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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