integral from 0 to pi of (9e^x+6sin(x))

解

\int_{0}^{π} (9 e^{x} + 6 sin (x)) d x

9 e^{π} + 3

十進法表記

211.26623 \dots

解答ステップ

\int_{0}^{π} 9 e^{x} + 6 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} 9 e^{x} d x + \int_{0}^{π} 6 sin (x) d x

\int_{0}^{π} 9 e^{x} d x = 9 (e^{π} - 1)

\int_{0}^{π} 6 sin (x) d x = 12

= 9 (e^{π} - 1) + 12

簡素化

9 (e^{π} - 1) + 12 : 9 e^{π} + 3

= 9 e^{π} + 3