Star
  • Star
what is the integral of xe^-x with limits 0 to infinity?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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myininaya
  • myininaya
\[\int\limits_{}^{}xe^{-x}dx=-xe^{-x}-\int\limits_{}^{}-e^{-x}dx=-xe^{-x}-e^{-x}\]
myininaya
  • myininaya
before we look at the limits lets check this
myininaya
  • myininaya
(-1)e^{-x}--x(-e^{-x})+e^{-x} =xe^{-x} YAY! now for the limists...

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myininaya
  • myininaya
\[\lim_{b \rightarrow \inf}(-be^{-b}-e^{-b}+e^0)\]
myininaya
  • myininaya
e^0=1 -e^{-b}->0 as b->inf now let's look at -be^{-b}
Star
  • Star
we are using limits because it is improper integral?
myininaya
  • myininaya
yes! :)
Star
  • Star
ahh i see. i was wondering where the e^0 came from?
myininaya
  • myininaya
i plugged in ther limits the bottom limit is 0 -[0e^0-e^0]=+e^0=+1
myininaya
  • myininaya
now we have
myininaya
  • myininaya
\[\lim_{b \rightarrow \inf}be^{-b}=\lim_{b \rightarrow \inf} \frac{b}{e^b}=\lim_{b \rightarrow \inf} \frac{1/e^b}\]
myininaya
  • myininaya
\[\lim_{b \rightarrow \inf}be^{-b}=\lim_{b \rightarrow \inf} \frac{b}{e^b}\]
myininaya
  • myininaya
use l'hospital's rule so we have \[\lim_{b \rightarrow \inf} \frac{1}{e^b}=\lim_{b \rightarrow \inf}e^{-b}=0\]
myininaya
  • myininaya
so we have -0-0+1=1
Star
  • Star
ah i see! thank you so much for your help :)
myininaya
  • myininaya
np

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