Question

If $\underset{\frac{1}{3}}{\overset{3}{\int }}\left|{\log }_{e}x\right|dx=\frac{m}{n}\log e\left(\frac{n^2}{v}\right)$, where $m$ and $n$ are coprime natural numbers, then $m^2+n^2-5$ is equal to _____

Answer 1

Answer: 20

$\underset{\frac{1}{3}}{\overset{3}{\int }}|\ell nx|dx=\underset{\frac{1}{3}}{\overset{1}{\int }}(-\ell nx)dx+\underset{1}{\overset{3}{\int }}(\ell nx)dx$
$=-[x\ell nx-x{]}_{1/3}^1+[x\ell nx-x{]}_1^3$
$=-\left[-1-\left(\frac{1}{3}\ell n\frac{1}{3}-\frac{1}{3}\right)\right]+[3\ell n3-3-(-1)]$
$=\left[-\frac{2}{3}-\frac{1}{3}\ell n\frac{1}{3}\right]+[3\ell n3-2]$
$=-\frac{4}{3}+\frac{8}{3}\ell n3$
$=\frac{4}{3}(2\ell n3-1)$
$=\frac{4}{3}\left(\ell n\frac{9}{e}\right)$
$\therefore m=4,n=3$
Now, $m^2+n^2-5=16+9-5=20$