<p>Integrate: \(\int_{-1.5}^{3.5} |x-1|\, dx\).</p>
Step-by-Step Solution
Key Concept: Split the integral at x=1 where the absolute value expression changes sign, since |x-1| = -(x-1) for x<1 and |x-1| = (x-1) for x≥1. This converts the absolute value integral into two standard polynomial integrals.
<p><strong>Step 1:</strong> Identify critical point where (x-1)=0, which is x=1. Since -1.5 < 1 < 3.5, split the integral.</p><p><strong>Step 2:</strong> For x ∈ [-1.5, 1]: |x-1| = -(x-1) = 1-x</p><p>For x ∈ [1, 3.5]: |x-1| = x-1</p><p><strong>Step 3:</strong> ∫₍₋₁.₅₎^(3.5) |x-1| dx = ∫₍₋₁.₅₎^1 (1-x) dx + ∫₁^(3.5) (x-1) dx</p><p><strong>Step 4:</strong> First integral: [x - x²/2]₍₋₁.₅₎^1 = (1 - 1/2) - (-1.5 - 2.25/2) = 0.5 - (-3) = 3.5</p><p><strong>Step 5:</strong> Second integral: [x²/2 - x]₁^(3.5) = (12.25/2 - 3.5) - (1/2 - 1) = 3.625 - (-0.5) = 4.125</p><p><strong>Step 6:</strong> Total = 3.5 + 4.125 = 7.625 or 61/8</p><p>∴ Answer: <strong>7.625 or 61/8</strong></p>
Correct Answer: 7