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

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

Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(y \right)} = y^{3}$$$:

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

Apply the power rule $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=3$$$:

$$\frac{{\color{red}{\int{y^{3} d y}}}}{2}=\frac{{\color{red}{\frac{y^{1 + 3}}{1 + 3}}}}{2}=\frac{{\color{red}{\left(\frac{y^{4}}{4}\right)}}}{2}$$

Therefore,

$$\int{\frac{y^{3}}{2} d y} = \frac{y^{4}}{8}$$

Add the constant of integration:

$$\int{\frac{y^{3}}{2} d y} = \frac{y^{4}}{8}+C$$

$$$\int \frac{y^{3}}{2}\, dy = \frac{y^{4}}{8} + C$$$A