<p>Area bounded by the curve \(y = x \sin x\) and X-axis between \(x = 0\) and \(x = 2\pi\) is</p>
Step-by-Step Solution
Key Concept: The curve y = x sin x oscillates around the x-axis, changing sign at x = π. We must split the integral into regions where the function is positive and negative, taking absolute values to find total area.
<p><strong>Step 1: Determine where y = x sin x changes sign</strong><br>For x ∈ [0, 2π]: sin x ≥ 0 when x ∈ [0, π] and sin x ≤ 0 when x ∈ [π, 2π]<br>Therefore: y ≥ 0 on [0, π] and y ≤ 0 on [π, 2π]</p><p><strong>Step 2: Set up the area integral with absolute values</strong><br>Area = ∫₀^π x sin x dx + |∫_π^{2π} x sin x dx| = ∫₀^π x sin x dx - ∫_π^{2π} x sin x dx</p><p><strong>Step 3: Evaluate ∫ x sin x dx using Integration by Parts</strong><br>Let u = x, dv = sin x dx<br>Then du = dx, v = -cos x<br>∫ x sin x dx = -x cos x + ∫ cos x dx = -x cos x + sin x + C</p><p><strong>Step 4: Evaluate the first integral ∫₀^π x sin x dx</strong><br>= [-x cos x + sin x]₀^π<br>= [(-π cos π + sin π) - (0 + sin 0)]<br>= [(-π)(-1) + 0 - 0]<br>= π</p><p><strong>Step 5: Evaluate the second integral ∫_π^{2π} x sin x dx</strong><br>= [-x cos x + sin x]_π^{2π}<br>= [(-2π cos 2π + sin 2π) - (-π cos π + sin π)]<br>= [(-2π)(1) + 0 - (π + 0)]<br>= -2π - π = -3π</p><p><strong>Step 6: Calculate total area</strong><br>Area = ∫₀^π x sin x dx - ∫_π^{2π} x sin x dx<br>= π - (-3π)<br>= π + 3π<br>= 4π sq units</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C