If the area of the region $\left{(x, y): 0 \leq y \leq \min \left{2 x, 6 x-x^{2}\right}\right}$ is $A$, then $12 \mathrm{~A}$ is equal to.
We have
$A=\frac{1}{2} \times 4 \times 8+\int_{4}^{6}\left(6 x-x^{2}\right) d x$
$A=\frac{76}{3}$
$12 A=304$
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