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Re: Sovling integrals: non-algebraic???

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  • Subject: [mg35376] Re: [mg35369] Sovling integrals: non-algebraic???
  • From: Andrzej Kozlowski <andrzej at>
  • Date: Wed, 10 Jul 2002 02:19:42 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

Your equation is:

Integrate[(c x^d), {x, a, b}] ==y

c*(-(a^(1 + d)/(1 + d)) + b^(1 + d)/(1 + d)) == y

Solve cannot solve this because it is not meant to. Solve can only solve 
equations which are "essentially algebraic", which means basically all 
for which they are general non-numerical methods.

Your other attempt using SolveAlways is just a case of complete 
misunderstanding of what this function does. SolveAlways is for 
"identities". For example:
SolveAlways[d + x == d, d]

means find x such that for all d, d + x == x, so of course the answer is

{{x -> 0}}

Your equation

SolveAlways[c*(-(a^(1 + d)/(1 + d)) + b^(1 + d)/(1 + d)) == y,d]

is saying: find c,a, b and y such that for all d the above relation 
holds. In fact the relation cannot hold for all d  whatever the values 
of the variables, but in any case this clearly has nothing to do with 
what you were trying to do.

In fact your equation can't be "solved" if by solving you mean 
expressing a "general solution" in terms of some known functions of the 
parameters. It is indeed, as you wrote, "mathematically impossible".

Andrzej Kozlowski
Toyama International University

On Tuesday, July 9, 2002, at 07:51  PM, Björn wrote:

> Hello!
> I can't make Mathematica solve an expression for a variable which is
> inside an integral. Integrals are of the form:
> Integrate[f, {x, xmin, xmax}]
> And my expression is of the kind:
> Solve[Integrate[(c x^d), {x, a, b}] == y, d]
> Now, Mathematica can solve for a, b and c. But it cannot solve for d.
> Why?
> It gives me the message:
> "The equations appear to involve the variables to be solved for in an
> essentially non-algebraic way."
> If I try SolveAlways[Integrate[c x^d, {x, a, b}] == y, d],
> I get the message:
> "The expression (a^(1 + d)) involves unknowns in more than one
> argument, so inverse functions cannot be used."
> (It isn't mathematically impossible, right?)
> Most thankful for any help!

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