This document contains C++ code and output to evaluate definite integrals using numerical integration methods like the Trapezoidal rule, Simpson's 1/3 rule, and Simpson's 3/8 rule. It includes 7 questions that provide code to integrate functions like exp(x), 1/(1+x^2), and sin(x)+cos(x) over different intervals. The code demonstrates how to implement the different numerical integration techniques to compute approximations of definite integrals and compare the results to exact solutions where available.

1. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
QUESTION 1
Write a C++ program to evaluate∫
using Trapezoidal rule up
to three significant figures.
#include <cmath>
#include <iostream>
#include <iomanip>
using namespace std;
float f(float x)
{
float func;
func = exp(x);
return func;
}
void main()
{
float a,b,h,sum=0, Trap;
int i, n;
cout <<"Enter the lower limit a : ";
cin >> a;
cout << "Enter the upper limit b: ";
cin >> b;
cout << "Enter the number of subintervals n: ";
cin >> n;
h=(b-a)/n;

2. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
for(i=1;i<n;i++)
/*or we can can write for(i=a+h;i<=b-h;i=i+h) { sum+=f(i);*/
{
sum= sum + f(a+i*h);
}
Trap= (h/2)*(f(a) + f(b) +2 *sum);
cout <<"nThe value of the integral is: " << setprecision(3) << Trap <<endl ;
}
//Output:
Enter the lower limit a : 0
Enter the upper limit b: 1.2
Enter the number of subintervals n: 8
The value of the integral is: 2.32
//Alternative way for question 1
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double trapezoidal(double a,double b,int n)
{

3. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
double h= (b-a)/n;
double z[9],I;
double x[9]= {0.0, 0.15, 0.3,0.45,0.6,0.75,0.9,1.05,1.2 };
for (int i=0; i<9 ; i++)
z[i]= exp(x[i]); //the function is exp(x)
I=h/2*( z[0]+z[8] + 2*(z[1]+z[2]+z[3]+z[4]+z[5]+z[6]+z[7]) );
return (I);
}
int main()
{
double J;
J=trapezoidal(0,1.2,8);
cout<< "The integral value is " << setprecision(3) << J << "nn";
return 0;
}
//Output:
The integral value is 2.32
QUESTION 2
Write a C++ function sub-program to evaluate ∫
using
Trapezoidal rule correct to four decimal places. Compare the
result with the exact solution 1.48766

4. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
#include <cmath>
#include <iostream>
#include <iomanip>
using namespace std;
float f(float x)
{
float func;
func = 1/(1+(x*x));
return func;
}
void main()
{
double a,b,h,sum=0, Trap, exact=1.48766;
// we can find the exact value by using the formula exact= atan(b) - atan(a);
int i, n;
cout <<"Enter the lower limit a : ";
cin >> a;
cout << "Enter the upper limit b: ";
cin >> b;
cout << "Enter the number of subintervals n: ";
cin >> n;
h=(b-a)/n;
for(i=1;i<n;i++)

5. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
/*or we can can write for(i=a+h;i<=b-h;i=i+h) { sum+=f(i);*/
{
sum= sum + f(a+i*h);
}
Trap= (h/2)*(f(a) + f(b) +2 *sum);
cout <<"nThe value of the integral is: " << setprecision(4) <<fixed << Trap <<endl ;
cout<<"The difference in Trapezoidal method = "<<setprecision(4) << fixed <<fabs(exact-
Trap)<<endl;
}
//Output:
Output 1:
Enter the lower limit a : 0
Enter the upper limit b: 12
Enter the number of subintervals n: 8
The value of the integral is: 1.5358
The difference in Trapezoidal method = 0.0482
Output 2:
Enter the lower limit a : 0
Enter the upper limit b: 12
Enter the number of subintervals n: 1000
The value of the integral is: 1.4877
The difference in Trapezoidal method = 0.0000

6. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
QUESTION 3
Write a C++ program to evaluate ∫ √
using Simpson’s
1/3 rule.
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double x)
{
double function;
function= pow(sin(x)+cos(x), 0.5);
return function;
}
void main()
{
double a,b, h,sum_even=0,sum_odd=0,Simpson_13;
int i,n;
cout << "Enter the lower limit a: ";
cin >> a;
cout << "Enter the upper limit b: ";
cin >> b;
cout << "Enter the number of subintervals: ";
cin >> n;
h=(b-a)/n;
for(i=1;i<n;i=i++)
{
if(i%2==0)

7. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
sum_even= sum_even+ f(a+i*h);
if(i%2!=0)
sum_odd= sum_odd+ f(a+i*h);
}
Simpson_13=(h/3)*(f(a)+f(b)+ 4*sum_odd +2*sum_even);
cout <<"The value of the integral is " << setprecision(4) <<fixed << Simpson_13 <<
"nn";
}
//Output:
Enter the lower limit a: 0
Enter the upper limit b: 1
Enter the number of subintervals: 8
The value of the integral is 1.1394
//Alternative way for question 3
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double simpson13(double a,double b,int n)
{
double z[9],Simp,h=(b-a)/n;
double x[9]={0,0.125,0.25,0.375,0.5,0.625,0.75,0.875,1.0};
for (int i=0; i<9 ; i++)
z[i]= pow(sin(x[i])+cos(x[i]),0.5);

8. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
Simp=(h/3)*((z[0]+z[8]) + 2*(z[2]+z[4]+z[6]) + 4*(z[1]+z[3]+z[5]+z[7]) );
return Simp;
}
int main()
{
double J;
J=simpson13(0,1,8);
cout<< "The integral value is: "<< setprecision(4) <<fixed << J<<endl;
return 0;
}
//Output:
The integral value is: 1.1394
QUESTION 4
Write a C++ function sub-program to evaluate∫
using
Simpson’s 1/3 correct to four decimal places.
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double x)
{
float function;
function= 1/(1+pow(x,2));
return function;
}

9. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
void main()
{
double a,b, h,sum_even=0,sum_odd=0,Simpson_13;
int i,n;
cout << "Enter the lower limit a: ";
cin >> a;
cout << "Enter the upper limit b: ";
cin >> b;
cout << "Enter the number of subintervals: ";
cin >> n;
h=(b-a)/n;
for(i=1;i<n;i=i++)
{
if(i%2==0)
sum_even= sum_even+ f(a+i*h);
if(i%2!=0)
sum_odd= sum_odd+ f(a+i*h);
}
Simpson_13=(h/3)*(f(a)+f(b)+ 4*sum_odd +2*sum_even);
cout <<"The value of the integral is " << setprecision(4) <<fixed << Simpson_13 << "nn";
}
//Output:
Enter the lower limit a: 0
Enter the upper limit b: 12
Enter the number of subintervals: 6
The value of the integral is 1.4020

10. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
//Alternative way for question 4
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double simpson13(double a,double b,int n)
{
double z[7],Simp,h=(b-a)/n;
double x[7]={0,2,4,6,8,10,12};
for (int i=0; i<7 ; i++)
z[i]= 1/(1+pow(x[i],2));
Simp=(h/3)*((z[0]+z[6]) + 2*(z[2]+z[4]) + 4*(z[1]+z[3]+z[5]) );
return Simp;
}
int main()
{
double J;
J=simpson13(0,12,6);
cout<< "The integral value is: "<< setprecision(4) <<fixed << J<<endl;
return 0;
}
//Output
The integral value is: 1.4020

11. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
QUESTION 5
Write a C++ program to evaluate∫
using Simpson’s 3/8
rule.
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
double f(double x)
{
double function;
function=exp(sin(x));
return function;
}
void main()
{
double a,b, h,sum_miss=0,sum_triple=0,Simpson_38;
int i,n;
cout << "Enter the lower limit a: ";
cin >> a;
cout << "Enter the upper limit b: ";
cin >> b;
cout << "Enter the number of subintervals: ";
//should be taken as multiples of 3
cin >> n;
h=(b-a)/n;

12. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
for (i=1;i<n;i++)
{
if ( i%3 != 0 )
sum_miss = sum_miss + f(a+i*h);
if(i%3==0)
sum_triple =sum_triple+ f(a+i*h);
}
Simpson_38= (3*h/8)*((f(a)+f(b))+ 3*sum_miss +2*sum_triple);
cout <<"The value of the integral is " << setprecision(4) <<fixed << Simpson_38 << "nn";
}
//Output:
Enter the lower limit a: 0
Enter the upper limit b: 1.570796
Enter the number of subintervals: 8
The value of the integral is 3.0381
QUESTION 6
Write a C++ program to evaluate∫
using Simpson’s 3/8
rule.
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;

13. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
double f(double x)
{
double function;
function=log10(x);
return function;
}
void main()
{
double a,b, h,sum_miss=0,sum_triple=0,Simpson_38;
int i,n;
cout << "Enter the lower limit a: ";
cin >> a;
cout << "Enter the upper limit b: ";
cin >> b;
cout << "Enter the number of subintervals: ";
//should be taken as multiples of 3
cin >> n;
h=(b-a)/n;
for (i=1;i<n;i++)
{
if ( i%3 != 0 )
sum_miss = sum_miss + f(a+i*h);
if(i%3==0)
sum_triple =sum_triple+ f(a+i*h);
}
Simpson_38= (3*h/8)*((f(a)+f(b))+ 3*sum_miss +2*sum_triple);

14. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
cout <<"The value of the integral is " << setprecision(4) <<fixed << Simpson_38 << "nn";
}
//Output:
Enter the lower limit a: 2
Enter the upper limit b: 6
Enter the number of subintervals: 8
The value of the integral is 2.2833
QUESTION 7
Write a C++ function program to evaluate∫
using
Trapezoidal, Simpson’s 1/3 and Simpson’s 3/8 rules. Compare the
results with the exact solution 2.32012
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
double f(double x)
{
return exp(x) ;
}
double simpson_38(double a,double b,double n)
{
double h=(b-a)/n, sum_miss=0,sum_triple=0,Simp38;
for (int i=1;i<n;i++)
{
if ( i%3 != 0 )
sum_miss = sum_miss + f(a+i*h);
if(i%3==0)
sum_triple =sum_triple+ f(a+i*h);
}

15. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
Simp38= (3*h/8)*((f(a)+f(b))+ 3*sum_miss +2*sum_triple);
return Simp38;
}
double simpson_13(double a,double b,double n)
{
double h=(b-a)/n, sum_odd=0,sum_even=0,Simp13;
for(int i=1; i<n; i++)
{
if(i%2==0)
sum_even =sum_even + f(a+i*h);
else
sum_odd=sum_odd + f(a+i*h);
}
Simp13 = (h/3)*( (f(a) + f(b)) + 4*sum_odd + 2*sum_even );
return Simp13;
}
double trapezoidal(double a,double b,double n)
{
double sum=0, Trap, h=(b-a)/n;
for(int i=1;i<n;i++)
sum = sum+f(a+i*h);
Trap = (h/2)*( f(a) + f(b) + 2*sum);
return Trap ;
}
int main()
{
double a,b,n, exact= 2.32012;
int ans;
char choice;
Pilihan:
cout << "1. Trapezoidaln2. Simpson's 1/3 rulen3. Simpson's 3/8 rulenn";
cout << "Which number you wanna choose : ";
cin >> ans;
cout << endl;

16. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
switch(ans)
{
case 1 : cout << "Please input a,b and n : ";
cin >> a >> b >> n;
cout<<"The Exact value = "<< setprecision(6) <<exact<<endl;
cout<<"The result for Trapezoidal method = "
<<setprecision(6)<<trapezoidal(a,b,n)<<endl;
cout<<"The difference in Trapezoidal method = "<<setprecision(6)<<
fabs(exact - trapezoidal(a,b,n))<<endl;
break;
case 2 : cout << "Please input a,b and n : ";
cin >> a >> b >> n;
cout<<"The Exact value = "<<setprecision(6)<<exact<<endl;
cout<<"The result for Simpson's 1/3 method = "<<setprecision(6)<<
simpson_13(a,b,n)<<endl;
cout<<"The difference in Simpson's 1/3
method="<<setprecision(6)<<fabs(exact- simpson_13(a,b,n))<<endl;
break;
case 3 : cout << "Please input a,b and n : ";
cin >> a >> b >> n;
cout<<"The Exact value = "<<setprecision(6)<<exact<<endl;
cout<<"The result for Simpson's 3/8 method = "<< setprecision(6)<<
simpson_38(a,b,n)<<endl;
cout<<"The difference in Simpson's 3/8 method
="<<setprecision(6)<<fabs(exact- simpson_38(a,b,n))<<endl;
break;
default: cout << "Invalid number. Try again : ";
cin >> ans;
goto Pilihan;
break;
}
if(fabs(exact-simpson_38(a,b,n)) <fabs(exact-simpson_13(a,b,n)) &&
fabs(exact-simpson_38(a,b,n))<fabs(exact-trapezoidal(a,b,n)))

17. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
cout<<"Simpson's 3/8 is better"<<endl;
else if(fabs(exact-simpson_13(a,b,n))<fabs(exact-simpson_38(a,b,n)) &&
fabs(exact-simpson_13(a,b,n))<fabs(exact-trapezoidal(a,b,n)))
cout<<"Simpson's 1/3 is better"<<endl;
else
cout<<"Trapezoidal rule is better"<<endl;
check:
cout << "Do u want to try again (Y/N): ";
cin >> choice;
if ( choice == 'N' || choice == 'n')
cout << "Thank You." << endl;
else if ( choice == 'Y' || choice == 'y')
{
goto Pilihan;
}
else goto check;
return 0;
}
//Output:
1. Trapezoidal
2. Simpson's 1/3 rule
3. Simpson's 3/8 rule
Which number you wanna choose : 1
Please input a,b and n : 0 1.2 8
The Exact value = 2.32012
The result for Trapezoidal method = 2.32447
The difference in Trapezoidal method = 0.00434551
Simpson's 1/3 is better
Do u want to try again (Y/N): y
1. Trapezoidal
2. Simpson's 1/3 rule
3. Simpson's 3/8 rule
Which number you wanna choose : 2
Please input a,b and n : 0 1.2 8

18. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
The Exact value = 2.32012
The result for Simpson's 1/3 method = 2.32012
The difference in Simpson's 1/3 method=3.43063e-006
Simpson's 1/3 is better
Do u want to try again (Y/N): y
1. Trapezoidal
2. Simpson's 1/3 rule
3. Simpson's 3/8 rule
Which number you wanna choose : 3
Please input a,b and n : 0 1.2 8
The Exact value = 2.32012
The result for Simpson's 3/8 method = 2.26695
The difference in Simpson's 3/8 method =0.0531698
Simpson's 1/3 is better
Do u want to try again (Y/N): n
Thank You.
//Alternative for question 7
#include<iostream>
#include<cmath>
#include<iomanip>
using namespace std;
#define f(x) exp(x)
double trap(double a,double b,double h,double n)
{
int i;
double x1,sumA=0,resultA;
x1=a;
for(i=1;i<n;i++)
{
x1=x1+h;
sumA=sumA+f(x1);
}
resultA=(h/2)*(f(a)+2*sumA+f(b));
return resultA;
}
double simp13(double a,double b,double h,double n)
{
int j;
double x2,sumB1=0,sumB2=0,resultB;
x2=a;
for(j=1;j<n;j++)

19. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
{
x2=x2+h;
if(j%2==0)
sumB1=sumB1+2*f(x2);
else
sumB2=sumB2+4*f(x2);
}
resultB=(h/3)*(f(a)+sumB1+sumB2+f(b));
return resultB;
}
double simp38(double a,double b,double h,double n)
{
double sum_miss=0,sum_triple=0,Simp38;
for (int i=1;i<n;i++)
{
if ( i%3 != 0 )
sum_miss = sum_miss + f(a+i*h);
if(i%3==0)
sum_triple =sum_triple+ f(a+i*h);
}
Simp38= (3*h/8)*(f(a)+f(b)+ 3*sum_miss +2*sum_triple);
return Simp38;
}
int main()
{
int n;
double a=0,b=1.2,h,exact= 2.32012;
cout<<"Enter the value of n: ";
cin>>n;
h=(b-a)/n;
cout<<"The Exact value = "<<setprecision(6)<<exact<<endl;
cout<<"The result for Trapezoidal method = "<<setprecision(6)<<trap(a,b,h,n)<<endl;
cout<<"The result for Simpson 1/3 method = "<<setprecision(6)<<simp13(a,b,h,n)<<endl;
cout<<"The result for Simpson 3/8 method = "<<setprecision(6)<<simp38(a,b,h,n)<<endl;
cout<<"The difference in Trapezoidal method = "<<setprecision(6) <<fabs(exact-trap(
a,b,h,n))<<endl;
cout<<"The difference in Simpson 1/3 method = "<<setprecision(6)<<fabs(exact-simp13(
a,b,h,n))<<endl;

20. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
cout<<"The difference in Simpson 3/8 method = "<<setprecision(6)<<fabs(exact-simp38(
a,b,h,n))<<endl;
if(fabs(exact-simp38(a,b,h,n)) <fabs(exact-simp13(a,b,h,n)) &&
fabs(exact-simp38(a,b,h,n))<fabs(exact-trap(a,b,h,n)))
cout<<"Simpson's 3/8 is better"<<endl;
else if(fabs(exact-simp13(a,b,h,n))<fabs(exact-simp38(a,b,h,n)) &&
fabs(exact-simp13(a,b,h,n))<fabs(exact-trap(a,b,h,n)))
cout<<"Simpson's 1/3 is better"<<endl;
else
cout<<"Trapezoidal rule is better"<<endl;
return 0;
}
//Output:
Enter the value of n: 8
The Exact value = 2.32012
The result for Trapezoidal method = 2.32447
The result for Simpson 1/3 method = 2.32012
The result for Simpson 3/8 method = 2.26695
The difference in Trapezoidal method = 0.00434551
The difference in Simpson 1/3 method = 3.43063e-006
The difference in Simpson 3/8 method = 0.0531698
Simpson's 1/3 is better
//Alternative for question 7
#include<iostream>
#include<cmath>
using namespace std;
double f(double a)
{
return (double) exp(a);
}
int main()

21. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
{
double a=0,b=1.2,n,h,Exact,sum3=0,sum=0,sumE=0,sumO=0;
double sumT=0,simps3per8,simp1per3,trapzoid;
cout<<"Enter the number of subintervals,n: ";
cin>>n;
h=(b-a)/n;
Exact=exp(b)-exp(a);
cout<<"Exact value="<<Exact<<endl;
for(int i=1; i<n; i++) //Simpson 3/8
{
if(i%3==0)
sum3=sum3 + f(a+i*h);
else
sum=sum + f(a+i*h);}
simps3per8=((3*h)/8)*(f(a)+f(b)+3*sum+2*sum3);
cout<<"Simpson's 3/8 ="<<simps3per8<<endl;
for(i=1; i<n; i++) //Simpson's 1/3
{
if(i%2==0)
sumE=sumE + f(a+i*h);
else

22. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
sumO=sumO + f(a+i*h);}
simp1per3=(h/3)*(f(a)+f(b)+4*sumO+2*sumE);
cout<<"Simpson's 1/3 ="<<simp1per3<<endl;
for(i=1;i<n;i++) //Trapezoidal
sumT=sumT+f(a+i*h);
trapzoid=(h/2)*(f(a)+f(b)+2*sumT);
cout<<"Trapezoidal= "<<trapzoid<<endl;
if(fabs(Exact-simps3per8)<fabs(Exact-simp1per3) &&
fabs(Exact-simps3per8)<fabs(Exact-trapzoid))
cout<<"Simpson's 3/8 is better"<<endl;
else if(fabs(Exact-simp1per3)<fabs(Exact-simps3per8) &&
fabs(Exact-simp1per3)<fabs(Exact-trapzoid))
cout<<"Simpson's 1/3 is better"<<endl;
else
cout<<"Trapezoidal rule is better"<<endl;
return 0;
}

23. SJEM2231: STRUCTURED PROGRAMMING (C++) TUTORIAL 8
//Output:
Enter the number of subintervals,n: 8
Exact value=2.32012
Simpson's 3/8 =2.26695
Simpson's 1/3 =2.32012
Trapezoidal= 2.32447
Simpson's 1/3 is better