Solution
= Solution
Since
$$
\int_0^a x^2(a-x)^2\,dx=\frac{a^5}{30},
$$
an explicitly normalized trial is
$$
\boxed{\widetilde\psi(x)=\sqrt{\frac{30}{a^5}}\,x(a-x),\qquad 0<x<a.}
$$
It is zero outside the box. The original expression $x(x-a)$ differs by a minus sign, which changes only the <global phase> and leaves <probabilities> and <expectation values> unchanged.