`f(x)=1/3x^3-2x^2+x-2019`
`*`
$\int f(x) dx = \int(\frac{1}{3}x^3-2x^2+x-2019) dx$
$= \int \frac{1}{3}x^3dx - \int 2x^2dx+ \int xdx - \int 2019$
$=\frac{1}{3} .\frac{x^4}{4}- 2. \frac{x^3}{3}+\frac{x^2}{2}-2019x +C$
$=\frac{x^4}{12}-\frac{2x^3}{3}+\frac{x^2}{2}-2019x+C$