14[anal add math cd]

1

Additional Mathematics SPM Chapter 14

© Penerbitan Pelangi Sdn. Bhd.

1. ∫ g(x) dx = f (x) + c

2. ∫ 3h(x) dx = 3 ∫ h(x) dx

= 3f (x) + c

3. ∫ 6(3x + 1) dx = 9x2 + 6x + c

4. ∫ 6(3x + 1) dx = 9x2 + 6x + c

6 ∫ (3x + 1) dx = 9x2 + 6x + c

\ ∫ (3x + 1) dx = 9—6 x2 + 6x–––

6 + c—6

= 3—2 x2 + x + c—

6

5. ∫ 2—3 h(x) dx = f (x) + c

2—3 ∫ h(x) dx = f (x) + c

∫ h(x) dx = 3—2 [f (x) + c]

6. (a) ∫ x 4 dx = x5–––

5 + c

(b) ∫ 2x5 dx = 2x 6––––6

+ c

= 1—3 x6 + c

(c) ∫ 3 dx = 3x + c

(d) ∫ 3––––5x2

dx = ∫ 3—5 x–2 dx

= 3—5 1 x –1

–––––1 2 + c

= – 3–––5x

+ c

(e) ∫ (8 – 3x2) dx = 8x – 3x3––––

3 + c

= 8x – x3 + c

(f) ∫ x(x2 – 4x) dx = ∫ (x3 – 4x2) dx

= 1—4 x4 – 4—

3 x3 + c

(g) ∫ 2(x2 – 1)(3x – 4) dx

= ∫ 2(3x3 – 4x2 – 3x + 4) dx

= ∫ (6x3 – 8x2 – 6x + 8) dx

= 6x4––––

4 – 8x3

––––3

– 6x2––––

2 + 8x + c

= 3—2 x4 – 8—

3 x3 – 3x2 + 8x + c

(h) ∫ 1 3x3 – 4x2 + 1–––––––––––x5 2 dx

= ∫ 1 3x3––––

x5 – 4x2

––––x5

+ 1–––x5 2 dx

= ∫ (3x–2 – 4x–3 + x–5) dx

= 3x–1–––––1

– 4x–2–––––2

+ x–4–––– 4

+ c

= – 3—x + 2–––

x2 – 1––––

4x4 + c

(i) ∫1 x2 – 16–––––––x – 4 2 dx = ∫ (x + 4)(x – 4)

––––––––––––(x – 4) dx

= ∫ (x + 4) dx

= 1—2 x2 + 4x + c

(j) ∫ x2 + 3x + 2––––––––––x + 2

dx = ∫ (x + 1)(x + 2)––––––––––––

(x + 2) dx

= ∫ (x + 1) dx

= 1—2 x2 + x + c

CHAPTER

14 Integration

2



7. dy

–––dx = 4x2 – x + 5

y = ∫ dy–––dx · dx

= ∫ (4x2 – x + 5) dx

y = 4—3 x3 – x2

–––2 + 5x + c ...................... 1

Substitute x = 1, y = –1 into 1,

–1 = 4—3 – 1—

2 + 5 + c

c = – 41–––6

Therefore, y = 4—3 x3 – 1—

2 x2 + 5x – 41–––6 .

8. dy

–––dx = x 4 – 1––––––

x2

= x4–––x2 – 1–––

x2

= x2 – x–2

y = ∫ (x2 – x–2) dx

= 1—3 x3 – x –1

––––1 + c

y = 1—3 x3 + 1—x + c .......................... 1

Substitute y = 3 and x = 2 into 1,

3 = 1—3 (2)3 + 1—

2 + c

= 8—3 + 1—

2 + c

c = – 1—6

Therefore, y = 1—3 x3 + 1—x – 1—

6 .

9. dy

–––dx = 4x3 – 8x + 1

y = ∫ (4x3 – 8x + 1) dx

y = x4 – 4x2 + x + c ........................ 1

Substitute x = 1, y = –1 into 1,–1 = 1 – 4 + 1 + c c = 1

Therefore, the equation of the curve isy = x4 – 4x2 + x + 1.

10. ∫ (2x + 3)5 dx

Let u = 2x + 3

du–––dx = 2

dx = du–––2

∫(2x + 3)5 dx = ∫ u5–––2 du

= u6––––6(2) + c

= u6–––12 + c

= (2x + 3)6–––––––

12 + c

11. ∫ u4 dx = ∫ (1 – 3x)4 dx

= (1 – 3x)5–––––––

5(–3) + c

= – (1 – 3x)5–––––––––

15 + c

12. (a) ∫ 2(5x + 3)4 dx

= 2 (5x + 3)5–––––––––

5(5) + c

= 2–––25 (5x + 3)5 + c

(b) ∫ 1–––––––(3x – 2)4 dx

= ∫ (3x – 2)–4 dx

= (3x – 2)–3––––––––

3(–3) + c

= – 1–––––––––9(3x – 2)3 + c

(c) ∫ 3–––––––(2 – 7x)2 dx

= ∫ 3(2 – 7x)–2 dx

= 3(2 – 7x)–1–––––––––

(–1)(–7) + c

= 3–––––––––7(2 – 7x) + c

(d) ∫ 2–––––––––3(4x – 1)3 dx

= ∫ 2—3 (4x – 1)–3 dx

= 2—3 3 (4x – 1)–2

–––––––––2(4) 4 + c

= – 1–––––––––12(4x – 1)2 + c

13. (a) ∫ 2

1 3 dx = 33x4

2

1 = (3 × 2) – (3 × 1) = 3

3



(b) ∫ 1

0 4x2 dx = 3 4—

3 x341

0

= 1 4—3 × 12 – 1 4—

3 × 02 = 4—

3

(c) ∫ 1

–1 3–––

x3 dx = ∫ 1

–1 3x–3 dx

= 3 3x –2–––––2 4

1

–1

= 3– 3–––2x2 4

1

–1

= 1– 3—2 2 – 1– 3—

2 2 = 0

(d) ∫ 1

–2 1–––2x2 dx = ∫ 1

–2 1—

2 x–2 dx

= 3 x –1–––––2(–1) 4

1

–2

= 3– 1–––2x 4

1

–2

= 1– 1—2 2 – 3– 1––––

(– 4) 4 = – 1—

2 – 1—4

= – 3—4

(e) ∫ 2

0 (3x2 – 4) dx = 3 3x3

––––3 – 4x4

2

0

= [23 – 4(2)] – (0) = 0

(f) ∫ 3

1 1—

2 (2x – 3)3 dx = 3 1—2 × (2x – 3)4

––––––––4(2) 4

3

1

= 3 1–––16 (2x – 3)44

3

1

= 3 1–––16 (3)44 – 3 1–––

16 (–1)44 = 5

(g) ∫ 2

1 1–––––––(2 – 3x)4 dx = ∫ 2

1 (2 – 3x)–4 dx

= 3 (2 – 3x)–3––––––––(–3)(–3) 4

2

1

= 3 1–––––––––9(2 – 3x)3 4

2

1

= 3 1––––––9(– 4)3 4 – 3 1––––––

9(–1)3 4 = 1–––––

–576 + 1—9

= 7–––64

(h) ∫ 1

–2 4x2 – 9–––––––

2x – 3 dx

= ∫ 1

–2 (2x – 3)(2x + 3)––––––––––––––

(2x – 3) dx

= ∫ 1

–2 (2x + 3) dx

= 3x2 + 3x41

–2

= (1 + 3) – (4 – 6) = 4 + 2 = 6

(i) ∫ 0

–2 2(3x – 4)(1 – 5x) dx

= ∫ 0

–2 2(3x – 15x2 – 4 + 20x) dx

= ∫ 0

–2 2(–15x2 + 23x – 4) dx

= 23–5x3 + 23–––2 x2 – 4x4

0

–2

= 2(0) – 2[–5(–2)3 + 23–––2 (–2)2 – 4(–2)]

= –188

14. (a) ∫ 2

1 4x dx = 3 4x2

–––2 4

2

1

= 32x242

1

= (2 × 22) – (2 × 12) = 6

(b) ∫ 1

2 4x dx = 32x24

1

2

= (2 × 12) – (2 × 22) = – 6

(c) 4 ∫ 2

1 x dx = 43 x2

–––2 4

2

1

= 41 22–––2 – 1—

2 2 = 412 – 1—

2 2 = 41 3—

2 2 = 6

(d) ∫ 1

2 4x dx = – ∫ 2

1 4x dx

(e) ∫ 2

1 4x dx = 4 ∫ 2

1 x dx

4



15. (a) Area = ∫ 3

1 y dx

= ∫ 3

1 (x2 + 1) dx

= 3 x3–––3 + x4

3

1

= 1 33–––3 + 32 – 1 1—

3 + 12 = 12 – 4—

3 = 10 2—

3 unit2

(b) Area = ∫ 4

0 y dx

= ∫ 4

0 x3 dx

= 3 x4–––4 4

4

0

= 1 44–––4 2 – (0)

= 64 unit2

(c) Area = ∫ 3

0 y dx

= ∫ 3

0 (x2 – 3x) dx

= 3 x3–––3 – 3—

2 x243

0

= 3 33–––3 – 3—

2 (3)24 – (0)

= 9 – 27–––2

= – 9—2

= 9—2 unit2

Therefore, the area of the shaded region is 9—2 unit2.

(d) Area = ∫ 3

1 1––––

3x2 dx

= ∫ 3

1 x–2–––3 dx

= 3 x–1–––––3(–1) 4

3

1

= 3 1–––––3x 4

3

1

= 1– 1—9 2 – 1– 1—

3 2 = – 1—

9 + 1—3

= 2—9 unit2

(e) Area = ∫ 1

–2 1–––––––

(x + 4)2 dx

= ∫ 1

–2 (x + 4)–2 dx

= 3 (x + 4)–1–––––––

–1 41

–2

= 3– 1–––––x + 4 4

1

–2

= 1– 1—5 2 – 1– 1—

2 2 = – 1—

5 + 1—2

= 3–––10 unit2

(f)

0

y

x = 1

x = –1

y = x 3

xP

Q

Area of region P = ∫ 0

–1 x3 dx

= 3 x4–––4 4

0

–1

= (0) – ( 1—4 )

= – 1—4

= 1—4 unit2

Area of region Q = ∫ 1

0 x3 dx

= 3 x4–––4 4

1

0

= 1 1—4 2 – (0)

= 1—4 unit2

Therefore, the area of the shaded region

= 1—4 + 1—

4

= 1—2 unit2

5



(g)

0

yy = 2x 2

y = –2x + 4

B(2, 0)

A(1, 2)

xP Q


0 2x2 dx

= 3 2—3 x34

1

0

= 1 2—3 2 – (0)

= 2—3 unit2

Area of region Q = 12

× 1 × 2

= 1 unit2

Area of the shaded region = 2—3 + 1

= 53

unit2

(h)

0

y

P

y = 4

y = x 2 + 3

3(1, 4)

x


0 y dx

= ∫ 1

0 (x2 + 3) dx

= 3 1—3 x3 + 3x4

1

0

= 1 1—3 + 32 – (0)

= 3 1—3 unit2

Area of rectangle = 1 × 4 = 4 unit2

Area of the shaded region = 4 – 3 1—3

= 2—3 unit2

16. (a) Area = ∫ 1

0 x dy

= ∫ 1

0 y2 dy

= 3 y3–––3 4

1

0

= 1 1—3 2 – (0)

= 1—3 unit2

(b) Area = ∫ 3

1 x dy

= ∫ 3

1 y3 dy

= 3 y4–––4 4

3

1

= 1 34–––4 2 – 1 14

–––4 2

= 81–––4 – 1—

4 = 20 unit2

(c) Area = ∫ 1

0 x dy

= ∫ 1

0 –y3 dy

= 3– y4–––4 4

1

0

= 1– 1—4 2 – (0)

= – 1—4

= 1—4 unit2

Therefore, the area of the shaded region

is 1—4 unit2.

(d)

0

y

PQ

B(0, 2)

A(1, 1)

x = 3y2

y = –x + 2

x

6



Area of region Q = ∫ 1

0 x dy

= ∫ 1

0 3y2 dy

= 3y341

0

= (1) – (0) = 1 unit2

Area of region P = 1—2 × 1 × 1

= 1—2 unit2

Area of the shaded region = 1 + 1—2

= 1 1—2 unit2

(e)

A�–, –�

0

1 1

3 31–3

yy = x

x = 3y 2

x

Area under the curve x = 3y2 with y-axis

= ∫ 1—3

0

x dy

= ∫ 1—3

0

3y2 dy

= 3y341—3

0

= 31 1—3 2

3

4 – (0)3

= 1–––27 unit2

Area under the straight line y = x with y-axis

= 1—2 × 1—

3 × 1—3

= 1–––18 unit2

Area of the shaded region = 1–––18 – 1–––

27

= 1–––54 unit2

17. ∫ a

0 f (x) dx + ∫ b

a g(x) dx

18. ∫ b

a x dy = ∫ b

a f –1(y) dy

19. (a) ∫ b

0 f (x)––––

2 dx = 1—

2 ∫ b

0 f (x) dx

= 1—2 × 12

= 6 cm2

(b) ∫ b

a f (x) dx = ∫ b

0 f (x) dx – ∫ a

0 f (x) dx

= 12 – 5 = 7 cm2

(c) ∫ b

a 2f (x) dx = 2 ∫ b

a f (x) dx

= 2 × 7 = 14 cm2

(d) ∫ c

a f(x) dx = ∫ c

0 f(x) dx – ∫ a

0 f(x) dx

= 18 – 5 = 13 cm2

(e) ∫ a

0 f (x) dx + ∫ b

a f (x) dx = ∫ b

0 f (x) dx

= 12 cm2

20. ∫ 2

1 f (x) dx + ∫ 5

3 f (x) dx = ∫ 5

1 f (x) dx – ∫ 3

2 f (x) dx

= 10 – 4 = 6

21. ∫ 5

1 [kx + 2f (x)] dx = 10

∫ 5

1 kx dx + ∫ 5

1 2f (x) dx = 10

3 kx2–––2 4

5

1+ 2 ∫ 5

1 f (x) dx = 10

1 25–––2 k – 1—

2 k2 + 2 × 2 = 10

12k = 6 k = 6–––

12

= 1—2

22. (a) y = 2x2 + 1 ................................... 1 y = –x + 4 .................................... 2

1 = 2, 2x2 + 1 = –x + 4 2x2 + x – 3 = 0 (2x + 3)(x – 1) = 0 x = 1

Substitute x = 1 into y = –x + 4, y = 3

Therefore, A(1, 3).

7



(b) Area of the shaded region

= ∫ 1

0 (2x2 + 1) dx

= 3 2—3 x3 + x4

1

0

= 1 2—3 + 12 – (0)

= 1 2—3 unit2

23.

0

y

y = 2

y = x 2 + 1

BA

DCx

y = x2 + 1 ............................................ 1y = 2 .................................................... 2

1 = 2, x2 + 1 = 2 x2 = 1 x = ±1

A(–1, 2) and B(1, 2).

Area of region under the curve y = x2 + 1 from A to B

= ∫ 1

–1 (x2 + 1) dx

= 3 x3–––3 + x4

1

–1

= 1 1—3 + 12 – 1– 1—

3 – 12= 1 4—

3 2 – 1– 4—3 2

= 8—3 unit2

Area of square ABCD = 2 × 2 = 4 unit2

Area of the shaded region = 4 – 8—3

= 4—3 unit2

24. x = y2 – 4When x = 0, y2 = 4 y = ±2

Therefore, A(0, 2).

Area of the shaded region = ∫ 2

0 x dy

= ∫ 2

0 (y2 – 4) dy

= 3 y3–––3 – 4y4

2

0

= 3 23–––3 – 4(2)4 – (0)

= 8—3 – 8

= – 16–––3

= 16–––3 unit2

25.

0

y

x = –y 2 + 4

y + x = 2

CE D(3, –1)

B(0, –2)

A

x

x = –y2 + 4 .................................... 1 y + x = 2.............................................. 2

Substitute 1 into 2, y – y2 + 4 = 2 y2 – y – 2 = 0(y – 2)(y + 1) = 0 y = 2, –1

Substitute y = 2 into 1,x = – 4 + 4 = 0

Substitute y = –1 into 1,x = –1 + 4 = 3

Therefore, A(0, 2) and D(3, –1).

Areaof∆ADE = 1—2 × 3 × 3

= 9—2 unit2

Area of region BDE = ∫ –1

–2 x dy

= ∫ –1

–2 (–y2 + 4) dy

= 3– y3–––3 + 4y4

–1

–2

= 3– (–1)3––––

3 + 4(–1)4 – 3– (–2)3––––

3 + 4(–2)4= 1 1—

3 – 42 – 1 8—3 – 82

= – 11–––3 – 1– 16–––

3 2= – 11–––

3 + 16–––3

= 5—3 unit2

8



Area of the shaded region = 9—2 + 5—

3

= 6 1—6 unit2

26. (a) Volumegenerated=π ∫ 2

0 y2 dx

=π ∫ 2

0 x2 dx

=π3 x3–––3 4

2

0

=π1 23–––3 – 02

= 8—3 πunit

3

(b) Volumegenerated=π ∫ 2

0 y2 dx

=π ∫ 2

0 x4 dx

=π3 x5–––5 4

2

0

=π1 25–––5 – 02

= 32–––5 πunit3

(c) Volume generated

=π ∫ 2

1 y2 dx

=π ∫ 2

1 (x2 + 1)2 dx

=π ∫ 2

1 (x4 + 2x2 + 1) dx

=π3 x5–––5 + 2—

3 x3 + x42

1

=π3 25–––5 + 2—

3 (2)3 + 24–π1 1—5 + 2—

3 + 12 = 11 13–––

15 πunit3

(d) Volume generated

=π ∫ 2

0 y2 dx

=π ∫ 2

0 3– 1––––––

(x + 1) 42 dx

=π ∫ 2

0 (x + 1)–2 dx

=π3 (x + 1)–1––––––––

–1 42

0

=π3– 1––––––(x + 1) 4

2

0

=π1– 1—3 2–π1– 1—

1 2 = 2—

3 πunit3

27. (a)

O

y

A(1, 1)

B

y = x

y = x2

x

y = x2................................1 y = x .................................2

1 = 2, x2 = x x2 – x = 0 x(x – 1) = 0 x = 0, 1

Therefore, A(1, 1)

The volume generated by the curve y = x2

=π ∫ 1

0 y2 dx

=π ∫ 1

0 x4 dx

=π3 x5–––5 4

1

0

= 1—5 πunit

3

The volume generated by the straight line OA

= 1—3 π(1)

2 × 1

= 1—3 πunit

3

Volume generated for the shaded region = 1—

3 π–1—5 π

= 2–––15 πunit

3

(b)

O

y

A

B

Cy = –x + 6

y = x 2

x

y = x2...................1 y = –x + 6 ...........2

1 = 2, x2 = –x + 6 x2 + x – 6 = 0 (x + 3)(x – 2) = 0 x = –3, 2

Therefore, A(2, 4)

x-coordinate of B = 6

9



Volume generated by the shaded region = Volume generated by AOC + Volume generated

by ABC =π ∫ 2

0 y2 dx + 1—

3 π(4)2 × (6 – 2)

=π ∫ 2

0 x4 dx + 64–––

3 π

=π3 x5–––5 4

2

0+ 64–––

3 π

=π1 32–––5 2 + 64–––

3 π

= 27 11–––15 πunit3

28. (a) Volumegenerated =π ∫ 2

0 x2 dy

=π3 x3–––3 4

2

0

=π1 23–––3 – 02

= 8—3 πunit

3

(b) Volumegenerated =π ∫ 2

1 x2 dy

=π ∫ 2

1 (y – 1) dy

=π3 y2–––2 – y4

2

1

=π1 22–––2 – 22–π1 1—

2 – 12 =π(0)–π1– 1—

2 2 = 1—

2 πunit3

(c) Volumegenerated=π ∫ 4

0 x2 dy

=π ∫ 4

0 (4 – y) dy

=π34y – y2–––2 4

4

0

=π34(4) – 42–––2 4–π(0)

=8πunit3

29. (a)

O

BA

D

C

y

y = –x + 2

x = y 2

x

x = y2............................... 1 y = –x + 2 ....................... 2

Substitute 1 into 2, y = –y2 + 2 y2 + y – 2 = 0 (y + 2)(y – 1) = 0 y = –2, 1

Substitute y = 1 into 1, x = 1

Therefore, A(1, 1)

y = –x + 2 When x = 0, y = 2 B(0, 2) When y = 0, x = 2 C(2, 0)

Volumegeneratedby∆ADB

= 1—3 π(1)

2 × (2 – 1)

= 1—3 πunit

3

Volume generated by AOD =π ∫ 1

0 x2 dy

=π ∫ 1

0 y4 dy

=π3 y5–––5 4

1

0

=π1 1—5 2–π(0)

= 1—5 πunit

3

Volume generated by the shaded region

= 1—3 π+

1—5 π

= 8–––15 πunit3

(b)

O

A

B

C

y

y = 2x

y = x 2

x

y = x2...................1 y = 2x ..................2

1 = 2, x2 = 2x x2 – 2x = 0 x(x – 2) = 0 x = 0, 2

10



When x = 2, y = 2(2) = 4

Therefore, A(2, 4).

Volume generated by the curve y = x2

=π ∫ 4

0 x2 dy

=π ∫ 4

0 y dy

=π3 y2–––2 4

4

0

=π1 42–––2 2–π(0)

=8πunit3

Volumegeneratedby∆AOC

= 1—3 π(2)

2 × 4

= 16–––3 πunit3


=8π– 16–––3 π

= 8—3 πunit

3

1. ∫ 3

p 2[ f (x) – x] dx = 15

∫ 3

p [2f (x) – 2x] dx = 15

2 ∫ 3

p f (x) dx – ∫ 3

p (2x) dx = 15

2 × 10 – 3x24 3p = 15 20 – (9 – p2) = 15 20 – 9 + p2 = 15 p2 = 4 p = ±2 Since p , 0, \ p = –2

2. ∫ 2

0 3f(x) dx = 3 4x

x2 – 1 42

0

∫ 2

0 f(x) dx = 1

3 3 4x

x2 – 1 42

0

= 13

1 83

– 02 = 8

9

3. ∫ 4 g(x) dx = 4 ∫ g(x) dx

= 4 [2 f(x)] + c = 8 f(x) + c

4. (a) ∫ 1

6 2g(x) dx = 2 ∫ 1

6 g(x) dx

= 2(–10) = –20

(b) ∫ 1

6 [kx – g(x)] dx = 20

∫ 1

6 kx dx – ∫ 1

6 g(x) dx = 20

3 kx2–––2 4

1

6 – (–10) = 20

3 k(1)2–––––

2 – k(6)2–––––

2 4 = 10

1—2 k – 36–––

2 k = 10

– 35–––2 k = 10

k = 10 × 1– 2–––35 2

= – 4—7

5. ∫ b

a 3f (x) dx = 3 ∫ b

a f(x) dx

= 3 × 8 = 24

6. ∫ 3

k 1–––

x2 dx = 2—3

∫ 3

k x–2 dx = 2—

3

3 x –1–––––1 4

3

k = 2—

3

1– 1—3 2 – 1– 1—

k 2 = 2—3

– 1—3 + 1—

k = 2—

3

1—k

= 1

k = 1

11



7. dS–––dt

= 4t3 – 5

s = ∫ (4t3 – 5) dt

s = t4 – 5t + c 10 = 24 – 5(2) + c = 6 + c c = 4 Therefore s = t4 – 5t + 4

8. (a) dy

–––dx

= kx2 + 5x .................. 1

y – 2x + 3 = 0 y = 2x – 3

Gradient of the straight line is 2.

Substitute x = 2 and dy

–––dx

= 2 into 1,

dy

–––dx = kx2 + 5x

2 = k(2)2 + 5(2) 4k = –8 k = –2

(b) y = ∫ (kx2 + 5x) dx

= ∫ (–2x2 + 5x) dx

= – 2x3––––

3 + 5x2––––

2 + c ................ 1

Substitute x = 2 and y = 1 into 1,

1 = – 2—3 (2)3 + 5—

2 (2)2 + c

1 = – 16–––3 + 10 + c

c = 1 + 16–––3 – 10

= – 11–––3

Therefore, the equation of the curve is

y = – 2—3 x3 + 5—

2 x2 – 11–––3 .

9. (a) dy

–––dx = 2x3 – 8x

y = ∫ (2x3 – 8x) dx

= 2—4 x4 – 8x2

––––2 + c

= 1—2 x4 – 4x2 + c .................. 1

Substitute x = –1 and y = 1—2 into 1,

1—2 = 1—

2 (–1)4 – 4(–1)2 + c

= 1—2 – 4 + c

c = 4

Therefore, the equation of the curve is y = 1—

2 x4 – 4x2 + 4.

(b) For turning points, dy

–––dx

= 0

2x3 – 8x = 0 2x(x2 – 4) = 0 2x(x – 2)(x + 2) = 0 x = 0, 2, –2

When x = 0, y = 1—2 (0)4 – 4(0)2 + 4

= 4

When x = 2, y = 1—2 (2)4 – 4(2)2 + 4

= – 4

When x = –2, y = 1—2 (–2)4 – 4(–2)2 + 4

= – 4 Therefore, the turning points are (0, 4), (2, – 4)

and (–2, – 4).

d 2y

––––dx2 = 6x2 – 8

When x = 0, d 2y

––––dx2 = –8 , 0

When x = 2, d 2y

––––dx2 = 6(2)2 – 8

= 16 . 0

When x = –2, d 2y

––––dx2 = 6(–2)2 – 8

= 16 . 0

Hence, the maximum point is (0, 4).

10. (a) y = 4–––––––(3x – 1)2

= 4(3x – 1)–2

dy

–––dx = 4(–2)(3x – 1)–3(3)

= – 24––––––––(3x – 1)3

12



The gradient of the tangent at the point P(1, 1)

= – 24–––––––––[3(1) – 1]3

= – 24–––8

= –3

Equation of the tangent is y – 1 = –3(x – 1) = –3x + 3 y = –3x + 4

(b) (i) Area = ∫ 3

1 4––––––––

(3x – 1)2 dx

= ∫ 3

1 4(3x – 1)–2 dx

= 34(3x – 1)–1–––––––––

(–1)(3) 43

1

= 3– 4––––––––3(3x – 1) 4

3

1

= 3– 4–––––––3(9 – 1) 4 – 3– 4–––––––

3(3 – 1) 4 = – 4–––

24 + 4—6

= 1—2 unit2

(ii) Volume generated

=π ∫ 3

1 y2 dx

=π ∫ 3

1 3 4––––––––

(3x – 1)2 42 dx

=π ∫ 3

1 16––––––––

(3x – 1)4 dx

=π ∫ 3

1 16(3x – 1)–4 dx

=16π3 (3x – 1)–3––––––––

(–3)(3) 43

1

=16π3– 1–––––––––9(3x – 1)3 4

3

1

=16π3– 1–––––9(8)3 4–16π3– 1–––––

9(2)3 4 =16π3– 1–––––

9(8)3 + 1–––––9(2)3 4

= 7–––32 πunit

3

11.

O

P(3, –2)R(0, –1)Q

–1

23

x = y2 – 1

y

y = – –x

x

x = y2 – 1 ................................ 1

y = – 2x–––3 ................................ 2

Substitute 1 into 2,

y = – 2—3 (y2 – 1)

3y = –2y2 + 2 2y2 + 3y – 2 = 0(2y – 1)(y + 2) = 0 y = 1—

2 , –2

Substitute y = –2 into x = y2 – 1,x = (–2)2 – 1 = 3

Therefore, P(3, –2).

Volumegeneratedby∆OPQ = 1—3 π(3)

2(2)

=6πunit3

Volume generated by region PRQ

=π ∫ –1

–2 x2 dy

=π ∫ –1

–2 (y2 – 1)2 dy

=π ∫ –1

–2 (y4 – 2y2 + 1) dy

=π3 y5

–––5 – 2—

3 y3 + y4–1

–2

=π3(–1)5––––

5 – 2—3 (–1)3 + (–1)4–π3(–2)5

––––5 – 2—

3 (–2)3 + (–2)4=π1– 1—

5 + 2—3 – 12–π1– 32–––

5 + 16–––3 – 22

= 2 8–––15 πunit

3


=6π–2 8–––15 π

= 3 7–––15 πunit

3

12. (a)

O k

BQ

P

A

y = 2x

y = –x 2 + 3x

y

x

y = 2x .....................................1 y = –x2 + 3x ...........................2

13



Substitute 1 into 2, 2x = –x2 + 3x x2 – x = 0 x(x – 1) = 0 x = 0, 1

Therefore, k = 1.

(b) y = –x2 + 3x When y = 0, –x2 + 3x = 0 x(–x + 3) = 0 x = 0, 3

Therefore, A = (3, 0).

(c) Substitute x = 1 into y = 2x, y = 2

Therefore, B(1, 2).

Areaof∆OBk = 1––2 × 1 × 2

= 1 unit2

Area of the region bounded by curve OB, x-axis and x = 1

= ∫ 1

0 y dx

= ∫ 1

0 (–x2 + 3x) dx

= 3– x3–––3 + 3x2

–––2 4

1

0

= 1– 1—3 + 3—

2 2 – (0)

= 7—6 unit2

Area of region Q = 7—6 – 1

= 1—6 unit2

(d) Volume generated by region P

=π ∫ 3

1 y2 dx

=π ∫ 3

1 (–x2 + 3x)2 dx

=π ∫ 3

1 (x4 – 6x3 + 9x2) dx

=π3 x5–––5 – 3—

2 x4 + 3x343

1

=π3 35–––5 – 3—

2 (3)4 + 3(3)34–π3 1—5 – 3—

2 + 34 = 6 2—

5 πunit3

13. (a) y = x3 + 2

dy

–––dx

= 3x2

Gradient of the tangent at A(–1, 1) = 3(–1)2

= 3 Gradient of the normal at A is – 1—

3 .

Gradient of AC = 0 – 1–––––k + 1

0 – 1–––––k + 1 = – 1—

3 k + 1 = 3 k = 2(b)

ODC(2, 0)

B

A(–1, 1)

y = x 3 + 2

2

y

x

Areaof∆ADC = 1—2 × 1 × 3

= 3—2

= 1 1—2 unit2

Area of the region under curve AB with the x-axis

= ∫ 2

–1 y dx

= ∫ 2

–1 (x3 + 2) dx

= 3 x4–––4 + 2x4

2

–1

= 3 24–––4 + 2(2)4 – 3 (–1)4

–––––4 + 2(–1)4

= 8 – 1– 7—4 2

= 9 3—4 unit2

Area of the shaded region = 9 3—4 – 1 1—

2 = 8 1—

4 unit2

14



(c) Volume generated by the shaded region

=π ∫ 2

–1 (x3 + 2)2 dx – 1—

3 ×π× 12 × 3

=π ∫ 2

–1 (x6 + 4x3 + 4) dx–π

=π3 x7–––7 + 4x4

–––4 + 4x4

2

–1–π

=π3 27–––7 + 24 + 4(2)4–π1 –1–––

7 + 1 – 42–π = 44 3—

7 πunit3

1. (a) ∫ 1

0 x(x + 2)–––––––

(x + 1)2 dx = 3 x2–––––x + 1 4

1

0

= 1 12–––––1 + 1 2 – (0)

= 1—2

(b) ∫ 1

0 2x(x + 2)–––––––––

(x + 1)2 dx = 2 ∫ 1

0 x(x + 2)–––––––

(x + 1)2 dx

= 2 × 1—2

= 1

(c) ∫ 1

0 2x(x + 2)–––––––––

3(x + 1)2 dx = 2—3 ∫ 1

0 x(x + 2)–––––––

(x + 1)2 dx

= 2—3 × 1—

2 = 1—

3

2. ∫ g(x)––––3 dx = 1—

3 ∫ g(x) dx

= 1—3 [2h(x) + c]

3. ∫ k

1 (2x + 1) dx = –2

3 2x2––––

2 + x4k

1 = –2

(k2 + k) – (1 + 1) = –2 k2 + k = 0 k(k + 1) = 0 k = 0, –1

4. ∫ 1

0 k–––––––(2x – 1)2 dx = 2k – 3

∫ 1

0 k(2x – 1)–2 dx = 2k – 3

3 k(2x – 1)–1–––––––––

(–1)(2) 41

0 = 2k – 3

1 k––––2 2 – 1 k—

2 2 = 2k – 3

–k = 2k – 3 2k + k = 3 3k = 3 k = 1

5. ∫ 3

0 1 x3 + 3kx––––––––x 2 dx = 36

∫ 3

0 (x2 + 3k) dx = 36

3 x3–––3 + 3kx4

3

0 = 36

3 33–––3 + 3k(3)4 – (0) = 36

9 + 9k = 36 9k = 27 k = 3

6. (a) ∫ 5

1 3f (x) dx = 3 ∫ 5

1 f(x) dx

= 3 × 10 = 30

(b) ∫ 7

1 f (x) dx – ∫ 1

5 kf (x) dx = 62

22 + k ∫ 5

1 f (x) dx = 62

k × 10 = 40 k = 4

7. (a) ∫ 5

2 g(x) dx

= ∫ 3

2 g(x) dx + ∫ 5

3 g(x) dx

= 4 + 10 = 14

(b) ∫ 5

3 3 2g(x) + k

––––––––3 4 dx = 4—

3

∫ 5

3 1 2—

3 g(x) + k—3 2 dx = 4—

3

2—3 ∫ 5

3 g(x) dx + ∫ 5

3 k—

3 dx = 4—3

2—3 × 10 + 3 k—

3 x45

3 = 4—

3

20–––3 + 1 5—

3 k – k2 = 4—3

2—3 k = 4—

3 – 20–––3

2—3 k = – 16–––

3

k = 1– 16–––3 21 3—

2 2 = –8

15



8. Area of the shaded region

=Areaof∆AOB – ∫ a

0 f (x) dx – ∫ 3

a g(x) dx

= 1—2 × 3 × 8 – 4 – 2

= 6 unit2

9.

O

P(3, 5)

A(3, 0)

B(0, 5)

y = f (x)

y = g(x)

y

x

Area of the shaded region = Area of rectangle AOBP – ∫ 3

0 g(x) dx – ∫ 5

0 f –1(y) dy

= 3 × 5 – 4 – 5= 6 unit2

10. dy

–––dx

= 3x2 + 4x – 5

y = ∫ (3x2 + 4x – 5) dx

y = x3 + 2x2 – 5x + c ....................... 1

Substitute x = 1 and y = 4 into 1,4 = 1 + 2 – 5 + cc = 6

Therefore, y = x3 + 2x2 – 5x + 6.

11. dy–––dx

= 3(1 – 2x)5

y = f (x) = ∫ 3(1 – 2x)5 dx

= 3(1 – 2x)6

–––––––––6(–2)

+ c

= – 1—4 (1 – 2x)6 + c ................ 1

Substitute x = 1 and f (x) = 3 into 1,

3 = – 1—4 (1 – 2)6 + c

= – 1—4 + c

c = 3 1—4

Therefore, f (x) = – 1—4 (1 – 2x)6 + 13–––

4 .

12. dy–––dx

= 2––––––––(3 + 5x)2

y = ∫ 2––––––––(3 + 5x)2 dx

= ∫ 2(3 + 5x)–2 dx

= 3 2(3 + 5x)–1––––––––––

(–1)(5) 4 + c

= – 2––––––––5(3 + 5x)

+ c ........................ 1

Substitute x = 1—5 , y = 3 into 1,

3 = – 2–––––5(4) + c

c = 3 1–––10

Therefore, y = – 2––––––––5(3 + 5x) + 31–––

10 .

13. dy

–––dx

= 3x2 – 4x + k

Gradient of the tangent at P(2, 5) = 3(2)2 – 4(2) + k = 4 + ky – 2x + 1 = 0 y = 2x – 1

The gradient of the straight line is 2.

Therefore, 4 + k = 2 k = –2dy

–––dx

= 3x2 – 4x – 2

y = ∫ (3x2 – 4x – 2) dx

= x3 – 2x2 – 2x + c ....................... 1

Substitute x = 2 and y = 5 into 1,5 = 23 – 2(2)2 – 2(2) + c = 8 – 8 – 4 + cc = 9

Hence, the equation of the curve is y = x3 – 2x2 – 2x + 9.

14. (a) 2y – x + 1 = 0 2y = x – 1 y = 1—

2 x – 1—2

The gradient of the normal at (3, –1) is 1—2 .

Therefore, the gradient of the tangent at (3, –1) is –2.

dy

–––dx

= 4k––––––––(3x + 1)2

–2 = 4k––––––––(3x + 1)2

(–2)(10)2 = 4k k = –200–––––

4 = –50

16



(b) dy

–––dx

= –200––––––––(3x + 1)2

y = ∫ –200(3x + 1)–2 dx

= –200(3x + 1)–1––––––––

(–1)(3) + c

= 200––––––––3(3x + 1) + c.................... 1

Substitute x = 3 and y = –1 into 1,

–1 = 200–––––3(10) + c

c = –1 – 20–––3

= – 23–––3

Therefore, the equation of the curve is y = 200––––––––

3(3x + 1) – 23–––3 .

15. π ∫ k

0 y2 dx=625π

π ∫ k

0 x4 dx=625π

3 x5–––5 4

k

0 = 625

1 k5–––

5 2 – (0) = 625

k5 = 625 × 5 = 55

k = 5

16. π ∫ k

1 x2 dy = 1—

2 π

∫ k

1 (y – 1) dy = 1—

2

3 y2–––2 – y4

k

1 = 1—

2

1 k2–––2 – k2 – 1 1—

2 – 12 = 1—2

k2–––2 – k = 0

k1 k—2 – 12 = 0

k = 0, 2

Based on the diagram, k . 0,therefore k = 2.

17. π ∫ k

0 y2 dx = 7–––

24 π

∫ k

0 1–––––––

(x – 2)4 dx = 7–––24

3 (x – 2)–3–––––––

–3 4k

0 = 7–––

24

3– 1––––––––3(x – 2)3 4

k

0 = 7–––

24

– 1––––––––3(k – 2)3 – 3– 1––––––

3(–2)3 4 = 7–––24

– 1––––––––3(k – 2)3 = 7–––

24 + 1–––24

= 1—3

3(k – 2)3 = –3 (k – 2)3 = –1 = (–1)3

k – 2 = –1 k = 1

18. y = x––––––x2 + 1

dy –––dx =

(x2 + 1) d –––dx (x) – x d –––

dx (x2 + 1)–––––––––––––––––––––––––

(x2 + 1)2

= (x2 + 1) – x(2x)–––––––––––––(x2 + 1)2

= 1 – x2––––––––(x2 + 1)2

1

0∫ 1 – x2

––––––––(x2 + 1)2 dx = 3 x–––––

x2 + 1 41

0

= 1 1—2 2 – (0)

= 1—2

∫ 1

0 x2 – 1––––––––3(x2 + 1)2 dx = – 1—

3 ∫ 1

0 1 – x2––––––––(x2 + 1)2 dx

= 1– 1—3 2 × 1 1—

2 2 = – 1—

6

19. (a) dy

–––dx = 3x2 + 2 ............................... 1

y = 5x – 2 ................................ 2

Substitute x = p, y = q and dy

–––dx = 5 into 1

and 2,

From 1, 5 = 3p2 + 2 3p2 = 3 p2 = 1 p = ±1 Since p . 0, then p = 1. From 2, q = 5 × 1 – 2 = 3

17



(b) dy

–––dx = 3x2 + 2

y = ∫ (3x2 + 2) dx

y = x3 + 2x + c ........................ 3

Substitute x = 1 and y = 3 into3, 3 = 1 + 2 + c c = 0 Therefore, the equation of the curve is

y = x3 + 2x.

20. (a) dy

–––dx = 2––––––––

(1 – 3x)2 .......................... 1

y + 2x + 5 = 0 y = –2x – 5 ................... 2

The gradient of the normal at point A is –2. Therefore, the gradient of the tangent at point A

is 1—2 .

Substitute x = r, y = t, dy

–––dx = 1—

2 into 1 and 2,

From 1,

1—2 = 2––––––––

(1 – 3r)2

(1 – 3r)2 = 4 1 – 3r = ±2 3r = 1 2 = 3, –1 r = 1, – 1—

3 Since r . 0, then r = 1.

From 2, t = –2(1) – 5 = –7

(b) y = ∫ 2––––––––(1 – 3x)2

dx

= ∫ 2(1 – 3x)–2 dx

= 2(1 – 3x)–1–––––––––

(–1)(–3) + c

= 2––––––––3(1 – 3x)

+ c ........................ 1

Substitute x = 1 and y = –7 into 1, –7 =

2–––––––3(1 – 3)

+ c

= – 2—6 + c

c = –7 + 1—3

= – 6 2—3

Therefore, the equation of the curve is

y = 2––––––––

3(1 – 3x) – 20–––

3 .

21. (a) dy

–––dx = 2x – 5

y = ∫ (2x – 5) dx

= x2 – 5x + c ........................ 1

Substitute x = 1 and y = –5 into 1, –5 = 1 – 5 + c c = –1 Therefore, y = x2 – 5x – 1

For minimum point, dy

–––dx = 0 and

d2y –––dx2 . 0

2x – 5 = 0 x = 5—

2 \ k = 5—

2 Substitute x = 5—

2 , y = p into y = x2 – 5x – 1,

p = 1 5—2 2

2 – 51 5—

2 2 – 1

= 25–––4 – 25–––

2 – 1

= –7 1—4

(b) Gradient of the tangent at A = 2(1) – 5 = –3 Gradient of the normal at A = 1—

3 Equation of the normal at A is y – (–5) = 1—

3 (x – 1)

y + 5 = 1—3 x – 1—

3

y = 1—3 x – 16–––

3

22. (a) Substitute x = p, y = q into y = – (x – 1)2 + 16 and y = –x + 15, q = – (p – 1)2 + 16 ....................... 1 q = – p + 15 ................................. 2

1 = 2, –p + 15 = – (p – 1)2 + 16 = –(p2 – 2p + 1) + 16 = –p2 + 2p – 1 + 16 p2 – 3p = 0 p(p – 3) = 0 p = 0, 3

Substitute p = 3 into 2, q = –3 + 15 = 12 Therefore, p = 3 and q = 12.

(b) When y = 0, y = –x + 15 0 = –x + 15 x = 15 C(15, 0)

18



0

15

A(3, 12)

y = –x + 15

y = –(x – 1)2 + 16

y

xC(15, 0)

B(5, 0)

D(3, 0)

When y = 0, y = – (x – 1)2 + 16 0 = – (x – 1)2 + 16 (x – 1)2 = 16 x – 1 = ±4 x = ±4 + 1 = 5, –3 B = (5, 0)

The area of the shaded region =Areaof∆ADC – Area of ADB

= 1—2 × 12 × 12 – ∫ 5

3 [–(x – 1)2 + 16] dx

= 72 – 316x – (x – 1)3

––––––3 4

5

3

= 72 – 316(5) – (5 – 1)3

––––––3 4 – 316(3) –

(3 – 1)3

––––––3 4

= 72 – 380 – 64–––3 – 148 – 8—

3 24 = 72 – 40–––

3 = 58 2—

3 unit2

23. (a) y = (2x – 3)2 ............. 1 y + x = 9 y = 9 – x ................... 2

1 = 2, 9 – x = (2x – 3)2

= 4x2 – 12x + 9 4x2 – 11x = 0 x(4x – 11) = 0 x = 0, 11–––

4

Substitute x = 11–––4 into 2,

y = 9 – 11–––4

= 25–––4

Therefore, B1 11–––4 , 25–––

4 2 For y + x = 9, when y = 0, 0 + x = 9 x = 9 Therefore, C(9, 0).

For y = (2x – 3)2, when y = 0, 0 = (2x – 3)2

x = 3—2

Therefore, D( 3—2 , 0)

(b)

OE

A(0, 9)

2

3

4

11

4

25

y = (2x – 3)2

y + x = 9

y

xC(9, 0)

D �–, 0�

B�–, –�

Area bounded by the curve and the straight line

= Trapezium AOEB – ∫ 11––4

0 (2x – 3)2 dx

= 1—2 × 11–––

4 × 19 + 25–––4 2 – 3(2x – 3)3

–––––––3(2) 4

11––4

0

= 3 11–––8 × 61–––

4 4 – 31 1 11–––2 – 32

3

–––––––––6 2 – 1 (–3)3

–––––6 24

= 671––––32 – 17 5–––

48 2 = 13 83–––

96 unit2

(c) Volume generated by the shaded region

=π ∫ 3—2

0 y2 dx

=π ∫ 3—2

0 (2x – 3)4 dx

=π3 (2x – 3)5––––––––

2(5) 43—2

0

=π31(3 – 3)5––––––

10 2 – 1 (–3)5–––––

10 24

=π30 + 243––––10 4

= 243––––10 πunit3

24. (a)

O

A(0, 4)

x = 3y 2

y

D

Ex

C(4, 0)

B(3, 1)

19



Volume generated by the shaded region = Volume generated by rectangle DOEB

– Volume generated by the curve =π(1)2 × 3–π ∫ 3

0 x—

3 dx

=3π–π3 x2—6 4

3

0

=3π–π1 9—6 – 02

=3π– 3—2 π

= 3—2 πunit

3

(b) Gradient of BC = 1 – 0–––––3 – 4

= 1––––1

= –1

y – 1–––––0 – 3 = –1

y – 1 = 3 y = 4

Therefore, A(0, 4).

(c) Area=Areaof∆ABD + Area of BOD = 1—

2 × 3 × 3 + ∫ 1

0 x dy

= 9—2 + ∫ 1

0 3y2 dy

= 9—2 + 3y34

1

0

= 9—2 + (1 – 0)

= 5 1—2 unit2

25. (a)

OQ

P

CD

A

y = k

y = x2 – 3

y

xB

π ∫ k

0 x2 dy = 7—

2 π

∫ k

0 (y + 3) dy = 7—

2

3 y2–––2 + 3y4

k

0 = 7—

2

k2–––2 + 3k = 7—

2

k2–––2 + 3k – 7—

2 = 0

k2 + 6k – 7 = 0 (k + 7)(k – 1) = 0 k = –7, 1 Since k . 0, then k = 1.

(b) When x = 0, y = x2 – 3 y = 0 – 3 = –3 Therefore, A(0, –3) When y = 0, y = x2 – 3 0 = x2 – 3 x = AB3 Therefore, B(AB3 , 0)

The area of shaded region Q

= ∫ AB3

0

y dx

= ∫ AB3

0

(x2 – 3) dx

= 3 x3–––3

– 3x4AB3

0

= 3 (AB3 )3

–––––3

– 3AB3 4 – 0

= –2AB3 = 2AB3 unit2

Hence, the area of shaded region Q is 2AB3 unit2.

26. (a) y = (x – 3)2 + 2 Therefore, A = (3, 2) y = x + 1 ................................1 y = (x – 3)2 + 2 ......................2

1 = 2, x + 1 = (x – 3)2 + 2 = x2 – 6x + 9 + 2 x2 – 7x + 10 = 0 (x – 2)(x – 5) = 0 x = 2, 5

Substitute x = 2 and x = 5 into1 respectively, When x = 2, When x = 5, y = 2 + 1 y = 5 + 1 = 3 = 6 Therefore, B(2, 3) and C(5, 6)

20



(b) The volume generated by the shaded region = The volume generated by trapezium BCDE

– The volume generated by the region under curve BAC with the x-axis

=π ∫ 5

2 (x + 1)2 dx–π ∫ 5

2 [(x – 3)2 + 2]2 dx

=π3 (x + 1)3–––––––

3 4 5

2–π ∫ 5

2 [(x – 3)4 + 4(x – 3)2 + 4] dx

=π1 63–––3

– 33–––3 2–π ∫ 5

2 [(x – 3)4 + 4(x – 3)2 + 4] dx

=63π–π3 (x – 3)5–––––––

5 + 4—

3 (x – 3)3 + 4x45

2

=63π–π31 25–––5

+ 4—3 (2)3 + 202 – 1 –1–––

5 – 4—

3 + 824 =63π–π130 3—

5 2 = 32 2—

5 πunit3

27. (a) y = x(x – 4) When y = 0, x(x – 4) = 0 x = 0, 4 Therefore, A(4, 0) y = x2 – 4x

dy–––dx

= 2x – 4

The gradient of the tangent AB = 2(4) – 4 = 4 The equation of the tangent AB is y – 0 = 4(x – 4) y = 4x – 16

(b) Substitute x = 0 into y = 4x – 16, y = –16 Therefore, B(0, –16)

Areaof∆AOB = 1—2 × 4 × 16

= 32 unit2

Area of the region between curve OA and the x-axis

= ∫ 4

0 (x2 – 4x) dx

= 3 x3–––3 – 2x24

4

0

= 3 43–––3

– 2(4)24 – (0)

= –10 2—3

= 10 2—3 unit2

Area of the shaded region = 32 – 10 2—3

= 21 1—3 unit2

28. (a) When x = 0, y = x3 + 8 y = 0 + 8 y = 8 Therefore, B = (0, 8)

When y = 0, 0 = x3 + 8 x3 = –8 x = –2 Therefore, A(–2, 0)

Area of region P = 0

–2∫ (x3 + 8) dx

= 3 x 4–––4

+ 8x4 0

–2

= (0) – 3 (–2)4–––––4 + 8(–2)4

= –(4 – 16) = 12 unit2

Area of region Q = 5—3 × Area of region P

= 5—3 × 12

= 20 unit2

(b) Volume generated by P = π—2 ∫ 0

–2 y2 dx

= π—2 ∫ 0

–2 (x3 + 8)2 dx

= π—2 ∫ 0

–2 (x 6 + 16x3 + 64) dx

= π—2 3 x 7–––7

+ 4x4 + 64x40

–2

= π—2 (0) – π—2 3 (–2)7–––––7 + 4(–2)4 + 64(–2)4

= π—2 1 576––––7 2 = 41 1—

7 πunit3

29. (a) Volume generated by region P = π—2 ∫ 4

0 x2 dy

= π—2 ∫ 4

0 y dy

= π—2 3 y2–––2 4

4

0

= π—2 1 42–––2

– 02 =4πunit3

21



(b) π ∫ 4

0 x2 dy =π ∫ 4

0 y dy

=π3 y2–––2 4

4

0

=π1 42–––2

– 02 =8πunit3

30. (a)

0

A

x = 1

y = R

y = –kx 2 – 3

y

x

B

P

E C

D

Q

Area of region P = 11–––3

∫ 1

0 (–kx2 – 3) dx = – 11–––

3

3– kx3–––3 – 3x4

1

0 = – 11–––

3

1– k—3 – 32 = – 11–––

3 – k—

3 = – 11–––3 + 3

= – 2—3

k = 2

Substitute x = 1 and y = R into y = –2x2 – 3, R = –2(1)2 – 3 R = –5 Therefore, k = 2 and R = –5.

(b) D = (0, –3) and B(1, –5) Volume generated by region Q about the y-axis

=π ∫ –3

–5 x2 dy

=π ∫ –3

–5 1– 3—

2 – 1—2 y2 dy

=π3– 3—2 y – 1—

4 y24–3

–5

=π1– 3—2 (–3) – 1—

4 (–3)22 –π1– 3—

2 (–5) – 1—4 (–5)22

=(2.25π)–(1.25π) =πunit3

(c) Let C = (x, 0) Areaof∆BEC = 26–––

3 – 11–––3

= 15–––3

= 5 unit2

1—2 × (x – 1) × 5 = 5

x – 1 = 5 × 2—5

= 2 x = 3

Therefore, the x-coordinate of point C is 3.

1. (a)

x

yy = f(x)

D

O C

A(0, 1)

B(3, 5)

∫ 3

0 f (x) dx = Area of trapezium OABC

= 1—2 × 3 × (1 + 5)

= 9 unit2

(b) ∫ 3

0 f (x) dx + ∫ 5

1 x dy

= Area of rectangle CODB = 3 × 5 = 15 unit2

2. Area of the shaded region

= ∫ b

a f (x) dx – ∫ b

a g(x) dx

= 10 – 6= 4 unit2

3. y – 3xy = x2

y(1 – 3x) = x2

y = x2––––––1 – 3x

dy

–––dx =

(1 – 3x) d–––dx

(x2) – x2 d–––dx

(1 – 3x)––––––––––––––––––––––––––––

(1 – 3x)2

= (1 – 3x)(2x) – x2(–3)––––––––––––––––––(1 – 3x)2

= 2x – 6x2 + 3x2––––––––––––

(1 – 3x)2

= 2x – 3x2––––––––(1 – 3x)2

22



∫ 1

0 2x – 3x2––––––––(1 – 3x)2 dx = 3 x2

––––––1 – 3x 4

1

0

∫ 1

0 x(2 – 3x)–––––––––

(1 – 3x)2 dx = 1 1–––––1 – 3

– 02 1—2 ∫ 1

0 2x(2 – 3x)–––––––––

(1 – 3x)2 dx = – 1—2

\ ∫ 1

0 2x(2 – 3x)–––––––––

(1 – 3x)2 dx = –1

4. d 2y

––––dx2 = 30x + 8

dy

–––dx = ∫(30x + 8) dx

= 15x2 + 8x + c

Substitute x = –1, dy

–––dx = 7 into the equation,

7 = 15(–1)2 + 8(–1) + c = 15 – 8 + c c = 0

dy

–––dx = 15x2 + 8x

y = ∫ (15x2 + 8x) dx

y = 5x3 + 4x2 + d

Substitute x = 1, y = 3 into the equation, 3 = 5 + 4 + d d = – 6

\ y = 5x3 + 4x2 – 6

5. (a) (3 – x)(4x – 9) = –1 12x – 27 – 4x2 + 9x = –1 – 4x2 + 21x – 27 = –1 4x2 – 21x + 26 = 0 (x – 2)(4x – 13) = 0 x = 2, x = 13–––

4 Since x = 2 is x-coordinate of point B, then

x-coordinate of point A is 13–––4 .

(b) dy

–––dx = 3 – x

y = ∫ (3 – x) dx

= 3x – x2–––2 + c

Substitute x = 2, y = 5 into the equation,

5 = 3(2) – 22–––2

+ c

5 = 6 – 2 + c c = 1

\ y = 3x – x2–––2

+ 1

dy

–––dx = 4x – 9

y = ∫ (4x – 9) dx

= 2x2 – 9x + d

Substitute x = 2, y = 5 into the equation, 5 = 2(2)2 – 9(2) + d d = 5 – 8 + 18 = 15

\ y = 2x2 – 9x + 15

Therefore, the equation of the curves are y = 3x – x2

–––2

+ 1 and y = 2x2 – 9x + 15.

6. (a) dy

–––dx = kx + 3

The gradient of the tangent at point (2, 2) = 2k + 3. The same tangent has gradient = 1 – 2––––––

3 – 2 = –1 \ 2k + 3 = –1 k = –2

(b) dy

–––dx = –2x + 3

y = ∫ (–2x + 3) dx

= –x2 + 3x + c

Substitute x = 2, y = 2 into the equation, 2 = –(2)2 + 3(2) + c c = 0 \ y = –x2 + 3x

When x = –1, y = –(–1)2 + 3(–1) = –1 – 3 = – 4

7. (a) For stationary point (1, 1), dy

–––dx = 0

\ dy

–––dx = –10x + k

0 = –10(1) + k k = 10

23



(b) dy

–––dx = –10x + 10

y = ∫(–10x + 10) dx

= –5x2 + 10x + c

Substitute x = 1, y = 1 into the equation, 1 = –5(1)2 + 10(1) + c c = – 4

Therefore, the equation of the curve is y = –5x2 + 10x – 4.

(c) At (1, 1),

dy

–––dx = –10(1) + 10

= 0 The tangent at (1, 1) is parallel to the x-axis. Therefore, the equation of the normal is x = 1.

8. (a) dy

–––dx = px + q

(2, –3) is a stationary point.

\ dy

–––dx = 0

px + q = 0 p(2) + q = 0 2p + q = 0 ...........................1

dy

–––dx = px + q

y = ∫ (px + q) dx

= p—2 x2 + qx + c

The curve passes through (0, 5) \5 = c \ y =

p—2 x2 + qx + 5

Substitute x = 2, y = –3 into the equation,

y = p—2 x2 + qx + 5

–3 = p—2 (2)2 + q(2) + 5

–3 = 2p + 2q + 5 2p + 2q = –8 ..........................2

2 – 1, q = –8

Substitute q = –8 into 1, 2p – 8 = 0 p = 4

Therefore, p = 4, q = –8.

(b) The equation of the curve is

y = 1 4—2 2x2 – 8x + 5

y = 2x2 – 8x + 5

9. (a) dh–––dt = –0.8t

h = ∫ (–0.8t) dt

= – 0.8–––2

t2 + c

= –0.4t2 + c

Given h = 10 when t = 0, \c = 10 \ h = –0.4t2 + 10

When t = 2, h = – 0.4(2)2 + 10 = 8.4 cm

(b) Whenwaterhasflownoutall,h = 0 \ h = – 0.4t2 + 10 0 = – 0.4t2 + 10

t2 = 10––––0.4

t = 5 s

10. (a) ds–––dt

= 4t

s = ∫ (4t) dt

= 2t2 + c

When t = 0, s = 0, \ c = 0 \ s = 2t2

After 3 seconds, t = 3 s = 2(3)2

= 18 cm

Therefore, the distance travelled in 3 seconds is 18 cm.

(b) For s = 32, 32 = 2t2

t2 = 16 t = 4 s

24



11. (a) Equation of straight line AB y = 3—

2 x ...................................1

Equation of curve y = (x + 1)(3 – x)...................2

1 = 2, 3—2 x = (x + 1)(3 – x)

3—2 x = 3x – x2 + 3 – x

x2 + 3—2 x – 2x – 3 = 0

x2 – 1—2 x – 3 = 0

× 2, 2x2 – x – 6 = 0 (x – 2)(2x + 3) = 0 2x + 3 = 0 x = 2 is for point B

x = – 3—2

x-coordinate of A is – 3—2 .

Substitute x = – 3—2 into 1,

y = 3—2 1– 3—

2 2 = – 9—

4

\ A(– 3—2 , – 9—

4 )

(b) Area of the shaded region

= ∫ 2

3– — 2

3(x + 1)(3 – x) – 3—2 x4 dx

= ∫ 2

3– — 2

13x – x2 + 3 – x – 3—2 x2 dx

= ∫ 2

3– — 2

1–x2 + 1—2 x + 32 dx

= 3– x3–––3 + 1—

4 x2 + 3x4 2

3– — 2

= 3– 23–––3 + 1—

4 (2)2 + 3(2)4

– 3– 1– 3—

2 23

––––––3

+ 1—4 1– 3—

2 22 + 31– 3—

2 24 = 4 1—

3 – 1–2 13–––16 2

= 7 7–––48 unit2

12. (a) f ′(x) = x2 + x – 2 For stationary points, f ′(x) = 0 x2 + x – 2 = 0 (x + 2)(x – 1) = 0 x = –2, 1 Therefore, the coordinates of A are (–2, 0).

(b) f (x) = ∫ f ′(x) dx

= ∫ (x2 + x – 2) dx

= 1—3 x3 + 1—

2 x2 – 2x + c

Substitute x = –2, f (x) = 0 into the equation,

0 = 1—3 (–2)3 + 1—

2 (–2)2 – 2(–2) + c

c = – 1—3 (–2)3 – 1—

2 (–2)2 + 2(–2)

= 8—3 – 2 – 4

= – 10–––3

\ f (x) = 1—3 x3 + 1—

2 x2 – 2x – 10–––3

(c) Area of the shaded region

= ∫ 1

–2 1 1—

3 x3 + 1—2 x2 – 2x – 10–––

3 2 dx

= 3 1–––12 x4 + 1—

6 x3 – x2 – 10–––3 x4

1

–2

= 1 1–––12 + 1—

6 – 1 – 10–––3 2

– 1 1–––12 (–2)4 + 1—

6 (–2)3 – (–2)2 – 10–––3 (–2)2

= 1– 4 1–––12 2 – 12 2—

3 2 = – 6 3—

4 = 6 3—

4 unit2

13.

x

x 2

4

y

y = –

0x = 3x = 1

Area of the region = ∫ 3

1 4–––

x2 dx

= ∫ 3

1 4x–2 dx

= 3 4x–1–––––1 4

3

1

= 3– 4—x 43

1

= 1– 4—3 2 – 1– 4—

1 2 = – 4—

3 + 4

= 2 2—3 unit2

25



14. x –2 –1 0 1y 7 0 1 2

0–1 1

2

1

7

–2

y = |x3 + 1|

y

x

∫ –1

–2 (x3 + 1) dx

= 3 x4–––4 + x4

–1

–2

= 3 (–1)4––––

4 + (–1)4 – 3 (–2)4––––

4 + (–2)4= 1 1—

4 – 12 – (2)

= – 3—4 – 2

= – 11–––4

Therefore, the area between the curve, the x-axis, x = –2 and x = –1 is 11–––

4 unit2.

Area between the curve, the x-axis, x = –1 and x = 1= ∫ 1

–1 (x3 + 1) dx

= 3 x4–––4 + x4

1

–1

= 1 14–––4 + 12 – 1 1—

4 – 12= 1 5—

4 2 – 1– 3—4 2

= 5—4 + 3—

4= 2 unit2

The total area = 11–––4 + 2

= 4 3—4 unit2

15.

O C

y = x 2

y

x

B(2, 4)

A(k, 0)

Area of BOC = ∫ 2

0 x2 dx

= 3 x3–––3 4

2

0

= 1 23–––3 2 – (0)

= 8—3 unit2

Area of BAC = 1—2 × 4 × (k – 2)

= 2(k – 2)

Given the total area = 10 2—3

8—3 + 2k – 4 = 32–––

3

2k = 32–––3 + 4 – 8—

3 2k = 12 k = 6

16. (a) Gradient of AB = 4 – 2–––––0 – 1

= –2 Equation of AB is y = –2x + 4

(b) Given Area of ABD–––––––––––––Area of BCOD

= 5—4

Area of AOCB = 1—2 × 1 × (2 + 4)

= 3 unit2

∫ 1

0 f (x) dx = Area of BCOD

= 4—9 × 3

= 4—3 unit2

17.

0

y = 2x + 4

x + 1 = y 2

y

B

A Cx

For A, substitute y = 0 into y = 2x + 4,0 = 2x + 4x = –2\ A(–2, 0)

26



For B, substitute x = 0 into y = 2x + 4,y = 0 + 4 = 4\ B(0, 4)

For C, substitute y = 0 into x + 1 = y2,x = –1\ C(–1, 0)


= 1—3 p42(2) – p ∫ 0

–1 (x + 1) dx

= 32–––3 p – p3 x2

–––2 + x4

0

–1

= 32–––3 p – p30 – 1 1—

2 – 124= 32–––

3 p – p1 1—2 2

= 32–––3 p – 1—

2 p

= 10 1—6 p unit3

18.

0

y = – + 1

y

B(0, t)C

A(0, 1)

3

x

x2

When x = 0, y = 0 + 1 = 1\ A(0, 1)

Volume generated

= p ∫ t

1 x2 dy

= p ∫ t

1 (3y – 3) dy

= p3 3—2 y2 – 3y4

t

1

= p31 3—2 t2 – 3t2 – 1 3—

2 – 324

= p1 3—2 t2 – 3t + 3—

2 2

Given the volume = 6p

p1 3—2 t2 – 3t + 3—

2 2 = 6p

3—2 t2 – 3t + 3—

2 = 6

21 3—2 t2 – 3t + 3—

2 2 = 2 × 6

3t2 – 6t + 3 – 12 = 0 3t2 – 6t – 9 = 0 t2 – 2t – 3 = 0 (t – 3)(t + 1) = 0 \ t = 3

19. x – 4 0y 0 16

0–4

16

y = (x + 4)2

y

x

Volume generated= p ∫ 1

–4 y2 dx

= p ∫ 1

–4 (x + 4)4 dx

= p3 (x + 4)5–––––––

5 41

–4

= p1 55–––5 – 02

= 54p= 625p unit3

27



20. When y = 2, 2 = (x – 2)2 + 1 x – 2 = ±1 x = 3, 1

0

(1, 2) (3, 2)

y = (x – 2)2 + 1

y

y = 2

x

Volume generated

= p ∫ 3

1 {2 – [(x – 2)2 + 1]}2 dx

= p ∫ 3

1 [1 – (x – 2)2]2 dx

= p ∫ 3

1 [1 – 2(x – 2)2 + (x – 2)4] dx

= p3x – 2—3 (x – 2)3 + 1—

5 (x – 2)543

1

= p313 – 2—3 + 1—

5 2 – 11 + 2—3 – 1—

5 24= p13 – 2—

3 + 1—5 – 1 – 2—

3 + 1—5 2

= 1 1–––15 p unit3

14[anal add math cd]

Documents

substitute x

x 15x2

f x c hx dx

x cd 35x2 dx

3x2 dx

dydx dx

3x3 dx

3x4 dx