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MathGroup Archive 2004

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Re: Constant function Integrate Assumption - More

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47268] Re: [mg47252] Constant function Integrate Assumption - More
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 2 Apr 2004 03:30:11 -0500 (EST)
  • References: <200404010503.AAA28320@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 1 Apr 2004, at 14:03, Reece, Daryl wrote:

> Many thanks to everyone offering solutions to my dilemma.  In trying to
> simplify the problem, I must have slipped up and oversimplified.
>
> Specifically I want to start with
>
> Integrate[50000 Units/Year UnitStep[(t-5 Month)/Month], t]
>
> and receive
>
> 50000 Units/Year (t-5 Month) UnitStep[(t-5 Month)/Month].
>
> I need the full answer so that the units work out correctly in the 
> answer.
> I am starting with a sale rate and want to derive the total sales as a
> function of time.  This time may then be input as Days, Months, 
> Years....
> If I perform the replace operation, as many of you suggest, this 
> ability
> disappears.
>
> A lister rightly pointed out that Mathematica is confused not because 
> it
> assumes that Month may be a function of t, but rather it does not know 
> the
> sign of Month.  Even using the assumption Month>0 does not fix the 
> problem.
>
> Other suggestions?
>

Only the same I have already made:


Integrate[50000*(Units/Year)*
    UnitStep[(s - 5*Month)/Month], {s, -Infinity, t},
   Assumptions -> Month > 0]


(50000*(t - 5*Month)*Units*UnitStep[t/Month - 5])/Year


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