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

Lösung

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

\frac{1962}{5}

Dezimale

392.4

Schritte zur Lösung

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

Multipliziere aus

(- 2 x^{2} + 2 x)^{2} : 4 x^{4} - 8 x^{3} + 4 x^{2}

= \int_{1}^{4} 4 x^{4} - 8 x^{3} + 4 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_{1}^{4} 4 x^{4} d x - \int_{1}^{4} 8 x^{3} d x + \int_{1}^{4} 4 x^{2} d x

\int_{1}^{4} 4 x^{4} d x = \frac{4092}{5}

\int_{1}^{4} 8 x^{3} d x = 510

\int_{1}^{4} 4 x^{2} d x = 84

= \frac{4092}{5} - 510 + 84

Vereinfache

\frac{4092}{5} - 510 + 84 : \frac{1962}{5}

= \frac{1962}{5}

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