Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>Let a function \(f: R \to R\) be defined as \(f(x) = x + \sin x\) and \(I = \displaystyle\int_0^{\pi} f^{-1}(x)\, dx\) then:</p>
<p>(a) \(I > \displaystyle\int_0^1 \frac{1}{1+x^3}\, dx\)</p>
<p>(b) \(I < \displaystyle\int_0^1 e^{x^2}\, dx\)</p>
<p>(c) \(2 < I < 3\)</p>
<p>(d) \(\dfrac{\pi}{4} < I < \dfrac{\pi}{2}\)</p>
Step-by-Step Solution
Key Concept: Use integration by parts with the substitution property: ∫₀^π f⁻¹(x)dx = πf⁻¹(π) - ∫₀^(f⁻¹(π)) f(t)dt. First verify f is bijective (f'(x) = 1 + cos x ≥ 0), then find f⁻¹(π) by solving f(t) = π.
<p><strong>Step 1:</strong> Verify f(x) = x + sin x is bijective on ℝ.</p><p>f'(x) = 1 + cos x ≥ 0 for all x ∈ ℝ (with equality only at isolated points), so f is strictly increasing and bijective. ✓</p><p><strong>Step 2:</strong> Use the inverse function integration formula: ∫₀^a f⁻¹(x)dx = af⁻¹(a) - ∫₀^(f⁻¹(a)) f(t)dt</p><p><strong>Step 3:</strong> Find f⁻¹(π). We need f(t) = π, so t + sin t = π.</p><p>At t = π: f(π) = π + sin π = π + 0 = π ✓</p><p>Therefore f⁻¹(π) = π</p><p><strong>Step 4:</strong> Apply the formula:</p><p>I = π·f⁻¹(π) - ∫₀^π f(t)dt</p><p>I = π·π - ∫₀^π (t + sin t)dt</p><p>I = π² - [t²/2 - cos t]₀^π</p><p>I = π² - [(π²/2 - cos π) - (0 - cos 0)]</p><p>I = π² - [π²/2 - (-1) - (-1)]</p><p>I = π² - [π²/2 + 2]</p><p>I = π² - π²/2 - 2</p><p>∴ I = <strong>π²/2 - 2</strong></p>
Correct Answer: A