The area of the region enclosed by the curves $y=x, x=e, y=\frac{1}{x}$ and the positive $X$-axis, is

• A

  $\frac{1}{2}$ sq unit
• B

  1 sq unit
• Correct answer
  C

  $\frac{3}{2}$ sq units
• D

  $\frac{5}{2}$ sq units

Solution

Required area $=\int_0^1 x d x+\int_1^e \frac{1}{x} d x$
$=\left[\frac{x^2}{2}\right]_0^1+[\log x]_1^\theta=\frac{1}{2}+1=\frac{3}{2}$

Hence, the correct answer is option: 'C'.