Mathematics - Area Under Curves Question with Solution | TestHub

We have $y=4x-x^2$ When $y=0$, we get $x(4-x)=0\Rightarrow x=0,4$ Required area is shaded. $\begin{array}{rl}\therefore A& ={\int }_0^4\left(4x-x^2\right)dx\\ & ={\left[\frac{4x^2}{2}-\frac{x^3}{3}\right]}_0^4={\left[2x^2-\frac{x^3}{3}\right]}_0^4\\ & =\left|2(16-0)-\frac{64-0}{3}\right|=\left|32-\frac{64}{3}\right|\\ & =\frac{32}{3}\end{array}$