Am-sln-20(e)

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Chapter 20 Applications of Definite Integrals

CHAPTER 20

5. By solving the simultaneous equations y = x 2 and y 2 = x , the points of intersection are (0, 0) and (1, 1).

Exercise 20A (p.225)

∴ The required area =

1. y = x 2 − 4 = ( x + 2)( x − 2) When y = 0 , x = ±2

2x 2 x3 1 − ]0 3 3 1 = 3

2

x3 2 ]−2 3

= [4 x − = 10

2 3

2. y 2 = 4 x y2 4 y2 dy −4 4 4 y2 = dy −4 4 y3 = [ ]4−4 12 2 = 10 3

∴ The required area =

∫ ∫

4

3  6. (a)  y = x ..................(1)  y = 4 x ................(2) Put (2) into (1), x 3 − 4 x = 0 x = 0 , 2, −2 y = 0 , 8, −8 ∴ The coordinates of A are (2, 8) . The coordinates of B are ( −2, − 8) . (b) The required area 2

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

π 2 π 2

∫0

2 sin x dx

x4 x4 2 − 2 x 2 ]0−2 + [2 x 2 − ]0 4 4 =8 =[

7. (a) y = ( x + 1)( x − 2)( x − 4) The coordinates of B are (2, 0) . The coordinates of C are ( 4, 0) .

2 sin xdx π

= 4[ − cos x ]02 =4

(b) When −1 < x < 2 , y > 0 When 2 < x < 4 , y < 0 ∴ The required area

4. When x 3 − x = 0, x = −1, 0, 1 When −1 < x < 0 , y > 0 When 0 < x < 1 , y < 0 ∴ The required area 0

3

− x )dx −

1

∫0 ( x

3

− x )dx

x4 x2 0 x4 x2 1 − ]−1 − [ − ]0 4 2 4 2 1 = 2 =[

2

∫−1 ( x + 1)( x − 2)( x − 4)dx 4 − ∫ ( x + 1)( x − 2)( x − 4)dx 2 2 = ∫ ( x 3 − 5 x 2 + 2 x + 8)dx −1 4 + ∫ ( − x 3 + 5 x 2 − 2 x − 8)dx 2 =

∫−1 ( x

3

∴ The coordinates of A are ( −1, 0) .

π 2

∫−

=2

=

x − x 2 )dx

=[

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

3. The required area =

1

∫0 ( 3

2

∴ The required area =

x=

141

x 4 5x 3 − + x 2 + 8 x ]2−1 4 3 − x 4 5x 3 +[ + − x 2 − 8 x ]24 4 3 63 16 = + 4 3 1 = 21 12 =[

142

Chapter 20 Applications of Definite Integrals

y

8.

R

A2 = 9 − 4

y2 = x

1 2

1 2 A1 : A2 = 1 : 1

=4

x =4



A x

O

11. (a) For the curve y 2 = 4 x ,

y = −1 Q

dy =4 dx dy 2 = dx y dy 2 = dx (1, 2 ) 2 =1 2y

P

Put x = 4 into y = x , then y = ±2 . Put y = −1 into y 2 = x , then x = 1 . 2

Therefore the coordinates of P, Q and R are P(4, −2), Q(1, −1) and R(4, 2). ∴ The required area =

2

∫−1 (4 − y

2

Equation of the tangent is

)dy

y − 2 = x −1 x − y +1 = 0

y3 = [ 4 y − ]2−1 3 =9

y2 − ( y − 1)]dy 0 4 y3 y 2 =[ − + y]20 12 2 2 = 3

(b) The required area =

9. Solving the simultaneous equations x = y 2 and x = 8 − y 2 , we obtain that y = ±2 . 2

∫−2 [(8 − y ) − y 2 = 4 ∫ ( 4 − y 2 )dy 0

∴ The required area =

2

2

]dy

1 = 4[ 4 y − y 3 ]20 3 1 = 21 3 10. (a) S o l v i n g t h e s i m u l t a n e o u s e q u a t i o n s y = 4 x − x 2 and y = x 2 − 2 x , we obtain that x = 0 or 3. ∴ The required area 3

∫0 [(4 x − x ) − ( x 3 = ∫ (6 x − 2 x 2 )dx 0 =

2

2





(b) Area of rectangle = 2( at 2 )(2 at ) = 4a 2t 3 8 a2t 3 shaded area = 3 23 area of rectangle 4 a t 2 = for any t > 0 3

2 = [3 x 2 − x 3 ]30 3 =9

3

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

A1 =

2

3 1 = [ x 2 − x 3 ]30 2 3 1 =4 2

[

12. (a) The required area = shaded area 2 at y2 = ( at 2 − )dy −2 at 4a 2 at y2 =2 ( at 2 − )dy 0 4a 1 3 2 at 2 = 2[at y − y ]0 12 a 8 = a2t 3 3

− 2 x )]dx

(b) The line joining the points of intersection is y = x.



2

13.

π

∫0 [sin x − k( x − π)

2

]dx = 4

k ( x − π )3 π ]0 = 4 3 kπ 3 1 − ( −1 + )=4 3

[ − cos x −

6 k=− 3 π

Chapter 20 Applications of Definite Integrals

14. (a)

dy = 3 − 6x dx y = (3 − 6 x ) dx = 3 x − 3 x 2 + C1

4 1 − 4) − (64 − 2 2 ) 3 2

= 30 + (



143

14 8 + 2 3 3 2 = (7 + 4 2 ) 3 =

Since (1, 6) lies on curve C, 6 = 3(1) − 3(1)2 + C1 C1 = 6 Hence the equation of curve C is y = −3 x 2 + 3 x + 6 .

y

17. (b) When y = 0 , x = 2 or −1

2

y =x −1

When −1 < x < 2 , y > 0

B

∴ The required area =

2

∫−1 (−3x

= [− x 3 + = 10 +

2

+ 3 x + 6)dx

3x 2 + 6 x ]2−1 2

7 2

A1

A x−y+5 =0

O

27 = 2

C A2

15. (a) 1 + sin α = 2 cos α (α is acute) 1 + 2 sin α + sin 2 α = 4 cos 2 α 5 sin 2 α + 2 sin α − 3 = 0 (sin α + 1)(5 sin α − 3) = 0 3 sin α = (Q α is acute) 5 cos α =

4 5

3π 2

∫α

The coordinates of C and D are (−1, 0) and (1, 0) respectively. Solving the simultaneous equations y = x 2 − 1 and y = x + 5, we obtain that the coordinates of A and B are (−2, 3) and (3, 8) respectively.

(1 + sin x − 2 cos x )dx

∴ The required area 4 4 dx − x 2 dx 2 2 x 4 1 = [2 x 2 ]14 + [ ]1 2 − [ x 3 ]4 2 x 3 4

x3 x2 + + 6 x ]3−2 3 2 5 = 20 6

A2 = −

of intersection are (1, 4), ( 2 , 2) and (4, 16).

∫1 4 xdx − ∫1

2

2

= [−

3π 2

16. By solving y = x 2 , y = 4 x and x 2 y = 4 , the points

=

3

∫−2 ( x + 5 − x + 1)dx 3 = ∫ ( − x 2 + x + 6)dx −2

A1 + A2 =

= [ x − cos x − 2 sin x ]α 3π 4 3 = + 2 − α + + 2( ) 2 5 5 3π −1 3 = + 4 − sin ( ) 2 5



x

Put y = 0 into y = x 2 − 1 , then x 2 − 1 = 0 x = 1 or −1

(b) The required area =

D

1

∫−1 ( x

2

− 1)dx

x3 − x ]1−1 3 1 =1 3 5 1 ∴ A1 = 20 − 1 6 3 1 = 19 2 = −[

144

Chapter 20 Applications of Definite Integrals

18. (a) A sketch of the graph of y = cos 2 x on the given figure is shown as follows.

π π π 3 ( − − ) 2 2 3 4 π2 3π = − 12 8 =

y y = cos 2x

π

π 2

O

x

y = cos x

= π[

π 4

∫0 πx dy = π ∫ tan 2 ydy 0 = π ∫ (sec 2 y − 1)dy 0

4. The required volume =

2

π 4

(c) The required area =

2

y3 + y 2 + y]12 3 19π = 3

(b) cos x = cos 2 x 2 cos 2 x − cos x − 1 = 0 (2 cos x + 1)(cos x − 1) = 0 1 or cos x = 1 cos x = − 2 2π or x=0 x= 3

2

∫1 πx dy 2 = π ∫ ( y + 1)2 dy 1 2 = π ∫ ( y 2 + 2 y + 1)dy 1

3. The required volume =

2π 3

∫0 (cos x − cos 2 x )dx π + ∫ (cos 2 x − cos x )dx

π 4

π

= π[tan y − y]04 π = π(1 − ) 4

2π 3

sin 2 x 23π sin 2 x ]0 + [ − sin x ]π2 π 3 2 2 3 3 3 3 = + 4 4 3 3 = 2 = [sin x −

1

∫0 πy dx 1 = ∫ πx 4 dx 0



=



x

2

]dx

π [2 x 2 + 4 x ]24 4 = 8π

2

x5 = π[ ]10 5 π = 5 2. The required volume =

2

=

Exercise 20B (p.233) 1. The required volume =

x+2

4

∫2 π[( 2 ) − ( 2 ) π 4 = ∫ ( 4 x + 4)dx 4 2

5. The required volume =

π 2 π 3 π 2 π 3



6. By solving the simultaneous equations y = x 2 and y = 2 − x 2 , the points of intersection are (1, 1) and (−1, 1). ∴ The required volume 1

π cos 2 xdx

∫−1[π(2 − x ) − π( x ) ]dx 1 = π ∫ ( 4 − 4 x 2 + x 4 − x 4 )dx −1 1 = π ∫ ( 4 − 4 x 2 )dx −1

π 2

= π[ 4 x −

πy 2 dx

1 = π π (1 + cos 2 x )dx 3 2 π sin 2 x π2 = [x + ]π 2 2 3

=

=

16 π 3

2 2

4x3 1 ]−1 3

2 2

Chapter 20 Applications of Definite Integrals

7. By solving the simultaneous equations y = x and

∫−4 ( y − 3 )dx 4 = π ∫ (16 − x 2 )dx −4

The required volume = π

y = x , the points of intersection are (0, 0) and (1, 1). 2

∴ The required volume =

1

∫0 π( y

2

− y 4 )dy

y3 y5 1 − ]0 3 5 2π = 15

12.

x 2 y2 + =1 a2 b2

x2 y 2 = b 2 (1 − 2 ) a 2 b = b2 − 2 x 2 a ∴ The required volume

9. Consider a circle with centre (0, 0) and radius r, then the equation of the circle is x 2 + y2 = r 2 y2 = r 2 − x 2 r

=

∫−r πy dx r = 2 π ∫ (r 2 − x 2 )dx 0

The volume of sphere =

2

= 2 π[r 2 x −

a

∫a − h πy dx 2

b2 (b 2 − 2 x 2 )dx a−h a 2 3 b x = π[b 2 x − 2 ]aa − h a 3 b2h2 π = (3a − h) 3a 2 =π

x3 r ]0 3

1 = 2 π(1 − )r 3 3 4 3 = πr 3

1 3 4 x ]0 3

256 π 3 ∴ The volume of the remaining part of the 256 π cm 3 . sphere is 3

∫0 π(4 y − y)dy

3y 2 1 ]0 2 3π = 2

2

=

1

= π[

2

= 2 π[16 x −

= π[

8. The required volume =

4

145



a

y

13.

10. B y s o l v i n g t h e s i m u l t a n e o u s e q u a t i o n s 4 x 2 + 9 y 2 = 36 a n d y = 0 , t h e p o i n t s o f intersection are (3, 0) and (−3, 0).

36 − 4 x 2 )dx −3 9 3 4x2 )dx = π (4 − 9 −3 3 4x2 )dx = 2π (4 − 9 0 4x3 3 ]0 = 2 π[ 4 x − 27 = 16 π

The required volume =



3



11. Consider the equation of the circle x 2 + y 2 = 25 . x 2 = 25 − 9 x = ±4

C

π(



When y = 3,

2

V1 1 1 2

A

V2 B x

O

(a) Solving the simultaneous equations y = x = 2 and y = x ,

∴ the coordinates of A are (1, 1) , 1 ), 2 the coordinates of C are (2, 2) .

the coordinates of B are (2,

1 , x

146

Chapter 20 Applications of Definite Integrals 2

∫1 [πx − π( x ) ]dx 2 1 = π ∫ ( x 2 − 2 )dx 1 x

(b) The required volume =

2

1

2

x3 1 2 + ]1 3 x 11π = 6

15. (a) (i) Capacity of the pot =

y2 8 ]0 2 = 32 π = π[

= π[

2

(ii) The required volume

= π[ 4 y − =

5π 3



2

y2 + y]8−1 − 32 π 2 1 = 40 π − 32 π 2 17π = 2

(b) The volume of liquid inside the pot at a height h units is given by

1



∴ The required volume = V1 + V2 =

V=

h

8π 3

∫0 x dy h = π ∫ (20 y − y 2 )dy 0 2

= π[10 y 2 −

h

∫0

πydy =

πh 2 2

dV π dh dh = (2 h) = πh = 2π dt 2 dt dt

14. (a) Let V cm 3 be the volume of water inside the bowl when the depth is h cm. V=π

∫−1 π( y + 1)dy − 32π

= π[

y3 2 ]1 3

1 π(2 2 − 2 )dy 1 y 2 1 1 = π 1 ( 4 − 2 )dy y 2 11 = π[ 4 y + ] 1 y 2 =π

V2 =

8

=

∫1 π(2 − y )dy 2 = π ∫ ( 4 − y 2 )dy 1

(c) V1 =

2

8

∫0 πydy

1 3 h y ]0 3

1 2 πh (30 − h) 3 ∴ The volume of water inside the bowl is 1 2 πh (30 − h) cm 3 . 3 1 (b) V = 10 πh 2 − πh 3 3 dV dh = (20 πh − πh 2 ) dt dt dV When h = 5 and = 10 , dt dh 10 = [20 π(5) − π(5)2 ] dt dh 2 = dt 15π ∴ The rate of increase of the water level is 2 cm / s . 15π



∫ hdh = 2∫ dt

h2 = 2t + C 2 When t = 0 , h = 0 . ∴ C = 0 ∴

h 2 = 4t

16 =4 4 ∴ The time taken to fill the pot with water up to a level of 4 units is 4 minutes. When h = 4 , t =

=

Revision Exercise 20 (p.236) 1. A sketch of the graph is shown as follows. y y = |(x − 1)(x + 1)|

A2 A1 −1

O

x 1

2

147

Chapter 20 Applications of Definite Integrals

(b) The required area

1

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

A1 =

= [x − =

3x + 4 x 2 − )dx −1 4 4 1 4 = (3 x + 4 − x 2 )dx 4 −1 x3 1 3x 2 = [ + 4 x − ]4−1 4 2 3 5 =5 24 =

2

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

 y = x 2 4. (a)  3  y = cx

x3 − x ]12 3 4 = 3 ∴ The required area = A1 + A2 8 = 3 =[

2.

(



x3 1 ]−1 3

4 3

A2 =



4



x=

1 c

1 1 ∴ The coordinates of A are ( , 2 ) . c c (b) The area of the shaded region =

x 4 − 4 x 2 = 5x 2

1 c

∫0 ( x

2

− cx 3 )dx

x 3 cx 4 1c − ] 3 4 0 1 1 = 3− 3 3c 4c 1 = 12c 3 2 = 3 1 1 ∴ c3 = , c = 2 8 =[

x 2 ( x 2 − 9) = 0

x = 0 , ±3 3

∫0 [5x − ( x − 4 x 3 = 2 ∫ (9 x 2 − x 4 ) dx 0

∴ The required area = 2

2

= 2[3 x 3 − = 64.8

4

1 5 3 x ]0 5

2

)] dx

π )=0 3 π 2x + = π 3 π x= 3

5. (a) sin(2 x + 3. (a) 3 x − 4 y + 4 = 0 ........................(1)  x 2 ...................................(2)   y = 4 Put (2) into (1),

π ∴ The coordinates of A are ( , 0) . 3

3x − x 2 + 4 = 0 x 2 − 3x − 4 = 0 ( x − 4)( x + 1) = 0 x = 4 or −1

Put x = −1 into (2), y =

π − 2x = 2x + x=

1 , 4

∴ The coordinates of A are ( −1,

1 ). 4

1 (16) , 4 ∴ The coordinates of B are ( 4, 4) .

Put x = 4 into (2), y =

π 6 y = sin 2 x π = sin 2( ) 6 3 = 2

π 3

π ∴ The coordinates of B are ( , 6

3 ). 2

148

Chapter 20 Applications of Definite Integrals

(b) The required area π 6

π = sin 2 xdx + π sin(2 x + )dx 3 0 6 π 1 1 π π = [ − cos 2 x ]06 + [ − cos(2 x + )] π3 2 2 3 6 1 1 = + 4 4 1 = 2





y  6. (a)  x = sin  2  x = − sin y y sin = − sin y 2 y sin + sin y = 0 2 y y y sin + 2 sin cos = 0 2 2 2 y y sin (1 + 2 cos ) = 0 2 2 y 1 y sin = 0 or cos = − 2 2 2 4 π 8π ∴ y = 0, 2 π, 4 π, L or y = , , L 3 3 4π 2π 3 When y = , x = sin = 3 3 2 3 4π , ). ∴ The coordinates of Q are ( 2 3 4π 3

∫0

y + sin y)dy 2 4π y = [ −2 cos − cos y]03 2 = 4.5

(b) The required area =

0

∫−1[ y( y − 1) − y]dy 0 = ∫ ( y 2 − 2 y)dy −1

A2 =

π 3

y3 − y 2 ]0−1 3 4 = 3 ∴ The required area = A1 + A2 8 = 3 =[

8. When x 2 − 6 = 0 , x = ± 6 When x 2 − 6 = x , x = −2, 3 ∴ The required area

=

−2

L

− xdx +

∫0

6

xdx

1 3 −2 1 x2 x ]− 6 + [ − x 2 ]0−2 + [ ]0 6 3 2 2 x2 x3 +[ − + 6 x ]3 6 2 3 ( 6 )3 8 27 ( 6 )3 = −12 + + 6 6 − +2+ + 3 3 2 3 −6 6 37 = 6 = [6 x −

(sin

y

9.

y 2 = 2x A1

y

7.

0

∫− 6 ∫−2 3 + ∫ x − ( x 2 − 6)dx 6 −( x 2 − 6)dx +

A

x = y(y −1)

(2, 2)

A2

A1 x

O

O

C

x

A2 y = −1

The equation of L is y = x . 2

∫0 [ y − y( y − 1)]dy 2 = ∫ (2 y − y 2 )dy 0

A1 =

= [y2 − =

4 3

y3 2 ]0 3

(a) By solving the simultaneous equations y2 = 2 x a n d x 2 + y2 − 4 x = 0 , x = 0 o r x=2 Put x = 2 into y 2 = 2 x , ∴ y = ±2

Chapter 20 Applications of Definite Integrals

The coordinates of A are (2, 2) .

The required area

4 The coordinates of centre C = ( −( − ), 0) 2 = (2, 0)

=

(b) Let A3 be the area of the quarter-circle OAC. A3 =

π( 2 ) 2 =π 4

A2 =

∫0

2

2 xdx

A1 = A 3 − A 2 = π −

∫0 ( x − 6 x + 8x − x + 4 x ) dx 4 + ∫ ( x 2 − 4 x − x 3 + 6 x 2 − 8 x ) dx 3 3 = ∫ ( x 3 − 7 x 2 + 12 x )dx 0 4 − ∫ ( x 3 − 7 x 2 + 12 x )dx 3 3

2

2

1 7 = [ x 4 − x 3 + 6 x 2 ]30 4 3 1 7 − [ x 4 − x 3 + 6 x 2 ]34 4 3 45 7 + = 4 12 5 = 11 6

3 2

2x 2 ]0 3 2 2 = ( 2 )3 3 8 = 3 = 2[

3

149

8 3

∴ The required area = 2 A1 = 2 π −

16 3

10. No solution is provided for the H.K.C.E.E. question because of the copyright reasons.

12. (a) y = (1 + x )(3 − x ) When y = 0 , x = −1 , 3 ∴ The coordinates of P are ( −1, 0) . ∴ The coordinates of Q are (3, 0) . y = 3 + 2x − x2 dy = 2 − 2x dx

11. (a) C1: y = x 3 − 6 x 2 + 8 x = x ( x − 2)( x − 4) C2 : y = x 2 − 4 x = x ( x − 4)

A sketch of the curves is shown as follows.

dy dy = 4, = −4 dx x = −1 dx x = 3 The equation of RP is y = 4( x + 1) .

y C 1: y = x 3 − 6 x 2 + 8 x

The equation of RQ is y = −4( x − 3) . By solving the equation of RP and RQ, the coordinates of R are (1, 8) .

(2, 0)

O

(4, 0)

x

(b) The required area

=

1 ( 4)(8) − 2

3

∫−1 (1 + x )(3 − x )dx

= 16 − [3 x + x 2 − 2

C 2: y = x − 4 x

(b) S o l v i n g t h e s i m u l t a n e o u s e q u a t i o n s

= 16 − =5

1 33 x ]−1 3

32 3

1 3

y = x 3 − 6 x 2 + 8 x and y = x ( x − 4) , x ( x − 2)( x − 4) = x ( x − 4) x ( x − 3)( x − 4) = 0 x = 0 , 3, 4

13 − 14. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.

150

Chapter 20 Applications of Definite Integrals

15. Capacity of the vessel 2π

∫0 πx dy 2π = π ∫ (1 + cos y)2 dy 0 2π 3 1 = π ∫ ( + 2 cos y + cos 2 y)dy 0 2 2 =

Put (1, 0) into (1),

2

A − 4 + B = 0 ....................(2) Put (0, 2) into (1), B=2

Put B = 2 into (2),

3 sin 2 y 2 π = π[ y + 2 sin y + ]0 2 4 = 3π 2

A=2 2  (ii)  y = 2 x − 4 x + 2 y = 2

16. (a) Solving the simultaneous equations y = cos x π π and y = sin 2 x , we obtain that x = or . 2 6 π 3 Put x = into y = cos x , y = 6 2 π 3 ). ∴ The coordinates of P are ( , 6 2 (b) Volume of the solid π 6

∫0

π sin 2 2 xdx +





π 2 π 6

π cos 2 xdx

π 6

1 1 = π ( − cos 4 x )dx 0 2 2 π 2 1 1 + π π ( + cos 2 x )dx 2 2 6 x sin 4 x π6 x sin 2 x π2 = π[ − ]0 + π[ + ]π 2 8 2 4 6 π 2 3 3π = − 4 16



1 17. (a) Put y = 2 into y = x 2 , x = ±2 2 ∴ The coordinates of B are (2, 2) . Put y = 2 into y = 2 x , x = 1



∴ The coordinates of P are (2, 2) . (b) Volume of the solid



y [(2 y) − ( )2 ]dy = π 2 2

8

∫2

(2 y −

= π[ y 2 − = 18π

2

∫0 π[2 − (2 x − 4 x + 2) ]dx 2 = − π ∫ ( 4 x 4 − 16 x 3 + 24 x 2 − 16 x )dx 0 =

2

2

2

4x5 − 4 x 4 + 8 x 3 − 8 x 2 ]20 5 32 π = 5 = − π[

19. (a) The equation of L is y − 2 = m( x − 1) . (b) Put y = x 2 into the equation of L,

x 2 − 2 = m( x − 1) x 2 − mx + ( m − 2) = 0 (i) α + β = m (ii) αβ = m − 2 (iii) β − α = m 2 − 4( m − 2)

Solving the simultaneous equations y = 2 x 1 and y = x 2 , we obtain that x = 0 or 4. 2 ∴ The coordinates of C are ( 4, 8) . 8

2x2 − 4x + 2 = 2 x = 0 or 2

When x = 2 , y = 2(2)2 − 4(2) + 2 = 2

∴ The coordinates of A are (1, 2) .

(b) π

y = Ax 2 − 4 x + B ...............(1)

18. (a) (i)

y2 )dy 4

y3 8 ]2 12

= m 2 − 4m + 8 (c) The required area = = = = =

β 1 2 (β + α 2 )(β − α ) − x 2 dx 2 α 1 2 1 3 2 (β + α )(β − α ) − (β − α 3 ) 2 3 1 2 2 (β − α )[3(β + α ) − 2(β 2 + αβ + α 2 )] 6 1 (β − α )3 6 3 1 2 ( m − 4 m + 8) 2 6



Chapter 20 Applications of Definite Integrals

(d) m 2 − 4 m + 8 = ( m − 2)2 + 4

dA dh = π( 4 h 3 + 4 h ) dt dt dh 2 = 4hπ(h + 1) dt dh 9 When h = 3 , (From (b)(i)), =− dt 20 dA 9 = 4(3)π(9 + 1)( − ) ∴ dt 20 = −54 π ∴ The rate of change of the surface area of the water at this moment is −54π

∴ When m = 2 , the area of the shaded region attains its minimum. ∴ Equation of L is y − 2 = 2( x − 1) y = 2x 20. (a) Let V cubic units be the volume of water. h

∫0 πx dy h = π ∫ ( y 2 + 1)2 dy 0 h = π ∫ ( y 4 + 2 y 2 + 1)dy 0

V=

2

y5 2 y3 + + y]0h 5 3 h 5 2h3 = π( + + h) 5 3 ∴ Volume of water h 5 2h3 = π( + + h) cubic units 5 3 = π[

h 5 2h3 + + h) 5 3 dV dh = π(h 4 + 2 h 2 + 1) dt dt dV = −5πh 2 (given) dt dV When h = 3 , = −5π(9) dt = −45π V = π(

π[34 + 2(3)2 + 1]

square units per minute. 21 − 25. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.

Enrichment 20 (p.244) Solving the simultaneous equations y = x 2 and y 2 = 8 x , we obtain that x = 0 , y = 0 or x = 2 , y = 4 . ∴ The volume of the solid

(b) (i) From (a),

dh = −45π dt dh −45 = dt 100 −9 = 20

2

∫0 x( 8x − x )dx 2 = 2 π ∫ ( 8 x − x 3 )dx 0 = 2π

2

3 2

5

4 2x 2 x4 2 − ]0 5 4 24 π = 5 = 2 π[

Classwork 1 (p.220) 1. Rewrite the equation y 2 = x as y = ± x . y

∴ The rate at which the level of the 9 water is falling when h = 3 is 20 unit per minute. (ii) Put y = h into y 2 = x − 1 , ∴ h 2 = x − 1 , or x = h 2 + 1 Let the surface area of water be A square units. A = πx 2 = π(h 2 + 1)2 = π(h 4 + 2 h 2 + 1)

151

y2 = x

A1 O

2

3

A2

x

152

Chapter 20 Applications of Definite Integrals

As shown in the diagram, the curve y 2 = x is split into two parts y = x and y = − x . The required area is the sum of A1 and A2 . Since the curve y 2 = x is symmetrical about the x-axis, A1 = A 2 .

A1 =

Classwork 3 (p.225) 1. By solving the simultaneous equations y 2 = 4 x and x + y = 8, the points of intersection are (4, 4) and (16, −8). The equations of the parabola and the line can be respectively written as x=

3

∫2

xdx

y2 and x = 8 − y . 4

y2 )dy 4 −8 y 2 y3 4 = [8 y − − ]−8 2 12 56 160 = + 3 3 = 72

∴ The required area =

3 2

2x 3 ]2 =[ 3 6 3−4 2 = 3 ∴ The required area = A1 + A 2 12 3 − 8 2 = 3

1

A2

−1

∴ The required area

∫0 − x + 3x − 2 x dx 1 = ∫ − x 3 + 3 x 2 − 2 x dx 0 2 + ∫ − x 3 + 3 x 2 − 2 x dx 1 2 1 = ∫ ( x 3 − 3 x 2 + 2 x )dx − ∫ ( x 3 − 3 x 2 + 2 x )dx 1 0

x

π

O

When 0 ≤ x ≤ 1 , y ≤ 0

3

y1 = sin x

P A1

= − x ( x − 3 x + 2) = − x ( x − 2)( x − 1) 2

2

(8 − y −

y

2.

2. y = − x 3 + 3 x 2 − 2 x

=



4

y2 = cos x

2

x4 x4 − x 3 + x 2 ]10 − [ − x 3 + x 2 ]12 4 4 1 1 = + 4 4 1 = 2 =[

The graphs of y1 = sin x and y2 = cos x on the interval 0 ≤ x ≤ π are shown. The two curves intersect at P. Solving for the x-coordinate of P, sin x = cos x tan x = 1 π x= 4

π . 4

The x-coordinate of P is ∴

π 4

∫0 ( y2 − y1 )dx = ∫ (cos x − sin x )dx 0

A1 =

π 4

π

= [sin x + cos x ]04 = 2 −1

Classwork 2 (p.221) 1 2

Rewrite the equation y = x 2 as x = ± y . ∴ The required area =

9

∫0 y

1 2

3

dy

2y 2 9 =[ ]0 3 = 18



π

∫ ( y1 − y2 )dx π = ∫ (sin x − cos x )dx

A2 =

π 4

π 4

= [ − cos x − sin x ]ππ 4

= 1+ 2 ∴ The required area = A1 + A2 =2 2

Chapter 20 Applications of Definite Integrals

(b) The required volume

Classwork 4 (p.231) 8

∫0 8 = ∫ πy 0

1. The required volume =

πx 2 dy 2 3

dy

5 3

3y 8 ]0 5 96 π = 5 = π[

4

∫0 πy dx 4 = ∫ πxdx 0

2. The required volume =

= π[ = 8π

2

x2 4 ]0 2

π

∫0 πx dy π = ∫ π sin 2 ydy 0 π1 = π∫ (1 − cos 2 y)dy 0 2

3. The required volume =

2

sin 2 y π π [y − ]0 2 2 π2 = 2 =

Classwork 5 (p.233) By solving the simultaneous equations y = 2 x 2 and y = 16 x − 30 , the points of intersection are (3, 18) and (5, 50). (a) The required volume 5

∫3 π[(16 x − 30) − 4 x ]dx 5 = π ∫ (256 x 2 − 960 x + 900 − 4 x 4 )dx 3 =

2

4

256 x 3 4x5 5 ]3 − 480 x 2 + 900 x − 3 5 2 656 π = 15 = π[

y + 30 2 ) ]dy 16 50 − y 2 17 y 225 =π ( + − )dy 18 256 64 64 − y 3 17 y 2 225 y 50 = π[ + − ]18 768 128 64 64 π = 3 =

50

∫18 π[ 2 − ( ∫

y

153

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