# Is the Fundamental Theorem of Calculus really applicable to the definition of work?

When the force $F$ on an object is not constant, then the work it performs is defined as $$W = \int_{x_0}^{x} F(X)dX.$$

Now, the Fundamental Theorem of Calculus states that

$$\text{If}\,\,\, f(x) = \int_{x_0}^{x} F(u)du, \,\,\,\text{then}\,\,\, \dfrac{d{f(x)}}{dx} = F(x).$$

If it is true (definitely it is), then I can say that the rate of change of work with respect to displacement is force or that the instantaneous rate of change of work with respect to displacement is force? I find this ridiculous! Can work be instantaneous? I can't imagine how work can be instantaneous. If the statement is wrong, how can the Fundamental Theorem of Calculus be true? Please help.

The Fundamental Theorem of Calculus is of course correct, and you are applying it correctly. The statements

• the rate of change of work with respect to displacement is force

and

• the instantaneous rate of change of work with respect to displacement is force

are correct. The thing to keep in mind is that it's not work that is instantaneous, but its rate of change. The work performed on the system is constantly changing, and it is performed continuously through time (and thence over displacement). Its derivative with respect do displacement is simply how fast it is changing, and this is a function of the particular instant as much as the work itself, and also the displacement, velocity, and so on.

There is an additional consequence to this interpretation: the force in an object is the work you would need to perform on an object to push it a unit displacement in the given direction. This is correct, though it is indeed a little mind-bending! If it's any help, this will get trumped by things like the principle of virtual work.

• +1 . Does instantaneous velocity exist? What about Zeno's paradox?? Mr. Feynman replied,Calculus gives the answer. And what about intuition?? ...if the instantaneous velocity is $v$ km/hr , it really does mean that if from that point, the motion were uniform, the body would have velocity $v$ after 1 hr. – user36790 Nov 24 '14 at 11:54
• Sir, just an example is needed! Can you please give?? I will be grateful to you. – user36790 Nov 24 '14 at 14:38
• You can directly translate: "If the instantaneous velocity is $v$ m/s, it really does mean that, if from that moment the motion were uniform, the body would have advanced $v$ m in 1s." This goes directly into "If the instantaneous-rate-of-change-of-work (i.e. force) is $F$ N, it really does mean that, if from that moment the motion were uniform, the external force would have performed $F$ joules of work over the next 1m." – Emilio Pisanty Nov 24 '14 at 14:58
• +1 . Really a great answer and, of course, the comment. Sir, in the same way can I apply FOTC in case of moment of inertia?? $$dI = r^2 .dm$$ . What will be the instantaneous rate of change of $I$ with respect to mass?? Thanks for clearing my confusion:) – user36790 Nov 24 '14 at 16:59
• That one is a bit harder. The interpretation of that integral goes something as "adding a unit mass of $m$ kg at squared-radius $r^2$ m$^2$ will increase the moment of inertia by $m r^2$ kg m$^2$". But in general it doesn't serve much to go around looking for every possible integral you might apply the FTC to. – Emilio Pisanty Nov 24 '14 at 17:49

No matter how ridiculous you may find it, it is true. The most general definition of work is indeed that the infinitesimal work done along an infinitesimal path is just force times the length of the path, i.e.

$$\mathrm{d}W = F\mathrm{d}s$$

Therefore, the amount of work done along a path in space $\gamma : [a,b] \to \mathbb{R}^3$ is the line integral

$$W = \int_\gamma \vec F(\vec x) \cdot \mathrm{d}\vec x$$

which, if the force is a conservative force, i.e. if there exists a function $U : \mathbb{R}^3 \to \mathbb{R}$ such that $\vec F = \vec \nabla U$, reduces to

$$W = U(\gamma(b)) - U(\gamma(a))$$

so that, for conservative forces in one dimension, indeed $\frac{\mathrm{d}W}{\mathrm{d}x} = F$.