integral from 20 to 40 of 3(3x+4)^2

Lösung

\int_{20}^{40} 3 (3 x + 4)^{2} d x

548160

Schritte zur Lösung

\int_{20}^{40} 3 (3 x + 4)^{2} d x

Entferne die Konstante:

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

= 3 \cdot \int_{20}^{40} (3 x + 4)^{2} d x

Multipliziere aus

(3 x + 4)^{2} : 9 x^{2} + 24 x + 16

= 3 \cdot \int_{20}^{40} 9 x^{2} + 24 x + 16 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

= 3 (\int_{20}^{40} 9 x^{2} d x + \int_{20}^{40} 24 x d x + \int_{20}^{40} 16 d x)

\int_{20}^{40} 9 x^{2} d x = 168000

\int_{20}^{40} 24 x d x = 14400

\int_{20}^{40} 16 d x = 320

= 3 (168000 + 14400 + 320)

Vereinfache

= 548160

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