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Star

  • 5 years ago

what is the integral of xe^-x with limits 0 to infinity?

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  1. myininaya
    • 5 years ago
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    \[\int\limits_{}^{}xe^{-x}dx=-xe^{-x}-\int\limits_{}^{}-e^{-x}dx=-xe^{-x}-e^{-x}\]

  2. myininaya
    • 5 years ago
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    before we look at the limits lets check this

  3. myininaya
    • 5 years ago
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    (-1)e^{-x}--x(-e^{-x})+e^{-x} =xe^{-x} YAY! now for the limists...

  4. myininaya
    • 5 years ago
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    \[\lim_{b \rightarrow \inf}(-be^{-b}-e^{-b}+e^0)\]

  5. myininaya
    • 5 years ago
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    e^0=1 -e^{-b}->0 as b->inf now let's look at -be^{-b}

  6. star
    • 5 years ago
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    we are using limits because it is improper integral?

  7. myininaya
    • 5 years ago
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    yes! :)

  8. star
    • 5 years ago
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    ahh i see. i was wondering where the e^0 came from?

  9. myininaya
    • 5 years ago
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    i plugged in ther limits the bottom limit is 0 -[0e^0-e^0]=+e^0=+1

  10. myininaya
    • 5 years ago
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    now we have

  11. myininaya
    • 5 years ago
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    \[\lim_{b \rightarrow \inf}be^{-b}=\lim_{b \rightarrow \inf} \frac{b}{e^b}=\lim_{b \rightarrow \inf} \frac{1/e^b}\]

  12. myininaya
    • 5 years ago
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    \[\lim_{b \rightarrow \inf}be^{-b}=\lim_{b \rightarrow \inf} \frac{b}{e^b}\]

  13. myininaya
    • 5 years ago
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    use l'hospital's rule so we have \[\lim_{b \rightarrow \inf} \frac{1}{e^b}=\lim_{b \rightarrow \inf}e^{-b}=0\]

  14. myininaya
    • 5 years ago
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    so we have -0-0+1=1

  15. star
    • 5 years ago
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    ah i see! thank you so much for your help :)

  16. myininaya
    • 5 years ago
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    np

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