**Problem:** Here we have to find integration for the (1/1+x*x)dx with lower limit =0 to upper limit = 6

**Algorithm:**

Step 1: input a,b,number of interval n

Step 2:h=(b-a)/n

Step 3:If(n%6==0)

Then , sum=sum+((3*h/10)*(y(a)+y(a+2*h)+5*y(a+h)+6*y(a+3*h)+y(a+4*h)+5*y(a+5*h)+y(a+6*h)));

a=a+6*h

and Weddle’s rule is applicable then go to step 6

Step 4: else, Weddle’s rule is not applicable

Step 5:Display output

**Code:**

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#include<stdio.h> float y(float x){ return 1/(1+x*x); //function of which integration is to be calculated } int main(){ float a,b,h,sum; int i,n,m; printf("Enter a=x0(lower limit), b=xn(upper limit), number of subintervals: "); scanf("%f%f%d",&a,&b,&n); h = (b-a)/n; sum=0; if(n%6==0){ sum=sum+((3*h/10)*(y(a)+y(a+2*h)+5*y(a+h)+6*y(a+3*h)+y(a+4*h)+5*y(a+5*h)+y(a+6*h))); a=a+6*h; printf("Value of integral is %f\n", sum); } else{ printf("Sorry ! Weddle rule is not applicable"); } } |

** Output:**