integral from 0 to 1 of (8x^e+3e^x)

解

\int_{0}^{1} (8 x^{e} + 3 e^{x}) d x

\frac{8}{e + 1} + 3 (e - 1)

十進法表記

7.30637 \dots

解答ステップ

\int_{0}^{1} 8 x^{e} + 3 e^{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}^{1} 8 x^{e} d x + \int_{0}^{1} 3 e^{x} d x

\int_{0}^{1} 8 x^{e} d x = \frac{8}{e + 1}

\int_{0}^{1} 3 e^{x} d x = 3 (e - 1)

= \frac{8}{e + 1} + 3 (e - 1)