= Solution
The Born <probability> is the <integral> of $|\psi_n|^2$ over the indicated <interval>. Changing variables to $\xi$ cancels the dimensional <normalization> factors and gives
$$
P_0(\ell)=\frac1{\sqrt\pi}\int_{-\ell}^{\ell}e^{-\xi^2}\,d\xi,\qquad
P_1(\ell)=\frac2{\sqrt\pi}\int_{-\ell}^{\ell}\xi^2e^{-\xi^2}\,d\xi.
$$
Since $\frac{d}{d\xi}(\xi e^{-\xi^2})=e^{-\xi^2}-2\xi^2e^{-\xi^2}$, integration gives
$$
\boxed{P_1(\ell)=P_0(\ell)-\frac2{\sqrt\pi}\ell e^{-\ell^2}.}
$$
As $\ell\to\infty$, the correction term tends to zero and $P_0(\ell)\to1$ by <normalization>. Hence $\boxed{P_1(\ell)\to1}$. This is expected because the expanding <interval> exhausts the whole position space and the state is normalized: the particle is somewhere on the <line> with total <probability> one.
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