Integral von $$$\frac{1}{\sqrt{y}}$$$

Bestimme $$$\int \frac{1}{\sqrt{y}}\, dy$$$.

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

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

Daher,

$$\int{\frac{1}{\sqrt{y}} d y} = 2 \sqrt{y}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sqrt{y}} d y} = 2 \sqrt{y}+C$$

$$$\int \frac{1}{\sqrt{y}}\, dy = 2 \sqrt{y} + C$$$A