⎥⎦⎤

⎢⎣⎡−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

=

−

−∫

x

x

dxx

10. ∫ −4

1)1( dxxx

227

21

31

216

364

23

)(

1

423

4

12

=

⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

−= ∫xx

dxxx

11. ∫ −+5

2)2)(1( dxxx

245

)2(22

232)5(2

25

35

223

)2(

2323

2

523

5

22

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=

−−= ∫

xxx

dxxx

12. 11|]|ln2[2 ee

xexdxx

e −=⎟⎠⎞

⎜⎝⎛ −∫

ee

eeee−−=

−−−= ••

2)1ln21()ln2(

2

13. dxx

dxx

xx∫∫ ⎟

⎠⎞

⎜⎝⎛ −=

− 4

2

4

2 2

2133

22ln3

)22ln3()44ln3(

]||ln3[ 24

−=−−−=

−= xx

14. dxxx

dxx

x∫∫ ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛+=

+ −4

121

4

1

11

24ln)12)1(ln()424(ln

]2 ||[ln

]2 ||[ln

41

142

1

+=−−−=

+=

+=

xx

xx

15. 0

3ln23ln

02

21

⎥⎦⎤

⎢⎣⎡=∫ xx edxe

4

)(21 03ln 2

=

−= ee

16. 0

2ln32ln

03

31)1( ⎥⎦

⎤⎢⎣⎡ +=+∫ xedxe xx

2ln37

312ln

38

0312ln

31 02ln 3

+=

−+=

⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ += ee

p.264

17. 1

2

2

2

1 3 )13(61

)13(1

⎥⎦

⎤⎢⎣

⎡

+−=

+∫ xdx

x

156811

]1)1(3[61

]1)2(3[61

22

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−−

+−=

18. 1535

1]1)1[()( −+=∫ xxxxf

432

112156 33

=−−−=



19. 10)0(3

1=+∫ dxax

2

82

8

1012

32

9

102 1

32

=

=

=⎟⎠⎞

⎜⎝⎛ +−⎟

⎠⎞

⎜⎝⎛ +

=⎥⎥⎦

⎤

⎢⎢⎣

⎡+

a

a

aa

xax

20. 3

38)2(1

2 =+∫b

dxx

3

3823 1

3=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

bxx

0)153)(3(

0456

1523

3382

312

3

2

3

3

3

=++−

=−+

=+

=⎟⎠⎞

⎜⎝⎛ +−⎟⎟

⎠

⎞⎜⎜⎝

⎛+

bbb

bb

bb

bb

∴ 3=b or )1(2

)15)(1(433 2 −±−=b

(rejected) 2

513 −±−=

21. 61)1(

0−=−∫

cdxxx

1or 21

013261

632

61

23

61

23

61)(

3

23

23

0

23

0

2

−=

=+−

−=−

−=−

−=⎥⎦

⎤⎢⎣

⎡−

−=−∫

c

cc

cc

cc

xx

dxxx

c

c

22. 221))(1(

5

2=++∫ dxpxp

221)1(

2)1(

221)]1()1[(

2

52

5

2

=⎥⎥⎦

⎤

⎢⎢⎣

⎡++

+

=+++∫

xppxp

dxppxp

29or 0

0)92(0927)1(2)1(7221)1(3)1(

221

221)]1(2)1(2[)1(5

2)1(5

2

22

−=

=+=+

=+++

=+++

=+++−⎥⎦

⎤⎢⎣

⎡++

+

p

pppp

ppp

ppp

pppppp

7.2 pp.269 – 270

p.269

1. ∫ −+−2

132 )254( dxxxx

20531

][23

545

4

254

12

1

22

1

24

1

25

2

1

2

1

2

132

14

=

−⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

−+−= ∫ ∫∫∫

xxxx

dxxdxdxxdxx

2. ∫− +++1

2

23 )123( dxxxx

415

][23

24

3

23

21

2

12

2

13

2

14

1

2

1

2

1

221

23

−=

+⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

+++=

−−−

− −−− ∫ ∫∫∫

xxxx

dxxdxdxxdxx

3. ∫ ++−1

0

2 )4476( dxxxe x

376

]4[2

43

7][6

4476

01

0

12

0

13

01

1

0

1

0

1

021

0

−=

+⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

++−= ∫ ∫∫∫

e

xxxe

dxxdxdxxdxe

x

x

4. ∫− −+0

232 )373( dxxxe x

241

23

23

47

213

373

4

2

02

2

04

2

02

0

2

0

230

22

−−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎥⎦

⎤⎢⎣⎡=

−+=

−−−

−−− ∫∫∫

e

xxe

xdxdxxdxe

x

x


Definite Integrals

5. ∫∫ =4

2

4

2|| xdxdxx

6

28

2 2

42

=−=⎥⎥⎦

⎤

⎢⎢⎣

⎡=

x

6. ∫∫∫ −−+−=

2

0

0

2

2

2)(|| xdxdxxdxx

4

22

22 0

22

2

02

=+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

−

xx

7. (a) ∫∫∫ +=6

2

2

1

6

1)()()( dxxfdxxfdxxf

734

=+=

(b) 0)(6

6=∫ dxxf

8. (a) ∫∫∫ +=6

3

3

0

6

0)()()( dxxgdxxgdxxg

321

=+=

(b) ∫∫ =6

3

6

3)(4)(4 dxxgdxxg

824

=×=

(c) ∫∫∫ −=−6

6

6

6

6

611)(3]11)(3[ dxdxxgdxxg

00)0(3

=−=

(d) ∫ −6

6])(2[ dxxxg

0

0)0(2

)(26

6

6

6

=−=

−= ∫∫ xdxdxxg

9. (a) ∫ −7

3)](2)([ dxxgxf

22

)8(26

)(2)(7

3

7

3

−=−−=

−= ∫∫ dxxgdxxf

(b) ∫ +7

3)](5)(4[ dxxgxf

164024

)8(5)6(4

)(5)(47

3

7

3

=+−=+−=

−= ∫∫ dxxgdxxf

10. (a) ∫ ∫ ∫− −+=

4

1

2

1

4


∫ ∫ ∫− −−=

2

1

4

1

4


4610

=−=

(b) ∫− +2

1)](3)(2[ dxxgxh

29

7342

)(3)(22

1

2

1

=×−×=

+= ∫∫ −−dxxgdxxh

11. (a) ∫ ∫ ∫+=7

0

5

0

7


∫ ∫ ∫−=7

5

7

0

5


3710

=−=

(b) ∫ ∫ ∫+=+7

0

7

0

7

022 )(2)](2[ dxxfdxxdxxfx

3403

63

343

)3(23 0

73

=

+=

+⎥⎥⎦

⎤

⎢⎢⎣

⎡=

x

(c) ∫ ∫ ∫−=−7

5

7

5

7

5)(])([ xdxdxxfdxxxf

9225

2493

23

5

72

−=

⎟⎠⎞

⎜⎝⎛ −−=

⎥⎦

⎤⎢⎣

⎡−=

x

(d) ∫ ∫=7

0

7

0)()( dxxfduuf

10=



p.270

12. ∫ ∫ •+ =2

1

2

1)(1)( edxedxe xfxf

ke

dxee xf

=

= ∫2

1)(

13. ∫−1

1)( dxxf

0

)()(

)()(

1

0

1

0

0

1

1

0

=

⎥⎦⎤

⎢⎣⎡−+=

+=

∫ ∫

∫∫ −

dxxfdxxf

dxxfdxxf

14. ∫−2

1)( dxxf

6

321

)()()(0

1

1

0

2

1

=++=

++= ∫∫∫ −dxxfdxxfdxxf

15. (a) ∫ −0

2)]()([ dxxfxg

451

)()(

)()(

0

2

2

0

0

2

0

2

=+−=

⎥⎦⎤

⎢⎣⎡−−−=

−=

∫ ∫

∫∫dxxfdxxg

dxxfdxxg

(b) ∫ +2

0)](5)(4[ dxxgxf

25)1(5)5(4

)(5)(4

)(5)(4

2

0

0

2

2

0

2

0

=+=

⎥⎦⎤

⎢⎣⎡−+=

+=

∫ ∫

∫∫dxxgdxxf

dxxgdxxf

16. (a) ∫ ++8

3 )()()()( dx

xgxfxgxf

538

][ 38

8

3

=−=

=

= ∫x

dx

(b) ∫ +

8

3 )()()( dx

xgxfxg

145

)()()(

)()()()(

)()()()()(

8

3

8

3

8

3

=−=

+−

++

=

+−+

=

∫∫

∫

dxxgxf

xfdxxgxfxgxf

dxxgxf

xfxgxf

17. 7)]()([2

1=+∫ dxxgxf

∫ ∫

∫

∫∫

=−

=−

=+

2

1

2

1

2

1

2

1

2

1

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

1)]()([

)1.......(..........7)()(

dxxgdxxf

dxxgxf

dxxgdxxf

:)2()1( +

4)(

8)(2

2

1

2

1

=

=

∫

∫dxxf

dxxf

:)2()1( −

3)(

6)(2

2

1

2

1

=

=

∫

∫dxxg

dxxg

18. 19)](3)(2[2

2=+∫ dxxgxf

∫ ∫

∫

∫∫

− −

−

−−

−=−

−=−

=+

2

2

2

2

2

2

2

2

2

2

)2......(..........7)(3)(4

7)](3)(4[

)1.......(..........19)(3)(2

dxxgdxxf

dxxgxf

dxxgdxxf

:)2()1( +

2)(

12)(6

2

2

2

2

=

=

∫

∫

−

−

dxxf

dxxf

Substituting 2)(2

2=∫− dxxf into (2),

7)(3)2(42

2−=− ∫− dxxg

∫− =2

215)(3 dxxg

∫− =2

25)( dxxg


Definite Integrals

7.3 pp.274 – 276

p.274

1. (a) 1

0100

19

10 −− ⎥

⎥⎦

⎤

⎢⎢⎣

⎡=∫

uduu

101

10)1(010

−=

−−=

(b) Let ,12 −= xu .2xdxdu =

When x = 0, .1−=u

When x = 1, u = 0.

∫∫ =−0

1

91

0

92

21)1( duudxxx

201

101

21

−=

⎟⎠⎞

⎜⎝⎛−=

2. (a) 188

1][ uu edue =∫

ee −= 8

(b) Let ,3xu = .3 2dxxdu =

When x = 1, u = 1. When x = 2, u = 8.

∴ ∫∫ =8

1

2

12

312

duedxex ux

)(31 8 ee −=

3. (a) ∫ ==2

1 12 2ln][ln1 udu

u

(b) Let ,15 += xu .5 4dxxdu =

When x = 0, u = 1. When x = 1, u = 2.

∴ ∫∫ =+

2

0

1

0 5

4 151

1du

udx

xx

2ln51

=

4. Let ,32 += xu .2dxdu =

When x = 0, u = 3. When x = 1, u = 5.

∫∫ =+5

341

04

21)32( duudxx

51441

535

21

521

55

3

55

=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

u

5. Let ,23 −= xu .3dxdu =

When x = 1, u = 1. When x = 2, u = 4.

∫∫ =−4

1

2

1 3123 duudxx

914

)14(92

32

31

23

23

1

423

=

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡= u

6. Let ,4+= xu .dxdu =

When x = 4, u = 8. When x = 23, u = 27.

∫∫ =+27

831

23

43 4 duudxx

4195

)827(43

43

34

34

8

2734

=

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡= u

7. Let ,2+= xu .dxdu = When ,1−=x u = 1. When ,2−= ex u = e.

∫∫−

−=

+

eedu

udx

x 1

2

1

12

1

1

1ln1

][ln 1

=−=

= eu



8. Let ,34 −= xu .4dxdu =

When x = 1, u = 1. When x = 2, u = 5.

∫∫ =−

5

1

2

1

141

341 du

udx

x

45ln

][ln41

15

=

= u

9. Let ,92 += xu .2xdxdu =

When x = 0, u = 9. When x = 4, u = 25.

∫∫ =+25

921

4

02

219 duudxxx

398

)925(31

32

21

23

23

9

2523

=

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡= u

10. Let ,13 += xu .3 2dxxdu =

When x = 0, u = 1. When x = 2, u = 9.

∫∫ =+9

121

2

032

311 duudxxx

952

)19(92

32

31

23

23

1

923

=

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡= u

11. Let ,4 xu −= .dxdu −=

When x = 2, u = 2. When x = 4, u = 0.

∴ ∫∫ −=− 0

2

4

2 4 duedxe ux

1

][

2

02

2

0

−=

=

= ∫

e

e

due

u

u

12. Let ,2xu = .2xdxdu =

When x = 0, u = 0. When x = 1, u = 1.

∫∫ =1

0

1

0 212

duedxxe ux

)1(

21

][21

01

−=

=

e

eu

13. Let ,12 −= xu .2xdxdu =

When x = 1, u = 0. When x = 2, u = 3.

∴ ∫∫ =−

3

0

2

1 2

121

1du

udx

x

x

3

]2[21

032

1

=

= u

14. Let ,4 2xu −= .2xdxdu −=

When x = 0, u = 4. When x = 2, u = 0.

∴ ∫∫ ×−=−

0

4

2

0 2

1221

4

2 duu

dxx

x

4

]2[ 04

4

021

=

=

= ∫−

u

duu

15. Let ,1x

u = .12 dx

xdu −=

When x = 2, .21

=u

When x = 3, .31

=u

∴ ∫∫ −= 31

21

3

2 2

1

duedxxe u

x

31

21

3121

21

31

][

ee

e

due

u

u

−=

=

= ∫


Definite Integrals

16. Let ,xu = .2

1 dxx

du =

When x = 1, u = 1. When x = 4, u = 2.

∴ ∫∫ =2

1

4

12 duedx

xe u

x

)(2

][2

2

12

ee

eu

−=

=

17. Let ,52 xu −= .5dxdu −=

When x = 0, u = 2. When x = 1, .3−=u

∴ ∫∫−

⎟⎠⎞

⎜⎝⎛×=

−

3

2 2

1

0 21

513

)52(3 du

udx

x

21

153

153

3

2

2

3 2

−=

⎥⎦

⎤⎢⎣

⎡−=

=

−

−∫

u

duu

18. Let ,31 2xu −= .6xdxdu −=

When x = 0, u = 1. When x = 2, .11−=u

∴ ∫∫−

⎟⎠

⎞⎜⎝

⎛−×=−

11

1 2

2

0 22

1614

)31(4 du

udx

xx

118

132

11

1

−=

⎥⎦⎤

⎢⎣⎡−=

−u

p.275

19. Let ,ln xu = .1 dxx

du −=

When x = e, u = 1. When ,2ex = u = 2.

∴ ∫∫ −=2

1

1ln12

duu

dxxx

e

e

2ln

|]|[ln 12

=

= u

20. Let ,1−= xu .dxdu =

x = u + 1. When x = 1, u = 0. When x = 2, u = 1.

∴ ∫∫ ++−=−+1

082

18 )21()1)(2( duuudxxx

2013

31

101

310

)3(

0

1910

1

089

=

+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

+= ∫uu

duuu

21. ∫ +++1

0 2

2

11 dx

xxx

∫∫ ++=

1

0 2

1

0 1dx

xxdx

For the second integral, let ,12 += xu .2xdxdu =

When x = 0, u = 1. When x = 1, u = 2.

∴ ∫ ∫∫ +=+++ 1

0

2

1

1

0 2

2 121

11 du

udxdx

xxx

22ln1

|]|[ln21][ 1

201

+=

+= ux

22. Let .22 xxu += .)1(2)22( dxxdxxdu +=+=

When x = 1, u = 3. When x = 2, u = 8.

∴ ∫∫ =++ 8

3

2

1 21

21

21 du

udx

xxx

)3ln8(ln

21

|]|[ln21

38

−=

= u

23. Let ,1+= xu ,dxdu = .1−= ux

When x = 0, u = 1. When x = 2, u = 3.

∴ ∫∫−−

=+− 3

1

2

0

1111 du

uudx

xx

3ln2213ln23

|]|ln2[

21

2

13

3

1

3

1

−=−−=

−=

⎟⎠⎞

⎜⎝⎛ −=

−=

∫

∫

uu

duu

duu

u



24. Let ,1+= xu ,dxdu = .1−= ux

When x = 0, u = 1. When x = 3, u = 4.

∴ ∫∫−

=+

4

1 2

3

0 21

)1(du

uudx

xx

434ln

1414ln

1||ln

11

1

4

4

1 2

−=

−+=

⎥⎦⎤

⎢⎣⎡ +=

⎟⎠⎞

⎜⎝⎛ −= ∫

uu

duuu

25. Let ,1−= xu ,dxdu = .1+= ux

When x = 2, u = 1. When x = 3, u = 2.

∴ ∫ −+3

2

2

12 dx

xx

2ln327

1ln3)1(22

12ln3)2(22

2

||ln322

32

32

2)1(

22

1

22

2

1

2

1

2

2

1

2

+=

⎥⎥⎦

⎤

⎢⎢⎣

⎡++−

⎥⎥⎦

⎤

⎢⎢⎣

⎡++=

⎥⎥⎦

⎤

⎢⎢⎣

⎡++=

⎟⎠⎞

⎜⎝⎛ ++=

++=

++=

∫

∫

∫

uuu

duu

u

duu

uu

duu

u

26. Let ,1+= xu ,dxdu = .1−= ux

When x = 0, u = 1. When x = 3, u = 4.

∴ ∫∫ −−=+4

1

3

0)1(1 duuudxxx

15116

32

524

324

52

32

52

)(

23

25

1

423

25

4

121

23

=

⎟⎠⎞

⎜⎝⎛ +−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛×−×=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

−= ∫

uu

duuu

27. Let ,12 −= xu ,2xdxdu = .12 += ux

When ,2=x u = 1.

When ,5=x u = 4.

∴ ∫∫−

=−

•5

2 2

25

2 2

3

11dx

x

xxdxx

x

310

232424

32

21

232

21

121

121

23

1

423

4

1

4

1

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+×=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+=

∫

∫

uu

duu

u

duu

u

28. (a) Let ,1−= xu ,dxdu = .1+= ux

When x = 1, u = 0. When x = 5, u = 4.

∴ ∫∫ +=4

0

5

1)1()( duufdxxf

∫ +=4

0)1( dxxf

(b) Let .)1()( 6−= xxf Then .)1( 6xxf =+

∴ ∫∫ =−4

065

16)1( dxxdxx

7384 16

7 0

47

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

x

29. (a) Let ,6 xu −= ,dxdu −= .6 ux −=

When x = 2, u = 4. When x = 4, u = 2.

∴ ∫∫ −−=2

4

4

2)6()( duufdxxf

∫

∫−=

−=

4

2

4

2

)6(

)6(

dxxf

duuf

(b) Let .6)( 3 xxf −= Then .)6( 3 xxf =−

∴ ∫∫ =−4

234

23 6 dxxdxx

)242(23

43

33

2

434

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡= x


Definite Integrals

30. (a) Let ,xu −= ,dxdu −= .ux −= When ,1−=x u = 1.

When x = 0, u = 0.

∴ ∫∫ −−=−

0

1

0

1)()( duufdxxf

∫

∫−=

−=

1

0

1

0

)(

)(

dxxf

duuf

(b) ∫− −−

1

1

2

xx eex

0

(a)) (from )(

1

0

1

0

22

1

0

1

0

22

1

0

1

0 )(

22

0

1

21

0

2

=−

−−

=

−+

−=

−−

+−

=

−+

−=

∫ ∫

∫ ∫

∫ ∫

∫∫

−−

−−

−−−−

− −−

dxee

xdxee

x

dxee

xxdxee

x

dxeexdx

eex

dxee

xdxee

x

xxxx

xxxx

xxxx

xxxx

31. (a) Let ,8+= xeu .dxedu x=

When x = 0, u = 9. When ,2ln=x u = 10.

∴ ∫∫ =+

20

9

2ln

0

18

duu

dxe

ex

x

9ln10ln

|]|[ln 910

−=

= u

(b) x

x

x

x

xx ee

ee

ee −

−

−

−•

+=

+=

+ 8181

181

∴ ∫∫ − −

−

− +=

+

0

2ln

0

2ln 8181 dx

ee

e x

x

x

Let ,xu −= .dxdu −= When ,2ln−=x .2ln=u

When x = 0, u = 0.

∴ ∫− +

0

2ln 181 dx

ex

9ln10ln8

82ln

0

0

2ln

−=+

=

+−=

∫

∫

due

e

due

e

u

u

u

u

7.4 pp.282 – 283

p.282

1. ⎟⎠⎞

⎜⎝⎛ ++++

+≈∫ 2464

2022.0)(

1

0dxxf

4.3=

2. ∫3

0)( dxxf

875.3

26.12.116.02

4.23.05.0

=

⎟⎠⎞

⎜⎝⎛ +++++

+≈

3. ∫4

1)( dxxf

195.3422.1)9.0()1.0(

3.14.28.12.1)1.0(2

43.03.0

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+++−+−+

++++−++−

≈

4. The width of each subinterval is .2.05

01=

−


∴ ∫ +1

031 dxx

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+++++++

≈3

333

8.01

6.014.012.012

012.0

1150.1= (cor. to 4 d. p.)


05.0=

−

The end points are 0, 0.1, 0.2, 0.3, 0.4 and 0.5.

∴ ∫ −5.0

0

2

dxe x

⎥⎥⎦

⎤

⎢⎢⎣

⎡++++

+≈ −−−−

−2222

22

4.03.02.01.05.00

22.0 eeeeee

4606.0= (cor. to 4 d. p.)


24=

−

The end points are 2, 2.4, 2.8, 2.8, 3.2, 3.6 and 4.

∴ ∫4

2 ln1 dxx

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

++

+++

≈

6.3ln1

2.3ln1

8.2ln1

4.2ln1

24ln

12ln

1

4.0

9344.1= (cor. to 4 d. p.)




12=

−


∴ ∫2

1dx

xex

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

+++++

≈8.16.14.12.12

22.08.16.14.12.1

2

eeeeee

0653.3= (cor. to 4 d. p.)


5.12=

−


∴ ∫+2

5.1

5 dxe

xx

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

++

++

++

++

++

+

≈

9.18.1

7.16.1

25.1

9.158.15

7.156.152

255.15

1.0

ee

eeee

5911.0= (cor. to 4 d. p.)


)1(1=

−−

The end points are ,1− ,6.0− ,2.0− 0.2, 0.6 and 1.

∴ ∫− +1

12169 dxx

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

++++

−++−++

++−+

≈

22

22

22

)6.0(169)2.0(169

)2.0(169)6.0(169

2)1(169)1(169

4.0

5574.7= (cor. to 4 d. p.)


5.01=

−


∴ ∫ +1

5.02 )1ln( dxx

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++++++

+++++

=

)19.0ln()18.0ln()17.0ln(

)16.0ln(2

)11ln()15.0ln(1.0

222

222

2252.0= (cor. to 4 d. p.)


12=

−


∴ ∫+

2

1 2 2

1 dxx

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

++

++

++

+++

++

≈

28.1

1

26.1

1

24.1

122.1

12

22

1

21

1

2.0

222

2

22

4879.0= (cor. to 4 d. p.)


23=

−


∴ ∫3

2ln dxxx

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+++

≈

8.2ln8.26.2ln6.2

4.2ln4.22.2ln2.22

3ln32ln22.0

3973.2= (cor. to 4 d. p.)

p.283

13. (a) The width of each subinterval is .25.04

01=

−

The end points are 0, 0.25, 0.5, 0.75 and 1. ∴ The approximation 1I

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++++

+++++

≈

)1()1(

)1(2

)1()1(25.0

75.05.0

25.00

ee

eee

7272.2= (cor. to 4 d. p.)

(b) xx eedxd

=+ )1(

∴ xx eedxd

=+ )1(2

2

(c) Since ,10 ≤≤ x

eex ≤≤1

∴ 0)1(2

2>+xe

dxd

∴ The approximation 1I is an overestimate

of I.


Definite Integrals


02=

−

The end points are 0, 0.5, 1, 1.5 and 2. ∴ The approximation 1I

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++++

+++++

≈44

444

5.1111

5.012

21015.0

7344.3= (cor. to 4 d. p.)

(b) )4(12

11 3

4

4 xx

xdxd

+=+

4

3

1

2

x

x

+=

∴ 42

21 x

dxd

+

23

4

42

4

4

3324

)1(

)3(21

1

22)6(1

x

xxx

x

xxxx

+

+=

+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+−+

=

(c) Since ,20 ≤≤ x

160

404

2

≤≤

≤≤

x

x

∴ 0)3(2 42 >+xx

∴ 0)1( 42

2>+ x

dxd


of I.


23=

−

The end points are 2, 2.2, 2.4, 2.6, 2.8 and 3. ∴ The approximation 1J

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

++++

++++

+++

≈

)38.2ln()36.2ln()34.2ln()32.2ln(

2)33ln()32ln(

2.022

22

22

2213.2= (cor. to 4 d. p.)

(b) )2(3

1)]3[ln( 22 x

xx

dxd

+=+

3

22 +

=x

x

∴ 22

22

2

2

)3()2(2)2)(3()]3[ln(

+−+

=+x

xxxxdxd

22

2

)3(26+

−=

xx

(c) Since ,32 ≤≤ x

122610

18216

94

2

2

2

−≥−≥−

−≥−≥−

≤≤

x

x

x

∴ 0)]3[ln( 22

2<+x

dxd

∴ The approximation 1J is an underestimate

of I.


01=

−

The end points are 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 and 1.

∴ The approximation 1I

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+++++

+++++

≈

29.0

28.0

27.0

26.0

25.0

24.0

23.0

22.0

21.02

120

22222

223222

21.0

eeeee

eeeeee

1963.1= (cor. to 4 d. p.)

(b) xeedxd xx

22

22

)( =

∴ 22222

2222

)(xxx

exeedxd

+=

)1( 22

2

xex

+=

(c) Since 10 ≤≤ x 10 2 ≤≤ x 211 2 ≤+≤ x

∴ 0)( 22

22

>x

edxd


of I. pp.287 – 289

p.287

1. ∫ ++−3

123 )243( dxxxx

3100

][223

44

3 13

1

32

1

33

1

34

=

+⎥⎥⎦

⎤

⎢⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

⎥⎥⎦

⎤

⎢⎢⎣

⎡= xxxx



2. 0ln0

0 2

2=∫ dx

exx

3. ∫−−−

1

1)32( dxee xx

e

e

ee xx

−=

+= −−

1

]32[ 11

4. ∫ ⎟⎠⎞

⎜⎝⎛ ++

3

2 2321 dxxx

23ln2

23

3||ln22

3

+=

⎥⎦⎤

⎢⎣⎡ −+=

xxx

5. ∫ ⎟⎠⎞

⎜⎝⎛ −⎟⎠⎞

⎜⎝⎛ +

2

1

1111 dxxx

21

1

11

1

2

2

1 2

=

⎥⎦⎤

⎢⎣⎡ +=

⎟⎠⎞

⎜⎝⎛ −= ∫

xx

dxx

6. ∫ ∫ ⎟⎠⎞

⎜⎝⎛ −=

−3

1

3

1

2 44 dxx

xdxx

x

3ln44

||ln42 1

32

−=⎥⎥⎦

⎤

⎢⎢⎣

⎡−= xx

7. Let ,42 −= xu .2xdxdu = When ,1−=x .3−=u

When x = 1, .3−=u

∴ ∫∫−

−−•=−

3

3

41

1

42

214)4(4 duudxxx

0=

8. Let ,13 += xu .3 2dxxdu =

When x = 0, u = 1. When x = 1, u = 2.

∴ ∫∫ =+2

1

1

032

311 duudxxx

)122(92

32

31

1

223

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡= u

9. Let ,122 ++= xxu .)1(2)22( dxxdxxdu +=+=

When x = 0, u = 1. When x = 1, u = 4.

∴ ∫∫ =++

+ 4

1

1

0 21

21

121 du

udx

xxx

2ln

2ln221

4ln21

|]|[ln21

14

=

×=

=

= u

10. Let ,2+= −xeu .dxedu x−−=

When x = 0, u = 3. When x = 2, .22 += −eu

∴ ∫∫+

−

−

−=+

− 2

3

2

0

2 12

e

x duu

dxe

x

⎟⎠⎞

⎜⎝⎛ +−=

=

=

+

+

−

−∫

21ln3ln

|]|[ln

1

2

23

3

2

2

2

e

u

duu

e

e

11. Let ),1ln( += xu .1

1 dxx

du+

=

When x = 2, .3ln=u When x = 3, .4ln=u

∴ ∫∫ =++ 4ln

3ln

3

2 1)1ln( ududx

xx

34ln12ln

21

)3ln4)(ln3ln4(ln21

2)3(ln

2)4(ln

222

3ln

4ln2

=

−+=

−=

⎥⎦

⎤⎢⎣

⎡=

u

12. Let ),ln( 2xu = .2 dxx

du =

When ,1e

x = .2−=u

When x = e, u = 2.

∴ ∫∫ −=

2

21

2

21)ln( ududx

xxe

e

0

221

2

22

=⎥⎥⎦

⎤

⎢⎢⎣

⎡=

−

u


Definite Integrals


02=

−


∴ ∫ +

2

0 2 41 dx

x

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

++

++

++

+++

++

≈

46.11

42.11

48.01

44.01

242

140

1

4.0

22

22

22

3919.0= (cor. to 4 d. p.)


12=

−


∴ ∫ +2

1)1ln( dxx

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++++

+++++

≈)18.1ln()16.1ln(

)14.1ln()12.1ln(2

3ln2ln2.0

9090.0= (cor. to 4 d. p.)


01=

−


∴ ∫1

0dxxex

⎥⎦⎤

⎢⎣⎡ ++++

+≈ 8.06.04.02.0 8.06.04.02.0

202.0 eeeee

0148.1= (cor. to 4 d. p.)


23=

−


∴ ∫3

0

2

dxex

⎥⎥⎦

⎤

⎢⎢⎣

⎡++++

+≈

2222 8.26.24.22.294

22.0 eeeeee

1011.1585= (cor. to 4 d. p.)


12=

−


∴ ∫2

12 ln xdxx

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++

+++

≈

8.1ln8.16.1ln6.1

4.1ln4.12.1ln2.12

2ln20ln02.0

22

2222

0832.1= (cor. to 4 d. p.)


)4(2=

−−−

The end points are ,4− ,6.3− ,2.3− ,8.2− 4.2− and .2−

∴ ∫−

− +

2

4 3 11 dx

x

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+−+

+−+

+−+

+−++−

++−

≈

333

3

33

14.21

18.21

12.31

16.31

212

114

1

4.0

6235.1−= (cor. to 4 d. p.)

19. (a) ∫− +4

1)](3)(2[ dxxgxf

43

)7(3)11(2

)(3)(24

1

4

1

=+=

+= ∫ ∫− −dxxgdxxf

(b) ∫ ∫ ∫− −−=

4

2

4

1

2


15)4(11

=−−=

(c) ∫ −4

2)](4)([ dxxgxf

92415

)17(415

)from(b)()()(415

)(4)(

4

1

2

1

4

2

4

2

−=−=

−−=

⎥⎦⎤

⎢⎣⎡ −−=

−=

∫ ∫

∫ ∫

− −dxxgdxxg

dxxgdxxf

p.288

20. Let ,ln xu = .1 dxx

du =

When x = 1, u = 0. When ,3ex = u = 3.

∴ ∫∫ =3

0

2

1

23 )(ln duudxxxe

9

3 0

33

=⎥⎥⎦

⎤

⎢⎢⎣

⎡=

u



21. Let ,1+= xu dxdu = and .1−= ux

When x = 0, u = 1. When ,1−= ex u = e.

∴ ∫∫−

=+

− eedu

uudx

xx

1

1

0

11

2

|]|ln[

11

1

1

−=

−=

⎟⎠⎞

⎜⎝⎛ −= ∫

e

uu

duu

e

e

22. Let ,42 −= xu .2xdxdu =

When x = 2, u = 0. When x = 3, u = 5.

∴ ∫∫ =−5

0

3

22

214 duudxxx

355

32

21

0

523

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡= u

23. Let ,12 2 −= xu ,4xdxdu = .2

12 +=

ux

When x = 1, u = 1. When x = 3, u = 17.

∴ ∫∫−

=−

•3

1 2

23

1 2

3

1212dx

x

xxdxx

x

31

3175

232

81

181

21

41

1

1723

17

1

17

1

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

+

=

∫

∫

uu

duu

u

duu

u

24. Let ,1+= xu ,dxdu = .1−= ux

When x = 0, u = 1. When x = 3, u = 4.

∴ duuudxxx ∫∫ −=+4

1

3

0)1(1

15116

32

52

)(

1

423

25

4

121

23

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

−= ∫

uu

duuu

25. Let ,21 xeu += .2 dxedu x=

When x = 0, u = 3. When ,3ln=x u = 7.

∴ ∫∫ =+

7

3

3ln

0

121

21du

udx

ee

x

x

37ln

21

|]|[ln21

37

=

= u

26. Let ,23 += xu ,3dxdu = .3

2−=

ux

When x = 0, u = 2. When x = 1, u = 5.

∴ ∫∫+

−

=++ 5

2 3

1

0 3

13

2

31

)23(1 du

u

u

dxxx

2009

211

91

1191

191

2

5

2

5

2 32

5

2 3

=

⎥⎦⎤

⎢⎣⎡ −−=

⎟⎠⎞

⎜⎝⎛ +=

+=

∫

∫

uu

duuu

duu

u

27. Let ,42 += xu ,2xdxdu = .42 −= ux

When x = 0, u = 4. When ,5=x u = 9.

∴ ∫ +5

023 4dxxx

15253

38

52

21

)4(21

)4(21

4

4

923

25

9

423

9

4

5

022

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

−=

−=

+=

∫

∫

∫

uu

duuu

duuu

xdxxx


Definite Integrals

28. Let ,12 += xet .2 dxetdt x=

222

2

)1(

112

−=

−=

+=

te

te

dxe

edt

x

x

x

x

When x = 0, .2=t When ,3ln=x t = 2.

∴ ∫+

3ln

0

3

1dx

e

ex

x

1521492

32

52

)12(2

)1(2

1

2

235

2

224

2

222

3ln

0

2

−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡+−=

+−=

−=

+=

∫

∫

∫

ttt

dttt

dtt

dxe

eex

xx

29. ,1x

u = .12 dx

xdu −=

,1u

x = ,12

2

ux = .11 2

22

uuu +

=+

∴ dxudu 2−=

2ududx−

=

When x = 10, u = 0.1. When x = 0.1, u = 10.

∴ ∫∫ +=

+

10

1.0 2

10

1.0 2 1

ln21

1ln

x

xdx

xdxx

∫

∫

∫

∫

+−=

+=

+=

+

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛

=

10

1.0 2

1.0

10 2

1.0

10 2

1.0

10

2

2

2

1ln

1ln

1ln

1

1ln21

dxx

x

dxx

x

duu

uu

uu

duu

∴ ∫ =+

10

1.0 2 01

ln2 dxx

x

∫ =+

10

1.0 2 01

ln dxx

x

30. (a) ⎟⎟⎠

⎞⎜⎜⎝

⎛

+

−

141

xx

dxd

23

)14(

3214

142)1(14

+

+=

+

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−−+

=

x

xx

xxx

(b) ∫+

+2

023

)14(

32 dx

x

x

34

(a)) (from 14

1

0

2

=

⎥⎦

⎤⎢⎣

⎡

+

−=

xx

31. (a) )2)(1(

221 −+

++−=

−+

+ xxBBxAAx

xB

xA

)2)(1(2)(

−+−++

=xx

ABxBA

Q )2)(1(12)(

)2)(1(1

−+−++

≡−+ xx

ABxBAxx

∴ ⎩⎨⎧

=−=+

)2(....................12)1..(....................0

ABBA

:)2()1( −

3113

−=

−=

A

A

Substituting31

−=A into (1)

01331

=+−

∴ 31

=B

(b) ∫ −+

6

3 )2)(1( xxdx

37ln2ln4

|]1|ln|2|[ln31

(a)) (from 2

11

131

36

6

3

−=

+−−=

⎟⎠⎞

⎜⎝⎛

−+

+−= ∫

xx

dxxx

32. (a) xx ++1

1

xx

xxxx

xxxx

xx

−+=

−+−+

=

−+

−+×

++=

1

)1(1

11

11



(b) ∫ ++

3

1 1 xxdx

322346

32)1(

32

(a)) (from )1(

1

323

23

3

1

−−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+=

−+= ∫

xx

dxxx

33. (a) 32 )1(1

)1(2

11

++

+−

+ xxx

3

2

3

2

3

2

)1(

)1(12212

)1(1)1(2)1(

+=

++−−++

=

+++−+

=

xx

xxxx

xxx

(b) ∫ +

1

0 3

2

)1(dx

xx

852ln

)1(21

12|1|ln

(a)) (from )1(

1)1(

21

1

0

1

2

1

0 32

−=

⎥⎦

⎤⎢⎣

⎡

+−

+++=

⎟⎟⎠

⎞⎜⎜⎝

⎛

++

+−

+= ∫

xxx

dxxxx

34. (a) Let ∫ += CxFdxxf )()( .

Then )()]([ xfxFdxd

= .

By the Fundamental Theorem of Calculus,

∫ −=x

FxFdttf0

)0()()(

∴ 23)0()( xFxF =−

xxf

xFdxdxF

dxd

xdxdFxF

dxd

6)(

6)]0([)]([

)3()]0()([ 2

=

=−

=−

(b) ∫∫ •=3

1

3

126)2( tdtdttf

48

212

12

1

32

3

1

=⎥⎥⎦

⎤

⎢⎢⎣

⎡=

= ∫t

tdt


01=

−


∴ ∫ +−1

0 11 dx

ee

x

x

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+−

++−

+

+−

++−

++−

++−

≈

8.0

8.0

6.0

6.0

4.0

4.0

2.0

2.00

0

11

11

11

11

211

11

2.0

ee

ee

ee

eee

eee

2399.0−= (cor. to 4 d. p.)

2

2

22

2

)1(2

)1(

)1())(1())(1(

11

x

x

x

xxxx

x

xxxx

x

x

ee

eeeee

eeeee

ee

dxd

+−=

++−−−

=

+−−−+

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

3

4

22

2

2

)1()1(2

)1())(1)(2()()1(2

11

x

x

x

xxxx

x

x

exe

eeeeee

ee

dxd

+−

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

++−+

−=⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

For ,10 ≤≤ x

.01≤−x

∴ The approximation is an underestimate. 36. (a) When 2)( =xf , xxf 3)( = and 24)( xxf = ,

I = G.

∴ ∫ ++=2

1)2()2()2(2 cbaxdx

)(2][ 21

2 cbax ++=

)1...(....................3222 =++ cba

∫ ×+×+×=2

12 )2(3)

23(3)1(33 cbadxx

cbax 6293][ 2

13 ++=

)2(....................141296 =++ cba

22

1223 )2(4)

23(4)1(44 ×+×+×=∫ cbadxx

cbax 1694][ 21

4 ++=

)3(....................151694 =++ cba

∴ (1), (2) and (3),

61

=a , 1=b ,31

=c

(b) )2(31)

23()1(

61 fffG ++=

∴ ∫ ++≈2

12ln

31

23ln1ln

61ln xdxx

6365.0=


Definite Integrals

37 – 38 (HKASLE Questions) Extended Questions p.290

39. (a) Let ,xu −= .dxdu −=

When kx −= , u = k. When x = k, ku −= .

∫−k

kdxxf )( ∫

−−−=

k

kduuf )(

∫

∫

∫

−

−

−

−=

−=

−=

k

k

k

k

k

k

dxxf

duuf

duuf

)(

)(

)(

∴ ∫− =k

kdxxf 0)(2

0)( =∫−k

kdxxf

(b) Let ).()( 4 xx eexxf −−=

Then

)()(

)(

)()()(

4

4

4

xfeex

eex

eexxf

xx

xx

xx

−=−−=

−=

−−=−

−

−

−

∴ ∫−− =−

2

24 0)( dxeex xx (from (a))

40. (a) ∫∫∫ −−+=

0

1

1

0

1


Consider ∫−0

1.)( dxxf

∫∫ −−+=

0

1

0

1)2()( dxxfdxxf

Let ,2+= xu dxdu =

When x = 0, u = 2. When ,1−=x u = 1.

∴ ∫∫− =+2

1

0

1)()2( duufdxxf

∫=2

1)( dxxf

∴ ∫∫∫ +=−

2

1

1

0

1


∫=2

0)( dxxf

(b) ∫ +2

0]2)([ dxxf

021

1

2

0

2

0

][2)(

2)(

xdxxf

dxdxxf

∫

∫∫

−+=

+= (from (a))

Since ),()( xfxf =− by using the result of

Question 38(a),

∴ 0)(1

1=∫− dxxf

∴ 4][2]2)([ 20

2

0==+∫ xdxxf

Open-ended Questions p.290

41. Consider .)( xxf =

∫=4

1xdxI

215

2 1

42

=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

x

The width of each subinterval is .3.0103

1014

==−

The end points are 1, 1.3, 1.6, 1.9, 2.2, 2.5, 2.8, 3.1, 3.4, 3.7 and 4. ∴ The approximation J of I

I=

=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++++

++++++

≈

215

7.34.31.38.2

5.22.29.16.13.12

413.0

(or other reasonable answers)

42. Let .21)( += xxg Then

1

11

1

2

21

221

−−∫ ⎥

⎥⎦

⎤

⎢⎢⎣

⎡+=⎟

⎠⎞

⎜⎝⎛ + xxdxx

101

=−=

(or other reasonable answers)

New Progress in Senior Mathematics Module 1 …...7 New Progress in Senior Mathematics Module 1 Book...

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