Question Number 64266 by Chi Mes Try last updated on 16/Jul/19
$$\:\underset{\mathrm{1}/{e}} {\overset{{e}} {\int}}\:\mid\:\mathrm{log}\:{x}\:\mid\:{dx}\:= \\ $$
Answered by ajfour last updated on 16/Jul/19
$${I}=−\int_{\mathrm{1}/{e}} ^{\:\:\mathrm{1}} \mathrm{ln}\:{xdx}+\int_{\mathrm{1}} ^{\:\:{e}} \mathrm{ln}\:{xdx} \\ $$$$\:\:\:=−\left({x}\mathrm{ln}\:{x}−{x}\right)\mid_{\mathrm{1}/{e}} ^{{e}} +\left({x}\mathrm{ln}\:{x}−{x}\right)\mid_{\mathrm{1}} ^{{e}} \\ $$$$\:\:\:=−\left({e}−{e}\right)+\left(−\frac{\mathrm{1}}{{e}}−\frac{\mathrm{1}}{{e}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({e}−{e}\right)−\left(\mathrm{0}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\int_{\mathrm{1}/{e}} ^{\:\:{e}} \mid\mathrm{ln}\:{x}\mid{dx}\:=\:\mathrm{1}−\frac{\mathrm{2}}{{e}}\:. \\ $$