1-e-e-ln-x-dx-

Question Number 88902 by M±th+et£s last updated on 13/Apr/20

\int_{\frac{1}{e}}^{e} l n ∣ x ∣ d x

Commented by abdomathmax last updated on 13/Apr/20

\int_{\frac{1}{e}}^{e} l n ∣ x ∣ d x = {[x l n x - x]}_{\frac{1}{e}}^{e} = (e - e) - (- \frac{1}{e} - \frac{1}{e})

= \frac{2}{e}

Commented by abdomathmax last updated on 13/Apr/20

a n o t h e r w a y l e t λ > 0 w e h a v e ∣ x ∣= x \Rightarrow

\int_{\frac{1}{λ}}^{λ} l n (x) d x = {[x l n x - x]}_{\frac{1}{λ}}^{λ} = λ l n λ - λ - (\frac{1}{λ} l n (\frac{1}{λ}) - \frac{1}{λ})

= λ l n (λ) - λ + \frac{l n (λ)}{λ} + \frac{1}{λ}

λ = e \Rightarrow \int_{\frac{1}{e}}^{e} l n (x) d x = e - e + \frac{1}{e} + \frac{1}{e} = \frac{2}{e}

Commented by M±th+et£s last updated on 13/Apr/20

t h a n k y o u s i r

Commented by turbo msup by abdo last updated on 13/Apr/20

y o u a r e w e l c o m e

Answered by TANMAY PANACEA. last updated on 13/Apr/20

t =∣ x ∣

d t = d x s i n c e x > 0

\int_{\frac{1}{e}}^{e} l n t d t

=∣ t l n t - t ∣_{\frac{1}{e}}^{e}

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

= - \frac{1}{e} \times - 1 + \frac{1}{e} = \frac{2}{e}

Commented by M±th+et£s last updated on 13/Apr/20

t h a n k y o u s i r

Commented by TANMAY PANACEA. last updated on 13/Apr/20

m o s t w e l c o m e s i r