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

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

Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=\frac{2}{3}$$$ and $$$f{\left(y \right)} = y^{2}$$$:

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

Apply the power rule $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

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

Therefore,

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

Add the constant of integration:

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

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