JEE Main · 2009 · 12 Apr 2009 · Offline · Area Under Curves

The tangent to $(y-2{)}^2=x-1$ at the point $(2,3)$ is $x-2y+4=0$. Treating the region by horizontal strips, the right boundary is the parabola $x=(y-2{)}^2+1$ and the left boundary is the tangent $x=2y-4$, running from the $x$-axis up to $y=3$. The area is ${\int }_0^3[(2y-4)-((y-2{)}^2+1)]dy={\int }_0^3-(y-3{)}^{2}dy=\left[\frac{(y-3{)}^3}{3}\right]$ evaluated from 0 to 3, which equals 9 square units.