area under curve, C/C++ Programming

Assignment Help:

write a program to find the area under the curve y=f(x) between x=a & x=b

Related Discussions:- area under curve

Explain the bit fields portable or not, Explain the  bit fields portable o...

Explain the  bit fields portable or not? - No, Bit fields aren't portable. - As Bit fields can't span machine words and number of bits in a machine word is different on diff

Program for greatest common divisor – c++ program, Greatest Common Divisor ...

Greatest Common Divisor (GCD) - The greatest common divisor (GCD) of two integers is the largest integer that will evenly divide both integers. The GCD algorithm involves intege

Compiler Design - Limit In The Method Instructions, Raj is a newbie to the ...

Raj is a newbie to the programming and while learning the programming language he came to know the following rules: · Each program must start with ''{'' and end with ''}''

Determine the canonical form, Rule: To determine the canonical form we s...

Rule: To determine the canonical form we should OR the min terms. A min term is defined as a Boolean equation of the input if the output is logic '1'. If the input is logic '1'

Search property from catalogue , Implement the search property from catalo...

Implement the search property from catalogue menu option.  After selecting this option the user should be asked to specify the property using the following sub-menu: 1. Specif

First line starts with T, Input Format: First line starts with T, which is...

Input Format: First line starts with T, which is the number of test cases. Each test case contains a string (S). Output Format: For each test case print the minimum number of ch

Hotel reservation, •Flow Chart and Pseudocode of Add module - Hotel booking...

•Flow Chart and Pseudocode of Add module - Hotel booking - Signup for new membership Delete module - Hotel reservation cancellation - Change of reservation

C++ Program Please see where i do mistake, #include #include #include ...

#include #include #include #include #include class Employee { private: char *Name; //Set them as pointers... int IdNumber; char *Department; char *Position; public: voi

diana

9/4/2012 4:19:48 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