Improper Integrals with the HP-15C LE & CE

[Werner]

Okay.. time to explain a little bit how the defining function used in integration (and solve, btw) works.

The integration routine calls the function to evaluate f(x). To that end, it fills the stack with the values of x - and your routine must produce f(x). Take your routine, I put the stack contents right next to it:

                X       Y       Z       T
 LBL 10         x       x       x       x
 1              1       x       x       x
 SQRT           1       x       x       x
 /              x       x       x       x
 RTN

There you are. As you demonstrated yourself as well, calling the routine with x=10, returns 10. So you essentially defined f(x)=x. That still doesn't solve your claim that, when integrating from 0 to 1, you obtain *2* instantly (you should get 0.5, as Namir said)

If instead you take Namir's definition:

                X       Y       Z       T
 LBL 10         x       x       x       x
 1/X           1/x      x       x       x
 SQRT        1/sqrt(x)  x       x       x
 RTN

This is the proper way to define f(x)=1/sqrt(x).

Cheers, Werner