integral from 0 to 1 of e^x(e^x-1)

Lösung

\int_{0}^{1} e^{x} (e^{x} - 1) d x

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

Dezimale

1.47624 \dots

Schritte zur Lösung

\int_{0}^{1} e^{x} (e^{x} - 1) d x

Wende U-Substitution an

= \int_{1}^{e} u - 1 d u

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}^{e} u d u - \int_{1}^{e} 1 d u

\int_{1}^{e} u d u = \frac{e ^{2} - 1}{2}

\int_{1}^{e} 1 d u = e - 1

= \frac{e ^{2} - 1}{2} - (e - 1)

Vereinfache

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

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