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# Integrate: $\int \limits_{1/e}^e | \log_e x| dx$

$\begin {array} {1 1} (a)\;\frac{1}{e} \\ (b)\;\frac{2}{e} \\ (c)\;\frac{e}{2} \\ (d)\;\frac{4}{e} \end {array}$
Can you answer this question?

$\log _e x=0$
at $x=1, \log _e n=0$
if it is negative x < 1
if it is positive x > 1
$\int \limits_{4e}^1 -(\log _e x)+\int \limits_1 ^e \log _e x dx$
=> $\large\frac{2}{e}$
Hence b is the correct answer.
answered Dec 12, 2013 by