Re: Definite integral from a to b of sqrt(x) (one-valued vs. multi-valued)

On Wed, 28 Nov 2007 15:21:44 -0800 (PST), pfntjux@xxxxxxxxx wrote:


Given that 0 < a < b, the definite integral from a to b of sqrt(x) yields
a positive value c. That's true if we treat sqrt(x) as a one-valued
function.

We do not have to "treat" sqrt(x) as a single-valued function, but x |->
sqrt(x) just IS a single-valued function (at least in "standard" math,
i.e., say, Real Analysis).

Moreover,

sqrt(x) >= 0 for all x e IR, x >= 0.


Now, I've sometimes seen the graph of sqrt(x) drawn as a
multi-valued function and it looks like a "C".

Well actually this figure isn't the graph of the function x |-> sqrt(x),
but consists of _two_ graphs, namely the graph of the function x |->
sqrt(x) and the graph of the function x |-> -sqrt(x).


___
^ V x
| /
2 + _ --
| _ ~ ´
1 + . ´
| ´
0 +----|----|----|----|---> x
| 1 2 3 4
-1+ `.
| ` - _
-2+ ` -- __
| \ ___
-V x



In such circumstances, is it correct to say that said definite integral
yields +/- c?

No. (Since sqrt(x) >= 0 for all x e IR, x >= 0.)


Does it make sense to use multi-valued functions in an integral, or
must one necessarily use one-valued functions?

Well, at least the definition of an /integral/ _I_ learned, only
involved "one-valued functions".


F.

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