integral from 0 to 10 of (2200e^{0.024t}-1460e^{0.018t})

解

\int_{0}^{10} (2200 e^{0.024 t} - 1460 e^{0.018 t}) d t

8867.98599 \dots

解答ステップ

\int_{0}^{10} 2200 e^{0.024 t} - 1460 e^{0.018 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

= \int_{0}^{10} 2200 e^{0.024 t} d t - \int_{0}^{10} 1460 e^{0.018 t} d t

\int_{0}^{10} 2200 e^{0.024 t} d t = 24864.50544 \dots

\int_{0}^{10} 1460 e^{0.018 t} d t = 15996.51945 \dots

= 24864.50544 \dots - 15996.51945 \dots

簡素化

= 8867.98599 \dots