Chapter 17 Indefinite Integral and Its Applications
CHAPTER 17
13.
∫x
2
∫
( x 2 + 4)dx = ( x 4 + 4 x 2 )dx
∫ 2 x dx = 2 4 x
2.
∫
dx 1 = x+C 3 3
3.
∫
dx = x4
4.
∫
3dx x3 9 2 − 13 ( )+C = x3 +C = 3 x dx = 3 2 3 x 2 3
5.
6.
7.
8.
9.
10.
1 3
11 4 1 + C = x4 + C 8
∫
1.2
1
3 2 25 x + 2x 2 + C 5
∫ 8 sin xdx = −8 cos x + C
17.
x 15 ) + C = x 2.2 + C 2.2 11
∫ (sin x + cos x )dx = − cos x + sin x + C
18.
∫ (4 cos x − 3 sin x )dx = 4 sin x + 3 cos x + C
x 0.6 2 ) = x 0.6 + C 0.6 3
19.
∫ (a cos x + b sin x )dx = a∫ cos xdx + b∫ sin xdx
2.2
∫
∫
x3 x2 (3 x − 2 x + 1)dx = 3( ) − 2( ) + x + C 3 2 = x3 − x2 + x + C
= a sin x − b cos x + C
2
∫ (2 x
3
x4 x2 ) − 7( ) + 6 x + C 4 2 1 4 7 2 = x − x + 6x + C 2 2
− 7 x + 6)dx = 2(
20.
∫ 5 sec
21.
∫ sin 2 x = ∫ csc
22.
∫ (3
2
xdx = 5 tan x + C
dx
− x − 6)dx
3
= 3(
x2 3 2
1
) + 2(
3
x2
23.
∫ (4 x
2 3
+
3 x
− 13
1
∫
2
5
∫
=
24.
∫
x3
2
∫
+ 3 x − 5) x 3 dx = ( x 5 + 3 x 4 − 5 x 3 )dx =
1 6 3 5 5 4 x + x − x +C 6 5 4
1
5 3
4
) + 3(
x3 4 3
)+C
12 53 9 43 x + x +C 5 4
3 1 1 3x 2 + 2 x − 1 dx = (3 x 2 + 2 x 2 − x − 2 )dx x
∫
5
∫ (x
)+C
)dx = ( 4 x 3 + 3 x 3 )dx = 4(
= 6(
12.
1 2
= 2x 2 + 4x 2 + C
(2 x + 4)(3 x − 8)dx = (6 x 2 − 4 x − 32)dx x3 x2 ) − 4( ) − 32 x + C 3 2 = 2 x 3 − 2 x 2 − 32 x + C
xdx = − cot x + C
∫
3
2
2
1 1 2 )dx = (3 x 2 + 2 x − 2 )dx x
x+
x6 x4 x2 ( 4 x − 2 x + 9 x )dx = 4( ) − 2( ) + 9( ) + C 6 4 2 2 6 1 4 9 2 = x − x + x +C 3 2 2 5
∫ ( x + 2)( x − 3)dx = ∫ ( x ∫
3
16.
2
1 1 = x3 − x2 − 6x + C 3 2
11.
∫
xdx = ( x 2 + 3 x 2 )dx
∫ 4 cos xdx = 4 sin x + C
1 x −3 + C = − x −3 + C −3 3
0.4 x −0.4 dx = 0.4(
∫
∫ ( x + 3)
15.
x −4 dx =
dx = 3(
14.
=
∫
∫ 3x
1 5 4 3 x + x +C 5 3
=
Exercise 17A (p.177) 1.
115
= 3( =
x2 5 2
3
) + 2(
x2 3 2
1
)−
x2 1 2
+C
1 6 25 4 23 x + x − 2x 2 + C 5 3
116
25.
Chapter 17 Indefinite Integral and Its Applications
∫ ( x 4 − 23 x )dx = ∫ ( x 1
1
−4
−
−3
1 − 13 x )dx 2
(b) From (a), 4
2 3
x 1x − +C −3 2 23 1 3 2 = − x −3 − x 3 + C 3 4 =
26.
∫ 2 sin 2 cos 2 dx = ∫ sin xdx = − cos x + C
27.
∫
28.
29.
x
4 sin 2
∫ 3 tan
∫
2
∴
x
∫
∫
1.
∫ = ∫ (sec 2 x + csc 2 x )dx
2.
= sec x + tan x − x + C
=
32. (a)
2 + 2x2 dx = dy = y + C (1 − x 2 )2
dy = 8x − 1 dx
∫
dy = 4x + 1 dx
∫
y = ( 4 x + 1)dx = 2x2 + x + C Put (−1, 3) into y = 2 x 2 + x + C , C = 2 Hence the equation of the curve is y = 2 x 2 + x + 2 .
2
∫ (1 − x 2 )2 + ∫ (1 − x 2 )2 ] dx
2 + 2x
y = 2x − 1
Put (−1, −1) into y = 4 x 2 − x + C ,
3.
dy 2(1 − x 2 ) − 2 x ( −2 x ) 2 + 2x2 = = 2 2 dx (1 − x ) (1 − x 2 )2
=
dx = 4 x + x + C
−1 = 4( −1)2 − ( −1) + C C = −6 ∴ y = 4x2 − x − 6
∫ = ∫ (sec x tan x + sec 2 x − 1)dx
2[
x x+ x
∫
= (sec x tan x + tan 2 x )dx
∴
2 x +1
dx
y = (8 x − 1)dx = 4 x 2 − x + C
(sec x + tan x ) tan xdx
∫
x x+ x
dy =2 dx y = 2x + C ∴
= 3(tan x − x ) + C = 3 tan x − 3 x + C
(b) From (a),
2 x +1
Put (1, 1) into y = 2 x + C , C = −1
∫
xdx = 3(sec 2 x − 1)dx
(tan x + cot x )2 dx = (tan 2 x + 2 + cot 2 x )dx
31. (a)
x x+ x
Exercise 17B (p.183)
x dx = 2(1 − cos x )dx 2 = 2( x − sin x ) + C = 2 x − 2 sin x + C
= tan x − cot x + C 30.
∫ 4dy = ∫
∫
2 x +1
dy = dx
x dx
2
∫ (1 − x 2 )2 dx 2x +C 1 − x2
1 dy 1 1 1 = ( x + x ) − 2 (1 + x − 2 ) dx 2 2 1+ 1 2 x = 2 x+ x 2 x +1 = 4 x x+ x
4. If Q
dy d2y = kx + 3, =k dx dx 2 d2y =2 dx 2 k=2
∴ dy = 2x + 3 dx y = (2 x + 3)dx
∫
= x 2 + 3x + C
Put (0, 4) into y = x 2 + 3 x + C , C = 4 Hence the equation of the curve is y = x 2 + 3 x + 4 .
Chapter 17 Indefinite Integral and Its Applications
5. Since
∫
dy = tan x sec x , dx
y = (tan x sec x )dx = sec x + C
Put (0, 1) into y = sec x + C , C = 0 Hence the equation of the curve is y = sec x .
6. Since
dy = 3 x + cos x , dx
∫
y = (3 x + cos x )dx 3 = x 2 + sin x + C 2 3 Put (0, 3) into y = x 2 + sin x + C , C = 3 2 Hence the equation of the curve is 3 y = x 2 + sin x + 3 . 2
d2y =4 7. dx 2 dy = 4 dx = 4 x + C1 dx dy When x = 1 , = 2 . ∴ C1 = −2 dx dy = 4x − 2 dx y = ( 4 x − 2)dx
∫
∫
= 2 x 2 − 2 x + C2
Put (2, 5) into y =
1 3 1 2 x − x − x + C2 , 6 2
8 − 2 − 2 + C2 6 2 C2 = 7 3 Hence the equation of the curve is 1 1 2 y = x3 − x2 − x + 7 . 6 2 3 5=
9. v = 2t + 1
∫
s = (2t + 1)dt = t 2 + t + C When t = 0 , s = 0 . ∴ C = 0 s = t2 + t
10. v = 3 t + 2
∫
3
s = (3 t + 2)dt = 2t 2 + 2t + C When t = 0 , s = 5 . ∴ C = 5 3
s = 2t 2 + 2t + 5 11. a = 6t 2 + 4t
∫
v = (6t 2 + 4t )dt = 2t 3 + 2t 2 + C When t = 0 , v = 4 . ∴ C = 4
v = 2t 3 + 2t 2 + 4 12. a = 12t 2 + 6t
∫
Put (1, 3) into y = 2 x 2 − 2 x + C2 , C2 = 3
v = (12t 2 + 6t )dt = 4t 3 + 3t 2 + C1
Hence the equation of the curve is
When t = 0 , v = −3 . ∴ C1 = −3
y = 2x2 − 2x + 3. d2y = x −1 8. dx 2 1 dy = ( x − 1)dx = x 2 − x + C1 2 dx
∫
When x = 2 , ∴
dy = −1. ∴ C1 = −1 dx
dy 1 2 = x − x −1 dx 2 1 y = ( x 2 − x − 1)dx 2 1 3 1 2 = x − x − x + C2 6 2
∫
117
v = 4t 3 + 3t 2 − 3
∫
s = ( 4t 3 + 3t 2 − 3)dt = t 4 + t 3 − 3t + C2
When t = 0 , s = 0 . ∴ C2 = 0
s = t 4 + t 3 − 3t 13. v = 4t + 2
∫
s = ( 4t + 2)dt = 2t 2 + 2t + C Let s = 0 when t = 0 , then C = 0 .
s = 2t 2 + 2t When t = 4 , s = 2( 4)2 + 2( 4) = 40 ∴ The distance that the particle moves in the first four seconds is 40 .
118
Chapter 17 Indefinite Integral and Its Applications
When s = 20 , 5t 2 = 20 t = 2 or −2 (rejected) v = 10(2) = 20
14. v = 2t − 4
∫
s = (2t − 4)dt = t 2 − 4t + C When t = 0, s = 0 2 − 4(0) + C = C
∴ The velocity is 20 m / s .
(a) When t = 2 , s = (2)2 − 4(2) + C = −4 + C ∴ Distance = C − ( −4 + C ) = 4 (b) When t > 2 , v > 0
17. (a) (b)
When t = 5 , s = (5) − 4(5) + C = 5 + C 2
∴ Distance = 2( 4) + 5 + C − C = 13 15. a = 48t − 24
∫
v = ( 48t − 24)dt = 24t 2 − 24t + C1 When t = 0 , v = 6 . ∴ C1 = 6 v = 24t 2 − 24t + 6
∫
2 (a) s = (24t − 24t + 6)dt
= 8t 3 − 12t 2 + 6t + C2
d ( x tan x ) = x sec 2 x + tan x dx dy 1 = x sec 2 x + tan x dx 2 1 dy = ( x sec 2 x + tan x )dx 2 1 ∴ y = x tan x + C 2
∫
∫
π Since the curve passes through ( , 1) , 4 1 π π 1 = ( ) tan( ) + C 2 4 4 π C = 1− 8 Hence the equation of the curve is 1 π y = x tan x + 1 − . 2 8
When t = 0 , s = −3 . ∴ C2 = −3 s = 8t 3 − 12t 2 + 6t − 3
(b) Put v = 0 into v = 24t 2 − 24t + 6 , 24t 2 − 24t + 6 = 0 4t 2 − 4t + 1 = 0 (2t − 1)2 = 0 1 t= 2 1 ∴ The particle is stationary when t = . 2
16. a = 10
∫
v = 10 dt = 10t + C1 When t = 0 , v = 0 ∴
C1 = 0
Then v = 10t
∫
s = 10tdt = 5t + C2 2
When t = 0 , s = 0 ∴
C2 = 0
Then s = 5t 2
18. (a) v = 60t − 30t 2 Put v = 0 , 60t − 30t 2 = 0 t (2 − t ) = 0 t = 0 or t = 2 ∴ His travelling time is 2 hours.
(b) v = 60t − 30t 2
∫
s = (60t − 30t 2 )dt = 30t 2 − 10t 3 + C When t = 0, s = 30(0)2 − 10(0)3 + C =C When t = 2 , s = 30(2)2 − 10(2)3 + C = 40 + C ∴ The distance between town A and town B is ( 40 + C − C ) km = 40 km . (c) v = 60t − 30t 2 = 30 − (30 − 60t + 30t 2 ) = 30 − 30(1 − t )2 Therefore, v reaches its maximum when t = 1 . ∴ The maximum velocity is 30 km / h .
Chapter 17 Indefinite Integral and Its Applications
19. (a) a = 8 − 15t 2
4.
∫
v = (8 − 15t 2 )dt = 8t − 5t 3 + C
∫ (x
−2
119
+ 2 x −3 + 3 x −4 ) dx
x −1 x −2 x −3 + 2( ) + 3( )+C −1 −2 −3 = − x −1 − x −2 − x −3 + C =
When t = 0 , v = 24 . ∴ C = 24 Put v = 0 , then −5t 3 + 8t + 24 = 0
5t 3 − 8t − 24 = 0 (t − 2)(5t 2 + 10t + 12) = 0 t=2 ∴ It stops once when t = 2 .
5.
∫ (x − x ) 2
2
∫
4 )dx x2 1 4 = x3 − 4x − + C x 3
dx = ( x 2 − 4 +
(b) (i) v = −5t 3 − 8t + 24
∫
s = ( −5t 3 + 8t + 24)dt 5 4 t + 4t 2 + 24t + C1 4 5 = − t 4 + 4t 2 + 24t (Put C1 = 0 ) 4 When t = 2 , 5 s = − (2) 4 + 4(2)2 + 24(2) = 44 4 ∴ Distance travelled is 44 m .
∫
7.
∫
8.
∫ (x
=−
(ii) When t = 4 , 5 s = − ( 4) 4 + 4( 4)2 + 24( 4) 4 = −160 When 2 < t < 4 , v < 0 ∴ Distance travelled = [2( 44) + 160] m = 248 m
9.
Revision Exercise 17 (p.186)
∫ (x
2
∫
+ 2)2 dx = ( x 4 + 4 x 2 + 4)dx =
2.
1 5 4 3 x + x + 4x + C 5 3
∫ ( x − a)( x − b)dx = ∫ [ x =
3.
2
− ( a + b) x + ab]dx
1 3 1 x − ( a + b) x 2 + abx + C 3 2
∫ (ax + b)( px + q) dx = ∫ [apx 2 + (bp + aq ) x + bq] dx =
1 1 apx 3 + (bp + aq ) x 2 + bqx + C 3 2
∫
x x x dx =
5
∫x
7 8
dx =
11.
12.
8 158 x +C 15
∫
+ 3 x 3 − 5 x ) x −3 dx = ( x 2 + 3 − 5 x −2 )dx =
10.
1.
x 4 + 2x3 + 9 dx = ( x 2 + 2 x + 9 x −2 )dx 2 x 1 = x 3 + x 2 − 9 x −1 + C 3
6.
1 3 x + 3 x + 5 x −1 + C 3
∫ (3sin θ + 4 cos θ)dθ = −3 cos θ + 4 sin θ + C
∫ cot
2
∫ (sec
∫
θ dθ = (csc 2 θ − 1) dθ = − cot θ − θ + C
2
θ + csc 2 θ)dθ = tan θ − cot θ + C
1 − cos 2θ 1 − (1 − 2 sin 2 θ) = 1 + cos 2θ 1 + (2 cos 2 θ − 1) 2 sin 2 θ = 2 cos 2 θ = tan 2 θ = sec 2 θ − 1
1 − cos 2θ
∫ 1 + cos 2θ dθ = ∫ (sec θ − 1)dθ = ∫ sec 2 θdθ − ∫ dθ 2
= tan θ − θ + C
120
Chapter 17 Indefinite Integral and Its Applications
θ θ + cos 4 4 4 θ θ θ 4θ = (sin + 2 sin 2 cos 2 + cos 4 ) 4 4 4 4 2 θ 2 θ − 2 sin cos 4 4 θ θ 2 θ 2 θ 2 1 = (sin + cos ) − (2 sin cos )2 4 4 2 4 4 1 2θ = 1 − sin 2 2 θ (b) cos θ = 1 − 2 sin 2 2 1 1 1 2θ cos θ = − sin 4 4 2 2 3 1 1 2θ + cos θ = 1 − sin 4 4 2 2 1 θ θ θ (sin 4 + cos 4 )dθ = (1 − sin 2 )dθ 4 4 2 2 3 1 = ( + cos θ)dθ 4 4 3 1 = θ + sin θ + C 4 4
13. (a) sin 4
∫
dy = 3 cot 2 θ( − csc 2 θ) − 3( − csc 2 θ) dθ = −3 cot 4 θ − 3 cot 2 θ + 3 cot 2 θ + 3
∫ (−3 cot
x4 − 3x + C , 2 C=0
Put (2, 2) into y =
Hence the equation of the curve is y = (b)
= cot 3 θ − 3 cot θ + C 1
∫ ∫ 1 4 3 ∫ cot θdθ = −3 [cot θ + −3 cot θ − 3θ] + C
−3 cot 4 θdθ = cot 3 θ − 3 cot θ + C1 − 3 dθ
cot 3 θ + cot θ + θ + C 3
x4 − 3x , 2 7 y= 2
Put x = −1 into y =
∴ The coordinates of the point are ( −1,
7 ). 2
17 − 18. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons. 19. (a)
dV = 3t 2 + 2t dt V = (3t 2 + 2t )dt = t 3 + t 2 + C C = 14 − 2 3 − 2 2 = 2 V = t3 + t2 + 2
(b) When t = 5 ,
V = (5)3 + (5)2 + 2 = 152 ∴ The volume of the bubble is 152 cm 3 . 20. (a) V1 = 40 − 10t V2 = t 2 − 10t + 24 When t = 4 ,
dy = 8 − 4x dx y = (8 − 4 x )dx
V1 = 40 − 10( 4) = 0
∫
V2 = ( 4)2 − 10( 4) + 24 = 0 ∴ When t = 4 , V1 = V2
= 8x − 2 x 2 + C = −2( x − 2) + C1 Q −2( x − 2)2 ≤ 0 ∴ ymax = C1 = 7 2
∫
(b) When 0 ≤ t ≤ 4 , h = V1dt
∫
h = ( 40 − 10t )dt
y = −2( x − 2) + 7 = −2 x + 8 x − 1 2
x4 − 3x . 2
dy = −5 dx 2 x 3 − 3 = −5 x = −1
∴
=−
x4 − 3x + C 2
When t = 2 , V = 14
θ + 3)dθ
= y + C1
15.
=
∫
= −3 cot 4 θ + 3 4
dy = 2x3 − 3 dx y = (2 x 3 − 3)dx
∫
∫ ∫
14. y = cot 3 θ − 3 cot θ
∴
16. (a)
2
= 40t − 5t 2 (constant = 0)
Hence the equation of the curve is
When t = 4 , h = 40( 4) − 5( 4)2 = 80
y = −2 x 2 + 8 x − 1 .
∴ The height of the particle is 80 m .
Chapter 17 Indefinite Integral and Its Applications
∫
(c) When t ≥ 4 , h = V2 dt
∫
121
(c) By (b), 1 + 2(2) + 3(2)2 + L + 10(2)9
1 − (10 + 1)210 + 10(2)10 +1 (1 − 2)2 = 9 217
1 h = (t − 10t + 24)dt = t 3 − 5t 2 + 24t + C 3 When t = 4 , V1 = V2 (from (a)),
=
2
∫ = ∫ V1dt
h = V2 dt
Enrichment 17 (p.188)
= 80 (from (b)) 2 ∴ C = 42 3 When t = 5 , 1 2 1 h = (5)3 − 5(5)2 + 24(5) + 42 = 79 3 3 3 1 ∴ The height of the particle is 79 m . 3 dV1 (d) Q = −10 ≠ 14 dt dV ∴ Here we only consider a = 2 = 2t − 10 . dt When a = 14 ,
∫ [ dx f ( x )]dx = x d
2. (a)
21. (a) Sn ( x ) = 1 + 2 x + 3 x 2 + L + nx n −1
∫ Sn ( x )dx = ∫ (1 + 2 x + 3 x 2 + L + nx n −1 )dx 2
(b)
∴
x (1 − x n ) Sn ( x )dx = + C ...............(3) 1− x
∫
(1 − x )(1 − nx n − x n ) + x (1 − x n ) Sn ( x ) = (1 − x )2 n 1 − (n + 1) x + nx n +1 = (1 − x )2
d sec 2 x = 2 sec x (sec x tan x ) = 2 sec 2 x tan x dx
x d
x − x n +1 1− x
(b) Differentiate (3) with respect to x, d Sn ( x )dx dx (1 − x )[1 − (n + 1) x n ] − ( x − x n +1 )( −1) = (1 − x )2
∫
∫
x d ( sec 2 x ) 2 dx
∫ 2 ( dx sec x )dx 1 d = [ ∫ ( x sec 2 x )dx − ∫ sec 2 xdx ] 2 dx
(1) − (2), (1 − x ) K = x − x n +1
∫
∫
x d 1 d ( sec 2 x ) = [ ( x sec 2 x ) − sec 2 x ] 2 dx 2 dx
xK = x 2 + x 3 + x 4 + L + x n +1 ................(2)
∴
+ sin x + C
By (a),
Let K = x + x 2 + x 3 + L + x n .................(1)
If x ≠ 1 , K =
2
d [ xf ( x )] = xf ' ( x ) + f ( x ) dx d xf ' ( x ) = [ xf ( x )] − f ( x ) dx d xf ' ( x )dx = [ xf ( x )] dx − f ( x ) dx dx = xf ( x ) − f ( x ) dx
(c) x sec 2 x tan x =
n
nx 2x 3x + +L+ +C n 2 3 = x + x2 + x3 + L + xn + C =x+
d f ( x ) = 2 x + cos x dx
(b)
∫
2
x3 − cos x + C 3
f ( x )dx =
d [ f ( x )dx ] = x 2 + sin x dx
2t − 10 = 14 t = 12
∴
∫ ∫
1. (a)
2
∫ x sec =
2
x tan xdx
1 1 x sec 2 x − tan x + C 2 2
Classwork 1 (p.175)
∫
x2 ) + C = 2x2 + C 2
1.
∫
4 xdx = 4 xdx = 4(
2.
∫
7 x 2 dx = 7 x 2 dx = 7(
∫
x3 7 ) + C = x3 + C 3 3
122
3.
4.
5.
∫
Chapter 17 Indefinite Integral and Its Applications
∫
∫ (2 x
∫ (4
3
Classwork 4 (p.180)
3
x2
1
15 xdx = 15 x 2 dx = 15(
3 2
3
) + C = 10 x 2 + C
∫
dy 3 = 4x − x2 , dx 2 3 2 y = ( 4 x − x )dx 2 x2 3 x3 = 4( ) − +C 2 2 3 1 = 2x2 − x3 + C 2
1. Since
∫
∫
∫
+ x 2 − 1)dx = 2 x 3 dx + x 2 dx − dx x4 x3 = 2( ) + − x+C 4 3 1 1 = x 4 + x3 − x + C 2 3
∫
∫
1 2
1 3
Put the point (2, 5) into y = 2 x 2 −
∫
x − 3 x + 2)dx = 4 x dx − x dx + 2 dx = 4( =
x
3 2
3 2
)−
x
4 3
4 3
1 5 = 2( 2 ) 2 − ( 2 ) 3 + C 2 C =1
+ 2x + C
Hence the equation of the curve is 1 y = − x3 + 2x2 + 1. 2
8 23 3 43 x − x + 2x + C 3 4
Classwork 2 (p.176) 1.
∫ 3 sin xdx = 3∫ sin xdx = −3 cos x + C
2.
∫ (3 sin x + 2 cos x )dx = 3∫ sin xdx + 2∫ cos xdx = −3 cos x + 2 sin x + C
3.
∫
4.
∫ 3 sin
∫
sin x dx = sec x tan xdx = sec x + C cos 2 x 2
∫ ∫
1 x ( )dx = 3 (1 − cos x )dx 2 2 3 3 = cos xdx dx − 2 2 3 3 = x − sin x + C 2 2
∫
2.
d2y = 2(1 − x ) dx 2 d dy ( ) = 2(1 − x ) dx dx dy = 2(1 − x )dx dx x2 = 2 x − 2( ) + C1 2 = − x 2 + 2 x + C1
∫
dy = 5 = −(1)2 + 2(1) dx C1 = 4
At (1, 3), ∴
Then
dy = −x2 + 2x + 4 dx y = ( − x 2 + 2 x + 4)dx
∫
Classwork 3 (p.177) (a) f ' ( x ) = − x sin x + cos x (b) Using the result of (a), we obtain
∫ (− x sin x + cos x ) dx = f ( x ) + C1 = x cos x + C1 ∴ ∫ − x sin xdx + ∫ cos xdx = x cos x + C1 ∫ x sin xdx = ∫ cos xdx − x cos x − C1
= sin x + C2 − x cos x − C1 = sin x − x cos x + C
1 = − x 3 + x 2 + 4 x + C2 3 Since (1, 3) is on the curve, 1 3 = − (1)3 + (1)2 + 4(1) + C2 3 5 C2 = − 3 Hence the equation of the curve is 1 5 y = − x3 + x2 + 4x − . 3 3
1 3 x +C, 2
Chapter 17 Indefinite Integral and Its Applications
Classwork 5 (p.183) 1. Suppose v m/s and s m are the velocity of the particle and its displacement from O after travelling t1 seconds.
∫
v = 10 dt = 10t + C1 When t = 0 , v = 2 ∴ C1 = 2 ∴
(c) When t = 2 , s =
∫
s = (10t + 2)dt = 5t 2 + 2t + C2
When t = 0 , s = 0 C2 = 0
∴
s = 5t 2 + 2t
∴ When t = t1 , s = 5t12 + 2t1 ∴ The displacement at time t1 is (5t12 + 2t1 ) m .
2. (a) Let s m be the distance of the stone from the ground at time t seconds after projection. ds = v = 4t − 2t 2 dt
∫
s = ( 4t − 2t 2 )dt = 2t 2 −
2 3 t +C 3
When t = 0 , s = 0 ∴
C=0
2 3 t 3 ∴ The distance at time t seconds after projection 2 is (2t 2 − t 3 ) m . 3 ∴
s = 2t 2 −
(b) When
8 3
When t = 2.5 , 2 s = 2(2.5)2 − (2.5)3 3 25 = 12
v = 10t + 2
∴
∴ The maximum height 2 = [ 2( 2 ) 2 − ( 2 ) 3 ] m 3 8 = m 3 = 2.67 m (corr. to 2 d.p.)
ds = v = 0 , 4t − 2t 2 = 0 dt t = 0 or 2
d 2s = 4 − 4t dt 2 d 2s = 4 − 4( 0 ) = 4 > 0 dt 2 t = 0 d 2s = 4 − 4(2) = −4 < 0 dt 2 t = 2
∴ s attains its maximum when t = 2 .
∴ The distance it travels from t = 2 to t = 2.5 8 25 =( − )m 3 12 = 0.58 m (corr. to 2 d.p.)
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