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:  ${(y-2)}^2=x-1\to 1$
$\Rightarrow y^2-4y-x+5=0$ 
tgt at (2,3) to the parabola is 
 $y\left(3\right)-2(y+3)-\frac{1}{2}(x+2)+5=0$
 $\Rightarrow x-2y+4=0$
From diagram

Reg.area  $=\underset{0}{\overset{3}{\int }}({(y-2)}^2+1-(2y-4\left)\right)dy$                 
$$\begin{gathered} =\underset{0}{\overset{3}{\int }}(y^2-6y+9)dy={\left(\frac{{(3-y)}^3}{-3}\right)}_0^3 \\ =(0+9)=9\mathrm{sq}.units \end{gathered}$$