Ex:
Formula:
Formula:
Here x0=0 and x0+nh=6 hence h= (6-0)/4 =1.5
therefor interval is 1.5 in each term
lower bound upper bound
x
|
0
|
1.5
|
3
|
4.5
|
6
|
f(x)
|
1
|
.30
|
.10
|
.04
|
.02
|
Now apply the trapezoidal formula :
= 1.5/2[(1+.02) +2(.30+.10+.04)]
= 1.5/2[1.027+2(.44)]
=1.5/2[1.027+.88)]
=1.5/2[1.907] =1.4
C Code:
#include<stdio.h>
float f(float);
void main()
{ float a,b,h,i,n,k=0;
printf("\nEnter the lower and upper limits for function");
scanf("%f%f",&a,&b);
printf("\nEnter the difference");
scanf("%f",&h);
//Calculate the integral
printf("\nx\tf(x)");
for(i=a;i<=b;i=i+h)
{ printf("\n%f\t%f",i,f(i));
k=k+f(i);
}
k=k-(f(a)+f(b));
k=(h/2.0)*(f(a)+f(b))+h*k;
printf("\nThe definite integral=%f",k);
}
float f(float x)
{ return(1/(1+x*x));
}
Output:
Enter the lower and upper limits for function
0
6
Enter the difference 1.5
x f(x)
0.000000 1.000000
1.500000 0.307692
3.000000 0.100000
4.500000 0.047059
6.000000 0.027027
The definite integral= 1.452397
No comments:
Post a Comment