Interesting that only integrals along the [tex]x[/tex]-axis are suggested when integrating along the [tex]y[/tex]-axis would be much simpler... Anyway, you have to split the interval of integration into two. The "height" of the region is not uniform over the entire interval.
When [tex]y=0[/tex], we have [tex]0^2=x-1\implies x=1[/tex]. When [tex]y=2[/tex], we have [tex]2^2=x-1\implies x=5[/tex]. Then the area we want is given by
[tex]\displaystyle\int_0^12\,\mathrm dx+\int_1^52-\sqrt{x-1}\,\mathrm dx[/tex]
which seems to agree with the last option.
