Integral de $$$b^{5} \sigma \sigma_{1}^{2}$$$ con respecto a $$$b$$$

Halla $$$\int b^{5} \sigma \sigma_{1}^{2}\, db$$$.

Aplica la regla del factor constante $$$\int c f{\left(b \right)}\, db = c \int f{\left(b \right)}\, db$$$ con $$$c=\sigma \sigma_{1}^{2}$$$ y $$$f{\left(b \right)} = b^{5}$$$:

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

Aplica la regla de la potencia $$$\int b^{n}\, db = \frac{b^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=5$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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