From: Jorge on
I would like to ask to you about this doubt I found "playing" a little
with my HP50G:

The Mode operating conditions are: ALG, STD, RAD, RECT. (Release 2.15)
Making the Integer Function in the "0" to "1" interval for SIN(X) dX,
the
result is:

"-COS(1)--1"

Making it in the interval from 0 to Pi the result is "2".


Thanks in advance for your answer.
From: Daniel on
On Jan 6, 6:19 pm, Jorge <jorge.po...(a)gmail.com> wrote:
> I would like to ask to you about this doubt I found "playing" a little
> with my HP50G:
>
> The Mode operating conditions are: ALG, STD, RAD, RECT. (Release 2.15)
> Making the Integer Function in the "0" to "1" interval for SIN(X) dX,
> the
> result is:
>
> "-COS(1)--1"
>
> Making it in the interval from 0 to Pi the result is "2".
>
> Thanks in advance for your answer.

That is correct. Confirm for yourself with Wolfram Alpha.

http://www.wolframalpha.com/input/?i=integrate+sin(x)+from+0+to+pi
http://www.wolframalpha.com/input/?i=integrate+sin(x)+from+0+to+1
From: Jorge on
On 6 ene, 19:30, Daniel <djpetro...(a)gmail.com> wrote:
> On Jan 6, 6:19 pm, Jorge <jorge.po...(a)gmail.com> wrote:
>
> > I would like to ask to you about this doubt I found "playing" a little
> > with my HP50G:
>
> > The Mode operating conditions are: ALG, STD, RAD, RECT. (Release 2.15)
> > Making the Integer Function in the "0" to "1" interval for SIN(X) dX,
> > the
> > result is:
>
> > "-COS(1)--1"
>
> > Making it in the interval from 0 to Pi the result is "2".
>
> > Thanks in advance for your answer.
>
> That is correct. Confirm for yourself with Wolfram Alpha.
>
> http://www.wolframalpha.com/input/?i=integrate+sin(x)+from+0+to+pihttp://www.wolframalpha.com/input/?i=integrate+sin(x)+from+0+to+1

Thanks for your clarifying answer. But the doubt on how HP50G gives
the output for the first case (even if mathematically correct -cos(1)
- - 1 is 1-cos(1)) remains. Just in how is made the output.
From: Dave Hayden on
On Jan 7, 2:48 pm, Jorge <jorge.po...(a)gmail.com> wrote:
> On 6 ene, 19:30, Daniel <djpetro...(a)gmail.com> wrote:
>
>
> Thanks for your clarifying answer. But the doubt on how HP50G gives
> the output for the first case (even if mathematically correct -cos(1)
> - - 1 is 1-cos(1)) remains. Just in how is made the output.

If you press EVAL, you'll get -(cos(1)-1), which is a little better.
You can get what you want by going to pressing MODE F3 and checking
"Incr Pow" to tell the CAS to display equations with the constant on
the left instead of the right. Now press OK twice to return to the
stack display and press EVAL to get 1-COS(X)

Dave



From: Han on
On Jan 7, 2:44 pm, Dave Hayden <d...(a)larou.com> wrote:
> On Jan 7, 2:48 pm, Jorge <jorge.po...(a)gmail.com> wrote:
>
> > On 6 ene, 19:30, Daniel <djpetro...(a)gmail.com> wrote:
>
> > Thanks for your clarifying answer. But the doubt on how HP50G gives
> > the output for the first case (even if mathematically correct -cos(1)
> > - - 1 is 1-cos(1)) remains. Just in how is made the output.
>
> If you press EVAL, you'll get -(cos(1)-1), which is a little better.
> You can get what you want by going to pressing MODE F3 and checking
> "Incr Pow" to tell the CAS to display equations with the constant on
> the left instead of the right.  Now press OK twice to return to the
> stack display and press EVAL to get 1-COS(X)
>
> Dave

I don't know how integration works for the HP50, but on the HP48, many
"basic" integrals are done using a method similar to using a look-up
table (via MATCH commands). Since the integrand is SIN(X), the
antiderivative is -COS(X). The output that you see is simply the
result of the Fundamental Theorem of Calculus. Using the Fundamental
Theorem of Calculus (which _IS_ implemented in the HP48), results in:

INT(0,1,SIN(X),X) = -COS(1) - -COS(0)

My guess is that the HP50, due to the settings, automatically
simplifies -COS(0) to -1.