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 method: if y = f(x), then ∫f⁻¹(x)dx = xf⁻¹(x) - ∫xf'(f⁻¹(x))dx. Alternatively, recognize that ∫₀^π f⁻¹(x)dx + ∫₀^(π) f(x)dx = π·f(π) using the geometric property of inverse functions.
<p><strong>Step 1:</strong> Recognize that f(x) = x + sin(x) is strictly increasing (f'(x) = 1 + cos(x) ≥ 0 for all x), so f⁻¹ exists.</p><p><strong>Step 2:</strong> Use the geometric property: For inverse functions, ∫₀^π f⁻¹(x)dx = π·f⁻¹(π) - ∫₀^(f⁻¹(π)) f(x)dx</p><p><strong>Step 3:</strong> Find f⁻¹(π): We need f(x) = π, so x + sin(x) = π. Since sin(x) ∈ [-1,1], we get x ≈ π (checking: f(π) = π + sin(π) = π + 0 = π). Thus f⁻¹(π) = π.</p><p><strong>Step 4:</strong> Apply the inverse function formula: I = π·π - ∫₀^π (x + sin x)dx = π² - [x²/2 - cos(x)]₀^π = π² - [π²/2 - cos(π) + cos(0)] = π² - [π²/2 + 1 + 1] = π²/2 - 2</p><p>∴ Answer: A</p>
Correct Answer: A