$$$b^{5} \sigma \sigma_{1}^{2}$$$ 對 $$$b$$$ 的積分

求$$$\int b^{5} \sigma \sigma_{1}^{2}\, db$$$。

套用常數倍法則 $$$\int c f{\left(b \right)}\, db = c \int f{\left(b \right)}\, db$$$，使用 $$$c=\sigma \sigma_{1}^{2}$$$ 與 $$$f{\left(b \right)} = b^{5}$$$：

$${\color{red}{\int{b^{5} \sigma \sigma_{1}^{2} d b}}} = {\color{red}{\sigma \sigma_{1}^{2} \int{b^{5} d b}}}$$

套用冪次法則 $$$\int b^{n}\, db = \frac{b^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=5$$$：

$$\sigma \sigma_{1}^{2} {\color{red}{\int{b^{5} d b}}}=\sigma \sigma_{1}^{2} {\color{red}{\frac{b^{1 + 5}}{1 + 5}}}=\sigma \sigma_{1}^{2} {\color{red}{\left(\frac{b^{6}}{6}\right)}}$$

因此，

$$\int{b^{5} \sigma \sigma_{1}^{2} d b} = \frac{b^{6} \sigma \sigma_{1}^{2}}{6}$$

加上積分常數：

$$\int{b^{5} \sigma \sigma_{1}^{2} d b} = \frac{b^{6} \sigma \sigma_{1}^{2}}{6}+C$$

$$$\int b^{5} \sigma \sigma_{1}^{2}\, db = \frac{b^{6} \sigma \sigma_{1}^{2}}{6} + C$$$A