integral from 0 to 2 of (4x-1)(2x^2-x)^4

Lösung

\int_{0}^{2} (4 x - 1) (2 x^{2} - x)^{4} d x

\frac{7776}{5}

Dezimale

1555.2

Schritte zur Lösung

\int_{0}^{2} (4 x - 1) (2 x^{2} - x)^{4} d x

Multipliziere aus

(4 x - 1) (2 x^{2} - x)^{4} : 64 x^{9} - 144 x^{8} + 128 x^{7} - 56 x^{6} + 12 x^{5} - x^{4}

= \int_{0}^{2} 64 x^{9} - 144 x^{8} + 128 x^{7} - 56 x^{6} + 12 x^{5} - x^{4} 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}^{2} 64 x^{9} d x - \int_{0}^{2} 144 x^{8} d x + \int_{0}^{2} 128 x^{7} d x - \int_{0}^{2} 56 x^{6} d x + \int_{0}^{2} 12 x^{5} d x - \int_{0}^{2} x^{4} d x

\int_{0}^{2} 64 x^{9} d x = \frac{32768}{5}

\int_{0}^{2} 144 x^{8} d x = 8192

\int_{0}^{2} 128 x^{7} d x = 4096

\int_{0}^{2} 56 x^{6} d x = 1024

\int_{0}^{2} 12 x^{5} d x = 128

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= \frac{32768}{5} - 8192 + 4096 - 1024 + 128 - \frac{32}{5}

Vereinfache

\frac{32768}{5} - 8192 + 4096 - 1024 + 128 - \frac{32}{5} : \frac{7776}{5}

= \frac{7776}{5}

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