Animated Solution for Mathematics - Definite Integration: If the area of the region $\{(x,y):0\le y\le \min \{2x,6x-x^2\}\}$ is $A$, then $12A$ is equal to.

The problem asks us to find the area of a region defined by $y\le \min \{2x,6x-x^2\}$. The notation $\min \{f(x),g(x)\}$ implies that for any given $x$, the boundary of our region is defined by the lower of the two curves.

To understand the behavior of these functions, we first identify their intersection points by setting them equal:

Rearranging the terms, we obtain:

The curves intersect at $x=0$ and $x=4$. At $x=4$, the height of both functions is $y=2(4)=8$.

We must determine which function is the "ceiling" in different intervals. For $0\le x\le 4$, the line $y=2x$ lies below the parabola $y=6x-x^2$. Thus, the line acts as the upper boundary in this interval.

For $x>4$, the parabola $y=6x-x^2$ dips below the line $y=2x$. The parabola intersects the $x$-axis at $6x-x^2=0$, which gives $x=6$. Therefore, the parabola acts as the upper boundary from $x=4$ to $x=6$.

Since the boundary function changes at $x=4$, we split the total area $A$ into two integrals:

The first integral represents the area of a triangle with base $4$ and height $8$:

For the second integral, we evaluate the parabola:

Evaluating at the limits:

Summing the two parts, we find the total area $A$:

The problem asks for the value of $12A$:

The final answer is 304.