1. We need the area

$A=\int_0^3 \min\{x^2+2,\,2x+2\}\,dx.$

So first, determine which of the two curves is smaller on $[0,3]$.

2. Compare the functions

$x^2+2 \quad \text{and} \quad 2x+2.$

Set them equal:

$x^2+2=2x+2$ $x^2-2x=0$ $x(x-2)=0.$

Thus, they intersect at

$x=0,\;2.$

3. Check which one is smaller in each interval.

Consider

$x^2+2-(2x+2)=x^2-2x=x(x-2).$

• For $0<x<2$, we have $x(x-2)<0$, so $x^2+2<2x+2.$ Hence, the minimum is $x^2+2$.

• For $2<x\le 3$, we have $x(x-2)>0$, so $x^2+2>2x+2.$ Hence, the minimum is $2x+2$.

Therefore,

$A=\int_0^2 (x^2+2)\,dx+\int_2^3 (2x+2)\,dx.$

4. Evaluate the integrals.

First,

$\int_0^2 (x^2+2)\,dx=\left[\frac{x^3}{3}+2x\right]_0^2=\frac{8}{3}+4=\frac{20}{3}.$

Second,

$\int_2^3 (2x+2)\,dx=\left[x^2+2x\right]_2^3=(9+6)-(4+4)=15-8=7.$

So,

$A=\frac{20}{3}+7=\frac{20}{3}+\frac{21}{3}=\frac{41}{3}.$

5. Now compute $12A$:

$12A=12\cdot \frac{41}{3}=4\cdot 41=164.$

So the required integer is

$\boxed{164}.$