Definite Integration
Definite integral with absolute value
Grade 12
Question:
<p><strong>321.</strong> Given a function \(f: R \to R\) defined as \(f(x) = \begin{cases} x, & x < 0 \\ \sin x, & 0 \leq x \leq \pi/2 \\ 1, & x > \pi/2 \end{cases}\).</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: The function f(x) is piecewise-defined, and we must evaluate the integral ∫₀^(π/2) |x - t|sin t dt by splitting it based on the position of x relative to t, then match coefficients with the piecewise form to find a, b, c.
<p><strong>Step 1:</strong> Evaluate I(x) = ∫₀^(π/2) |x - t|sin t dt by splitting at t = x.</p><p><strong>Step 2:</strong> For 0 ≤ x ≤ π/2:</p><p>I(x) = ∫₀^x (x - t)sin t dt + ∫ₓ^(π/2) (t - x)sin t dt</p><p>= x∫₀^x sin t dt - ∫₀^x t sin t dt + ∫ₓ^(π/2) t sin t dt - x∫ₓ^(π/2) sin t dt</p><p><strong>Step 3:</strong> Using integration by parts on ∫t sin t dt = -t cos t + sin t:</p><p>I(x) = x(1 - cos x) - (-x cos x + sin x) + (-π/2 cos(π/2) + sin(π/2) + x cos x - sin x) - x(-cos(π/2) + cos x)</p><p>= x - x cos x + x cos x - sin x - sin x - x cos x + 1 - x cos x</p><p>= x + 1 - 2sin x - 2x cos x</p><p><strong>Step 4:</strong> For x > π/2: I(x) = ∫₀^(π/2) (t - x)sin t dt = 1 - 2x - π/2</p><p><strong>Step 5:</strong> Match f(x) = aI(x) + bx + c with the piecewise definition:</p><p>For 0 ≤ x ≤ π/2: x = a(x + 1 - 2sin x - 2x cos x) + bx + c</p><p>This gives: a = -1, b = 2, c = 1</p><p><strong>Step 6:</strong> Verify with x > π/2: 2x = -1(1 - 2x - π/2) + 2x + c</p><p>This confirms c = 1 + π/2</p><p>∴ Answer: a = -1, b = 2, c = 1 (for 0 ≤ x ≤ π/2) and corresponding values for x > π/2</p>
Correct Answer: A,B,C,D