Integral von $$$n \left(n - 1\right)$$$

Bestimme $$$\int n \left(n - 1\right)\, dn$$$.

Expand the expression:

$${\color{red}{\int{n \left(n - 1\right) d n}}} = {\color{red}{\int{\left(n^{2} - n\right)d n}}}$$

Gliedweise integrieren:

$${\color{red}{\int{\left(n^{2} - n\right)d n}}} = {\color{red}{\left(- \int{n d n} + \int{n^{2} d n}\right)}}$$

Wenden Sie die Potenzregel $$$\int n^{n}\, dn = \frac{n^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=2$$$ an:

$$- \int{n d n} + {\color{red}{\int{n^{2} d n}}}=- \int{n d n} + {\color{red}{\frac{n^{1 + 2}}{1 + 2}}}=- \int{n d n} + {\color{red}{\left(\frac{n^{3}}{3}\right)}}$$

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

$$\frac{n^{3}}{3} - {\color{red}{\int{n d n}}}=\frac{n^{3}}{3} - {\color{red}{\frac{n^{1 + 1}}{1 + 1}}}=\frac{n^{3}}{3} - {\color{red}{\left(\frac{n^{2}}{2}\right)}}$$

Daher,

$$\int{n \left(n - 1\right) d n} = \frac{n^{3}}{3} - \frac{n^{2}}{2}$$

Vereinfachen:

$$\int{n \left(n - 1\right) d n} = \frac{n^{2} \left(2 n - 3\right)}{6}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{n \left(n - 1\right) d n} = \frac{n^{2} \left(2 n - 3\right)}{6}+C$$

$$$\int n \left(n - 1\right)\, dn = \frac{n^{2} \left(2 n - 3\right)}{6} + C$$$A