Integral of $$$- 4 t$$$

Find $$$\int \left(- 4 t\right)\, dt$$$.

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

$${\color{red}{\int{\left(- 4 t\right)d t}}} = {\color{red}{\left(- 4 \int{t d t}\right)}}$$

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

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

Therefore,

$$\int{\left(- 4 t\right)d t} = - 2 t^{2}$$

Add the constant of integration:

$$\int{\left(- 4 t\right)d t} = - 2 t^{2}+C$$

$$$\int \left(- 4 t\right)\, dt = - 2 t^{2} + C$$$A