integral from 0 to 1/2 of 6/7 (2x^2+x)

Lösung

\int_{0}^{\frac{1}{2}} \frac{6}{7} (2 x^{2} + x) d x

\frac{5}{28}

Dezimale

0.17857 \dots

Schritte zur Lösung

\int_{0}^{\frac{1}{2}} \frac{6}{7} (2 x^{2} + x) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{6}{7} \cdot \int_{0}^{\frac{1}{2}} 2 x^{2} + 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

= \frac{6}{7} (\int_{0}^{\frac{1}{2}} 2 x^{2} d x + \int_{0}^{\frac{1}{2}} x d x)

\int_{0}^{\frac{1}{2}} 2 x^{2} d x = \frac{1}{12}

\int_{0}^{\frac{1}{2}} x d x = \frac{1}{8}

= \frac{6}{7} (\frac{1}{12} + \frac{1}{8})

Vereinfache

\frac{6}{7} (\frac{1}{12} + \frac{1}{8}) : \frac{5}{28}

= \frac{5}{28}

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