Question

The area bounded by the curves $ y = \log_e x $ and $ y = (\log_e x)^2 $ is:

• A. \( (3 - e) \) sq. units
• B. \( (e - 3) \) sq. units
• C. \( \frac{1}{2} (3 - e) \) sq. units
• D. \( \frac{1}{2} (e - 3) \) sq. units

Hint:

When calculating areas between curves, first find the points of intersection and then integrate the difference between the functions over the given range.

Answer 1

The Correct Option is C

We are asked to find the area between the curves \( y = \log_e x \) and \( y = (\log_e x)^2 \).
Step 1: Find the points of intersection.
To find the points where the curves intersect, set: \[ \log_e x = (\log_e x)^2 \] Let \( u = \log_e x \). Then the equation becomes: \[ u = u^2 \quad \Rightarrow \quad u^2 - u = 0 \quad \Rightarrow \quad u(u - 1) = 0. \] Thus, \( u = 0 \) or \( u = 1 \). Since \( u = \log_e x \), the corresponding values of \( x \) are \( x = 1 \) and \( x = e \).
Step 2: Set up the integral.
The area between the curves is
given by the integral:
\[ \text{Area} = \int_1^e \left( \log_e x - (\log_e x)^2 \right) \, dx. \]
Step 3: Simplify and integrate.
Let \( u = \log_e x \), so \( du = \frac{1}{x} dx \). When \( x = 1 \), \( u = 0 \), and when \( x = e \), \( u = 1 \). The integral becomes: \[ \text{Area} = \int_0^1 (u - u^2) \, du = \left[ \frac{u^2}{2} - \frac{u^3}{3} \right]_0^1 = \left( \frac{1}{2} - \frac{1}{3} \right) - (0) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}. \] Thus, the total area is: \[ \text{Area} = \frac{1}{2} (3 - e). \]