Question

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

• A. $\frac{2}{e}\text{sq. units}$
• B. $1-\frac{2}{e}\text{sq. units}$
• C. $\frac{1}{e}\text{sq. units}$
• D. $1-\frac{1}{e}\text{sq. units}$

Answer 1

The correct option is A $\frac{2}{e}\text{sq. units}$

$y=x{e}^x$
Let $\frac{dy}{dx}=0\Rightarrow e^x+x{e}^x=0\Rightarrow x=-1$

Also, at $x=-1,\frac{dy}{dx}$ changes sign from $-ve$ to $+ve$.
Hence, $x=-1$ is a point of minima.

When $x\to \infty ,y\to \infty$
Also,
$\underset{x\to -\infty }{lim}x{e}^x=\underset{x\to -\infty }{lim}\frac{x}{e^{-x}}$
$=\underset{x\to -\infty }{lim}\frac{1}{-e^{-x}}$ (Applying L'Hospital Rule)
$=0$

Similarly, we can draw the graph of $y=x{e}^{-x}.$



Required area
$=\underset{0}{\overset{1}{\int }}x{e}^{x}dx-\underset{0}{\overset{1}{\int }}x{e}^{-x}dx$
$=[x{e}^x{]}_0^1-\underset{0}{\overset{1}{\int }}{e}^{x}dx-([-x{e}^{-x}{]}_0^1+\underset{0}{\overset{1}{\int }}{e}^{-x}dx)$
$=e-(e-1)-(-e^{-1}-(e^{-1}-1\left)\right)$
$=\frac{2}{e}\text{sq. units.}$