= Solution
Parametrize any actual particle trajectory by time, with $\mathbf v=\dot{\mathbf r}$ and acceleration $\mathbf F=\dot{\mathbf v}$. The <chain rule> gives
$$
\frac{d}{dt}\left(\frac12|\mathbf v|^2\right)=\mathbf v\cdot\dot{\mathbf v}=\mathbf F\cdot\dot{\mathbf r}.
$$
Integrating from the origin to the endpoint gives
$$
I=\int\mathbf F\cdot d\mathbf r=K_A-K_O,\qquad\boxed{\Delta K=1.}
$$
Here $K$ is <specific kinetic energy>, because the source defines it without a mass factor. Equivalently, $|\mathbf v_A|^2-|\mathbf v_O|^2=2$ in the chosen units. The individual initial or final speeds are not fixed by this result. For a particle of mass $m$, the change of ordinary <kinetic energy> would be $mI$.
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