Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1994
*January
*February
*March
*April
*May
*June
*July
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1994

[Date Index] [Thread Index] [Author Index]

Search the Archive

RE: Negative Area

  • To: mathgroup at christensen.cybernetics.net
  • Subject: [mg220] RE: [mg213] Negative Area
  • From: Chip Sample <sample at shire.ac.arknet.edu>
  • Date: Wed, 23 Nov 1994 11:12:10 CST

On integrating a +ve function (A>B) I obtain a -ve

function.  Would you consider this to be a bug?

In[347]:=
Integrate[1/(A+B Cos[x])^2,{x,0,2 Pi}]/.{A->2,B->1}
Out[347]=
  -4 Pi
---------
3 Sqrt[3]

---
Yours,

Roger
(rmj at llewelyn.slac.stanford.edu)
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Since the form of the indefinite integral depends on the values of A and B, the 
proper thing for Mma to do is leave the integral unevaluated.  

Maybe this is related to other complaints that MMa is incapable of keeping 
track of such constraints as A is integer or B >0.  It would seem to need this 
capability to recognize that a pattern match is conditional on such a 
constraint.

At any rate,  it seems like a good idea to give Mma all the inputs you have up 
front  if possible to avoid this "bug".  I tried:
 
Integrate[1/(A+B Cos[x])^2/.{A->2,B->1},{x,0,2 Pi}]

Out[4]=
  4 Pi
---------
3 Sqrt[3]

apparently the right answer.

Chip





  • Prev by Date: RE: Negative Area: Definitive Answer (not)
  • Next by Date: Re: recognition puzzle
  • Previous by thread: Negative Area
  • Next by thread: Re: Negative Area