Question

Evaluate: \[ \int_{1}^{3} (|x - 1| + |x - 2|) dx. \]

Hint:

For absolute value, split at points where expression inside $| \cdot |$ is zero.

Answer 1

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