Question

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

Hint:

To integrate functions involving absolute values, split the integral at points where the expressions inside the absolute values change sign, and evaluate the integral piecewise.

Answer 1

The integrand includes absolute value functions. The solution requires analyzing the behavior of \(|x - 1|\), \(|x - 2|\), and \(|x - 3|\) over the interval \([1, 3]\).

Step 1: Partition the interval \([1, 3]\) at the points \(x = 1\), \(x = 2\), and \(x = 3\). The resulting sub-intervals are: \[ [1, 2], \quad [2, 3]. \]

Step 2: Determine the value of each absolute value expression within each sub-interval.
- For \(x \in [1, 2]\): The expressions simplify to \(|x - 1| = x - 1\), \(|x - 2| = 2 - x\), and \(|x - 3| = 3 - x\).
The integrand becomes: \[ |x - 1| + |x - 2| + |x - 3| = (x - 1) + (2 - x) + (3 - x) = 4 - x. \]
- For \(x \in [2, 3]\): The expressions simplify to \(|x - 1| = x - 1\), \(|x - 2| = x - 2\), and \(|x - 3| = 3 - x\).
The integrand becomes: \[ |x - 1| + |x - 2| + |x - 3| = (x - 1) + (x - 2) + (3 - x) = x. \]

Step 3: Calculate the definite integral over each sub-interval.
1. For \(x \in [1, 2]\): The integral is \[ \int_{1}^{2} (4 - x) \, dx = \left[4x - \frac{x^2}{2}\right]_{1}^{2}. \]
Evaluation yields: \[ \left[4(2) - \frac{(2)^2}{2}\right] - \left[4(1) - \frac{(1)^2}{2}\right] = (8 - 2) - (4 - 0.5) = 6 - 3.5 = 2.5. \]
2. For \(x \in [2, 3]\): The integral is \[ \int_{2}^{3} x \, dx = \left[\frac{x^2}{2}\right]_{2}^{3}. \]
Evaluation yields: \[ \left[\frac{(3)^2}{2}\right] - \left[\frac{(2)^2}{2}\right] = \frac{9}{2} - \frac{4}{2} = \frac{5}{2}. \]

Step 4: Sum the results from the sub-intervals.
\[ \int_{1}^{3} \left(|x - 1| + |x - 2| + |x - 3|\right) \, dx = 2.5 + 2.5 = 5. \]

Final Answer: \[ \int_{1}^{3} \left(|x - 1| + |x - 2| + |x - 3|\right) \, dx = 5. \]