2. The following definite integral represents the area of a region in the $x y$-plane. Draw a well-labeled picture of the region, and use known geometric formulas to find the area. \[ \int_{0}^{6}|x-4| d x \]

Step 1 ：The integral represents the area under the curve of the function \(|x-4|\) from \(x=0\) to \(x=6\).

Step 2 ：The function \(|x-4|\) is a V-shaped function with the vertex at \((4,0)\).

Step 3 ：The area under the curve can be divided into two triangles, one from \(x=0\) to \(x=4\) and the other from \(x=4\) to \(x=6\).

Step 4 ：The area of a triangle is given by the formula \(\frac{1}{2} \times \text{base} \times \text{height}\).

Step 5 ：The base of the first triangle is 4 and the height is 4. The base of the second triangle is 2 and the height is 2.

Step 6 ：We can calculate the area of each triangle and add them together to get the total area. The area of the first triangle is \( \frac{1}{2} \times 4 \times 4 = 8 \) and the area of the second triangle is \( \frac{1}{2} \times 2 \times 2 = 2 \).

Step 7 ：The total area is \( 8 + 2 = 10 \).

Step 8 ：Final Answer: The area of the region represented by the integral \(\int_{0}^{6}|x-4| d x\) is \(\boxed{10}\).