//Simpson's (1/3)rd Rule
deff('y=f(x)','y=x^4')
a=input("Enter Lower Limit: ")
b=input("Enter Upper Limit: ")
n=input("Enter number of sum intervals: ")
h=(b-a)/n
add1=0
add2=0
add3=0
for i=0:n
x=a+i*h
y=f(x)
disp([x y])
if (i==0)|(i==n) then
add1=add1+y
else if (modulo(i,2)==0) then
add2=add2+y
else
add3=add3+y
end
end
end
I=(h/3)*(add1+2*add2+4*add3)
disp(I,"Integration by Simpsons (1/3)rd Rule is:")
Output
-->exec('D:\Scilab prog by me\Simpson's (1-3)rd Rule 1.sce', -1)
Enter Lower Limit: -3
Enter Upper Limit: 3
Enter number of sum intervals: 12
- 3. 81.
- 2.5 39.0625
- 2. 16.
- 1.5 5.0625
- 1. 1.
- 0.5 0.0625
0. 0.
0.5 0.0625
1. 1.
1.5 5.0625
2. 16.
2.5 39.0625
3. 81.
Integration by Simpsons (1/3)rd Rule is:
97.25
deff('y=f(x)','y=x^4')
a=input("Enter Lower Limit: ")
b=input("Enter Upper Limit: ")
n=input("Enter number of sum intervals: ")
h=(b-a)/n
add1=0
add2=0
add3=0
for i=0:n
x=a+i*h
y=f(x)
disp([x y])
if (i==0)|(i==n) then
add1=add1+y
else if (modulo(i,2)==0) then
add2=add2+y
else
add3=add3+y
end
end
end
I=(h/3)*(add1+2*add2+4*add3)
disp(I,"Integration by Simpsons (1/3)rd Rule is:")
Output
-->exec('D:\Scilab prog by me\Simpson's (1-3)rd Rule 1.sce', -1)
Enter Lower Limit: -3
Enter Upper Limit: 3
Enter number of sum intervals: 12
- 3. 81.
- 2.5 39.0625
- 2. 16.
- 1.5 5.0625
- 1. 1.
- 0.5 0.0625
0. 0.
0.5 0.0625
1. 1.
1.5 5.0625
2. 16.
2.5 39.0625
3. 81.
Integration by Simpsons (1/3)rd Rule is:
97.25
thank you!!!!!
ReplyDeletewhat is add1,add2 & add3?
ReplyDeleteI=(h/3)*(add1+2*add2+4*add3+addb)
ReplyDeletehey
ReplyDeletehow can i define
y=1/1+5x