Integral von $$$x^{3} \left(- x - 2\right)$$$

Bestimme $$$\int x^{3} \left(- x - 2\right)\, dx$$$.

Den Integranden vereinfachen:

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=-1$$$ und $$$f{\left(x \right)} = x^{3} \left(x + 2\right)$$$ an:

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

Expand the expression:

$$- {\color{red}{\int{x^{3} \left(x + 2\right) d x}}} = - {\color{red}{\int{\left(x^{4} + 2 x^{3}\right)d x}}}$$

Gliedweise integrieren:

$$- {\color{red}{\int{\left(x^{4} + 2 x^{3}\right)d x}}} = - {\color{red}{\left(\int{2 x^{3} d x} + \int{x^{4} d x}\right)}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=4$$$ an:

$$- \int{2 x^{3} d x} - {\color{red}{\int{x^{4} d x}}}=- \int{2 x^{3} d x} - {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=- \int{2 x^{3} d x} - {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=2$$$ und $$$f{\left(x \right)} = x^{3}$$$ an:

$$- \frac{x^{5}}{5} - {\color{red}{\int{2 x^{3} d x}}} = - \frac{x^{5}}{5} - {\color{red}{\left(2 \int{x^{3} d x}\right)}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=3$$$ an:

$$- \frac{x^{5}}{5} - 2 {\color{red}{\int{x^{3} d x}}}=- \frac{x^{5}}{5} - 2 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- \frac{x^{5}}{5} - 2 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

Daher,

$$\int{x^{3} \left(- x - 2\right) d x} = - \frac{x^{5}}{5} - \frac{x^{4}}{2}$$

Vereinfachen:

$$\int{x^{3} \left(- x - 2\right) d x} = \frac{x^{4} \left(- 2 x - 5\right)}{10}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{x^{3} \left(- x - 2\right) d x} = \frac{x^{4} \left(- 2 x - 5\right)}{10}+C$$

$$$\int x^{3} \left(- x - 2\right)\, dx = \frac{x^{4} \left(- 2 x - 5\right)}{10} + C$$$A