integral from 1 to 2 of 2r^4ln(r)

解

\int_{1}^{2} 2 r^{4} ln (r) d r

2 (- \frac{31}{25} + \frac{32}{5} ln (2))

十進法表記

6.39228 \dots

解答ステップ

\int_{1}^{2} 2 r^{4} ln (r) d r

定数を除く:

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

= 2 \cdot \int_{1}^{2} r^{4} ln (r) d r

部分積分を適用する

= 2 [\frac{1}{5} r^{5} ln (r) - \int \frac{r ^{4}}{5} d r]_{1}^{2}

\int \frac{r ^{4}}{5} d r = \frac{r ^{5}}{25}

= 2 [\frac{1}{5} r^{5} ln (r) - \frac{r ^{5}}{25}]_{1}^{2}

限度を計算する:

- \frac{31}{25} + \frac{32}{5} ln (2)

= 2 (- \frac{31}{25} + \frac{32}{5} ln (2))