To find the area of the region bounded by the two functions $y = |x - 4|$ and $y = \frac{x}{3}$, we need to identify the points of intersection and then integrate the absolute difference between the functions over the interval defined by these intersection points.

Step 1: Finding the Points of Intersection

The points of intersection occur when $|x - 4| = \frac{x}{3}$. This gives two cases due to the absolute value function.

Case 1: $x - 4 = \frac{x}{3}$

1. Multiply both sides by 3 to clear the fraction: $3(x - 4) = x$ $3x - 12 = x$
2. Solve for $x$: $2x = 12 \Rightarrow x = 6$
3. Substitute $x = 6$ into either function to find $y$: $y = \frac{6}{3} = 2$ So, one intersection point is $(6, 2)$.

Case 2: $-(x - 4) = \frac{x}{3}$

1. Simplify the left side: $-x + 4 = \frac{x}{3}$
2. Multiply both sides by 3: $-3x + 12 = x$
3. Solve for $x$: $-4x = -12 \Rightarrow x = 3$
4. Substitute $x = 3$ into either function to find $y$: $y = \frac{3}{3} = 1$ So, the other intersection point is $(3, 1)$.

Step 2: Setting Up the Integral

The area of the region is given by integrating the absolute difference between the two functions from $x = 3$ to $x = 6$: $\text{Area} = \int_{3}^{6} \left(|x - 4| - \frac{x}{3}\right) \, dx$ On the interval $[3, 6]$, $|x - 4|$ becomes $4 - x$ when $x < 4$ and $x - 4$ when $x \geq 4$.

Thus, we split the integral at $x = 4$: $\text{Area} = \int_{3}^{4} \left((4 - x) - \frac{x}{3}\right) \, dx + \int_{4}^{6} \left((x - 4) - \frac{x}{3}\right) \, dx$

Step 3: Calculating Each Integral

1. First Integral from $x = 3$ to $x = 4$: $\int_{3}^{4} \left(4 - x - \frac{x}{3}\right) \, dx = \int_{3}^{4} \left(4 - \frac{4x}{3}\right) \, dx$

2. Second Integral from ( x =