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Articles Home » C# » C# Program to Find the Value of 1/(1+x2) using Simpson?s 1/3 Rule

C# Program to Find the Value of 1/(1+x2) using Simpson?s 1/3 Rule

C# Program to Find the Value of 1/(1+x2) using Simpson’s 1/3 Rule

This C# Program Finds Value of 1/(1+x2) using Simpsons 1/3 Rule. Here Simpson’s 1/3 Rule Numerical Integration is used to estimate the value of a definite integral. It works by creating an even number of intervals and fitting a parabola in each pair of intervals.

Here is source code of the C# Program to Find Value of 1/(1+x2) using Simpsons 1/3 Rule. The C# program is successfully compiled and executed with Microsoft Visual Studio. The program output is also shown below.

```
/*
```

 * C# Program to Find Value of 1/(1+x2) using Simpsons 1/3 Rule

```
 */
```
```
using System;
```
```
 
```
```
class simpson
```
```
{
```
```
    float a, b;
```
```
    int n;
```
```
    public void readdata()
```
```
    {
```

        Console.WriteLine("Enter the Lower Limit Value : ");

        a = Convert.ToSingle(Console.ReadLine());

        Console.WriteLine("Enter the Upper Limit Value : ");

        b = Convert.ToSingle(Console.ReadLine());

        Console.WriteLine("Enter the Number of Intervals : ");

        n = Convert.ToInt32(Console.ReadLine());

```
    }
```
```
    public void simp()
```
```
    {
```
```
        int i;
```
```
        float x, sum = 0.0f, h;
```
```
        float[] y = new float[n + 1];
```
```
        h = (b - a) / n;
```
```
        x = a;
```
```
        for (i = 0; i <= n; i++)
```
```
        {
```
```
            y[i] = 1.0f / (1 + x * x);
```
```
            x = x + h;
```
```
        }
```
```
        sum = y[0] + y[n];
```
```
        for (i = 1; i < n - 1; i += 2)
```
```
        {
```

            sum += 4 * y[i]+2* y[i + 1];

```
 
```
```
        }
```
```
        sum = sum * h / 3.0f;
```

        Console.WriteLine("Integral Value : {0} ", sum);

```
        Console.ReadLine();
```
```
    }
```
```
    public static void Main()
```
```
    {
```
```
        simpson obj = new simpson();
```
```
        obj.readdata();
```
```
        obj.simp();
```
```
    }
```
```
}
```

Here is the output of the C# Program:

Enter the Lower Limit Value : 1
Enter the Upper Limit Value : 10
Enter the Number of Intervals : 100
Integral Value : 0.6845199

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