<p>The area enclosed between the curve \(y = \log_e(x + e)\) and the coordinate axes is</p>
Step-by-Step Solution
Key Concept: The area is bounded by the curve y = logₑ(x + e), the y-axis (x = 0), and the x-axis (y = 0). Find where the curve meets the axes, then integrate: Area = ∫₀^(1-e) logₑ(x + e) dx using substitution and integration by parts.
<p><strong>Step 1:</strong> Find intersection points with axes.</p><p>When x = 0: y = logₑ(e) = 1 (y-axis intersection)</p><p>When y = 0: logₑ(x + e) = 0 ⟹ x + e = 1 ⟹ x = 1 - e ≈ -1.718 (x-axis intersection)</p><p><strong>Step 2:</strong> Set up the integral. The area enclosed is:</p><p>A = ∫₍₁₋ₑ₎⁰ logₑ(x + e) dx</p><p><strong>Step 3:</strong> Use substitution u = x + e, so du = dx. When x = 1 - e, u = 1; when x = 0, u = e.</p><p>A = ∫₁ᵉ logₑ(u) du</p><p><strong>Step 4:</strong> Apply integration by parts: ∫logₑ(u) du = u·logₑ(u) - u + C</p><p>A = [u·logₑ(u) - u]₁ᵉ = [e·logₑ(e) - e] - [1·logₑ(1) - 1]</p><p>A = [e·1 - e] - [0 - 1] = 0 + 1 = 1</p><p>∴ Answer: A (Area = 1)</p>
Correct Answer: A