Re: Integrate with unknow function

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• Subject: [mg130231] Re: Integrate with unknow function
• From: "shan@mitbbs" <shan.pub at gmail.com>
• Date: Sun, 24 Mar 2013 04:17:25 -0400 (EDT)
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Hi Bob,

It works great!

But I actually can't define g as the indefinite integral of function f[x].
Can mathematica handle something like the following problem?

Two functions f1 and f2, i.e.,
f1=Integrate[f11[x],{x,0,a1}]+
Integrate[f12[x],{x,a1,k}]
f2=Integrate[f21[x],{x,0,a2}]+Integrate[f22[x],{x,a2,k}]
where f11,f12,f21,f22 have no exact forms, and 0<a1<a2<k.

Solve f1-f2 and get result as [image:
[;\int_0^{a1}(f11[x]-f21[x])dx+\int_{a1}^{a2}(f12[x]-f21[x])dx+\int_{a2}^{k}(f12[x]-f22[d])dx;]]

Thanks a lot!

Shan

On Sat, Mar 23, 2013 at 8:04 AM, Bob Hanlon <hanlonr357 at gmail.com> wrote:

> If g is the indefinite integral of a contrinuous function f[x] then
>
> Using a replacement rule
>
> ClearAll[f, g, expr];
>
> expr = Integrate[f[d], {d, 0, 3}] -
>    Integrate[f[d], {d, 0, 2}];
>
> expr /. Integrate[f[x_], {x_, a_, b_}] ->
>   g[b] - g[a]
>
> -g[2] + g[3]
>
> Alternatively, defining an upvalue for f (TagSet)
>
> ClearAll[f, g, expr];
>
> expr = Integrate[f[d], {d, 0, 3}] -
>    Integrate[f[d], {d, 0, 2}];
>
> f /: Integrate[f[x_], {x_, a_, b_}] =
>   g[b] - g[a];
>
> expr
>
> -g[2] + g[3]
>
>
> Bob Hanlon
>
>
> On Sat, Mar 23, 2013 at 3:25 AM, Shan <shan.pub at gmail.com> wrote:
> > Hi,
> >
> > I have a very rookie question as follows:
> >
> > Integrate[f[d], {d, 0,3}] - Integrate[f[d], {d, 0,2}]
> >
> > How can I get the result as f[d]? Thanks very much for any help!
> >
> > shan
> >
>



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