Mathematics - Area Under Curves Question with Solution | TestHub

The required area is shaded The point of intersection of the curves are $x^2=x\Rightarrow x(x-1)=0$ i.e. $O(0,0)$ and $P(1,1)$
$\begin{array}{rl}& \therefore A={\int }_0^1\left(x-x^2\right)dx\\ & ={\left[\frac{x^2}{2}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\end{array}$