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人気のある微分積分 >

integral from 0 to 1 of (r^3)/(4+r^2)

解

\int_{0}^{1} \frac{r ^{3}}{4 + r ^{2}} d r

解

\frac{1}{2} (1 - 4 (ln (5) - 2 ln (2)))

+1

十進法表記

0.05371 \dots

解答ステップ

\int_{0}^{1} \frac{r ^{3}}{4 + r ^{2}} d r

u置換積分法を適用する

= \int_{4}^{5} \frac{u - 4}{2 u} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{4}^{5} \frac{u - 4}{u} d u

拡張

\frac{u - 4}{u} : 1 - \frac{4}{u}

= \frac{1}{2} \cdot \int_{4}^{5} 1 - \frac{4}{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

= \frac{1}{2} (\int_{4}^{5} 1 d u - \int_{4}^{5} \frac{4}{u} d u)

\int_{4}^{5} 1 d u = 1

\int_{4}^{5} \frac{4}{u} d u = 4 (ln (5) - 2 ln (2))

= \frac{1}{2} (1 - 4 (ln (5) - 2 ln (2)))

グラフ

人気の例

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