Integral von $$$b^{5} \sigma \sigma_{1}^{2}$$$ nach $$$b$$$

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

Wende die Konstantenfaktorregel $$$\int c f{\left(b \right)}\, db = c \int f{\left(b \right)}\, db$$$ mit $$$c=\sigma \sigma_{1}^{2}$$$ und $$$f{\left(b \right)} = b^{5}$$$ an:

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

Wenden Sie die Potenzregel $$$\int b^{n}\, db = \frac{b^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=5$$$ an:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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