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why the output is wrong?

–2 votes
asked May 7, 2020 by meghna kattekola (130 points)
#include<math.h>
float f(float x)
{
    return 1/(1+x*x);
}
int main()
{
  float a,b,h,x,sum=0.0;    
  int i,n;
  printf("Enter lower limit and upper limit:\n");
  scanf ("%f%f",&a,&b);
  printf("Enter number of intervals:\n");
  scanf("%d",&n);
  h=(b-a)/n;
  for(i=1;i<=n-1;i++)
    {
      x=a+(i*h);
      if(i%2==0)
      sum=sum+2*f(x);
      else
        sum=sum+4*f(x);
    }
  sum=(h/3)*(sum+f(a)+f(b));
  printf("The value of integral:%f\n",sum);
  return 0;
 }                               //????//

4 Answers

0 votes
answered May 7, 2020 by Abhinav Challa (140 points)

add this line at the top of the code

#include<stdio.h> 

0 votes
answered May 7, 2020 by Raj (150 points)
You have asked a wrong question
0 votes
answered May 8, 2020 by Abhishek Nalinisathiamurthi (190 points)

#include<stdio.h> // <--  a small error
#include<math.h>
float f(float x)
{
    return 1/(1+x*x);
}
int main()
{
  float a,b,h,x,sum=0.0;    
  int i,n;
  printf("Enter lower limit and upper limit:\n");
  scanf ("%f%f",&a,&b);
  printf("Enter number of intervals:\n");
  scanf("%d",&n);
  h=(b-a)/n;
  for(i=1;i<=n-1;i++)
    {
      x=a+(i*h);
      if(i%2==0)
      sum=sum+2*f(x);
      else
        sum=sum+4*f(x);
    }
  sum=(h/3)*(sum+f(a)+f(b));
  printf("The value of integral:%f\n",sum);
  return 0;
 }                               //????//

0 votes
answered May 12, 2020 by LiOS (6,380 points)
#include<stdio.h>
#include<math.h>
float f(float x)
{
    return 1/(1+x*x);
}
int main()
{
  float a,b,h,x,sum=0.0;    
  int i,n;
  printf("Enter lower limit and upper limit:\n");
  scanf ("%f%f",&a,&b);
  printf("Enter number of intervals:\n");
  scanf("%d",&n);
  h=(b-a)/n;
  for(i=1;i<=n-1;i++)
    {
      x=a+(i*h);
      if(i%2==0)
      sum=sum+2*f(x);
      else
        sum=sum+4*f(x);
    }
  sum=(h/3)*(sum+f(a)+f(b));
  printf("The value of integral:%f\n",sum);
  return 0;
 }
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