Solution to 2008 Problem 85

We first find the normalization constant A, which we can take to be real and positive.
\begin{align*}\int_0^L \left|\psi\right|^2dx &= A^2\int_0^L\sin^2\left(\frac{3 \pi x}{L} \right)dx \\&= A^2 \int_0^L ...
So, in order for \psi to be normalized, A must equal \sqrt{2/L}. So,the desired probability is
\begin{align*}\int_{L/3}^{2L/3} \left|\psi\right|^2dx &= A^2\int_{L/3}^{2L/3}\sin^2\left(\frac{3 \pi x}{L} \right)dx \\&a...
Therefore, answer (B) is correct.

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