Integral of $$$\frac{a c \rho v^{2}}{2}$$$ with respect to $$$v$$$

Find $$$\int \frac{a c \rho v^{2}}{2}\, dv$$$.

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{a c \rho}{2}$$$ and $$$f{\left(v \right)} = v^{2}$$$:

$${\color{red}{\int{\frac{a c \rho v^{2}}{2} d v}}} = {\color{red}{\left(\frac{a c \rho \int{v^{2} d v}}{2}\right)}}$$

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

$$\frac{a c \rho {\color{red}{\int{v^{2} d v}}}}{2}=\frac{a c \rho {\color{red}{\frac{v^{1 + 2}}{1 + 2}}}}{2}=\frac{a c \rho {\color{red}{\left(\frac{v^{3}}{3}\right)}}}{2}$$

Therefore,

$$\int{\frac{a c \rho v^{2}}{2} d v} = \frac{a c \rho v^{3}}{6}$$

Add the constant of integration:

$$\int{\frac{a c \rho v^{2}}{2} d v} = \frac{a c \rho v^{3}}{6}+C$$

$$$\int \frac{a c \rho v^{2}}{2}\, dv = \frac{a c \rho v^{3}}{6} + C$$$A