integral from 1 to-1 of (6r+5)^2

Lösung

\int_{1}^{- 1} (6 r + 5)^{2} d r

- 74

Schritte zur Lösung

\int_{1}^{- 1} (6 r + 5)^{2} d r

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{1} (6 r + 5)^{2} d r

Multipliziere aus

(6 r + 5)^{2} : 36 r^{2} + 60 r + 25

= - \int_{- 1}^{1} 36 r^{2} + 60 r + 25 d r

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}^{1} 36 r^{2} d r + \int_{- 1}^{1} 60 r d r + \int_{- 1}^{1} 25 d r)

\int_{- 1}^{1} 36 r^{2} d r = 24

\int_{- 1}^{1} 60 r d r = 0

\int_{- 1}^{1} 25 d r = 50

= - (24 + 0 + 50)

Vereinfache

= - 74

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