The area of the region enclosed by the curves $y=x, x=e, y=\frac{1}{x}$ and the positive $x$-axis is
- 1 square units
- $\frac{3}{2}$ square units
- $\frac{5}{2}$ square units
- $\frac{1}{2}$ square units
Solution

Area $=\int_0^1 x d x+\int_1^e \frac{1}{x} d x=\frac{1}{2}+1=\frac{3}{2}$
Asked in: JEE Main 2011