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
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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

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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

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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)

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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




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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]

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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
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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
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     
[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

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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









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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)]

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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)

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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)

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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

=π

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