This document presents a numerical methods group project on Simpson's rule and Weddle's rule. It introduces the group members and their student IDs. It then provides examples of using Simpson's 1/3 rule and Weddle's rule to evaluate definite integrals. For each integral, it shows the setup including limits, intervals, and function values at each point. It then applies the appropriate formula to calculate the integral value to three decimal places.

1. WELCOME TO OUR
PRESENTATION
OF
NUMERICAL METHOD

2. GROUP MEMBERS
NAME
1.A.K.M. ASADUZZAMAN.
2.KOUSHIK ROY.
3.MOSTAKIM ISLAM.
4.MD. ZAHID HASAN.
5.MD. ASIF-AL-FAHAD.

3. TOPIC
1.SIMPSON’S 1/3 RULE.
2.WEDDLE’S RULE.

4. A.K.M. ASADUZZAMAN
ID NO: 142-15-4004

5. SIMPSON’S 1/3 RULE
QUESTION NO-1:
Evaluate 𝟐
𝟖 𝟏
𝟏+𝒙
𝒅𝒙 correct up to
three decimal place Where N=6
by using Simpson’s
𝟏
𝟑
rule or
simply Simpson’s rule.

6. Solution:
Here,
Upper limit (𝑋 𝑛) =8
Lower limit (𝑋0) =2
No. of intervals (n) =6
Interval gap =h
Now,
𝑋 𝑛 = 𝑋0 + 𝑛ℎ
h =
𝑋 𝑛−𝑋0
𝑛
h =
8 − 2
6
h= 1

7. Simpson’s
1
3
Rule:
I =
ℎ
3
𝑦0 + 4(𝑦1 + 𝑦3 + 𝑦5) + 2(𝑦2 + 𝑦4) + 𝑦6]
I = 1
3
0.3333 + 4 0.25 + 0.1666 + 0.125 + 2 0.2 + 0.1428 + ]0.1111
I = 1.0988 (Answer)
X 2 3 4 5 6 7 8
Y=
1
1+𝑥
0.3333 0.25 0.2 0.1666 0.1428 0.125 0.1111

8. KOUSHIK ROY
ID NO: 142-15-4061

9. SIMPSON’S 1/3 RULE
QUESTION NO:2
Evaluate 𝟐
𝟖 𝟏
(𝟏+𝒙) 𝟐 𝒅𝒙 correct up to
three decimal place Where N=6
by using Simpson’s
𝟏
𝟑
rule or
simply Simpson’s rule.

10. Solution:
Here,
Upper limit (𝑋 𝑛) =8
Lower limit (𝑋0) =2
No. of intervals (n) =6
Interval gap =h
Now,
𝑋 𝑛 = 𝑋0 + 𝑛ℎ
h =
𝑋 𝑛−𝑋0
𝑛
h =
8 − 2
6
h= 1

11. Simpson’s
1
3
Rule:
I =
ℎ
3
𝑦0 + 4(𝑦1 + 𝑦3 + 𝑦5) + 2(𝑦2 + 𝑦4) + 𝑦6]
I = 1
3
0.1111 + 4 0.625 + 0.027 + 0.015 + 2 0.04 + 0.020 + ]0.012
I = 0.970366 (Answer)
X 2 3 4 5 6 7 8
Y=
1
(1+𝑥)2
0.1111 0.0625 0.04 0.0277 0.020 0.015 0.012

12. MOSTAKIM ISLAM
ID NO: 141-15-3337

13. SIMPSON’S 1/3 RULE
QUESTION NO-3:
Evalute I= 𝟎
𝟏 𝟏
𝟏+𝒙 dx,correct to
three decimal places
where,h=0.125 by simpsons
Rule.

14. Solution:
We know,
xn= x0+nh
1=0+n*0.125
n=
1−0
𝑜.125
n = 8
Now,
x0 = 0 , y0 =
1
1+0
=1
x1=0.125, y1=
1
1+0.125
= 0.8888
x2=0.25, y2=
1
1+0.25
= 0.8

15. x3=0.375, y3=
1
1+0.5
= 0.7272
x4=0.5, y4=
1
1+0.5
= 0.6666
x5=0.625, y5=
1
1+0.625
= 0.6153
x6=0.75 , y6=
1
1+75
= 0.5714
x7=0.875, y7=
1
1+0.875
= 0.5333
x8=1, y8=
1
1+1
= 0.5

16. Simpsons 1/3 Rule :
I=
ℎ
3
y0+4(y1+y3+y5+y7)+2(y2+y4+y6)+y8]
I = 0.125
3
[1+4(0.8888+0.7272+0.6153+0.5333)+2(0.8+0.6666+0.571)+0.5]
I = 0.6931 (Answer)

17. MD. ZAHID HASAN
ID NO: 142-15-3466

18. WEDDLE’S RULE
QUESTION NO-1:
Evaluate 𝟎.𝟐𝟓
𝟏.𝟕𝟓 𝟏
𝟏+𝒙
𝒅𝒙 correct up to
three decimal place Where N=6
by using Weddle’s rule.

19. Solution:
Here,
Upper limit (𝑋 𝑛) =1.75
Lower limit (𝑋0) =0.25
No. of intervals (n) =6
Interval gap =h
Now,
𝑋 𝑛= 𝑋0 + 𝑛ℎ
h =
𝑋 𝑛−𝑋0
𝑛
h =
1.75 − 0.25
6
h = 0.25

20. Weddle’s Rule: It’s a 7-point quadrature formula, I .e n=6
I =
3ℎ
10
𝑦0 + 5𝑦1 + 𝑦2 + 6𝑦3 + 𝑦5 + 5𝑦5 + 𝑦6]
I = 3∗0.25
10
0.8 + 5 ∗ 0.6666 + 0.5714 + 6 ∗ 0.5 + 0.4444 + 5 ∗ 0.4 + ]0.3636
I = 0.7884 (Answer).
X 0.25 0.5 0.75 1.0 1.25 1.5 1.75
Y=
1
1+𝑥
0.8 0.6666 0.5714 0.5 0.4444 0.4 0.3636

21. MD. ASIF-AL-FAHAD
ID NO: 142-15-3659

22. WEDDLE’S RULE
QUESTION NO-2:
Evaluate 𝟒
𝟓.𝟐
𝒍𝒏𝒙 𝒅𝒙 correct up to
three decimal place Where N=6
by using Weddle’s rule.

23. Solution:
Here,
Upper limit (𝑋 𝑛) =5.2
Lower limit (𝑋0) =4
No. of intervals (n) =6
Interval gap =h
Now,
𝑋 𝑛 = 𝑋0 + 𝑛ℎ
h =
𝑋 𝑛−𝑋0
𝑛
h =
5.2−4
6
h= 0.2

24. Weddle’s Rule: It’s a 7-point quadrature formula, I .e n=6
I =
3ℎ
10
𝑦0 + 5𝑦1 + 𝑦2 + 6𝑦3 + 𝑦5 + 5𝑦5 + 𝑦6]
I = 3∗0.2
10
1.3862 + 5 ∗ 1.4350 + 1.4816 + 6 ∗ 1.5260 + 1.5686 + 5 ∗ 1.6094 + ]1.6486
I = 1.8277 (Answer)
X 4 4.2 4.4 4.6 4.8 5.0 5.2
Y= 𝑙𝑛𝑥 1.3862 1.4350 1.4816 1.5260 1.5686 1.6094 1.6486