Approximate the integral [0 10] using Simpson’s Rule N = 10

`Solution:`

```
public class Integral {
public static void main(String[] args) {
float a = 0.0f; // lower Bound
float b = 10.0f; // high Bound
int N = 10; // define N
float x[] = new float[N+1]; // reservation for last and first value
//calculate x_i
x[0] = a;
x[N] = b;
float solution = x[0]*x[0] + x[N]*x[N];
float function = 1;
float h = (b-a)/(N*3);
for(int i = 1;i<=x.length-2;i++) {
x[i] = (a + i*(b-a)/N) * (a + i*(b-a)/N) ; // function of x^2 = x*x
if(i%2 == 0) {
solution += 2*x[i];
}else {
solution += 4*x[i];
}
//System.out.println(x[i]);
}
System.out.printf("%.2f",solution*h); // In conclusion we make an output
}
}
```

Solution: 333,33

` `

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