= Solution
The factor $2x-x^3=x(2-x^2)$ is positive on $(0,\sqrt2)$ and negative on $(\sqrt2,2)$. For every nonzero $\alpha$, the proposed <probability density function> is therefore negative on an interval. The choice $\alpha=0$ has total integral zero. In fact,
$$
\int_0^2(2x-x^3)\,dx=\left[x^2-\frac{x^4}{4}\right]_0^2=0,
$$
so no scaling could normalise it even without the sign obstruction. Consequently
$$
\boxed{\text{No value of }\alpha\text{ gives a probability density function; no corresponding expectation exists}.}
$$
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