If the distance is given by 5t, then it's true that the integral of 5t (starting at 0) would be (5/2)t\^2, but the integral of distance wouldn't tell you the total distance. The units would be off.

The integral of velocity (measured in miles per hour, say) with respect to time (measured in hours) would give you a distance or displacement (measured in miles).

The integral of distance with respect to time would give you something whose units are your distance units multiplied by your time units. It might not be obvious what that would mean physically, but that's how the units would work out.