2. Find the area of the figure defined on the coordinate plane by the system

$$
\left\{\begin{array}{l}
\sqrt{1-x}+2 x \geqslant 0 \\
-1-x^{2} \leqslant y \leqslant 2+\sqrt{x}
\end{array}\right.
$$
Answer: 4.

Solution. The system is defined under the condition $x \in[0,1]$, under which the first inequality of the system is satisfied. Since the functions $y=x^{2}$ and $y=\sqrt{x}$ are inverses of each other for $x \in[0,1]$, the area of the figure defined by the conditions $x \in[0,1],-1-x^{2} \leqslant y \leqslant 2+\sqrt{x}$, consists of the same parts as the area of the rectangle defined by the conditions $x \in[0,1], y \in[-2,2]$, i.e., it is equal to 4.