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人気のある微分積分 >

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

解

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

解

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

+1

十進法表記

0.75796 \dots

解答ステップ

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

u置換積分法を適用する

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

拡張

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

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

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

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

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

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

簡素化

= 3 + \frac{e ^{2} - 1}{2} - 2 e

グラフ

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