Integrals are so much, euhh, integral to this discussion that that Newton guy made them up* to make sense of this stuff.
Speed is the integral of acceleration, displacement the integral of speed. You cannot do any math in this field without the mathematics.
*yes, there were other people working in that field.
Re: pushing a box in a circle.
While it is certainly possible to make the the sum total of the work 0, it is also possible (and more correct) to calculate the work done underway.
Work is not really a vector, you cannot really subtract the work going one way from the work going back: you should add them.
To express this more clearly:
Work really is not a vector.
@Cayuga, in case you haven’t got back to your high school freshman on this, @naita is correct, work has been done in this case, and that work is due to the force you have to apply to the box to overcome the friction between it and your kitchen floor. Unfortunately, you do need to calculate the integral, but as long as you and the freshman know that the circumference of a circle is 2(pi)R, where R is the radius of the circle, then if F is the (assumed constant) force applied, the total work done is 2(pi)FR because you pushed the box exerting force F over a distance equal to the circumference of the circle. Note that this also assumes that you were very precisely adjusting your push so that the box traveled in a perfect circle and the constant force F applied was always tangential to the circle.
Sorry for using the inelegant (pi) but I can’t remember how to insert symbols…
Even if your floor were frictionless, you’d still be doing work. In that case, the box would be moving faster and faster as you pushed it, and the energy you were putting in would be becoming kinetic energy.
I left out the time dependency of the force applied, and assumed that the initial acceleration would be compensated for by the final deceleration to bring the box to a dead stop once the circle was complete. I also neglected to account for the additional work done in rotating the box around it’s center if we assume that the pusher kept the same side facing her/him as they traversed the circle.
You can in some circumstances. Think of the typical pendulum, at the top of its arc. As it swings down, gravity is doing positive work on it, causing it to speed up. As it swings up, gravity is doing negative work on it, causing it to slow down.
(Potential energy is defined by the amount of work needed to bring something to its current state.)
I can relate to the way moes_lotion is thinking. In problems like in the OP, we often start by developing a simple mathematical model such as w = f times d to describe a situation. However, our simple mathematical model will almost always be based on lots and lots of assumptions. As a practical matter, these assumptions are often very sensible. For example, if the force we are talking about in the OP is the force necessary to overcome friction (and not a force to constantly accelerate a frictionless box as Chronos * suggested) then we probably assumed that the friction can be a described as a constant coefficient of friction times the normal force. But it might be that the friction is not constant; it might be a function of speed, temperature, time (maybe because the box heats up the longer it is pushed), and a few other things I haven’t thought of yet. And it could very well be that those effects are so small we really don’t have to worry about them.
The point is, no matter how well we refine our mathematical models to describe real-world phenomena, Mother Nature is actually much more complicated. It just depends on how far down the rabbit hole we want to go. As for myself, I really enjoy rabbit holes.
_* I still like Chronos’s idea of a frictionless box.
It should also be pointed out that the displacement calculated relative to the inertial reference frame of the pushed object. Leaning on the wall of an truck barrelling down the street also doesn’t do any work, even though there is a lot of displacement relative to the people on the ground.
Actually, leaning on the wall of a speeding truck would do work. It’s just that your feet on the truck’s floor would be doing negative that amount of work, so the net work would be zero.
You think you can explain frames of reference when integrals are deemed too complex?
Good luck. 
If you break the problem up into a bunch of smaller, easier-to-understand, straight line segments and add them all up, it’s pretty much the same as an integral. I’d take this as an opportunity to de-mystify calculus. Get young students believing it’s really not so bad (and it isn’t).
I’m picturing a curling rink situation but with totally friction-less ice. What kind of shoes would I need to be able to even get to the box in the first place - skates maybe?