Using integration, find the area of the region $\left\{(\mathrm{x},\mathrm{y}):{\mathrm{y}}^2\le \mathrm{x}\le \mathrm{y}\right\}$

Given: the region $\left\{(\mathrm{x},\mathrm{y}):{\mathrm{y}}^2\le \mathrm{x}\le \mathrm{y}\right\}$
Notice that the region is the intersection of the following region,
${\mathrm{R}}_1=\left\{(\mathrm{x},\mathrm{y}):{\mathrm{y}}^2\le \mathrm{x}\right\}\text{ and }{\mathrm{R}}_2=\left\{\right(\mathrm{x},\mathrm{y}):\mathrm{x}\le \mathrm{y}\}$
Now, considering the equation,
$\mathrm{x}={\mathrm{y}}^2......\left(1\right)$
$\mathrm{x}=\mathrm{y}$
Observe the below graph,

From the equation (1) and (2), it follows
$\therefore \mathrm{y}={\mathrm{y}}^2\Rightarrow {\mathrm{y}}^2-\mathrm{y}=0\Rightarrow \mathrm{y}(\mathrm{y}-1)=0\Rightarrow \mathrm{y}=0,1$
Therefore, curve (1) and lin (2) will meet at point O(0, 0) and A(1, 1).
For required region, 
Area of the shaded region in graph $={\int }_0^1 \mathrm{xdy}$
$\begin{array}{l}={\int }_0^1 \left(\mathrm{y}-{\mathrm{y}}^2\right)\mathrm{dy}\\ ={\int }_0^1 \mathrm{ydy}-{\int }_0^1 {\mathrm{y}}^2\mathrm{dy}\\ ={\left[\frac{{\mathrm{y}}^2}{2}-\frac{{\mathrm{y}}^3}{3}\right]}_0^1=\left[\left(\frac{1}{2}-\frac{1}{3}\right)-(0-0)\right]=\frac{1}{6}{\text{ unit }}^2\end{array}$
Therefore, the area of the region for $\left\{(\mathrm{x},\mathrm{y}):{\mathrm{y}}^2\le \mathrm{x}\le \mathrm{y}\right\}\text{ is }\frac{1}{6}{\text{ unit }}^2$