<p>The value of \(\int_{-2}^{3} |1 - x^2| \, dx\) is</p>
Step-by-Step Solution
Key Concept: The absolute value function |1 - x²| changes sign when 1 - x² = 0, i.e., at x = ±1. Split the integral at these critical points and remove the absolute value by determining the sign of (1 - x²) in each region.
<p><strong>Step 1: Find critical points</strong></p><p>Set 1 - x² = 0 → x = ±1. These points lie in [-2, 3], so we must split the integral.</p><p><strong>Step 2: Determine sign in each region</strong></p><p>• For x ∈ [-2, -1]: x² > 1, so 1 - x² < 0, thus |1 - x²| = x² - 1<br/>• For x ∈ [-1, 1]: x² ≤ 1, so 1 - x² ≥ 0, thus |1 - x²| = 1 - x²<br/>• For x ∈ [1, 3]: x² > 1, so 1 - x² < 0, thus |1 - x²| = x² - 1</p><p><strong>Step 3: Split and integrate</strong></p><p>∫₋₂³ |1 - x²| dx = ∫₋₂⁻¹ (x² - 1) dx + ∫₋₁¹ (1 - x²) dx + ∫₁³ (x² - 1) dx</p><p><strong>Step 4: Evaluate each integral</strong></p><p>∫₋₂⁻¹ (x² - 1) dx = [x³/3 - x]₋₂⁻¹ = (-1/3 + 1) - (-8/3 + 2) = 2/3 + 2/3 = 4/3</p><p>∫₋₁¹ (1 - x²) dx = [x - x³/3]₋₁¹ = (1 - 1/3) - (-1 + 1/3) = 2/3 + 2/3 = 4/3</p><p>∫₁³ (x² - 1) dx = [x³/3 - x]₁³ = (9 - 3) - (1/3 - 1) = 6 + 2/3 = 20/3</p><p><strong>Step 5: Add results</strong></p><p>Total = 4/3 + 4/3 + 20/3 = 28/3</p><p>∴ Answer: <strong>28/3</strong></p>
Correct Answer: A