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Re: piecewise integration


load[x_]=-9*10^3*
    DiracDelta[x]+Piecewise[{{x*10*(10^3/3),0â?¤xâ?¤3}},0]-6*10^3*DiracDelta[x-5];

If a bound of the integration is on a DiracDelta then Mathematica integrates that DiracDelta to 1/2 of its coefficient

Integrate[DiracDelta[x],{x,0,1}]

1/2

Integrate[DiracDelta[x],{x,-1,0}]

1/2

Integrate[DiracDelta[x],{x,-1,1}]

1

This impacts both ends of your integral

Integrate[#,{x,0,5}]&/@load[x]

7500

Integrate[#,{x,0,6}]&/@load[x]

4500

Integrate[#,{x,-1,5}]&/@load[x]

3000

Integrate[#,{x,-1,6}]&/@load[x]

0


Bob Hanlon

---- Chris Chiasson <chris at chiasson.name> wrote: 
> The Integrate result seems pretty weak. Is there any way to obtain a
> more explicit exact answer besides manually converting the Piecewise
> function to two UnitStep functions? Can the same be done if the final
> limit of integration is a variable?
> 
> in
> 
> load[x_]=-9*10^3*DiracDelta[x]+Piecewise[{{x*10*(10^3/3),0<=x<=3}}]-6*10^3*DiracDelta[x-5]//InputForm
> 
> out
> 
> -6000*DiracDelta[-5 + x] - 9000*DiracDelta[x] +
> Piecewise[{{(10000*x)/3, 0 <= x <= 3}}, 0]
> 
> in
> 
> Integrate[load[x],{x,0,5}]//InputForm
> 
> out
> 
> Integrate[InputForm[-6000*DiracDelta[-5 + x] - 9000*DiracDelta[x] +
> Piecewise[{{(10000*x)/3, 0 <= x <= 3}}, 0]], {x, 0, 5}]
> 
> -- 
> http://chris.chiasson.name/
> 

--

Bob Hanlon
hanlonr at cox.net



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