<p>The value of \(\int_{-2}^{3} |1-x^2|\, dx\) is:</p>
Step-by-Step Solution
Key Concept: Split the integral at points where the expression inside absolute value changes sign (at x = ±1), then evaluate each piece by removing the absolute value based on the sign of (1-x²) in that region.
<p><strong>Step 1:</strong> Find critical points where |1-x²| changes behavior. Set 1-x² = 0 → x² = 1 → x = ±1</p><p><strong>Step 2:</strong> Determine the sign of (1-x²) in each interval:</p><ul><li>For x ∈ [-2, -1]: x² > 1, so 1-x² < 0 → |1-x²| = x²-1</li><li>For x ∈ [-1, 1]: x² ≤ 1, so 1-x² ≥ 0 → |1-x²| = 1-x²</li><li>For x ∈ [1, 3]: x² > 1, so 1-x² < 0 → |1-x²| = x²-1</li></ul><p><strong>Step 3:</strong> Split and evaluate:</p><p>∫₋₂³ |1-x²| dx = ∫₋₂⁻¹ (x²-1) dx + ∫₋₁¹ (1-x²) dx + ∫₁³ (x²-1) dx</p><p><strong>Step 4:</strong> Compute each integral:</p><ul><li>∫₋₂⁻¹ (x²-1) dx = [x³/3 - x]₋₂⁻¹ = (-1/3 + 1) - (-8/3 + 2) = 2/3 - (-2/3) = 4/3</li><li>∫₋₁¹ (1-x²) dx = [x - x³/3]₋₁¹ = (1 - 1/3) - (-1 + 1/3) = 2/3 - (-2/3) = 4/3</li><li>∫₁³ (x²-1) dx = [x³/3 - x]₁³ = (9 - 3) - (1/3 - 1) = 6 - (-2/3) = 20/3</li></ul><p><strong>Step 5:</strong> Add the results: 4/3 + 4/3 + 20/3 = 28/3</p><p>∴ Answer: <strong>28/3</strong></p>
Correct Answer: A