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Write functions to calculate the definite integral of a function (f(x)) using rectangular (left, right, and midpoint), trapezium, and Simpson's methods. Your functions should take in the upper and lower bounds (a and b) and the number of approximations to make in that range (n). Assume that your example already has a function that gives values for f(x).

Simpson's method is defined by the following pseudocode:

h := (b - a) / n
sum1 := f(a + h/2)
sum2 := 0

loop on i from 1 to (n - 1)
    sum1 := sum1 + f(a + h * i + h/2)
    sum2 := sum2 + f(a + h * i)

answer := (h / 6) * (f(a) + f(b) + 4*sum1 + 2*sum2)

What it will be doing is computing integrals for multiple quadratics (like the one shown in Wikipedia) and summing them.

See also

Active object for integrating a function of real time.

ActionScript

Integration functions: <lang ActionScript>function leftRect(f:Function, a:Number, b:Number, n:uint):Number { var sum:Number = 0; var dx:Number = (b-a)/n; for (var x:Number = a; n > 0; n--, x += dx) sum += f(x); return sum * dx; }

function rightRect(f:Function, a:Number, b:Number, n:uint):Number { var sum:Number = 0; var dx:Number = (b-a)/n; for (var x:Number = a + dx; n > 0; n--, x += dx) sum += f(x); return sum * dx; }

Ada

This solution creates a generic package into which the function F(X) is passed during generic instantiation. The first part is the package specification. The second part is the package body.

<lang ada>generic

  with function F(X : Long_Float) return Long_Float;

package Integrate is

  function  Left_Rect(A, B, N : Long_Float) return Long_Float;
  function Right_Rect(A, B, N : Long_Float) return Long_Float;
  function   Mid_Rect(A, B, N : Long_Float) return Long_Float;
  function  Trapezium(A, B, N : Long_Float) return Long_Float;
  function    Simpson(A, B, N : Long_Float) return Long_Float;

end Integrate;

package body Integrate is

  ---------------
  -- Left_Rect --
  ---------------

  function Left_Rect (A, B, N : Long_Float) return Long_Float is
     H : constant Long_Float := (B - A) / N;
     Sum : Long_Float := 0.0;
     X : Long_Float := A;
  begin
     while X <= B - H loop
        Sum := Sum + (H * F(X));
        X := X + H;
     end loop;
     return Sum;
  end Left_Rect;

  ----------------
  -- Right_Rect --
  ----------------

  function Right_Rect (A, B, N : Long_Float) return Long_Float is
     H : constant Long_Float := (B - A) / N;
     Sum : Long_Float := 0.0;
     X : Long_Float := A + H;
  begin
     while X <= B - H loop
        Sum := Sum + (H * F(X));
        X := X + H;
     end loop;
     return Sum;
  end Right_Rect;

  --------------
  -- Mid_Rect --
  --------------

  function Mid_Rect (A, B, N : Long_Float) return Long_Float is
     H : constant Long_Float := (B - A) / N;
     Sum : Long_Float := 0.0;
     X : Long_Float := A;
  begin
     while X <= B - H loop
        Sum := Sum + (H / 2.0) * (F(X) + F(X + H));
        X := X + H;
     end loop;
     return Sum;
  end Mid_Rect;

  ---------------
  -- Trapezium --
  ---------------

  function Trapezium (A, B, N : Long_Float) return Long_Float is
     H : constant Long_Float := (B - A) / N;
     Sum : Long_Float := F(A) + F(B);
     X : Long_Float := 1.0;
  begin
     while X <= N - 1.0 loop
        Sum := Sum + 2.0 * F(A + X * (B - A) / N);
        X := X + 1.0;
     end loop;
     return (B - A) / (2.0 * N) * Sum;
  end Trapezium;

  -------------
  -- Simpson --
  -------------

  function Simpson (A, B, N : Long_Float) return Long_Float is
     H : constant Long_Float := (B - A) / N;
     Sum1 : Long_Float := 0.0;
     Sum2 : Long_Float := 0.0;
     Limit : Integer := Integer(N) - 1;
  begin
     for I in 0..Limit loop
        Sum1 := Sum1 + F(A + H * Long_Float(I) + H / 2.0);
     end loop;
     for I in 1..Limit loop
        Sum2 := Sum2 + F(A + H * Long_Float(I));
     end loop;
     return H / 6.0 * (F(A) + F(B) + 4.0 * Sum1 + 2.0 * Sum2);
  end Simpson;

end Integrate;</lang>

ALGOL 68

<lang algol68>MODE F = PROC(LONG REAL)LONG REAL;

2. left rect ##

PROC left rect = (F f, LONG REAL a, b, INT n) LONG REAL: BEGIN

  LONG REAL h= (b - a) / n;
  LONG REAL sum:= 0;
  LONG REAL x:= a;
  WHILE x <= b - h DO
     sum := sum + (h * f(x));
     x +:= h
  OD;
  sum

END # left rect #;

2. right rect ##

PROC right rect = (F f, LONG REAL a, b, INT n) LONG REAL: BEGIN

  LONG REAL h= (b - a) / n;
  LONG REAL sum:= 0;
  LONG REAL x:= a + h;
  WHILE x <= b - h DO
     sum := sum + (h * f(x));
     x +:= h
  OD;
  sum

END # right rect #;

2. mid rect ##

PROC mid rect = (F f, LONG REAL a, b, INT n) LONG REAL: BEGIN

  LONG REAL h= (b - a) / n;
  LONG REAL sum:= 0;
  LONG REAL x:= a;
  WHILE x <= b - h DO
     sum := sum + (h / 2) * (f(x) + f(x + h));
     x +:= h
  OD;
  sum

END # mid rect #;

2. trapezium ##

PROC trapezium = (F f, LONG REAL a, b, INT n) LONG REAL: BEGIN

  LONG REAL h= (b - a) / n;
  LONG REAL sum:= f(a) + f(b);
  LONG REAL x:= 1;
  WHILE x <= n - 1 DO
     sum := sum + 2 * f(a + x * h );
     x +:= 1
  OD;
  (b - a) / (2 * n) * sum

END # trapezium #;

2. simpson ##

PROC simpson = (F f, LONG REAL a, b, INT n) LONG REAL: BEGIN

  LONG REAL h= (b - a) / n;
  LONG REAL sum1:= 0;
  LONG REAL sum2:= 0;
  INT limit:= n - 1;
  FOR i FROM 0 TO limit DO
     sum1 := sum1 + f(a + h * LONG REAL(i) + h / 2)
  OD;
  FOR i FROM 1 TO limit DO
     sum2 +:= f(a + h * LONG REAL(i))
  OD;
  h / 6 * (f(a) + f(b) + 4 * sum1 + 2 * sum2)

END # simpson #; SKIP</lang>

AutoHotkey

ahk discussion <lang autohotkey>MsgBox % Rect("fun", 0, 1, 10,-1) ; 0.45 left MsgBox % Rect("fun", 0, 1, 10) ; 0.50 mid MsgBox % Rect("fun", 0, 1, 10, 1) ; 0.55 right MsgBox % Trapez("fun", 0, 1, 10) ; 0.50 MsgBox % Simpson("fun", 0, 1, 10) ; 0.50

Rect(f,a,b,n,side=0) { ; side: -1=left, 0=midpoint, 1=right

  h := (b - a) / n
  sum := 0, a += (side-1)*h/2
  Loop %n%
     sum += %f%(a + h*A_Index)
  Return h*sum

}

Trapez(f,a,b,n) {

  h := (b - a) / n
  sum := 0
  Loop % n-1
     sum += %f%(a + h*A_Index)
  Return h/2 * (%f%(a) + %f%(b) + 2*sum)

}

Simpson(f,a,b,n) {

  h := (b - a) / n
  sum1 := sum2 := 0, ah := a - h/2
  Loop %n%
     sum1 += %f%(ah + h*A_Index)
  Loop % n-1
     sum2 += %f%(a + h*A_Index)
  Return h/6 * (%f%(a) + %f%(b) + 4*sum1 + 2*sum2)

}

fun(x) { ; linear test function

  Return x

}</lang>

BASIC

Works with: QuickBasic version 4.5

Translation of: Java

<lang qbasic>FUNCTION leftRect(a, b, n)

    h = (b - a) / n
    sum = 0
    FOR x = a TO b - h STEP h
       sum = sum + h * (f(x))
    NEXT x
    leftRect = sum

END FUNCTION

FUNCTION rightRect(a, b, n)

    h = (b - a) / n
    sum = 0
    FOR x = a + h TO b - h STEP h
       sum = sum + h * (f(x))
    NEXT x
    rightRect = sum

END FUNCTION

FUNCTION midRect(a, b, n)

    h = (b - a) / n
    sum = 0
    FOR x = a TO b - h STEP h
       sum = sum + (h / 2) * (f(x) + f(x + h))
    NEXT x
    midRect = sum

END FUNCTION

FUNCTION trap(a, b, n)

    h = (b - a) / n
    sum = f(a) + f(b)
    FOR i = 1 TO n-1
       sum = sum + 2 * f((a + i * h))
    NEXT i
    trap = h / 2 * sum

END FUNCTION

FUNCTION simpson(a, b, n)

    h = (b - a) / n
    sum1 = 0
    sum2 = 0

    FOR i = 0 TO n-1
       sum1 = sum + f(a + h * i + h / 2)
    NEXT i

    FOR i = 1 TO n - 1
       sum2 = sum2 + f(a + h * i)
    NEXT i

    simpson = h / 6 * (f(a) + f(b) + 4 * sum1 + 2 * sum2)

END FUNCTION</lang>

C

Translation of: Java

<lang c>#include <math.h>

double int_leftrect(double from, double to, double n, double (*func)()) {

  double h = (to-from)/n;
  double sum = 0, x;
  for(x=from; x <= (to-h); x += h)
     sum += func(x);
  return h*sum;

}

double int_rightrect(double from, double to, double n, double (*func)()) {

  double h = (to-from)/n;
  double sum = 0, x;
  for(x=from+h; x <= (to-h); x += h)
     sum += func(x);
  return h*sum;

}

double int_midrect(double from, double to, double n, double (*func)()) {

  double h = (to-from)/n;
  double sum = 0.0, x;
  for(x=from; x <= (to-h); x += h)
     sum += (func(x) + func(x + h));
  return h*sum/2.0;

}

double int_trapezium(double from, double to, double n, double (*func)()) {

  double h = (to - from) / n;
  double sum = func(from) + func(to);
  int i;
  for(i = 1;i < n;i++)
      sum += func(from + i * h);
  return  h * sum;

}

double int_simpson(double from, double to, double n, double (*func)()) {

  double h = (to - from) / n;
  double sum1 = 0.0;
  double sum2 = 0.0;
  int i;

  for(i = 0;i < n;i++)
     sum1 += func(from + h * i + h / 2.0);

  for(i = 1;i < n;i++)
     sum2 += func(from + h * i);

  return h / 6.0 * (func(from) + func(to) + 4.0 * sum1 + 2.0 * sum2);

}</lang>

Usage example:

<lang c>#include <stdio.h>

include <stdlib.h>

double f1(double x) {

  return 2.0*x*log(x*x+1.0);

}

double f1a(double x) {

  double y = x*x+1;
  return y*(log(y)-1.0);

}

double f2(double x) {

 return cos(x);

}

double f2a(double x) {

 return sin(x);

}

typedef double (*pfunc)(double, double, double, double (*)()); typedef double (*rfunc)(double);

define INTG(F,A,B) (F((B))-F((A)))

int main() {

    int i,j;
    double ic;
    
    pfunc f[5] = { &int_leftrect, &int_rightrect,
                        &int_midrect,  &int_trapezium,
                        &int_simpson };
    const char *names[5] = {"leftrect", "rightrect", "midrect",
                      "trapezium", "simpson" };
    rfunc rf[2] = { &f1, &f2 };
    rfunc If[2] = { &f1a, &f2a };
    
    for(j=0; j<2; j++)
    {
      for(i=0; i < 5 ; i++)
      {
        ic = (*f[i])(0.0, 1.0, 100.0, rf[j]);
        printf("%10s [ 0,1] num: %+lf, an: %lf\n",
               names[i], ic, INTG((*If[j]), 0.0, 1.0));
      }
      printf("\n");
    }

}</lang>

Output of the test (with a little bit of enhancing by hand):

  leftrect [ 0,1] num: +0.365865, an: 0.386294
 rightrect [ 0,1] num: +0.365865
   midrect [ 0,1] num: +0.372628
 trapezium [ 0,1] num: +0.393254
   simpson [ 0,1] num: +0.386294

  leftrect [ 0,1] num: +0.838276, an: 0.841471
 rightrect [ 0,1] num: +0.828276
   midrect [ 0,1] num: +0.836019
 trapezium [ 0,1] num: +0.849165
   simpson [ 0,1] num: +0.841471

C++

Due to their similarity, it makes sense to make the integration method a policy. <lang cpp>// the integration routine template<typename Method, typename F, typename Float>

double integrate(F f, Float a, Float b, int steps, Method m)

{

 double s = 0;
 double h = (b-a)/steps;
 for (int i = 0; i < steps; ++i)
   s += m(f, a + h*i, h);
 return h*s;

}

// methods class rectangular { public:

 enum position_type { left, middle, right };
 rectangular(position_type pos): position(pos) {}
 template<typename F, typename Float>
  double operator()(F f, Float x, Float h)
 {
   switch(position)
   {
   case left:
     return f(x);
   case middle:
     return f(x+h/2);
   case right:
     return f(x+h);
   }
 }

private:

 position_type position;

};

class trapezium { public:

 template<typename F, typename Float>
  double operator()(F f, Float x, Float h)
 {
   return (f(x) + f(x+h))/2;
 }

};

class simpson { public:

 template<typename F, typename Float>
  double operator()(F f, Float x, Float h)
 {
   return (f(x) + 4*f(x+h/2) + f(x+h))/6;
 }

};

// sample usage double f(double x) { return x*x; )

// inside a function somewhere: double rl = integrate(f, 0.0, 1.0, 10, rectangular(rectangular::left)); double rm = integrate(f, 0.0, 1.0, 10, rectangular(rectangular::middle)); double rr = integrate(f, 0.0, 1.0, 10, rectangular(rectangular::right)); double t = integrate(f, 0.0, 1.0, 10, trapezium()); double s = integrate(f, 0.0, 1.0, 10, simpson());</lang>

Common Lisp

<lang lisp>(defun left-rectangle (f a b n &aux (d (/ (- b a) n)))

 (* d (loop for x from a below b by d summing (funcall f x))))

(defun right-rectangle (f a b n &aux (d (/ (- b a) n)))

 (* d (loop for x from b above a by d summing (funcall f x))))

(defun midpoint-rectangle (f a b n &aux (d (/ (- b a) n)))

 (* d (loop for x from (+ a (/ d 2)) below b by d summing (funcall f x))))

(defun trapezium (f a b n &aux (d (/ (- b a) n)))

 (* (/ d 2)
    (+ (funcall f a)
       (* 2 (loop for x from (+ a d) below b by d summing (funcall f x)))
       (funcall f b))))

(defun simpson (f a b n)

 (loop with h = (/ (- b a) n)
       with sum1 = (funcall f (+ a (/ h 2)))
       with sum2 = 0
       for i from 1 below n
       do (incf sum1 (funcall f (+ a (* h i) (/ h 2))))
       do (incf sum2 (funcall f (+ a (* h i))))
       finally (return (* (/ h 6)
                          (+ (funcall f a)
                             (funcall f b)
                             (* 4 sum1)
                             (* 2 sum2))))))</lang>

D

The function f(x) is Func1/Func2 in this program. <lang d>module integrate ; import std.stdio ; import std.math ;

alias real function(real) realFn ;

class Sigma{

 int n ;
 real a, b , h ;
 realFn fn ; 
 string desc ;
 enum Method {LEFT = 0, RGHT, MIDD, TRAP, SIMP} ;
 static string[5] methodName = 
 ["LeftRect ", "RightRect", "MidRect  ", "Trapezium", "Simpson  "] ;
 static Sigma opCall() { 
   Sigma s = new Sigma() ; 
   return s ; 
 }
 static Sigma opCall(realFn f, int n, real a, real b) { 
   Sigma s = new Sigma(f, n, a, b) ; 
   return s ; 
 } 
 static real opCall(realFn f, int n, real a, real b, Method m) { 
   return Sigma(f,n,a,b).getSum(m) ;
 }
 private this() {} ;
 this(realFn f, int n, real a, real b) {
   setFunction(f) ;
   setStep(n) ;
   setRange(a,b) ;   
 } 
 Sigma opCall(Method m) { 
   return doSum(m) ; 
 }
 Sigma setFunction(realFn f) { 
   this.fn = f ; 
   return this ; 
 }
 Sigma setRange(real a, real b) { 
   this.a = a ; this.b = b ; 
   setInterval() ;
   return this ; 
 }
 Sigma setStep(int n) { 
   this.n = n ; 
   setInterval() ;
   return this ; 
 }
 Sigma setDesc(string d) { 
   this.desc = d ; 
   return this ; 
 }
 private void setInterval() { 
   this.h = (b - a) / n ; 
 }
 private real partSum(int i, Method m) {
   real x = a + h * i ;
   switch(m) {
     case Method.LEFT:
       return fn(x) ;
     case Method.RGHT:
       return fn(x + h) ;
     case Method.MIDD:
       return fn(x + h/2) ;
     case Method.TRAP:
       return (fn(x) + fn(x + h))/2 ;
     default:
   }
   //case SIMPSON:
   return (fn(x) + 4 * fn(x + h/2) + fn(x + h))/6 ;
 } 
 real getSum(Method m) {
   real sum = 0 ;
   for(int i = 0; i < n ; i++)
     sum += partSum(i, m) ;
   return sum * h ;
 } 
 Sigma doSum(Method m) {
   writefln("%10s = %9.6f", methodName[m], getSum(m)) ; 
   return this ; 
 }
 Sigma showSetting() {
   writefln("\n%s\nA = %9.6f, B = %9.6f, N = %s, h = %s", desc, a, b, n, h) ;
   return this ;
 } 
 Sigma doLeft() { return doSum(Method.LEFT) ; }
 Sigma doRight() { return doSum(Method.RGHT) ; }
 Sigma doMid() { return doSum(Method.MIDD) ; }
 Sigma doTrap() { return doSum(Method.TRAP) ; }
 Sigma doSimp() { return doSum(Method.SIMP) ; }
 Sigma doAll() {
   showSetting() ;
   doLeft() ; doRight() ; doMid() ; doTrap() ; doSimp() ;
   return this ; 
 }

} real Func1(real x) {

 return cos(x) + sin(x)  ;

} real Func2(real x) {

 return 2.0L/(1 + 4*x*x)  ;

} void main() {

 // use as a re-usable and manageable object
 Sigma s = Sigma(&Func1, 10, -PI/2, PI/2).showSetting()
 .doLeft().doRight().doMid().doTrap()(Sigma.Method.SIMP) ;     
 s.setFunction(&Func2).setStep(4).setRange(-1.0L,2.0L) ; 
 s.setDesc("Function(x) = 2 / (1 + 4x^2)").doAll() ;
 // use as a single function call 
 writefln("\nLeftRect Integral of FUNC2 =") ;
 writefln("%12.9f (%3dstep)\n%12.9f (%3dstep)\n%12.9f (%3dstep).",
   Sigma(&Func2,   8, -1.0L, 2.0L, Sigma.Method.LEFT),  8,
   Sigma(&Func2,  64, -1.0L, 2.0L, Sigma.Method.LEFT), 64,
   Sigma(&Func2, 512, -1.0L, 2.0L, Sigma.Method.LEFT),512) ;
 writefln("\nSimpson Integral of FUNC2 =") ;
 writefln("%12.9f (%3dstep).",
   Sigma(&Func2, 512, -1.0L, 2.0L, Sigma.Method.SIMP),512) ;

}</lang> Parts of the output:

Function(x) = 2 / (1 + 4x^2)
A = -1.000000, B =  2.000000, N = 4, h = 0.75
 LeftRect  =  2.456897
 RightRect =  2.245132
 MidRect   =  2.496091
 Trapezium =  2.351014
 Simpson   =  2.447732

One can verifies his result with this applet

E

Translation of: Python

<lang e>pragma.enable("accumulator")

def leftRect(f, x, h) {

 return f(x)

}

def midRect(f, x, h) {

 return f(x + h/2)

}

def rightRect(f, x, h) {

 return f(x + h)

}

def trapezium(f, x, h) {

 return (f(x) + f(x+h)) / 2

}

def simpson(f, x, h) {

 return (f(x) + 4 * f(x + h / 2) + f(x+h)) / 6

}

def integrate(f, a, b, steps, meth) {

  def h := (b-a) / steps
  return h * accum 0 for i in 0..!steps { _ + meth(f, a+i*h, h) }

}</lang>

<lang e>? integrate(fn x { x ** 2 }, 3.0, 7.0, 30, simpson)

value: 105.33333333333334

? integrate(fn x { x ** 9 }, 0, 1, 300, simpson)

value: 0.10000000002160479</lang>

Forth

<lang forth>fvariable step

defer method ( fn F: x -- fn[x] )

left execute ;

right step f@ f+ execute ;

mid step f@ 2e f/ f+ execute ;

trap

 dup fdup  left
     fswap right f+  2e f/ ;

simpson

 dup fdup  left
 dup fover mid 4e f* f+
     fswap right f+  6e f/ ;

set-step ( n F: a b -- n F: a )

 fover f- dup 0 d>f f/ step f! ;

integrate ( xt n F: a b -- F: sigma )

 set-step
 0e
 0 do
   dup fover method f+
   fswap step f@ f+ fswap
 loop
 drop fnip
 step f@ f* ;
\ testing similar to the D example

test

 ' is method  ' 4 -1e 2e integrate f. ;

fn1 fsincos f+ ;

fn2 fdup f* 4e f* 1e f+ 2e fswap f/ ;

7 set-precision test left fn2 \ 2.456897 test right fn2 \ 2.245132 test mid fn2 \ 2.496091 test trap fn2 \ 2.351014 test simpson fn2 \ 2.447732</lang>

Fortran

In ISO Fortran 95 and later if function f() is not already defined to be "elemental", define an elemental wrapper function around it to allow for array-based initialization: <lang fortran>elemental function elemf(x)

  real :: elemf, x
  elemf = f(x)

end function elemf</lang>

Use Array Initializers, Pointers, Array invocation of Elemental functions, Elemental array-array and array-scalar arithmetic, and the SUM intrinsic function. Methods are collected into a single function in a module. <lang fortran>module Integration

 implicit none

contains

 ! function, lower limit, upper limit, steps, method
 function integrate(f, a, b, in, method)
   real :: integrate
   real, intent(in) :: a, b
   integer, optional, intent(in) :: in
   character(len=*), intent(in), optional :: method
   interface
      elemental function f(ra)
        real :: f
        real, intent(in) :: ra
      end function f
   end interface

   integer :: n, i, m
   real :: h
   real, dimension(:), allocatable :: xpoints
   real, dimension(:), target, allocatable :: fpoints
   real, dimension(:), pointer :: fleft, fmid, fright

   if ( present(in) ) then
      n = in
   else
      n = 20
   end if

   if ( present(method) ) then
      select case (method)
      case ('leftrect')
         m = 1
      case ('midrect')
         m = 2
      case ('rightrect')
         m = 3
      case ( 'trapezoid' )
         m = 4
      case default
         m = 0
      end select
   else
      m = 0
   end if

   h = (b - a) / n

   allocate(xpoints(0:2*n), fpoints(0:2*n))

   xpoints = (/ (a + h*i/2, i = 0,2*n) /)

   fpoints = f(xpoints)
   fleft  => fpoints(0 : 2*n-2 : 2)
   fmid   => fpoints(1 : 2*n-1 : 2)
   fright => fpoints(2 : 2*n   : 2)

   select case (m)
   case (0) ! simpson
      integrate = h / 6.0 * sum(fleft + fright + 4.0*fmid)
   case (1) ! leftrect
      integrate = h * sum(fleft)
   case (2) ! midrect
      integrate = h * sum(fmid)
   case (3) ! rightrect
      integrate = h * sum(fright)
   case (4) ! trapezoid
      integrate = h * sum(fleft + fright) / 2
   end select

   deallocate(xpoints, fpoints)
 end function integrate

end module Integration</lang>

Usage example: <lang fortran>program IntegrationTest

 use Integration
 use FunctionHolder
 implicit none

 print *, integrate(afun, 0., 3**(1/3.), method='simpson')
 print *, integrate(afun, 0., 3**(1/3.), method='leftrect')
 print *, integrate(afun, 0., 3**(1/3.), method='midrect')  
 print *, integrate(afun, 0., 3**(1/3.), method='rightrect')
 print *, integrate(afun, 0., 3**(1/3.), method='trapezoid')

end program IntegrationTest</lang>

The FunctionHolder module:

<lang fortran>module FunctionHolder

 implicit none

contains

 pure function afun(x)
   real :: afun
   real, intent(in) :: x

   afun = x**2
 end function afun

end module FunctionHolder</lang>

Haskell

Different approach from most of the other examples: First, the function f might be expensive to calculate, and so it should not be evaluated several times. So, ideally, we want to have positions x and weights w for each method and then just calculate the approximation of the integral by

<lang haskell>approx f xs ws = sum [w * f x | (x,w) <- zip xs ws]</lang>

Second, let's to generalize all integration methods into one scheme. The methods can all be characterized by the coefficients vs they use in a particular interval. These will be fractions, and for terseness, we extract the denominator as an extra argument v.

Now there are the closed formulas (which include the endpoints) and the open formulas (which exclude them). Let's do the open formulas first, because then the coefficients don't overlap:

<lang haskell>integrateOpen :: Fractional a => a -> [a] -> (a -> a) -> a -> a -> Int -> a integrateOpen v vs f a b n = approx f xs ws * h / v where

 m = fromIntegral (length vs) * n
 h = (b-a) / fromIntegral m
 ws = concat $ replicate n vs
 c = a + h/2
 xs = [c + h * fromIntegral i | i <- [0..m-1]]</lang>

Similarly for the closed formulas, but we need an additional function overlap which sums the coefficients overlapping at the interior interval boundaries:

<lang haskell>integrateClosed :: Fractional a => a -> [a] -> (a -> a) -> a -> a -> Int -> a integrateClosed v vs f a b n = approx f xs ws * h / v where

 m = fromIntegral (length vs - 1) * n
 h = (b-a) / fromIntegral m
 ws = overlap n vs
 xs = [a + h * fromIntegral i | i <- [0..m]]

overlap :: Num a => Int -> [a] -> [a] overlap n [] = [] overlap n (x:xs) = x : inter n xs where

 inter 1 ys     = ys
 inter n []     = x : inter (n-1) xs
 inter n [y]    = (x+y) : inter (n-1) xs
 inter n (y:ys) = y : inter n ys</lang>

And now we can just define

<lang haskell>intLeftRect = integrateClosed 1 [1,0] intRightRect = integrateClosed 1 [0,1] intMidRect = integrateOpen 1 [1] intTrapezium = integrateClosed 2 [1,1] intSimpson = integrateClosed 3 [1,4,1]</lang>

or, as easily, some additional schemes:

<lang haskell>intMilne = integrateClosed 45 [14,64,24,64,14] intOpen1 = integrateOpen 2 [3,3] intOpen2 = integrateOpen 3 [8,-4,8]</lang>

Some examples:

*Main> intLeftRect  (\x -> x*x) 0 1 10  
0.2850000000000001
*Main> intRightRect (\x -> x*x) 0 1 10  
0.38500000000000006
*Main> intMidRect   (\x -> x*x) 0 1 10 
0.3325
*Main> intTrapezium (\x -> x*x) 0 1 10 
0.3350000000000001
*Main> intSimpson   (\x -> x*x) 0 1 10 
0.3333333333333334

J

Solution: <lang j>integrate=: adverb define

 'a b steps'=. 3{.y,128
 size=. (b - a)%steps
 size * +/ 2 u\ a + size * i.>:steps

)

rectangle=: adverb def 'u -: +/ y'

trapezium=: adverb def '-: +/ u y'

simpson =: adverb def '6 %~ +/ 1 1 4 * u y, -:+/y'</lang> Example usage
Integrate square (*:) from 0 to π in 10 steps using various methods. <lang j> *: rectangle integrate 0 1p1 10 10.3095869962

  *: trapezium integrate 0 1p1 10

10.3871026879

  *: simpson integrate 0 1p1 10

10.3354255601</lang> Integrate sin from 0 to π in 10 steps using various methods. <lang j> sin=: 1&o.

  sin rectangle integrate 0 1p1 10

2.00824840791

  sin trapezium integrate 0 1p1 10

1.98352353751

  sin simpson integrate 0 1p1 10

2.00000678444</lang> Note that J has a primitive verb p.. for integrating polynomials. For example the integral of $x^{2}$ (which can be described in terms of its coefficients as 0 0 1) is: <lang j> 0 p.. 0 0 1 0 0 0 0.333333333333

  0 p.. 0 0 1x        NB. or using rationals

0 0 0 1r3</lang> That is: $0x^{0}+0x^{1}+0x^{2}+{\tfrac {1}{3}}x^{3}$
So to integrate $x^{2}$ from 0 to π : <lang j> 0 0 1 (0&p..@[ -~/@:p. ]) 0 1p1 10.3354255601</lang>

Java

The function in this example is assumed to be f(double x). <lang java5>public class Integrate{

  public static double leftRect(double a, double b, double n){
     double h = (b - a) / n;
     double sum = 0;
     for(double x = a;x <= b - h;x += h)
        sum += f(x);
     return h * sum;
  }

  public static double rightRect(double a, double b, double n){
     double h = (b - a) / n;
     double sum = 0;
     for(double x = a + h;x <= b - h;x += h)
        sum += f(x);
     return h * sum;
  }

  public static double midRect(double a, double b, double n){
     double h = (b - a) / n;
     double sum = 0;
     for(double x = a;x <= b - h;x += h)
        sum += (f(x) + f(x + h));
     return (h / 2) * sum;
  }

  public static double trap(double a, double b, double n){
     double h = (b - a) / n;
     double sum = f(a) + f(b);
     for(int i = 1;i < n;i++)
        sum += f(a + i * h);
     return  h * sum;
  }

  public static double simpson(double a, double b, double n){
     double h = (b - a) / n;
     double sum1 = 0;
     double sum2 = 0;

     for(int i = 0;i < n;i++)
        sum1 += f(a + h * i + h / 2);

     for(int i = 1;i < n;i++)
        sum2 += f(a + h * i);

     return h / 6 * (f(a) + f(b) + 4 * sum1 + 2 * sum2);
  }
  //assume f(double x) elsewhere in the class

}</lang>

Logo

<lang logo>to i.left :fn :x :step

 output invoke :fn :x

end to i.right :fn :x :step

 output invoke :fn :x + :step

end to i.mid :fn :x :step

 output invoke :fn :x + :step/2

end to i.trapezium :fn :x :step

 output ((i.left :fn :x :step) + (i.right :fn :x :step)) / 2

end to i.simpsons :fn :x :step

 output ( (i.left :fn :x :step)
        + (i.mid :fn :x :step) * 4
        + (i.right :fn :x :step) ) / 6

end

to integrate :method :fn :steps :a :b

 localmake "step (:b - :a) / :steps
 localmake "sigma 0
 ; for [x :a :b-:step :step] [make "sigma :sigma + apply :method (list :fn :x :step)]
 repeat :steps [
   make "sigma :sigma + (invoke :method :fn :a :step)
   make "a :a + :step ]
 output :sigma * :step

end

to fn2 :x

 output 2 / (1 + 4 * :x * :x)

end print integrate "i.left "fn2 4 -1 2 ; 2.456897 print integrate "i.right "fn2 4 -1 2 ; 2.245132 print integrate "i.mid "fn2 4 -1 2 ; 2.496091 print integrate "i.trapezium "fn2 4 -1 2 ; 2.351014 print integrate "i.simpsons "fn2 4 -1 2 ; 2.447732</lang>

MATLAB

f.m (The funtion you are wanting to approximate an integral too) <lang MATLAB> function f=f(x) f=x; </lang>

Numerical_Integration.m <lang MATLAB> function Numerical_Integration(a,b,n) format long; h=(b-a)/n; x1=linspace(a,b,n+1); LftRecRl=h.*sum(f(x1(1:n)))

x2=linspace(a,b,n+1); RgtRecRl=h.*sum(f(x2(2:n+1)))

x3=linspace(a,b,n+1); MdPtRl=h.*sum(f((x3(1:n)+x3(2:n+1))/2))

x4=linspace(a,b,n+1); TrpRl=h.*sum((f(x3(1:n))+f(x4(2:n+1)))/2)

x5=linspace(a,b,n+1); SmpRl=h/6.*sum(f(x5(1:n))+4*f((x5(1:n)+x5(2:n+1))/2)+f(x5(2:n+1))) </lang>

OCaml

<lang ocaml>let integrate f a b steps meth =

 let h = (b -. a) /. float_of_int steps in
 let rec helper i s =
   if i >= steps then s
   else helper (succ i) (s +. meth f (a +. h *. float_of_int i) h)
 in
 h *. helper 0 0.

let left_rect f x _ =

f x

let mid_rect f x h =

 f (x +. h /. 2.)

let right_rect f x h =

 f (x +. h)

let trapezium f x h =

 (f x +. f (x +. h)) /. 2.

let simpson f x h =

 (f x +. 4. *. f (x +. h /. 2.) +. f (x +. h)) /. 6.

let square x = x *. x

let rl = integrate square 0. 1. 10 left_rect let rm = integrate square 0. 1. 10 mid_rect let rr = integrate square 0. 1. 10 right_rect let t = integrate square 0. 1. 10 trapezium let s = integrate square 0. 1. 10 simpson</lang>

Pascal

<lang pascal>function RectLeft(function f(x: real): real; xl, xr: real): real;

begin
 RectLeft := f(xl)
end;

function RectMid(function f(x: real): real; xl, xr: real) : real;

begin
 RectMid := f((xl+xr)/2)
end;

function RectRight(function f(x: real): real; xl, xr: real): real;

begin
 RectRight := f(xr)
end;

function Trapezium(function f(x: real): real; xl, xr: real): real;

begin
 Trapezium := (f(xl) + f(xr))/2
end;

function Simpson(function f(x: real): real; xl, xr: real): real;

begin
 Simpson := (f(xl) + 4*f((xl+xr)/2) + f(xr))/6
end;

function integrate(function method(function f(x: real): real; xl, xr: real): real;

                  function f(x: real): real;
                  a, b: real;
                  n: integer);
var
 integral, h: real;
 k: integer;
begin
 integral := 0;
 h := (b-a)/n;
 for k := 0 to n-1 do
  begin
   integral := integral + method(f, a + k*h, a + (k+1)*h)
  end;
 integrate := integral
end;</lang>

PL/I

<lang PL/I> integrals: procedure options (main);

/* The function to be integrated */ f: procedure (x) returns (float);

  declare x float;
  return (3*x**2 + 2*x);

end f;

  declare (a, b) float;
  declare (rect_area, trap_area, Simpson) float;
  declare (d, dx) fixed decimal (10,2);
  declare (l, r) float;
  declare (S1, S2) float;

  l = 0; r = 5;
  a = 0; b = 5; /* bounds of integration */
  dx = 0.05;

  /* Rectangle method */
  rect_area = 0;
  do d = a to b by dx;
     rect_area = rect_area + dx*f(d);
  end;
  put skip data (rect_area);

  /* trapezoid method */
  trap_area = 0;
  do d = a to b by dx;
     trap_area = trap_area + dx*(f(d) + f(d+dx))/2;
  end;
  put skip data (trap_area);

  /* Simpson's */
  S1 = f(a+dx/2);
  S2 = 0;
  do d = a to b by dx;
     S1 = S1 + f(d+dx+dx/2);
     S2 = S2 + f(d+dx);
  end;
  Simpson = dx * (f(a) + f(b) + 4*S1 + 2*S2) / 6;
  put skip data (Simpson);

end integrals; </lang>

Python

<lang python>def left_rect(f,x,h):

 return f(x)

def mid_rect(f,x,h):

 return f(x + h/2)

def right_rect(f,x,h):

 return f(x+h)

def trapezium(f,x,h):

 return (f(x) + f(x+h))/2.0

def simpson(f,x,h):

 return (f(x) + 4*f(x + h/2) + f(x+h))/6.0

def square(x):

 return x*x

def integrate( f, a, b, steps, meth):

  h = (b-a)/steps
  ival = h * sum(meth(f, a+i*h, h) for i in xrange(steps))
  return ival

t = integrate( square, 3.0, 7.0, 30, simpson )</lang> A faster Simpson's rule integrator is <lang python>def faster_simpson(f, a, b, steps):

  h = (b-a)/steps
  a1 = a+h/2
  s1 = sum( f(a1+i*h) for i in range(0,steps))
  s2 = sum( f(a+i*h) for i in range(1,steps))
  return (h/6.0)*(f(a)+f(b)+4.0*s1+2.0*s2)</lang>

Ruby

Translation of: Tcl

<lang ruby>def leftrect(f, left, right)

 f.call(left)

end

def midrect(f, left, right)

 f.call((left+right)/2.0)

end

def rightrect(f, left, right)

 f.call(right)

end

def trapezium(f, left, right)

 (f.call(left) + f.call(right)) / 2.0

end

def simpson(f, left, right)

 (f.call(left) + 4*f.call((left+right)/2.0) + f.call(right)) / 6.0

end

def integrate(f, a, b, steps, method)

 delta = 1.0 * (b - a) / steps
 total = 0.0
 steps.times do |i|
   left = a + i*delta
   right = left + delta
   total += delta * send(method, f, left, right)
 end
 total

end

def square(x)

 x**2

end

def def_int(f, a, b)

 l = case f.to_s
     when /sin>/
       lambda {|x| -Math.cos(x)}
     when /square>/
       lambda {|x| (x**3)/3.0}
     end
 l.call(b) - l.call(a)

end

a = 0 b = Math::PI steps = 10

for func in [method(:square), Math.method(:sin)]

 puts "integral of #{func} from #{a} to #{b} in #{steps} steps"
 actual = def_int(func, a, b)
 for method in [:leftrect, :midrect, :rightrect, :trapezium, :simpson]
   int = integrate(func, a, b, steps, method)
   diff = (int - actual) * 100.0 / actual
   printf "   %-10s  %s\t(%.1f%%)\n", method, int, diff
 end

end</lang> outputs

integral of #<Method: Object#square> from 0 to 3.14159265358979 in 10 steps
   leftrect    8.83678885388545	(-14.5%)
   midrect     10.3095869961997	(-0.2%)
   rightrect   11.9374165219154	(15.5%)
   trapezium   10.3871026879004	(0.5%)
   simpson     10.3354255600999	(0.0%)
integral of #<Method: Math.sin> from 0 to 3.14159265358979 in 10 steps
   leftrect    1.98352353750945	(-0.8%)
   midrect     2.00824840790797	(0.4%)
   rightrect   1.98352353750945	(-0.8%)
   trapezium   1.98352353750945	(-0.8%)
   simpson     2.0000067844418	(0.0%)

Scheme

<lang scheme>(define (integrate f a b steps meth)

 (define h (/ (- b a) steps))
 (* h
    (let loop ((i 0) (s 0))
      (if (>= i steps)
          s
          (loop (+ i 1) (+ s (meth f (+ a (* h i)) h)))))))

(define (left-rect f x h) (f x)) (define (mid-rect f x h) (f (+ x (/ h 2)))) (define (right-rect f x h) (f (+ x h))) (define (trapezium f x h) (/ (+ (f x) (f (+ x h))) 2)) (define (simpson f x h) (/ (+ (f x) (* 4 (f (+ x (/ h 2)))) (f (+ x h))) 6))

(define (square x) (* x x))

(define rl (integrate square 0 1 10 left-rect)) (define rm (integrate square 0 1 10 mid-rect)) (define rr (integrate square 0 1 10 right-rect)) (define t (integrate square 0 1 10 trapezium)) (define s (integrate square 0 1 10 simpson))</lang>

Standard ML

<lang sml>fun integrate (f, a, b, steps, meth) = let

 val h = (b - a) / real steps
 fun helper (i, s) =
   if i >= steps then s
   else helper (i+1, s + meth (f, a + h * real i, h))

in

 h * helper (0, 0.0)

end

fun leftRect (f, x, _) = f x fun midRect (f, x, h) = f (x + h / 2.0) fun rightRect (f, x, h) = f (x + h) fun trapezium (f, x, h) = (f x + f (x + h)) / 2.0 fun simpson (f, x, h) = (f x + 4.0 * f (x + h / 2.0) + f (x + h)) / 6.0

fun square x = x * x

val rl = integrate (square, 0.0, 1.0, 10, left_rect ) val rm = integrate (square, 0.0, 1.0, 10, mid_rect ) val rr = integrate (square, 0.0, 1.0, 10, right_rect) val t = integrate (square, 0.0, 1.0, 10, trapezium ) val s = integrate (square, 0.0, 1.0, 10, simpson )</lang>

Tcl

<lang tcl>package require Tcl 8.5

proc leftrect {f left right} {

   $f $left

} proc midrect {f left right} {

   set mid [expr {($left + $right) / 2.0}]
   $f $mid

} proc rightrect {f left right} {

   $f $right

} proc trapezium {f left right} {

   expr {([$f $left] + [$f $right]) / 2.0}

} proc simpson {f left right} {

   set mid [expr {($left + $right) / 2.0}]
   expr {([$f $left] + 4*[$f $mid] + [$f $right]) / 6.0}

}

proc integrate {f a b steps method} {

   set delta [expr {1.0 * ($b - $a) / $steps}]
   set total 0.0
   for {set i 0} {$i < $steps} {incr i} {
       set left [expr {$a + $i * $delta}]
       set right [expr {$left + $delta}]
       set total [expr {$total + $delta * [$method $f $left $right]}]
   }
   return $total

}

interp alias {} sin {} ::tcl::mathfunc::sin proc square x {expr {$x*$x}} proc def_int {f a b} {

   switch -- $f {
       sin    {set lambda {x {expr {-cos($x)}}}}
       square {set lambda {x {expr {$x**3/3.0}}}}
   }
   return [expr {[apply $lambda $b] - [apply $lambda $a]}]

}

set a 0 set b [expr {4*atan(1)}] set steps 10

foreach func {square sin} {

   puts "integral of ${func}(x) from $a to $b in $steps steps"
   set actual [def_int $func $a $b]
   foreach method {leftrect midrect rightrect trapezium simpson} {
       set int [integrate $func $a $b $steps $method]
       set diff [expr {($int - $actual) * 100.0 / $actual}]
       puts [format "   %-10s  %s\t(%.1f%%)" $method $int $diff]
   }

}</lang>

integral of square(x) from 0 to 3.141592653589793 in 10 steps
   leftrect    8.836788853885448	(-14.5%)
   midrect     10.30958699619969	(-0.2%)
   rightrect   11.93741652191543	(15.5%)
   trapezium   10.387102687900438	(0.5%)
   simpson     10.335425560099939	(0.0%)
integral of sin(x) from 0 to 3.141592653589793 in 10 steps
   leftrect    1.9835235375094544	(-0.8%)
   midrect     2.0082484079079745	(0.4%)
   rightrect   1.9835235375094544	(-0.8%)
   trapezium   1.9835235375094546	(-0.8%)
   simpson     2.0000067844418012	(0.0%)

TI-89 BASIC

TI-89 BASIC has built-in numerical integration with the ∫ operator, but no control over the method used is available so it doesn't really correspond to this task.

$\overbrace {{\textstyle \int }(\mathrm {expr} ,x,a,b)} ^{\mathrm {TI-89} }=\overbrace {\int _{a}^{b}\mathrm {expr} \ dx} ^{\text{Math notation}}$

An explicit numerical integration program should be written here.

Ursala

A higher order function parameterized by a method m returns a function that integrates by that method. The method m is meant to specify whether it's rectangular, trapezoidal, etc.. The integrating function constructed from a given method takes a quadruple (f,a,b,n) containing the integrand f, the bounds (a,b), and the number of intervals n.

<lang Ursala>#import std

import nat
import flo

(integral_by "m") ("f","a","b","n") =

iprod ^(* ! div\float"n" minus/"b" "a",~&) ("m" "f")*ytp (ari successor "n")/"a" "b"</lang> An alternative way of defining this function shown below prevents redundant evaluations of the integrand at the cost of building a table-driven finite map in advance. <lang Ursala>(integral_by "m") ("f","a","b","n") =

iprod ^(* ! div\float"n" minus/"b" "a",~&) ^H(*+ "m"+ -:"f"+ * ^/~& "f",~&ytp) (ari successor "n")/"a" "b"</lang> As mentioned in the Haskell solution, the latter choice is preferable if evaluating the integrand is expensive. An integrating function is defined for each method as follows. <lang Ursala>left = integral_by "f". ("l","r"). "f" "l" right = integral_by "f". ("l","r"). "f" "r" midpoint = integral_by "f". ("l","r"). "f" div\2. plus/"l" "r" trapezium = integral_by "f". ("l","r"). div\2. plus "f"~~/"l" "r" simpson = integral_by "f". ("l","r"). div\6. plus:-0. <"f" "l",times/4. "f" div\2. plus/"l" "r","f" "r"></lang> As shown above, the method passed to the integral_by function is itself a higher order function taking an integrand f as an argument and returning a function that operates on the pair of left and right interval enpoints. Here is a test program showing the results of integrating the square from zero to pi in ten intervals by all five methods. <lang Ursala>#cast %eL

examples = <.left,midpoint,rignt,trapezium,simpson> (sqr,0.,pi,10)</lang> output:

<
   8.836789e+00,
   1.030959e+01,
   1.193742e+01,
   1.038710e+01,
   1.033543e+01>

(The GNU Scientific Library integration routines are also callable in Ursala, and are faster and more accurate.)