= Solution
{c}
For a <position measurement>, the <Born rule> integrates the <probability density> over the chosen interval. The three terms from part (b) have integrals
$$
\int_0^{\pi/4}1\,dx=\frac\pi4,\qquad
\int_0^{\pi/4}\cos^2(2x)\,dx=\left[\frac x2+\frac{\sin4x}{8}\right]_0^{\pi/4}=\frac\pi8,
$$
and
$$
\int_0^{\pi/4}2\cos(2x)\cos(\Omega t)\,dx=\left[\sin(2x)\cos(\Omega t)\right]_0^{\pi/4}=\cos(\Omega t).
$$
Thus
$$
\boxed{\Pr\left(0\leq x\leq\frac\pi4\right)=\frac18+\frac1{3\pi}\cos\left(\frac{2\hbar t}{m}\right).}
$$
The oscillating term comes from <quantum interference> between the zero and positive <energy> components. The <probability> lies between $1/8-1/(3\pi)$ and $1/8+1/(3\pi)$, both within $[0,1]$; integrating the same <probability density> over the whole circle gives one.
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