integral from 0 to 20 of-0.4x^2+1000-840

解

\int_{0}^{20} - 0.4 x^{2} + 1000 - 840 d x

2133.33333 \dots

解答ステップ

\int_{0}^{20} - 0.4 x^{2} + 1000 - 840 d x

簡素化

= \int_{0}^{20} - 0.4 x^{2} + 160 d x

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}^{20} 0.4 x^{2} d x + \int_{0}^{20} 160 d x

\int_{0}^{20} 0.4 x^{2} d x = 1066.66666 \dots

\int_{0}^{20} 160 d x = 3200

= - 1066.66666 \dots + 3200

簡素化

= 2133.33333 \dots