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We need to consider the intervals where the expressions inside the absolute values change signs. These points are $x=1$, $x=2$, and $x=3$. Since the integral is from 1 to 3, we break it into intervals [1, 2] and [2, 3].
For $1 \le x \le 2$:\r\n$|x-1| = x-1$\r\n$|x-2| = 2-x$\r\n$|x-3| = 3-x$\r\nSo, $|x-1|+|x-2|+|x-3| = (x-1) + (2-x) + (3-x) = 4-x$\r\nTherefore, $\int_{1}^{2}(|x-1|+|x-2|+|x-3|)dx = \int_{1}^{2}(4-x)dx$
For $2 \le x \le 3$:\r\n$|x-1| = x-1$\r\n$|x-2| = x-2$\r\n$|x-3| = 3-x$\r\nSo, $|x-1|+|x-2|+|x-3| = (x-1) + (x-2) + (3-x) = x$\r\nTherefore, $\int_{2}^{3}(|x-1|+|x-2|+|x-3|)dx = \int_{2}^{3}x dx$
$\int_{1}^{2}(4-x)dx = [4x - \frac{x^2}{2}]_{1}^{2} = (4(2) - \frac{2^2}{2}) - (4(1) - \frac{1^2}{2}) = (8-2) - (4-\frac{1}{2}) = 6 - \frac{7}{2} = \frac{12-7}{2} = \frac{5}{2}$\r\n$\int_{2}^{3}x dx = [\frac{x^2}{2}]_{2}^{3} = \frac{3^2}{2} - \frac{2^2}{2} = \frac{9}{2} - \frac{4}{2} = \frac{5}{2}$
$\int_{1}^{3}(|x-1|+|x-2|+|x-3|)dx = \int_{1}^{2}(4-x)dx + \int_{2}^{3}x dx = \frac{5}{2} + \frac{5}{2} = 5$
Final Answer: 5
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