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# Chaos

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Submitted By huajui
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CHAPTER 9 INTEGRATION

Focus on Exam 9
1

3 + sin x dx = cos2 x

∫cos

3
2

x

+ sin2 x dx cos x

= ∫[3 sec2 x + (cos x)−2 sin x] dx = ∫[3 sec2 x − (cos x)−2 (−sin x)] dx (cos x)−1 = 3 tan x − +c −1 = 3 tan x + 1 + c cos x 2 ∫sin2 x cos2 x dx = ∫(sin x cos x)2 dx = = = = 1 ∫2 sin 2x
2

dx

sin 2x = 2 sin x cos x 1 sin x cos x = sin 2x 2

∫ 4 sin
1 4

1

2

2x dx

∫

1 − cos 4x dx 2

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

∫ 1 1 = x − sin 4x + c 8 4
3

1 (1 − cos 4x) dx 8

du x 2 dx = 1 − u2 1 − x4

= = =

1 du 1 − u2 2

u = x2 du = 2x dx du x dx = 2

1 1 1+u × ln +c 2 2 1−u

1 1 + x2 ln +c 1 − x2 4

∫ a − x = − 2a dx 1
2 2

ln

a − x + c
© Oxford Fajar Sdn. Bhd. (008974-T) 2012

a+x

2

ACE AHEAD Mathematics (T) Second Term

4 Let x = sin θ dx = cos θ dθ dx = cos θ dθ

∫x

2

dx cos θ dθ = 2 2 1−x sin θ 1 − sin2 θ

= =

∫ sin θ ∫ sin dθ 2

cos θ dθ 2 cos2 θ

θ q 1

x

= ∫cosec2 θ dθ = − cot θ + c 1 − x2 =− x +c 5 Let u = ln x du 1 = dx x dx = du x dx 1 dx ∫ x ln x = ∫  x ln x =

1 − x2

[Shown]

∫u

du

= ln |u| + c = ln |ln x| + c 6 d (tan3 x) = 3 tan2 x sec2 x dx = 3(sec2 x − 1) sec2 x = (3 sec2 x − 3) sec2 x = 3 sec4 x − 3 sec2 x
4 2 3

[Shown]

∫(3 sec x − 3 sec x) dx = tan x + c ∫3 sec4 x dx − ∫3 sec2 x dx = tan3 x + c ∫3 sec4 x dx = ∫3 sec2 x dx + tan3 x + c = 3 tan x + tan3 x + c 1 c ∫ sec4 x dx = tan x + tan3 x + 3 3 1 ∫sec4 x dx = tan x + tan3 x + c′ 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

3

7 ∫x sin x cos x dx = x =

1 ∫ 2 sin 2x dx 1

∫ 2 x sin 2x dx 1 1 1 1 = ∫ cos 2x   x − ∫ − cos 2x    dx 2 2 2 2
1 1 = − x cos 2x + cos 2x dx 4 4

1 1 = − x cos 2x + sin 2x + c 8 4 8

x 1 dx = 2 2 0 1 + x 1 = 2 1 = 2 1 = 2 1 = 2
1 2 0

2x dx 2 0 1 + x
1

[ln |1 + x2|]1 0 [ln (1 + 12) − ln (1 + 02)] (ln 2 − ln 1) ln 2

9

x2 dx 16 − x2

Let x = 4 sin θ dx = 4 cos θ dθ For the lower limit, when x = 0, 0 = 4 sin θ θ=0 For the upper limit, when x = 2, 2 = 4 sin θ 1 sin θ = 2 π θ= 6

π 6

0

16 sin2 θ (4 cos θ dθ) = 16 − 16 sin2 θ = =

π 6

0

16 sin2 θ (4 cos θ dθ) 4 1 − sin2 θ 16 sin2 θ dθ 16 π 6

π 6 π 6

0

0

1 − cos 2θ dθ 2

=8

0

(1 − cos 2θ) dθ

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

4

ACE AHEAD Mathematics (T) Second Term

1 = 8 θ − sin 2θ 2

3

4

π 6

0

=8 =

 π − 1 sin π − 0 6 2 3

3 4 π−4 2 3 4 = π− 2 3 3

 
[Shown]

10

Let t = tan

x 2

dt 1 x = sec2 dx 2 2 dt 1 x = 1 + tan2 dx 2 2 dt 1 = (1 + t2) dx 2 2 dt dx = 1 + t2

For the lower limit, when x = 0, t = tan 0 =0 π For the upper limit, when x = , 2 t = tan =1

π 4

1 + t2 x 1 − t2

2t

π 2
0

5 dx =5 3 sin x + 4 cos x =5 =5

∫
1

3

∫ ∫

0 1 0 1 0

2t 1 − t2 2 + 4 1+t 1 + t2 2 dt 6t + 4 − 4t2

2 dt 1 + t2

 

dt 3t + 2 − 2t2
1 0 1 0

= −5 = −5

∫ ∫

dt 2t2 − 3t − 2 dt (2t + 1)(t − 2)

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

5

Let

A 1 B = + (2t + 1)(t − 2) 2t + 1 t − 2 1 ≡ A(t − 2) + B(2t + 1)

Letting t = 2, 1 = B(5) 1 B= 5 1 5 Letting t = − , 1 = A − 2 2 A=− \ −5 2 5

 
−2 1 ∫ 35(2t + 1) + 5(t − 2)4 dt
1 0

dt = −5 0 (2t + 1)(t − 2)
1

= −5 −

3

1 1 ln 2t + 1 + ln t − 2 5 5
1

4

1 0

= 3ln 2t + 1 − ln t − 240 = ln

3 

2t + 1 t−2

4

1 0

= ln −3 − ln − = ln

 1 2


3 1 2 [Shown]

= ln 6 11

dx 0 1 − sin x x Let t = tan 2 2 tan x 2 x 2

2π 3

tan x = =

1 − tan2

1 + t2 x 1 − t2

2t

2t 1 − t2 x t = tan 2 dt 1 x = sec2 dx 2 2 1 x = 1 + tan2 2 2 1 = (1 + t2) 2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

6

ACE AHEAD Mathematics (T) Second Term

2 dt = (1 + t2) dx dx = 2 dt 1 + t2 =0 When x = 0 2

When x = 0, t = tan

2π π , t = tan 3 3 = 3 2 dt 1 + t2 2t 1− 1 + t2

2π 3
0

dx = 1 − sin x =

∫ ∫

3
0

3
0

2 dt 1 + t2 − 2t
3

=2 = = = = = = =

0

dt (t − 1)2

3

2(t − 1)−1 3 −1 0

4

3t −214 −

3
0

−2 −2 − −1 3−1 −2 −2 3−1

 

−2 − 2 3 − 1 3−1 −2 3 3−1 2 3 1− 3 [Shown]

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

7

12

Let t = tan

dt 1 x = sec2 dx 2 2 dt 1 x = 1 + tan2 dx 2 2 dt 1 = (1 + t2) dx 2 2 dt dx = 1 + t2

x 2

For the lower limit, when x = 0, t = tan 0 =0 For the upper limit, when x = 2π π , t = tan 3 2 = 3
2

1 + t2 x 1 − t2

2t

12+dtt  dx 3 ∫ 5 + 4 cos x dx = ∫ 5 + 4 1 − t 1 + t 
2 π 3

3

3

0

0

2

2

=

3
0

6 dt 5 + 5t + 4 − 4t2
2

=6

3
0

dt 9 + t2

=6×

1 t 3 tan−1 3 3 0

3

4

∫a + x dx 2

2

=

1 x tan−1 + c a a

= 2 tan−1 =2 =

3 − tan−1 0 3

π − 0 6
[Shown]

π 3

13 sin x ≡ A(3 sin x + 4 cos x) + B(3 cos x − 4 sin x) sin x ≡ (3A − 4B)sin x + (4A + 3B)cos x Equating the coefficients of sin x, … 1 = 3A − 4B Equating the coefficients of cos x, … 0 = 4A + 3B Solving  and  simultaneously, we have 3 4 A= and B = − 25 25
© Oxford Fajar Sdn. Bhd. (008974-T) 2012

8

ACE AHEAD Mathematics (T) Second Term

π 2
0

sin x dx = 3 sin x + 4 cos x = = =

π 2
0

3 4 (3 sin x + 4 cos x) − (3 cos x − 4 sin x) 25 25 dx 3 sin x + 4 cos x 3 cos x − 4 sin x

3 4 ∫ 3 25 − 25 3 sin x + 4 cos x4 dx
0

π 2

3

3 4 x− ln |3 sin x + 4 cos x| 25 25

4

π 2

0

3 π 4 [ln −0 − |3 + 0| − ln |0 + 4|] 25 2 25

= 0.235 14 A B 17 + x ≡ + (4 − 3x)(1 + 2x) 4 − 3x 1 + 2x 17 + x ≡ A(1 + 2x) + B(4 − 3x) 1 1 1 Letting x = − , 16 = B 4 − 3 − 2 2 2 1 1 16 = B 5 2 2 B=3 4 Letting x = , 3 1 4 =A 1+2 3 3 1 2 18 = A 3 3 3 A=5 5 3 17 + x \ = + (4 − 3x)(1 + 2x) 4 − 3x 1 + 2x 18
1 2 1 3

3  4   3  4  

17 + x dx = (4 − 3x)(1 + 2x) =

5 3 ∫ 4 − 3x + 1 + 2x dx 5 −3

1 2 1 3

1 2 1 3

−3 3 dx + 4 − 3x 2
1

1 2 1 3

2 dx 1 + 2x
1

3 5 = − [ln |4 − 3x|]2 + [ln |1 + 2x|] 2 1 1 2 3 3 3 =− 3 3 3 2 2 5 ln 4 − − ln 4 − + ln 1 + − ln 1 + 2 3 2 2 3 3 5 5 3 5 = − ln − ln 3 + ln 2 − ln 3 2 2 3

3  

4

3  

4

5 2 5 = − ln 3 3

2 3 ln 2 5 3 5 5 3 6 = − ln + ln 3 6 2 5 + = 0.577
© Oxford Fajar Sdn. Bhd. (008974-T) 2012

     
[Shown]

Fully Worked Solution

9

15

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

Letting x = 1,

\

−2 6x − 6 2x ≡ + (x + 3)(x2 + 3) x + 3 x2 + 3 −2 2x ∫ x + 3 + x + 3 dx
2 1 2

6x − 6 dx = 2 1 (x + 3)(x + 3)
2

= −2 [ln |x + 3|]2 + [ln |x2 + 3|]2 1 1 = −2 (ln 5 − ln 4) + ln 7 − ln 4 = −2 ln 5 7 + ln 4 4 7 4 5 4

= ln

3  4
2

= ln 16

28 25

[Shown]

A B C 13 − 11x + 6x2 ≡ + + (x + 3)(x − 2)2 x + 3 (x − 3)2 x − 2 13 − 11x + 6x2 ≡ A(x − 2)2 + B(x + 3) + C(x + 3)(x − 2) Letting x = 2, Letting x = −3, Letting x = 0, 15 = 5B B=3 100 = 25A A=4 13 = 4A + 3B − 6C 13 = 4(4) + 3(3) − 6C C=2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

10

ACE AHEAD Mathematics (T) Second Term

\

4 13 − 11x + 6x2 3 2 = + + (x + 3)(x2 + 3) x + 3 (x − 2)2 x − 2

13 − 11x + 6x2 = 2 3 (x + 3)(x + 3)
4

∫ x + 3 dx + ∫
3

4

4

3 dx + 2 3 (x − 2)
4

4 3

2 dx x−2

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

4 3

4 (x − 2)−2 dx + 2[ln |x − 2|]3

3

3 x−2

4

4 3

4 + 2[ln |x − 2|]3

= 4 (ln 7 − ln 6) − = 4 ln

7 3 + + 2 ln 2 6 2 = 3.50 [Shown] 17 1 x + 2x − 15 x2 + 2x − 14 x2 + 2x − 15 1 x2 + 2x − 14 1 =1+ 2 x2 + 2x − 15 x + 2x − 15 1 =1+ (x − 3)(x + 5) A 1 B Let ≡ + (x − 3)(x + 5) x − 3 x + 5
2

3 − 3  + 2(ln 2 − ln 1) 2 1

1 ≡ A(x + 5) + B(x − 3) Letting x = −5, 1 = −8B 1 B=− 8 Letting x = 3, 1 = 8A 1 A= 8 1 x2 + 2x − 14 1 \ 2 =1+ − 8(x − 3) 8(x + 5) x + 2x − 15

x2 + 2x − 14 dx = 2 4 x + 2x − 15
5

1 1 ∫ 1 + 8(x − 3) − 8(x + 5) dx
5 4

= x+ =5+

3

1 1 ln |x − 3| − ln |x + 5| 8 8 ln 2 −

4

5 4

1 8 1 =5+ 8 1 =1+ 8 1 =1+ 8

1 1 1 ln 10 − 4 + ln 1 − ln 9 8 8 8 1 1 ln 2 − ln 10 − 4 + ln 9 8 8 ln 2 × 9 10 9 ln [Shown] 5

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

11

18

x 0

x cos2 x dx = = =

1 + cos 2x ∫ x 2  dx x 0

cos 2x = 2 cos2 x − 1 cos2 x =

1 (x + x cos 2x) dx 0 2 x x

 1 + cos 2x  2

∫ 2 x dx + ∫ 2 x cos 2x dx 1 x 1 1 = 3  4 + 3 sin 2x   x4 − ∫ 2 2 2 2
0 0 2 x x 0 0

1

x

1

x 0

1 1 sin 2x  dx 2 2



Integrating by parts.

=

3 4 3 x2 4 +

x

x 1 1 = + x sin 2x + cos 2x 4 0 4 8 0 0 π2 1 1 1 = − 0 + π sin 2π − 0 + cos 2π − cos 0 4 4 8 8 2 π 1 1 = + − 4 8 8 π2 = [Shown] 4 19 y = 4x2 − 5x ⋅⋅⋅ y = 5x − 6x2 ⋅⋅⋅ Substituting  into , 4x2 − 5x = 5x − 6x2 10x2 − 10x = 0 10x(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 = 4x2 − 5x = x(4x − 5) The curve cuts the x-axis at the dy 1 = 8x − 5 points (0, 0) and 1 , 0 . dx 4 Since a > 0, the curve has a minimum point. dy At minimum point, =0 dx 8x − 5 = 0 5 x= 8 2 5 5 5 When x = , y = 4 −5 8 8 8 9 = −1 16

3 4 3

0 2 x

1 x sin 2x − 4 0 x 4 ∫ 4 3

x

1 sin 2x dx 0 4 x 4

x





© Oxford Fajar Sdn. Bhd. (008974-T) 2012

12

ACE AHEAD Mathematics (T) Second Term

Hence, the minimum point of the curve is y = 5x − 6x2 = x(5 − 6x) dy = 5 − 12x dx dy =0 dx 5 − 12x = 0 5 x= 12 5 5 5 When x = , y = 5 −6 12 12 12 1 =1 24 At maximum point,

9 5, −1 16. 8
The curve cuts the x-axis at the points 5 (0, 0) and , 0 . 6

 

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

   

2

Hence, the minimum point of the curve is

5 1 12 , 1 24.

The graphs of y = 4x2 − 5x and y = 5x − 6x2 are as shown in the following diagram. y 2 5 , 1 1 12 24 1 A1 O −1 y = 5x − 6x
2

y = 4x 2 − 5x

x 1 A2 (1, −1) y = −x 2

−2

5, 9 −1 8 16

The equation of the chord joining the points of intersection (0, 0) and (1, −1) is y = −x. A1 = =

∫ ∫

1 0 1 0

35x − 6x2 − (−x)4 dx
(6x − 6x2) dx
1

= 33x2 − 2x34 0 = 3(1)2 − 2(1)3 − 0 = 1 unit2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

13

A2 =

∫ =∫

1 0 1 0

[−x − (4x2 − 5x)]dx (4x − 4x2) dx

x3 1 3 0 1 = 2(1)2 − 4 −0 3 2 = units2 3 2 A1 : A2 = 1 : = 3: 2 3 = 2x2 − 4

3

 4 

[Shown]

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

O

2

3

4

x

Required area = =

1

3 1

y dx
To be integrated.

3

3 ln |x − 2| dx
To be kept. Copy back

= [3x ln |x − 2|]3 −

4

4 3

3x 

1 dx x−2

Differentiate

= 33x ln |x − 2|4 3 − 3 = 33x ln |x − 2|4 3 − 3
4 3 4

4

4 3

x dx x−2

2 ∫ 1 + x − 2 dx
4 3 4

= 33x ln |x − 2|4 − 33x + 2 ln |x − 2|4 3 = 12 ln 2 − 12 − 6 ln 2 − 9(0) + 9 + 6(0) = 6 ln 2 − 3 [Shown]

1 x − 2 x x−2 2 x 2 \ =1+ x−2 x−2

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

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

14

ACE AHEAD Mathematics (T) Second Term

21 y = −x2 y = x2 − x3

… …

Substituting  into , −x2 = x2 − x3 x3 − 2x2 = 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).

dy = 2x − 3x2 dx d2y = 2 − 6x dx2 At turning points, dy = 0. dx 2x − 3x2 = 0 x = 0 or When x = 0, y = 0 and 2 3

x(2 − 3x) = 0

d2y = 2 − 6(0) dx2 = 2 (x > 0)

Thus, (0, 0) is a minimum point. 2 22 23 When x = , y = − 3 3 3 4 = 27 and d2y 2 =2−6 dx2 3 = −2 (x < 0)

  

Thus,

4 2, 27 is a maximum point. 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

15

The graphs of y = − x2 and y = x2 − x3 are as shown in the following diagram. y y2 = x 2 − x 3 2, 4 3 27 O x

y1 = −x 2 (2, −4)

Required area =

∫ =∫

2

0 2

( y2 − y1) dx 3(x2 − x3 − (−x2)4 dx
2 3

0 2

∫ (2x − x ) dx 2x x =3 − 4 3 4
=
0 3 4 2 0

2 16 = (8) − −0 3 4 1 = 1 units2 [Shown] 3 22 y = 4 x 2 y = 4(x − 1) … …

Substituting  into : 42 = 4(x − 1) x 16 = 4(x − 1) x2 4 =x−1 x2 4 = x3 − x2 3 2 x −x − 4=0 By inspection, x = 2 satisfies the equation.



\ (x − 2)(x2 + x + 2) = 0 x−2=0 x=2 or x2 + x + 2 = 0 No real roots because b2 − 4ac = 12 − 4(1)(2) = −7 (< 0)

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

16

ACE AHEAD Mathematics (T) Second Term

When x = 2, y =

4 2 =2

Hence, the point of intersection of the curves is (2, 2). 4 The graphs of the curves y = and y2 = 4 (x − 1) is as shown in the following diagram. x y 3 2 A2 A1 O 1 (2, 2)

y 2 = 4(x − 1)

4 y =x x

y 2 = 4(x − 1)

Required area = A1 + A2 = =

∫  4 +1 dy + ∫
2 0

y2

3 2

4 dy y
3

2 y3 +y 12 0 8 = +2− 12 8 = + 4 ln 3

3  

4

+ [4 ln y]2

 3 units 2

0 + (4 ln 3 − 4 ln 2)
2

[Shown]

23 y = x(x + 2)(x − 3) = x3 − x2 − 6x …  y = x(x − 3) = x2 − 3x … Substituting  into , x3 − x2 − 6x = x2 − 3x 3 x − 2x2 − 3x = 0 x(x2 − 2x − 3) = 0 x(x + 1)(x − 3) = 0 x = 0, −1 or 3 From : When x = −1, y = (−1)2 − 3(−1) =4 When x = 0, y = 0 When x = 3, y = 32 − 3(3) =0 Hence, the points of intersection are (−1, 4), (0, 0) and (3, 0). y = x(x + 2)(x − 3) = x3 − x2 − 6x
The curve cuts the x-axis at the points (−2, 0), (0, 0) and (3, 0).

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

17

dy = 3x2 − 2x − 6 dx d2y = 6x − 2 dx2 dy At turning points, =0 dx 3x2 − 2x − 6 = 0 2 ± (−2)2 − 4(3)(−6) x= 2(3) x = 1.79 or −1.12 When x = 1.79, y = 1.79(1.79 + 2)(1.79 − 3) = −8.21 and d2y = 6(1.79) − 2 dx2 = 8.74 (> 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 d2y = 6(−1.12) − 2 dx2 = −8.72 (< 0) \ (−1.12, 4.06) is a maximum point. y = 1.5(1.5 − 3) 0+3 y = x(x − 3) x= 2 = x2 − 3x 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 8 6 4 A1 2 y = x(x − 3) y = x(x + 2)(x − 3)

(−1.12, 4.06)

−2 −1 O −2 −4 −6 −8

1 A2

2

3

x

(1.5, −2.25)

(1.79, −8.21)

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

18

ACE AHEAD Mathematics (T) Second Term

Required area = Area A1 + Area A2

∫ =∫
= =

0 −1 0 −1

3x3 − x2 − 6x − (x2 − 3x)4 dx + (x3 − 2x2 − 3x) dx +
3 0 3

2

3 0

3x2 − 3x − (x3 − x2 − 6x)4 dx

∫ (−x + 2x + 3x) dx x 4 2x3 3x2 + + 4 3 2
2

3

x 4 2x3 3x2 − − 4 3 2

4 3
−1 3

0

+ −

4

3 0

=0 − =−

3

(−1) 2(−1) 3(−1) 3 2(3)3 3(3)2 − − + − + + 4 3 2 4 3 2
4

4 3

4

4

3 0

−0

1 + 2 − 3 + − 81 + 18 + 27 4 3 2 4 2
[Shown]

5 = 11 units2 6 24 y = e x When x = 0, y = e0 =1 When x → + ∞, y → + ∞

When x → − ∞, y → 0 y = 2 + 3e−x 3 =2+ x e 3 When x = 0, y = 2 + 0 = 5 e 3 When x → + ∞, x → 0 and thus y → 2 e When x → − ∞, y → + ∞ y y = 2 + 3e−x y = ex

5

2 1 O In 3 x

y = ex y = 2 + 3e−x

… …

Substituting  into , we have: e x = 2 + 3e−x 3 ex = 2 + x e (e x)2 = 2e x + 3
© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

19

(e x)2 − 2e x − 3 = 0 (e x − 3)(e x + 1) = 0 ex = 3 or e x = −1 x = ln 3 (No solution) Hence, the x-coordinate of the point of intersection of the curves y = e x and y = 2 + 3e−x is ln 3. Area of the shaded region =

∫ 32 + 3e  − e 4 dx 1 = 32x + 3  e − e 4 −1 ln 3 0 −x x −x x ln 3 0

3 − ex ex 0 3 3 = 2 ln 3 − ln3 − e ln 3 − 2(0) − 0 − e0 e e 3 = 2 ln 3 − − 3 − 0 + 3 + 1 3 = 2.20 units2 = 2x − ln 3

3

4

25 (a) y 2 = x(x − 4)2 y = ± x (x − 4) Hence, the axis of symmetry is the x-axis. (b) Since y 2 ≥ 0, then x(x − 4)2 ≥ 0. Because (x − 4)2 ≥ 0, x(x − 4)2 ≥ 0 if and only if x ≥ 0. Hence, the curve exists only for x ≥ 0. (c) y 2 = x(x − 4)2 = x(x2 − 8x + 16) = x3 − 8x2 + 16x dy 2y = 3x2 − 16x + 16 dx dy 3x2 − 16x + 16 = dx 2y dy At turning points, = 0. dx 3x2 − 16x + 16 =0 2y 3x2 − 16x + 16 = 0 (3x − 4)(x − 4) = 0 x = 4 is not accepted because when x = 4, y = 0 4 dy 0 x= = (undefined). and = 3 dx 0 2 4 4 4 When x = , y 2 = −4 3 3 3 13 =9 27 y = ±3.08 1 1 Hence, the turning points are 1 , 3.08 and 1 , −3.08 . 3 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

20

ACE AHEAD Mathematics (T) Second Term

(d) The curve y 2 = x(x − 4)2 is as shown in the following diagram. y 1 1 , 3.08 3

O

4

x

1 1 , −3.08 3

(e) Volume generated = π

∫ y dx = π ∫ x(x − 4) dx = π ∫ (x − 8x + 16x) dx x 8x = π3 − + 8x 4 4 3
2 0 4 0 2 4 0 3 2 4 3 4 0 2

4

= π 64 −

3

8 (64) + 128 3

4

1 = 21 π units3 3 26 y 2 = 6x y = −2x + 6 ...  ... 

Substituting  into , (−2x + 6)2 = 6x 4x2 − 24x + 36 = 6x 4x2 − 30x + 36 = 0 2x2 − 15x + 18 = 0 (2x − 3)(x − 6) = 0 3 x = or 6 2 3 3 From : When x = , y = −2 +6 2 2 =3



When x = 6, y = −2(6) + 6 = −6 Hence, the points of intersection of the curve y 2 = 6x and the straight line y = −2x + 6 are 3, 3 2 and (6, − 6).
© Oxford Fajar Sdn. Bhd. (008974-T) 2012

 

Fully Worked Solution

21

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

y2 = −2x + 6 y12 = 6x V1

3

3 O 3 2 V2

6

x

(6, − 6)

V1 = π =π =π

∫ 
3 0

y2 2 dy + π 6

∫
6 3 6 3

6−y 2

2

dy

3

0

y4 dy + 1 π 36 4
3 0

(36 − 12y + y 2) dy

3 4
 

y5 180

y3 + 1 π 36y − 6y 2 + 3 4

3

4

6 3

= π 243 + 1 π 36(6) − 6(6)2 + 216 − 36(3) − 6(3)2 + 27 180 4 3 3 = 27 π + 1 π(72 − 63) 20 4 = 18 π units3 5 V2 = π

3

4

6

0

6x dx − π

6

3

(−2x + 6)2 dx

6 = π [3x2]0 − π

6

3

(4x2 − 24x + 36) dx

= 3π (36 − 0) − π

34x 3

3

− 12x2 + 36x

4

6 3

4 4 = 108π − π (6)3 − 12(6)2 + 36(6) − (3)3 − 12(3)2 + 36(3) 3 3 = 108π − π [72 − 36] = 72π units3 18 π V1 : V2 = 5 72π 1 = 20 = 1:20 [Shown] 27 y = x(4 − x) = 4x − x2 4 y = −1 x … …

3

4

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

22

ACE AHEAD Mathematics (T) Second Term

Substituting  into , 4 4x − x2 = −1 x 4x2 − x3 = 4 − x x3 − 4x2 − x + 4 = 0 By inspection, x = 1 satisfies the equation. (x − 1)(x2 − 3x − 4) = 0 (x − 1)(x + 1)(x − 4) = 0 x = 1, −1 or 4 x = −1 is not accepted \ x = 1 or 4 4 From : When x = 1, y = − 1 1 =3 When x = 4, y = 4 −1 4 =0 x2 − 3x − 4 x − 1 x3 − 4x2 − x + 4 x3 − x2 −3x2 − x −3x2 + 3x − 4x + 4 − 4x + 4 0

Hence, the points of intersection of the curves are (1, 3) and (4, 0) for x > 0. 4 The graphs of y = x(4 − x) and y = − 1 for x ≥ 0 are as shown in the following diagram. x y y2 = 4 − 1 x (2, 4)

(1, 3)

y1 = x(4 − x)

O −1

4

x

Volume generated = π =π

∫ ∫ ∫

4

1 4

y22 dx − π

4

1

y12 dx

1 4

x2 (4 − x)2 dx − π

∫
4 1

4 − 1 dx x 16 8 ∫  x − x + 1 dx
4 1 2

2

= π x2(16 − 8x + x2) dx − π
1

=π =π

4

1

(16x2 − 8x3 + x4) dx − π

16 8 ∫  x − x + 1 dx
4 1 2

3

16x3 x5 − 2x4 + 3 5

4

4 1

−π −

3

16 − 8 ln x + x x

4

4 1

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

23

3

16(4)3 45 16 1 − 2(4)4 + − −2+ 3 5 3 5

4
4

−π−

3

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

3 = 30 π − π(15 − 8 ln 4) 5 3 = 15 π + 8π ln 22 5 3 = 15 π + 16π ln 2 5 3 = 15 + 16 ln 2 π [Shown] 5

28 y = e x When x = 0, y = e0 = 1 When x → + ∞, y → + ∞ When x → − ∞, y → 0 y = 2 + 3e−x 3 y=2+ x e 3 When x = 0, y = 2 + 0 = 5 e 3 When x → + ∞, x → 0 and thus y → 2 e When x → − ∞, y → + ∞ y y = ex 5 y = 2 + 3e−x 2 1 O In 3 x

y = ex … y = 2 + 3e−x … Substituting  into , we have: e x = 2 + 3e−x 3 ex = 2 + x e (e x)2 = 2e x + 3 x 2 x (e ) − 2e − 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.
© Oxford Fajar Sdn. Bhd. (008974-T) 2012

24

ACE AHEAD Mathematics (T) Second Term

Area of the shaded region =

∫ 32 + 3e  − e 4 dx ln 3 0 −x x

= 2x + 3 = 2x −

3 3

1 −1 e

−x

− ex

4

ln 3 0

3 − ex ex

4

ln 3 0

= 2 ln 3 − = 2 ln 3 −

3 3 − e ln 3 − 2(0) − 0 − e0 eln3 e 3 − 3 −0 + 3 + 1 3

= 2.20 units2 29 2x + 1 Ax + B C ≡ + (x2 + 1)(2 − x) x2 + 1 2 − x 2x + 1 ≡ (Ax + B)(2 − x) + C(x2 + 1) Letting x = 2, Letting x = 0, 5 = C(5) C=1 1 = 2B + C 1 = 2B + 1 B=0 3 = (A + B) + 2C 3 = (A + 0) + 2(1) A=1

Letting x = 1,

\

2x + 1 1 = 2x + (x + 1)(2 − x) x + 1 2 − x
2

1 0

2x + 1 dx = (x + 1)(2 − x)
2

1 2

1 0

x dx + x +1
2 1 0 2

1 0

1 dx 2−x
1 0

= =

2x dx − x +1
2

−1 dx 2−x

32 ln (x

1

+ 1)

4

1 0

− [ln (2 − x)]1 0

= 1 (ln 2 − ln 1) − (ln 1 − ln 2) 2 = 1.04 30 (a) 1 x + 2 x2 + x + 2 x2 +2 x x2 + x + 2 x \ 2 =1+ 2 x +2 x +2
2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

25

x2 + x + 2 dx = x2 + 2 =

∫ 1 + x

2

x dx +2
2

∫ 31 + 2 x

1

2x +2

4 dx

=x+ (b)

1 ln |x2 + 2| + c 2

∫e

x x+1 dx = xe− (x + 1) dx 1 − (x + 1) e  x − −e− (x + 1)  1 dx −1 x = − x + 1 + e− (x + 1) dx e x 1 − (x + 1) = − x+1 + e +c e −1 =

=− =− 31 (a) dy 3x − 5 = 2 x dx y= y=

x e x+1 −

1 e x+1 +c

x+1 +c ex + 1

3x − 5 ∫  2 x  dx 3 ∫ 2 x
1 2

5 − − x 2 dx 2
1

1

3 x2 5 y= − 2 3 2 2
3 1

  
3

x2 1 2

+c

y = x 2 − 5x 2 + c Since the curve passes through the point (1, − 4), then − 4 = (1)2 − 5(1)2 + c −4 = 1 − 5 + c c=0 Hence, the equation of the curve is y = x2 − 5x2 = x2 (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.
© Oxford Fajar Sdn. Bhd. (008974-T) 2012
1 3 1 3 1

26

ACE AHEAD Mathematics (T) Second Term

At a turning point,

dy = 0. dx

3x − 5 =0 2 x 3x − 5 = 0 x= 5 3

5 5 5 When x = , y = −5 3 3 3 = −4.30 dy 3x − 5 = dx 2 x

3 5 − = x2 − x 2 2 2 d2y 3 − 1 5 − 3 = x 2+ x 2 4 dx2 4 = 31 + 53 4x 2 4x 2 5 d2y When x = , 2 = 3 dx 3 45 3

1

1



1 2

+

5 45 3



3 2

(> 0)

2 Hence, 1 , −4.30 is a minimum point. 3 Then curve of y = x (x − 5) is as shown below. y 

O −2 −4

x 1 2 3 4 5

2 1 , −4.30 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

27

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

∫ y dx = ∫ x − 3x  dx
5 0 5 0

3 2

1 2

= =

3

5

2x 2 − 5 2x 2 5 3
5 2

 4 
3
0

5

2 (5) − 10 (5) − 0 5 3 2 10 =  ( 5) − ( 5)  5 3
3 2
5 3

=

2 (25 5  

5) −

10 (5 5 ) 3

= 10 5 − = − = 32 20 5 3

50 5 3

20 5 units2 3

3 2

(x − 2)2 dx = x2 =

x ∫ 
3 2 3 2

2

− 4x + 4 dx x2

4 ∫ 1 − x + 4x  dx
−2

−1 = x − 4 ln |x| + 4 x −1

3 3

 4
3 2

3 2

= x − 4 ln |x| − = 3 − 4 ln 3 − = = = 33 y = 6 − e x On the x-axis, y = 0. 6 − ex = 0 ex = 6 x = ln 6

4 x

4

4 4 − 2 − 4 ln 2 − 3 2

5 + 4 ln 2 − 4 ln 3 3 5 + 4 (ln 2 − ln 3) 3 5 2 + 4 ln 3 3



[Shown]

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

28

ACE AHEAD Mathematics (T) Second Term

Thus, the curve y = 6 − e x intersects the x-axis at (ln 6, 0). On the y-axis, x = 0. y = 6 − e0 y=5 Thus, the curve y = 6 − e x intersects the y-axis at (0, 5). As x → ∞, y → − ∞ As x → − ∞, y → 6 y 6

5 y = 6 − ex

y = 5e−x (In 5, 1) O In 6 x

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 − ex …  y = 5e−x …  Substituting  into , 6 − e x = 5e−x x 6e − (e x)2 = 5 Letting e x = p, 6p − p2 = 5 p2 − 6p + 5 = 0 (p − 1)(p − 5) = 0 p = 1 or 5 When p = 1, ex = 1 x = ln 1 x=0
© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

29

When x = 0, y = 6 − e0 = 5 When p = 5, ex = 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 =

∫ 36 − e  − 5e 4 dx ln 5 0 x −x

= 6x − e x − = 6x − e x +

3 3

5 −x e (−1) 5 ex

4 

ln 5 0

4

ln 5 0

= 6 ln 5 − eln 5 = 6 ln 5 − 5 +

5 5 − 0 − e0 + 0 eln 5 e

5 − (−1 + 5) 5 = 6 ln 5 − 5 + 1 + 1 − 5 = (6 ln 5 − 8) units 2 Volume of the solid generated = π

∫ 36 − e  − 5e  4 dx = π∫ 336 − 12e + e − 25e 4 dx ln 5 0 x −x ln 5 0 x 2x −2x

2

2

1 25 −2x = π 36x − 12e x + e2x − e 2 (−2) 1 25 = π 36x − 12e x + e2x − 2x 2 2e

3 3 3 3

4

ln 5 0

4

ln 5 0

1 25 1 25 = π 36 ln 5 − 12eln 5 + e 2 ln 5 + 2 ln 5 − 0 − 12e0 + e0 + 0 2 2e 2 2e 1 25 1 25 = π 36 ln 5 − 12(5) + (25) + − −12 + + 2 2 2 2(25) = π (36 ln 5 − 48) = 12(3 ln 5 − 4)π units3 34 Let u = 1 − x du = − 1 dx dx = − du When x = 0, u = 1. When x = 1, u = 0.

4

4

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

30

ACE AHEAD Mathematics (T) Second Term

1 0

x2 (1 − x) 3 dx = = = = =

1

∫ ∫ ∫

0 1 0 1 0 1 0 1

(1 − u)2 u 3 (−du) − u 3 (1 − u)2 du − u 3 (1 − 2u + u2) du
1 3 1 1

1

∫ − u

+ 2u 3 − u 3 du
7 3 10 3

4

7

3
3

u 2u u + − 4 7 10 3 3 3
4 3 7 3

4 3

4
7 3

0

1

3 6 3 = − u + u − u 4 7 10

10 3

4

0

1

3 6 3 = 0 − − (1) + (1) − (1) 3 4 7 10 3 6 3 − + 4 7 10 27 = 140 = 35 (a) y 3

4 3

10

4

−1 −2 O 2 R −2 −3 −4

y2 = x − 2 x

y1 = x 2 − 4

(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).
© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

31

(c) Area of R =

∫ (y − y ) dx = ∫ 3(x − 2) − (x − 4)4 dx = ∫ (− x + x + 2) dx x x = 3− + + 2x4 3 2
−1 2 2 1 2 −1 2 2 −1 3 2 2 −1

2

(−1)3 (−1)2 23 22 + + 2(2) − − + + 2(−1) 3 2 3 2 10 7 = − − 3 6 9 = units2 2 =−

3

4

 

(d) Volume generated = π =π =π

∫ ∫

2

−1 2

(y12 − y22) dx 3(x2 − 4)2 − (x − 2)24 dx
4 2 2

−1 2

∫ 3(x − 8x + 16) − (x − 4x + 4)4 dx = π ∫ 3(x − 9x + 4x + 12) dx x = π 3 − 3x + 2x + 12x4 5
−1 2 4 2 −1 5 2 3 2 −1

=π =π =

32 − 3(2) 5
5

3

+ 2(2)2 + 12(2) −

 

(−1)5 − 3(−1)3 + 2(−1)2 + 12(−1) 5

4

372 − − 364 5 5

108 π units3 5

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

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...INTRODUCTION TO CHAOS THEORY BY T.R.RAMAMOHAN C-MMACS BANGALORE -560037 SOME INTERESTING QUOTATIONS * “PERHAPS THE NEXT GREAT ERA OF UNDERSTANDING WILL BE DETERMINING THE QUALITATIVE CONTENT OF EQUATIONS; TODAY WE DO NOT KNOW WHETHER THE EQUATIONS OF FLUID FLOW CONTAIN THE BARBER POLE STRUCTURE OF TURBULENCE; TODAY WE DO NOT KNOW WHETHER THE SCHRODINGER EQUATION CONTAINS FROGS, MUSICAL COMPOSERS OR MORALITY OR WHETHER SOMETHING BEYOND IT LIKE GOD IS NEEDED OR NOT “ PARAPHRASED FROM R.P.FEYNMAN * “ONLY A CORNER OF THE VEIL HAS BEEN LIFTED, BUT PERHAPS WE HAVE STARTED ON THE ABOVE JOURNEY ” PARAPHRASED FROM FLORIS TAKENS WHAT IS CHAOS THEORY? * CHAOS THEORY IS BASED ON THE OBSERVATION THAT SIMPLE RULES WHEN ITERATED CAN GIVE RISE TO APPARENTLY COMPLEX BEHAVIOR. * EG. LET US CONSIDER X0 RULE X1 X2 RULE | | | | X1 XN RULE X N+1 THE BASIC OBSERVATION IS ; IF IN THE IMPLEMENTATION OF THE ABOVE RULE X 0 IS CHANGED BY A SMALL AMOUNT, THE RESULTING SEQUENCE WILL BE VERY DIFFERENT. AFTER SOME ITERATIONS MATHEMATICIANS HAVE STILL NOT AGREED ON A DEFINITION OF CHAOS THAT IS ACCEPTABLE TO ALL HOW EVER AN OPERATIONAL (MEANING ONE THAT CAPTURES MOST ESSENTIAL FEATURES) DEFINITION OF A CHAOTIC SYSTEM IS ; 1) THE SOLUTION MUST BE APERIODIC 2) THE SOLUTION MUST BE BOUNDED 3) THE SOLUTION MUST BE EXPONENTIALLY SENSITIVE TO INITIAL CONDITIONS THE RULE THAT GOVERNS A CHAOTIC SYSTEM MUST SATISFY CERTAIN PROPERTIES 1) THE RULE MUST BE NONLINEAR 2) IN THE...

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#### Divorce Chaos

...Divorce Chaos Rebecca H. Fischer, Esq. Howard Chusid, Ed.D, NCC, LMHC The scenario is one that is known to many. Your spouse and you are continually arguing. You can’t agree on anything. Everything, be it little or large, becomes an argument. The children hear it, your friends hear it in your voice, and your parents hear it in your interactions with them. You decide to visit a counselor, together and individually, but it doesn’t help. You still are not getting along and are left with few options. So, your friends tell you of a great attorney who has done great for many. You visit with the attorney and are told that you need a \$5000 retainer but you have no idea where that is going to come from. The attorney also tells you that the cost for the divorce will be in the \$50,000 – \$80,000 range by the time you get through. You don’t have that kind of money to throw away on attorneys’ fees. If you are feeling this way, there is a high probability your spouse is feeling the exact same way. The exclusion of happiness is always felt by both partners. Problems and irritations are not just a unitary function of a marriage; both partners experience it, usually at the same time. Moreover, we should not forget the kids because they always know what is going on. There are three models of way to get through a divorce: the litigated model, collaborative model and the mediation option. The last two are similar in ideology but are different in approach. The litigated model is the one......

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#### Chaos and Fractals

...Chaos and Fractals What is a Fractal A fractal is a never ending pattern. Fractals are never ending complex patterns that are self-similar across different scales. They are created by by being repeated over and over in a feedback loop. Fractals are everywhere. They can be found in nature, algebra, and math. Nature Algebra Geometry Mandelbrot Set This was created in 1980 shortly after the first personal computer in order to calculate numbers thousands and sometimes millions of times. Equation (old)Z=(new)Z2+C We start by plugging a value for the variable C into the simple equation below. Each complex number is actually a point in a 2-dimensional plane. The equation gives an answer, Znew . We plug this back into the equation, as Zold and calculate it again. We are interested in what happens for different starting values of C. When you square a number it gets bigger. Then you square the answer and it get even bigger.. Eventually, it goes to infinity. This is the fate of most starting values of C. Some values of C do not get bigger, but get smaller, or alternate between a set of fixed values. These are the points inside the Mandelbrot Set, which we color black. Outside the Set, all the values of C cause the equation to go to infinity, and the colors are proportional to the speed at which they expand. Ju Fractals in Nature Natural fractals can be found anywhere in nature. Even in our body. It can be the blood vessels in our arms, or...

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#### Chaos in the Caribbean

...Chaos in the Caribbean Strayer University Forensic Accounting and Fraud Examination ACC 571 Dr. Timothy Brown ------------------------------------------------- 1. Evaluate Avey’s role as an expert witness for the Jamaican government. Avey and his firm were hired by the Jamaican government starting in the early in the 90s to investigate accusations of fraud and mismanagement and prepare reports outlining his findings starting with the Blaise Merchant Bank and Trust Co which spread to two similar but larger cases involving Century National Bank and its related financial entities and Eagle Merchant Bank. As an expert witness, Avey was hired by the Jamaican government to provide forensic investigation and audit support. He utilized specialized investigative skills in carrying out an inquiry conducted in such a manner that the outcome would be applicable to a court of law. In addition he examined evidence regarding assertions to determine its correspondence to established criteria carried out in a manner suitable to the court. Avey conducted his investigation grounded in sound forensic accounting principles where he quickly discovered self dealing in the Blaise Case where money was lent from one Blaise financial entity to companies controlled by its principal shareholders. In the Century Case, dishonesty was the main cause of the problems where the use of depositors’ funds to acquire such assets as real estate (which had also been the case with Blaise) for the...

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#### Order and Chaos in Jurassic Park

...Film Analysis 1 Fine Arts Film- Spring 2014 Order and Chaos in Jurassic Park First time watching Jurassic Park the theme that stands out most is Man vs. Nature but this is not the best fit. The true theme is almost shown straight forward during the helicopter ride to the island when Dr. Malcolm states, “John doesn't subscribe to Chaos, particularly what it has to say about his little science project!” (Koepp 19) pitting him against Hammond. These two are the most obvious players in the chaos game, but Dr. Sattler against Dr. Grant and the Tyrannosaurus Rex (T. rex) against the Velociraptors (Raptors) are also players. The other characters do not have as defined roles but still play into the game. They all have a common theme: Chaos vs. Order. Most of the main characters arrive on the island on the same helicopter. The first view inside shows Hammond, Dr. Sattler, and Dr. Grant on the right with Dr. Malcolm and Gennaro on the left. This scene shows who is on which side of the line, but Hammond’s and Dr. Malcolm’s conflict with each other is most prevalent. Hammond is wearing all white while Dr. Malcolm is, “dressed all in black” (Koepp 18) adding visual representation of the polar difference between chaos and order. Chaos and order is the root topic during the Velociraptor birth and lunch scene. Dr. Sattler and Dr. Grant debate which will win in the end but Dr. Malcolm and Hammond have already made up their minds. Chaos is thrown its first blow when Dr. Malcolm is hurt......

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#### A Solution to the Urban Chaos of Cairo

...A Solution: Urban Chaos in Cairo What is known today as the Greater Cairo Region is in fact a combination of a number of neighbouring cities and governorates (including Cairo, Giza, Helwan, Qalyubia and 6th of October. Since the first establishment in the region, Al- Fustat in 641 AD, Cairo was in continuous growth. The numerous and different reigns over Egypt since the establishment of Cairo, different Islamic periods, Ottoman, French, English until the revolution in 1952, have led to the inconsistency in development and urban planning which gave rise to cumulative problems. During the course of the last fifty years, the population of Cairo has increased dramatically as a result of high birth rates and large rural-urban migration from other parts of the country (Yousry and Aboul Atta). Currently Cairo is considered one of the biggest cities in the world with a population averaging at 14-15 million (Sims, 2003). The fast and continued increase in the population has been a major urban challenge that the city encountered. Today, the urban landscape of the Greater Cairo Region (GCR) is somewhat unclear: main built up, dense areas are surrounded by a number of satellite settlements at the periphery, but most importantly occupied by infinite unplanned and random constructions in the core and surroundings of the city. Informal settlements, that are named after their informal legal condition, gradually, but rapidly, took over a large area of Cairo’s landscape and crawled over......

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#### Chaos Is Come Again

...Chaos is Come Again Is it easy to turn a sane, loving man into a murderer? The famous play Othello, The Moor of Venice by William Shakespeare introduces the story of a brave black general in a predominantly white society, who has fallen in love with Desdemona, and struggles to see the true character of those around him. Iago his close friend was driven by jealousy to manipulate everyone he came in contact with. Othello is accepted by the Venetian leaders because his dedication to their military. When the story starts Othello’s strengths can be described as self-sacrificing and loving Desdemona wholeheartedly. Desdemona is quickly enveloped by this and expresses her love for Othello; “My heart’s subdued/Even to the very quality of my lord. / I saw Othello’s visage in his mind, /And to his honors and his valiant parts” (1.3.253-256). Desdemona is in love with Othello’s commitment to war and even points out that she was truly in love when she knew his thoughts. Othello is too trusting, insecure about himself, and he makes snap judgments. All of these weaknesses are used against him by Iago. “Chaos is come again” transforms Othello from a strong black warrior, loving husband, and trusting man into a cynical shell of his former self. Iago uses ethos to make Othello believe that he is trustworthy which helps him continue to make Othello doubt his wife. Othello's inclination to trust Iago is easily perceived, “The Moor already changes with my poison" (3.3.325), saying this shows......

Words: 1096 - Pages: 5