Study Material for Numerical Differentiation and Integration

1. Page | 1
VIVEKANANDA COLLEGE, TIRUVEDAKAM WEST
(Residential & Autonomous – A Gurukula Institute of Life-Training)
(Affiliated to Madurai Kamaraj University)
PART III: PHYSICS MAJOR – FOURTH SEMESTER-CORE SUBJECT PAPER-II
NUMERICAL METHODS – 06CT42
(For those who joined in June 2018 and after)
Reference Text Book: Numerical Methods – P.Kandasamy, K.Thilagavathy & K.Gunavathi,
S.Chand & Company Ltd., New Delhi, 2014.
UNIT – IV Numerical Differentiation and Integration
Numerical Differentiation
Introduction
We found the polynomial curve y = f (x), passing through the (n+1) ordered pairs (xi, yi), i=0, 1,
2…n. Now we are trying to find the derivative value of such curves at a given x = xk (say), whose x0 <
xk < xn.
To get derivative, we first find the curve y = f (x) through the points and then differentiate and get its
value at the required point.
If the values of x are equally spaced. We get the interpolating polynomial due to Newton-Gregory.
• If the derivative is required at a point nearer to the starting value in the table, we use
Newton’s forward interpolation formula.
• If we require the derivative at the end of the table, we use Newton’s backward interpolation
formula.
• If the value of derivative is required near the middle of the table value, we use one of the
central difference interpolation formulae.
Newton’s forward difference formula to get the derivative
Newton’s forward difference interpolation formula is
𝑦 (𝑥0 + 𝑢ℎ) = 𝑦𝑢 = 𝑦0 + 𝑢∆𝑦0 +
𝑢(𝑢 − 1)
2!
∆2
𝑦0 +
𝑢(𝑢 − 1)(𝑢 − 2)
3!
∆3
𝑦0 + ⋯
where 𝑦 (𝑥) is a polynomial of degree 𝑛 𝑖𝑛 𝑥 𝑎𝑛𝑑 𝑢 =
𝑥 − 𝑥0
ℎ
where ℎ 𝑖s interval of differencing
The values of first and second derivative at the starting value 𝑥 = 𝑥0 for given by
(
𝑑𝑦
𝑑𝑥
)
𝑥=𝑥0
=
1
ℎ
[∆ 𝑦0 −
1
2
∆2
𝑦0 +
1
3
∆3
𝑦0 − ⋯ ]
(
𝑑2
𝑦
𝑑𝑥2
)
𝑥=𝑥0
=
1
ℎ2
[∆2
𝑦0 − ∆3
𝑦0 +
11
12
∆4
𝑦0 − ⋯ ]

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Problems:
1. The table given below revels the velocity v of a body during the time ‘t’ specified. find its
acceleration at t = 1.1
t : 1.0 1.1 1.2 1.3 1.4
v : 43.1 47.7 52.1 56.4 60.8
𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧. 𝑣 is dependent on time 𝑡 𝑖. 𝑒. , 𝑣 = 𝑣(𝑡). we require acceleration =
𝑑𝑣
𝑑𝑡
.
therefore, we have to find 𝑣′(1.1).
𝐴𝑠
𝑑𝑣
𝑑𝑡
𝑎𝑡 𝑡 = 1.1 is require, (nearer to beginning value), we use 𝑁𝑒𝑤𝑡𝑜𝑛 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.
𝑣(𝑡) = 𝑣 (𝑥0 + 𝑢ℎ) = 𝑣0 + 𝑢∆𝑣0 +
𝑢(𝑢 − 1)
2!
∆2
𝑣0 +
𝑢(𝑢 − 1)(𝑢 − 2)
3!
∆3
𝑣0 + ⋯
𝑑𝑣
𝑑𝑡
=
1
ℎ
.
𝑑𝑣
𝑑𝑡
=
1
ℎ
[∆𝑣0 +
2𝑢 − 1
2
∆2
𝑣0 +
3𝑢2
− 6𝑢 + 2
6
∆3
𝑣0 + ⋯ ]
where 𝑢 =
𝑡 − 𝑡0
ℎ
=
1.1 − 1.0
0.1
= 1
(
𝑑𝑣
𝑑𝑡
)
𝑡=1.1
= (
𝑑𝑣
𝑑𝑡
)
𝑛=1
=
1
0.1
[4.6 +
1
2
(−0.2) +
1
6
(0.1) +
1
12
(0.1)]
= 10[4.6 − 0.1 − 0.0166 + 0.0083]
= 𝟒𝟒. 𝟗𝟏𝟕
t
1.0
1.1
1.2
1.3
1.4
v
43.1
47.7
52.1
56.4
60.8
∆𝑣
4.6
4.4
4.3
4.4
∆2
𝑣
-0.2
-0.1
0.1
∆3
𝑣
0.1
0.2
∆4
𝑣
0.1

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2. Derive the Newton’s forward difference formula to get the derivative.
We are given (𝑛 + 1)ordered pairs (𝑥𝑖, 𝑦𝑖)𝑖 = 0, 1, … 𝑛. we want to find the derivative of
𝑦 = 𝑓(𝑥) passing through the (𝑛 + 1) points, at a point near to the startinng value 𝑥 = 𝑥0
Newton’s forward difference interpolation formula is
𝑦 (𝑥0 + 𝑢ℎ) = 𝑦𝑢 = 𝑦0 + 𝑢∆𝑦0 +
𝑢(𝑢 − 1)
2!
∆2
𝑦0 +
𝑢(𝑢 − 1)(𝑢 − 2)
3!
∆3
𝑦0 + ⋯ … (1)
where 𝑦 (𝑥) is a polynomial of degree 𝑛 𝑖𝑛 𝑥 𝑎𝑛𝑑 𝑢 =
𝑥 − 𝑥0
ℎ
Differentiating 𝑦(𝑥) w. r. t. 𝑥,
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑢
.
𝑑𝑢
𝑑𝑥
=
1
ℎ
.
𝑑𝑦
𝑑𝑢
𝑑𝑦
𝑑𝑥
=
1
ℎ
[∆𝑦0 +
2𝑢 − 1
2
∆2
𝑦0 +
3𝑢2
− 6𝑢 + 2
6
∆3
𝑦0 +
(4𝑢3
− 18𝑢2
+ 22𝑢 − 6)
24
∆4
𝑦0] … . (2)
Equation (2) gives the value of
𝑑𝑦
𝑑𝑥
at general 𝑥 which may be anywhere in the interval.
In special case like 𝑥 = 𝑥0, 𝑖. 𝑒. , 𝑢 = 0 𝑖𝑛 (2)
(
𝑑𝑦
𝑑𝑥
)
𝑥=𝑥0
= (
𝑑𝑦
𝑑𝑥
)
𝑢=0
=
1
ℎ
[∆𝑦0 +
1
2
∆2
𝑦0 +
1
3
∆3
𝑦0 −
1
4
∆4
𝑦0 + ⋯ ] … (3)
Differentiating (2) again w. r. t. 𝑥,
𝑑2
𝑦
𝑑𝑥2
=
𝑑
𝑑𝑢
(
𝑑𝑦
𝑑𝑥
) .
𝑑𝑢
𝑑𝑥
=
𝑑
𝑑𝑢
(
𝑑𝑦
𝑑𝑥
) .
1
ℎ
𝑑2
𝑦
𝑑𝑥2
=
1
ℎ2
[∆2
𝑦0 + (𝑢 − 1)∆3
𝑦0 +
(6𝑢2
− 18𝑢 + 11)
12
∆4
𝑦0 + ⋯ ] … (4)
Equation (4) give the second derivative value at 𝑥 = 𝑥.
setting 𝑥 = 𝑥0 𝑖. 𝑒. , 𝑢 = 0 𝑖𝑛 (4)
(
𝑑2
𝑦
𝑑𝑥2
)
𝑥=𝑥0
=
1
ℎ2
[∆2
𝑦0 − ∆3
𝑦0 +
11
12
∆4
𝑦0 + ⋯ ] … (5)
This equation (5) give the value of second derivative at the starting value 𝑥 = 𝑥0
3. 𝐓𝐡𝐞 𝐭𝐚𝐛𝐥𝐞 𝐛𝐞𝐥𝐨𝐰 𝐠𝐢𝐯𝐞𝐬 𝐭𝐡𝐞 𝐫𝐞𝐬𝐮𝐥𝐭𝐬 𝐨𝐟 𝐚𝐧 𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧: 𝛉 𝐢𝐬 𝐭𝐡𝐞 𝐨𝐛𝐬𝐞𝐫𝐯𝐞𝐝 𝐭𝐞𝐦𝐩𝐞𝐫𝐚𝐭𝐮𝐫𝐞 𝐢𝐧
𝐝𝐞𝐠𝐫𝐞𝐞𝐬 𝐜𝐞𝐧𝐭𝐫𝐢𝐠𝐫𝐚𝐝𝐞 𝐨𝐟𝐚 𝐯𝐞𝐬𝐬𝐞𝐥 𝐨𝐟 𝐜𝐨𝐨𝐥𝐢𝐧𝐠 𝐰𝐚𝐭𝐞𝐫; 𝐭 𝐢𝐬 𝐭𝐡𝐞 𝐭𝐢𝐦𝐞 𝐢𝐧 𝐦𝐢𝐧𝐮𝐭𝐞𝐬 𝐟𝐫𝐨𝐦 𝐭𝐡𝐞
𝐛𝐞𝐠𝐢𝐧𝐧𝐢𝐧𝐠 𝐨𝐟 𝐨𝐛𝐬𝐞𝐫𝐯𝐚𝐭𝐢𝐨𝐧.
𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐚𝐩𝐩𝐫𝐨𝐱𝐢𝐦𝐚𝐭𝐞 𝐫𝐚𝐭𝐞 𝐨𝐟 𝐜𝐨𝐨𝐥𝐢𝐧𝐠 𝐚𝐭 𝒕 = 𝟑 𝒂𝒏𝒅 𝟑. 𝟓
t :
𝜽 :
1
85.3
3
74.5
5
67.0
7
60.5
9
54.3

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Solution. we form below the difference table
t
1
3
5
7
9
𝜃
85.3
74.5
67.0
60.5
54.3
∆𝜃
-10.8
-7.5
-6.5
-6.2
∆2
𝜃
3.3
1.0
0.3
∆3
𝜃
-2.3
-0.7
∆4
𝜃
1.6
𝑑𝜃
𝑑𝑡
represents the rate of cooling
𝑢 =
𝑡 − 𝑡0
ℎ
=
𝑡 − 1
2
𝐴𝑡 𝑡 = 3, 𝑢 = 1 𝐴𝑡 𝑡 = 3.5, 𝑢 = 1.25 ℎ = 2
(i) Putting 𝑢 = 1 𝑖𝑛
𝑑𝑦
𝑑𝑥
(
𝑑𝜃
𝑑𝑡
)
𝑡=3
= (
𝑑𝜃
𝑑𝑡
)
𝑢=1
=
1
2
[−10.8 −
1
2
(3.3) −
1
6
(−2.3) +
1
12
(1.6)]
=
1
2
[−10.8 + 1.65 + 0.38333 + 0.13333]
= −𝟒. 𝟑𝟏𝟔𝟔𝟕
(ii)Putting 𝑢 = 1.25 𝑖𝑛
𝑑𝑦
𝑑𝑥
(
𝑑𝜃
𝑑𝑡
)
𝑡=3.5
= (
𝑑𝜃
𝑑𝑡
)
𝑢=1.25
=
1
2
[−10.8 + 0.75(3.3) − (0.1354)(−2.3) + (0.04948)(1.6)]
=
1
2
[−10.8 + 2.475 + 0.31142 + 0.079168]
= −𝟑. 𝟗𝟔𝟕𝟏𝟖
Newton’s backward difference formula to compute the derivative
Consider Newton’s backward difference interpolation formula.
𝑦(𝑥) = 𝑦 (𝑥0 + 𝑣ℎ) = 𝑦𝑛 + 𝑣∆𝑦𝑛 +
𝑣(𝑣 + 1)
2!
∇2
𝑦𝑛 +
𝑣(𝑣 + 1)(𝑣 + 2)
3!
∇3
𝑦𝑛 + ⋯

5. Page | 5
where 𝑣 =
𝑥 − 𝑥0
ℎ
h is interval of differencing.
The value of first and second derivative at the ending value 𝑥 = 𝑥 𝑛 given by
(
𝑑𝑦
𝑑𝑥
)
𝑥=𝑥 𝑛
=
1
ℎ
[∇𝑦𝑛 +
1
2
∇2
𝑦𝑛 +
1
3
∇3
𝑦𝑛 +
1
4
∇4
𝑦𝑛 + ⋯ ]
(
𝑑2
𝑦
𝑑𝑥2
)
𝑥=𝑥 𝑛
=
1
ℎ2
[∇2
𝑦𝑛 + ∇3
𝑦𝑛 +
11
12
∇4
𝑦𝑛 + ⋯ ]
4. 𝐀 𝐫𝐨𝐝 𝐢𝐬 𝐫𝐨𝐭𝐚𝐭𝐢𝐧𝐠 𝐢𝐧 𝐚 𝐩𝐥𝐚𝐧𝐞. 𝐓𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐧𝐠 𝐭𝐚𝐛𝐥𝐞 𝐠𝐢𝐯𝐞𝐬 𝐭𝐡𝐞 𝐚𝐧𝐠𝐥𝐞 𝛉 (𝐢𝐧 𝐫𝐚𝐝𝐢𝐚𝐧𝐬) 𝐭𝐡𝐫𝐨𝐮𝐠𝐡
𝐰𝐡𝐢𝐜𝐡 𝐭𝐡𝐞 𝐫𝐨𝐝 𝐡𝐚𝐬 𝐭𝐮𝐫𝐧𝐞𝐝 𝐟𝐨𝐫 𝐯𝐚𝐫𝐢𝐨𝐮𝐬 𝐯𝐚𝐥𝐮𝐞𝐬 𝐨𝐟 𝐭𝐢𝐦𝐞 𝐭 (𝐬𝐞𝐜𝐨𝐧𝐝𝐬). 𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞 𝐭𝐡𝐞 𝐚𝐧𝐠𝐮𝐥𝐚𝐫
𝐯𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐚𝐧𝐝 𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐚𝐜𝐜𝐞𝐥𝐞𝐫𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐫𝐨𝐝 𝐚𝐭 = 𝟎. 𝟔 𝐬𝐞𝐜𝐨𝐧𝐝𝐬.
t : 0 0.2 0.4 0.6 0.8 1.0
𝜽 : 0 0.12 0.49 1.12 2.02 3.20
Solution. We form the difference table below:
t
0
0.2
0.4
0.6
0.8
1.0
𝜃
0
0.12
0.49
1.12
2.02
3.20
∇𝜃
0.12
0.37
0.63
0.90
1.18
∇2
𝜃
0.25
0.26
0.27
0.28
∇3
𝜃
0.01
0.01
0.01
∇4
𝜃
0
0
𝑥 = 0.6 is toward the end, we will use backward difference formula. ℎ = 0.2
(
𝑑𝑦
𝑑𝑥
)
𝑥
=
1
ℎ
[∆𝑦𝑛 +
2𝑣 +
2
∇2
𝑦𝑛 +
3𝑣2
+ 6𝑣 + 2
6
∇3
𝑦𝑛 +
4𝑣3
+ 18𝑣2
+ 22𝑣 + 6
24
∇4
𝑦𝑛 + ⋯ ] … (1)
𝐻𝑒𝑟𝑒 𝑣 =
𝑥 − 𝑥 𝑛
ℎ
=
0.6 − 1.0
0.2
= −2
Using in (1),
𝑑𝜃
𝑑𝑡
represents the angular velocity

6. Page | 6
(
𝑑𝜃
𝑑𝑡
)
𝑡=0.6
=
1
0.2
[1.18 −
3
2
(0.28) +
1
3
(0.01)]
= 5[1.18 − 0.42 + 0.00333]
= 𝟑. 𝟖𝟏𝟔𝟔𝟓
𝒓𝒂𝒅
𝒔𝒆𝒄
𝑑2
𝜃
𝑑𝑡2
represents the angular acceleration
Also, (
𝑑2
𝑦
𝑑𝑡2
) =
1
ℎ2
[∇2
𝑦𝑛 + (v + 1)∇3
𝑦𝑛 + ⋯ ]
(
𝑑2
𝜃
𝑑𝑡2
)
𝑡=0.6
=
1
0.04
[0.28 − 0.01]
= 𝟔. 𝟕𝟓
𝒓𝒂𝒅
𝒔𝒆𝒄 𝟐
5. 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐟𝐢𝐫𝐬𝐭 𝐭𝐰𝐨 𝐝𝐞𝐫𝐢𝐯𝐚𝐭𝐢𝐞𝐬 𝐨𝐟 (𝒙)
𝟏
𝟑 𝒂𝒕 𝒙 = 𝟓𝟎 𝐚𝐧𝐝 𝒙 = 𝟓𝟔 𝐠𝐢𝐯𝐞𝐧 𝐭𝐡𝐞 𝐭𝐚𝐛𝐥𝐞 𝐛𝐞𝐥𝐨𝐰
x : 50 51 52 53 54 55 56
y =(𝒙)
𝟏
𝟑 :
3.6840 3.7084 3.7325 3.7563 3.7798 3.8030 3.8259
Solution.
We require 𝑓′(𝑥) 𝑎𝑡 𝑥 = 50 we use 𝑁𝑒𝑤𝑡𝑜𝑛′
𝑠 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 and
𝑡o get 𝑓′(𝑥) 𝑎𝑡 𝑥 = 56 we use 𝑁𝑒𝑤𝑡𝑜𝑛′
𝑠 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.
By 𝑁𝑒𝑤𝑡𝑜𝑛′
𝑠 𝑓𝑜𝑟𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑚𝑢𝑙𝑎,
ℎ = 1 𝑢 =
𝑥 − 𝑥0
ℎ
=
50 − 50
1
= 0
𝐃𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐜𝐞 𝐓𝐚𝐛𝐥𝐞
x
50
51
52
53
54
55
56
y
3.6840
3.7084
3.7325
3.7563
3.7798
3.8030
3.8259
∆𝑦
0.0244
0.0241
0.0238
0.0235
0.0232
0.0229
∆2
𝑦
-0.0003
-0.0003
-0.0003
-0.0003
-0.0003
∆3
𝑦
0
0
0
0

7. Page | 7
(
𝑑𝑦
𝑑𝑥
)
𝑥=𝑥0
= (
𝑑𝑦
𝑑𝑥
)
𝑢=0
=
1
ℎ
[∆ 𝑦0 −
1
2
∆2
𝑦0 +
1
3
∆3
𝑦0 − ⋯ ]
=
1
1
[0.0244 −
1
2
(−0.0003) +
1
3
(0)]
= 𝟎. 𝟎𝟐𝟒𝟓𝟓
(
𝑑2
𝑦
𝑑𝑥2
)
𝑥=50
= (
𝑑2
𝑦
𝑑𝑥2
)
𝑢=0
=
1
ℎ2
[∆2
𝑦0 − ∆3
𝑦0 + ⋯ ]
= 1[−0.0003]
= −𝟎. 𝟎𝟎𝟎𝟑.
𝐵𝑦 𝑁𝑒𝑤𝑡𝑜𝑛′
𝑠 𝑏𝑎𝑐𝑘𝑤𝑎𝑟𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎,
(
𝑑𝑦
𝑑𝑥
)
𝑥=𝑥 𝑛
= (
𝑑𝑦
𝑑𝑥
)
𝑣=0
=
1
ℎ
[∇𝑦𝑛 +
1
2
∇2
𝑦𝑛 +
1
3
∇3
𝑦𝑛 + ⋯ ]
(
𝑑𝑦
𝑑𝑥
)
𝑥=56
=
1
1
[0.0229 +
1
2
(−0.0003) + 0 ]
= 𝟎. 𝟎𝟐𝟐𝟕𝟓
(
𝑑2
𝑦
𝑑𝑥2
)
𝑥=𝑥 𝑛
=
1
ℎ2
[∇2
𝑦𝑛 + ∇3
𝑦𝑛 + ⋯ ]
=
1
1
[−0.0003]
= −𝟎. 𝟎𝟎𝟎𝟑
NUMERICAL INTEGRATION
Introduction
We know that ∫ 𝑓(𝑥)𝑑𝑥 represents the area between 𝑦 = 𝑓(𝑥), 𝑥 − axis and the
𝑏
𝑎
ordinates 𝑥 = 𝑎 and 𝑥 = 𝑏. This integration is possible only if the 𝑓(𝑥)is explicitly given and
if it is integrable. The problem of numerical integration can be stated as follows: Given a
set of (𝑛 + 1) paird values (𝑥𝑖, 𝑦𝑖), 𝑖 = 0, 1, 2, … 𝑛 𝑜f the function 𝑦 = 𝑓(𝑥), where 𝑓(𝑥) is
not known explicitly, it is required to compute ∫ 𝑦 𝑑𝑥.
𝑥 𝑛
𝑥0
As we did in the case of interpolation
or numerical differentation, we replace 𝑓(𝑥) by an interpolating polynomial 𝑃𝑛(𝑥) 𝑎nd obtain
∫ 𝑃𝑛(𝑥)𝑑𝑥 which is approximately taken as the value for ∫ 𝑓(𝑥)𝑑𝑥.
𝑥 𝑛
𝑥0
𝑥0
𝑥 𝑛

8. Page | 8
A general quadrature formula for equidistant ordinates ( or Newton-Cote’s formula)
For equally spaced intervals, we have Newton’s forward difference formula as
𝑦 (𝑥0 + 𝑢ℎ) = 𝑦𝑢 = 𝑦0 + 𝑢∆𝑦0 +
𝑢(𝑢 − 1)
2!
∆2
𝑦0 +
𝑢(𝑢 − 1)(𝑢 − 2)
3!
∆3
𝑦0 + ⋯
Here, 𝑢 =
𝑥 − 𝑥0
ℎ
where h is interval of differencing
𝑠ince 𝑥 𝑛 = 𝑥0 + 𝑛ℎ and , 𝑢 =
𝑥 − 𝑥0
ℎ
we have
𝑥 − 𝑥0
ℎ
= 𝑛 = 𝑢.
∫ 𝑓(𝑥) 𝑑𝑥 ≈ ℎ[
𝑥 𝑛
𝑥0
𝑛𝑦0 +
𝑛2
2
∆𝑦0 +
1
2
(
𝑛3
3
−
𝑛2
2
) ∆2
𝑦0 +
1
6
(
𝑛4
4
− 𝑛3
+ 𝑛2
) ∆3
𝑦0 + ⋯
Above the equation is called 𝑁𝑒𝑤𝑡𝑜𝑛 − 𝐶𝑜𝑡𝑒′
𝑠 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑢𝑟𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 and is a general
𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑢𝑟𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎. Giving various values for 𝑛, we get a number of special formula.
Trapezoidal rule
𝐵𝑦 𝑝𝑢𝑡𝑡𝑖𝑛𝑔 𝑛 = 1, 𝑖𝑛 𝑡ℎ𝑒 𝑁𝑒𝑤𝑡𝑜𝑛 − 𝐶𝑜𝑡𝑒′
𝑠 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑢𝑟𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 (𝑖. 𝑒. , 𝑡ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑜𝑛𝑙𝑦 𝑡𝑤𝑜
𝑝𝑎𝑖𝑟𝑒𝑑𝑣𝑎𝑙𝑢𝑒𝑠 𝑎𝑛𝑑 𝑖𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑛𝑔 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 𝑖𝑠 𝑙𝑖𝑛𝑒𝑎𝑟).
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥 + ⋯ + ∫ 𝑓(𝑥) 𝑑𝑥
𝑥0+𝑛ℎ
𝑥0+(𝑛−1)ℎ
𝑥0+2ℎ
𝑥0+ℎ
𝑥0+ℎ
𝑥0
𝑥0+𝑛ℎ
𝑥0
𝑥 𝑛
𝑥0
=
ℎ
2
[(𝑦0 + 𝑦𝑛) + 2(𝑦1 + 𝑦2 + 𝑦3 + ⋯ + 𝑦 𝑛−1)]
=
ℎ
2
[(𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑙𝑎𝑠𝑡 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠)
+ 2(𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠)
This is known as Trapezoidal Rule.
Simpson’s one-third rule
Setting n =2 in Newton-Cotes quadrature formula,
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ⋯ + ∫ 𝑓(𝑥) 𝑑𝑥
𝑥 𝑛
𝑥 𝑛−2
𝑥4
𝑥2
𝑥2
𝑥0
𝑥 𝑛
𝑥0
=
ℎ
3
[(𝑦0 + 𝑦𝑛) + 2(𝑦2 + 𝑦4 + ⋯ ) + 4(𝑦1 + 𝑦3 + ⋯ )]
=
ℎ
3
[𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑙𝑎𝑠𝑡 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠
+ 2(𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑜𝑑𝑑 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 + 4(𝑠𝑢𝑚 𝑜𝑓 𝑒𝑣𝑒𝑛 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠)]

9. Page | 9
Simpson’s three-eighths rule
Putting n = 3 in Newton-Cote’s quadrature formula,
𝐼𝑓 𝑛 𝑖𝑠 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3
∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥 + ⋯ + ∫ 𝑓(𝑥) 𝑑𝑥
𝑥0+𝑛ℎ
𝑥0+(𝑛−3)ℎ
𝑥0+6ℎ
𝑥0+3ℎ
𝑥0+3ℎ
𝑥0
𝑥0+𝑛ℎ
𝑥0
=
3ℎ
8
[(𝑦0 + 𝑦𝑛) + 3(𝑦1 + 𝑦2 + 𝑦4 + 𝑦5 + ⋯ + 𝑦 𝑛−1) + 2(𝑦3 + 𝑦6 + 𝑦9 + ⋯ + 𝑦𝑛)]
Above the equation is called 𝑆𝑖𝑚𝑝𝑠𝑜𝑛′
𝑠 𝑡ℎ𝑟𝑒𝑒 − 𝑒𝑖𝑔ℎ𝑡𝑠𝑠 𝑟𝑢𝑙𝑒 which is applicable only when
𝑛 𝑖s multiple of 3.
6. Evaluate ∫ 𝒙 𝟒
𝒅𝒙 𝐮𝐬𝐢𝐧𝐠 𝐢) 𝐓𝐫𝐚𝐩𝐞𝐳𝐨𝐢𝐝𝐚𝐥 𝐫𝐮𝐥𝐞 𝐢𝐢) 𝐒𝐢𝐦𝐩𝐬𝐨𝐧′
𝐬 𝐫𝐮𝐥𝐞
𝟑
−𝟑
Solution. Here 𝑦(𝑥) = 𝑥4
. Interval length (𝑏 − 𝑎) = 6. So, we divide 6 equal intervals with
ℎ =
3 − (−3)
6
=
6
6
= 1. we form below the table
x -3 -2 -1 0 1 2 3
y 81 16 1 0 1 16 81
i) By Trapezoidal rule,
∫ 𝑥4
𝑑𝑥 ≈
ℎ
2
[(𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑙𝑎𝑠𝑡 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 + 2(𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠)
3
−3
≈
1
2
[(81 + 81) + 2(16 + 1 + 0 + 1 + 16)
≈ 𝟏𝟏𝟓
𝑖𝑖) 𝐵𝑦 𝑠𝑖𝑚𝑝𝑠𝑜𝑛′
𝑠 𝑜𝑛𝑒 − 𝑡ℎ𝑖𝑟𝑑 𝑟𝑢𝑙𝑒
∫ 𝑦 𝑑𝑥 ≈
1
3
[(81 + 81) + 2(1 + 1) + 4(16 + 0 + 16)]
3
−3
≈ 𝟗𝟖
𝑖𝑖𝑖) 𝑆𝑖𝑛𝑐𝑒 𝑛 = 6, (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 𝑡ℎ𝑟𝑒𝑒), we can also use 𝑆𝑖𝑚𝑝𝑠𝑜𝑛′
𝑠 𝑡ℎ𝑟𝑒𝑒 − 𝑒𝑖𝑔ℎ𝑡ℎ𝑠 𝑟𝑢𝑙𝑒.
∫ 𝑦 𝑑𝑥 ≈
3
8
[(81 + 81) + 3(16 + 1 + 1 + 16) + 2(0)] ≈ 𝟗𝟗.
3
−3
7. 𝐄𝐯𝐚𝐥𝐮𝐭𝐞 ∫
𝐝𝐱
𝟏+𝐱
𝐮𝐬𝐢𝐧𝐠 𝐢) 𝐓𝐫𝐚𝐩𝐞𝐳𝐨𝐢𝐝𝐚𝐥 𝐫𝐮𝐥𝐞 𝐢𝐢) 𝐒𝐢𝐦𝐩𝐬𝐨𝐧′
𝐬 𝐫𝐮𝐥𝐞
𝟔
𝟎
Solution. Take the number of intervals as 6

10. Page | 10
ℎ =
6 − 0
6
= 1
x 0 1 2 3 4 5 6
𝑦 =
1
1 + 𝑥
1 0.5 1/3 1/4 1/5 1/6 1/7
𝑖) 𝐵𝑦 𝑇𝑟𝑎𝑝𝑒𝑧𝑜𝑖𝑑𝑎𝑙 𝑟𝑢𝑙𝑒,
∫
𝑑𝑥
1 + 𝑥
6
0
=
ℎ
2
[(𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑙𝑎𝑠𝑡 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 + 2(𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠)
=
1
2
[(1 +
1
7
) + 2 (0.5 +
1
3
+
1
4
+
1
5
+
1
6
)]
= 𝟐. 𝟎𝟐𝟏𝟒𝟐𝟖𝟓𝟕
𝑖𝑖) 𝐵𝑦 𝑆𝑖𝑚𝑝𝑠𝑜𝑛′
𝑠 𝑜𝑛𝑒 − 𝑡ℎ𝑖𝑟𝑑 𝑟𝑢𝑙𝑒,
𝐼 =
1
3
[(1 +
1
7
) + 2 (
1
3
+
1
5
) + 4 (
1
2
+
1
4
+
1
6
)]
=
1
3
(1 +
1
7
+
16
15
+
22
6
) = 𝟏. 𝟗𝟓𝟖𝟕𝟑𝟎𝟏𝟔
𝑖𝑖𝑖) 𝐵𝑦 𝑆𝑖𝑚𝑝𝑠𝑜𝑛′
𝑠 𝑡ℎ𝑟𝑒𝑒 − 𝑒𝑖𝑔ℎ𝑡ℎ𝑠 𝑟𝑢𝑙𝑒,
𝐼 =
3
8
[(1 +
1
7
) + 3 (0.5 +
1
3
+
1
5
+
1
6
) + 2 (
1
4
)]
= 𝟏. 𝟗𝟔𝟔𝟎𝟕𝟏𝟒𝟑
8. 𝐀 𝐫𝐢𝐯𝐞𝐫 𝐢𝐬 𝟖𝟎 𝐦𝐞𝐭𝐫𝐞𝐬 𝐰𝐢𝐝𝐞. 𝐓𝐡𝐞 𝐝𝐞𝐩𝐭𝐡 ′d'𝐢𝐧 𝐦𝐞𝐭𝐫𝐞𝐬𝐚𝐭 𝐚 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝒙 𝒎𝒆𝒕𝒓𝒆𝒔 𝐟𝐫𝐨𝐦 𝐨𝐧𝐞 𝐛𝐚𝐧𝐤 𝐢𝐬
𝐠𝐢𝐯𝐞𝐧 𝐛𝐲 𝐭𝐡𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐢𝐧𝐠 𝐭𝐚𝐛𝐥𝐞. 𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐞 𝐭𝐡𝐞 𝐚𝐫𝐞𝐚 𝐨𝐟 𝐜𝐫𝐨𝐬𝐬 𝐬𝐞𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐭𝐡𝐞 𝐫𝐢𝐯𝐞𝐫 𝐮𝐬𝐢𝐧𝐠
𝐒𝐢𝐦𝐩𝐬𝐨𝐧′
𝐬 𝐫𝐮𝐥𝐞.
x : 0 10 20 30 40 50 60 70 80
d : 0 4 7 9 12 15 14 8 3
Solution. Here h =10. Area of cross section is ∫ 𝑦 𝑑𝑥
80
0
𝐴 =
10
3
[(0 + 3) + 2(7 + 12 + 14) + 4(4 + 9 + 15 + 8)]
=
10
3
[3 + 66 + 144]
= 𝟕𝟏𝟎 𝒔𝒒. 𝒎𝒆𝒕𝒓𝒆𝒔

11. Page | 11
9. 𝐄𝐯𝐚𝐥𝐮𝐭𝐞 ∫
𝒅𝒙
𝟏+𝒙 𝟐 𝐮𝐬𝐢𝐧𝐠 𝐓𝐫𝐚𝐩𝐞𝐳𝐨𝐢𝐝𝐚𝐥 𝐫𝐮𝐥𝐞 𝒘𝒊𝒕𝒉 𝒉 = 𝟎. 𝟐. 𝐇𝐞𝐧𝐜𝐞 𝐨𝐛𝐭𝐚𝐢𝐧 𝐚𝐩𝐩𝐫𝐨𝐱𝐢𝐦𝐚𝐭𝐞 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝝅.
𝟏
𝟎
𝐂𝐚𝐧 𝐲𝐨𝐮 𝐮𝐬𝐞 𝐨𝐭𝐡𝐞𝐫 𝐟𝐨𝐫𝐦𝐮𝐥𝐚𝐞 𝐢𝐧 𝐭𝐡𝐢𝐬 𝐜𝐚𝐬𝐞.
Solution. 𝐿𝑒𝑡 𝑦(𝑥) =
𝑑𝑥
1+𝑥2
Interval is (1 − 0) = 1 ℎ = 0.2
x 0 0.2 0.4 0.6 0.8 1.0
𝑦 =
𝑑𝑥
1 + 𝑥2
1 0.96154 0.86207 0.73529 0.60976 0.50000
∫
𝑑𝑥
1 + 𝑥2
1
0
=
ℎ
2
[(𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑎𝑛𝑑 𝑙𝑎𝑠𝑡 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠 + 2(𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒𝑠)
=
0.2
2
[(1 + 0.5) + 2(0.96154 + 0.86207 + 0.73529 + 0.60976)
= (0.1)[1.5 + 6.33732]
= 𝟎. 𝟕𝟖𝟑𝟕𝟑𝟐
𝐵𝑦 𝑎𝑐𝑡𝑢𝑎𝑙 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛,
∫
𝑑𝑥
1 + 𝑥2
= (tan−1
𝑥)0
1
=
𝜋
4
1
0
∴
𝜋
4
≈ 0.783732
∴ 𝜋 ≈ 𝟑. 𝟏𝟑𝟒𝟗𝟑 (𝒂𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒆𝒍𝒚)
10. 𝐄𝐯𝐚𝐥𝐮𝐭𝐞 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐠𝐫𝐚𝐥 𝑰 = ∫ 𝒍𝒐𝒈 𝒆 𝒙 𝒅𝒙 𝐮𝐬𝐢𝐧𝐠 𝐓𝐫𝐚𝐩𝐞𝐳𝐨𝐢𝐝𝐚𝐥 𝐚𝐧𝐝 𝐒𝐢𝐦𝐩𝐬𝐨𝐧′
𝐬 𝐫𝐮𝐥𝐞𝐬
𝟓.𝟐
𝟒
Solution. Here b - a = 5.2 – 4 = 1.2
Hence, ℎ =
1.2
6
= 0.2
x : 4 4.2 4.4 4.6 4.8 5.0 5.2
𝑓(𝑥)
= 𝑙𝑜𝑔 𝑒 𝑥:
1.3862944 1.4350845 1.4816045 1.5260563 1.5686159 1.6094379 1.6486586
𝑖) 𝐵𝑦 𝑇𝑟𝑎𝑝𝑒𝑧𝑜𝑖𝑑𝑎𝑙 𝑟𝑢𝑙𝑒,
∫ 𝑙𝑜𝑔 𝑒 𝑥 𝑑𝑥 =
0.2
2
[(1.3862944 + 1.6486586
5.2
4
+ 2(1.4350845 + 1.4816045 + 1.5260563 + 1.5686159 + 1.6094379)]
= 𝟏. 𝟖𝟐𝟕𝟔𝟓𝟓𝟏𝟐

12. Page | 12
ii) Since n = 6, we can use Simpson′
s rule
𝐵𝑦 𝑆𝑖𝑚𝑝𝑠𝑜𝑛′
𝑠 𝑜𝑛𝑒 − 𝑡ℎ𝑖𝑟𝑑 𝑟𝑢𝑙𝑒
𝐼 =
0.2
3
[(1.3862944 + 1.6486586 + 2(1.4816045 + 1.5686159)
+ 4(1.4350845 + 1.5260563)]
= 𝟏. 𝟖𝟐𝟕𝟖𝟒𝟕𝟐𝟒
𝑖𝑖𝑖) 𝐵𝑦 𝑆𝑖𝑚𝑝𝑜𝑛𝑠𝑜𝑛′
𝑠 𝑡ℎ𝑖𝑟𝑑 − 𝑒𝑖𝑔ℎ𝑡ℎ𝑠 𝑟𝑢𝑙𝑒,
𝐼 =
3(0.2)
8
[(1.3862944 + 1.6486586)
+ 3(1.4350845 + 1.4816045 + 1.5686159 + 1.6094379 + 2(1.5260563)]
= 𝟏. 𝟖𝟐𝟕𝟖𝟒𝟕𝟐𝟒
Short Answer:
1. Write down the Newton-Cote’s quadrature formula.
∫ 𝑓(𝑥) 𝑑𝑥 ≈ ℎ [𝑛𝑦0 +
𝑛2
2
∆𝑦0 +
1
2
(
𝑛3
3
−
𝑛2
2
) ∆2
𝑦0 +
1
6
(
𝑛4
4
− 𝑛3
+ 𝑛2
) ∆3
𝑦0 + ⋯ ]
𝑥 𝑛
𝑥0
This equation is called 𝑁𝑒𝑤𝑡𝑜𝑛 − 𝐶𝑜𝑡𝑒′
𝑠 𝑞𝑢𝑎𝑑𝑟𝑎𝑡𝑢𝑟𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎.
2. What is the nature of y (x) in the case of trapezoidal rule?
In trapezoidal rule, y (x) is a linear function of x.
3. State the nature of y (x) and number of intervals in the case of Simpson’s one-third rule?
In Simpson’s one-third rule, y (x) is a polynomial of degree two. To apply this rule n, the number of
intervals must be even.
4. What is the nature of y (x) in the case of Simpson’s three-eighths rule and when it is
applicable?
In Simpson’s third-eighths rule, y (x) is a polynomial of degree three. This rule is applicable if n, the
number of intervals is a multiple of 3.
5. Differentiate between Simpson’s one-third rule and Simpson’s three-eighths rule.
S.No Simpson’s one-third rule Simpson’s three-eighths rule
1 y (x) is a polynomial of degree two y (x) is a polynomial of degree three
2 The number of intervals must be even. The number of intervals is a multiple of 3.