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

Arrays, how to declare arrays

how to declare arrays

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

Pro, em Ipsum is that it has a more-or-less normal distribution of letters,...

em Ipsum is that it has a more-or-less normal distribution of letters, as opposed to using ''Content here, content here'', making it look like readable English. Many desktop publis

Expression, i need expression and its types with example programs in c++

i need expression and its types with example programs in c++

Define generic bitwise operations, Define Generic Bitwise Operations? ...

Define Generic Bitwise Operations? Bitwise operators only work on an incomplete number of types int and char. It turns out there's above one kind of int. In particular there's

Implementation of the stack class in c++, Implementation of the Stack class...

Implementation of the Stack class in C++: How to implement stack class in c++. int Stack::push(int elem) {    if (top    {       list[top++] = elem;       r

How does c++ help with the tradeoff of security vs.usability, A: In C, enca...

A: In C, encapsulation was completed by making things static in a compilation unit or module. It prevented another module from accessing the static stuff. (Incidentally, now static

C program to display the contents of file on a screen, Program is to displa...

Program is to display the contents of file on a screen: Write a program to display the contents of file on a screen void main()   {  clrscr();  ifstream fin("ascii

Programs that perform the identical recursive algorithm, Goals For this ass...

Goals For this assignment you will write programs in C and LC-3 assembly code. Both programs will perform the identical recursive algorithm. The goals of this programming assignmen

Hungarian notation, describe how identifiers of different data type are def...

describe how identifiers of different data type are defined using this notaion?

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