Integraal van $$$b^{5} \sigma \sigma_{1}^{2}$$$ met betrekking tot $$$b$$$

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

Pas de constante-veelvoudregel $$$\int c f{\left(b \right)}\, db = c \int f{\left(b \right)}\, db$$$ toe met $$$c=\sigma \sigma_{1}^{2}$$$ en $$$f{\left(b \right)} = b^{5}$$$:

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

Pas de machtsregel $$$\int b^{n}\, db = \frac{b^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=5$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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