Question

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

• A. $1\mathrm{sq}\mathrm{units}$
• B. $\frac{3}{2}\mathrm{sq}\mathrm{units}$
• C. $\frac{5}{2}\mathrm{sq}\mathrm{units}$
• D. $\frac{1}{2}\mathrm{sq}\mathrm{units}$

Answer 1

The correct option is B

$\frac{3}{2}\mathrm{sq}\mathrm{units}$

Explanation for the correct option:

Given,

$$\begin{gathered} \mathrm{y}=\mathrm{x}, \\ \mathrm{x}=\mathrm{e}, \\ \mathrm{y}=\frac{1}{\mathrm{x}} \end{gathered}$$

Area,

$$\begin{gathered} =\mathrm{A}\left(\mathrm{ODA}\right)+\mathrm{A}\left(\mathrm{ABCD}\right) \\ ={\int }_0^1\mathrm{x}d\mathrm{x}+{\int }_1^{\mathrm{e}}\frac{1}{\mathrm{x}}d\mathrm{x} \\ =\frac{1}{2}+\left[\log \left(\mathrm{e}\right)-\log \left(1\right)\right] \\ =\frac{1}{2}+1-0 \\ =\frac{3}{2}\mathrm{sq}\mathrm{units} \end{gathered}$$

Hence, option $\left(\mathrm{B}\right)$ is the correct answer.