Area Ques 56
Question
- Compute the area of the region bounded by the curves $y= e x \log x$ and $y=\frac{\log x}{e x}$, where $\log e=1$.
$(1990,4\ M)$
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Answer:
Correct Answer: 56.$(\frac{e^{2}-5}{4 e})$ sq units
Solution:
Formula:
- Both the curves are defined for $x>0$.
Both are positive when $x>1$ and negative when $0<x<1$. We know that, $\lim _{x \rightarrow 0^{+}}(\log x) = -\infty$
Hence, $\lim _{x \rightarrow 0^{+}} \frac{\log x}{e x} \rightarrow-\infty$. Thus, $Y$-axis is asymptote of first curve.
$ \begin{aligned} & \text { And } \lim _{x \rightarrow 0^{+}} x \log x \ & =\lim _{x \rightarrow 0^{+}} \frac{e \log x}{1 / x} \\ & =\lim _{x \rightarrow 0^{+}} \frac{e (\frac{1}{x})}{(-\frac{1}{x^{2}})}=-\infty \end{aligned} $
Thus, the first curve starts from $(0,0)$ but does not include the point $(0,0)$.
Now, the given curves intersect, hence
$\therefore \quad$ The required area is calculated.
$=\int _{1 / e}^{1} (\frac{(\log x)}{e x}-e x \log x) d x$
$=\frac{1}{e}{[\frac{(\log x)^{2}}{2}]} _{1 / e}^{1}-e [\frac{x^{2}}{4}(2 \log x-1)] _{1 / e}^{1}=(\frac{e^{2}-5}{4 e})$ sq units
BETA
