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anonymous

  • 5 years ago

how do you integrate t times e to the negative t? (te^-t)?

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  1. anonymous
    • 5 years ago
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    by parts

  2. anonymous
    • 5 years ago
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    (te^-t) just and one to 1then put te over -t+1 te^-t+1/-t+1

  3. anonymous
    • 5 years ago
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    \[t e ^{-1/}/-1 +\int\limits e^t dt\]

  4. amistre64
    • 5 years ago
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    very carefully; you dont wana scare it

  5. anonymous
    • 5 years ago
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    \[-t/e^t+ e^t +C\]

  6. amistre64
    • 5 years ago
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    parts is good ..... e^-t +t -e^-t -1 e^-t +0 ............. right?

  7. amistre64
    • 5 years ago
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    -t e^-t + e^-t + C ??

  8. anonymous
    • 5 years ago
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    so u=t? and dv=e^-t?

  9. anonymous
    • 5 years ago
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    oh missed it :)

  10. amistre64
    • 5 years ago
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    yes

  11. anonymous
    • 5 years ago
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    so is v=-e^-t?

  12. anonymous
    • 5 years ago
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    \[\int\limits_{}^{}te^{-t}dt=-te^{-t}+\int\limits_{}^{}e^{-t}dt=-te^{-t}-e^{-t}+C\]

  13. anonymous
    • 5 years ago
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    u=t, du=1 dv=e^-t, v=-e^-t

  14. amistre64
    • 5 years ago
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    Lets see if this reverts back... Dt(-t e^-t + e^-t + C) t e^-t - e^-t e^-t (t-1) ... not quite, unless i forgot how to do that too ;)

  15. anonymous
    • 5 years ago
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    thanks i got it :)

  16. amistre64
    • 5 years ago
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    yay!!

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