The points of extremum of $\phi(x)=\int\limits_1^x e^{-t^2 / 2}\left(1-t^2\right) d t$ are

$x=1,-1$ $x=-1,2$ $x=2,1$ $x=-2,1$

$x=1,-1$

We have, $\phi(x)=\int\limits_1^x e^{-t^2 / 2}\left(1-t^2\right) d t \Rightarrow \phi'(x)=e^{-x^2 / 2}\left(1-x^2\right)$ ∴  $\phi'(x)=0 \Rightarrow x= \pm 1$ Hence, the points of extremum are $x= \pm 1$