Question

Find the area enclosed the curves : $y=ex\log x$ and $y=\frac{\log x}{ex}$ where $\log e=1$

• A. $\frac{e^2-5}{4e}$
• B. $\frac{e^2+5}{4e}$
• C. $\frac{e^2-3}{2e}$
• D. $\frac{e^2+3}{2e}$

Answer 1

The correct option is A $\frac{e^2-5}{4e}$

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
$ex\log x=\frac{\log x}{ex}\Rightarrow (e^2{x}^2-1)\log x=0$
$\Rightarrow x=1,\frac{1}{e}(sincex>0)$
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$
$=\frac{e^2-5}{4e}$