Why does the definite integral of $F\left(x\right)$ with respect to $x$ represent the work done by a variable force from $x=a$ to $x=b$?

• A. Integrating force with respect to position yields momentum rather than work.
• B. The force remains constant over the motion, so multiplying average force by time yields exact work.
• C. The interval $[a,b]$ is partitioned into tiny displacements $\Delta x$, where the work done across each small subinterval is approximately $F\left(x\right)\Delta x$, and taking the limit of their sum yields the exact total work.
• D. The slope of the force-position graph over the interval is equal to the mechanical work done.

Answer: (C)