chapter 9 integration

8/12/2019 Chapter 9 Integration

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O xford Fajar Sdn. Bhd. (008974-T) 2012

CHAPTER 9 INTEGRATION

Focus on Exam 9

1 3 + sin xcos2 x d x = 3cos2 x + sin xcos2 x d x = [3 sec2 x + (cos x)2 sin x] d x = [3 sec2 x (cos x)2(sin x)] d x

= 3 tan x (cos x)1

1 + c

= 3 tan x + 1cos x

+ c

2 sin2 x cos 2 x d x = (sin x cos x)2 d x

= 12 sin 2 x2

d x

= 14 sin2 2 x d x = 14 1 cos 4 x2 d x = 18 (1 cos 4 x) d x = 1

8 x 1

4 sin 4 x + c

3 x1 x4 d x = d u2

1 u2

=12 d u1 u2 = 12

12

ln 1 + u1 u + c

= 14

ln 1 + x 2

1 x 2 + c

d xa2 x2 = 12aln a + x

a x +

c

sin 2 x = 2 sin x cos x

sin x cos x = 12

sin 2 x

cos 4 x = 1 2 sin2 2 x sin2 2 x

= 1 cos 4 x2

u = x2d ud x

= 2 x x d x = d u2


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ACE AHEAD Mathematics (T) Second Term2

4 Let x = sin

d xd

= cos

d x = cos d

d x

x2 1 x 2 =

cos d

sin2 1 sin 2

= cos d sin2 cos2 = d sin2 = cosec2 d = cot + c

= 1 x 2

x + c [Shown]

5 Let u = ln x

d ud x

= 1 x

d x x

= d u

d x x ln x = d x x 1ln x = d uu = ln |u| + c = ln |ln x | + c

6 d d x

(tan3 x) = 3 tan2 x sec2 x = 3(sec 2 x 1) sec 2 x = (3 sec 2 x 3) sec2 x =3 sec 4 x 3 sec 2 x [Shown] (3 sec4 x 3 sec2 x) d x = tan 3 x + c 3 sec4 x d x 3 sec2 x d x = tan 3 x + c 3 sec4 x d x = 3 sec2 x d x + tan3 x + c = 3 tan x + tan3 x + c sec4 x d x = tan x +13 tan

3 x + c3 sec4 x d x = tan x + 1

3 tan 3 x + c

1 x

q

1 x 2


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Fully Worked Solution 3

7 x sin x cos x d x = x 12 sin 2 x d x = 12 x sin 2 x d x =

1

2 cos 2 x 1

2 x

1

2 cos 2 x 1

2 d x

= 14 x cos 2 x + 14 cos 2 x d x = 1

4 x cos 2 x + 1

8 sin 2 x + c

8 1

0

x1 + x2 d x =

12

1

0

2 x1 + x2 d x

= 12 [ln |1 + x2|]10

= 12 [ln (1 + 12) ln (1 + 02)] = 12 (ln 2 ln 1)

= 12

ln 2

9 2

0

x2

16 x2 d x

Let x = 4 sin d x = 4 cos d For the lower limit, when x = 0, 0 = 4 sin = 0 For the upper limit, when x = 2, 2 = 4 sin

sin = 12 = 6

06 16 sin

2 16 16 sin 2

(4 cos d ) = 06 16 sin

2

4 1 sin2 (4 cos d )

= 06 16 sin2 d

= 06 16 1 cos 2

2 d = 8 0

6 (1 cos 2 ) d


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Let 1(2t + 1)(t 2) =

A2t + 1 +

Bt 2

1 A(t 2) + B(2t + 1) Letting t = 2, 1 = B(5)

B =15

Letting t = 12, 1 = A 52

A = 25

5 1

0

d t (2t + 1)(t 2) = 5

1

0 25(2t + 1) + 15(t 2) d t = 5 1

5 ln 2t + 1 + 1

5 ln t 2

1

0

= ln 2t + 1 ln t 2 10

= ln 2t + 1t 2

1

0

= ln 3 ln 12 = ln 31

2

= ln 6 [Shown]

11 023 d x1 sin x

Let t = tan x2

tan x =2 tan x

21 tan2 x

2 = 2t 1 t 2

t = tan x2

d t d x

= 12 sec2 x

2

= 12 1 + tan2 x

2

= 12 (1 + t 2)

x

1 + t 2

1 t 2

2 t


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2 d t = (1 + t 2) d x

d x = 2 d t 1 + t 2

When x = 0, t = tan 02 = 0 When x = 23 , t = tan

3

= 3

023 d x

1 sin x = 03

2 d t 1 + t 2

1 2t 1 + t 2

= 03

2 d t

1 + t 2 2t

= 2 03 d t (t 1)2

= 2(t 1)1

1 03

= 2

t 1 0

3

= 23 1

21

= 23 1

2

= 2 2 3 1

3 1

= 2 33 1

= 2 31 3

[Shown]


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12 Let t = tan x2 d t

d x = 12 sec

2 x2

d t d x

= 12 1 + tan2 x

2

d t d x

= 12 (1 + t 2)

d x = 2 d t 1 + t 2

For the lower limit, when x = 0, t = tan 0 = 0

For the upper limit, when x = 23 , t = tan2

= 3

023

35 + 4 cos x d x = 0

3

3 2 d t 1 + t 2

5 + 41 t 2

1 + t 2 d x

= 0 3 6 d t 5 + 5t 2 + 4 4t 2 = 6 0 3 d t 9 + t 2 = 6 13 tan

1 t 3 0

3

d xa2 + x2 = 1a tan1 xa + c = 2 tan1 3 3 tan

1 0

= 2 6

0

= 3

[Shown]

13 sin x A(3 sin x + 4 cos x) + B(3 cos x 4 sin x) sin x (3 A 4 B)sin x + (4 A + 3 B)cos x Equating the coefficients of sin x, 1 = 3 A 4 B Equating the coefficients of cos x, 0 = 4 A + 3 B Solving and simultaneously, we have

A =

325

and B =

425

x

1 + t 2

1 t 2

2 t


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02 sin x

3 sin x + 4 cos x d x = 02

325

(3 sin x + 4 cos x) 425 (3 cos x 4 sin x)

3 sin x + 4 cos x d x

= 02 3

25 425

3 cos x 4 sin x3 sin x + 4 cos x d x

= 325 x 425

ln |3 sin x + 4 cos x|0

2

= 325 2

0 425 [ln |3 + 0| ln |0 + 4|]

= 0.235

14 17 + x(4 3 x)(1 + 2 x) A

4 3 x + B

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

Letting x = 12, 16 12 = B 4 312 16 1

2 = B 512

B = 3

Letting x = 43, 1813 = A 1 + 2

43

18 13 = A 3

23

A = 5 17 + x(4 3 x)(1 + 2 x) =

54 3 x +

31 + 2 x

1213

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

1213

54 3 x +

31 + 2 x d x

= 53 1213

34 3 x d x +

32

1213

21 + 2 x d x

= 53 [ln |4 3 x|]

1213

+ 32 [ln |1 + 2 x|]1213

= 53 ln 4

32 ln 4

33 +

32 ln 1 +

22 ln 1 +

23

= 53 ln52 ln 3 + 32 ln 2 ln

53

= 53 ln523 + 32 ln 25

3

= 53 ln56 +

32 ln

65

= 0.577 [Shown]


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15 6 x 6( x + 3)( x2 + 3)

A x + 3 +

Bx + C x2 + 3

6 x 6 A( x2 + 3) + ( Bx + C )( x + 3) Letting x = 3, 18 6 = 12 A 24 = 12 A A = 2 Letting x = 0, 6 = 3 A + 3C 6 = 6 + 3C C = 0 Letting x = 1, 0 = 4 A + ( B + C )(4) 0 = 8 + 4 B B = 2

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

2 x + 3 +

2 x x2 + 3

2

1

6 x 6

( x + 3)( x2 + 3) d x = 2

1 2 x + 3 + 2 x x2 + 3 d x = 2 [ln | x + 3|]12 + [ln | x2 + 3|]12

= 2 (ln 5 ln 4) + ln 7 ln 4

= 2 ln 54 + ln74

= ln7454

2

= ln 2825

[Shown]

16 13 11 x + 6 x2

( x + 3)( x 2)2 A

x + 3 + B

( x 3)2 +C

x 2 13 11 x + 6 x2 A( x 2)2 + B( x + 3) + C ( x + 3)( x 2)

Letting x = 2, 15 = 5 B B = 3 Letting x = 3, 100 = 25 A A = 4 Letting x = 0, 13 = 4 A + 3 B 6C 13 = 4(4) + 3(3) 6C C = 2


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13 11 x + 6 x2

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

x + 3 +3

( x 2)2 +2

x 2

4

3

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

4

3

4 x + 3 d x +

4

3

3( x 2)2 d x +

4

3

2 x 2 d x

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

4

3

( x 2)2 d x + 2[ln | x 2|]34

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

x 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 1

x2 + 2 x 15 x2 + 2 x 14 x2 + 2 x 15

1 x

2 + 2 x 14 x2 + 2 x 15 = 1 +

1 x2 + 2 x 15

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

( x 3)( x + 5) A

x 3 + B

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

Letting x = 5, 1 = 8 B B = 18

Letting x = 3, 1 = 8 A A = 18 x

2 + 2 x 14 x2 + 2 x 15 = 1 +

18( x 3)

18( x + 5)

5

4

x2 + 2 x 14 x2 + 2 x 15 d x =

5

4 1 + 18( x 3) 18( x + 5) d x = x + 18 ln | x 3|

18 ln | x + 5| 4

5

= 5 + 18 ln 2 18 ln 10 4 +

18 ln 1

18 ln 9

= 5 + 18 ln 2 18 ln 10 4 +

18 ln 9

= 1 + 18 ln2 9

10 = 1 +

18 ln

95 [Shown]


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18 x

0 x cos2 x d x =

x

0 x 1 + cos 2 x

2 d x =

x

0

12 ( x + x cos 2 x) d x

cos 2 x = 2 cos2 x 1

cos2 x = 1 + cos 2 x2

= x

0 12 x d x + x

0 12 x cos 2 x d x

= 12 x

2

2

x

0 + 12 sin 2 x

12 x

x

0

x

0

12 sin 2 x

12 d x Integrating by parts.

= x2

4 x

0 + 14 x sin 2 x

x

0

x

0

14 sin 2 x d x

= x2

4 x

0 + 14 x sin 2 x

x

0 + 18 cos 2 x

x

0

= 2

4 0 + 1

4 sin 2 0 + 18 cos 2

18 cos 0

= 2

4 + 18

18

= 2

4 [Shown]

19 y = 4 x2 5 x y = 5 x 6 x2 Substituting into ,

4 x2 5 x = 5 x 6 x2 10 x2 10 x = 0 10 x( x 1) = 0 x = 0 or 1 From : When x = 0, y = 0. When x = 1, y = 1. Hence, the points of intersection are (0, 0) and (1, 1) .

y = 4 x2 5 x = x(4 x 5) The curve cuts the x-axis at the points (0, 0) and 1 1

4, 0 .

d yd x

= 8 x 5

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

At minimum point, d yd x

= 0 8 x 5 = 0 x = 58 When x = 58, y = 4

58

2 5 58

= 1 916


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Hence, the minimum point of the curve is 58

, 1 916 .

y = 5 x 6 x2 = x(5 6 x)

d yd x

= 5 12 x

The curve cuts the x-axis at the points

(0, 0) and 56

, 0 .

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

At maximum point, d yd x

= 0

5 12 x = 0 x = 512

When x = 512 , y = 55

12 6512

2

= 11

24

Hence, the minimum point of the curve is 512

, 1 124

.

The graphs of y = 4 x2 5 x and y = 5 x 6 x2 are as shown in the following diagram.

y = 4x 2 5x

y = x

(1, 1)

x O

y

1

1

1

2

2

2

y = 5x 6x 2

124

512

, 1

,

A1

A2

1 916

58

The equation of the chord joining the points of intersection (0, 0) and (1,

1) is y=

x.

A1 = 1

05 x 6 x2 ( x) d x

= 1

0(6 x 6 x2) d x

= 3 x2 2 x3 10 = 3(1)2 2(1)3 0 = 1 unit2


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A2 = 1

0[ x (4 x2 5 x)]d x

= 1

0 (4 x 4 x2) d x = 2 x2 4 x

3

3

1

0

= 2(1)2 4 13 0

= 23 units2

A1 : A2 = 1 :23

= 3:2 [Shown]

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

x O 2 43

y = 3 ln ( x 2)

y

Required area = 1

3 y d x =

1

3 3 ln | x 2| d x Copy back

= [3 x ln | x 2|]43 4

33 x 1

x 2 d x

Differentiate

= 3 x ln | x 2| 43 3 4

3

x x 2 d x

= 3 x ln | x 2| 43 3 4

3 1 + 2 x 2 d x = 3 x ln | x 2| 43 3 x + 2 ln | x 2|

43

1 x 2 x x 2 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 2 3 [Shown]

To be kept.

To be integrated.


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21 y = x2 y = x2 x3 Substituting into ,

x2 = x2 x3 x3 2 x2 = 0 x3( x 2) = 0 x = 0 or 2 From : When x = 0, y = 0

When x = 2, y = 22 = 4 Hence, the points of intersection of the curves are (0, 0) and (2, 4) . y = x2 x3

= x2

(1 x)

The curve intersects the x-axis at the points (0, 0) and (1, 0).

d yd x

= 2 x 3 x2

d 2 y

d x2 = 2 6 x

At turning points,d yd x = 0.

2 x 3 x2 = 0 x(2 3 x) = 0 x = 0 or 23 When x = 0, y = 0 and d

2 yd x2

= 2 6(0)

= 2 ( x > 0) Thus, (0, 0) is a minimum point.

When x = 23, y = 23

2 233

= 427 and

d 2 y

d x2 = 2 6 23

= 2 ( x


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The graphs of y = x2 and y = x2 x3 are as shown in the following diagram.

x

y

O

y 1 = x 2

y 2 = x 2

x 3

(2, 4)

427

23

,

Required area =

2

0

( y2 y1) d x

= 2

0 ( x2 x3 ( x2) d x =

2

0 (2 x2 x3) d x = 2 x

3

3 x

4

4

2

0

= 23 (8) 164

0

= 113 units

2

[Shown]

22 y = 4 x

y 2 = 4( x 1) Substituting into :

4 x

2

= 4( x 1)

16

x2 = 4( x 1)

4 x2

= x 1 4 = x3 x2 x3 x2 4 = 0 By inspection, x = 2 satisfies the equation. ( x 2)( x2 + x + 2) = 0 x 2 = 0 or x2 + x + 2 = 0 x = 2 No real roots because b2 4ac = 12 4(1)(2)

=

7 ( 0) (1.79, 8.21) is a minimum point. When x = 1.12, y = 1.12(1.12 + 2)(1.12 3) = 4.06 and

d 2 y

d x2 = 6(1.12) 2

= 8.72 (< 0) (1.12, 4.06) is a maximum point. y = x( x 3) = x2 3 x The curve cuts the x-axis at the points (0, 0) and (3, 0). Its minimum point is (1.5, 2.25). The graphs of y = x( x + 2)( x 3) and y = x( x 3) is as shown in the following diagram.

y

x O

2

4

6

8

1 2

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 + 32 y = 1.5(1.5 3)


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Required area = Area A1 + Area A2 =

0

1 x3 x2 6 x ( x2 3 x) d x +

3

0 x2 3 x ( x3 x2 6 x) d x

= 0

1 ( x3 2 x2 3 x) d x +

3

0 ( x3 + 2 x2 + 3 x) d x

= x 44 2 x3

3 3 x22

0

1 + x 44 + 2 x

3

3 + 3 x22

3

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 = 11 5

6 units 2 [Shown]

24 y = e x When x = 0, y = e0 = 1 When x + , y + When x , y 0 y = 2 + 3e x

= 2 + 3e x When x = 0, y = 2 + 3e0 = 5

When x + , 3e x

0 and thus y 2 When x , y +

x

y

O In 3

1

2

5

y = 2 + 3e x

y = e x

y = e x y = 2 + 3e x Substituting into , we have: e x = 2 + 3e x

e x = 2 + 3

e x

(e x)2 = 2e x + 3


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(d) The curve y 2 = x( x 4)2 is as shown in the following diagram.

x

y

O 4

13

1 , 3.08

13

1 , 3.08

(e) Volume generated =

4

0 y 2 d x

= 4

0 x( x 4)2 d x

= 4

0 ( x3 8 x2 + 16 x) d x

= x4

4 8 x

3

3 + 8 x2

4

0

= 64 83 (64) + 128

= 2113 units

3

26 y 2 = 6 x ... y = 2 x + 6 ... Substituting into , (2 x + 6)2 = 6 x 4 x2 24 x + 36 = 6 x 4 x2 30 x + 36 = 0 2 x2 15 x + 18 = 0

(2 x 3)( x 6) = 0 x = 32 or 6

From : When x = 32, y = 232

+ 6

= 3 When x = 6 , y = 2(6) + 6 = 6 Hence, the points of intersection of the curve y 2 = 6 x and the straight line y = 2 x + 6 are 3

2, 3

and (6, 6) .


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The graphs of y 2 = 6 x and y = 2 x + 6 are as shown in the following diagram.

x

y

O

y 12 = 6x

y 2 = 2x + 6

(6, 6)

6

3

3 6

V 1

V 2

32

V 1 = 3

0 y2

6

2

d y + 6

3 6 y2 2

d y

= 30 y 436 d y + 14 6

3 (36 12 y + y 2) d y

= y5

180

3

0 + 1

4 36 y 6 y 2 + y

3

3

6

3

= 243180

+ 14

36(6) 6(6)2 + 2163

36(3) 6(3)2 + 273

= 2720

+ 14

(72 63)

= 185 units3

V 2 = 6

0 6 x d x

6

3 (2 x + 6)2 d x

= [3 x2]60 6

3 (4 x2 24 x + 36) d x

= 3(36 0) 4 x3

3 12 x2 + 36 x

6

3

= 108 43 (6)3 12(6)2 + 36(6) 43 (3)

3 12(3) 2 + 36(3)

= 108 [72 36] = 72 units3

V 1 : V 2 = 185 72

= 120 = 1:20 [Shown] 27 y = x(4 x) = 4 x x2 y = 4

x 1


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23/31



= 16(4)3

3 2(4)4 + 4

5

5 163 2 +

15

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

= 30 35

(15 8 ln 4)

= 15 35 + 8 ln 2 2

= 15 35 + 16 ln 2

= 15 35

+ 16 ln 2 [Shown]

28 y = e x When x = 0, y = e0 = 1 When x + , y + When x , y 0 y = 2 + 3e x y = 2 + 3e x When x = 0, y = 2 + 3e0 = 5

When x + , 3e x 0 and thus y 2

When x , y +

x

y

O In 3

1

2

5

y = e x

y = 2 + 3e x

y = e x

y = 2 + 3e x Substituting into , we have: e x = 2 + 3e x

e x = 2 + 3e x (e x)2 = 2e x + 3 (e x)2 2e x 3 = 0 (e x 3)(e x + 1) = 0 e x = 3 or e x = 1 x = ln 3 (Not possible) Hence, the x-coordinate of the point of intersection of the curves y = e x and y = 2 + 3e x is ln 3.


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Area of the shaded region = ln 3

0 2 + 3e x e x d x = 2 x + 3 11 e

x e xln 3

0

= 2 x 3e x e

xln 3

0

= 2 ln 3 3eln3 eln 3 2(0) 3e0 e

0

= 2 ln 3 33 3 0 + 3 + 1

= 2.20 units 2

29 2 x + 1( x2 + 1)(2 x)

Ax + B x2 + 1 +

C 2 x

2 x + 1 ( Ax + B)(2 x) + C ( x2 + 1) Letting x = 2, 5 = C (5) C = 1 Letting x = 0, 1 = 2 B + C 1 = 2 B + 1 B = 0 Letting x = 1, 3 = ( A + B) + 2C 3 = ( A + 0) + 2(1) A = 1

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

x 2 + 1 + 1

2 x

1

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

1

0 x x2 + 1 d x +

1

0 12 x d x

= 12 1

0 2 x x2 + 1 d x

1

0 12 x d x

= 12 ln ( x2 + 1)

1

0 [ln (2 x)]10

= 12

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

= 1.04

30 (a)

x2 + x + 2 x2 + 2 = 1 +

x x2 + 2

1 x2 + 2 x2 + x + 2

x2 + 2 x


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x2 + x + 2 x2 + 2 d x = 1 +

x x2 + 2 d x

= 1 + 12 2 x x2 + 2 d x = x + 1

2 ln | x 2 + 2| + c

(b) xe x + 1 d x = xe( x + 1) d x = 11 e

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

11 e

( x + 1) + c

= xe x + 1 1e x + 1 + c

= x + 1e x + 1

+ c

31 (a) d yd x

= 3 x 52 x

y = 3 x 52 x d x

y = 32 x

12

52 x

12 d x

y = 32 x

32

32

52 x

12

12

+ c

y = x32 5 x

12 + c

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

4 = (1)32 5(1)

12 + c

4 = 1 5 + c c = 0

Hence, the equation of the curve is y = x32 5 x

12

= x12 ( x 5)

= x ( x 5) (b) At the x-axis, y = 0 x( x 5) = 0 x = 0 or 5 x = 0 is ignored because it is given that x > 0. Therefore, x = 5.


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At a turning point, d yd x

= 0.

3 x 52 x

= 0

3 x 5 = 0

x = 53

When x = 53, y = 53

53

5

= 4.30

d yd x

= 3 x 52 x

= 32

x12 5

2 x

12

d 2 y

d x2 = 34 x

12 + 54 x

32

= 34 x

12

+ 54 x

32

When x = 53,d 2 yd x2

= 34 5

3

12

+ 54 5

3

32

(> 0)

Hence, 1 23

, 4.30 is a minimum point.

Then curve of y = x( x 5) is as shown below.

y

x O 4 5321

2

4

23

1 , 4.30


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28/31



Thus, the curve y = 6 e x intersects the x-axis at (ln 6, 0). On the y-axis, x = 0. y = 6 e 0 y = 5 Thus, the curve y = 6 e x intersects the y-axis at (0, 5). As x , y As x , y 6

(In 5, 1)

In 6

5

6

y = 6 e x

y = 5e x

O x

y

y = 5e x On the y-axis, x = 0. y = 5(e 0) y = 5 Therefore, the curve y = 5e x intersects the y-axis at (0, 5). As x , y 0. As x , y The curves y = 6 e x and y = 5e x are as shown. y = 6 e x y = 5e x Substituting into , 6 e x = 5e x 6e x (e x)2 = 5 Letting e x = p , 6 p p2 = 5 p2 6 p + 5 = 0 ( p 1)( p 5) = 0 p = 1 or 5 When p = 1, e x = 1 x = ln 1 x = 0


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When x = 0, y = 6 e0 = 5 When p = 5, e x = 5 x = ln 5 When x = 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 e x 5e x d x = 6 x e x 5(1) e

x ln 5

0

= 6 x e x + 5e xln 5

0

= 6 ln 5 e ln 5 5eln 5 0 e0 + 5e0

= 6 ln 5 5 + 55 (1 + 5) = 6 ln 5 5 + 1 + 1 5 = (6 ln 5 8) units 2

Volume of the solid generated = ln 5

0 6 e x

2

5e x2

d x

= ln 5

0 36 12e x + e 2 x 25e 2 x d x

= 36 x 12e x + 12 e2 x 25(2) e

2 xln 5

0

= 36 x 12e x + 12 e2 x 252e2 x

ln 5

0

= 36 ln 5 12e ln 5 + 12 e 2 ln 5 + 252e 2 ln 5 0 12e

0 + 12 e0 + 252e0

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

2(25) 12 + 12 +

252

= (36 ln 5 48) = 12(3 ln 5 4) units 3

34 Let u = 1 x

d ud x

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

When x = 1 , u = 0.


30/31



1

0 x2 (1 x)

13 d x =

0

1 (1 u)2 u

13 (d u)

= 0

1 u

13 (1 u)2 d u

=

0

1

u13 (1 2u + u2) d u

= 0

1 u

13 + 2u

43 u

73 d u

= u43

43

+ 2u73

73

u103

103

0

1

= 34 u43 + 67 u

73 310 u

103

0

1

= 0 34 (1)43 + 67 (1)

73 310 (1)103

= 34 67

+ 310 = 27

140

35 (a)

O

y 1 = x 2 4

y 2=

x

2

4

2

3

2

1

2x

y

R

(b) y = x 2 y = x2 4 Substituting into , x2 4 = x 2 x2 x 2 = 0 ( x 2)( x + 1) = 0 x = 2 or 1 When x = 2, y = 2 2 = 0 When x = 1, y = 1 2

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


31/31


(c) Area of R = 2

1 ( y2 y1) d x

= 2

1 ( x 2) ( x2 4) d x

=

2

1 ( x2 + x + 2) d x

= x3

3 + x

2

2 + 2 x

2

1

= 23

3 + 2

2

2 + 2(2) (1)

3

3 + (1)

2

2 + 2(1)

= 103 76

= 92

units 2

(d) Volume generated = 2

1 ( y12

y22

) d x

= 2

1 ( x2 4)2 ( x 2)2 d x

= 2

1 ( x4 8 x2 + 16) ( x2 4 x + 4) d x

= 2

1 ( x4 9 x2 + 4 x + 12) d x

= x5

5 3 x3 + 2 x2 + 12 x

2

1

= 25

5 3(2)3 + 2(2)2 + 12(2) (1)

5

5 3(1)3 + 2(1)2 + 12(1)

= 725 365

= 1085

units 3

chapter 9 integration

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