Given, curve $y =
\begin{cases}
x^2, & x < 0\\
x, & x \ge 0
\end{cases}$
and line $y=4$
Area of OABO,
$A_{1} =\int\limits_{y=0}^{4} \sqrt{y} d y$
$=\left[\frac{2}{3} y^{3 / 2}\right]_{0}^{4}$
$t=\frac{2}{3}\left[(4)^{3 / 2}-0\right]$
$=\frac{2}{3} \times(2)^{3}$
$=\frac{16}{3}$
and area of $OBCO$,
$A_{2} =\int\limits_{0}^{4} y d y=\left[\frac{y^{2}}{2}\right]_{0}^{4}$
$=\frac{1}{2}\left[(4)^{2}-0\right]=\frac{16}{2}=8$
Hence, area of $O A B O=A_{1}+A_{2}$
$=\frac{16}{3}+8=\frac{40}{3}$ sq unit
