Chapter 09 FWS

8/9/2019 Chapter 09 FWS

1/31

Oxford Fajar Sdn. Bhd. (008974-T) 2014

CHAPTER 9 INTEGRATION

Focus on Exam 9

1 3 +sinxcos2x dx= 3

cos2x+ sinx

cos2xdx =[3 sec2x+(cosx)2sinx] dx

=[3 sec2x(cosx)2(sinx)] dx

=3 tanx(cosx)1

1+c

=3tanx+ 1cosx

+c

2 sin2xcos2xdx=(sinxcosx)2dx

=12sin 2x2

dx

=14sin22xdx

=14

1 cos 4x2 dx

=18

(1 cos 4x) dx

=18

x14sin4x+c

3 x1 x4dx=du2

1 u2

= 12

du1 u2

=12

12

ln 1 +u1 u+c

=1

4ln

1 +x2

1 x2

+c

dxa2x2=1

2aln a+xax +c

sin 2x=2 sinxcosx

sinxcosx=1

2sin 2x

cos 4x=1 2 sin22x

sin22x =1 cos 4x

2

u=x2dudx

=2x

x dx =du2


2/31


ACE AHEADMathematics (T) Second Term2

4 Let x=sin

dxd

=cos

dx=cos d

dx

x2 1 x2=

cos d

sin2 1 sin2

= cos dsin2 cos2 = dsin2 =csc2d

=cot +c

= 1x2

x +c [Shown]

5 Let u=lnx

dudx

=1x

dxx

= du

dxx lnx= dxx

1lnx

= duu =ln |u| +c

= ln |lnx| +c

6ddx

(tan3x)=3 tan2x sec2x

=3(sec2x1) sec2x =(3 sec2x3) sec2x = 3sec4x3sec2x [Shown]

(3 sec4x3 sec2x) dx=tan3x+c

3 sec4xdx 3 sec2xdx=tan3 x+c

3 sec4 xdx=3 sec2xdx+tan3x+c

=3 tanx+tan3x+c

sec4xdx= tanx+ 13tan3x+ c

3

sec4xdx=tanx+13tan3x+c

1 x

q

1 x2


3/31


Fully Worked Solutions 3

7 xsinxcosxdx= x12sin 2xdx

=12xsin 2xdx

=1

2

cos 2x

1

2

x

1

2

cos 2x

1

2dx

=14

x cos 2x+14cos 2xdx

=14

xcos2x+18

sin2x+c

8 1

0

x

1 +x2dx =1

2

1

0

2x1 +x2

dx

=12

[ln |1 +x2|]10

=12[ln (1 +12) ln (1 +02)]

=12

(ln 2 ln 1)

=12

ln2

9 2

0

x2

16 x2dx

Letx=4 sin dx=4 cos d

For the lower limit, whenx=0, 0 =4 sin =0

For the upper limit,

whenx=2, 2 =4 sin

sin =12

=6

06 16 sin

2

16 16 sin2(4 cos d) = 0

6

16 sin2

4 1 sin2(4 cos d)

= 0616 sin2 d

= 06161 cos 22 d

=806(1 cos 2) d


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=8 12sin 206

=8612

sin3

0

=43 4

3

2 =4

3 2 3 [Shown]

10 Let t=tan x2

dtdx

=12

sec2x

2

dtdx

=12

1 +tan2x2 dt

dx=1

2(1 +t2)

dx = 2 dt1+t2

For the lower limit, whenx=0, t=tan 0 =0

For the upper limit, whenx= 2

,

t= tan 4 = 1

11

x

t2

1 t2

+ 2t

0

2 5 dx3 sinx +4 cosx= 5

1

0

2 dt

1+t2

3 2t1+t2+ 41t21+t2

= 5 1

0

2 dt6t+4 4t2

= 5 1

0

dt3t+2 2t2

= 51

0

dt2t23t2

= 51

0 dt

(2t+1)(t2)


5/31



Let1

(2t +1)(t2)= A

2t+1+ B

t 2

1 A(t 2) +B(2t +1)

Letting t=2, 1 =B(5)

B=15

Letting t= 12

, 1 =A 52 A= 2

5

51

0

dt(2t +1)(t2)

= 51

0 25(2t +1)+

15(t 2)dt

= 5

1

5

ln 2t +1+ 1

5

ln t 2

1

0

= ln 2t +1 ln t 21

0

= ln 2t +1t 2 1

0

=ln 3 ln 12 =ln

31

2

=ln 6 [Shown]

11 023

dx1sinx

Let t=tan x2

tanx=2 tan

x

2

1 tan2x

2

= 2t1t2

t =tan x2

dtdx

=12

sec2x

2

=12

1 +tan2x2

=12(1 +t2)

x

1 + t2

1 t2

2t


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2 dt =(1 +t2) dx

dx = 2 dt1+t2

Whenx=0, t=tan 02

=0

Whenx=23

, t=tan 3

= 3

023

dx1sinx

=03

2 dt1+t2

1 2t1+t2

=03

2 dt

1+t22t

= 203

dt

(t1)2

= 2(t 1)1

1 03

= 2t 1

0

3

= 23 1

21

= 23 1

2

=2 2 31

3 1

=2 3

3 1

=2 3

1 3 [Shown]


7/31



12 Let t=tan x2

dtdx

=12

sec2x

2

dtdx

=12

1 +tan2x2

dtdx

=12

(1 +t2)

dx = 2 dt1+t2

For the lower limit, whenx=0, t=tan 0 =0

For the upper limit, whenx= 23

, t= tan 2

= 3

023

35+4 cosx

dx =03

3 2 dt1+t25 + 41 t

2

1+t2dx

=03

6 dt5+5t2+4 4t2

= 603

dt9 +t2

= 6 13

tan1 t303

dxa2+x2=1

atan1

x

a+c

= 2tan1 33 tan10 = 260 =

3 [Shown]

13 sinxA(3 sinx+4 cosx) +B(3 cosx4 sinx) sinx (3A4B)sinx+(4A+3B)cosx

Equating the coefficients of sinx,

1 =3A4B

Equating the coefficients of cosx,

0 =4A+3B Solvingandsimultaneously, we have

A= 3

25andB= 4

25

x

1 + t2

1 t

2

2t


8/31



02 sinx

3 sinx+4 cosxdx =0

2

325

(3 sinx+4 cosx) 425

(3 cosx4 sinx)

3 sinx+4 cosxdx

=02 325

425

3 cosx4 sinx3 sinx+4 cosxdx

= 325x 425ln |3 sinx+4 cosx|02

= 325

2 0425

[ln |3 +0| ln |0 +4|]

=0.235

1417 +x

(4 3x)(1 +2x)A

4 3x+ B

1 + 2x 17 +xA(1 + 2x) +B(4 3x)

Lettingx= 12, 16 12= B4 3 12 16

12

= B5 12 B=3

Lettingx= 43

, 1813

= A1 + 243 18

13

= A323 A=5

17 +x(4 3x)(1 +2x)

= 54 3x

+ 31 + 2x

1

21

3

17 +x

(4 3x)(1 +2x)dx=

1

21

3

54 3x+3

1 + 2xdx

= 53

1

21

3

3

4 3xdx +3

2

1

21

3

2

1 + 2xdx

= 53

[ln |4 3x|]1

21

3

+32

[ln |1 + 2x|]1

21

3

=

5

3ln 4 3

2 ln 4 3

3+3

2ln1 +2

2 ln1 +2

3 = 5

3ln 52 ln 3+

32

ln 2 ln 53

= 53

ln 523+3

2ln 253

= 53

ln56

+32

ln65

=0.577 [Shown]


9/31



156x6

(x +3)(x2+3) A

x+3+ Bx +C

x2+ 3

6x6 A(x2+ 3) +(Bx + C)(x +3) Lettingx=3, 18 6 =12A 24 =12A

A =2 Lettingx=0, 6 =3A +3C 6 =6 +3C C =0

Lettingx=1, 0 =4A + (B+C)(4) 0 =8 +4B B =2

6x6(x +3)(x2+3)

2x+3 +

2x

x2+ 3

2

1

6x6

(x +3)(x2+3)dx=

2

1 2x+3+

2xx2+ 3dx

= 2 [ln |x+ 3|]12+[ln |x2+ 3|]

12

= 2 (ln 5 ln 4) +ln7 ln 4

= 2 ln 54

+ln 74

= ln

7

4

542

= ln 2825

[Shown]

1613 11x+6x2(x +3)(x2)2

Ax+3

+ B(x 3)2

+ Cx 2

13 11x+6x2A(x 2)2+B(x + 3) +C(x +3)(x 2)

Lettingx=2, 15 =5B B =3

Lettingx=3, 100 =25A A =4

Lettingx=0, 13 =4A + 3B6C 13 =4(4) + 3(3) 6C C =2


10/31



13 11x+6x2

(x +3)(x2+3)= 4

x+3+ 3

(x 2)2+ 2

x 2

4

3

13 11x+6x2(x +3)(x2+3)

=4

3

4x+3

dx+4

3

3(x2)2

dx + 4

3

2x 2

dx

= 4 [ln |x+ 3|]34+3

4

3

(x 2)2dx+2[ln |x 2|]3

4

= 4 [ln |x+ 3|]3

4 3x 24

3+2[ln |x 2|]

3

4

= 4 (ln 7 ln 6) 32 31+2(ln 2 ln 1)

= 4 ln 76

+32

+2 ln 2

=3.50 [Shown]

17 1x2+2x 15x2+2x 14

x2+2x 151

x2+2x 14x2+2x 15

=1 + 1x2+2x 15

=1 + 1(x 3)(x +5)

Let1

(x 3)(x+5) A

x 3+ B

x+ 5 1 A(x+ 5) +B(x 3)

Lettingx=5, 1 =8BB= 1

8

Lettingx=3, 1 =8A

A= 18

x2+2x 14

x2+2x 15=1 + 1

8(x 3) 1

8(x+ 5)

5

4

x2+2x 14x2+2x 15

dx = 5

41 + 18(x 3)

18(x+ 5)dx

= x+ 18 ln |x3| 18

ln |x+ 5|4

5

= 5 + 18

ln2 18

ln 10 4 + 18 ln 1 18

ln 9 = 5 + 1

8ln2 1

8ln 10 4 + 1

8ln 9

= 1 + 18

ln 2 910

=1

+

1

8ln

9

5 [Shown]


11/31



18

0x cos2xdx =

p

0x1 +cos 2x2 dx

=p

0

12

(x +xcos 2x) dx

cos 2x=2 cos2x1

cos2x = 1 +cos 2x2

=p

012 x dx +

p

012 x cos 2xdx

=12x2

2 p

0

+12 sin 2x 12

xp

0

p

0

12

sin 2x12dx Integrating by parts.

=x2

4 p

0

+14x sin 2xp

0

p

0

14

sin 2xdx

=x2

4 p

0

+14x sin 2xp

0

+18cos 2x

0

=2

4

0 +1

4

sin 20+1

8cos 21

8cos 0

=2

4+1

81

8

= 2

4 [Shown]

19y =4x25x y =5x6x2

Substitutinginto,

4x25x=5x6x2

10x2

10x=0 10x(x 1) =0 x =0or1 From: Whenx=0, y=0. Whenx =1,y=1. Hence, the points of intersection are (0, 0) and (1,1).

y =4x25x=x(4x5) The curve cuts thex-axis at the

points (0, 0) and1 14, 0.dy

dx=8x5

Since a > 0, the curve has a minimum point.

At minimum point,dy

dx=0

8x5 =0

x=58

Whenx=58

,y=4582

558 =1 9

16


12/31



Hence, the minimum point of the curve is 58, 1916.

y =5x6x2=x(5 6x)

dy

dx=5 12x

The curve cuts thex-axis at the points

(0, 0) and 56, 0.

Since a < 0, the curve has a maximum point.

At maximum point,dy

dx=0

5 12x=0

x= 512

Whenx= 512

,y=5 512 6512

2

=11

24

Hence, the minimum point of the curve is 512 , 1124.

The graphs ofy=4x25xandy=5x6x2are as shown in the following diagram.

y= 4x2 5x

y= x

(1, 1)

xO

y

1

1

1

2

2

2

y= 5x 6x2

124512 , 1

,

A1

A2

1 91658

The equation of the chord joining the points of intersection (0, 0) and (1, 1) isy =x.

A1=

1

05x6x2(x)dx

=1

0(6x 6x2) dx

=3x22x310 =3(1)2 2(1)3 0 =1 unit2


13/31



A2=

1

0[x(4x25x)] dx

=1

0(4x 4x2) dx =2x24x

3

3 1

0

=2(1)2 413 0 =2

3units2

A1: A

2=1 : 2

3

=3:2 [Shown]

20 The graph ofy=3 ln (x 2) is as shown in the following diagram.

xO 2 43

y= 3 ln (x2)

y

Required area =4

3 y dx

=4

3 3 ln |x 2| dx Copy back

=[3xln |x2|]43

4

33x 1

x 2dx

Differentiate

=3xln |x2|43 3

4

3

x

x 2dx

=3xln |x2|43 3

4

31 + 2x 2dx

=3xln |x2|43 3x+2 ln |x2|4

3

1

x2 x x2 2

x

x 2 =1 +2

x 2

=3(4) ln 2 3(4 +2 ln 2)[3(3) ln 1 3(3 + 2 ln 1)]

=12 ln 2 12 6 ln 2 9(0) +9 +6(0)

=6 ln 23 [Shown]

To be kept.

To be integrated.


14/31



21y=x2 y =x2x3

Substitutinginto,

x2=x2x3

x32x2=0 x3(x 2) =0 x =0or2

From: Whenx=0, y=0

Whenx =2,y=22

=4

Hence, the points of intersection of the curves are (0, 0) and (2,4).

y =x2x3

=x2

(1 x)

The curve intersects thex-axis at the

points (0, 0) and (1, 0).

dy

dx=2x3x2

d2y

dx2=2 6x

At turning points,

dy

dx=0. 2x3x2=0

x(23x) =0

x=0 or 23

Whenx=0, y=0 andd2y

dx2=2 6(0)

=2 (>0)

Thus, (0, 0) is a minimum point.

Whenx=23

,y=232 23

3

= 427

and

d2y

dx2=2 6 23

=2 (< 0)

Thus, 23,

427is a maximum point.


15/31



The graphs ofy=x2andy=x2x3are as shown in the following diagram.

x

y

O

y1= x2

y2= x2 x3

(2, 4)

42723 ,

Required area =2

0 (y2y1) dx =

2

0 (x2x3(x2)dx

=2

0 (2x2x3) dx

=2x3

3x

4

4 2

0

=23

(8) 164

0

=11

3units2

[Shown]

22 y =4x

y2=4(x 1)

Substitutinginto:

4x2

=4(x 1)

16

x

2=4(x 1)

4

x2=x 1

4 =x3x2

x3x2 4 =0 By inspection, x=2 satisfies the equation.

(x2)(x2+x+2) =0

x2 =0 or x2+x+2 =0 x=2 No real roots because b24ac =12 4(1)(2)

=7 (< 0)


16/31



Whenx=2, y =42

=2

Hence, the point of intersection of the curves is (2, 2).

The graphs of the curvesy =4x

andy2=4 (x 1) is as shown in the following diagram.

y

xO

(2, 2)

y=4x

y2= 4(x 1)

y2

=

4(x

1)

3

2

1

A2

A1

Required area =A1+A

2

=2

0y

2

4+1dy+

3

2

4y

dy

=y3

12+y

2

0

+[4 lny]32

= 812+ 2 0+(4 ln 3 4 ln 2)

=8

3+ 4 ln 3

2units2

[Shown]

23 y=x(x+2)(x3) =x3x2 6x y=x(x3) =x23x

Substitutinginto, x3x2 6x =x23x x32x2 3x =0 x(x22x3) =0

x(x+1)(x3) =0 x=0, 1 or 3 From: Whenx=1,y=(1)2 3(1) =4 Whenx=0, y=0 Whenx=3, y=32 3(3) =0 Hence, the points of intersection are (1, 4), (0, 0)and(3, 0).

y=x(x+2)(x3) =x3x2 6x The curve cuts thex-axis at the points (2, 0), (0, 0) and (3, 0).


17/31



dy

dx=3x22x 6

d2y

dx2=6x2

At turning points,dy

dx=0

3x2

2x 6 =0 x=

2 (2)2 4(3)(6)2(3)

x=1.79 or 1.12 Whenx=1.79,y=1.79(1.79 +2)(1.79 3) =8.21 and

d2y

dx2=6(1.79) 2

=8.74 (> 0)

(1.79, 8.21) is a minimum point. Whenx=1.12,y=1.12(1.12 + 2)(1.12 3) =4.06 and

d2y

dx2=6(1.12) 2

=8.72 (< 0) (1.12, 4.06) is a maximum point. y=x(x3) =x23x The curve cuts thex-axis at the points (0, 0) and (3, 0). Its minimum point is (1.5, 2.25).

The graphs ofy=x(x+2)(x3) andy=x(x3) is as shown in the following diagram.y

xO

2

4

6

8

12

2

4

6

8

321

y= x(x 3)

y= x(x+ 2)(x 3)(1.12, 4.06)

(1.5, 2.25)

(1.79, 8.21)

A2

A1

x =0 +3

2

y =1.5(1.5 3)


18/31



Required area =AreaA1+AreaA

2

=0

1x3x26x(x23x)dx+

3

0x23x(x3x26x) dx

=0

1(x32x23x) dx +

3

0(x3+2x2+3x) dx

=x4

42x

3

33x

2

20

1+x

4

4+2x

3

3+3x

2

23

0

=0 (1)4

4

2(1)3

3

3(1)2

2 +34

4+2(3)

3

3+3(3)

2

2 3

0

0

=14+23

32+

814

+18 +272

=1156

units2 [Shown]

24y=ex

Whenx=0, y=e0 =1

Whenx + ,y +

Whenx ,y0 y=2 +3ex

=2 + 3ex

Whenx=0, y=2 + 3e0

=5

Whenx + , 3

ex 0 and thusy 2

Whenx ,y +

x

y

O In 3

1

2

5

y= 2 + 3ex

y= ex

y=ex y=2 +3ex

Substitutinginto, we have:

ex=2 +3ex

ex=2 + 3

e

x

(ex)2=2ex+3


19/31



(ex)2 2ex 3 =0 (ex 3)(ex+ 1) =0 ex=3 or ex=1 x=ln 3 (No solution)

Hence, thex-coordinate of the point of intersection of the curves y=exandy=2 +3exis ln 3.

Area of the shaded region =ln 3

0 2 +3ex

ex

dx =2x+3 11ex ex

ln 3

0

=2x 3ex exln 3

0

=2 ln 3 3eln3

eln 32(0) 3e0e0 =2 ln 3 3

3 3 0 + 3 + 1

=2.20 units2

25 (a) y 2=x(x4)2

y = x (x4) Hence, the axis of symmetry is thex-axis.

(b) Sincey 20, thenx(x4)20. Because (x4)20, x(x4)20 if and only ifx0. Hence, the curve exists only forx0.

(c) y 2=x(x4)2

=x(x2

8x+16) =x3 8x2+16x

2ydy

dx=3x216x+16

dy

dx=3x

216x+162y

At turning points,dy

dx=0.

3x216x+16

2y=0

3x216x+16 =0 (3x4)(x4) =0

x=43

Whenx=43

,y 2=43

4342

=9 1327

y=3.08

Hence, the turning points are 113, 3.08and 1

13,3.08.

x=4 is not accepted because whenx=4,y=0

anddy

dx=

0

0(undefined).


20/31



(d) The curvey 2=x(x4)2is as shown in the following diagram.

x

y

O 4

131 , 3.08

131 , 3.08

(e) Volume generated =

4

0

y 2dx

=4

0x(x4)2dx

=4

0(x3 8x2+16x) dx

= x4

4 8x

3

3+ 8x2

4

0

= 64 83(64) +128

=2113 units

3

26y 2=6x ... y =2x +6 ...

Substitutinginto,

(2x +6)2=6x 4x224x +36=6x 4x230x +36 =0 2x215x +18=0

(2x 3)(x 6)=0 x =3

2or 6

From: Whenx =32

,y=2 32+6 =3

Whenx =6,y =2(6) +6 = 6

Hence, the points of intersection of the curvey 2=6xand the straight liney=2x+6 are 32

, 3and (6, 6).


21/31



The graphs ofy 2=6x andy=2x+6 are as shown in the following diagram.

x

y

O

y12= 6x

y2= 2x+ 6

(6, 6)

6

3

3 6

V1

V2

32

V1=

3

0 y 2

6 2

dy+6

3 6 y

2 2

dy

=3

0y

4

36dy+1

4

6

3(36 12y+y 2) dy

= y5

1803

0

+14

36y6y 2+y3

3 6

3

= 243180+14

36(6) 6(6)2+2163 36(3) 6(3)2+273

=2720

+14

(72 63)

=185 units

3

V2=

6

06x dx

6

3(2x+6)2dx

= [3x2]60 6

3(4x224x+36) dx

=3 (36 0) 4x3

3 12x2+36x

6

3

=108 43(6)312(6)2+36(6) 43

(3)312(3)2+36(3) =108 [72 36] =72 units3

V1: V

2=

185

72

= 120

=1: 20 [Shown]

27y =x(4 x) =4x x2

y =4x

1


22/31



Substitutinginto,

4x x2=4x

1

4x2x3=4 x x34x2x +4=0

By inspection,x =1 satisfies the equation. (x 1)(x23x 4) =0

(x1)(x+1)(x4) =0 x =1, 1 or 4 x =1is not accepted x =1or 4

From: Whenx =1, y =41

1

=3

Whenx=4, y=44

1

=0

Hence, the points of intersection of the curves are (1, 3) and (4, 0) forx>0.

The graphs ofy=x(4 x) andy=4x

1 forx0 are as shown in the following diagram.

y

xO

(2, 4)

(1, 3)

y2=4x

y1= x(4 x)

1

4

1

Volume generated =4

1y

22dx

4

1y

12dx

=4

1x2 (4 x)2dx

4

14x 1

2

dx

=4

1x2(16 8x +x2) dx

4

116x2

8x

+ 1dx =

4

1(16x28x3+x4) dx

4

116x2

8x

+ 1dx

=16x3

3 2x4

+x5

5 4

1

16x 8 lnx+x

4

1

x2 3x 4

x 1 x34x2x+4x3x2

3x2x3x2+3x

4x+ 44x+4

0


23/31



=16(4)3

32(4)4+4

5

5163 2 +

15

164 8 ln 4 + 4 (16 8 ln 1 +1) =303

5(15 8 ln 4)

=1535

+8ln 22

=1535

+16ln 2

=1535+16ln2 [Shown]

282x+1

(x2+1)(2 x)Ax+B

x2+1+ C

2 x

2x+1 (Ax+B)(2 x) +C(x2

+1) Lettingx=2, 5 =C(5) C=1

Lettingx=0, 1 =2B+C 1 =2B +1 B =0

Lettingx=1, 3 =(A+B) +2C 3 =(A+0) +2(1) A=1

2x+1(x2+1)(2 x)

= xx2+1

+ 12x

1

0

2x+1(x2+1)(2 x)

dx=1

0 xx2+1

dx+1

0

12 x

dx

=12

1

0

2xx2+1

dx1

0

12 x

dx

=12ln (x2+1)1

0[ln (2 x)]1

0

=12

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

=1.04

29 (a)

x2+x+2

x2

+2=1+ x

x2

+2

1

x2+2 x2+x+2x2 +2

x


24/31



x2+x+2x2+2

dx =1+ xx2+2dx

=1+122x

x2+2dx =x+1

2

ln|x2+2| +c

(b) x

ex+1dx =xe (x+1)dx

= 11

e(x+1)xe (x+1)1 dx = x

ex+1+e (x+1)dx

= xex+1

+ 11

e (x+1)+c

= xex+1 1ex+1+c

=x+1ex+1

+c

30 (a)dy

dx=3x5

2 x

y=3x52 xdx

y=3

2x

12

5

2x

12

dx y=3

2x

32

3252x

12

12+c

y=x32 5x

12 +c

Since the curve passes through the point (1, 4), then

4 =(1)3

2 5(1)1

2+c

4 =1 5 +c c=0

Hence, the equation of the curve isy=x32 5x

12

=x12(x5)

= x (x5)

(b) At thex-axis,y=0

x(x5) =0 x=0 or 5

x=0 is ignored because it is given thatx>0. Therefore,x=5.


25/31



At a turning point,dy

dx=0.

3x52 x

=0

3x5 =0

x=53

Whenx=53

,y= 53

535 =4.30

dy

dx=3x5

2 x

=3

2

x125

2

x 1

2

d2y

dx2=3

4x

12+5

4x

32

= 3

4x12

+ 5

4x32

Whenx=53

,d2y

dx2= 3

45312

+ 5

45332

(>0)

Hence, 1 23, 4.30is a minimum point.

Then curve ofy= x(x5) is as shown below.

y

xO 4 5321

2

4

231 , 4.30

(c) Area of the region bounded by the curve and thex-axis =5

0ydx

=5

0 x32

3x

12

dx


26/31



=2x52

55 2x

32

3

5

0

=25(5)5210

3(5)

320

=25( 5)

5

103 ( 5)

3

=25(25 5)

103

(5 5) =10 5503 5 =203 5 =20

3

5units2

31 3

2(x2)2

x2dx =

3

2x

24x+4x2 dx

=3

21 4x+4x2 dx

=x 4 ln |x| +4 x1

13

2

=x 4 ln |x| 4x3

2

=3 4 ln 3 43

2 4 ln 2 42 =5

3+4 ln 2 4 ln 3

=53

+4 (ln 2 ln 3)

=53

+4 ln 23 [Shown]

32y=6 ex

On thex-axis,y=0. 6 ex=0 ex=6 x=ln 6

Thus, the curvey=6 exintersects thex-axis at (ln 6, 0).

On they-axis,x=0. y=6 e0

y=5


27/31



Thus, the curvey=6 ex intersects they-axis at (0, 5).

Asx ,y Asx ,y6

(In 5, 1)

In 6

5

6

y= 6 ex

y= 5ex

O x

y

y=5ex

On they-axis,x=0. y=5(e0) y=5

Therefore, the curvey=5exintersects they-axis at (0, 5). Asx ,y 0. Asx

,y

The curvesy=6 exandy=5exare as shown. y=6 ex y=5ex

Substitutinginto,

6 ex=5ex 6ex(ex)2=5 Letting ex=p, 6pp2=5

p

2

6p+5 =0 (p1)(p5) =0 p=1 or 5 Whenp=1, ex=1 x=ln 1 x=0 Whenx=0, y=6 e0=5 Whenp=5, ex=5 x=ln 5


28/31



Whenx=ln 5,y=6 e ln 5=6 5 =1 Hence, the points of intersection are (0, 5) and (ln 5, 1).

Area of the shaded region =ln 5

0 6 ex 5exdx

=6xex

5

(1)ex

ln 5

0

=6xe x+ 5exln 5

0

=6 ln 5 eln 5 5eln 5

0 e0+ 5e0 =6 ln 5 5 +5

5 (1 + 5)

=6 ln 5 5 +1 +1 5

=(6 ln 58) units2

Volume of the solid generated =ln 5

06 e x

2

5ex2

dx

=ln 5

036 12ex+e2x25e2xdx

=36x12ex+12e2x25

(2)e2x

ln 5

0

=36x12ex+12e2x252e2x

ln 5

0

=36 ln 5 12eln 5+12e2 ln 5+25

2e2 ln 50 12e0+12e0+

252e0

=36 ln 5 12(5) +12(25) +

252(25)

12+12+252

=(36 ln 5 48)

=12(3 ln 54)units3

33 Let u=1 x

dudx

=1

dx =du

Whenx =0,u =1.

Whenx =1, u =0.

1

0x2(1 x)

1

3dx =0

1(1 u)2u

1

3(du)

=0

1u

1

3

(1 u)2

du


29/31



=0

1u

1

3(1 2u+u2) du

=0

1u

1

3+2u4

3u7

3du

=u43

43

+2u73

73

u103

103

0

1

=34u4

3+67

u7

3 310

u10

3 0

1

=0 34(1)4

3+67

(1)7

3 310

(1)10

3

=3

46

7+3

10

= 27140

34 (a)

O

y1= x2 4

y2= x2

4

2

3

2

1

2

x

y

R

(b) y=x2 y=x24

Substitutinginto,

x24 =x2

x2

x2 =0 (x2)(x+1) =0 x=2 or 1

Whenx=2, y=2 2 =0

Whenx=1,y=1 2 =3

Hence, the coordinates of the points of intersection are (2, 0) and (1,3).


30/31



(c) Area ofR =2

1(y

2y

1) dx

=2

1(x2) (x2 4)dx

=

2

1(x2+x+2) dx

=x3

3+x

2

2+2x

2

1

=23

3+2

2

2+2(2) (1)

3

3+

(1)2

2+2(1)

=103

76 =9

2units2

(d) Volume generated =

2

1(y12

y22

) dx

=2

1(x24)2(x 2)2dx

=2

1(x48x2+16) (x24x+4)dx

=2

1(x49x2+4x +12) dx

=x

5

53x3+2x2+12x

2

1

=25

53(2)3+2(2)2+12(2) (1)

5

53(1)3+2(1)2+12(1)

=725 365

=1085

units3

35 (a)

x

y

y =

y = x2 1

12

x2

1 O 1

(b) y=x2 1

y

x=

12

2


31/31


x4x2= 12

x4x2 12 = 0

(x2+ 3)(x2 4) = 0 x2= 4 x= 2

Whenx= 2, y = 4 1 = 3

Hence, the points of intersection are (2, 3) and (2, 3).

(c)

xx

2

21

12 =

Volume = y yy

y y+( ) + 112

10

32

0

12

3

12

d d d

= +

+ [ ] [ ] y

y y y

2

0

3

3

12

0

12

2 (12) ln

= +

+ ( ) ( )

3

23 6 12

2

ln ln12 3-

= +15

212 12ln 4

= +9

224 ln 2

= 24 l 2

9

2units

3n

Chapter 09 FWS

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Transcript of Chapter 09 FWS