The area bounded by the curve \(y=\left\{\begin{array}{cl}x^{2}, & x < 0 \\ x, & x \geq 0\end{array}\right.\) and the line \(y=4\), is

Given, curve \(y=\left\{\begin{array}{c}x^{2}, x < 0 \\ x, x \geq 0\end{array}\right.\) and line \(y=4\)
Area of \(\mathrm{OABO}\),


\[
\begin{aligned}
\mathrm{A}_{1} &=\int_{\mathrm{y}=0}^{4} \sqrt{\mathrm{y}} \mathrm{dy} \\
&=\left[\frac{2}{3} \mathrm{y}^{3 / 2}\right]_{0}^{4} \\
\mathrm{t} &=\frac{2}{3}\left[(4)^{3 / 2}-0\right] \\
&=\frac{2}{3} \times(2)^{3} \\
&=\frac{16}{3}
\end{aligned}
\]
and area of \(\mathrm{OBCO}\),
\[
\begin{aligned}
A_{2} &=\int_{0}^{4} y d y=\left[\frac{y^{2}}{2}\right]_{0}^{4} \\
&=\frac{1}{2}\left[(4)^{2}-0\right]=\frac{16}{2}=8
\end{aligned}
\]
Hence, area of \(\mathrm{OABO}=\mathrm{A}_{1}+\mathrm{A}_{2}\) \(=\frac{16}{3}+8=\frac{40}{3}\) sq unit