Re: Integration Problem

• To: mathgroup at smc.vnet.net
• Subject: [mg102026] Re: Integration Problem
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Mon, 27 Jul 2009 05:53:40 -0400 (EDT)

```On 7/26/09 at 3:55 AM, liquidsolids at hotmail.com (JerrySpock) wrote:

>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?

When you want a numeric result use NIntegrate instead of
N[Integrate. That is

In[2]:= NIntegrate[
Sqrt[(2 Exp[2*m])^2 + (1.5 Exp[1.5*m])^2], {m, 1, 2}]

Out[2]= 49.7621

IMO, this is the simplest, most efficient way to deal with the
issue. The other choice would be:

In[3]:= N[
Integrate[Sqrt[(2 Exp[2*m])^2 + (3/2 Exp[3/2*m])^2], {m, 1, 2}]]

Out[3]= 49.7621

where I've replaced all of the machine precision values with
exact values. Integrate is intended to give exact symbolic
solutions to integrals. You cannot get exact symbolic solutions
if some of the constants are machine precision numbers. If you
have machine precision numbers use NIntegrate which intended for
this use.

```

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