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MathGroup Archive 1994

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RE: Negative Area: Definitive Answer (not)

  • To: mathgroup at christensen.cybernetics.net
  • Subject: [mg223] RE: [mg213] Negative Area: Definitive Answer (not)
  • From: Chip Sample <sample at shire.ac.arknet.edu>
  • Date: Wed, 23 Nov 1994 12:38:21 CST

I've figured it out!  Your example gives the wrong answer as shown here,

In[10]:=
Integrate[1/(A+B Cos[x])^2,{x,0,2 Pi}]/.{A->2,B->1}
Out[10]=
  -4 Pi
---------
3 Sqrt[3]

but if you reverse the roles of A and B you get the right answer.

In[11]:=
Integrate[1/(B+A Cos[x])^2,{x,0,2 Pi}]/.{A->1, B->2}
Out[11]=
  4 Pi
---------
3 Sqrt[3]

So clearly, you just put your A's and B's in the wrong places!  ;o)

Chip






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