# Problem with a wavefunction in Quantum Mechanics (math) (Book solution possibly wrong?)

Well there is a problem in my book which lists this problem:

Calculate the probability that a particle will be found at $0.49L$ and $0.51L$ in a box of length $L$ when it has (a) $n = 1$. Take the wave function to be constant in this range.

The answer in the solutions manual for this specific problem is given as:

However when I workout the equation myself I get this:

So i don't understand why the solutions book says that: $$\frac{2}{L}\int \sin^2 \left(\frac{n \pi x}{L} \right) \approx \frac{2 \Delta x}{L} \sin^2 \left(\frac{n \pi x}{L} \right)$$

What am I doing wrong here??

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Hi, any feedback on my answer would be nice... – draks ... Apr 3 '14 at 11:26

## 2 Answers

You might have missed a factor $2$ in your calculation. Wolfram confirms that $$\frac2L \int_{.49L}^{.51L} \sin^2\left(\frac{n\pi x}L\right) dx= 0.399,$$ which is close to the approximation given in your solution book: $$\frac{2 \Delta x}{L} \sin^2 \left(\frac{n \pi x}{L} \right)=\frac{2 \cdot 0.02L}{L} \sin^2 \left(\frac{.5L \pi}{L} \right)=0.04\sin^2\left(\frac\pi 2\right)=0.04$$

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Thanks for the answer, sorry about delay in feedback. You're right the book does give this approximation, but why use this when you can use the method I used?? That doesn't make sense to me – Biochemist_HK Apr 4 '14 at 14:28

Because the function is very close to constant over that range it can be pulled outside the integral. It gives a very good approximation to the correct answer without the need to do anything but simple multiplication

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