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

Lösung

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

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

Dezimale

0.62011 \dots

Schritte zur Lösung

\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

Wende die Summenregel an:

\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

Vereinfache

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

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