WELCOME TO OUR PRESENTATION

OF NUMERICAL METHOD

GROUP MEMBERS

NAME

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

TOPIC

1.SIMPSON’S 1/3 RULE.

2.WEDDLE’S RULE.

A.K.M. ASADUZZAMAN

ID NO: 142-15-4004

SIMPSON’S 1/3 RULEQUESTION NO-1:

Evaluate correct up to three decimal place Where N=6 by using Simpson’s rule or simply Simpson’s rule.

Solution: Here, Upper limit () =8Lower limit () =2 No. of intervals (n) =6Interval gap =h Now,

h =

h= 1

Simpson’s Rule: I =

I =

I = 1.0988 (Answer)

X 2 3 4 5 6 7 8

Y= 0.3333 0.25 0.2 0.1666 0.1428 0.125 0.1111

KOUSHIK ROY

ID NO: 142-15-4061

SIMPSON’S 1/3 RULEQUESTION NO:2

Evaluate correct up to three decimal place Where N=6 by using Simpson’s rule or simply Simpson’s rule.

Solution: Here, Upper limit () =8Lower limit () =2No. of intervals (n) =6 Interval gap =h Now,

h =

h= 1

Simpson’s Rule: I =

I =

I = 0.970366 (Answer)

X 2 3 4 5 6 7 8

Y= 0.1111 0.0625 0.04 0.0277 0.020 0.015 0.012

MOSTAKIM ISLAM

ID NO: 141-15-3337

SIMPSON’S 1/3 RULEQUESTION NO-3:

Evalute I=dx,correct to three decimal places where,h=0.125 by simpsons Rule.

Solution:We know,xn= x0+nh 1=0+n*0.125 n= n = 8 Now, x0 = 0 , y0 = =1

x1=0.125, y1= = 0.8888 x2=0.25, y2= = 0.8

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

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

I = 0.6931 (Answer)

MD. ZAHID HASAN

ID NO: 142-15-3466

WEDDLE’S RULEQUESTION NO-1:

Evaluate correct up to three decimal place Where N=6 by

using Weddle’s rule.

Solution: Here, Upper limit () =1.75 Lower limit () =0.25 No. of intervals (n) =6 Interval gap =h

Now,

h =

h = 0.25

Weddle’s Rule: It’s a 7-point quadrature formula, I .e n=6

I =

I =

I = 0.7884 (Answer).

X 0.25 0.5 0.75 1.0 1.25 1.5 1.75

Y= 0.8 0.6666 0.5714 0.5 0.4444 0.4 0.3636

MD. ASIF-AL-FAHAD

ID NO: 142-15-3659

WEDDLE’S RULEQUESTION NO-2:

Evaluate correct up to three decimal place Where N=6 by using Weddle’s rule.

Solution: Here, Upper limit () =5.2Lower limit () =4 No. of intervals (n) =6Interval gap =h Now,

h = h h= 0.2

Weddle’s Rule: It’s a 7-point quadrature formula, I .e n=6

I =

I =

I = 1.8277 (Answer)

X 4 4.2 4.4 4.6 4.8 5.0 5.2

Y= 1.3862 1.4350 1.4816 1.5260 1.5686 1.6094 1.6486