The area bounded by the curves $y=x{e}^x,y=x{e}^{-x}$ and the line $x=1$ is 

Curve tracing $:y=x{e}^x$
 $\text{Let }\frac{dy}{dx}=0\Rightarrow e^x+x{e}^x=0\text{ or }x=-1$

Also, at $x=-1,\frac{dy}{dx}$ changes sign from –ve to +ve.

Hence, $x=-1$ is a point of minima.

$\text{ When }x\to \mathrm{\infty },y\to \mathrm{\infty }$

$\text{ Also }\underset{x\to -\mathrm{\infty }}{lim} x{e}^x=\underset{x\to -\mathrm{\infty }}{lim} \frac{x}{e^{-x}}=\underset{x\to -\mathrm{\infty }}{lim} \frac{1}{-e^{-x}}=0$

With similar types of arguments, we can draw the graph of $y=x{e}^{-x}$

Required area $={\int }_0^1 x{e}^{x}dx-{\int }_0^1 x{e}^{-x}dx$

$={\left[x{e}^x\right]}_0^1-{\int }_0^1 e^{x}dx-\left({\left[-x{e}^{-x}\right]}_0^1+{\int }_0^1 e^{-x}dx\right)$

$=e-(e-1)-\left(-e^{-1}-\left(e^{-1}-1\right)\right)=\frac{2}{e}\text{ sq.units }$