integral from-1 to 1 of pi(2-2x^2)^2

Lösung

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

π \frac{64}{15}

Dezimale

13.40412 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

= π \cdot \int_{- 1}^{1} (2 - 2 x^{2})^{2} d x

Multipliziere aus

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

= π \cdot \int_{- 1}^{1} 4 - 8 x^{2} + 4 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

= π (\int_{- 1}^{1} 4 d x - \int_{- 1}^{1} 8 x^{2} d x + \int_{- 1}^{1} 4 x^{4} d x)

\int_{- 1}^{1} 4 d x = 8

\int_{- 1}^{1} 8 x^{2} d x = \frac{16}{3}

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

= π (8 - \frac{16}{3} + \frac{8}{5})

Füge

8 - \frac{16}{3} + \frac{8}{5}

zusammen:

\frac{64}{15}

= π \frac{64}{15}

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