Evaluate the definite integral $\int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x$.

Let $I = \int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x$.
$\Rightarrow I = \int_{1}^{4}|x-1| d x + \int_{1}^{4}|x-2| d x + \int_{1}^{4}|x-3| d x$.
$I = I_{1} + I_{2} + I_{3}$ ... $(1)$
where $I_{1} = \int_{1}^{4}|x-1| d x$,$I_{2} = \int_{1}^{4}|x-2| d x$,and $I_{3} = \int_{1}^{4}|x-3| d x$.
For $I_{1} = \int_{1}^{4}|x-1| d x$:
Since $(x-1) \geq 0$ for $1 \leq x \leq 4$,
$I_{1} = \int_{1}^{4}(x-1) d x = \left[\frac{x^{2}}{2} - x\right]_{1}^{4} = (8 - 4) - (\frac{1}{2} - 1) = 4 + \frac{1}{2} = \frac{9}{2}$. ... $(2)$
For $I_{2} = \int_{1}^{4}|x-2| d x$:
Since $x-2 \leq 0$ for $1 \leq x \leq 2$ and $x-2 \geq 0$ for $2 \leq x \leq 4$,
$I_{2} = \int_{1}^{2}(2-x) d x + \int_{2}^{4}(x-2) d x = \left[2x - \frac{x^{2}}{2}\right]_{1}^{2} + \left[\frac{x^{2}}{2} - 2x\right]_{2}^{4} = (4 - 2 - (2 - \frac{1}{2})) + ((8 - 8) - (2 - 4)) = \frac{1}{2} + 2 = \frac{5}{2}$. ... $(3)$
For $I_{3} = \int_{1}^{4}|x-3| d x$:
Since $x-3 \leq 0$ for $1 \leq x \leq 3$ and $x-3 \geq 0$ for $3 \leq x \leq 4$,
$I_{3} = \int_{1}^{3}(3-x) d x + \int_{3}^{4}(x-3) d x = \left[3x - \frac{x^{2}}{2}\right]_{1}^{3} + \left[\frac{x^{2}}{2} - 3x\right]_{3}^{4} = (9 - \frac{9}{2} - (3 - \frac{1}{2})) + ((8 - 12) - (\frac{9}{2} - 9)) = 2 + \frac{1}{2} = \frac{5}{2}$. ... $(4)$
From equations $(1), (2), (3),$ and $(4)$,we obtain:
$I = \frac{9}{2} + \frac{5}{2} + \frac{5}{2} = \frac{19}{2}$.