Integralen av $$$\frac{1}{\sqrt{y}}$$$

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

Tillämpa potensregeln $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=- \frac{1}{2}$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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