Let the area of the region $\left{(x, y): 0 \leq x \leq 3,0 \leq y \leq \min \left{x^{2}+2,2 x+2\right}\right}$ be $A$. Then $12 \mathrm{~A}$ is equal to
$\mathrm{y}=2 \mathrm{x}+2$
$A=\int_{0}^{2}\left(x^{2}+2\right) d x+\int_{2}^{3}(2 x+2) d x$
$A=\frac{41}{3}$
$12 A=41 \times 4=164$
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