integral from 0 to 1 of x^2(4x^3-1)^4

Lösung

\int_{0}^{1} x^{2} (4 x^{3} - 1)^{4} d x

\frac{61}{15}

Dezimale

4.06666 \dots

Schritte zur Lösung

\int_{0}^{1} x^{2} (4 x^{3} - 1)^{4} d x

Multipliziere aus

x^{2} (4 x^{3} - 1)^{4} : 256 x^{14} - 256 x^{11} + 96 x^{8} - 16 x^{5} + x^{2}

= \int_{0}^{1} 256 x^{14} - 256 x^{11} + 96 x^{8} - 16 x^{5} + x^{2} 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} 256 x^{14} d x - \int_{0}^{1} 256 x^{11} d x + \int_{0}^{1} 96 x^{8} d x - \int_{0}^{1} 16 x^{5} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} 256 x^{14} d x = \frac{256}{15}

\int_{0}^{1} 256 x^{11} d x = \frac{64}{3}

\int_{0}^{1} 96 x^{8} d x = \frac{32}{3}

\int_{0}^{1} 16 x^{5} d x = \frac{8}{3}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= \frac{256}{15} - \frac{64}{3} + \frac{32}{3} - \frac{8}{3} + \frac{1}{3}

Vereinfache

\frac{256}{15} - \frac{64}{3} + \frac{32}{3} - \frac{8}{3} + \frac{1}{3} : \frac{61}{15}

= \frac{61}{15}

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