Originally Posted by

**angel.white** velocity is the derivative of distance, Mr F says: No, not in general. Only true when the direction of the velocity does not change. In general, velocity is the derivative of **displacement** (with respect to time).

so to get distance, we integrate velocity. We want to go from zero seconds to 4 seconds, so we set those values:

$\displaystyle \int_0^4 (3t^2+1) ~dt$

$\displaystyle =[t^3+t+C]_0^4$

$\displaystyle =[4^3+4+C]-[0^4+0+C]$

$\displaystyle =4^3+4+C-C$

$\displaystyle =68 ~metres$

You can see that the value of the constant is not relevant, because it will always be added and then subtracted. So when calculating definite integrals, we usually ignore it.