(delete)
- Thread starter vstypicals
- Start date
You're kidding right? How about you try doing the assignment and we'll give you hints when you get stuck.
I'm doing it as we speak. =/ I know how to do most of the stuff. I just don't know how to put them together.. Sorry if that's confusing. Heh.
make an attempt at putting them together and if it doesn't work we can try to point out how to fix it 
Well, you would prompt the user... and if you know calculus, then it shoudl be easy to figure out a basic algorithm. HINT: It involves a loop.
Final product... (dont mind the commenting; the professor wanted that stuff.)
--------------------------------------------------------
#include <stdio.h>
/* variable declarations */
double func();
double h, sum;
int i, n;
main()
{
/* prompt the user for the number of iterations (n) */
printf("How many steps? ");
scanf("%d", &n);
sum = 0; /* initialize the sum to 0 */
h = 2.0/n; /* formula for height */
/* loop to add up of the sums */
for(i = 0; i < n; i++)
{
sum += ((h/2.0)*(func(i*h) + func(i*h + h)));
}
/* print results */
printf("The integral approximation from 0 to 2 with %d steps is %f\n", n, sum);
}
/* func - this function computes the square of x
*
* inputs:
* x - the value to be squared
*
* outputs:
* the square of x
*
* side effects:
* none
*/
double func(double x)
{
return x*x;
}
--------------------------------------------------------
#include <stdio.h>
/* variable declarations */
double func();
double h, sum;
int i, n;
main()
{
/* prompt the user for the number of iterations (n) */
printf("How many steps? ");
scanf("%d", &n);
sum = 0; /* initialize the sum to 0 */
h = 2.0/n; /* formula for height */
/* loop to add up of the sums */
for(i = 0; i < n; i++)
{
sum += ((h/2.0)*(func(i*h) + func(i*h + h)));
}
/* print results */
printf("The integral approximation from 0 to 2 with %d steps is %f\n", n, sum);
}
/* func - this function computes the square of x
*
* inputs:
* x - the value to be squared
*
* outputs:
* the square of x
*
* side effects:
* none
*/
double func(double x)
{
return x*x;
}
Uh.. It prompts the user for the number of steps.. then exits once it's given.. Or was that a trojan? =/
Originally posted by: vstypicals
Uh.. It prompts the user for the number of steps.. then exits once it's given.. Or was that a trojan? =/
It probably prompts and the prints the result and immediately closes. It goes by too fast for you to see.
Originally posted by: igowerf
Originally posted by: vstypicals
Uh.. It prompts the user for the number of steps.. then exits once it's given.. Or was that a trojan? =/
It probably prompts and the prints the result and immediately closes. It goes by too fast for you to see.
seriously. I didn't write a GUI app to do this, run it from a command line and you'll see the result.
Works just fine on two different windows machines for me.
Here's the source if you want to compile it yourself.
I just directly ran it.. and it did close too fast right after printing the results for me to see. I didn't run it from a command. Sorry.
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