integral from-1 to 0 of 1/(1+e^t)

解

\int_{- 1}^{0} \frac{1}{1 + e ^{t}} d t

- ln (2) + ln (1 + \frac{1}{e}) + 1

十進法表記

0.62011 \dots

解答ステップ

\int_{- 1}^{0} \frac{1}{1 + e ^{t}} d t

\frac{1}{1 + e ^{t}} = - \frac{e ^{t}}{1 + e ^{t}} + 1

= \int_{- 1}^{0} - \frac{e ^{t}}{1 + e ^{t}} + 1 d t

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{- 1}^{0} \frac{e ^{t}}{1 + e ^{t}} d t + \int_{- 1}^{0} 1 d t

\int_{- 1}^{0} \frac{e ^{t}}{1 + e ^{t}} d t = ln (2) - ln (1 + \frac{1}{e})

\int_{- 1}^{0} 1 d t = 1

= - (ln (2) - ln (1 + \frac{1}{e})) + 1

簡素化

= - ln (2) + ln (1 + \frac{1}{e}) + 1