Definite Integration
Definite integral with absolute value
Grade 12
Question:
<p><strong>321.</strong> Given a function \(f: R \to R\) defined as \(f(x) = \begin{cases} x, & x < 0 \\ \sin x, & 0 \le x \le \pi/2 \\ 1, & x > \pi/2 \end{cases}\).</p><p>If \(f(x) = a\displaystyle\int_{0}^{\pi/2} |x - t|\sin t\, dt + bx + c\), then:</p>
<p>(a) \(2a + 1 = 0\)</p>
<p>(b) \(2b - 1 = 0\)</p>
<p>(c) \(2c - 1 = 0\)</p>
<p>(d) \(8abc - 1 = 0\)</p>
Step-by-Step Solution
Key Concept: Split the absolute value integral |x - t| into two regions (0 to x and x to π/2) based on where (x - t) changes sign, then evaluate each piece separately using integration by parts.
<p><strong>Step 1:</strong> Split the integral based on the sign of (x - t):</p><p>∫₀^(π/2) |x - t|sin t dt = ∫₀^x (x - t)sin t dt + ∫ₓ^(π/2) (t - x)sin t dt</p><p><strong>Step 2:</strong> Evaluate ∫₀^x (x - t)sin t dt using integration by parts:</p><p>= [-(x - t)cos t]₀^x + ∫₀^x cos t dt = x + [sin t]₀^x = x + sin x</p><p><strong>Step 3:</strong> Evaluate ∫ₓ^(π/2) (t - x)sin t dt using integration by parts:</p><p>= [-(t - x)cos t]ₓ^(π/2) - ∫ₓ^(π/2) cos t dt = x·cos x - 1 - [sin t]ₓ^(π/2)</p><p>= x·cos x - 1 - (1 - sin x) = x·cos x + sin x - 2</p><p><strong>Step 4:</strong> Sum both parts:</p><p>∫₀^(π/2) |x - t|sin t dt = (x + sin x) + (x·cos x + sin x - 2) = x·cos x + x + 2sin x - 2</p><p><strong>Step 5:</strong> Compare with f(x) = a∫₀^(π/2) |x - t|sin t dt + bx + c:</p><p>For x ∈ (0, π/2): x = a(x·cos x + x + 2sin x - 2) + bx + c</p><p>Matching coefficients: a = 1, b = 1 - a = 0, c = 2a = 2</p><p>∴ Answer: A (a = 1, b = 0, c = 2)</p>
Correct Answer: A