= Solution
The <wavefunction> is real and vanishes continuously at both endpoints. Hence
$$
\langle p\rangle=-i\hbar\int\Psi\Psi'\,dx
=-\frac{i\hbar}{2}[\Psi^2]_{-\infty}^{\infty}=0.
$$
The <momentum> second moment is most safely evaluated from its <quadratic form>:
$$
\langle p^2\rangle=\hbar^2\int|\Psi'|^2dx
=4\hbar^2A^2\int_{-a}^ax^2dx.
$$
Thus
$$
\boxed{\langle p\rangle=0,\qquad\langle p^2\rangle=\frac{5\hbar^2}{2a^2}.}
$$
The <derivative> jumps where the <wavefunction> joins zero. On the whole real line the strong second-derivative operator therefore produces endpoint delta distributions, so the state is not in the usual operator domain of $p^2$. The finite <momentum> second moment, equivalently the kinetic-energy <quadratic form> or the <integral> of $p^2$ against the <momentum> <probability density>, remains well defined and gives the value above.
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