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Question: Can't figure this out

Question:Can't figure this out

Rouben Rostamian 7938 Maple
April 06 2015
1

Consider

> z := Int(f(t-s), s=0..1);

Int(f(t-s), s = 0 .. 1)

The result is clearly a function of t (and definitely not a function of s, which is a dummy variable).  Let's define

> F := unapply(z, t);

proc (t) options operator, arrow; Int(f(t-s), s = 0 .. 1) end proc

Then

> F(q);

Int(f(q-s), s = 0 .. 1)

as expected.  However:

> F(s);

Int(f(0), s = 0 .. 1)

which is not correct, as this confuses the argument s of F with the dummy integration variable s.  How would you salvage the situation?

 
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