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人気のある微分積分 >

integral from 0 to 1 of (-1+2/(1+t))

解

\int_{0}^{1} (- 1 + \frac{2}{1 + t}) d t

解

- 1 + 2 ln (2)

+1

十進法表記

0.38629 \dots

解答ステップ

\int_{0}^{1} - 1 + \frac{2}{1 + t} d t

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} 1 d t + \int_{0}^{1} \frac{2}{1 + t} d t

\int_{0}^{1} 1 d t = 1

\int_{0}^{1} \frac{2}{1 + t} d t = 2 ln (2)

= - 1 + 2 ln (2)

グラフ

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