Integral de $$$\frac{2 y^{2}}{3}$$$

Halla $$$\int \frac{2 y^{2}}{3}\, dy$$$.

Aplica la regla del factor constante $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ con $$$c=\frac{2}{3}$$$ y $$$f{\left(y \right)} = y^{2}$$$:

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

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

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

Por lo tanto,

$$\int{\frac{2 y^{2}}{3} d y} = \frac{2 y^{3}}{9}$$

Añade la constante de integración:

$$\int{\frac{2 y^{2}}{3} d y} = \frac{2 y^{3}}{9}+C$$

$$$\int \frac{2 y^{2}}{3}\, dy = \frac{2 y^{3}}{9} + C$$$A