Integral of $$$- 2 t^{2}$$$

Find $$$\int \left(- 2 t^{2}\right)\, dt$$$.

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

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

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

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

Therefore,

$$\int{\left(- 2 t^{2}\right)d t} = - \frac{2 t^{3}}{3}$$

Add the constant of integration:

$$\int{\left(- 2 t^{2}\right)d t} = - \frac{2 t^{3}}{3}+C$$

$$$\int \left(- 2 t^{2}\right)\, dt = - \frac{2 t^{3}}{3} + C$$$A