<p>Let a function f: ℝ → ℝ be defined as f(x) = x + sin x. The value of ∫₀^{2π} f⁻¹(x)dx will be:</p>
Step-by-Step Solution
Key Concept: Use the property that ∫₀ᵃ f⁻¹(x)dx + ∫₀ᶠ⁽ᵃ⁾ f(x)dx = a·f(a). Here we need to find the range of f to determine the upper limit, then apply this integration property.
<p><strong>Step 1: Verify f is invertible</strong></p><p>f(x) = x + sin x, so f'(x) = 1 + cos x ≥ 0 for all x ∈ ℝ, with f'(x) = 0 only at isolated points. Since f'(x) > 0 almost everywhere, f is strictly increasing and invertible on ℝ.</p><p><strong>Step 2: Find the range of f on [0, 2π]</strong></p><p>At x = 0: f(0) = 0 + sin(0) = 0</p><p>At x = 2π: f(2π) = 2π + sin(2π) = 2π + 0 = 2π</p><p>Since f is strictly increasing, the range of f on [0, 2π] is [0, 2π].</p><p><strong>Step 3: Apply the inverse function integral formula</strong></p><p>For a strictly monotonic function with f(a) = b, we have:</p><p>∫₀ᵇ f⁻¹(x)dx = a·b - ∫₀ᵃ f(x)dx</p><p>Here: a = 2π, b = f(2π) = 2π</p><p>∫₀^{2π} f⁻¹(x)dx = 2π · 2π - ∫₀^{2π} (x + sin x)dx</p><p><strong>Step 4: Evaluate ∫₀^{2π} (x + sin x)dx</strong></p><p>∫₀^{2π} (x + sin x)dx = [x²/2 - cos x]₀^{2π}</p><p>= [(2π)²/2 - cos(2π)] - [0 - cos(0)]</p><p>= [4π²/2 - 1] - [0 - 1]</p><p>= 2π² - 1 + 1 = 2π²</p><p><strong>Step 5: Calculate the final answer</strong></p><p>∫₀^{2π} f⁻¹(x)dx = 4π² - 2π² = 2π²</p><p><strong>∴ Answer:</strong> a</p>
Correct Answer: a