Integral of $$$\frac{n^{2}}{4}$$$

Find $$$\int \frac{n^{2}}{4}\, dn$$$.

Apply the constant multiple rule $$$\int c f{\left(n \right)}\, dn = c \int f{\left(n \right)}\, dn$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(n \right)} = n^{2}$$$:

$${\color{red}{\int{\frac{n^{2}}{4} d n}}} = {\color{red}{\left(\frac{\int{n^{2} d n}}{4}\right)}}$$

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

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

Therefore,

$$\int{\frac{n^{2}}{4} d n} = \frac{n^{3}}{12}$$

Add the constant of integration:

$$\int{\frac{n^{2}}{4} d n} = \frac{n^{3}}{12}+C$$

$$$\int \frac{n^{2}}{4}\, dn = \frac{n^{3}}{12} + C$$$A