area of 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. The area under a curve between two points can be found by doing a definite integral between the two points.

Related Discussions:- area of curve

Functions, write a program to calculate gross salary and net salary using h...

write a program to calculate gross salary and net salary using hra da pf in c++

#tit, 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

Describe how can i allocate/unallocate an array of things?, A: Use p = new ...

A: Use p = new T[n] and delete[] p:   Fred* p = new Fred[100]; ... delete[] p; Any time you allocate an array of objects through new (generally with the [n] in the n

Artificial block, What if I cannot wrap the local in an artificial block? n...

What if I cannot wrap the local in an artificial block? need help on Artificial Block in c++.

Create a program for decision structure , Create the following program or p...

Create the following program or propose one of your own. The program must include at least one of each of the following. Decision structure Repetitive structure fu

Recursive procedure to computes the number of digits, (a) Write a recursive...

(a) Write a recursive procedure (digits n) that computes the number of digits in the integer n using a linear recursive process. For example, (digits 42) should return 2 and (digit

What is precedence and order of evaluation, Precedence and Order of evaluat...

Precedence and Order of evaluation The languages follow a standard precedence for basic operators. Precedence rules help in deleting ambiguity of the order of operations perfor

String, A string is said to be "Beautiful"€, if it contains only non repet...

A string is said to be "Beautiful"€, if it contains only non repetitive alphabets

diana

9/4/2012 4:19:46 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