Integral von $$$x \left(3 - x\right)$$$

Bestimme $$$\int x \left(3 - x\right)\, dx$$$.

Expand the expression:

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

Gliedweise integrieren:

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

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

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

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

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

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

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

Daher,

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

Vereinfachen:

$$\int{x \left(3 - x\right) d x} = \frac{x^{2} \left(9 - 2 x\right)}{6}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{x \left(3 - x\right) d x} = \frac{x^{2} \left(9 - 2 x\right)}{6}+C$$

$$$\int x \left(3 - x\right)\, dx = \frac{x^{2} \left(9 - 2 x\right)}{6} + C$$$A