integral from 0 to 4 of (2v+5)(3v-1)

解

\int_{0}^{4} (2 v + 5) (3 v - 1) d v

212

解答ステップ

\int_{0}^{4} (2 v + 5) (3 v - 1) d v

拡張

(2 v + 5) (3 v - 1) : 6 v^{2} + 13 v - 5

= \int_{0}^{4} 6 v^{2} + 13 v - 5 d v

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_{0}^{4} 6 v^{2} d v + \int_{0}^{4} 13 v d v - \int_{0}^{4} 5 d v

\int_{0}^{4} 6 v^{2} d v = 128

\int_{0}^{4} 13 v d v = 104

\int_{0}^{4} 5 d v = 20

= 128 + 104 - 20

簡素化

= 212