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anonymous
 4 years ago
Integration help!
anonymous
 4 years ago
Integration help!

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{\infty}dx/(x^5(e^{1/x}1))\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first year university

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1Doesn't look like an analytic solution. It can be solved by series, or numerical. Are you in any of these?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0show that \[e^t \ge 1+t\] for \[t \ge0\] hence explain briefly why the integral must converge.

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1By 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 \( e^t1 \ge \) 1+t1 =t for t\( \gt 0 \) or \( 1/(e^t1) \le \) 1/t for t\( \gt 0 \) therefore I < integral of 1/x^6

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1and \[\int\limits_{1}^{\infty} \frac{dx}{x^6} \ = \ \frac{1}{5}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont think we can use MacLaurin's expansion though

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1Then you can show that e^t=1 when t=0. and d(e^t)/dt >1 for t>0, therefore e^t1 >0 for t>0.
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