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 involves absolute values. To solve, we analyze the behavior of \(|x - 1|\), \(|x - 2|\), and \(|x - 3|\) over the given interval \([1, 3]\). 

Step 1: Break the interval \([1, 3]\) at the critical points \(x = 1\), \(x = 2\), and \(x = 3\). The intervals are: \[ [1, 2], \quad [2, 3]. \] 

Step 2: Evaluate the expressions for each interval. - For \(x \in [1, 2]\): \[ |x - 1| = x - 1, \quad |x - 2| = 2 - x, \quad |x - 3| = 3 - x. \] 

Thus, the integrand becomes: \[ |x - 1| + |x - 2| + |x - 3| = (x - 1) + (2 - x) + (3 - x) = 4 - x. \] - For \(x \in [2, 3]\): \[ |x - 1| = x - 1, \quad |x - 2| = x - 2, \quad |x - 3| = 3 - x. \] Thus, the integrand becomes: \[ |x - 1| + |x - 2| + |x - 3| = (x - 1) + (x - 2) + (3 - x) = x. \] 

Step 3: Compute the integral over each sub-interval. 1. For \(x \in [1, 2]\): \[ \int_{1}^{2} (4 - x) \, dx = \left[4x - \frac{x^2}{2}\right]_{1}^{2}. \] 

Evaluate: \[ \left[4(2) - \frac{(2)^2}{2}\right] - \left[4(1) - \frac{(1)^2}{2}\right] = \left(8 - 2\right) - \left(4 - 0.5\right) = 6 - 3.5 = 2.5. \] 2. For \(x \in [2, 3]\): \[ \int_{2}^{3} x \, dx = \left[\frac{x^2}{2}\right]_{2}^{3}. \] 

Evaluate: \[ \left[\frac{(3)^2}{2}\right] - \left[\frac{(2)^2}{2}\right] = \frac{9}{2} - \frac{4}{2} = \frac{5}{2}. \] Step 4: Add the results. \[ \int_{1}^{3} \left(|x - 1| + |x - 2| + |x - 3|\right) \, dx = 2.5 + 2.5 = 5. \] 

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