Integral von $$$x^{3} - 3 x^{2}$$$

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

Gliedweise integrieren:

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

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

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

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

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

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

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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