Question

(b) 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 integral contains absolute value functions. To evaluate, the behavior of \(|x - 1|\), \(|x - 2|\), and \(|x - 3|\) over the interval \([1, 3]\) is analyzed. Step 1: The interval \([1, 3]\) is divided at the critical points \(x = 1\), \(x = 2\), and \(x = 3\), resulting in the sub-intervals \([\1, 2]\) and \([2, 3]\). Step 2: The expressions are evaluated for each sub-interval. For \(x \in [1, 2]\), \(|x - 1| = x - 1\), \(|x - 2| = 2 - x\), and \(|x - 3| = 3 - x\). The integrand simplifies to \((x - 1) + (2 - x) + (3 - x) = 4 - x\). For \(x \in [2, 3]\), \(|x - 1| = x - 1\), \(|x - 2| = x - 2\), and \(|x - 3| = 3 - x\). The integrand simplifies to \((x - 1) + (x - 2) + (3 - x) = x\). Step 3: Integration is performed over each sub-interval. For \(x \in [1, 2]\), \(\int_{1}^{2} (4 - x) \, dx = \left[4x - \frac{x^2}{2}\right]_{1}^{2} = \left(8 - 2\right) - \left(4 - 0.5\right) = 6 - 3.5 = 2.5\). For \(x \in [2, 3]\), \(\int_{2}^{3} x \, dx = \left[\frac{x^2}{2}\right]_{2}^{3} = \frac{9}{2} - \frac{4}{2} = \frac{5}{2} = 2.5\). Step 4: The results from the sub-intervals are summed. \(\int_{1}^{3} \left(|x - 1| + |x - 2| + |x - 3|\right) \, dx = 2.5 + 2.5 = 5\). Final Answer: \[\int_{1}^{3} \left(|x - 1| + |x - 2| + |x - 3|\right) \, dx = 5.\]