integral from 0 to 1 of (2-2x^3)^3

Lösung

\int_{0}^{1} (2 - 2 x^{3})^{3} d x

\frac{162}{35}

Dezimale

4.62857 \dots

Schritte zur Lösung

\int_{0}^{1} (2 - 2 x^{3})^{3} d x

Multipliziere aus

(2 - 2 x^{3})^{3} : 8 - 24 x^{3} + 24 x^{6} - 8 x^{9}

= \int_{0}^{1} 8 - 24 x^{3} + 24 x^{6} - 8 x^{9} 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 d x - \int_{0}^{1} 24 x^{3} d x + \int_{0}^{1} 24 x^{6} d x - \int_{0}^{1} 8 x^{9} d x

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

\int_{0}^{1} 24 x^{3} d x = 6

\int_{0}^{1} 24 x^{6} d x = \frac{24}{7}

\int_{0}^{1} 8 x^{9} d x = \frac{4}{5}

= 8 - 6 + \frac{24}{7} - \frac{4}{5}

Vereinfache

8 - 6 + \frac{24}{7} - \frac{4}{5} : \frac{162}{35}

= \frac{162}{35}

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