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

Lösung

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

- \frac{1}{e} + 2

Dezimale

1.63212 \dots

Schritte zur Lösung

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

Multipliziere aus

\frac{e ^{x} + 1}{e ^{x}} : 1 + \frac{1}{e ^{x}}

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

Wende die Summenregel an:

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

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

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

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

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

Vereinfache

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

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