Integral de $$$9 d t$$$ con respecto a $$$t$$$

Halla $$$\int 9 d t\, dt$$$.

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=9 d$$$ y $$$f{\left(t \right)} = t$$$:

$${\color{red}{\int{9 d t d t}}} = {\color{red}{\left(9 d \int{t d t}\right)}}$$

Aplica la regla de la potencia $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=1$$$:

$$9 d {\color{red}{\int{t d t}}}=9 d {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}=9 d {\color{red}{\left(\frac{t^{2}}{2}\right)}}$$

Por lo tanto,

$$\int{9 d t d t} = \frac{9 d t^{2}}{2}$$

Añade la constante de integración:

$$\int{9 d t d t} = \frac{9 d t^{2}}{2}+C$$

$$$\int 9 d t\, dt = \frac{9 d t^{2}}{2} + C$$$A