Solution:
Area of the region enclosed by the parabola, $x^{2}=y$, the line, $y=x+2$, and $x$-axis is represented bv the shaded reaion $\mathrm{OABCO}$ as

Point of intersection of the parabola, $x^{2}=y$, and the line, $y=x+2$, is A $(-1,1)$
$\therefore$ Area $\mathrm{OABCO}=$ Area $(\mathrm{BCA})+$ Area COAC
$\begin{array}{l}
=\int_{-2}^{-1}(x+2) d x+\int_{-1}^{0} x^{2} d x \\
=\left[\frac{x^{2}}{2}+2 x\right]_{-2}^{-1}+\left[\frac{x^{3}}{3}\right]_{-1}^{0} \\
=\left[\frac{(-1)^{2}}{2}+2(-1)-\frac{(-2)^{2}}{2}-2(-2)\right]+\left[-\frac{(-1)^{3}}{3}\right] \\
=\left[\frac{1}{2}-2-2+4+\frac{1}{3}\right] \\
=\frac{5}{6} \text {sq. units }
\end{array}$