Mathematics - Area Under Curves Question with Solution | TestHub

Refer figure, point of intersection of given curves are ${\mathrm{x}}^2=4\mathrm{x}$ $\Rightarrow \mathrm{x}(\mathrm{x}-4)=0$
$$\therefore \mathrm{O}\equiv (0,0)\text{ and }\mathrm{P}\equiv (4,4)$$ Required area is shaded. $$\begin{array}{rl}& \therefore A={\int }_0^4(\sqrt{4x}-x)dx=2{\int }_0^4{x}^{\frac{1}{2}}dx-{\int }_0^{4}xdx\\ & =2{\left[\frac{x^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}\right]}_0^4-{\left[\frac{x^2}{2}\right]}_0^4-\left(\frac{4}{3}\right)(4\times 2)-\frac{16}{2}\\ & =\frac{32}{3}-\frac{16}{2}=\frac{16}{6}=\frac{8}{3}\text{ sq. units. }\end{array}$$