integral from 0 to 1 of 8xe^{4x}+2e^{4x}

Lösung

\int_{0}^{1} 8 x e^{4 x} + 2 e^{4 x} d x

2 e^{4}

Dezimale

109.19630 \dots

Schritte zur Lösung

\int_{0}^{1} 8 x e^{4 x} + 2 e^{4 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} 8 x e^{4 x} d x + \int_{0}^{1} 2 e^{4 x} d x

\int_{0}^{1} 8 x e^{4 x} d x = \frac{3 e ^{4} + 1}{2}

\int_{0}^{1} 2 e^{4 x} d x = \frac{e ^{4} - 1}{2}

= \frac{3 e ^{4} + 1}{2} + \frac{e ^{4} - 1}{2}

Vereinfache

\frac{3 e ^{4} + 1}{2} + \frac{e ^{4} - 1}{2} : 2 e^{4}

= 2 e^{4}

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