integral from 0 to 1 of (3x^e+7e^x)

解

\int_{0}^{1} (3 x^{e} + 7 e^{x}) d x

\frac{3}{e + 1} + 7 (e - 1)

十進法表記

12.83479 \dots

解答ステップ

\int_{0}^{1} 3 x^{e} + 7 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} 3 x^{e} d x + \int_{0}^{1} 7 e^{x} d x

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

\int_{0}^{1} 7 e^{x} d x = 7 (e - 1)

= \frac{3}{e + 1} + 7 (e - 1)