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

.testvita, answer for tcs is working question

answer for tcs is working question

Wap to print the largest number from any 10 numbers, WAP TO PRINT THE LARGE...

WAP TO PRINT THE LARGEST NUMBER FROM ANY 10 NUMBERS #include stdio.h> #include conio.h>   void main()   {                    int a[10],i,max;

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

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

., what is polymorphism

what is polymorphism

Area of curve, Write a program to find the area under the curve y = f(x) be...

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

Last fibonnaci standing, i need a program for finding the last fibonacci nu...

i need a program for finding the last fibonacci number

C++ code, write c++ programm calculate electricity bill with person name,us...

write c++ programm calculate electricity bill with person name,use ,id

KRPano, I am seeking an expert to create a 360 degree virtual tour in KRPan...

I am seeking an expert to create a 360 degree virtual tour in KRPano for use on PCs, tablets and smartphones. You will be given with panoramas in order to create an advanced KRPano

Android - iphone mobile game, I am seeking a developer to create a mobile g...

I am seeking a developer to create a mobile game for me. I want an endless runner game with simple controls. The game needs to be a 3D game in a 2D view. The graphics and animation

Introduction to C language, Design and implement a library system that does...

Design and implement a library system that does the following: ? Takes details of a student/library users: first name, last name, other names, user ID number, and nationality. Th

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