integral from 0 to 4 of 1/(e^x+1)

解

\int_{0}^{4} \frac{1}{e ^{x} + 1} d x

- ln (e^{4} + 1) + ln (2) + 4

十進法表記

0.67499 \dots

解答ステップ

\int_{0}^{4} \frac{1}{e ^{x} + 1} d x

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

= \int_{0}^{4} - \frac{e ^{x}}{e ^{x} + 1} + 1 d x

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_{0}^{4} \frac{e ^{x}}{e ^{x} + 1} d x + \int_{0}^{4} 1 d x

\int_{0}^{4} \frac{e ^{x}}{e ^{x} + 1} d x = ln (e^{4} + 1) - ln (2)

\int_{0}^{4} 1 d x = 4

= - (ln (e^{4} + 1) - ln (2)) + 4

簡素化

= - ln (e^{4} + 1) + ln (2) + 4