<p><strong>95.</strong> Let a function \(f(x)\) be defined in \([-2, 2]\) as \(f(x) = \begin{cases} \{x\}, & -2 \leq x < -1 \\ |\text{sgn }x|, & -1 \leq x \leq 1 \\ \{-x\}, & 1 < x \leq 2 \end{cases}\), where \(\{x\}\) denotes fractional part, then area bounded by graph of \(f(x)\) and \(x\)-axis is:</p>
Step-by-Step Solution
Key Concept: The fractional part function {x} = x - ⌊x⌋ is periodic with period 1 and ranges from 0 to 1. The area under {x} from -2 to 2 requires careful handling of how {x} behaves across integer boundaries, particularly the shift in floor function values at negative integers.
<p><strong>Step 1: Understand the fractional part function</strong></p><p>For any real x: {x} = x - ⌊x⌋, where 0 ≤ {x} < 1</p><p><strong>Step 2: Identify {x} on each subinterval of [-2, 2]</strong></p><p>• For x ∈ [-2, -1): ⌊x⌋ = -2, so {x} = x - (-2) = x + 2, ranges from [0, 1)</p><p>• For x ∈ [-1, 0): ⌊x⌋ = -1, so {x} = x - (-1) = x + 1, ranges from [0, 1)</p><p>• For x ∈ [0, 1): ⌊x⌋ = 0, so {x} = x, ranges from [0, 1)</p><p>• For x ∈ [1, 2): ⌊x⌋ = 1, so {x} = x - 1, ranges from [0, 1)</p><p>• At x = 2: {2} = 0</p><p><strong>Step 3: Calculate the area on each subinterval</strong></p><p>∫₋₂⁻¹ (x+2)dx = [x²/2 + 2x]₋₂⁻¹ = (1/2 - 2) - (2 - 4) = -3/2 + 2 = 1/2</p><p>∫₋₁⁰ (x+1)dx = [x²/2 + x]₋₁⁰ = 0 - (1/2 - 1) = 1/2</p><p>∫₀¹ x dx = [x²/2]₀¹ = 1/2</p><p>∫₁² (x-1)dx = [x²/2 - x]₁² = (2 - 2) - (1/2 - 1) = 1/2</p><p><strong>Step 4: Sum all areas</strong></p><p>Total area = 1/2 + 1/2 + 1/2 + 1/2 = 2</p><p>∴ Answer: B</p>
Correct Answer: B