The area enclosed by the curves xy = 4, y = x, \(x = e^2\) and the x-axis, is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(10-2\log_e 4\)
The line y = x meets the curve xy = 4 where
\(x^2 = 4,\quad x = 2\) (first quadrant, where the line \(x = e^2\) lies)
Since \(e^2 \approx 7.39 \gt 2\), the region above the x-axis has the line y = x as its top from x = 0 to x = 2, and the curve \(y = \frac{4}{x}\) as its top from x = 2 to \(x = e^2\). So the area is the sum of two integrals:
\( \displaystyle A = \int_{0}^{2} x\,dx + \int_{2}^{e^2}\frac{4}{x}\,dx \)
\( \displaystyle = \left[\frac{x^2}{2}\right]_{0}^{2} + 4\Big[\log_e x\Big]_{2}^{e^2} \)
\( = 2 + 4\left(2 - \log_e 2\right) \)
\( = 10 - 4\log_e 2 \)
Since \(4\log_e 2 = 2\log_e 2^2 = 2\log_e 4\):
\( A = 10 - 2\log_e 4 \)