From: Jorge on 6 Jan 2010 19:19 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 6 Jan 2010 19:30 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 7 Jan 2010 14:48 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 1cos(1)) remains. Just in how is made the output.
From: Dave Hayden on 7 Jan 2010 15:44 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 1cos(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 1COS(X) Dave
From: Han on 7 Jan 2010 20:39 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 1cos(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 1COS(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 lookup 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.

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