anonymous
  • anonymous
Integration help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{1}^{\infty}dx/(x^5(e^{1/x}-1))\]
anonymous
  • anonymous
wut grade r u in
anonymous
  • anonymous
first year university

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mathmate
  • mathmate
Doesn't look like an analytic solution. It can be solved by series, or numerical. Are you in any of these?
anonymous
  • anonymous
show that \[e^t \ge 1+t\] for \[t \ge0\] hence explain briefly why the integral must converge.
mathmate
  • mathmate
By MacLaurin's expansion, \( e^t \) = 1+t/1!+t^2/2!+t^3/3! +... => \( e^t \ge \) 1+t for t\( \gt 0 \) This means that \( e^t-1 \ge \) 1+t-1 =t for t\( \gt 0 \) or \( 1/(e^t-1) \le \) 1/t for t\( \gt 0 \) therefore I < integral of 1/x^6
mathmate
  • mathmate
and \[\int\limits_{1}^{\infty} \frac{dx}{x^6} \ = \ \frac{1}{5}\]
anonymous
  • anonymous
i dont think we can use MacLaurin's expansion though
mathmate
  • mathmate
Then you can show that e^t=1 when t=0. and d(e^t)/dt >1 for t>0, therefore e^t-1 >0 for t>0.
anonymous
  • anonymous
ok thanks

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