Integral of $$$\frac{z}{3}$$$

Find $$$\int \frac{z}{3}\, dz$$$.

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

$${\color{red}{\int{\frac{z}{3} d z}}} = {\color{red}{\left(\frac{\int{z d z}}{3}\right)}}$$

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

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

Therefore,

$$\int{\frac{z}{3} d z} = \frac{z^{2}}{6}$$

Add the constant of integration:

$$\int{\frac{z}{3} d z} = \frac{z^{2}}{6}+C$$

$$$\int \frac{z}{3}\, dz = \frac{z^{2}}{6} + C$$$A