Solution Manual - Textbook Add Math BM F5 KSSM

3. y = x3 + 3x2 – 9x + 2

dy = 0
dx

3x2 + 6x – 9 = 0

x2 + 2x – 3 = 0

( x + 3)(x – 1) = 0

x = –3 atau x = 1

Apabila x = –3, y = (–3)3 + 3(–3)2 – 9(–3) + 2

= –27 + 27 + 27 + 2

= 29

Apabila x = 1, y = 13 + 3(1)2 – 9(1) + 2

=1+3–9+2

= –3

Jadi, koordinat titik A yang mungkin ialah (–3, 29) dan (1, –3).

d 2y = 6x + 6
dx2
d 2y
Apabila x = –3, dx2 = 6(–3) + 6
= –18 + 6

d 2y = –12
Apabila x = 1, dx2 = 6(1) + 6
=6+6

= 12

Latihan Formatif 2.3

1. xy – 2x2 = 3

xy = 3 + 2x2
3 + 2x2
y = x

= 2x + 3
x
= 2x + 3x –1

dy = 2 – 3x –2
dx
3
= 2 – x2

d 2y = 6x –3
dx2 6
= x3

( ) ( )x2d 2y+ x ddyx = x2 6 + x 2 – 3
dx2 x3 x2
= 6 3
x + 2x – x

= 2x + 3  (Tertunjuk)
x
2. (a) f (x) = 3x – 2x3

f (x) = 3 – 6x2

f (x) = –12x

f (1) = 3 – 6(1)2

= –3

f (1) = –12(1)

= –12

(b) f (x) = x2(5x – 3)

= 5x3 – 3x2

f (x) = 15x2 – 6x

f (x) = 30x – 6

f (1) = 15(1)2 – 6(1)

=9

28

f (1) = 30(1) – 6

= 24
x3 + x
(c) f (x) = x2

= x + 1
x
= x + x–1

f (x) = 1 – x –2
1
= 1 – x2

f (x) = 2x –3
2
= x3

f (1) = 1 – 1
=0 12

f (1) = 2
13
=2

3. f (x) = ! x2 – 5

( )f (x) = 1 (x2 – 5)– 12(2x)
2
x
= ! x2 – 5

5)– 12(2x)
( [ ) ]f (x) =! x2 – 5 (1) – x 1 (x2 –
2
! x2 – 5 2

! x2 – 5 – x2 5
! x2 –
=
x2 – 5

= x2 – 5 – x2
(x2 – 5)! x2 – 5

= –  – 5 – 5
(x2 5)! x2

= – ! (x25– 5)3

f (3) = 3
! 32 – 5

= 3
! 4

= 3
2

f (–3) = –  5 – 5]3
! [(–3)2

= –  5
! 43

= – 58

4. a = t3 + 2t2 + 3t + 4

da = 3t2 + 4t + 3
dt
d 2a
dt2 = 6t + 4

29

da = d 2a
dt dt2

3t2 + 4t + 3 = 6t + 4

3t2 – 2t – 1 = 0

(3t + 1)(t – 1) = 0
t = – 31   atau t = 1
5. g(x) = hx3 – 4x2 + 5x

g(x) = 3hx2 – 8x + 5

g(x) = 6hx – 8

Diberi g(1) = 4

6h(1) – 8 = 4

6h = 12

h=2

6. f (x) = x3 – x2 – 8x + 9
f (x) = 0
(a)

3x2 – 2x – 8 = 0

(3x + 4)(x – 2) = 0
– 43
x = atau x = 2
(b) f (x) 2
= 6x –
(c) f (x) = 0

6x – 2 = 0

6x = 2
1
x= 3

(d) f (x) , 0
6x – 2 , 0
6x , 2

x , 1
3

Latihan Kendiri 2.7

1. (a) (i) y = 9x + 1
x
= 9x + x–1

dy = 9 – x –2
dx

= 9 – 1
x2

( ) Apabila x = 1 , dy = 9 – 1
4 dx 12
4

= 9 – 16

= –7

Apabila x = 1, dy = 9 – 1
dx 12

=9–1

=8

(ii) Pada x = 1 , kecerunan tangennya adalah negatif, iaitu –7 (, 0). Jadi, garis tangen condong ke kiri.
4

Pada x = 1 pula, kecerunan tangennya adalah positif, iaitu 8 (. 0). Jadi, garis tangen condong ke kanan.

30

(b) dy = 0
dx

9 – 1 = 0
x2
1
9 = x2

9x2 = 1

x2 = 1
9
x = ± 13
( ) ( ) ( ) Apabila x = 1 1 1 – 31 , – 13 1
3 , y = 9 3 + 1 , Apabila x = y = 9 + – 31

3 = –3 – 3
= 3 + 3

= 6 = – 6

1 – 13 , – 6
( ) ( ) º 3 , 6 dan

2. (a) y = ax2 + b
x

= ax2 + bx–1
ddyx
= 2ax – bx –2
= 2ax – b
x2
dy dy
Apabila x = 1 , dx = –14 , Apabila x = 2, dx = 7
2
( ) ( ) 1 b b
2a 2 – 1 2 = –14 , 2a(2) – 22 = 7

2 1
4
a – 4b = –14 …1 , 4a – b = 7

16a – b = 28 … 2
2 × 4: 64a – 4b = 112 …3
3 – 1: 63a = 126

a = 126
63

= 2

Gantikan a = 2 ke dalam 1: 2 – 4b = –14

4b = 16

b=4

º a = 2, b = 4

(b) dy = 0
dx
4
2(2)(x) – x2 = 0

4x – 4 = 0
x2
4
4x = x2

x3 = 1

x=1 4
1
Apabila x = 1, y = 2(1)2 +
= 2+4

= 6

º (1, 6)

31

Latihan Kendiri 2.8
1. (a) f (x) = 5x2 – 7x – 1

f (x) = 10x – 7

Apabila x = 1, f (1) = 10(1) – 7
=3

Persamaan tangen: y – (–3) = 3(x – 1)

y + 3 = 3x – 3

y = 3x – 6
––x 31+(x1–
Persamaan normal: y – (–3) = 1)
3y + 9 =

3y + x + 8 = 0

(b) f (x) = x3 – 5x + 6

f (x) = 3x2 – 5

Apabila x = 2, f (2) = 3(2)2 – 5

= 12 – 5
=7

Persamaan tangen: y – 4 = 7(x – 2)

y – 4 = 7x – 14

y = 7x – 10

Persamaan normal: y – 4 = – 71 (x – 2)
7y – 28 = –x + 2

7y + x = 30
(c) f (x) = ! 2x + 1
1)– 12(2)
f (x) = 1 (2x +
2

= 1
! 2x + 1

Apabila x = 4, f (4) = 1
! 2(4) + 1

= 1
! 9

= 1
3
1
Persamaan tangen: y – 3 = 3 (x – 4)

3y – 9 = x – 4

3y – x = 5

Persamaan normal: y – 3 = –3(x – 4)

y – 3 = –3x + 12

y = –3x + 15
x+1
(d) f (x) = x–1

f (x) = (x – 1)(1) – (x + 1)(1)
(x – 1)2

= x –1–x– 1
(x – 1)2

= – (x 2 1)2
–

Apabila x = 3, f (3) = – (3 2 1)2
–

= – 21

32

Persamaan tangen: y – 2 = – 21 (x – 3)
2y – 4 = –x + 3
2y = –x + 7

Persamaan normal: y – 2 = 2(x – 3)
y – 2 = 2x – 6

y = 2x – 4

2. (a) y = 2x3 – 4x + 3

dy = 6x2 – 4
dx

Apabila x = 1, y = 2(1)3 – 4(1) + 3

=2–4+3

=1
dy
dan dx = 6(1)2 – 4

=6–4

=2

Persamaan tangen: y – 1 = 2(x – 1)

y – 1 = 2x – 2

y = 2x – 1

Persamaan normal: y – 1 = – 12 (x – 1)
2y – 2 = –x + 1

2y + x = 3
1
(b) y = ! x – ! x

= 1 – x– 21

x2

dy = 1 x– 21 + 1 x– 32
dx 2 2
1 1
= 2! x + 2! x3

Apabila x = 4, y = ! 4 – 1
! 4
1
=2– 2

= 3
2
dy 1 1
dan dx = 2! 4 + 2! 43

= 1 + 1
4 16
5
= 16

Persamaan tangen: y – 3 = 156(x – 4)
2

16y – 24 = 5x – 20

16y – 5x = 4

Persamaan normal: y– 3 = – 156(x – 4)
2
15
5y – 2 = –16x + 64

10y – 15 = –32x + 128

10y = –32x + 143

33

(c) y = ! x + 1

1

y = (x + 1)2

dy = 1 (x + 1)– 12
dx 2
1
= 2! x + 1

Apabila x = 3, y = ! 3 + 1
= ! 4
=2

dan dy = 1
dx 2! 3 + 1

= 1
4
1
Persamaan tangen: y – 2 = 4 (x – 3)

y= 1 x – 3 +2
4 4
1 5
y= 4 x + 4

Persamaan normal: y – 2 = – 4(x – 3)

y = – 4x + 12 + 2

y = – 4x + 14
5
(d) y = x2 + 1

dy = (x2 + 1)(0) – 5(2x)
dx (x2 + 1)2

= – (x21+0x1)2

Apabila x = –2, y = 5 + 1
(–2)2

=1

dan dy = –  10(–2)
dx [(–2)2 + 1]2

= 4
5

Persamaan tangen: y – 1 = 4 (x + 2)
5
5y – 5 = 4x + 8

5y – 4x = 13

Persamaan normal: y – 1 = – 45 (x + 2)

4y – 4 = –5x – 10

4y + 5x + 6 = 0
1
(e) y = 2 + x

dy = – x12
dx

Apabila x = –1, y = 2 + 1
–1
=1

dan dy = – (–11)2
dx

= –1

Persamaan tangen: y – 1 = –[x – (–1)]
y – 1 = –x – 1
y = –x

34

Persamaan normal: y – 1 = x – (–1)

y=x+1+1

y=x+2
x2 + 3
(f) y = x+1

dy = (x + 1)(2x) – (1)(x2 + 3)
dx (x + 1)2

= 2x2 + 2x – x2 – 3
(x + 1)2

= x2 + 2x – 3
(x + 1)2

Apabila x = 3, y = 32 + 3
3+1

=3

dan dy = 32 + 2(3) – 3
dx (3 + 1)2
3
= 4

Persamaan tangen: y – 3 = 3 (x – 3)
4
3 9
y = 4 x – 4 + 3

y = 3 x + 3
4 4
Persamaan normal: y – 3 = – 34 (x – 3)
y = – 43 x + 4 + 3
y = – 34 x + 7
3. (a) y = x! 1 – 2x

2x)– 21
[ ]dy = x 1 (1 – (–2) + ! 1 – 2x (1)
2
dx
x + ! 1 – 2x
= – ! 1 – 2x

= –x + 1 – 2x
! 1 – 2x

= 1 – 3x
! 1 – 2x

Apabila x = – 4, dy = 1 – 3(– 4)
dx ! 1 – 2(– 4)

= 13
! 9

= 13
3

(b) Apabila x = – 4, y = – 4! 1 – 2(– 4)
= – 4(3)

= –12

Persamaan tangen: y + 12 = 13 (x + 4)
3

3y + 36 = 13x + 52

3y – 13x = 16

(c) Persamaan normal: y + 12 = – 133(x + 4)

13y + 156 = –3x – 12

13y + 3x + 168 = 0

35

4. (a) y = (x – 2)2

= x2 – 4x + 4
dy
dx = 2x – 4

Apabila x = 3, dy = 2(3) – 4
dx
=2

Persamaan tangen: y – 1 = 2(x – 3)

y – 1 = 2x – 6

y = 2x – 5

Pada titik (k, 7): 7 = 2k – 5

2k = 12

6 k=6
x
(b) y = 7x –

= 7x – 6x –1 6
1
Apabila x = 1, y = 7(1) –

=1

dan dy = 7 + 6x –2
dx
6
= 7 + x2

= 7 + 6
12
= 13
––x 113+(x1
Persamaan normal: y – 1 = – 1)
13y – 13 =

13y + x = 14

Pada paksi-x, y=0

13(0) + x = 14

x = 14

º A(14, 0)

Latihan Kendiri 2.9

1. (a) y = x2 – 3x + 4

dy = 2x – 3
dx
dy
Apabila x = 1, dx = 2(1) – 3

= 2 – 3

= –1

Persamaan tangen di titik A(1, 2) ialah

y – 2 = –1(x – 1)

y – 2 = –x + 1

y+x=3

(b) Apabila x = 3, dy = 2(3) – 3
dx
= 6 – 3

= 3 – 13

Persamaan normal di titik B(3, 4) dengan kecerunan ialah
y – 4 = – 31 (x – 3)

3y – 12 = –x + 3

3y + x = 15
(c) y + x = 3 …1
3y + x = 15 …2
2 – 1: 2y = 12

y=6

36

Gantikan y = 6 ke dalam 1:
6 + x = 3
x = –3
º C(–3, 6)

2. (a) y = 2x2 – 5x – 2
dy
dx = 4x – 5

= 4(1) – 5

= –1

Persamaan normal: y + 5 = x – 1

y=x–6

(b) x – 6 = 2x2 – 5x – 2

2x2 – 6x + 4 = 0

2(x2 – 3x + 2) = 0

2(x – 2)(x – 1) = 0

x = 2 atau x = 1

Apabila x = 2, y = 2 – 6

= – 4

º B(2, – 4) –5 + (– 4)
2
( )(c) Titik tengah AB, M = 1 + 2,
2
( )= 3 – 29
2 ,

3. (a) mPQ = 1–0
1
2 – 1 2

= 2

y = ax 3 − 4x + b

dy = 3ax2 − 4
dx
dy
Pada titik P(2, 1), dx = 2

3a(2)2 − 4 = 2

12a = 6

a = 1
2
dan 1
1 = 2 (2)3 − 4(2) + b

1=4−8+b

1 = −4 + b

1 b=5
2
Maka, a = dan b = 5.

(b) P(2, 1) dan m = − 21 − 1 −−x 21+(x2− 2)
Persamaan normal: y − 2 =
=
2y

2y + x = 4

(c) Pada paksi-x, 2(0) + x = 4

x=4

Jadi, koordinat R ialah (4, 0).

(d) Luas ∆ PQR = 1   2 3 4 2
2 1 2 1

00
 4  
= 1 – 3
2 2
( )=1 5
2 1 2
4
= 1 unit2

37

4. (a) 3y − x = 14

3y = x + 14
yke=ce13ruxn+an13t4angen
b ialah −3.
Jadi, x

y = ax +

= ax + bx−1
dy b
dx = a − x2

Pada titik P(1, 5), dy = –3
dx
b
a − 12 = –3

a – b = –3 ……1

dan 5 = a(1) + b
1

a + b = 5 ……2
1 + 2: 2a = 2

a=1
Gantikan a = 1 ke dalam 1:

1 – b = –3

b = 4

Maka, a = 1 dan b = 4.

(b) P(1, 5) dan m = –3

Persamaan tangen: y – 5 = –3(x – 1)

y – 5 = –3x + 3

y + 3x = 8

(c) 3y − x = 14

y= x + 14 ……1
3
4
y=x+ x ……2

Gantikan 1 ke dalam 2:

x + 14 = x + 4
3 x
x(x + 14) = 3x2 + 12

x2 + 14x = 3x2 + 12

2x2 – 14x + 12 = 0

x2 – 7x + 6 = 0

(x – 1)(x – 6) = 0

x = 1 atau x = 6

Gantikan x = 6 ke dalam 1: y = 6 + 14
3
= 6 32
( ) Maka, koordinat Q ialah 6, 6 32 .
2
( )(d) MPQ = 1 + 6 , 5 + 6 3
2
2

( )= 3 21 , 5 56

38

5. (a) y = ! 2x + 1

dy = 1 (2x + 1)– 12 (2)
dx 2
1
= ! 2x + 1

Pada titik A(4, 3), dy = 1
dx ! 2(4) + 1

= 1
3

Persamaan tangen: y – 3 = 1 (x – 4)
3
3y – 9 = x – 4

3y = x + 5

Pada paksi-x, 3(0) = x + 5

x = –5

Jadi, koordinat B ialah (–5, 0).
Maka, dAB = ! (–5 –  4)2 + (0 – 3)2
= ! 81 + 9

= ! 90

= 3! 10 unit
(b) y = x2 + 6x + 4

dy = 2x + 6
dx
dy
Pada titik (−2, − 4), dx = 2(−2) + 6

= 2

Pada titik (1, 1 ), dy = − 21
2 dx

3h(1)2 + k = − 12

6h + 2k = −1 ……1

dan 1 = h(1)3 + k(1) + 2
2
1
2 =h+k+2

h + k = − 32

2h + 2k = −3 ……2
1 − 2: 4h = 2
1
h= 2

( ) Gantikan h = 1 1
2 ke dalam 1: 6 2 + 2k = −1

3 + 2k = –1

2k = – 4

1 k = –2
2
Maka, h = dan k = −2.

39

Aktiviti Penerokaan 7 (Halaman 57)

4. Koordinat-x bagi titik P –1 0 123
4 2 0 –2 – 4
Kecerunan lengkung pada titik P, dy
dx + + 0––
/ / –\ \
Tanda bagi dy
dx

Lakaran tangen

Lakaran graf

5. Koordinat-x bagi titik P –3 –2 –1 0 1
2 4
Kecerunan lengkung pada titik P, dy – 4 –2 0
dx + +
/ /
Tanda bagi dy ––0
dx

Lakaran tangen \ \–

Lakaran graf

6. Koordinat-x bagi titik P –2 –1 0 12
12 3 0 3 12
Kecerunan lengkung pada titik P, dy
dx + + 0 ++
/ / – //
Tanda bagi dy
dx

Lakaran tangen

Lakaran graf

7. (a) (i) (1, 4)

(ii) Nilai dy berubah daripada 4 kepada 0 dan kemudian kepada – 4 apabila x menokok melalui titik pegun (1, 4)
dx
(iii) Tandanya berubah daripada positif kepada negatif apabila x menokok melalui titik pegun (1, 4)

(iv) Titik pusingan maksimum

(b) (i) (–1, – 4)

(ii) Nilai dy berubah daripada – 4 kepada 0 dan kemudian kepada 4 apabila x menokok melalui titik pegun
dx
(–1, – 4)

(iii) Tandanya berubah daripada negatif kepada positif apabila x menokok melalui titik pegun (–1, – 4)

(iv) Titik pusingan minimum

(c) (i) (0, 4) dy
dx
(ii) Nilai berubah daripada 12 kepada 0 dan kemudian kepada 12 sekali lagi apabila x menokok melalui

titik pegun (0, 4)

(iii) Tandanya tidak berubah apabila x menokok melalui titik pegun (0, 4)

(iv) Titik lengkok balas

Perbincangan (Halaman 62)
y = x3 + 6x2 + 12x + 7, y = x3 – 6x2 + 12x – 5 dan lain-lain lagi.

40

Latihan Kendiri 2.10

1. (a) y = x3 – 12x

dy = 3x2 – 12
dx
dy
Untuk titik pusingan, dx = 0

3x2 – 12 = 0

3x2 = 12

x2 = 4

x = ±2

Apabila x = –2, y = (–2)3 – 12(–2)

= –8 + 24

= 16

Apabila x = 2, y = 23 – 12(2)

= 8 – 24

= –16

Maka, titik pusingan ialah (–2, 16) dan (2, –16).

Apabila x = –2, d 2y = 6x
dx2 = 6(–2)

= –12 (, 0)

Maka, (–2, 16) ialah titik maksimum.

Apabila x = 2, d 2y = 6x
dx2 = 6(2)

= 12 (. 0)

Maka, (2, –16) ialah titik minimum.

(b) y = x(x – 6)2

= x(x2 – 12x + 36)

= x3 – 12x2 + 36x
dy
dx = 3x2 – 24x + 36

Untuk titik pusingan, dy = 0
dx

3x2 – 24x + 36 = 0

x2 – 8x + 12 = 0

(x – 2)(x – 6) = 0

x = 2 atau x = 6

Apabila x = 2, y = 23 – 12(2)2 + 36(2)

= 8 – 48 + 72

= 32

Apabila x = 6, y = 63 – 12(6)2 + 36(6)

= 216 – 432 + 216

= 0

Maka, titik pusingan ialah (2, 32) dan (6, 0).

Apabila x = 2, d 2y = 6x – 24
dx2 = 6(2) – 24

= –12 (, 0)

Maka, (2, 32) ialah titik maksimum.

Apabila x = 6, d 2y = 6x – 24
dx2 = 6(6) – 24

= 12 (. 0)

Maka, (6, 0) ialah titik minimum.

41

(c) y = x! 18 – x2

x2)– 21
[ ]dy=x 1 + ! 18 – x2(1)
2
dx
(18 – (–2x)

= – ! 18x2– x2 + ! 18 – x2

= –x2 + 18 – x2
! 18 – x2

= 18 – 2x2
! 18 – x2

= 2(9 – x2)
! 18 – x2

= 2(3 + x)(3 – x)
! 18 – x2

Untuk titik pusingan, dy = 0
dx

2(3 + x)(3 – x) =0
! 18 – x2

2(3 + x)(3 – x) = 0
x = –3 atau x = 3

Apabila x = –3, y = –3! 18 – (–3)2

= –3! 9
= –9

Apabila x = 3, y = 3! 18 – 32

= 3! 9

= 9

Maka, titik pusingan ialah (–3, –9) dan (3, 9).
– x2)– 12(–2x)
[ ]d 2y= ! 18 – x2(– 4x) – (18 – 2x2) 1 (18
2
( )dx2
! 18 – x2 2

– 4x! 18 – x2 + 18x – 2x3
! 18 – x2
=
18 – x2

= – 4x(18 – x2) + 18x – 2x3

3
(18 – x2)2

= –72x + 4x3 + 18x – 2x3
! (18 – x2)3

= 2x3 – 54x
! (18 – x2)3

= 2x(x2 – 27)
! (18 – x2)3

Apabila x = –3, d 2y = 2(–3)[(–3)2 – 27]
dx2 ! [18 – (–3)2]3

= 108
27

= 4 (. 0)

Maka, (–3, –9) ialah titik minimum.

42

Apabila x = 3, d 2y = 2(3)[(3)2 – 27]
dx2 ! [18 – (3)2]3

= – 12078

= – 4 (, 0)

Maka, (3, 9) ialah titik maksimum.

(d) y = (x – 6)(4 – 2x)

= 4x – 2x2 – 24 + 12x

= –2x2 + 16x – 24
dy
dx = – 4x + 16

Untuk titik pusingan, dy = 0
dx

– 4x + 16 = 0

x=4

Apabila x = 4, y = (4 – 6)[4 – 2(4)]

= (–2)(– 4)

= 8

Maka, titik pusingan ialah (4, 8).

Apabila x = 4, d 2y = – 4 (, 0)
dx2

Maka, (4, 8) ialah titik maksimum.

(e) y = x + 4
x
= x + 4x–1

dy = 1 – 4x –2
dx
4
= 1 – x2

Untuk titik pusingan, dy = 0
dx

1 – 4 = 0
x2
4
1 = x2

x2 = 4

x = ±2

Apabila x = –2, y = –2 + 4
(–2)
= – 4

Apabila x = 2, y = 2 + 4
2
= 4

Maka, titik pusingan ialah (–2, – 4) dan (2, 4).

Apabila x = –2, d 2y = 8x –3
dx2 8
= x3

= 8
(–2)3

= –1 (, 0)

Maka, (–2, – 4) ialah titik maksimum.

Apabila x = 2, d 2y = 8
dx2 23

= 1 (. 0)

Maka, (2,  4) ialah titik minimum.

43

(f) y = x2 + 1
x2
= x2 + x –2

dy = 2x – 2x –3
dx
2
= 2x – x3

Untuk titik pusingan, dy = 0
dx

2x – 2 = 0
x3
2
2x = x3

x 4 = 1

x=1

Apabila x = 1, y = 12 + 1
12
= 2

Maka, titik pusingan ialah (1, 2).

Apabila x = 1, d 2y = 2 + 6x– 4
dx2 =
2+ 6
x4
6
= 2 + 14

= 8 (. 0)

Maka, (1, 2) ialah titik minimum.

(g) y = x + x 1 1
–
= x + (x – 1)–1

dy = 1 – 1(x – 1)–2(1)
dx
1
= 1 – (x – 1)2

Untuk titik pusingan, dy = 0
dx

1 – (x 1 = 0
– 1)2
1
1 = (x – 1)2

(x – 1)2 = 1

x – 1 = ±1

x = ±1 + 1

x = –1 + 1  atau  x = 1 + 1

= 0 = 2

Apabila x = 0, y = 0 + 0 1 1
–
= –1

Apabila x = 2, y = 2 + 2 1 1
–
= 3

Maka, titik pusingan ialah (0, –1) dan (2, 3).

Apabila x = 0, d 2y = (0 2
dx2 – 1)3

= –2 (, 0)

Maka, (0, –1) ialah titik maksimum.

Apabila x = 2, d 2y = (2 2
dx2 – 1)3

= 2 (. 0)

Maka, (2, 3) ialah titik minimum.

44

(h) y = (x – 3)2
x

= x2 – 6x + 9
x

= x – 6 + 9x–1

dy = 1 – 9x–2
dx
9
= 1 – x2

Untuk titik pusingan, dy = 0
dx

1 – 9 = 0
x2
9
1 = x2

x2 = 9

x = ±3

Apabila x = –3, y = (–3 – 3)2
–3
= –12

Apabila x = 3, y = (3 – 3)2
3
= 0

Maka, titik pusingan ialah (–3, –12) dan (3, 0).

Apabila x = –3, d 2y = 18
dx2 x3

= 18
(–3)3

= – 32 (, 0)

Maka, (–3, –12) ialah titik maksimum.

Apabila x = 3, d 2y = 18
dx2 33

= 2 (. 0)
3

Maka, (3, 0) ialah titik minimum.

2. (a) y = x(x – 2)3

dy = x[3(x – 2)2(1)] + (x – 2)3(1)
dx
= 3x(x – 2)2 + (x – 2)3

= (x – 2)2[3x + (x – 2)]

= (4x – 2)(x – 2)2

= 2(2x – 1)(x – 2)2

(b) Untuk titik pegun, dy = 0
dx

2(2x – 1)(x – 2)2 = 0

2x – 1 = 0  atau  (x – 2)2 = 0
1
x= 2 x=2

( ) Apabila x = 1 , y = 1 1 – 2 3
2 2 2

( )=1 – 287
2

= – 1276

Apabila x = 2, y = 2(2 – 2)3

= 0
( ) 1 – 1267
ºP 2 , dan Q(2, 0)

45

(c) Q ialah titik lengkok balas.

Kuiz Pantas (Halaman 64)

Boleh.
!Gantikan j =
 5π1 t2 ke dalam L = 8j2 + 2πjt. Kemudian, tentukan nilai t apabila dL = 0.
dt

!Seterusnya, gantikan nilai t = 10.186 yang diperoleh ke dalam j =  5π1 t2 untuk mencari nilai j.

Latihan Kendiri 2.11

1. (a) sPQ = 80 – 2j

jq = 80 – 2j

q = 80 – 2j
j
1
Luas sektor POQ, A = 2 j2q

( ) = 1 j2 80 – 2j
2 j

1 = 1 j(80 – 2j) (Tertunjuk)
2 2
(b) A = j(80 – 2j)

= 40j – j2

Untuk luas maksimum, dA = 0
dj

40 – 2j = 0

2j = 40

j = 20

Apabila j = 20, A = 1 (20)[80 – 2(20)]
2

= 400
d 2A
dj2 = –2 (. 0)

Jadi, A adalah maksimum.

Maka, luas maksimum bagi sektor POQ ialah 400 cm2.

2. (a) 2y + 2(13x) + 24x = 240

2y + 50x = 240

y + 25x = 120

y = 120 – 25x

(b) L = 24xy + 1 (24x)(5x)
2
= 24x(120 – 25x) + 60x2

= 2 880x – 600x2 + 60x2

= 2 880x – 540x2 (Tertunjuk)

(c) (i) Untuk nilai maksimum, dL = 0
dx

2 880 – 1 080x = 0

1 080x = 2 880
2
x = 2 3

( ) Apabila x = 2 2 , y = 120 – 25 2 2
3 3
1
= 53 3

dan d 2L = –1 080 (, 0)
dx2
2 1
Maka, L mempunyai nilai maksimum apabila x = 2 3 cm dan y = 53 3 cm.

46

( ) ( )(ii) L Maks2 880 2 2 2 2 2
= 3 – 540 3

= 7 680 – 3 840

= 3 840

Maka, luas maksimum rantau ialah 3 840 cm2.

3. (a) Katakan jejari dan tinggi silinder masing-masing ialah j cm dan t cm.

πj2t = 32π

t= 32π
πj2

= 32 …1
j2

Fungsi kos, C = 2πj2(2) + 2πjt(1)

( ) = 4πj2 + 2πj 32
j2

= 4πj2 + 64jπ (Tertunjuk)

(b) Untuk C minimum, dC = 0
dj

8πj – 64πj–2 = 0

8πj = 64π
j2

j3 = 8
j = 3! 8

= 2 cm
32 32
dan t = j2 = 22 = 8 cm

dd 2jC2 = 8π + 128π
j3

Apabila j = 2, dd 2jC2 = 8π + 128π
23

= 8π + 16π
= 24π (. 0)

Jadi, C adalah minimum apabila j = 2. Maka, kilang itu mesti mengeluarkan tin berbentuk silinder dengan
jejari 2 cm dan tinggi 8 cm untuk memperoleh kos minimum.

Aktiviti Penerokaan 8 (Halaman 65)

2. 3π cm3 ˜ 1 saat

48π cm3 ˜ 48π = 16 saat
3π
Jadi, masa yang diambil untuk memenuhkan air di dalam setiap bekas ialah 16 saat.

3.

Bekas Silinder Kon

h cm h cm

9 9

Graf kedalaman-masa

0 ts 0 ts
16 16

4. (b) Luas permukaan air di dalam bekas berbentuk silinder sentiasa seragam apabila air diisikan ke dalamnya. Jadi,
aras air meningkat secara seragam terhadap masa dan kedalaman air dikatakan meningkat pada kadar malar.

Luas permukaan air di dalam bekas berbentuk kon pula bertambah apabila aras air meningkat. Jadi, kadar
perubahan kedalaman aras air berubah-ubah apabila air diisikan, iaitu kadar perubahan kedalaman air menyusut
terhadap masa.

47

Latihan Kendiri 2.12

1. (a) y = 3x2 – 4

( ) dy = 6x dy
dx dx
Apabila x = 1 , =6 1
2 2

ddyt dy dx = 3
dx dt
= ×

dy = 3 × 2 = 6 unit s–1
dt

(b) y = 2x2 + 1 = 2x2 + x–1
x = 4(1)
dy 1 dy
dx = 4x – x2 dx

Apabila x = 1, – 1
12

= 4 – 1

ddyt dy dx = 3
dx dt
= ×

dy = 3 × 2 = 6 unit s–1
dt
2
(c) y= (3x – 5)3 = 2(3x – 5)–3 5)4
= – 6(3x – 5)– 4(3) = – (3x1–8
dy
dx dy
dx
Apabila x = 2, = –  18 5]4
[3(2) –

ddyt dy dx = –18
dx dt
= ×

dy = –18 × 2 = –36 unit s–1
dt

(d) y = (4x – 3)5

dy = – 3)4(4) = 3)4
[ ( ) ] dx 5(4x = = 20(4x – –3

Apabila x 1 , dy 20 4 1 4
2 dx 2

ddyt dy dx = 20
dx dt
= ×

dy = 20 × 2 = 40 unit s–1
dt

(e) y = x x 1
+
dy (x + 1)(1) – x(1) 1
dx = (x + 1)2 = (x + 1)2

Apabila y = 2, 2 = x x 1
+

2x + 2 = x

x = –2
dy
dan dx = (–2 1 1)2
+

ddyt dy dx =1
dx dt
= ×

dy = 1 × 2 = 2 unit s–1
dt

48

(f) y = x3 + 2

dy = 3x2
dx

Apabila y = 10, 10 = x3 + 2

x3 = 8

x=2

dan dy = 3(2)2
dx
ddyt = 12
dy dx
= dx × dt

dy = 12 × 2 = 24 unit s–1
dt
2. (a) y = x3 – 2x2

dy = 3x2 – 4x
dx
dy
Apabila x = 1, dx = 3(1)2 – 4(1)
= 3–4


ddyt dy dx = –1
dx dt
= ×

6 = –1 × dx
dt
dx
dt = – 6 unit s–1

(b) y = x2 + 4
x

= x2 + 4x–1
dy
dx = 2x – 4
x2

Apabila x = 2, dy = 2(2) – 4
dx 22
= 4 – 1

ddyt dy dx =3
dx dt
= ×

6 = 3 × dx
dt

dx = 2 unit s–1
dt
2x2
(c) y = x–1

dy = (x – 1)(4x) – 2x2(1)
dx (x – 1)2

= 4x2 – 4x – 2x2
(x – 1)2

= 2x2 – 4x
(x – 1)2

= 2x(x – 2)
(x – 1)2

Apabila x = 3, dy = 2(3)(3 – 2)
dx (3 – 1)2

= 6
4
3
= 2

49

dy = dy × dx
dt dx dt
dx
6 = 3 × dt
2
dx
( ) dt = 6 2 = 4 unit s–1
3
(d) y = (x – 6)! x – 1

[ ]dy = (x – 6) 1 (x – 1)– 12 (1) + ! x – 1
2
dx
x–6 + ! x – 1
= 2! x – 1

= x – 6 + 2(x – 1)
2! x – 1

= x – 6 + 2x – 2
2! x – 1

= 3x – 8
2! x – 1

Apabila x = 2, dy = 3(2) – 8
dx 2! 2 – 1

= – 22

= –1

dy = dy × dx
dt dx dt
dx
6 = –1 × dt
= – 6
dx unit s–1
dt
2x – 1
(e) y = x+1

dy = (x + 1)(2) – (2x – 1)(1)
dx (x + 1)2

= 2x + 2 – 2x + 1
(x + 1)2

= (x 3
+ 1)2

Apabila y = 3, 3 = 2x – 1
x+1

3x + 3 = 2x – 1
x = – 4

dan dy = 3 1)2
dx (– 4 +

= 3
9
1
= 3

dy = dy × dx
dt dx dt
dx
6 = 1 × dt
3
dx
dt = 18 unit s–1

(f) y = ! 2x + 7

dy = 1 (2x + 7)– 21 (2)
dx 2
1
= ! 2x + 7

50

Apabila y = 3, 3 = ! 2x + 7

9 = 2x + 7

2x = 2

x=1

dan dy = 1 + 7
dx ! 2(1)

= 1
! 9

= 1
3
dy dy dx
dt = dx × dt

6 = 1 × dx
= 3 dt
dx 18
dt unit s–1

3. (a) y = (x – 8)! x + 4

[ ]dy = (x – 8) 1 (x + 4)– 12 (1) + ! x + 4(1)
2
dx
x–8 + ! x + 4
= 2! x + 4

= x – 8 + 2(x + 4)
2! x + 4

= x –8+ 2x + 8
2! x +4

= 3x
2! x + 4

(b) Apabila x = 5, dy = 3(5)
dx 2! 5 + 4

= 15
2(3)

= 15
6
5
= 2

dy = dy × dx
dt dx dt
ddyt =
5 ×6
2

= 15 unit s–1

Perbincangan (Halaman 68)
1
(a) V = 3  π r2h …1

8r = h
16
h
r = 16 × 8

= h …2
2

Gantikan 2 ke dalam 1:

( )V =1  π  h 2h
3 2

= 1  π h3
12

51

Kadar perubahan V diberi oleh:

dV = dV × dh (petua rantai)
dt dh dt

( )=ddh  1 πh3 × dh
12 dt

= 1  πh2 × dh
4 dt
dV
Apabila h = 8 dan dt = 64π, kita peroleh

64π = 1  π(8)2 × dh
4 dt
dh
64π = 16π × dt

dh = 4
dt

Jadi, kadar perubahan kedalaman air dalam bekas itu ialah 4 cms–1.
(b) L = πr2 …3
Gantikan 2 ke dalam 3:
( )L h 2
= π  2

= 1  πh2
4

Kadar perubahan L diberi oleh:

dL = dL × dh (petua rantai)
dt dh dt

( )=d 1 πh2 × dh
dh 4 dt

= 1  πh × 4
2
dh
Apabila h = 8 dan dt = 4, kita peroleh

dL = 1  π(8) × 4
dt 2

= 16π

Jadi, kadar perubahan luas permukaan mengufuk ialah 16π cm2s–1.

Latihan Kendiri 2.13

1. y = 1  x2
8
dy
dx = 1  x
4
dy 1
Apabila x = 4, dx = 4 (4) = 1

dy = dy × dx
dt dx dt
dy
dt = 1 × 3

= 3 unit s–1

2. L = x2
dL
dx = 2x

Apabila L = 4,

x2 = 4
x2 = ! 4

dL = = 2 (. 0)
dan dx 2(2) = 4

52

dL = dL × dx
dt dx dt
dx
8 = 4 × dt

ddxt = 8
4

= 2 cms–1

3. I = x3
dI
dx = 3x2

Apabila x = 10, dI = 3(10)2
dx
= 300
dI dI dx
dt = dx × dt

–10.5 = 300 × dx
dt

dx = – 1300.05 = – 2070 cmmin–1
dt

4. (a) V = π r2h

= π(3)2h

= 9πh
dV
(b) dt = 9π × – 0.6

= –5.4π cm3min–1

5. Katakan panjang bayang-bayang dan panjang hujung bayang-bayang dari kaki tiang lampu masing-masing ialah

s m dan l m.
6 1.8
x+s = s

6s = 1.8(x + s)

6s = 1.8x + 1.8s 6m
4.2s = 1.8x
3
s = 7 x 1.8 m

(a) s= 3 x xm sm
7 lm
ds 3
dx = 7

Jadi, ds = ds × dx
dt dx dt

= 3 × 3.5
7

= 1.5

Maka, kadar perubahan panjang bayang-bayang ialah 1.5 ms–1.

(b) l = x + s
3
= x + 7 x

= 170 x
dl 10
dx = 7

Jadi, dl = dl × dx
dt dx dt

= 10 × 3.5
7

= 5

Maka, kadar perubahan hujung bayang-bayang yang bergerak ialah 5 ms–1.

Perbincangan (Halaman 71)
Boleh digunakan tetapi jawapannya tidak tepat dan bukan penghampiran yang terbaik.

53

Latihan Kendiri 2.14

1. (a) y = 4x3 – 3x2

dy = 12x2 – 6x
dx

Apabila x = 1, dx = 1.05 – 1

= 0.05
dy
dan dx = 12(1)2 – 6(1)
= 12 – 6

=6

Jadi, dy ≈ dy × dx
dx

= 6 × 0.05

= 0.3 unit
(b) y = 4! x + 3x2

( )dy = 4 1 x– 12 + 6x
2
dx
2
= ! x + 6x

Apabila x = 4, dx = 3.98 – 4

= – 0.02

dan dy = 2 + 6(4)
dx ! 4

= 1 + 24

= 25

Jadi, dy ≈ dy × dx
dx

= 25 × – 0.02

= – 0.5 unit
3
2. (a) y = 2x2
dy 1
dx =
3x2

= 3! x 3

Apabila y = 16, 16 = 2x2

3

x2 = 8

( ) x= 23 2
3

=4

dy = 15.7 – 16

= – 0.3

dan dy = 3! 4
dx
= 3(2)

=6
dy
Jadi, dy ≈ dx × dx

– 0.3 = 6 × dx

dx = – 06.3
= – 0.05 unit
x + 2
(b) y = 2

= 1 x + 1
2
dy
dx = 1
2

54

Apabila y = 2, 2 = x + 2
2

4=x+2

x=2

dy = 2 + p – 2

= p

Jadi, dy ≈ dy × dx
dx
1
p= 2 × dx

dx = 2p unit
16
3. y = x2 = 16x–2

dy = –32x–3 = – 3x23
dx 2, y= 16
22
Apabila x =

y = 16
4
=4

dx = 2.02 – 2

= 0.02
dy – 3223
dan dx =

= – 4 dy
dx
f (x + dx) ≈ y +  dx

(2 16 = 4 + (– 4)(0.02)
+ 0.02)2
= 4 – 0.08

= 3.92

5
4. y = x4
dy 5 1
dx = 4 x 4

dy = 4 y
100
= 0.04y
5
= 0.04x4

dy ≈ dy × dx
dx

5 = 5 1 × dx
4
0.04x4 x4

5
0.04x4
dx = 1
5
4 x4

= 0.032x

Peratus perubahan hampir dalam x = dx × 100%
x

= 0.032x × 100%
x

= 3.2%

55

Latihan Kendiri 2.15
!  1.
T = 2π  l
10
l – 21
[ ( ) ( )] ddTl 1 10 1
= 2π 2 10

!  = π

10 l
10
dT π
! Apabila l = 9, dl =
9
10 10

= π! 10
30

dan dl = 9.05 – 9 = 0.05

Maka, dT ≈ dT × dl
dl
π! 10
= 30 × 0.05

= π! 10 saat
600
2. L = π j2

dL = 2π j
dj
L = 4π

π j2 = 4π

j2 = 4

j = 2 (. 0)

Apabila j = 2, ddLj = 2π(2)
= 4π

dL = 4.01π – 4π
= 0.01π

dan dL ≈ dL × dj
dj

0.01π = 4π × dj

dj = 0.01π
4π

= 0.0025 cm

3. V = x3

dV = 3x2
dx

Apabila x = 2, dx = 1.99 – 2

dV = – 0.01
dx
dan = 3(2)2
= 12

Maka, dV = dV × dx
dx

= 12 × – 0.01

= – 0.12 cm3
4
4. I = 3 π j3

dI = 4π j2
dj

Apabila j = 5, dj = 4.98 – 5

= – 0.02

56

dan dI = 4π (5)2
dj = 100π

Maka, dI ≈ dI × dj
dj

= 100π × – 0.02

= –2π cm3

Latihan Formatif 2.4

1. (a) y = ! x + 1

dy = 1 (x + 1)– 12 (1)
dx 2
1
= 2! x + 1

Apabila x = 0, ddyx = 1
2! 0 + 1

= 1
2
1
Persamaan tangen: y–1= 2 (x – 0)

2y – 2 = x
2y – x = 2

Pada paksi-x, y = 0

2(0) – x = 2

x = –2

º Q(–2, 0)

(b) Persamaan normal: y – 1 = –2(x – 0)

y – 1 = –2x

y = –2x + 1

Pada paksi-x, y = 0

0 = –2x + 1

2x = 1
1
( ) 1 x= 2
2
ºR , 0

(c) Luas ∆ PQR = 1   0 –2 1 0
2 1 0 2 1
0

  =1 1
2 2 – (–2) 

= 1  21 + 2 
2

= 5
4
1
= 1 4 unit2

2. (a) 2y = 4 – x 1
2
y = 2 – x

y = x2 – 4x + 1

dy = 2x – 4
dx

Jadi, 2x – 4 = 2
2x = 2 + 4
2x = 6

x=3

57

Apabila x = 3, y = 32 – 4(3) + 1

= 9 – 12 + 1

= –2

º a = 3, b = –2

(b) Persamaan tangen: y + 2 = 2(x – 3)

y + 2 = 2x – 6

y = 2x – 8

Pada paksi-x, y = 0

0 = 2x – 8

2x = 8

x=4

º B(4, 0)

(c) Persamaan normal: y + 2 = – 12 (x – 3)
2y + 4 = –x + 3

2y + x + 1 = 0

Pada paksi-x, y = 0

2(0) + x + 1 = 0

x = –1

º C(–1, 0)

(d) Luas ∆ BPC = 1   4 –1 3 4
2 0 0 –2 0

 = 1  2 – (–8) 
2
 10 
= 1
2

= 5 unit2

3. (a) Luas = 75

x2 + 4hx = 75

4hx = 75 – x2

h = 75 – x2
4x

Isipadu, V = x2h
( ) = x2
1 75 – x2
4 4x

= x(75 – x2)

= 1 (75x – x3) (Tertunjuk)
4
(b) Untuk V maksimum, ddVx = 0

75 – 3 x2 = 0
4 4
75 3
4 = 4 x2

3x2 = 75

x2 = 25
x = 5 (. 0)
1
4
V = [75(5) – 53]

= 1 (375 – 125)
4
1
= 4 (250)

= 62.5 cm3

d 2V = – 32 x
dx2

58

Apabila x = 5, d 2V = – 23 (5)
dx2
= –7.5 (, 0) Ú V adalah maksimum

Maka, V mempunyai nilai maksimum apabila x = 5 cm dan isi padu maksimum kotak ialah 62.5 cm3.

4. (a) x2 + y2 = 102

y2 = 100 – x2
y = ! 100 – x2

dy = 1 (100 – x2)– 21 (–2x)
dx 2
x
= –  ! 100 – x2

dy = dy × dx
dt dx dt

= –  x – x2 ×3
! 100
3x
= –  ! 100 – x2

Apabila x = 8, dy = –  3(8) 82
dt ! 100 –

= – 4

Jadi, kadar perubahan hujung kayu A ialah – 4 ms–1.

(b) x2 + y2 = 102

Apabila y = 6, x2 + 62 = 102

x2 = 100 – 36

x = ! 64

= 8
dy 8
Apabila x = 8, dx = –  ! 100 – 82

= – 43
dy dy dx
dan dt = dx × dt

–2 = – 43 × dx
dt
dx
dt = 1.5

Jadi, kadar perubahan hujung kayu B ialah 1.5 ms–1.

5. Katakan x m ialah jarak mengufuk antara helikopter dengan budak lelaki dan z m ialah jarak antara helikopter

dengan budak lelaki pada masa t.

z2 = x2 + 1352 xm
z = ! x2 + 18 225

dz = 1 (x2 + 18 225)– 21(2x)
dx 2 x
=
! x2 + 18 225 135 m zm

Jadi, dz = dz × dx
dt dx dt
dx ddzt 72
Apabila x = 72 dan dt = –17, = ! 722 + 18 225 × –17

= 72 × –17
153

= –8

º –8 ms–1

59

Latihan Sumatif
( ) [ ] 1. (a)
had   8 + 2x – x2 = had (4 – x)(2 + x)
8 – 2x2 2(4 – x2)
x ˜ –2 x ˜ –2

[ ]=
had (4 – x)(2 + x)
2(2 + x)(2 – x)
x ˜ –2

[ ]= had
x ˜ –2
(4 – x)
2(2 – x)

= 4 – (–2)
2[2 – (–2)]

= 6
8

= 3
4
(b) had

( ) ( )( )x˜0
! 1 +x+ x2 –1 = had ! 1 + x + x2 – 1 ! 1 + x + x2 + 1
x x ! 1 + x + x2 + 1
x˜0

[ ]( )= had
x˜0
1 + x + x2 – 1
x ! 1 + x + x2 + 1

[ ]( )= had
x˜0
x + x2
x ! 1 + x + x2 + 1

[ ]( )= had
x˜0
x(1 + x)
x ! 1 + x + x2 + 1

( )= had
x˜0
1+x
! 1 + x + x2 + 1

= 1 1
! 1 +

= 1
2

(c) had 9 – x 2 =8
4 – ! x 2 + 7
x˜k

9 – k 2 =8
4 – ! k 2 + 7
( )( )
9 – k 2 4 + ! k 2 + 7 =8
4 – ! k 2 + 7 4 + ! k 2 + 7
( )(9 – k 2) 4 + ! k 2 + 7
=8
16 – (k2 + 7)
( )(9 – k 2) 4 + ! k 2 + 7
=8
9 – k2

4 + ! k 2 + 7 = 8
! k 2 + 7 = 4
k2 + 7 = 16

k2 = 9

k = ±3

( ) 2. hada–5 = –3
x ˜ –1x+4
a–5
–1 + 4 = –3

a–5 = –3
3
a – 5 = –9

a = –9 + 5

a = – 4

60

( ) 3. d d
(a) dx 1 = dx [(2x + 1)–1]
2x + 1

= –1(2x + 1)–2(2)

= – (2x 2 1)2
+

( b) ddx[4x(2x – 1)5] = 4x[5(2x – 1)4(2)] + (2x – 1)5(4)
= 40x(2x – 1)4 + 4(2x – 1)5

= 4(2x – 1)4[10x + (2x – 1)]

= 4(12x – 1)(2x – 1)4
[ ](c)
d (2 6 = d [6(2 – x)–2]
dx – x)2 dx

= 6(–2)(2 – x)–3(–1)

= 12
(2 – x)3
( )(d) + 3)– 21(1) + ! x +
d (x! x + 3) = x 1 (x 3(1)
dx 2
x
= 2! x + 3 + ! x + 3

= x + 2(x + 3)
2! x + 3

= 3x + 6
2! x + 3

= 3(x + 2)
2! x + 3

4. (a) y = x(3 – x)

= 3x – x2
ddxy = 3 – 2x
d 2y
dx2 = –2

y dd x2y2 + x ddyx + 12 = (3x – x2)(–2) + x(3 – 2x) + 12
= –6x + 2x2 + 3x – 2x2 + 12

= 12 – 3x

(b) 12 – 3x = 0

3x = 12

x = 12
3

x = 4

5. y = ax + b
x2
= ax + bx–2

ddxy = a – 2bx–3
2b
= a – x3

( )Pada titik , – 72 b
–1, – 27 = a(–1) + (–1)2

– 27 = –a + b
7
a–b= 2

2a – 2b = 7 …1

61

dan dy = 2
dx
2b
a – (–1)3 = 2

a + 2b = 2 …2

1 + 2: 3a = 9
a=3

Gantikan a = 3 ke dalam 1: 2(3) – 2b = 7

6 – 2b = 7

2b = –1

– 21 b = – 12

ºa = 3, b =
V= 4 π r3
6. 3

dV = 4π r2
dr

dV = dV × dr
dt dr dt

20π = 4π r2 × 0.2

20 = 0.8r2

r2 = 25

r = 5 (. 0)

º r = 5 cm = 14(6x3 + 1)– 21

7. (a) y = 14
! 6x3 + 1

( ) dy – 21 (6x3 + 1)– 23(18x2)
dx = 14

= –  126x2
3
(6x3 + 1)2

Apabila x = 2, dx = 2.05 – 2

= 0.05

dan dy = –  126(2)2
dx 3

[6(2)3 + 1]2

= – 350443

= – 4729

Jadi, dy ≈ dy × dx
dx
= – 4729 × 0.05

= –0.0735 unit
14
(b) Apabila x = 2, y = ! 6(2)3 + 1

= 14
7

= 2

dx = 2.05 – 2 = 0.05

dan dy = – 7429
dx

14
! 6(2.05)3 + 1
( ) – 7492 (0.05)
=2+

= 2 – 0.0735

= 1.927

62

8. y = 1
! x

= x– 12

dy = – 21 x– 23
dx

= – 2!1 x3

Apabila x = 4, y = 1
! 4

= 1
2

dx = 2 × 4 = 0.08
100
dy
dan dx = – 2!1 43 = – 116

Jadi, dy ≈ dy × dx
dx
= – 116 × 0.08

= – 0.005

Maka, peratus perubahan hampir dalam y = dy × 100%
y

= – 0.0105 × 100%
2

= –1%

9. y = 3x2 – 4x + 6

dy = 6x – 4
dx

Apabila x = 2, y = 3(2)2 – 4(2) + 6

= 12 – 8 + 6

dy = 10
dx
= 6(2) – 4

=8

dan dx = p × 2
100
= 0.02p

dy = dy × dx
dx

= 8 × 0.02p

= 0.16p

Maka, peratus perubahan dalam y ialah 0.16p × 100 = 1.6p%
10
10. (a) Titik maksimum ialah (–1, 6) dan titik minimum ialah (1, 2)

(b)

y

(–1, 6) y = f (x)

(1, 2)
0x


63

11. (a) y = 3x3 – 4x + 2

dy = 9x2 – 4
dx

Pada titik A(2, 1), dy = 9(2)2 – 4
dx
= 36 – 4

= 32

Persamaan tangen di titik A(2, 1) ialah:

y – 1 = 32(x – 2)

y – 1 = 32x – 64

dy y = 32x – 63
dx
(b) = 32

9x2 – 4 = 32

9x2 = 36

x2 = 4

x = ±2

Apabila x = –2, y = 3(–2)3 – 4(–2) + 2

= –24 + 8 + 2

= –14

º (–2, –14)

( ) 12. (a) j2 = 6! 3 2 – t2 A
t cm
j = ! 108 – t2 …1 D

I = 1  πj2t …2
3
Gantikan 1 ke dalam 2: 6� 3 cm
B j cm
I = 1  π (! 108 – )t2 2t
3
1
= 3  π (108 – t2)t

= 36π t – 1  π t3 ddIt  
3
Untuk isi padu maksimum, = 0

36π – π t2 = 0

36π = πt2

36 = t2
t = 6 (. 0)


º t = 6 cm

(b) I = 36π t – 1  π t3
3
1
Apabila t = 6, I = 36π(6) – 3  π (6)3

= 216π – 72π

= 144π

º Isi padu kon ialah 144π cm3.

13. AC = ! 302 + x2 = ! 900 + x2

Jumlah masa yang diambil dari A ke D ialah

T= ! 900 + x2 + 400 – x
40 50

T = 1 ! 900 + x2 + 8 – 1  x
40 50

64

Untuk nilai pegun T, dT  = 0
dx
(900 + x2)– 12(2x) –
( )11 1 = 0
2 50
40 x 1
40! 900 + x2 50
=

50x = 40! 900 + x2

5x = 4! 900 + x2

25x2 = 16(900 + x2)

25x2 = 14 400 + 16x2

9x2 = 14 400

x2 = 1 600

x = ! 1 600

= 40

º Jarak dari B ke C ialah 40 m.

14. I = 8

x3 = 8

x = 2

L = 6x2
AddLxpab=il1a2xx
= 2, dL   = 12(2)
dx = 24

dL = dL × dx
dt dx dt
dL
dt = 24 × 2

= 48

º Kadar perubahan jumlah luas permukaan kubus ialah 48 cm2s–1.

15. (a) Luas, A = 1 xy
2
1
= 2 x(6x – x2)

= 1 (6x2 – x3) (Tertunjuk)
2

(b) (i) A= 1 (6x2 – x3) = 3x2 – 1 x3
2 2
dA 3
dx = 6x – 2 x2

dA = dA × dx
dt dx dt
( ) dA 3
dt = 6x – 2 x2 (2)

[ ] Apabila = 2, dA = 3 (2)2 (2)
x dt 6(2) – 2

= (12 – 6)(2)

= 12

Kadar tokokan A ialah 12 unit2 s–1

[ ](ii) Apabila x = 5, dA = 3 (5)2 (2)
dt 6(5) – 2

= (30 – 37.5)(2)

= –15

º Kadar susutan A ialah 15 unit2 s–1

65

16. (a) r = h
12 20

r = 12  h
20

r = 3  h …1
5

V = 1 π r2h …2
3
( )Gantikan
1 ke dalam 2: V = 1  π  3  h 2h
3 5
( ) 1 9
= 3  π  25 h2 h

= 3  π h3 (Tertunjuk)
3 25
(b) (i) V= 25  π h3

dV = 9  π h2
dh 25

Apabila h = 5, dh = 4.99 – 5 = – 0.01

dan ddVh = 9  π (5)2
25

= 9π

d V ≈ dV × d h
dh

d V = 9π × – 0.01

= – 0.09π

º Perubahan kecil dalam isi padu air ialah – 0.09π cm3.

(ii) Jika h menyusut sebanyak p%,
d h = – 1p00(h) = – 1p0h0
dV
d V ≈ dh × d h

d V = 9  π h2 × – 1p0h0
25
9
= – 2 500 π ph3

Jadi, dV × 100% = – 2 9 π ph3 × 100%
V 500
3
25  π h3

= –3p

º Isi padu menyusut sebanyak 3p%.

66

Jawapan

BAB 3 PENGAMIRAN

Aktiviti Penerokaan 1 (Halaman 82)

∫ 3. (a) Graf fungsi g(x) = f (x) dx adalah sama dengan graf fungsi f (x).
∫(b) Graf fungsi k(x) = h(x) dx adalah sama dengan graf fungsi h(x).
∫(c) Graf fungsi n(x) = m(x) dx adalah sama dengan graf fungsi m(x).

Kuiz Pantas (Halaman 83)

1. Mencari isi padu air dalam baldi daripada kadar pengaliran air sebuah pili.

2. Mencari jarak yang dilalui oleh suatu objek daripada kadar perubahan jarak atau laju.

3. Mencari luas pembiakan kulat daripada kadar pertambahan luas kulat pada roti.

Latihan Kendiri 3.1

∫ 1. (15x2 + 4) dx = 5x3 + 4x

∫ 2. 24x2 dx = 8x3

3. (a) dJ = 300t2 + 60t
dt

(b) dJ = 1 500t2 + 300t
dt
= 5(300t2 + 60t)

J = 5(100t3 + 30t2)

= 500t 3 + 150t2

Apabila t = 2, J = 500(2)3 + 150(2)2

= 4 600 liter

Latihan Formatif 3.1

1. dy = 3(3)(2)(2x + 2)2
dx
= 18(2x + 2)2
∫ 18(2x + 2)2 dx = 3(2x + 2)3

2. f (x) = (2 – 3x)(5) – (–3)(5x + 2)
(2 – 3x)2
16
= (2 – 3x)2

∫ ∫ f (x) dx = (2 16 dx
– 3x)2

= 5x + 2
2 – 3x

3. dy = 15(x + 2)2
dx
Bandingkan dengan ddyx,
h = 15 dan k = 2

h + k = 15 + 2

= 17

1

∫ ( ) ∫1 dy dx = 1 15(x + 2)2 dx
dx 10
10
= 110[5(x + 2)3]
(x + 2)3
= 2

∫ ( )Apabila x = 2, nilai bagi 1 dy dx = (2 + 2)3
10 dx 2

= 32

4. f (x) = 3x(2x + 1)2
f (x) = 3(2x + 1)2 + 2(2)(2x + 1)(3x)

= (2x + 1)[3(2x + 1) + 12x]

= (2x + 1)(6x + 3 + 12x)

= (2x + 1)(18x + 3)

= 36x2 + 6x + 18x + 3

= 36x2 + 24x + 3

= 3(12x2 + 8x + 1)

∫ ∫(12x2 + 3x + 1) dx = 1  f (x) dx
3

∫ = 1 f (x) dx
3
1
af(x) = 3  f(x)

a = 1
(a) 3
5. dA = 200t + 150t2
dt
dA
Apabila t = 5, dt = 200(5) + 150(5)2

= 4 750

Kadar keuntungan harian hasil jualan tiket bas bagi syarikat K ialah RM4 750 sehari.

(b) Fungsi keuntungan harian hasil jualan tiket bas bagi syarikat H

∫= 30t2 + 40t dt
∫= 1
5   200t + 150t2 dt

= 1 [100t2 + 50t3]
5
= 20t2 + 10t3

Maka, syarikat K mendapat keuntungan 5 kali ganda lebih daripada syarikat H.

Aktiviti Penerokaan 2 (Halaman 85)
Kes 1

Fungsi f (x) ∫ f (x) dx

f (x) = 2x + c 2 2x + c
f (x) = 3x + c 3 3x + c
f (x) = 0.5x + c 0.5 0.5x + c
f (x) = –7x + c –7 –7x + c
f (x) = – 4x + c – 4 – 4x + c

∫Maka, a dx = ax + c.

2

Kes 2

Fungsi ∫f (x) f (x) dx Pola

f (x) = 2x2 + c 4x 4x2 + c = 2x2 + c 4x1 + 1 + c
6x2 2 1+1
8x3
f (x) = 2x3 + c 10x4 6x3 + c = 2x3 + c 6x2 + 1 + c
12x5 3 2+1

f (x) = 2x4 + c 8x4 + c = 2x4 + c 8x3 + 1 + c
4 3+1

f (x) = 2x5 + c 10x5 + c = 2x5 + c 10x4 + 1 + c
5 4+1

f (x) = 2x6 + c 12x6 + c = 2x6 + c 12x5 + 1 + c
6 5+1

∫Maka, axn dx = axn + 1 .
n+1

Kuiz Pantas (Halaman 86)

∫ ∫(a) dx = 1 dx

= x + c

∫(b) 0 dx = 0 + c
= c
1
2 x2 + c jika x > 0

∫(c)  x  dx = – 12 x2 + c jika x , 0



Perbincangan (Halaman 87)

Walaupun kamiran bagi fungsi yang melibatkan penambahan atau penolakan akan melibatkan beberapa pemalar
pengamiran, namun hasil tambah semua pemalar ini masih lagi merupakan suatu pemalar. Oleh itu, kamiran bagi suatu
fungsi yang melibatkan penambahan dan penolakan sebutan-sebutan algebra boleh diwakilkan dengan satu pemalar
pengamiran sahaja.

Latihan Kendiri 3.2

∫ 1. (a) 2 dx = 2x + c

∫(b) 5 dx = 5  x + c
6 6

∫(c) –2 dx = –2x + c
∫(d) π π
3 dx = 3  x + c

∫ 2. (a) 3x2 dx = 3x3 + c
3
= x3 + c

∫ ( )(b) 4  x3 dx = 4 x4 +c
3 3 4

= x4 + c
3

∫(c) –x dx = – x22 + c

∫ ∫(d) – x22 dx = –2x –2 dx
–2x –1
= –1 + c

= 2 +c
x

3

∫ ∫(e) 3 dx = 3x –3 dx
x3
3x –2
= –2 +c

= – 23x2 + c
1
∫ ∫(f)
3! x dx = 3x2 dx

3
3x2
= 3 + c

2
3

= 2x2 + c

= 2! x3 + c

∫ ∫(g) 2 2x – 31 dx
3! x dx =
2
2x3
= 2 + c

3
2

= 3x3 + c

= 33! x2 + c

∫ ( ) ∫(h) – !3 x 3 dx = –27x – 32 dx

= –27x– 12 +c
– 21

= 54 +c
! x

∫ 3. (a) 2x + 3 dx = 2x2 + 3x + c
2

= x2 + 3x + c
∫(b) 4x3 5x2
4x2 + 5x dx = 3 + 2 + c

= 4 x3 + 5 x2 + c
3 2

∫ ( )(c) 1  x3 + 5x – 2 dx = 1   x4 + 5x2 – 2x + c
2 2 4 2

= 1 x4 + 5 x2 – 2x + c
8 2

∫ ∫(d) 3 + 4x – 2 dx = 3x –2 + 4x – 2 dx
x2

= 3x –1 + 4x2 – 2x + c
–1 2

= – 3x + 2x2 – 2x + c

∫ ∫ 4. (a) (x + 2)(x – 4) dx = x2 + 2x – 4x – 8 dx

∫= x2 – 2x – 8 dx

= x3 – 2x2 – 8x + c
3 2
x3
= 3 – x2 – 8x + c

∫ ∫(b) x2(3x2 + 5x) dx = 3x4 + 5x3 dx
3x5 5x4
= 5 + 4 +c

= 3 x5 + 5 x4 + c
5 4

4

1

5x2 – 3x2 dx
∫ ∫ ( )(c) (5x2 – 3! x ) dx =
3
5x3 3x2
= 3 – 3 + c

2

= 5 x3 – 3 + c
3
2x2

= 5 x3 – 2! x3 + c
3

∫ ∫(d) (5x – 3)2 dx = 25x2 – 30x + 9 dx
25x3 30x2
= 3 – 2 + 9x + c

= 25 x3 – 15x2 + 9x + c
3
∫ ( ) ∫(e)
5x2 – 3x dx = 5x – 3 dx
x
5x2
= 2 – 3x + c

= 5 x2 – 3x + c
2
3
∫ ∫( )(f)
x + ! x 2 dx = (x2 + 2x2 + x) dx

5
x3 2x2 x2
= 3 + 5 + 2 + c

2

= 1 x3 + 4 5 + 1 x2 + c
3 5 2
x2

Perbincangan (Halaman 88)
Tidak boleh. Kaedah penggantian hanya boleh digunakan untuk fungsi dalam kurungan yang berbentuk linear sahaja.

Kita perlu mengembangkan fungsi dalam kurungan yang bukan linear bagi mendapatkan kamiran.

Latihan Kendiri 3.3
∫ 1. (a) (x – 3)2 dx
Katakan u = x – 3 dan du = 1
dx

∫ ∫ (x – 3)2 dx = u2 du
u3
= 3 + c

= (x – 3)3 + c
3

(b) Katakan u = 3x – 5 dan du = 3
dx

∫ ∫ u9
(3x – 5)9 dx = 3 du

= u10 + c
30
(3x – 5)10
= 30 + c

(c) Katakan u = 5x – 2 dan du = 5
dx

∫ ∫ 4u5
4(5x – 2)5 dx = 5 du

= 4u6 + c
30

= 125(5x – 2)6 + c

5

(d) Katakan u = 7x – 3 dan du = 7
dx
(7x – 3)4 u4
∫ ∫ 3 dx = 21 du

= u5 + c
105

= (7x – 3)5 + c
105

(e) Katakan u = 2x – 6 dan du = 2
dx
12 12u–3
(2x – 6)3 2
∫ ∫ dx = du

= 6u–2 + c
–2

= – (2x 3 6)2 + c
–

(f) Katakan u = 3x – 2 dan du = 3
dx
2 2u–2
3(3x – 9
∫ ∫ dx = du
2)2

= 2u–1 + c
–9
= – 9(3x2– 2) + c

∫ 2. (a) (4x + 5)4 dx = (4x + 5)5 + c
5(4)

= (4x + 5)5 + c
20

∫(b) 2(3x – 2)3 dx = 2(3x – 2)4 + c
4(3)

= (3x – 2)4 + c
6

∫(c) (5x – 11)4 dx = (5x – 11)5 + c
5(5)

= (5x – 11)5 + c
25

∫(d) (3x – 2)5 dx = (3x – 5)6 +c
5 5(3)(6)

= (3x – 5)6 + c
90

∫ ∫(e) 5 5(6x – 3)–6 dx
(6x – 3)6 dx =

= 5(6x – 3)–5 + c
–5(6)

= – 6(6x1– 3)5 + c

12
(3x – 5)8
∫ ∫(f) 12(3x – 5)–8 dx
dx =

= 12(3x – 5)–7 + c
–7(3)

= – 7(3x4– 5)7 + c

Latihan Kendiri 3.4

∫ 1. (a) y = 4x – 2 dx
4x2
y = 2 – 2x + c

y = 2x2 – 2x + c …1

6

Gantikan x = –1 dan y = 7 ke dalam persamaan 1:

7 = 2(–1)2 – 2(–1) + c

c=3

∫(b) y = – 6x – 6 dx
x3
∫y = – 6x – 6x –3 dx
– 62x2 6x –2
y = – –2 + c

y = –3x2 + 3 + c…1
x2
Gantikan x = –1 dan y = 6 ke dalam persamaan 1:
3
6 = –3(–1)2 + (–1)2 + c
c = 6

∫ 2. y = 20x3 – 6x2 – 6 dx
20x4 6x3
y= 4 – 3 – 6x + c

y = 5x4 – 2x3 – 6x + c …1

Gantikan x = 1 dan y = 2 ke dalam persamaan 1:

2 = 5(1)4 – 2(1)3 – 6(1) + c

c=5
y = 5x4 – 2x3 – 6x + 5 …2

Gantikan x = 1 ke dalam persamaan 2:
2

( ) ( ) ( )y = 514–2 1 3 –6 1 +5
2 2 2

y = 33
16

∫ 3. (a) y = 9x2 – 2 dx
9x3
y = 3 – 2x + c

y = 3x3 – 2x + c …1
Gantikan titik (1, 6) ke dalam persamaan 1:

6 = 3(1)3 – 2(1) + c

c=5

Persamaan lengkung ialah y = 3x3 – 2x + 5
∫(b) y = 10x – 2 dx

y = 10x2 – 2x + c
2
y = 5x2 – 2x + c …1

Gantikan titik (2, 13) ke dalam persamaan 1:
13 = 5(2)2 – 2(2) + c

c = –3
Persamaan lengkung ialah y = 5x2 – 2x – 3

∫(c) y = 24x2 – 5 dx

y = 24x3 – 5x + c
3
y = 8x3 – 5x + c …1
Gantikan titik (1, 1) ke dalam persamaan 1:

1 = 8(1)3 – 5(1) + c

c = –2

Persamaan lengkung ialah y = 8x3 – 5x – 2

7

∫(d) y = 18x2 + 10x dx

y = 18x3 + 10x2 + c
3 2
y = 6x3 + 5x2 + c …1

Gantikan titik (–2, –10) ke dalam persamaan 1:
–10 = 6(–2)3 + 5(–2)2 + c

c = 18
Persamaan lengkung ialah y = 6x3 + 5x2 + 18

Latihan Formatif 3.2

∫ 1. (a) 1 dx = 1 x + c
2 2

∫ ∫(b) 5 dx = 5x –3 dx
3x3 3

= 5x –2 + c
– 6
= – 65x2 + c
x – 12 dx
∫ ∫(c) 1
! x dx =

1
x2
= 1 +c

2 1

= 2x2 + c

∫ ( ) ∫(d) 2 – 3 dx = 2x –3 – 3x – 4 dx
x3 x4
2x –2 3x –3
= –2 – –3 + c

= – x12 + 1 + c
x3

5x2 – 3x3
∫ ∫ 2. (a) x dx = 5x – 3x2 dx

= 5x2 – 3x3 + c
2 3
5
= 2 x2 – x3 + c

6x3 + 2x2
2x2
∫ ∫(b) dx = 3x + 1 dx

= 3x2 + x + c
2
3
= 2 x2 + x + c

∫(c) (5 – 6x)3 dx = (5 – 6x)4 + c
–24

= – (5 – 6x)4 + c
24

∫ ∫(d) 1 (5 – 2x)– 41 dx
 4! 5 – 2x dx =
3
(5 – 2x)4
( )= 3 +c
4 (–2)

3
– 2(5 – 2x)4
= 3 + c

8

3. dy = 10x + p …1
dx x2

Gantikan dy = 20 1 dan x = 2 ke dalam persamaan 1:
dx 2
p
41 = 10(2) + (2)2
2
p=2

∫y = 10x + 2 dx
x2

∫y = 10x + 2x –2 dx

y = 10x2 + 2x–1 + c
2 –1

y = 5x2 – 2 + c …2
x
Gantikan y = 19 dan x = 2 ke dalam persamaan 2:

19 = 5(2)2 – 2 +c
2
c = 0
2
y = 5x2 – x …3

Gantikan x = –2 ke dalam persamaan 3:
2
y = 5(–2)2 – –2
= 21

∫ 4. (a) y = 4x3 – 15x2 + 6 dx

y = 4x4 – 15x3 + 6x + c
4 3
y = x4 – 5x3 + 6x + c …1

Gantikan x = 3 dan y = –20 ke dalam persamaan 1:
–20 = (3)4 – 5(3)3 + 6(3) + c

c = 16
y = x 4 – 5x3 + 6x + 16 …2

Gantikan x = –2 ke dalam persamaan 2:

y = (–2)4 – 5(–2)3 + 6(–2) + 16

y = 60
∫(b) y = 2x + 2 dx
2x2
y = 2 + 2x + c

y = x2 + 2x + c …1

Gantikan x = 2 dan y = 2 ke dalam persamaan 1:
2 = (2)2 + 2(2) + c

c = – 6
y = x2 + 2x – 6 …2
Gantikan y = – 6 ke dalam persamaan 2:
– 6 = x2 + 2x – 6
x2 + 2x = 0

x(x + 2) = 0

x = 0 atau x = –2

∫ 5. y = 3x2 – 8x dx
3x3 8x2
y = 3 – 2 + c

y = x3 – 4x2 + c …1

9

Gantikan titik (1, –1) ke dalam persamaan 1:

–1 = (1)3 – 4(1)2 + c

c = 2

Persamaan lengkung tersebut ialah y = x3 – 4x2 + 2.

6. dy = – (6x – 2)
dx

∫ y = 2 – 6x dx
6x2
y = 2x – 2 + c

y = 2x – 3x2 + c …1

Gantikan titik (2, 2) ke dalam 1:

2 = 2(2) – 3(2)2 + c

c = 10

Maka, y = 2x – 3x2 + 10

7. dy = ax + b
dx
dy
Pada titik (–2, 8), dx = –7

a(–2) + b = –7
–2a + b = –7 …1

Pada titik (0, 6), dy = 5
dx

a(0) + b = 5

b = 5
Gantikan b = 5 ke dalam 1,

–2a + 5 = –7

–2a = –12

a=6

∫ y = 6x + 5 dx
6x2
= 2 + 5x + c

= 3x2 + 5x + c …2

Gantikan titik (–2, 8) ke dalam 2:
8 = 3(–2)2 + 5(–2) + c
c = 8 – 12 + 10
=6
Maka, persamaan lengkung ialah y = 3x2 + 5x + 6.

∫ 8. s = 10t – 2 dt

s = 10t2 – 2t + c
2
s = 5t2 – 2t + c …1
Gantikan s = 8 dan t = 1 ke dalam 1:

8 = 5(1)2 – 2(1) + c

c=5
s = 5t2 – 2t + 5 …2
Gantikan t = 3 ke dalam 2:

s = 5(3)2 – 2(3) + 5

= 44 m

Kuiz Pantas (Halaman 92)

∫ [ ](a) 2 1 dx = x 2
1 1

= 2 – 1

= 1

10

∫(b) 2 0 dx = 0
1

Aktiviti Penerokaan 3 (Halaman 93)

4.

∫  2 ∫  6
2 3x2 dx 2 3x2 dx

∫  6 ∫  6
2 3x2 dx
3 2 3x2 dx
∫  6
2 3(3x2) dx ∫ ∫   6 6
2 3x2 dx + 2 6x dx
∫ ∫   43x2 dx + 6 ∫  2
1 –  6 3x2 dx
4 3x2 dx 0

∫  6 11
2 (3x2 + 6x) dx

Latihan Kendiri 3.5

∫ [ ] 1. (a) 4 x3 dx = x4 4
2 4
2

= 44 – 24
4 4
= 60

∫  ∫ (b) 42 4 2x –2
1 x2 dx = 1 dx

[ ]= 2x –1 4
–1 1
( –2 ) (– 12 )
=  4 –

= – 21 + 2
3
= 2

∫  [ ](c) 5(2x2 + 3x) dx = 2x3 + 3x2 5
1 3 2 1

[ ] [ ]=2(5)3 + 3(5)2 – 2(1)3 + 3(1)2
3 2 3 2

= 356
3
∫  ( ) ∫ (d)
6 1 – 2x dx = 6 (x –3 – 2x) dx
2 x3 2

[ ]= x –2 – 2x2 6
–2 2 2

[ ] [ ]= – 2(16)2 – 62 – – 2(12)2 – 22

= – 2897
1

3x – x2
∫   ∫  ( ) ( )(e) 3 3
1 3x – ! x dx = 1 dx

[ ]2 1 33
3x2 x2
= 2 – 3

[ ] [ ]2 3 3
3(3)2 32 3(1)2 12
= 2 – 3 – 2 – 3

= 9.203 2