The area enclosed by the curves $xy = 4, y=x, x=e^2$ and the x-axis, is :

$4-log_e2$ $10-2log_e4$ $6-2log_e4$ $10-log_e8$

$10-2log_e4$

The correct answer is Option (2) → $10-2log_e4$ $xy = 4, y=x, x=e^2$ $x=y$ and $xy=4$ intersect at $x=4$ required area = $\frac{1}{2}×2×2+\int\limits_2^{e^2}\frac{4}{x}dx$ $=2+4[\log x]_2^{e^2}$ $=2+4(2-\log 2)$ $=10-2\log 4$