1. We need the area of the region $\{(x,y): 0\le y\le \min\{2x,\,6x-x^2\}\}.$ This means for each $x$, the height of the region is $y=\min\{2x,\,6x-x^2\},$ provided this minimum is nonnegative.

2. First, find where the two curves intersect: $2x=6x-x^2$ $x^2-4x=0$ $x(x-4)=0$ So the intersection points are at $x=0,\ 4.$

3. Compare the two functions on the interval where they are relevant. Let $f(x)=2x,\qquad g(x)=6x-x^2.$ Then $g(x)-f(x)=6x-x^2-2x=4x-x^2=x(4-x).$ Hence:

   • for $0<x<4$, $x(4-x)>0$, so $g(x)>f(x)$, thus $\min\{2x,6x-x^2\}=2x;$
   • for $x>4$, $x(4-x)<0$, so $g(x)<f(x)$, thus $\min\{2x,6x-x^2\}=6x-x^2.$

4. Also, since $y\ge 0$, we need the minimum to be nonnegative. Now $2x\ge 0 \iff x\ge 0,$ and $6x-x^2=x(6-x)\ge 0 \iff 0\le x\le 6.$ Therefore the region exists only for $0\le x\le 6.$

5. So the required area is $A=\int_0^4 2x\,dx+\int_4^6 (6x-x^2)\,dx.$

6. Compute each integral: $\int_0^4 2x\,dx=\left[x^2\right]_0^4=16.$

   Next, $\int_4^6 (6x-x^2)\,dx=\left[3x^2-\frac{x^3}{3}\right]_4^6.$ At $x=6$: $3(36)-\frac{216}{3}=108-72=36.$ At $x=4$: $3(16)-\frac{64}{3}=48-\frac{64}{3}=\frac{80}{3}.$ Therefore, $\int_4^6 (6x-x^2)\,dx=36-\frac{80}{3}=\frac{28}{3}.$

7. Thus, $A=16+\frac{28}{3}=\frac{48+28}{3}=\frac{76}{3}.$

8. Hence, $12A=12\cdot \frac{76}{3}=4\cdot 76=304.$

Therefore the required integer is $\boxed{304}.$