The area of the region bounded by the parabola $(y-2{)}^2=x-1$, the tangent to the parabola at the point $(2,3)$ and the x-axis is:

Parabola: $(y-2{)}^2=x-1$. Point $P(2,3)$.Differentiating: $2(y-2)y^{\prime }=1⟹y^{\prime }=\frac{1}{2(y-2)}$.Slope at $(2,3)$: $m=\frac{1}{2(3-2)}=1/2$.Equation of tangent: $y-3=\frac{1}{2}(x-2)⟹2y-6=x-2⟹x=2y-4$.Area is better integrated w.r.t $y$ from $0$ to $3$.The x-boundary of the parabola is $x=(y-2{)}^2+1$.The x-boundary of the tangent is $x=2y-4$.Area $={\int }_0^3[(y-2{)}^2+1-(2y-4)]dy$ $={\int }_0^3[y^2-4y+4+1-2y+4]dy={\int }_0^3[y^2-6y+9]dy$ $={\int }_0^3(y-3{)}^{2}dy=[\frac{(y-3{)}^3}{3}{]}_0^3=0-(-27/3)=9$.