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Using my understanding of definite integrals, $I_i$ would be the inertia of the mass that you are not including in your calculation, while $I_f$ is the inertia of all the masses, including your mass of interest. $I_f$ - $I_i$ therefore is the inertia of the mass you are interested in. It appears as if $I_f(m)$ and $I_i(m)$ are not a functions of mass per se ...

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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 ...

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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 $$... 1 Where are you getting these books? Here it makes absolutely no sense to mention indefinite integrals and I have never seen such confusing statements. You need definite integral, usually in multiple dimensions (ideally 3D, as we live in 3D space, but you can simplify in some cases). The integration goes over the entire volume of the object, you have ... 1 I think the book authors might have the Huygens-Steiner theorem in mind when saying that. The theorem says that if the body has moment of inertia I_{cm} with respect to axis crossing its center of mass, then moving the reference axis to another parallel axis gives a new moment of inertia I', related to original one by$$I'=I_{cm}+md^2, where $d$ is ...

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Unless I have the rest of the book, it is difficult to understand what he meant. But many books have equations that could make not much sense or whose explanations are very unclear (no author is perfect). The only thing you really need to know is what you already seem to know. In an actual calculation you use a definite integral, and the constant $C$ doesn't ...

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As it looks like another question I've supplied an answer to might be duplicated here (and hence closed), I am going to provide a similar but not identical answer here. In words - divergence is the flux of something into or out of a closed volume, per unit volume. The best visual picture I have of this is a fluid flow. Imagine water spewing out of a tap - ...

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