integral from 0 to pi of (2e^x+3sin(x))

解

\int_{0}^{π} (2 e^{x} + 3 sin (x)) d x

2 e^{π} + 4

十進法表記

50.28138 \dots

解答ステップ

\int_{0}^{π} 2 e^{x} + 3 sin (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}^{π} 2 e^{x} d x + \int_{0}^{π} 3 sin (x) d x

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

\int_{0}^{π} 3 sin (x) d x = 6

= 2 (e^{π} - 1) + 6

簡素化

2 (e^{π} - 1) + 6 : 2 e^{π} + 4

= 2 e^{π} + 4