8.1. Find the area of the figure defined on the coordinate plane by the inequality $2(2-x) \geq\left|y-x^{2}\right|+\left|y+x^{2}\right|$.
Answer: 15. Solution. In the region above the graphs of the functions $y=x^{2}$ and $y=-x^{2}$, the original inequality takes the form $2-x \geq y$. In the region below the graphs of the functions $y=x^{2}$ and $y=-x^{2}$, the original inequality takes the form $2-x \geq -y$, i.e., $y \geq x-2$. In the region lying above the graph of the function $y=-x^{2}$ and below the graph of the function $y=x^{2}$, the original inequality has the form $2-x \geq x^{2}$.

Therefore, the points $(x, y)$ satisfying the given inequality form a trapezoid $A B C D$ on the coordinate plane, bounded by the lines $y=-x+2$, $y=x-2, x=-2$ and $x=1$ (see figure), with vertices at the points $(-2 ;-4),(-2 ; 4),(1 ; 1)$ and $(1 ;-1)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_adf4a392ca1a490a8c79g-08.jpg?height=471&width=603&top_left_y=1872&top_left_x=1366)

Since $A B=8, C D=2$, and the height of the trapezoid is 3, the desired area is $\frac{8+2}{2} \cdot 3=15$