Step 1: Find breakpoints where expressions change sign:
\[
x - 1 = 0 \implies x = 1, x - 2 = 0 \implies x = 2.
\text{So, split: } [1, 2], [2, 3].
\]
Step 2:
- For $1 \leq x<1$: not needed.
- For $1 \leq x<2$:
\[
|x - 1| = x - 1, |x - 2| = -(x - 2) = 2 - x.
\]
So,
\[
I_1 = \int_{1}^{2} [(x - 1) + (2 - x)] dx = \int_{1}^{2} [1] dx = [x]_{1}^{2} = 1.
\]
- For $2 \leq x \leq 3$:
\[
|x - 1| = x - 1, |x - 2| = x - 2.
\]
So,
\[
I_2 = \int_{2}^{3} [(x - 1) + (x - 2)] dx = \int_{2}^{3} (2x - 3) dx.
\]
\[
= [x^2 - 3x]_{2}^{3} = [(9 - 9) - (4 - 6)] = 0 - (-2) = 2.
\]
Therefore,
\[
I = I_1 + I_2 = 1 + 2 = 3.
\]
Final Answer: 3