Question

If the area of the region bounded by the curve $y=ex\ln x$ and $y=\frac{\left(\ln x\right)}{ex}$ is $\frac{e^2-m}{4e}$.Find $m$

Answer 1

Given curves are

$y=ex\ln x-\left(1\right)$

$y=\ln x/ex$ is $\frac{e^2-m}{4e}-\left(2\right)$

$m=?$

Points of intersection of (1) and (2)

are $P\left(1/e\right)-1)$ and $Q(1,0)$

for curves $\left(1\right),y<0$ for $0<x<1$ and

$y\ge 0$ for $x\ge 1$

$y\to 0$ when $x\to 0$

Area $\left|{\int }_{1/e}^{1}ex\ln xdx\right|-\left|{\int }_{1/e}^1\frac{\ln x}{ex}dx\right|$

$=\left|\frac{1}{4e}(3-e^2)\right|-\left|\frac{-1}{2e}\right|$

$\Rightarrow \frac{e^2-3}{4e}-\frac{1}{2e}=\frac{e^2-5}{4e}sq\cdot units$

So, $m=5$ answer.