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Hunus Group Title

Question on integration by parts

  • 2 years ago
  • 2 years ago

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  1. Hunus Group Title
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    When integrating \[\int\limits_{0}^{1} xe^{-x}dx\] is the answer \[-2/e -1\] or\[1-2/e\]

    • 2 years ago
  2. nitishanand Group Title
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    1-2/e

    • 2 years ago
  3. Hunus Group Title
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    I keep getting \[-2/e-1\] but Wolfram says its the opposite

    • 2 years ago
  4. amistre64 Group Title
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    if you did the int by parts correctly, you should only arrive at one solution ... how did you go about this?

    • 2 years ago
  5. amistre64 Group Title
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    e^-x x -e^-x: -x e^-x -1 e^-x : -e^-x 0 -e^-x : 0 I get: -x e^(-x) - e^(-x) if i did it right

    • 2 years ago
  6. amistre64 Group Title
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    oh, then 0 to 1 lol

    • 2 years ago
  7. DHASHNI Group Title
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    let u =x dv=e^(-x)dx du=dx v=-e^(-x) =>-xe^(-x)-e^(-x) on limits the ans is -1/e-1/e-1=-2/e-1

    • 2 years ago
  8. amistre64 Group Title
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    -1 e^(-1) - e^(-1) + 0 e^(-0) + e^(-0) \[-\frac{1}{e}-\frac{1}{e}+0+1\] \[-\frac{2}{e}+1\]

    • 2 years ago
  9. amistre64 Group Title
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    the algebra is the killer in most cases

    • 2 years ago
  10. Hunus Group Title
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    \[u=x\] \[du=dx\] \[v=-e^{-x}\] \[dv=e^{-x}\] \[\int\limits_{0}^{1} xe^{-x}dx=-xe^{-x}|_{0}^{1} + \int_{0}^{1}e^{-x}dx\] \[\int\limits_{0}^{1} xe^{-x}dx=-1e^{-1} + -e^{-x}|_{0}^{1}\] \[\int\limits_{0}^{1} xe^{-x}dx=-e^{-1} + -e^{-1}+1\] \[\int\limits_{0}^{1} xe^{-x}dx=-2e^{-1}+1\]

    • 2 years ago
  11. Zarkon Group Title
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    @Hunus your integrand is positive (on (0,1)) ...your answer is negative...this tells you you are not correct.

    • 2 years ago
  12. Hunus Group Title
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    Yea, I see where I went wrong on the \[-e^{-x}|_{0}^{1}\] I put -e^-1 - 1 instead of -e^-1 -(-1)

    • 2 years ago
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