Integral von $$$\frac{y^{2}}{4} - 1$$$

Bestimme $$$\int \left(\frac{y^{2}}{4} - 1\right)\, dy$$$.

Gliedweise integrieren:

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

Wenden Sie die Konstantenregel $$$\int c\, dy = c y$$$ mit $$$c=1$$$ an:

$$\int{\frac{y^{2}}{4} d y} - {\color{red}{\int{1 d y}}} = \int{\frac{y^{2}}{4} d y} - {\color{red}{y}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(y \right)} = y^{2}$$$ an:

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

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

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

Daher,

$$\int{\left(\frac{y^{2}}{4} - 1\right)d y} = \frac{y^{3}}{12} - y$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(\frac{y^{2}}{4} - 1\right)d y} = \frac{y^{3}}{12} - y+C$$

$$$\int \left(\frac{y^{2}}{4} - 1\right)\, dy = \left(\frac{y^{3}}{12} - y\right) + C$$$A