Re: Integration Problem

• To: mathgroup at smc.vnet.net
• Subject: [mg102040] Re: [mg102004] Integration Problem
• From: Patrick Scheibe <pscheibe at trm.uni-leipzig.de>
• Date: Mon, 27 Jul 2009 05:56:16 -0400 (EDT)
• References: <200907260755.DAA18985@smc.vnet.net>

Hi,

at a first glance I ignored your solution and just wrote my solution
down

arcl = Sqrt[#.#] &[D[#, t] & /@ {Exp[2 t], Exp[3/2 t]}]
Integrate[arcl, {t, 1, 2}]
N[%]

This gives your predicted solution, but please note that I substituted
the 1.5 with 3/2.
After checking your solution, I couldn't find any mistake since I just
didn't know what "bad thing" should happen when using 1.5 except of 3/2.
Now check following

arcl1 = Sqrt[#.#] &[D[#, t] & /@ {Exp[2 t], Exp[3/2 t]}]
arcl2 = Sqrt[#.#] &[D[#, t] & /@ {Exp[2 t], Exp[1.5 t]}]
Integrate[arcl1, {t, 1, 2}] // N
NIntegrate[arcl1, {t, 1, 2}]
Integrate[arcl2, {t, 1, 2}] // N
NIntegrate[arcl2, {t, 1, 2}]

{49.7621, 49.7621, 79.6261, 49.7621}

But on the other hand, if you integrate and *then* substitute the
integration interval with the *bad itegral* you get

(# /. t -> 2) - (# /. t -> 1) &@Integrate[arcl2, t]

49.7621

And finally taking the absolute value

Integrate[Abs[arcl2], {t, 1, 2}]

49.7621

gives the right result. Anyone who knows what's happening here exactly?

Cheers
Patrick

On Sun, 2009-07-26 at 03:55 -0400, JerrySpock wrote:
> Hello, everyone.
>
> I'm having a problem integrating to find an arc length.
>
> I have two parametric equations:
>
> x=e^(2t)
>
> and
>
> y=e^(1.5t)
>
> I'm looking for the arc length from 1 to 2.
>
> N[
> Integrate[
> Sqrt[
> (2Exp[2*m])^2 + (1.5Exp[1.5*m])^2
> ],{m, 1, 2}]]
>
> I keep getting the answer 79.6, but my TI-83 says the answer is 49.8.  I've been playing with this for hours, and I can't get it to work.  Any ideas what I'm doing wrong?
>
>     [Edited by: admin on Jul 25, 2009 7:22 AM]
>

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