The area (in square units) of the region enclosed by the curves $y = x, x =e, y=\frac{1}{x}$ and the positive x-axis is
Answer & explanation
Correct answer: option 3
Required area = $\int\limits_0^1x\,dx+\int\limits_1^e\frac{1}{x}dx$
$=\left[\frac{x^2}{2}\right]_0^1+[\log x]_1^e=\frac{1}{2}+1=\frac{3}{2}$
