integral from 0 to 1 of e^{2x}-e^x

解

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

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

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

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

簡素化

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