Let $S = \left\{(x,y): \frac{y(3x-1)}{x(3x-2)} < 0\right\}$, $S' = \{(x,y): A \times B: -1 \leq A \leq 1, -1 \leq B \leq 1\}$, then the area of the region enclosed by all points in $S \cap S'$ is____.
Step-by-Step Solution
Key Concept: Analyze the inequality $\frac{y(3x-1)}{x(3x-2)} < 0$ by finding critical points at $x=0, x=\frac{1}{3}, x=\frac{2}{3}$ and determining sign changes in each interval, then find the intersection with the square region $S' = [-1,1] \times [-1,1]$ and compute the enclosed area using integration.
The shaded region represents $S \setminus S'$ where two regions are bounded by curves. By carefully analyzing the geometry shown in the diagram with critical points at $(-1,0)$, $(0,1)$, $(1/3, 1)$, and $(1,0)$, the enclosed area between the curves is computed using integration. The total area of the clearly enclosed region is $2$ square units.
Correct Answer: 2