integral from 0 to r of (br+c)2pir

解

\int_{0}^{r} (b r + c) 2 π r d r

\frac{2 πb r ^{3}}{3} + π c r^{2}

解答ステップ

\int_{0}^{r} (b r + c) \cdot 2 π r d r

定数を除く:

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

= 2 π \cdot \int_{0}^{r} (b r + c) r d r

部分積分を適用する

= 2 π [\frac{1}{2} r^{2} (b r + c) - \int \frac{b r ^{2}}{2} d r]_{0}^{r}

\int \frac{b r ^{2}}{2} d r = \frac{b r ^{3}}{6}

= 2 π [\frac{1}{2} r^{2} (b r + c) - \frac{b r ^{3}}{6}]_{0}^{r}

簡素化

2 π [\frac{1}{2} r^{2} (b r + c) - \frac{b r ^{3}}{6}]_{0}^{r} : 2 π [\frac{1}{6} (2 b r^{3} + 3 c r^{2})]_{0}^{r}

= 2 π [\frac{1}{6} (2 b r^{3} + 3 c r^{2})]_{0}^{r}

限度を計算する:

\frac{1}{6} (2 b r^{3} + 3 c r^{2})

= 2 π \frac{1}{6} (2 b r^{3} + 3 c r^{2})

簡素化

= \frac{2 πb r ^{3}}{3} + π c r^{2}