calculate-0-6-e-x-x-1-e-x-dx-

Question Number 37280 by abdo.msup.com last updated on 11/Jun/18

c a l c u l a t e \int_{0}^{6} \frac{e^{x - [x]}}{1 + e^{x}} d x .

Commented by prof Abdo imad last updated on 16/Jun/18

I = \sum_{k = 0}^{5} \int_{k}^{k + 1} \frac{e^{x - k}}{1 + e^{x}} d x

= \sum_{k = 0}^{5} e^{- k} \int_{k}^{k + 1} \frac{e^{x}}{1 + e^{x}} d x

= \sum_{k = 0}^{5} e^{- k} {[l n (1 + e^{x})]}_{k}^{k + 1}

= \sum_{k = 0}^{5} e^{- k} {l n (1 + e^{k + 1}) - l n (1 + e^{k})}

= \sum_{k = 0}^{5} e^{- k} l n (\frac{1 + e^{k + 1}}{1 + e^{k}})

= l n (\frac{1 + e}{2}) + e^{- 1} l n (\frac{1 + e^{2}}{1 + e}) + e^{- 2} l n (\frac{1 + e^{3}}{1 + e^{2}})

+ e^{- 3} l n (\frac{1 + e^{4}}{1 + e^{3}}) + e^{- 4} l n (\frac{1 + e^{5}}{1 + e^{4}}) + e^{- 5} l n (\frac{1 + e^{6}}{1 + e^{5}}) .