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Re: Sheer frustration with integration of piecewise continuous functions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40911] Re: Sheer frustration with integration of piecewise continuous functions
  • From: bobhanlon at aol.com (Bob Hanlon)
  • Date: Thu, 24 Apr 2003 05:24:22 -0400 (EDT)
  • References: <b85m4u$ace$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

This will work much better if you use UnitStep

f[x_,L_,fpeak_]:= 
    fpeak/L(5x/3*UnitStep[x]+
          (L-5x/3)*UnitStep[x-3L/5]+
          (10(L-x)-L)*UnitStep[x-9L/10]-
          10(L-x)*UnitStep[x-L]);

Plot[f[x,1,1],{x,-0.1,1.1}, 
    PlotStyle -> {AbsoluteThickness[2], RGBColor[1, 0, 0]}];

FullSimplify[Integrate[f[x,L,fpeak],{x,0,L}], L>=0]

(-(13/20))*fpeak*L*(-1 + UnitStep[-L])

However, since

Simplify[%, L==0]

0

The result is just

FullSimplify[%%, L>0]

(13*fpeak*L)/20


Bob Hanlon

In article <b85m4u$ace$1 at smc.vnet.net>, Madhusudan Singh
<spammers-go-here at yahoo.com> wrote:

<< 
Subject:	Sheer frustration with integration of piecewise continuous
functions
From:		Madhusudan Singh <spammers-go-here at yahoo.com>
To: mathgroup at smc.vnet.net
Date:		Wed, 23 Apr 2003 09:24:46 +0000 (UTC)

Clear["'*"];
(*ClearAttributes[Which, HoldAll];*)

f[x_, L_, fpeak_] := (fpeak /L) Which[((0 <= x) && (x < 0.6 L)), x/(0.6 
),((0.6 L <= x) && (x <= 0.9 L)), L, ((0.9 L < x) && (x <= L)) , 10 (L 
-x)];
Plot[f[x, 1, 1], {x, 0, 1}];
Print[Integrate[f[x, L, fpeak], {x, 0, L}, Assumptions -> {L >= 0, fpeak >= 
0, x >= 0, x <= L}]];

I have struggled with the above integration (believe me, its just a test 
case, I have a much more complicated function) for an hour now.

What is missing above ? The plot evaluates but the integral does not. Why 
does Mathematica make it so damned difficult to work with piecewise 
continuous functions ?












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