If the area of the region ${(x, y): 0 \leq y \leq \min {2 x, 6 x-x^{2} } }$ is $A$, then $12 A$ 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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