integral from 0 to 1 of (3x^e+4e^x)

Lösung

\int_{0}^{1} (3 x^{e} + 4 e^{x}) d x

\frac{3}{e + 1} + 4 (e - 1)

Dezimale

7.67995 \dots

Schritte zur Lösung

\int_{0}^{1} 3 x^{e} + 4 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} 3 x^{e} d x + \int_{0}^{1} 4 e^{x} d x

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

\int_{0}^{1} 4 e^{x} d x = 4 (e - 1)

= \frac{3}{e + 1} + 4 (e - 1)

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