integral from 0 to 1 of 2^xe^x

Lösung

\int_{0}^{1} 2^{x} e^{x} d x

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

Dezimale

2.62030 \dots

Schritte zur Lösung

\int_{0}^{1} 2^{x} e^{x} d x

Wende Exponentenregel an:

a^{n} \cdot b^{n} = (a \cdot b)^{n}

2^{x} e^{x} = (2 e)^{x}

= \int_{0}^{1} (2 e)^{x} d x

\int a^{x} d x = \frac{a ^{x}}{ln a}

= [\frac{( 2 e ) ^{x}}{ln ( 2 e )}]_{0}^{1}

Berechne die Grenzen:

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

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

\int_{- 1}^{0} 2 x - 1 d x

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