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Basic integration trouble

  1. Mar 6, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine the probability of finding a particle of mass m between x=0 and x=L/10, if it is in n=3 state of an infinite well.


    2. Relevant equations
    [tex]P = \int_a^b\left |\psi\left(x\right)\right|^2 dx[/tex]
    [tex]\left|\psi\right|^2 = \frac{2}{L}Sin^2\left(\frac{nx\pi}{L}\right)[/tex]


    3. The attempt at a solution
    I'm trying to integrate...

    [tex]\int_0^{\frac{L}{10}}\frac 2 L \sin^{2}\left(\frac{3x\pi}{L}\right)dx[/tex]



    step (1)
    [tex]\frac{1}{L}\left[ \int_0^\frac{L}{10}1-\int_0^\frac{L}{10}Cos\left(\frac{6x\pi}{L}\right)\right]dx[/tex]


    step (2)

    [tex]\frac{1}{L}\left[\frac{L}{10} - Sin\left(\frac{3\pi}{5}\right)\right] [/tex]

    When I finish solving, I end up with an L in the answer...
    which I know I'm not suppose to have, did I mess up my integration somewhere?
    sorry about my pooooor pooor latex
     
    Last edited: Mar 6, 2008
  2. jcsd
  3. Mar 6, 2008 #2
    [tex]\int_0^{\frac{L}{10}}\frac 2 L \sin^{2}\left(\frac{3x\pi}{L}\right)dx[/tex]

    Yes?
     
  4. Mar 6, 2008 #3
    yep that's right sorry =(
     
  5. Mar 6, 2008 #4
    Sorry, PF was like broke last night. Use this trig identity:

    [tex]\cos{2x}=\cos^{2}x-\sin^{2}x[/tex]

    [tex]\sin^{2}x=\frac 1 2(1+\cos{2x})[/tex]

    But I think that's what you did, so it's good? Sorry don't have time to work it out myself.
     
  6. Mar 6, 2008 #5
    Yes thank you!
     
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