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人気のある微分積分 >

integral from 0 to 1 of (x^2)/((1+x))

解

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

解

ln (2) - \frac{1}{2}

+1

十進法表記

0.19314 \dots

解答ステップ

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

u置換積分法を適用する

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

拡張

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

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

\int_{1}^{2} u d u = \frac{3}{2}

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

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

= \frac{3}{2} - 2 + ln (2)

分数を組み合わせる

- 2 + \frac{3}{2} : - \frac{1}{2}

= ln (2) - \frac{1}{2}

グラフ

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