SM025 Self-Directed Learning

NSO/KKW/CCW

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Self-Directed Learning 2019/2020

NSO/KKW/CCW

Table of Contents

Chapter 1: Integration............................................................................................................................ 1
1.1 Integration of functions ................................................................................................................ 1
1.2 Integration of trigonometric functions .......................................................................................11
1.3 Technique of Integration ............................................................................................................14
1.4 Definite Integrals.........................................................................................................................21

Chapter 2: First Order Differential Equations .....................................................................................31
2.1 Separable Variable ......................................................................................................................31
2.2 First Order Linear Differential Equations ....................................................................................34
2.3 Applications of First Order Linear Differential Equations ...........................................................37

Chapter 3: Numerical Method .............................................................................................................42
3.1 Newton-Raphson Method ..........................................................................................................42
3.2 Trapezoidal Rule..........................................................................................................................43

Chapter 4: Conics..................................................................................................................................49
4.1 Circle ...........................................................................................................................................49
4.2 Ellipses.........................................................................................................................................51
4.3 Parabolas.....................................................................................................................................55

Chapter 5: Vectors................................................................................................................................59
5.1 Vectors in two and three dimensions .........................................................................................59
5.2 Scalar Product (Dot Product) ......................................................................................................61
5.3 Vector Product (Cross Product) ..................................................................................................64
5.4 Application of Vectors in Geometry............................................................................................67

Chapter 6: Data Description.................................................................................................................71
A. Stem-and-Leaf diagrams...............................................................................................................71
B. Ungrouped Data............................................................................................................................71
C. Grouped Data................................................................................................................................75
D. Pearson’s Coefficient Of Skewness...............................................................................................78

Chapter 7: Permutations and Combinations.......................................................................................81
A. Permutations ................................................................................................................................81
B. Combinations................................................................................................................................88

NSO/KKW/CCW

Chapter 8: Probability ..........................................................................................................................89
Chapter 9: Random Variable................................................................................................................97

9.1 Discrete Random Variable ..........................................................................................................97
9.2 Continuous Random Variable ...................................................................................................107
Chapter 10: Special Probability Distributions ...................................................................................123
10.1 Binomial Distribution ~ ( , )...........................................................................................123
10.2 Poisson Distribution ~ 0( ) ...............................................................................................125
10.3 Normal Distribution ~ ( , 2) ..........................................................................................129
10.4 Normal Approximation to the Binomial Distribution .............................................................133

NSO/KKW/CCW

Chapter 1: Integration

1.1 Integration of functions

a) ∫ = +

b) ∫ = +1 + , ≠ −1
+1

c) ∫( + ) = ( + ) +1 + , ≠ −1
( +1)

d) ∫ + = 1 + +
ln

e) ∫ 1 = 1 ln| + | +
+


f) ∫ ′( ) = ln| ( )| +
( )

g) ∫ ′( ) ( ) = ( ) +

h) ∫ ′( )[ ( )] = [ ( )] +1 +
+1

a) ∫ = +

Find the following integrals:

i) ∫ 5 ii) ∫ 5

iii) ∫ ln 2 iv) ∫ ln 2

v) ∫ 2 vi) ∫ 2

vii) ∫ viii) ∫
2 2

ix) ∫ x) ∫

Solution:

i) ∫ 5 = 5 + ii) ∫ 5 = 5 +

iii) ∫ ln 2 = (ln 2) + iv) ∫ ln 2 = (ln 2) +

v) ∫ 2 = 2 + vi) ∫ 2 = 2 +

vii) ∫ = + viii) ∫ = +
2 2
2 2

ix) ∫ = + x) ∫ = +

1

NSO/KKW/CCW

b) ∫ = +1 + , ≠ −1
+1

Find the following integrals: ii) ∫ 5
b)
iv) ∫ √x3 ln 2
i) ∫ 5

iii) ∫ x2 ln 2

v) ∫ 2√ vi) ∫ 2 3√

vii) ∫ viii) ∫ 3
2 2 √

ix) ∫( − 1)( + 2) x) ∫ 3+2


Solution:

i) ∫ 5 ii) ∫ 5

= 5 ∫ = 5 ∫

2 2
= 5 [ 2 ] + = 5 [ 2 ] +

5 2 5 2
= 2 + = 2 +

iii) ∫ x2 ln 2 iv) ∫ √x3 ln 2

= ln 2 ∫ 2 3

= ln 2 ∫ 2

3 5
= ln 2 [ 3 ] +
2
= ln 2 3 + = ln 2 [ 5 ] +
3
2

2 ln 2 5
= 5 2 +

v) ∫ 2√ vi) ∫ 2 3√

1 1

= 2 ∫ 2 = 2 ∫ 3

3 4
2 3
= 2 [ 3 ] + = 2 [ 4 ] +

2 3

= 2 3 + = 3 2 4 +
3 4 3
2 2

2

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vii) ∫ viii) ∫ 3
2 2 √

1 1
= 2 ∫ 2 = 3 ∫

√

= ∫ −2 = 3 ∫ −21
2

−1 1
= 2 [ −1 ] +
2
= 3 [ 1 ] +
= − 2 +
2

= 6√ +

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

= ∫( 2 + − 2) 3 2
3 2 = ∫ ( + )

= 3 + 2 − 2 + = ∫( 2 + 2)

3
= 3 + 2 +

c) ∫( + ) = ( + ) +1 + , ≠ −1
( +1)

Find the following integrals: ii) ∫ 3(2 − 1)4
i) ∫( + 3)5

iii) ∫ 9 ( − 1)5 iv) ∫ 3√ − 5

3 vi) ∫ 2
√4 +1
v) ∫ 2(√1 − )

vii) ∫ viii) ∫ 3
(3− )2 √3−2

3

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Solution: ii) ∫ 3(2 − 1)4

i) ∫( + 3)5 = 3 ∫(2 − 1)4
( + 3)6
(2 − 1)5
= (1)(6) + = 3 [ (2)(5) ] +
( + 3)6

= 6 +

= 3 ( − 1)5 +
10

iii) ∫ 9( + 1)5 iv) ∫ 3√ − 5

= 9 ∫( + 1)5 1

( + 1)6 = 3 ∫( − 5)2
= 9 [ (1)(6) ] +
3
= 3 ( + 1)6 +
2 ( − 5)2
= 3 [ (1) (32) ] +

3

= 2( − 5)2 +

3 vi) ∫ 2
√4 +1
v) ∫ 2(√1 − )
= 2∫ 1
3
√4 + 1
= 2 ∫(1 − )2

5 = 2 ∫(4 + 1)−21
(1 − )2
= 2 [ (52)] +
(−1)
1
45
= − 5 (1 − )2 + (4 + 1)2
= 2 [ (4) (12) ] +
= √4 + 1

vii) ∫ viii) ∫ 3
(3− )2 √3−2

1 = 3∫ 1
= ∫ (3 − )2
√3 − 2

= ∫(3 − )−2 = 3 ∫(3 − 2 )−12

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

= −3√3 − 2 +

4

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d) ∫ + = 1 + +
ln

Find the following integrals: ii) ∫ −4
i) ∫ 2

iii) ∫ 52 −3 iv) ∫ 1−

v) ∫ 1 vi) ∫ 2
2
−3

vii) ∫( + 1)( − 3) viii) ∫ 2 −3


Solution: ii) ∫ −4
−4
i) ∫ 2
2 = (1)(ln ) +
−4
= (1)(ln 2) +
2 = ln +

= ln 2 + iv) ∫ 1−
1−
iii) ∫ 52 −3
52 −3 = (−1)(ln ) +
= − 1− +
= (2)(ln 5) +
52 −3

= 2 ln 5 +

v) ∫ 1 vi) ∫ 2
2
−3

1 1
= ∫ 2 = 2 ∫ −3

= ∫ −2 = 2 ∫ − +3

−2 − −3
= (−2)(ln ) + = 2 [(−1)(ln )] +

1 2
= − 2 2 + = − +3 +

5

NSO/KKW/CCW

vii) ∫( + 1)( − 3) viii) ∫ 2 −3
= ∫( 2 − 3 + − 3)

2 3
= ∫ ( − )

= ∫( 2 − 2 − 3) = ∫( − 3 − )

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

e) ∫ 1 = 1 ln| + | +
+


Find the following integrals:

i) ∫ 1 ii) ∫ 1
2 −3

iii) ∫ 2 iv) ∫ 5
4 +3 1−3

v) ∫ 1 vi) ∫ 3+
3− 9−

vii) ∫ (1 + 1 ) ( − 3) viii) ∫ 2−3


Solution:

i) ∫ 1 ii) ∫ 1
2 −3

= ln| | + = 1 ln|2 − 3| +
2

iii) ∫ 2 iv) ∫ 5
4 +3 1−3

1 1
= 2 ∫ 4 + 3 = 5 ∫ 1 − 3

1 = 5 1 ln|1 − 3 |] +
= 2 [4 ln|4 + 3|] + [−3

= 1 ln|4 + 3| + 5
2 = − 3 ln|1 − 3 | +

6

NSO/KKW/CCW

v) ∫ 1 vi) ∫ 3+
3− 9− 2

= 1 ln|3 − | + 3 +
−1 = ∫ (3 + )(3 − )

= − ln|3 − | + 1
= ∫ 3 −

= 1 ln|3 − | +
−1

= − ln|3 − | +

vii) ∫ (1 + 1) ( − 3) viii) ∫ 2−3



3 2 3
= ∫ ( − 3 + 1 − ) = ∫ ( − )

3 3
= ∫ ( − 2 − ) = ∫ ( − )

2 2
= 2 − 2 − 3 ln| | + = 2 − 3 ln| | +

f) ∫ ′( ) = ln| ( )| +
( )

Find the following integrals:

i) ∫ 2 ii) ∫ 1−2
1+2 − 2

iii) ∫ iv) ∫ 3
2+3 2−1

v) ∫ vi) ∫ 1
+1 ln

Solution:

i) ∫ 2 ii) ∫ 1−2
1+2 − 2

( ) = 1 + 2 , ′( ) = 2 ( ) = − 2, ′( ) = 1 − 2

⇒ ∫ 2 = ln|1 + 2 | + ⇒ ∫ 1−2 = ln| − 2| +
1+2 − 2

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iii) ∫ iv) ∫ 3
2+3 2−1

( ) = 2 + 3, ′( ) = 2 ( ) = 2 − 1, ′( ) = 2

⇒ ∫ = 1 ∫ 2 ⇒ ∫ 3 = 3 ∫
2+3 2 2+3 2−1 2−1

= 1 ln| 2 + 3| + 3 2
2 = 2 ∫ 2 − 1

= 3 ln| 2 − 1| +
2

v) ∫ vi) ∫ 1
+1 ln

( ) = + 1, ′( ) = ( ) = ln , ′( ) = 1


⇒ ∫ = ln| + 1| + ⇒ ∫ 1 = ∫ 1/
+1 ln ln


= ln|ln | +

g) ∫ ′( ) ( ) = ( ) + ii) ∫ 5 2 3
Find the following integrals: iv) ∫ 1− 2
vi) ∫ cos sin
i) ∫ 2 2

iii) ∫(3 − 4 ) 3 −2 2
v) ∫(1 − ) 2 − 2

Solution: ii) ∫ 5 2 3
i) ∫ 2 2 ( ) = 3, ′( ) = 3 2
( ) = 2, ′( ) = 2
⇒ ∫ 5 2 3 = 5 ∫ 2 3
⇒ ∫ 2 2 = 2 +

= 5 ∫ 3 2 3
3

= 5 3 +
3

8

NSO/KKW/CCW

iii) ∫(3 − 4 ) 3 −2 2 iv) ∫ 1− 2

( ) = 3 − 2 2, ′( ) = 3 − 4 ( ) = 1 − 2, ′( ) = −2

⇒ ∫(3 − 4 ) 3 −2 2 = 3 −2 2 + ⇒ ∫ 1− 2 = 1 ∫ −2 1− 2
v) ∫(1 − ) 2 − 2 −2

= − 1 1− 2 +
2

vi) ∫ cos sin

( ) = 2 − 2, ′( ) = 2 − 2 ( ) = sin , ′( ) = cos
= 2(1 − ) ⇒ ∫ cos sin = sin +

⇒ ∫(1 − ) 2 − 2

= 1 ∫ 2(1 − ) 2 − 2
2

= 1 2 − 2 +
2

h) ∫ ′( )[ ( )] = [ ( )] +1 +
+1

Find the following integrals: ii) ∫ 2( 3 + 4)5
i) ∫ 2 ( 2 + 4)5
iv) ∫ 2
iii) ∫(2 + 1) 2+ ( 2+3)2

v) ∫ ( + 1)3 vi) ∫ sin2 cos

9

Solution: NSO/KKW/CCW
i) ∫ 2 ( 2 + 4)5 ii) ∫ 2( 3 + 4)5

( ) = 2 + 4, ′( ) = 2 , = 5 ( ) = 3 + 4, ′( ) = 3 2, = 5

⇒ ∫ 2 ( 2 + 4)5 = ( 2 + 4)6 + ⇒ ∫ 2( 3 + 4)5
6

= 1 ∫ 3 2( 3 + 4)5
3

1 ( 3 + 4)6
= 3 [ 6 ] +

= 1 ( 3 + 4)6 +
18

iii) ∫(2 + 1)( 2 + )3 iv) ∫ 2
( ) = 2 + , ′( ) = 2 + 1, = 3 ( 2+3)2

= ∫ 2 ( 2 + 3)−2

⇒ ∫(2 + 1)( 2 + )3 ( ) = 2 + 3, ′( ) = 2 , = −2
( 2 + )4
2 ( 2 + 3)−1
= 4 + ⇒ ∫ ( 2 + 3)2 = −1 +

v) ∫ ( + 1)3 −1
( ) = + 1, ′( ) = , = 3 = 2 + 3 +

vi) ∫ cos sin2
( ) = sin , ′( ) = cos , = 2

⇒ ∫ ( + 1)3 = ( + 1)4 + ⇒ ∫ cos sin2 = sin3 +
4 3

10

NSO/KKW/CCW

1.2 Integration of trigonometric functions

i) ∫ sin( + ) = − 1 cos( + ) +


j) ∫ cos( + ) = 1 sin ( + ) +


k) ∫ sec2( + ) = 1 tan( + ) +


Note: Compound
sin( ± ) = sin cos ± cos sin angle formula

cos( ± ) = cos cos ∓ sin sin

sin2 = 1 − cos 2
2

cos2 = 1 + cos 2 Double angle
2 formula

sin 2 = 2 sin cos

2 cos cos = cos( + ) + cos( − ) Product to
−2 sin sin = cos( + ) − cos( − ) sum formula
2 sin cos = sin( + ) + sin( − )
2 cos sin = sin( + ) − sin( − )

Exercise 2. ∫ cos3 −cos2 +1
Find the following integrals: cos2

1. ∫(sin 2 + cos ) 4. ∫ tan 2 cos 2
3. ∫( 2 + sec2 3 )
5. ∫ (scions2 ) 6. ∫(sin + cos )2
7. ∫ cos2 8. ∫ cos2 3
9. ∫ sin 3 cos
11. ∫(cos 3 cos + sin 3 sin ) 10. ∫ sin 3 sin

12. ∫(sin 4 cos + cos 4 sin )

11

NSO/KKW/CCW

Solution: 2. ∫ cos3 −cos2 +1
cos2
1. ∫(sin 2 + cos )
cos 2 cos3 cos2 1
= ∫ (cos2 − cos2 + cos2 )
= − 2 + sin +

= ∫(cos − 1 + sec2 )

= sin − + tan +

3. ∫( 2 + sec2 3 ) 4. ∫ tan 2 cos 2
2 tan 3 sin 2

= 2 + 3 + = ∫ cos 2 cos 2

= ∫ sin 2
cos 2

= − 2 +

5. ∫ sin 2 6. ∫(sin + cos )2
cos = ∫(sin2 + 2 sin cos + cos2 )

Double angle formula = ∫(1 + 2 sin cos )

2 sin cos = ∫(1 + sin 2 )
= ∫ cos cos 2

= ∫ 2 sin = − 2 +

= −2 cos +

7. ∫ cos2 8. ∫ cos2 3

Double angle formula Double angle formula

1 + cos 2 1 + cos 2(3 )
= ∫ ( 2 ) = ∫ ( 2 )

= 1 ∫(1 + cos 2 ) 1
2 = 2 ∫(1 + cos 6 )

1 sin 2 1 sin 6
= 2 ( + 2 ) + = 2 ( + 6 ) +

12

NSO/KKW/CCW

9. ∫ sin 3 cos 10. ∫ sin 3 sin

Product to sum formula Product to sum formula

= 1 ∫[sin(3 + ) + sin(3 − )] = − 1 ∫[cos(3 + ) − cos(3 − )]
2 2

= 1 ∫[sin(4 ) + sin(2 )] = − 1 ∫[cos(4 ) − cos(2 )]
2 2

1 cos 4 cos 2 1 sin 4 sin 2
= 2 (− 4 − 2 ) + = − 2 ( 4 − 2 ) +

= − 1 (cos 4 + 2 cos 2 ) + = − 1 (sin 4 − 2 sin 2 ) +
8 8

11. ∫(cos 3 cos + sin 3 sin ) 12. ∫(sin 4 cos + cos 4 sin )

Compound angle formula Compound angle formula

= ∫ cos(3 − ) = ∫ sin(4 + )

= ∫ cos 2 = ∫ sin 5
sin 2 cos 5

= 2 + = − 5 +

13

NSO/KKW/CCW

1.3 Technique of Integration

Substitution
The purpose in using substitution is to rewrite the integration problem in terms of new
variable so that one or more of the basic integration formulas can then be applied.
The first technique to try when basic rules of integration doesn’t work.

Example: Solve ∫ . Steps:
2+1 a) Identify

Solution:

Let = 2 + 1 b) Compute


1 c) Substitute and into the integral
= 2 ⇒ = 2
- the new integral should be in terms of

1 d) Integrate with respect to
∫ 2 + 1 = ∫ 2
e) Convert back to

11
= 2 ∫

1
= 2 ln| | +

= 1 ln| 2 + 1| +
2

Integration By Parts

➢ Usually involve two different functions.
➢ ∫ = − ∫
➢ Choose according to the priority of LPET : L- or , P- polynomials, E- Exponent, T-

Trigonometric

Example: Solve ∫ .

Solution: =
Let =

= 1 ⇒ = = Steps:

a) Identify and

∫ = − ∫ b) Differentiate to get , integrate to get

= − +
c) Substitute into the formula

d) Integrate

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NSO/KKW/CCW

Partial Fractions

 Fraction with two or more factors at the denominator.
 Must check the fraction is proper fraction or improper fraction.
 If improper fraction, must use the long division method to get the proper fraction first.

Partial fraction:

Example: Solve ∫ 4 +5 . 4 + 5
( +2)( −1) ( + 2)( − 1) = ( + 2) + ( − 1)

Solution: 4 + 5 = ( − 1) + ( + 2)

When = −2: − 3 = −3 When = 1: 3 = 9
= 3
∫ 4 +5 = ∫ 1 + 3 = 1

( +2)( −1) ( +2) ( −1) 4 + 5 13

= | + 2| + 3 | − 1| + ∴ ( + 2)( − 1) = ( + 2) + ( − 1)

Exercise (Part A) Exercise (Part C)
Find the following integrals:
Find the following integrals:
1. ∫ sin
1
1. ∫ −1

2. ∫ sin

2. ∫ 1
2−1
3. ∫ sin 2

4. ∫ sin 3. ∫
cos 2−1

5. ∫ 2 4. ∫
√ 2−1

5. ∫ √ 2 − 1

Exercise (Part B) 6. ∫ √ − 1
Find the following integrals:
7. ∫ 1
1. ∫ ln ( −1)2

2. ∫ ln

3. ∫ 1 ln 8. ∫ −1
+2

4. ∫ 1 (ln )2 9. ∫ −1


5. ∫ x (ln 2)

6. ∫ 1
ln

15

NSO/KKW/CCW

Solution (Part A): Basic rules
1. ∫ sin = − +

2. ∫ sin = sin Integration by parts
Let = = −
=

∫ sin = ( )(− cos ) − ∫ − cos
= − cos + sin +

3. ∫ sin 2

Let = 2 Substitution

= 2



=

2

∫ 2 = ∫ sin
2

= 1 (− cos ) +
2

= − 1 (cos 2) +

2

4. ∫

Substitution
Let =

= −



=
−1

∫ = ∫ −1


= − | | +

= − | | +

5. ∫ 2 = ∫( )2 Double angle

∫ 2 = ∫ 1− 2 2 = 1 − 2 2
2
2 1 − 2
= 2
= 1 ∫ 1 − 2
2

= 1 [ − 2 2 ] +
2

= 1 − 2 +

24

16

NSO/KKW/CCW

Solution (Part B):

1. ∫

Let = = Integration by parts
=
du = 1
dx x

∫ = ( )( ) − ∫ (1 )



= − ∫ 1

= − +

2. ∫ = Integration by parts
Let = 2
du = 1
= 2
dx x

= 1



∫ = ( ) ( 2) − ∫ 2 ( 1 )
2
2

= 1 2 − 1 ∫
2 2

= 2 − 1 ( 2) +

2 22

= 2 − 2 +

24

3. ∫ 1 Substitution


Let =

du = 1 dx
x

∫ 1 = ∫


= 2 +

2

= ( )2 +

2

4. ∫ 1 ( )2 Substitution
17

Let =

NSO/KKW/CCW

du = 1 dx

x

∫ 1 ( )2 = ∫ 2


= 3 +

3

= ( )3 +

3

5. ∫ x ( 2) = ∫ (2 ) ln = ln
Integration by parts
= 2 ∫

Let = =
2
du = 1
= 2
dx x

= 1



∫ x ( 2) = 2 [( ) ( 2 2 − ∫ 2 ( 1 )]
2
)

= 2 − ∫

= 2 − 2 +

2

1 1

6. ∫ = ∫



= ( ) +

Solution (Part C):

1. ∫ 1 = | − 1| + Basic rules
−1

2. ∫ 1 = ∫ 1 − 1 Partial fraction:
2−1 2( −1) 2( +1)
11
= 1 | − 1| − 1 | + 1| + 2 − 1 = ( − 1)( + 1)
2 2
1
= 1 [ | − 1| − | + 1|] + ( − 1)( + 1) = ( − 1) + ( + 1)

2 1 = ( + 1) + ( − 1)

= 1 [ +−11] + When = 1: 2 = 1 When = −1: − 2 = 1
2
= 1 = − 1

2 2

1 11
∴ ( − 1)( + 1) = 2( − 1) − 2( + 1)

18

NSO/KKW/CCW

3. ∫ Substitution
2−1

Let = 2 − 1

= 2


=

2

∫ = ∫ 1
2−1 2

= 1 | | +

2

= 1 | 2 − 1| +
2

4. ∫
√ 2−1
Substitution
Let = 2 − 1

= 2



=

2

∫ = ∫ 1
√ 2−1 √ 2

= 1 ∫ −12
2

1

= 1 [ 2 ]+c
2
1

2

1

= 2 +

= √ 2 − 1 +

5. ∫ √ 2 − 1 Substitution
Let = 2 − 1
= 2



=

2

∫ √ 2 − 1 = ∫ √
2

= 1 [ 3⁄2 ] +
2
3

2

= 1 ( 2 − 1)3⁄2 +

3

19

NSO/KKW/CCW

6. ∫ √ − 1 = ∫( − 1)1⁄2 Basic rules

= ( −1)3⁄2 +

3

2

= 2 ( − 3 +

3 1)2

7. ∫ 1 = ∫( − 1)−2 Basic rules
( −1)2

= ( −1)−1 +

−1

= −1 +
( −1)

8. ∫ −1 Substitution
+2 = − 2

Let = + 2

=

∫ −1 =∫ −2−1
+2

= ∫ −3


= ∫ 1 − 3 ∫ 1


= − 3 | | +

= ( + 2) − 3 | + 2| +

9. ∫ −1 = ∫ − ∫ 1 Simplify to use basic rule


= ∫ 1 − | | +
= − | | +

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NSO/KKW/CCW

1.4 Definite Integrals 3

Area = ∫( − )

1 2 3

= ∫( − ) = [2 − ]

1
32 12
Example: = 2 − 3 − ( 2 − 1)

= 2 2
=

=
1 3

3

= ∫( − ) = ∫( − )

1 2 3

Example:


3 =

= [3 − 2 ]

1
32 12
= 3(3) − 2 − (3(1) − 2 )

1 = 2 2



=

21

NSO/KKW/CCW

Volume Axis of rotation:
x-axis

3
= ∫( − )
= ∫( − )

1
Example: 3

= = ∫( 2 − 1)

= 1 3 3
1 3
= [3 − ]

1
33 13
= [ 3 − 3 − ( 3 − 1)]

= 20 3
3

Axis of rotation:
y-axis
= ∫( − )



Example: 3

= ∫( − )
3
= 1
3

= ∫( − )

1 3 3

= [9 − 3 ]

1 1
33 13
= [9(3) − 3 − (9(1) − 3 )]

= 28 3
3
=

22

NSO/KKW/CCW

Exercise
1.
a) Find the area of the region bounded by = 2 + 2, the − , = 0 and = 4.
b) Find the volume of the solid formed in part (a) by rotating completely about the − .

2.
a) Find the area of the region bounded by 2 = 2 − , the − = 0 and = 1.
b) Find the volume of the solid formed in part (a) by rotating completely about the − .

3.
a) Find the area of the region bounded by = 3 + 1, the − , = 0 and = 1.
b) Find the volume of the solid formed in part (a) by rotating completely about the − .

4. Given the region R bounded by the curve = − , the − , the − and = 1.
a) Sketch the region R
b) Determine the area of the region R
c) Find the volume of solid obtained by rotating the region R about the − .

5. Given the curve = 2√ + 1 and the line − + 1 = 0.
a) Sketch the region bounded by the curve and the line
b) Find the area of the region.
c) Determine the volume of solid obtained when the region is revolved completely about
the − .

6. Sketch the region bounded by the curve = √ − 2 , the − , the − and the line
+ = 4.
a) Find the area of the region.
b) Calculate the volume of the solid when the region is revolved completely about the −
.

7. The region R is bounded by the curve = 2 + 6, = −1, = 1 and = 3.
a) Calculate the area of the region.
b) Find the volume of the solid formed by revolving R through 360° about the x-axis.

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Solution:
1. = +

R =
04

a) = ∫04( + − ) Upper - lower

= 3 4
[3 + 2 ]

0

43 02
= 3 + 2(4) − ( 3 + 2(0))

= 88 2
3

b) = ∫04 ( + )2 − ( )2

4

= ∫( 4 + 4 2 + 4)

0

5 4 3 4

= [ 5 + 3 + 4 ]

0

45 4(4)3
= [ 5 + 3 + 4(4) − 0]

= 4590 3
15

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2.
=

= −

R =
02

a) = ∫01( − − ) Right - Left

3 1
= [2 − ]
3
0

13
= 2(1) − 3 − (0)

= 5 2
3

b) = ∫01( − )2 −

1

= ∫ 4 − 4 2 + 4

0

= [4 − 4 3 5 1
+ ]
3 5
0

4(1)3 15
= [4(1) − 3 + 5 − 0]

= 43 3
15

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3.


= +

1
R =

01

a) = ∫01( + − ) Upper - lower

4 2
= [4 + ]

0

14
= 4 + 1 − (0)

= 5 2
4

b) = ∫01( + )2 −

1

= ∫( 6 + 2 3 + 1)

0

7 2 4 1

= [ 7 + 4 + ]

0

17 2(1)4
= [ 7 + 4 + 1 − 0]

= 23 3
14

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4. a)
= −

1 =

R
01

b) = ∫01( − − ) Upper - lower

= [− − ]10

= − 1 + 0
1

= 0.632 2

c) = ∫01( − )2 −

1

= ∫ −2

0

= [ −2 1
]
−2
0

−2 0
= [ −2 + 2 ]

= 0.432 3

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5. Make ‘x’ as the subject

a) − + = Point of intersection:

= √ − 2
R 4 + 1 = + 1

01 2
4 − = 0

(4 − 1) = 0

= 0, = 4

b) = ∫04 + − ( + ) Right - Left
Expand this


4 2
= ∫ ( − 4 )

0

2 3 4
= [ 2 − 12]0

42 43
= 2 − 12 − (0)

= 8 2
3

c) = ∫04( + )2 − ( + 2

)

= 4 ( 2 + 2 + 1 − 4 − 2 − 1)
16 2
∫

0

4 4 2
= ∫ (− 16 + 2 + 2 )

0

= [− 5 + 3 + 4
80 6
2]

0

= [− 45 + 43 + 42 − 0]
80 6

= 208 3
15

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6. Points of intersection:

+ = = √ − 2 + 2 = 4 −
4 2 + − 2 = 0
( + 2)( − 1) = 0

R = −2, = 1


0 24

a) = ∫01( + − ) + ∫14( − − ) Right - Left

3 1 2 4
= [3 + 2 ] + [4 − ]
2
0 1

13 42 12
= [ 3 + 2(1) − 0] + [4(4) − 2 − 4(1) + 2 ]

= 41 2
6

b) = ∫01( + )2 + ∫14( − )

14

= ∫( 4 + 4 2 + 4) + ∫(4 − )2

01

5 4 3 1 (4 − )3 4
= [ 5 + + 4 ] + [ ]
3 −3
0 1

= 15 + 4(1)3 + 4(1) − 0] + [16(4) − 4(4)2 + 43 − 16(1) + 4(1)2 − 13
[5 3 3 3]

83
= 15 + 9

= 218 3
15

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7.


= +

6 =
R 1

-1

a) = ∫−11( + − ) Upper - lower

1

= ∫( 2 + 3)

−1

3 1
= [3 + 3 ]

−1

13 (−1)3
= 3 + 3(1) − 3 − 3(−1)

= 20 2
3

b) = ∫−11( + )2 −

1

= ∫( 4 + 12 2 + 36 − 9)

0

= 5 + 4 3 + 1
[5
27 ]

−1

= 15 + 4(1)3 + 27(1) − (−1)5 − 4(−1)3 − 27(−1)]
[5 5

= 312 3
5

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Chapter 2: First Order Differential Equations

2.1 Separable Variable

Note:
Differential equation (DE) ➔ = 7 −

Order: Order of the highest derivative, so the above DE is first order.
Degree: Power of the highest derivative, so the above DE is degree one.

Solving separable variable differential equations:

Step 1: Consider the first order differential equations

= ( ). ( )



Step 2: Separate the variables

= ( )

( )

Step 3: Integrate both sides of the equation and the general solution will be obtained.

∫ = ∫ ( )
( )

Example

1. 3 2 + 5 − 2 = 0
2

Order: 2
Degree: 1

2. Solve ( + 1) = 2.


Solution:

( + 1) = 2


( + 1) = 2 Separate the variables

∫( + 1) = ∫ 2 Integrate both sides

2 3
2 + = 3 +

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3. Find the general solution of 1 = 2+1 2 .


Solution:

1 = 2+1 2


1 = 2+1 Separate the variables
2

∫ 2 = 1 ∫ 2 2+1 Integrate both sides
2

tan = 1 2+1 +

2

4. Solve = ln given (1) = 10.
( −2)

Solution:

( − 2) = ln Separate the variables


Substitution: ∫ ln = ∫

Integrate both sides Let = ln
ln = 1
∫( − 2) = ∫ = 2 +
Substitution
2
= 1
2 (ln )2
2 − 2 = 2 +

When = 1, = 10

(10)2 (ln(1))2
2 − 2(10) = 2 +

= 30

2 (ln )2
2 − 2 = 2 + 30

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Exercise
1. Determine the order and degree of the following differential equations:

(a) ( 2 2 )2 + ( )3 − = 0 (b) 2 + 2 ( )2 = sin
2


2. Find the general solution of the following equations:

(a) = 2 (b) = 2
+


3. Solve the following initial value problem:

(a) (1 + 2) + 3 = 0; (2) = 2 (b) = 1 ; (−1) = 0
−2( +1)

Solution:

1(a) Order: 2 Degree: 2 1(b) Order: 2 Degree: 1

2(a) = 2 2(b) = 2
+

1 = 1 = 2
2 +

∫ 1 = ∫ 1 = 2
2


∫ 2 = ∫ sin ∫ = 2 ∫ −

tan = − cos + = 2 ( − +

)
−1

= −2 +


= −2 − +

3(a) (1 + 2) + 3 = 0 3(b) = 1
−2( +1)

(1 + 2) = − 3 ( + 1) = − 1
2

∫(1 + 2) = ∫ − 3 ∫( + 1) = ∫ − 1
2

+ 3 = − 4 + 2 + = − 1 +

34 22

When = 2; = 2

(2) + (2)3 = − (2)4 + When = −1; = 0

34 (0)2 + (0) = − 1 (−1) +

= 26 22

3

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+ 3 = − 4 + 26 = − 1
2
3 43
2 + = − 1 − 1
3 + 3 = − 3 4 + 26
2 22
4

2 + 2 = − − 1

= − 2 − 2 − 1

2.2 First Order Linear Differential Equations

Note:
Integrating Factor

Technique RIP

R ➔ Rearrange the equation, to get ( ) and ( ).

I ➔ Integrating factor, ( ) = ∫ ( )

P ➔ Plug-in. ( ) ∙ = ∫ ( ) ( )

Example:

1. Find the general solution of = 3 .


Solution:

− 3 = 0

∴ ( ) = −3, ( ) = 0

( ) = ∫ ( )

( ) = ∫ −3

( ) = −3

( ) ∙ = ∫ ( ) ( )

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−3 ∙ = ∫ −3 . (0)
−3 ∙ = 0 +

=
−3

= 3

2. Solve + 2 = sin , given (0) = 1 .



Solution:
+ 2 = sin



+ 2 = sin
2

2 sin
∴ ( ) = , ( ) = 2

( ) = ∫ ( )

( ) = ∫ 2

( ) = 2 ln

( ) = ln 2

( ) = 2

( ) ∙ = ∫ ( ) ( )

2 ∙ = ∫ 2 sin
2

2 ∙ = ∫ sin

2 ∙ = − cos +
When = 0; = 1
(0)(1) = − cos(0) +
= 1
2 ∙ = − cos + 1

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1 − cos
= 2

Exercise
1. Find the general solution of the following differential equations

(a) − cot = cos (b) ( 2 + 1) + 2 = 1



2. Solve the following differential equations subject to the given initial condition.

(a) ( + 1) + = 2 − 1; (0) = 4 (b) = 2 + 2 ; (0) = 4



Solution:

1(a) − cot = cos 1(b) ( 2 + 1) + 2 = 1



∴ ( ) = − cot , ( ) = cos + 2 = 1
( 2+1) ( 2+1)


( ) = ∫ ( ) ∴ ( ) = 2 ; ( ) = 1
( 2+1) ( 2+1)

( ) = ∫ −csoins ( ) = ∫ ( 22 + 1)

( ) = − ln sin

( ) = ln(sin )−1 ( ) = ln( 2+1)

( ) = 1 ( ) = ( 2 + 1)

sin ( ) ∙ = ∫ ( ) ( )

( ) ∙ = ∫ ( ) ( ) ( 2 ∫( 2 1
( 2+1)
+ 1) ∙ = + 1)

1 ∙ = ∫ 1 cos
sin sin
( 2 + 1) ∙ = ∫ 1

1 ∙ = ln sin + ( 2 + 1) ∙ = +

sin

= sin ln|sin | + sin = +
( 2+1) ( 2+1)

2(a) ( + 1) + = 2 − 1 2(b) = 2 + 2


+ 1 = 2−1 − 2 = 2
( +1) ( +1)

+ 1 = ( −1)( +1) ( ) = 2
( +1) ( +1) ∴ ( ) = −2;

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

∴ ( ) = 1 ; ( ) = ( − 1) ( ) ∙ = ∫ ( ) ( )
( +1)

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( ) = ∫( +11) −2 ∙ = ∫ −2 2
−2 ∙ = ∫ 1
( ) = ln( +1) −2 ∙ = +
When = 0, = 4
( ) = ( + 1) −2(0) ∙ (4) = (0) +
= 4
( ) ∙ = ∫ ( ) ( ) −2 ∙ = + 4
= 2 ( + 4)
( + 1) ∙ = ∫( + 1)( − 1)

( + 1) ∙ = ∫ 2 − 1

( + 1) = 3 − +

3

When = 0, = 4

((0) + 1)(4) = (0)3 − (0) +

3

= 4

( + 1) = 3 − + 4

3

= 33− +4
( +1)

2.3 Applications of First Order Linear Differential Equations

Population growth

Question:

Prof A starts an experiment using bacteria with a population of 200. It is found that the rate of the

population growth of the bacteria, is given by = 0.04 where is the population at the time


minutes after the experiment starts.

(a) Find an expression for , where is a function of .

(b) Estimate the population of the bacteria 50 minutes after the start of the experiment.

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NSO/KKW/CCW

Solution:

(a) = 0.04



1
∫ = ∫ 0.04
ln = 0.04 +
= 0.04 +
General solution: = 0.04

= 0.04
Since = 0, = 200
∴ 200 = 0

= 200
Particular solution: = 200 0.04

(b) When = 50, = 200 0.04(50) = 1477.81
After 50 minutes, the population of the bacteria is approximately 1478.

Radioactive decay

Question:

Radium decomposes at a rate = − where is the mass of Radium at anytime years. If 10%


decomposes in 200 years, what percentage of the original amount of Radium will remain after 1000

years?

Solution:

= −


∫ = ∫ −
ln = − +
= − +
= −
= − , where =

Let the original amount of Radium , at = 0 be that is
= − (0)

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NSO/KKW/CCW

∴ =

= 0, = − (0)

=

= 200, = 0.9 10% decompose ⇒ 90% remain
0.9 = − (200)

− (200) = 0.9

= ln 0.9

−200

= −[l−n200.09]

= [ln2000.9]

When = 1000, = [ln2000.9](1000) = 0.59

∴ 59% will remain after 1000 years.

Newton’s Law of Cooling

Question:
The rate of change of the temperature of the body ( ) at any given time ( ) is proportional to the
difference between its temperature and the temperature of the surrounding medium.


= − ( − 60)
A body whose temperature is 180℃ is cooled by immersing in a liquid at 60℃. In one minute, the
temperature of the body has fallen to 120℃. How long will it take for the temperature of the body
to fall to 90℃?

Solution:

= − ( − 60)

1
∫ ( − 60) = ∫ −
ln| − 60| = − +
− 60 = − +
− 60 = −
− 60 = − where =

When = 0, = 180
180 − 60 = − (0)
= 120

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− 60 = 120 −

When = 1, = 120

120 − 60 = 120 − (1)

− = 120 − 60 = 1
120 2

1
− = ln 2

= −0.69

∴ − 60 = 120 −0.69

If = 90, 90 − 60 = 120 −0.69

−0.69 = 30 = 1

120 4

= ln14

−0.69

= 2 minutes

It will take 2 minutes for the temperature to fall to 90℃.

Electric Circuits

Question:

In an electric circuit, the current , in amperes, after seconds is given by the equation + 4 =

5

12. At time = 0, the current = 0.

(a) Express in terms of and hence, deduce the current after long periods of time.

(b) Find the difference in the current flowing in the circuit from = 5 to = 10.

Solution:

(a) + 4 = 12

5

4
= 12 − 5

4
= 5 (15 − )

14
∫ (15 − ) = ∫ 5

− ln|15 − | = 4 +
5

When = 0, = 0

− ln 15 =

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∴ − ln|15 − | = 4 − ln 15
5

ln|15 − | − ln 15 = − 4
5

|15 − | 4
ln 15 = − 5

15 − = −45
15

15 − = 15 −54

= 15 − 15 −45

= 15 (1 − −45 )

When → ∞, −54 → 0
= 15(1 − 0)
= 15
(b) When = 5, = 15 [1 − −54(5)]

= 14.7253
When = 10, = 15 [1 − −54(10)]

= 14.9950

Difference in current = 14.9950 − 14.7253
= 0.2697

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Chapter 3: Numerical Method

3.1 Newton-Raphson Method

To approximate a root of an equation.

Example:

Show that the equation 2 − 1 + 4 = 0 has a root between 0.2 and 0.3. Hence, evaluate this


root using the Newton-Raphson method taking 0.2 as the first approximation correct to

three significant figures.

Solution:

Let ( ) = 2 − 1 + 4


(0.2) = 0.22 − 1 + 4 = −0.96 < 0 Use numerical method to
0.2 check for a sign change

(0.3) = 0.32 − 1 + 4 = 0.76 > 0
0.3

Since (0.2) and (0.3) have opposite signs, there is a root between 0.2 and 0.3.

′ ( ) = 2 + 1 Find ′( )
2

0 = 0.2 Find the initial value (if not given in question)

1 = 0.2 − 0.22 − 1 + 4 = 0.2378 Substitute into formula
0.2
( )
2(0.2) + 1 +1 = − ′ ( )
0.22

2 = 0.2378 − 0.23782 − 1 + 4 = 0.2460
0.2378

2(0.2378) + 1
0.23782

3 = 0.2460 − 0.24602 − 1 + 4 = 0.2463
0.2460

2(0.2460) + 1
0.24602
Stop when get 2
4 = 0.2463 − 0.24632 − 1 + 4 = 0.2463 consecutive numbers
0.2463
Note: The decimal for
2(0.2463) + 1
0.24632 working steps > the

∴ = 0.246 Final
answer

decimal for final answer

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3.2 Trapezoidal Rule

To approximate a root of an equation.

Example:
Use the trapezoidal rule with 5 sub-intervals to estimate ∫010 √1 + 2 , giving your
answer correct to three decimal places.

Solution: = 5 Identify , and
= 0, = 10,

− 10 − 0 Compute ℎ
ℎ = = 5 = 2

Lower limit 0 = 0 = √1 + 2 Note:
Upper limit 1 = 2 1.0000 5 sub-intervals
2 = 4 = 5 strips
3 = 6 2.2361 = 5 equal parts
4 = 8 4.1231 = 6 ordinates
5 = 10 6.0828
Total 8.0623
10.0499
11.0499 20.5043

First and Remaining ordinates
last

∫010 √1 + 2 = 2 [11.0499 + 2(20.5043)] Substitute into formula
2
ℎ [( & ) + 2( )]
= 52.059 2

43

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Exercise

1. Show that 3 + − 6 = 0 has a root between = 1 and = 2, and find this root correct to
three significant figures.

2. Show that the equation ln + − 3 = 0 has a root between = 1 and = 3. Using 1 as
the first approximation, use Newton-Raphson method to find this root correct to three
decimal places.

3. Show that = ln( + 2) has a root between = 1 and = 2. Using the Newton-Raphson,
find this root correct to three significant figures.

1

4. By using Newton-Raphson method, find an approximate value of 72 correct to three decimal
places.

5. Use the trapezoidal rule to obtain approximate values of the following integrals giving your
answer correct to three decimal places:
a) ∫01 √sin by using 5 ordinates
b) ∫01 √1 − 2 by using 4 sub-intervals
c) ∫02 2 by using 4 strips
d) ∫13 ln by using sub-intervals of 0.5

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Solution

1. Let ( ) = 3 + − 6
(1) = 13 + 1 − 6 = −4 < 0
(2) = 23 + 2 − 6 = 24 > 0
Since (1) and (2) have opposite signs, there is a root between 1 and 2.

′( ) = 3 2 + 1

0 = 1.5
1.53 + 1.5 − 6

1 = 1.5 − 3(1.5)2 + 1 = 1.6452

1.64523 + 1.6452 − 6
2 = 1.6452 − 3(1.6452)2 + 1 = 1.6344

1.63443 + 1.6344 − 6
3 = 1.6344 − 3(1.6344)2 + 1 = 1.6344

∴ = 1.63 Final answer in 3 significant figures

2. Let ( ) = ln + − 3
(1) = ln 1 + 1 − 3 = −2 < 0
(3) = ln 3 + 3 − 3 = 1.0986 > 0
Since (1) and (3) have opposite signs, there is a root between 1 and 3.

′ ( ) = 1 + 1


0 = 1

1 = 1 − ln 1 + 1 − 3 =2

1 + 1
1

2 = 2 − ln 2 + 2 − 3 = 2.2046

1 + 1
2

3 = 2.2046 − ln 2.2046 + 2.2046 − 3 = 2.2079

1 + 1
2.2046

4 = 2.2079 − ln 2.2079 + 2.2079 − 3 = 2.2079

1 + 1
2.2079

∴ = 2.208 Final answer in 3 decimal places

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3. Let ( ) = − ln( + 2)
(1) = 1 − ln(1 + 2) = −0.0986 < 0
(2) = 2 − ln(2 + 2) = 0.6137 > 0
Since (1) and (2) have opposite signs, there is a root between 1 and 2.

′ ( ) = 1 − 1 2
+

0 = 1.5

1 = 1.5 − 1.5 − ln(1.5 + 2) = 1.154

1 − 1 2
1.5 +

2 = 1.154 − 1.154 − ln(1.154 + 2) = 1.146

1 − 1 + 2
1.154

3 = 1.146 − 1.146 − ln(1.146 + 2) = 1.146

1 − 1 + 2
1.146

∴ = 1.15 Final answer in 3 significant figures

4. Let ( ) = 2 − 7
(2) = 22 − 7 = −3 < 0
(3) = 32 − 7 = 2 > 0
Since (2) and (3) have opposite signs, there is a root between 2 and 3.

′( ) = 2

0 = 2.5
2.52 − 7

1 = 2.5 − 2(2.5) = 2.65

2.652 − 7
2 = 2.65 − 2(2.65) = 2.6458

2.64582 − 7
3 = 2.6458 − 2(2.6458) = 2.6458

∴ = 2.646 Final answer in 3 decimal places

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5. 5 ordinates = 4 sub-intervals
a) = 0, = 1, = 4

ℎ = − = 1−0 = 0.25

4

= √sin Use radian instead of degree
0.0000
0 = 0.00
1 = 0.25 0.4974
2 = 0.50 0.6924
3 = 0.75 0.8256
4 = 1.00 0.9173
0.9173 2.0154
Total

∫01 √sin = 0.25 [0.9173 + 2(2.0154)]
2

= 0.619

b) = 0, = 1, = 4
ℎ = − = 1−0 = 0.25

4

0 = 0.00 = √1 − 2 4 decimal places for working
1 = 0.25 1.0000
2 = 0.50
3 = 0.75 0.9682
4 = 1.00 0.8660
0.6614
Total 0.0000
1.0000 2.4956

∫01 √1 − 2 = 0.25 [1.0000 + 2(2.4956)]
2

= 0.749 3 decimal places for final answer

47