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

解

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

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

十進法表記

1.08616 \dots

解答ステップ

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

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

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

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

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

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

簡素化

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