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

Lösung

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

\frac{128}{15}

Dezimale

8.53333 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int_{- 1}^{1} (4 - 4 x^{2})^{2} d x

Multipliziere aus

(4 - 4 x^{2})^{2} : 16 - 32 x^{2} + 16 x^{4}

= \frac{1}{2} \cdot \int_{- 1}^{1} 16 - 32 x^{2} + 16 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

= \frac{1}{2} (\int_{- 1}^{1} 16 d x - \int_{- 1}^{1} 32 x^{2} d x + \int_{- 1}^{1} 16 x^{4} d x)

\int_{- 1}^{1} 16 d x = 32

\int_{- 1}^{1} 32 x^{2} d x = \frac{64}{3}

\int_{- 1}^{1} 16 x^{4} d x = \frac{32}{5}

= \frac{1}{2} (32 - \frac{64}{3} + \frac{32}{5})

Vereinfache

\frac{1}{2} (32 - \frac{64}{3} + \frac{32}{5}) : \frac{128}{15}

= \frac{128}{15}

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