The area (in sq. units) bounded by the curve $f\left(x\right)=\left\{\begin{array}{ll}{e}^{-\left|x\right|}& \left|x\right|\ge 1\\ \left|x\right|-1+\frac{1}{e}& \left|x\right|<1\end{array}\right.$ and the $x$-axis is

• A

  $2+\frac{2}{e^2}$
• B

  $2+\frac{1}{e^2}$
• C

  $\frac{1}{e^2}$
• Correct answer
  D

  $1+\frac{2}{e^2}$

Solution

$\because$ Graph of $f\left(x\right)$ is symmetric about $x$-axis, hence,
Area $=2\underset{0}{\overset{\mathrm{\infty }}{\int }}\left|f\left(x\right)\right|dx$
$=2\left[\frac{1}{2}\times \left(1-\frac{1}{e}\right)\left(1-\frac{1}{e}\right)+\frac{1}{2}\times \frac{1}{e}\times \frac{1}{e}+\underset{1}{\overset{\mathrm{\infty }}{\int }}{e}^{-x}dx\right]$
$=2\left[\frac{1}{2}\left[1-\frac{2}{e}+\frac{2}{e^2}\right]+{\left[-e^{-x}\right]}_1^{\mathrm{\infty }}\right]$
$=\left[1-\frac{2}{e}+\frac{2}{e^2}+\frac{2}{e}\right]=1+\frac{2}{e^2}$ sq. units

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