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Basic integration trouble

  • Thread starter dangsy
  • Start date
15
0
1. Homework Statement
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. Homework 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:
1,750
1
[tex]\int_0^{\frac{L}{10}}\frac 2 L \sin^{2}\left(\frac{3x\pi}{L}\right)dx[/tex]

Yes?
 
15
0
yep that's right sorry =(
 
1,750
1
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.
 
15
0
Yes thank you!
 

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