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Homework Help: Area under Integral.

  1. Oct 10, 2007 #1
    This question is quite hard for me.
    1. The problem statement, all variables and given/known data
    Find the area between the curve [tex]y=e^{-x}\left|sinx\right|[/tex] and the straight line y=0 for [tex]x\geq 0[/tex]

    3. The attempt at a solution

  2. jcsd
  3. Oct 10, 2007 #2


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    Because sine and cosine cross the x-axis, you can't just take the absolute value thorough the integral like that.
  4. Oct 10, 2007 #3
    Than ,actually, i should divide it to pieces.
    Last edited: Oct 10, 2007
  5. Oct 10, 2007 #4
    Is it equals to
    [tex](-1)^k\sum_{k=1}^{\infty}\frac{e^{-k\pi}}{2}[/tex] ?
  6. Oct 10, 2007 #5
  7. Oct 10, 2007 #6
    from this one I've found [tex]\frac{(1-e^{-\pi})}{2(1+e^{-\pi})}[/tex]

    Is this answer true?Please,check.
  8. Oct 10, 2007 #7

    Gib Z

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    Posts 3 and 5 are definitely correct, though either you or I have made a sign error on the sum of the geometric series, check that again just to be safe.
  9. Oct 11, 2007 #8
    I did it in this way
  10. Oct 11, 2007 #9

    Gib Z

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    Your last post makes no sense to me :( Maybe I'm doing the mistake, so heres how I'm getting it

    [tex]\frac{1}{2}+\sum_{k=1}^{\infty}(-1)^ke^{-k\pi} = \frac{1}{2} + \frac{-e^{-\pi}}{1+e^{-\pi}} = \frac{1-3e^{-\pi}}{2(1-e^{-\pi})}[/tex]
  11. Oct 11, 2007 #10
    Hey,Gib Z
    Should't be your answer be same as mine.
  12. Oct 11, 2007 #11

    Gib Z

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    lol well i dont know if your making the mistake or me, but i cant see mine, so unless u do, check ur answer again
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