integral from 3 to-3 of (6r+5)^2

解

\int_{3}^{- 3} (6 r + 5)^{2} d r

- 798

解答ステップ

\int_{3}^{- 3} (6 r + 5)^{2} d r

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

= - \int_{- 3}^{3} (6 r + 5)^{2} d r

拡張

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

= - \int_{- 3}^{3} 36 r^{2} + 60 r + 25 d r

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int_{- 3}^{3} 36 r^{2} d r + \int_{- 3}^{3} 60 r d r + \int_{- 3}^{3} 25 d r)

\int_{- 3}^{3} 36 r^{2} d r = 648

\int_{- 3}^{3} 60 r d r = 0

\int_{- 3}^{3} 25 d r = 150

= - (648 + 0 + 150)

簡素化

= - 798