Mathematics - Area Under Curves Question with Solution | TestHub

Given curves are $x^2=y$ and $y=4x$ Intersection points are $(0,0)$ and $(4,16)$ $\begin{array}{rl}\therefore \text{ Required area }& ={\int }_0^4\left(4x-x^2\right)dx\\ & ={\left[\frac{4x^2}{2}-\frac{x^3}{3}\right]}_0^4\\ & =\left[32-\frac{64}{3}\right]\\ & =\frac{32}{3}\mathrm{sq}\text{ unit }\end{array}$