Help with Array: Evaluating Integral of sin(x)

[999iscool]

You will write a program that evaluates the integral of sin(x) using the left-hand rectangle rule with
2000 subintervals, over 10 intervals. The intervals to test are [0, 1), [1, 2), …, [8, 9), [9, 10). You will
declare an array of type double that can hold 10 elements, and you will use this array to hold all 10
results you get from evaluating each interval.

What I need help with is the output.

What should be our output?
For example, [0, 1) ?

 

[Mark44]

The output should be the 10 elements in the array. Each array element is the number you get from integrating sin(x) on one of the intervals.

For example, result[0] should have a number that is close to -.5403.
Edit: that should be .4597.

 

[999iscool]

The output should be the 10 elements in the array. Each array element is the number you get from integrating sin(x) on one of the intervals.

For example, result[0] should have a number that is close to -.5403.

How did you get that number? I thought plain riemann sum (in this case, the right hand rule)

Here is the short version (with no array), tested [0,1)

#include <iostream>
#include <cmath>
   using namespace std;
   
double integrand(double);

int main()
{
	const int SIZE = 10;
	const double WIDTH = 0.0005;  // (b-a)/n
	
	double sum_array[SIZE];
	double a = 0.0;
	double b = 1.0;
	double sum = 0.00;
	double y1 = 0;

	for (double k = a; k < b; k += WIDTH)
	{
		y1 = integrand(k);
		sum += y1;
	}
	cout << (sum*WIDTH) << endl;
	return 0;	
}

double integrand(double x)
{
	double y;
	y = sin(x);
	return y;
}

edited the code again (added (sum*WIDTH). I forgot about this)

I got 0.45968 for [0, 1) using left endpoint

 

[Mark44]

Forgot to add 1. Make that .4597.

 

[999iscool]

Forgot to add 1. Make that .4597.

LOL
OKay, so it is 1-cos(1) for [0, 1)

Just really need to confirm that. Otherwise, even if I have perfect codes, with no REAL solution i still can't test it.
My other classmates are giving me weird numbers.

Thanks

 

[Mark44]

How did you get that number? I thought plain riemann sum (in this case, the right hand rule)

Your first post says to use the left endpoints.

Here is the short version (with no array), tested [0,1)

#include <iostream>
#include <cmath>
   using namespace std;
   
double integrand(double);

int main()
{
	const int SIZE = 10;
	const double WIDTH = 0.0005;  // (b-a)/n
	
	double sum_array[SIZE];
	double a = 0.0;
	double b = 1.0;
	double sum = 0.00;
	double y1 = 0;

	for (double k = a; k < b; k += WIDTH)
	{
		y1 = integrand(k);
		sum += y1;
	}
	cout << (sum*WIDTH) << endl;
	return 0;	
}

double integrand(double x)
{
	double y;
	y = sin(x);
	return y;
}

edited the code again (added (sum*WIDTH). I forgot about this)

I got 0.45968 for [0, 1) using left endpoint

Yeah, that's what you should get.

Some suggestions. Your integrand function can be shortened.

double integrand(double x)
{
    return sin(x);
}

Also, you could write a function that does essentially what you do in main. It should take three parameters: left endpoint, right endpoint, number of subintervals, and should return the computed value for that interval. The 10-element array would still be in main -- just loop through each element in the array and call your calculating function for each of the intervals 0 - 1, 1 - 2, ... 9 -10.