Integral de $$$2 \sqrt{y}$$$

Halla $$$\int 2 \sqrt{y}\, dy$$$.

Aplica la regla del factor constante $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ con $$$c=2$$$ y $$$f{\left(y \right)} = \sqrt{y}$$$:

$${\color{red}{\int{2 \sqrt{y} d y}}} = {\color{red}{\left(2 \int{\sqrt{y} d y}\right)}}$$

Aplica la regla de la potencia $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=\frac{1}{2}$$$:

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

Por lo tanto,

$$\int{2 \sqrt{y} d y} = \frac{4 y^{\frac{3}{2}}}{3}$$

Añade la constante de integración:

$$\int{2 \sqrt{y} d y} = \frac{4 y^{\frac{3}{2}}}{3}+C$$

$$$\int 2 \sqrt{y}\, dy = \frac{4 y^{\frac{3}{2}}}{3} + C$$$A