Mathematics - Area Under Curves Question with Solution | TestHub

Given curves are $y=\sqrt{x}$ ...(i) and $2y-x+3=0$ ...(ii)
On solving eqs. (i) and (ii), we get $\begin{array}{rl}& 2\sqrt{\mathrm{x}}-(\sqrt{\mathrm{x}}{)}^2+3=0\\ & \Rightarrow (\sqrt{\mathrm{x}}{)}^2-2\sqrt{\mathrm{x}}-3=0\\ & \Rightarrow (\sqrt{\mathrm{x}}-3)(\sqrt{\mathrm{x}}+1)=0\\ & \Rightarrow \sqrt{\mathrm{x}}=3(\because \sqrt{\mathrm{x}}=-1\text{ is not possible })\\ & \therefore \mathrm{y}=3\\ & \therefore \text{ Required area }\\ & ={\int }_0^3\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)\mathrm{dy}={\int }_0^3\left\{(2\mathrm{y}+3)-{\mathrm{y}}^2\right\}\mathrm{dy}\\ & ={\left[{\mathrm{y}}^2+3\mathrm{y}-\frac{{\mathrm{y}}^3}{3}\right]}_0^3=9+9-9=9\end{array}$