Question

Compute the area of the region bounded by the curves $y=ex\log x$ and $y=\frac{\log x}{ex}$, where $\log e=1$ is

Answer 1

Both the curves are defined for $x>0$
Both are positive when $x>1$ and negative when $0<x<1$
We know $\underset{x\to 0^{+}}{lim}\left(\log x\right)\to -\infty$
Hence, $\underset{x\to 0^{+}}{lim}\left(\log x\right)\to -\infty$ .Thus $y$-axis is asymptote of second curve.
And $\underset{x\to 0^{+}}{lim}ex\log x........[\left(0\right)\times \infty form]$
$=\underset{x\to 0^{+}}{lim}\frac{e\log x}{\frac{1}{x}}(-\frac{\infty }{\infty }form)$$=\underset{x\to 0^{+}}{lim}\frac{e\left(\frac{1}{x}\right)}{(-\frac{1}{x^2})}=0$ (using L'hopital rule)
Thus, the first curve starts from $(0,0)$ but does not include $(0,0)$
Now, the given curves intersect.

Therefore $ex\log x=\frac{\log x}{ex}$

$\Rightarrow (e^2{x}^2-1)\log x=0$

$\Rightarrow x=1,\frac{1}{e}$ .............. Since $(x>0)$
Therefore required area $={\int }_{\frac{1}{e}}^1(\frac{\left(\log x\right)}{ex}-ex\log x)dx$

$=\frac{1}{e}{\left[\frac{{\left(\log x\right)}^2}{2}\right]}_{\frac{1}{e}}^1-e{\left[\frac{x^2}{4}(2\log x-1)\right]}_{\frac{1}{e}}^1$
$=\frac{e^2-5}{4e}$