integral from 0 to 1 of e^{4x}-e^{2x}

解

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

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

十進法表記

10.20500 \dots

解答ステップ

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

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

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

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

簡素化

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

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