integral from 0 to 1 of e^x(e^x-1)

解

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

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

十進法表記

1.47624 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{e} u - 1 d u

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_{1}^{e} u d u - \int_{1}^{e} 1 d u

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

\int_{1}^{e} 1 d u = e - 1

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

簡素化

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