1-e-e-log-x-dx-

Question Number 64266 by Chi Mes Try last updated on 16/Jul/19

\underset{1 / e}{\int^{e}} ∣ \log x ∣ d x =

Answered by ajfour last updated on 16/Jul/19

I = - \int_{1 / e}^{1} \ln x d x + \int_{1}^{e} \ln x d x

= - (x \ln x - x) ∣_{1 / e}^{e} + (x \ln x - x) ∣_{1}^{e}

= - (e - e) + (- \frac{1}{e} - \frac{1}{e})

+ (e - e) - (0 - 1)

\Rightarrow \int_{1 / e}^{e} ∣ \ln x ∣ d x = 1 - \frac{2}{e} .

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