<p>Sketch the region bounded by the curves \(y = \log_e x\) and \(y = (\log_e x)^2\). Also find the area of the region.</p>
Step-by-Step Solution
Key Concept: The curves y = ln(x) and y = (ln x)² intersect when ln(x) = (ln x)², which occurs at x = 1 and x = e. The key is recognizing which curve is above the other in the interval (1, e) by testing a point, then integrating the difference.
<p><strong>Step 1: Find intersection points</strong></p><p>Set y = ln(x) equal to y = (ln x)²:</p><p>ln(x) = (ln x)²</p><p>Let u = ln(x): u = u²</p><p>u² - u = 0 → u(u - 1) = 0</p><p>So u = 0 or u = 1</p><p>Thus x = 1 or x = e</p><p><strong>Step 2: Determine which curve is above</strong></p><p>Test x = √e (where ln(√e) = 1/2):</p><p>y = 1/2 and y = (1/2)² = 1/4</p><p>Since 1/2 > 1/4, the curve y = ln(x) is above y = (ln x)² on (1, e)</p><p><strong>Step 3: Set up and evaluate the integral</strong></p><p>Area = ∫₁ᵉ [ln(x) - (ln x)²] dx</p><p><strong>For ∫ln(x)dx:</strong> Using integration by parts (u = ln x, dv = dx):</p><p>∫ln(x)dx = x·ln(x) - x + C</p><p><strong>For ∫(ln x)²dx:</strong> Using integration by parts twice (u = (ln x)², dv = dx):</p><p>∫(ln x)²dx = x(ln x)² - 2x·ln(x) + 2x + C</p><p><strong>Step 4: Apply limits</strong></p><p>At x = e: [e·ln(e) - e] - [e(ln e)² - 2e·ln(e) + 2e] = (e - e) - (e - 2e + 2e) = 0 - e = -e</p><p>At x = 1: [1·ln(1) - 1] - [1(ln 1)² - 2·1·ln(1) + 2·1] = (0 - 1) - (0 - 0 + 2) = -1 - 2 = -3</p><p>Area = [-e] - [-3] = 3 - e</p><p>∴ <strong>Answer: 3 - e</strong></p>
Correct Answer: 3 - e