integral from 0 to 2 of 5e^x+9cos(x)

解

\int_{0}^{2} 5 e^{x} + 9 cos (x) d x

5 (e^{2} - 1) + 9 sin (2)

十進法表記

40.12895 \dots

解答ステップ

\int_{0}^{2} 5 e^{x} + 9 cos (x) 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}^{2} 5 e^{x} d x + \int_{0}^{2} 9 cos (x) d x

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

\int_{0}^{2} 9 cos (x) d x = 9 sin (2)

= 5 (e^{2} - 1) + 9 sin (2)