Integral von $$$\frac{\sqrt{x}}{8}$$$

Bestimme $$$\int \frac{\sqrt{x}}{8}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{8}$$$ und $$$f{\left(x \right)} = \sqrt{x}$$$ an:

$${\color{red}{\int{\frac{\sqrt{x}}{8} d x}}} = {\color{red}{\left(\frac{\int{\sqrt{x} d x}}{8}\right)}}$$

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

$$\frac{{\color{red}{\int{\sqrt{x} d x}}}}{8}=\frac{{\color{red}{\int{x^{\frac{1}{2}} d x}}}}{8}=\frac{{\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{8}=\frac{{\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}}{8}$$

Daher,

$$\int{\frac{\sqrt{x}}{8} d x} = \frac{x^{\frac{3}{2}}}{12}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\sqrt{x}}{8} d x} = \frac{x^{\frac{3}{2}}}{12}+C$$

$$$\int \frac{\sqrt{x}}{8}\, dx = \frac{x^{\frac{3}{2}}}{12} + C$$$A