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

**References**:**Constant function Integrate Assumption - More***From:*"Reece, Daryl" <Daryl.Reece@goody.com>