$$W = \int_{x_i}^{x_f} F_x dx + \int_{y_i}^{y_f} F_y dy \int_{z_i}^{z_f} F_z dz$$ same as saying $$E = m\int_{0}^{x_f} a_x dx + m\int_{0}^{y_f} a_y dy + m\int_{0}^{z_f} a_z dz$$ Using definitions of $a_{avg} = \frac{1}{2}\frac{\Delta {v^2}}{\Delta x}$ and maybe $a_{avg} = \frac{\Delta{v}}{\Delta{t}}$
Not sure how to do the integration to prove that this leads to $E = \frac{1}{2}m{v^2}$
Does the integral simply resolve to $m a x_f$ where $\Delta x$ and $x_f$ reduce leaving $\frac{1}{2}m\Delta {V^2}$? Does it make sense that one could integrate something like that? I wouldn't do $\int x dx = xx = x^2$ Maybe I'm missing something with fundamental definitions (I derived $a_{avg}$ from Galilean Kinematic Equations for uniform acceleration)
edit: I guess an easier question is: Does $\int{a}{dx}$ resolve to the same expression as $\int{a}{dt}$? If not, how are the integrations performed using formal notation?
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$\begingroup$$$ \int a\mathrm dx=\int a\frac{\mathrm dx}{\mathrm dt}\mathrm dt=\int\frac{\mathrm d^2x}{\mathrm dt^2}\frac{\mathrm dx}{\mathrm dt}\mathrm dt=\frac12\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\text{const}\;. $$
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