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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

    [tex]\int_{0}^{\infty}e^{-x}\left|sinx\right|dx=-e^{-x}\left|sinx\right|+\int_{0}^{\infty}e^{-x}\left|cosx\right|dx[/tex]
    [tex]=-e^{-x}\left|sinx\right|-e^{-x}\left|cosx\right|-\int_{0}^{\infty}e^{-x}\left|sinx\right|dx[/tex]
    [tex]\int_{0}^{\infty}e^{-x}\left|sinx\right|dx=\frac{-e^{-x}\left|sinx\right|-e^{-x}\left|cosx\right|}{2}[/tex]
     
  2. jcsd
  3. Oct 10, 2007 #2

    NateTG

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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.
    [tex]\int_{0}^{\infty}e^{-x}\left|sinx\right|dx=\int_{0}^{\pi}e^{-x}sinxdx+\int_{\pi}^{2\pi}e^{-x}(-sinx)dx+\cdots[/tex]
    [tex]=\sum_{k=0}^{\infty}(-1)^k\int_{k\pi}^{(k+1)\pi}e^{-x}sinxdx[/tex]
     
    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
    Oh,sorry
    [tex]\frac{1}{2}+\sum_{k=1}^{\infty}(-1)^ke^{-k\pi}[/tex]
     
  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
    [tex]\frac{1}{2}-\frac{e^{-\pi}}{1-e^{-2\pi}}+\frac{e^{-2\pi}}{1-e^{-2\pi}}[/tex]
     
  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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