Numerical Integration using Simpson's 1/3 Rule

Program in Fortran 95

===================================================================================
!Program to find Integral of a function using Simpson's 1/3 rule
!LANGUAGE :: FORTRAN 95
!Compiler :: GNU Fortran
!===================================================================================
PROGRAM simpson1_3rd
    INTEGER j,n
    REAL x,upper,lower,I,summ,h
    PRINT*,'PROGRAM TO FIND THE INTEGRAL OF 1/(1+X) BY SIMPSON''S 1/3 METHOD'
    PRINT*,('-',j=1,65)
    PRINT*,'Enter the lower limit'
    READ*,lower
    PRINT*,'Enter the upper limit'
    READ*,upper
    PRINT*,'Enter the no of intervals'
    READ*,n
    h=(upper-lower)/n
    x=lower
    summ=0
    WRITE(*,10)
10  FORMAT (8X,"x",18x,"f(x)")
    WRITE(*,*)('-',j=1,32)
    DO j=0,n
        IF ((j.EQ.0).OR.(j.EQ.n)) THEN
            summ=summ+func(x)
        ELSE IF (MOD(j,2).NE.0) THEN
            summ=summ+(4*func(x))
        ELSE IF (MOD(j,2).EQ.0) THEN
            summ=summ+(2*func(x))
        END IF
        PRINT*,x,'  ',func(x)
        x=x+h
    END DO
    I=h*summ/3
    PRINT*,'Integral is : ',I
    STOP
END PROGRAM
!------------------FUNCTION SUBPROGRAM----------------
    REAL FUNCTION func(x1)
    func=1/(1+x1)
    RETURN
    END

 ------------------------------OUTPUT-----------------------------
  PROGRAM TO FIND THE INTEGRAL OF 1/(1+X) BY SIMPSON'S 1/3 METHOD
 -----------------------------------------------------------------
 Enter the lower limit
0
 Enter the upper limit
1
 Enter the no of intervals
10
        x                  f(x)
 --------------------------------
   0.00000000          1.00000000
  0.100000001         0.909090936
  0.200000003         0.833333313
  0.300000012         0.769230783
  0.400000006         0.714285731
  0.500000000         0.666666687
  0.600000024         0.625000000
  0.700000048         0.588235259
  0.800000072         0.555555522
  0.900000095         0.526315749
   1.00000012         0.499999970
 Integral is :   0.693150222
  

Program in C

==================================================================================================
Program to find Integral of a function using Simpson's 1/3 rule 
LANGUAGE :: C
Compiler :: GNU GCC
Program By:: G.R. Mohanty , www.numericalmethods.in
==================================================================================================


#include <stdio.h>

float func(float x) {
    return 1.0/(1.0 + x);
}

int main() {
    int j, n;
    float x, upper, lower, I, summ, h;

    printf("PROGRAM TO FIND THE INTEGRAL OF 1/(1+X) BY SIMPSON'S 1/3 METHOD\n");
    for (j = 1; j <= 65; j++) {
        printf("-");
    }
    printf("\n");
    printf("Enter the lower limit: ");
    scanf("%f", &lower);
    printf("Enter the upper limit: ");
    scanf("%f", &upper);
    printf("Enter the number of intervals: ");
    scanf("%d", &n);
    h = (upper - lower)/n;
    x = lower;
    summ = 0.0;
    
    /*Simpson's 1/3 Rule Calculation: Uses a loop to apply Simpson's 1/3 rule. 
    Depending on whether j is at the boundaries (j == 0 or j == n) or is odd/even, 
    it adds func(x) multiplied by appropriate coefficients (1, 4, or 2) to summ.*/
    
    printf("\nx           f(x)\n");
    for (j = 0; j <= n; j++) {
        if (j == 0 || j == n) {
            summ += func(x);
        } else if (j % 2 != 0) {
            summ += 4*func(x);
        } else {
            summ += 2*func(x);
        }
        printf("%8.6f   %8.6f\n", x, func(x));
        x += h;
    }
    I = h*summ/3.0;
    printf("\nIntegral is : %8.6f\n", I);
    return 0;
}

========================= OUTPUT ============================
PROGRAM TO FIND THE INTEGRAL OF 1/(1+X) BY SIMPSON'S 1/3 METHOD
-----------------------------------------------------------------
Enter the lower limit: 0
Enter the upper limit: 1
Enter the number of intervals: 10

x           f(x)
0.000000   1.000000
0.100000   0.909091
0.200000   0.833333
0.300000   0.769231
0.400000   0.714286
0.500000   0.666667
0.600000   0.625000
0.700000   0.588235
0.800000   0.555556
0.900000   0.526316
1.000000   0.500000

Integral is : 0.693252

Program in Scilab

======================================================================
 Program to find Integral of a function using Simpson's 1/3 rule										
 LANGUAGE  : SCILAB 
 Program By:: G.R. Mohanty , www.numericalmethods.in
=======================================================================

function f = func(x)
    f = 1/(1+x);
endfunction

printf("PROGRAM TO FIND THE INTEGRAL OF 1/(1+X) BY SIMPSON''S 1/3 METHOD\n");
printf("----------------------------------------------------------------");
lower = input("Enter the lower limit:");
upper = input("Enter the upper limit:");
n = input("Enter the number of intervals:");
h = (upper-lower)/n;
x = lower;
summ = 0.0;


disp("x       f(x)");
for j = 0:n
    if (j == 0 | j == n) then
        summ = summ + func(x);
    elseif modulo(j,2) <> 0 then
        summ = summ + 4*func(x);
    else
        summ = summ + 2*func(x);
    end
    printf("%8.6f %8.6f\n", x, func(x));
    x = x + h;
end

I = h*summ/3.0;
printf("Integral is :%f ",I);

========================= OUTPUT ============================
PROGRAM TO FIND THE INTEGRAL OF 1/(1+X) BY SIMPSON'S 1/3 METHOD
----------------------------------------------------------------
Enter the lower limit:0

Enter the upper limit:1

Enter the number of intervals:10


  "x       f(x)"
0.000000 1.000000
0.100000 0.909091
0.200000 0.833333
0.300000 0.769231
0.400000 0.714286
0.500000 0.666667
0.600000 0.625000
0.700000 0.588235
0.800000 0.555556
0.900000 0.526316
1.000000 0.500000

Integral is :0.693150