Mathematics - Area Under Curves Question with Solution | TestHub

Equation of tangent at $(2,3)$ to $$\begin{array}{rl}& (y-2{)}^2=x-1\text{ is }{S}_1=0\\ & \Rightarrow x-2y+4=0\end{array}$$ Required Area $=$ Area of $△OCB+$ Area of OAPD - Area of $△PCD$ $$\begin{array}{rl}& =\frac{1}{2}(4\times 2)+{\int }_0^3\left(y^2-4y+5\right)dy-\frac{1}{2}(1\times 2)\\ & =4+{\left[\frac{y^3}{3}-2y^2+5y\right]}_0^3-1=4-9-18+15-1\\ & =28-19=9\text{ sq. units }\end{array}$$
$$\text{ (or) }$$ $$\text{ Area }={\int }_0^3\left(2y-4-y^2+4y-5\right)dy={\int }_0^3\left(-y^2+6y-5\right)dy=-{\int }_0^3(3-y{)}^{2}dy={\left[\frac{(y-3{)}^3}{3}\right]}_0^3=\frac{27}{3}=9\text{ sq.units }$$