Area under curve, C/C++ Programming

Assignment Help:
Write a program to find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b.

Related Discussions:- Area under curve

Program for basically a small store, ¦It should ask customers to enter deta...

¦It should ask customers to enter details such as their names and addresses. ¦It should print a welcome message that includes the customer’s name. ¦It should provide a list of avai

.Change to palindrome, A palindrome is a string that reads the same from bo...

A palindrome is a string that reads the same from both the ends. Given a string S convert it to a palindrome by doing character replacement. Your task is to convert S to palindrome

C++, Write a program to find the area under the curve y = f(x) between x = ...

Write a program to find the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between the limits of a and b. The area under a curve between two points can b

Required, requiredrequiredrequiredrequiredrequiredrequiredrequiredrequired

requiredrequiredrequiredrequiredrequiredrequiredrequiredrequired

Define how passing arrays to a function, Define How Passing Arrays to a Fun...

Define How Passing Arrays to a Function? A complete array can be passed to a function as an argument. The manner in which the array is passed be different from that of an ordin

C pgms, find area under the curve y=f(x) between x=a &y=b integrate y=f(x)

find area under the curve y=f(x) between x=a &y=b integrate y=f(x)

Write a program that reads data from array, Write a program that reads data...

Write a program that reads data into an array of type int. You must use an array. The numbers input are attendance numbers at a conference. Keep a running average, number of events

Decoding smugglers message, Smugglers are becoming very smart day by day. N...

Smugglers are becoming very smart day by day. Now they have developed a new technique of sending their messages from one smuggler to another. In their new technology, they are send

Playback pattern sequences, This task involves creating a new class(es) tha...

This task involves creating a new class(es) that will allow the system to playback pattern sequences using the Neo  Freerunner  hardware. The first thing to note is that the vibrat

What do you mean by inheritance, Inheritance is the process of forming new ...

Inheritance is the process of forming new classes, called derived classes, from existing classes or base classes. The derived class inherits all the capabilities of the base class,

diana

9/4/2012 4:20:01 AM

#include
float start_point, /* GLOBAL VARIABLES */
end_point,
total_area;
int numtraps;
main( )
{
void input(void);
float find_area(float a,float b,int n); /* prototype */
print("AREA UNDER A CURVE");
input( );
total_area = find_area(start_point, end_point, numtraps);
printf("TOTAL AREA = %f", total_area);
}
void input(void)
{
printf("\n Enter lower limit:");
scanf("%f", &start_point);
printf("Enter upper limit:");
scanf("%f", &end_point);
printf("Enter number of trapezoids:");
scanf("%d", &numtraps);
}
float find_area(float a, float b, int n)
{
floatbase, lower, h1, h2; /* LOCAL VARIABLES */float function_x(float x); /* prototype */float trap_area(float h1,float h2,floatbase);/*prototype*/base = (b-1)/n;
lower = a;
for(lower =a; lower <= b-base; lower = lower + base)
{
h1 = function_x(lower);
h1 = function_x(lower + base);
total_area += trap_area(h1, h2, base);
}
return(total_area);
float trap_area(float height_1,float height_2,floatbase)
{
float area; /* LOCAL VARIABLE */
area = 0.5 * (height_1 + height_2) * base;
return(area);
}
float function_x(float x)
{
/* F(X) = X * X + 1 */return(x*x + 1);
}

Output
AREA UNDER A CURVE
Enter lower limit: 0
Enter upper limit: 3
Enter number of trapezoids: 30
TOTAL AREA = 12.005000
AREA UNDER A CURVE
Enter lower limit: 0
Enter upper limit: 3
Enter number of trapezoids: 100
TOTAL AREA = 12.000438

Solution in java ::

// hackerx sasi kamaraj college of engineering and technology 2910007 java Program


//The answer to be precise... although the type was a double, it rounds off the answer. Any help would be //appreciated...
//java code: 1. :: try this or the another one below this one
//Program code ::

public class Reimann
{
private static double integral(String s, double[] descriptors, double lb, double ub)
{

double area = 0; // Area of the rectangle
double sumOfArea = 0; // Sum of the area of the rectangles
double oldSumOfArea = 0;
double width = ub - lb;
boolean firstPass = true;

while ( (Math.abs((oldSumOfArea - sumOfArea) / sumOfArea) > .0001) || firstPass )
{

System.out.println((Math.abs((oldSumOfArea - sumOfArea) / sumOfArea) > .0001) || firstPass);
if (s.equals("poly"))
{
for (int i = 1; i <= ((ub - lb) / width); i++) // represents # of rectangles
{
for (int j = 0; j < descriptors.length; j++) // Goes through all the coefficients
{
area = width * descriptors[j] * Math.pow ( (double)( (i * width + lb + (i -1.0) * width + lb) / 2.0 ), j);
/*Above code computes area of each rectangle */

sumOfArea += area;

}
}
}
width = width / 2;
firstPass = false;
oldSumOfArea = sumOfArea;
}
return sumOfArea;
}

/*private static void runMyTests()
{
assert ( integral() <= 48.00001 ) && ( integral() >= 47.99999 );
}*/

public static void main (String [] args)
{

double lb = Double.parseDouble(args[args.length -2]);
double ub = Double.parseDouble(args[args.length -1]);

double[] coefficients = new double[args.length - 3];

if (args[0].equals("poly"))
{
for (int i = 1; i < args.length - 2; i++)
{
coefficients[i-1] = Double.parseDouble(args[i]);
}

System.out.println(integral("poly", coefficients, lb, ub));
}
}
}



Java Program 2 ::

public class Riemann
{
private static double integral(String s, double[] descriptors, double lb, double ub)
{

double area = 0; // Area of the rectangle
double sumOfArea = 0; // Sum of the area of the rectangles
double oldSumOfArea = 0;
double width = ub - lb;
boolean firstPass = true;

while ( (Math.abs((oldSumOfArea - sumOfArea) / sumOfArea) > .0001) || firstPass )
{

System.out.println((Math.abs((oldSumOfArea - sumOfArea) / sumOfArea) > .0001) || firstPass);
if (s.equals("poly")) // Statement for polynomial
{
for (int i = 1; i <= ((ub - lb) / width); i++) // represents # of rectangles
{
for (int j = 0; j < descriptors.length; j++) // Goes through all the coefficients
{
area = width * descriptors[j] * Math.pow ( (double)( (i * width + lb + (i -1.0) * width + lb) / 2.0 ), j);
/*Above code computes area of each rectangle */

sumOfArea += area;

}
}
}

else if (s.equals("sin")) // Statement for sin
{
for (int i = 1; i <= ((ub - lb) / width); i++) // represents # of rectangles
{
for (int j = 0; j < descriptors.length; j++) // Goes through all the coefficients
{
area = width * descriptors[j] * Math.sin(Math.toRadians(( (double)( (i * width + lb + (i -1.0) * width + lb) / 2.0 ))));
/*Above code computes area of each rectangle */

sumOfArea += area;

}
}
}

else if (s.equals("cos")) // Statement for cos
{
for (int i = 1; i <= ((ub - lb) / width); i++) // represents # of rectangles
{
for (int j = 0; j < descriptors.length; j++) // Goes through all the coefficients
{
area = width * descriptors[j] * Math.cos(Math.toRadians(( (double)( (i * width + lb + (i -1.0) * width + lb) / 2.0 ))));
/*Above code computes area of each rectangle */

sumOfArea += area;

}
}
}

width = width / 2;
firstPass = false;
oldSumOfArea = sumOfArea;
}

return sumOfArea;
}

/*private static void runMyTests()
{
assert ( integral() <= 48.00001 ) && ( integral() >= 47.99999 );
}*/

public static void main (String [] args)
{
double lb = Double.parseDouble(args[args.length -2]);
double ub = Double.parseDouble(args[args.length -1]);

double[] coefficients = new double[args.length - 3];

if (args[0].equals("poly"))
{
for (int i = 1; i < args.length - 2; i++)
{
coefficients[i-1] = Double.parseDouble(args[i]);
}

System.out.println(integral("poly", coefficients, lb, ub));
}

else if (args[0].equals("sin"))
{
for (int i = 1; i < args.length - 2; i++)
{
coefficients[i-1] = Double.parseDouble(args[i]);
}

System.out.println(integral("sin", coefficients, lb, ub));
}

else if (args[0].equals("cos"))
{
for (int i = 1; i < args.length - 2; i++)
{
coefficients[i-1] = Double.parseDouble(args[i]);
}

System.out.println(integral("cos", coefficients, lb, ub));
}
}
}



Question ::
Area Under Curve

diana

9/4/2012 4:20:21 AM

#include
float start_point, /* GLOBAL VARIABLES */
end_point,
total_area;
int numtraps;
main( )
{
void input(void);
float find_area(float a,float b,int n); /* prototype */
print("AREA UNDER A CURVE");
input( );
total_area = find_area(start_point, end_point, numtraps);
printf("TOTAL AREA = %f", total_area);
}
void input(void)
{
printf("\n Enter lower limit:");
scanf("%f", &start_point);
printf("Enter upper limit:");
scanf("%f", &end_point);
printf("Enter number of trapezoids:");
scanf("%d", &numtraps);
}
float find_area(float a, float b, int n)
{
floatbase, lower, h1, h2; /* LOCAL VARIABLES */float function_x(float x); /* prototype */float trap_area(float h1,float h2,floatbase);/*prototype*/base = (b-1)/n;
lower = a;
for(lower =a; lower <= b-base; lower = lower + base)
{
h1 = function_x(lower);
h1 = function_x(lower + base);
total_area += trap_area(h1, h2, base);
}
return(total_area);
float trap_area(float height_1,float height_2,floatbase)
{
float area; /* LOCAL VARIABLE */
area = 0.5 * (height_1 + height_2) * base;
return(area);
}
float function_x(float x)
{
/* F(X) = X * X + 1 */return(x*x + 1);
}

Output
AREA UNDER A CURVE
Enter lower limit: 0
Enter upper limit: 3
Enter number of trapezoids: 30
TOTAL AREA = 12.005000
AREA UNDER A CURVE
Enter lower limit: 0
Enter upper limit: 3
Enter number of trapezoids: 100
TOTAL AREA = 12.000438

Solution in java ::

// hackerx sasi kamaraj college of engineering and technology 2910007 java Program


//The answer to be precise... although the type was a double, it rounds off the answer. Any help would be //appreciated...
//java code: 1. :: try this or the another one below this one
//Program code ::

public class Reimann
{
private static double integral(String s, double[] descriptors, double lb, double ub)
{

double area = 0; // Area of the rectangle
double sumOfArea = 0; // Sum of the area of the rectangles
double oldSumOfArea = 0;
double width = ub - lb;
boolean firstPass = true;

while ( (Math.abs((oldSumOfArea - sumOfArea) / sumOfArea) > .0001) || firstPass )
{

System.out.println((Math.abs((oldSumOfArea - sumOfArea) / sumOfArea) > .0001) || firstPass);
if (s.equals("poly"))
{
for (int i = 1; i <= ((ub - lb) / width); i++) // represents # of rectangles
{
for (int j = 0; j < descriptors.length; j++) // Goes through all the coefficients
{
area = width * descriptors[j] * Math.pow ( (double)( (i * width + lb + (i -1.0) * width + lb) / 2.0 ), j);
/*Above code computes area of each rectangle */

sumOfArea += area;

}
}
}
width = width / 2;
firstPass = false;
oldSumOfArea = sumOfArea;
}
return sumOfArea;
}

/*private static void runMyTests()
{
assert ( integral() <= 48.00001 ) && ( integral() >= 47.99999 );
}*/

public static void main (String [] args)
{

double lb = Double.parseDouble(args[args.length -2]);
double ub = Double.parseDouble(args[args.length -1]);

double[] coefficients = new double[args.length - 3];

if (args[0].equals("poly"))
{
for (int i = 1; i < args.length - 2; i++)
{
coefficients[i-1] = Double.parseDouble(args[i]);
}

System.out.println(integral("poly", coefficients, lb, ub));
}
}
}



Java Program 2 ::

public class Riemann
{
private static double integral(String s, double[] descriptors, double lb, double ub)
{

double area = 0; // Area of the rectangle
double sumOfArea = 0; // Sum of the area of the rectangles
double oldSumOfArea = 0;
double width = ub - lb;
boolean firstPass = true;

while ( (Math.abs((oldSumOfArea - sumOfArea) / sumOfArea) > .0001) || firstPass )
{

System.out.println((Math.abs((oldSumOfArea - sumOfArea) / sumOfArea) > .0001) || firstPass);
if (s.equals("poly")) // Statement for polynomial
{
for (int i = 1; i <= ((ub - lb) / width); i++) // represents # of rectangles
{
for (int j = 0; j < descriptors.length; j++) // Goes through all the coefficients
{
area = width * descriptors[j] * Math.pow ( (double)( (i * width + lb + (i -1.0) * width + lb) / 2.0 ), j);
/*Above code computes area of each rectangle */

sumOfArea += area;

}
}
}

else if (s.equals("sin")) // Statement for sin
{
for (int i = 1; i <= ((ub - lb) / width); i++) // represents # of rectangles
{
for (int j = 0; j < descriptors.length; j++) // Goes through all the coefficients
{
area = width * descriptors[j] * Math.sin(Math.toRadians(( (double)( (i * width + lb + (i -1.0) * width + lb) / 2.0 ))));
/*Above code computes area of each rectangle */

sumOfArea += area;

}
}
}

else if (s.equals("cos")) // Statement for cos
{
for (int i = 1; i <= ((ub - lb) / width); i++) // represents # of rectangles
{
for (int j = 0; j < descriptors.length; j++) // Goes through all the coefficients
{
area = width * descriptors[j] * Math.cos(Math.toRadians(( (double)( (i * width + lb + (i -1.0) * width + lb) / 2.0 ))));
/*Above code computes area of each rectangle */

sumOfArea += area;

}
}
}

width = width / 2;
firstPass = false;
oldSumOfArea = sumOfArea;
}

return sumOfArea;
}

/*private static void runMyTests()
{
assert ( integral() <= 48.00001 ) && ( integral() >= 47.99999 );
}*/

public static void main (String [] args)
{
double lb = Double.parseDouble(args[args.length -2]);
double ub = Double.parseDouble(args[args.length -1]);

double[] coefficients = new double[args.length - 3];

if (args[0].equals("poly"))
{
for (int i = 1; i < args.length - 2; i++)
{
coefficients[i-1] = Double.parseDouble(args[i]);
}

System.out.println(integral("poly", coefficients, lb, ub));
}

else if (args[0].equals("sin"))
{
for (int i = 1; i < args.length - 2; i++)
{
coefficients[i-1] = Double.parseDouble(args[i]);
}

System.out.println(integral("sin", coefficients, lb, ub));
}

else if (args[0].equals("cos"))
{
for (int i = 1; i < args.length - 2; i++)
{
coefficients[i-1] = Double.parseDouble(args[i]);
}

System.out.println(integral("cos", coefficients, lb, ub));
}
}
}



Question ::
Area Under Curve

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd