Key Concepts and Formulas

• Area under a curve: The area under the curve $y=f(x)$ from $x=a$ to $x=b$ is given by the definite integral $\int_a^b f(x) \, dx$.
• Finding the minimum of two functions: The function $\min\{f(x), g(x)\}$ returns the smaller of the two function values at each $x$. To find the area under $\min\{f(x), g(x)\}$, we need to determine the intervals where $f(x) \le g(x)$ and $g(x) \le f(x)$.
• Intersection of curves: To find the intersection points of two curves $y=f(x)$ and $y=g(x)$, we solve the equation $f(x) = g(x)$.

Step-by-Step Solution

Step 1: Identify the Bounding Functions and Their Properties

We are given the functions $f(x) = 2x$ and $g(x) = 6x - x^2$.

• $f(x) = 2x$ is a straight line passing through the origin with a slope of 2.
• $g(x) = 6x - x^2$ is a downward-opening parabola. We can find its x-intercepts by setting $g(x) = 0$: $x(6-x) = 0$, so $x=0$ and $x=6$. The vertex is at $x = -\frac{b}{2a} = -\frac{6}{2(-1)} = 3$, and $g(3) = 6(3) - 3^2 = 18 - 9 = 9$. Thus, the vertex is at $(3, 9)$.
• The region is defined by $0 \leq y \leq \min\{2x, 6x-x^2\}$. The condition $y \geq 0$ limits the x values to where both $2x$ and $6x - x^2$ are non-negative.

Step 2: Find the Intersection Points of the Two Functions

To find where the functions intersect, we set $f(x) = g(x)$: $2x = 6x - x^2$ $x^2 - 4x = 0$ $x(x-4) = 0$ So, $x = 0$ or $x = 4$. The corresponding y-values are $f(0) = 0$ and $f(4) = 8$. The intersection points are $(0, 0)$ and $(4, 8)$.

Step 3: Determine the Intervals Where Each Function is the Minimum

We need to determine which function is smaller in the intervals $[0, 4]$ and $[4, 6]$.

• For $x \in (0, 4)$, let's test $x = 2$: $f(2) = 2(2) = 4$ and $g(2) = 6(2) - 2^2 = 12 - 4 = 8$. Since $4 < 8$, $f(x) \le g(x)$ in this interval. So $\min\{2x, 6x-x^2\} = 2x$.
• For $x \in (4, 6)$, let's test $x = 5$: $f(5) = 2(5) = 10$ and $g(5) = 6(5) - 5^2 = 30 - 25 = 5$. Since $5 < 10$, $g(x) \le f(x)$ in this interval. So $\min\{2x, 6x-x^2\} = 6x - x^2$.

Therefore, $\min\{2x, 6x - x^2\} = \begin{cases} 2x, & 0 \le x \le 4 \\ 6x - x^2, & 4 \le x \le 6 \end{cases}$

Step 4: Set Up the Definite Integral for the Area

The area $A$ is given by the sum of the integrals over the intervals $[0, 4]$ and $[4, 6]$: $A = \int_0^4 2x \, dx + \int_4^6 (6x - x^2) \, dx$

Step 5: Evaluate the Integrals

• $\int_0^4 2x \, dx = [x^2]_0^4 = 4^2 - 0^2 = 16$
• $\int_4^6 (6x - x^2) \, dx = [3x^2 - \frac{x^3}{3}]_4^6 = (3(6^2) - \frac{6^3}{3}) - (3(4^2) - \frac{4^3}{3}) = (108 - 72) - (48 - \frac{64}{3}) = 36 - 48 + \frac{64}{3} = -12 + \frac{64}{3} = \frac{-36 + 64}{3} = \frac{28}{3}$

Step 6: Calculate the Total Area A

$A = 16 + \frac{28}{3} = \frac{48 + 28}{3} = \frac{76}{3}$

Step 7: Calculate 12A

$12A = 12 \times \frac{76}{3} = 4 \times 76 = 304$

Common Mistakes & Tips

• Careless algebraic errors when solving for intersection points. Always double-check your work.
• Forgetting to consider the condition $y \geq 0$, which restricts the region to the area above the x-axis.
• Incorrectly identifying the minimum function in each interval. Testing a point within the interval is crucial.

Summary

The area of the region is found by identifying the minimum of the two functions $2x$ and $6x-x^2$, splitting the integral at their intersection point, and summing the two resulting definite integrals. Finally, we multiply the total area by 12 to obtain the final answer.

The final answer is $\boxed{304}$.