integral from 0 to 1 of (2+e^x)^2

解

\int_{0}^{1} (2 + e^{x})^{2} d x

4 e + \frac{e ^{2} - 1}{2}

十進法表記

14.06765 \dots

解答ステップ

\int_{0}^{1} (2 + e^{x})^{2} d x

拡張

(2 + e^{x})^{2} : 4 + 4 e^{x} + e^{2 x}

= \int_{0}^{1} 4 + 4 e^{x} + e^{2 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} 4 d x + \int_{0}^{1} 4 e^{x} d x + \int_{0}^{1} e^{2 x} d x

\int_{0}^{1} 4 d x = 4

\int_{0}^{1} 4 e^{x} d x = 4 (e - 1)

\int_{0}^{1} e^{2 x} d x = \frac{e ^{2} - 1}{2}

= 4 + 4 (e - 1) + \frac{e ^{2} - 1}{2}

簡素化

4 + 4 (e - 1) + \frac{e ^{2} - 1}{2} : 4 e + \frac{e ^{2} - 1}{2}

= 4 e + \frac{e ^{2} - 1}{2}