Integral de $$$\frac{2 g r^{2} \sigma v}{9}$$$ em relação a $$$v$$$

Encontre $$$\int \frac{2 g r^{2} \sigma v}{9}\, dv$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ usando $$$c=\frac{2 g r^{2} \sigma}{9}$$$ e $$$f{\left(v \right)} = v$$$:

$${\color{red}{\int{\frac{2 g r^{2} \sigma v}{9} d v}}} = {\color{red}{\left(\frac{2 g r^{2} \sigma \int{v d v}}{9}\right)}}$$

Aplique a regra da potência $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=1$$$:

$$\frac{2 g r^{2} \sigma {\color{red}{\int{v d v}}}}{9}=\frac{2 g r^{2} \sigma {\color{red}{\frac{v^{1 + 1}}{1 + 1}}}}{9}=\frac{2 g r^{2} \sigma {\color{red}{\left(\frac{v^{2}}{2}\right)}}}{9}$$

Portanto,

$$\int{\frac{2 g r^{2} \sigma v}{9} d v} = \frac{g r^{2} \sigma v^{2}}{9}$$

Adicione a constante de integração:

$$\int{\frac{2 g r^{2} \sigma v}{9} d v} = \frac{g r^{2} \sigma v^{2}}{9}+C$$

$$$\int \frac{2 g r^{2} \sigma v}{9}\, dv = \frac{g r^{2} \sigma v^{2}}{9} + C$$$A