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The function is given by $y = |x - 6|$. Based on the definition of the absolute value function, we can split it into two parts: $$y = \begin{cases} -(x-6) & \text{if } x < 6 \\ x-6 & \text{if } x \geq 6 \end{cases}$$
The area $A$ is the integral of $y$ with respect to $x$ from $4$ to $8$. We must split the integral at the critical point $x=6$: $$A = \int_{4}^{6} -(x-6) \, dx + \int_{6}^{8} (x-6) \, dx$$
First part: $$\int_{4}^{6} (-x+6) \, dx = \left[ -\frac{x^2}{2} + 6x \right]_{4}^{6} = (-18 + 36) - (-8 + 24) = 18 - 16 = 2$$ Second part: $$\int_{6}^{8} (x-6) \, dx = \left[ \frac{x^2}{2} - 6x \right]_{6}^{8} = (32 - 48) - (18 - 36) = -16 - (-18) = 2$$
Summing the two parts: $$A = 2 + 2 = 4 \text{ square units}$$
Final Answer: 4 square units