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

Find $$$\int \frac{t^{2}}{4}\, dt$$$.

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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