integral from-1 to 2 of (2x-1)(4x+5)

Lösung

\int_{- 1}^{2} (2 x - 1) (4 x + 5) d x

18

Schritte zur Lösung

\int_{- 1}^{2} (2 x - 1) (4 x + 5) d x

Multipliziere aus

(2 x - 1) (4 x + 5) : 8 x^{2} + 6 x - 5

= \int_{- 1}^{2} 8 x^{2} + 6 x - 5 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}^{2} 8 x^{2} d x + \int_{- 1}^{2} 6 x d x - \int_{- 1}^{2} 5 d x

\int_{- 1}^{2} 8 x^{2} d x = 24

\int_{- 1}^{2} 6 x d x = 9

\int_{- 1}^{2} 5 d x = 15

= 24 + 9 - 15

Vereinfache

= 18

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