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Re: Integration Problem

  • To: mathgroup at
  • Subject: [mg102036] Re: Integration Problem
  • From: mark mcclure <mcmcclur at>
  • Date: Mon, 27 Jul 2009 05:55:32 -0400 (EDT)
  • References: <h4h1vg$ibr$>

On Jul 26, 3:52 am, JerrySpock <liquidsol... at> wrote:
> 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.

Definitely, a little strange.  Could be that there's a little
numerical instability in the integrand, but I don't see it.  Of
course, you should always analyze your answer from multiple
perspectives.  In this case, a simple plot shows that 49.8 is closest
to the correct answer.  The most straightforward way that I can think
to do the problem in Mathematica is as follows

p[t_] = {Exp[2 t], Exp[3 t/2]};
NIntegrate[Norm[p'[t]], {t, 1, 2}]

Note that I used NIntegrate, rather than N[Integrate[...]] as you
did.  The problem seems to arise only when you try to apply Integrate
to an inexact integrand.  If your integrand is inexact, then why not
apply NIntegrate in the first place?  It should be faster and,
perhaps, more accurate.  Also,

Integrate[Sqrt[((3/2) Exp[3 t/2])^2 + (2 Exp[2 t])^2], {t, 1, 2}]

produces the correct result, as you can check by passing the result to
N.  On the other hand,

Integrate[Sqrt[((3/2.) Exp[3 t/2])^2 + (2. Exp[2 t])^2], {t, 1, 2}]

produces a totally bogus negative result, while

Integrate[Sqrt[((3/2.) Exp[3 t/2.])^2 + (2. Exp[2. t])^2], {t, 1, 2}]

refuses to integrate!

Mark McClure

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