Troubleshooting Integration for Particle Probability in an Infinite Well

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The discussion focuses on calculating the probability of finding a particle in the n=3 state of an infinite well between x=0 and x=L/10. The integral setup is presented as P = ∫(0 to L/10) (2/L) sin²(3πx/L) dx. The user encounters an issue with an extraneous L in the final answer, indicating a potential mistake in integration. Suggestions include using the trigonometric identity sin²(x) = 1/2(1 - cos(2x)) to simplify the integration process. The conversation concludes with acknowledgment of the user's efforts and a note of appreciation for the assistance provided.
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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.

Homework Equations


P = \int_a^b\left |\psi\left(x\right)\right|^2 dx
\left|\psi\right|^2 = \frac{2}{L}Sin^2\left(\frac{nx\pi}{L}\right)

The Attempt at a Solution


I'm trying to integrate...

\int_0^{\frac{L}{10}}\frac 2 L \sin^{2}\left(\frac{3x\pi}{L}\right)dx
step (1)
\frac{1}{L}\left[ \int_0^\frac{L}{10}1-\int_0^\frac{L}{10}Cos\left(\frac{6x\pi}{L}\right)\right]dxstep (2)

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

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:
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\int_0^{\frac{L}{10}}\frac 2 L \sin^{2}\left(\frac{3x\pi}{L}\right)dx

Yes?
 
yep that's right sorry =(
 
Sorry, PF was like broke last night. Use this trig identity:

\cos{2x}=\cos^{2}x-\sin^{2}x

\sin^{2}x=\frac 1 2(1+\cos{2x})

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