Mathematics - Area Under Curves Question with Solution | TestHub

Required area is shaded. Point of intersection of given curves are $(0,0)$ and $(4,16)$ $$\begin{array}{rl}& \therefore A={\int }_0^4\left(4x-x^2\right)dx\\ & =4{\int }_0^{4}xdx-{\int }_0^4{x}^{2}dx=4{\left[\frac{x^2}{2}\right]}_0^4-{\left[\frac{x^3}{3}\right]}_0^4\\ & =2(16)-\frac{(4{)}^2}{3}=32-\frac{64}{3}=\frac{32}{3}\text{ sq. units }\end{array}$$