<p>Evaluate \(\displaystyle\int_{-1}^{3} |\sin \pi x|\, dx\).</p>
Step-by-Step Solution
Key Concept: Split the integral at points where sin(πx) = 0 within [-1,3], which occur at x = -1, 0, 1, 2, 3. The absolute value removes the negative portions, making each sub-interval contribution positive.
<p><strong>Step 1:</strong> Identify zeros of sin(πx) in [-1,3].</p><p>sin(πx) = 0 when πx = nπ, so x = n where n ∈ ℤ. Within [-1,3]: x = -1, 0, 1, 2, 3.</p><p><strong>Step 2:</strong> Determine sign of sin(πx) in each interval.</p><p>• [-1, 0]: sin(πx) ≤ 0, so |sin(πx)| = -sin(πx)</p><p>• [0, 1]: sin(πx) ≥ 0, so |sin(πx)| = sin(πx)</p><p>• [1, 2]: sin(πx) ≤ 0, so |sin(πx)| = -sin(πx)</p><p>• [2, 3]: sin(πx) ≥ 0, so |sin(πx)| = sin(πx)</p><p><strong>Step 3:</strong> Evaluate integral piece by piece.</p><p>∫₋₁⁰ -sin(πx) dx = [cos(πx)/π]₋₁⁰ = (1/π)[cos(0) - cos(-π)] = (1/π)[1 - (-1)] = 2/π</p><p>∫₀¹ sin(πx) dx = [-cos(πx)/π]₀¹ = (-1/π)[cos(π) - cos(0)] = (-1/π)[-1 - 1] = 2/π</p><p>∫₁² -sin(πx) dx = [cos(πx)/π]₁² = (1/π)[cos(2π) - cos(π)] = (1/π)[1 - (-1)] = 2/π</p><p>∫₂³ sin(πx) dx = [-cos(πx)/π]₂³ = (-1/π)[cos(3π) - cos(2π)] = (-1/π)[-1 - 1] = 2/π</p><p><strong>Step 4:</strong> Sum all contributions.</p><p>∴ Answer: 2/π + 2/π + 2/π + 2/π = <strong>8/π</strong></p>
Correct Answer: 2