integral from 0 to 0.2 of 5(1-e^{-2t})

解

\int_{0}^{0.2} 5 (1 - e^{- 2 t}) d t

0.17580 \dots

解答ステップ

\int_{0}^{0.2} 5 (1 - e^{- 2 t}) d t

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int_{0}^{0.2} 1 - e^{- 2 t} d t

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

= 5 (\int_{0}^{0.2} 1 d t - \int_{0}^{0.2} e^{- 2 t} d t)

\int_{0}^{0.2} 1 d t = 0.2

\int_{0}^{0.2} e^{- 2 t} d t = 0.16483 \dots

= 5 (0.2 - 0.16483 \dots)

簡素化

= 0.17580 \dots