1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Area under Integral.
Loading...