Integral of $$$- \rho t + 1$$$ with respect to $$$t$$$

Find $$$\int \left(- \rho t + 1\right)\, dt$$$.

Integrate term by term:

$${\color{red}{\int{\left(- \rho t + 1\right)d t}}} = {\color{red}{\left(\int{1 d t} - \int{\rho t d t}\right)}}$$

Apply the constant rule $$$\int c\, dt = c t$$$ with $$$c=1$$$:

$$- \int{\rho t d t} + {\color{red}{\int{1 d t}}} = - \int{\rho t d t} + {\color{red}{t}}$$

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

$$t - {\color{red}{\int{\rho t d t}}} = t - {\color{red}{\rho \int{t d t}}}$$

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

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

Therefore,

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

Simplify:

$$\int{\left(- \rho t + 1\right)d t} = \frac{t \left(- \rho t + 2\right)}{2}$$

Add the constant of integration:

$$\int{\left(- \rho t + 1\right)d t} = \frac{t \left(- \rho t + 2\right)}{2}+C$$

$$$\int \left(- \rho t + 1\right)\, dt = \frac{t \left(- \rho t + 2\right)}{2} + C$$$A