Question

The area of the region bounded by the curves $\mathrm{y}={\mathrm{e}}^{\mathrm{x}},\mathrm{y}={\log }_{\mathrm{e}}\mathrm{x}$ and lines $\mathrm{x}=1,\mathrm{x}=2$is

• A. ${(\mathrm{e}–1)}^2$
• B. ${\mathrm{e}}^2–\mathrm{e}+1$
• C. ${\mathrm{e}}^2–\mathrm{e}+1–2{\log }_{\mathrm{e}}2$
• D. ${\mathrm{e}}^2+\mathrm{e}–2{\log }_{\mathrm{e}}2$

Answer 1

The correct option is C

${\mathrm{e}}^2–\mathrm{e}+1–2{\log }_{\mathrm{e}}2$

Explanation for correct option

Calculating the area of the region bounded by the three curves:

Given, equation of the three curves are $\mathrm{y}={\mathrm{e}}^{\mathrm{x}},\mathrm{y}={\log }_{\mathrm{e}}\mathrm{x},\mathrm{x}=1,\mathrm{x}=2$

Visualizing the curves on a graph we get:

$\mathrm{y}={\mathrm{e}}^{\mathrm{x}}$ and $\mathrm{y}=\log x$ are symmetrical

Hence, the required area is given by:

$$\begin{gathered} A={\int }_1^2\left(e^x-{\log }_{e}x\right)dx \\ ={\left[e^x\right]}_1^2-{\left[x{\log }_{e}x-\int dx\right]}_1^2 \\ =e^2-e-[x{\log }_{e}x-x{]}_1^2 \\ =e^2-e-2{\log }_{e}2+1 \end{gathered}$$

Therefore, the correct answer is Option (C).