Derivation for kinetic energy. Justification There is a derivation for kinetic energy using calculus:
\begin{align}
\Delta E_k&=\int_{x_0}^{x_1} F \ {\rm d}x \\
&= \int_{x_0}^{x_1} ma \ {\rm d}x \\
&= m \int_{x_0}^{x_1} \frac{{\rm d}v}{{\rm d}t} \ {\rm d}x \\
&= m \int_{x_0}^{x_1} \frac{{\rm d}v}{{\rm d}x}\frac{{\rm d}x}{{\rm d}t} \ {\rm d}x \\
&= m \int_{v_0}^{v_1} \frac{{\rm d}x}{{\rm d}t} \ {\rm d}v \\
&= m \frac{v^2}{2}\bigg\vert_{v_0}^{v_1}
\end{align}
How to justify change of limits in this case: $t_1 \rightarrow v_1$ and $t_0 \rightarrow v_0$? I suppose it's a simple change of variables, but I'm not sure how to do it.
 A: I think it would be more helpful if you would go back to "integration by parts". When you change the $dx$ to $dv$, you're changing what you have "respect to" in terms of the integral. Going back to basic calculus, remember how when you let $u$ equal some variables and input it into an integral, you had to change the bounds. This is the same situation.
Let $T = \frac{1}{2}mv^2$,
Therefore, $$\frac{dT}{dt} = \frac{1}{2}m\frac{d}{dt}(v^2)$$
$$=\frac{1}{2}m[\dot{v} \cdot v + v \cdot \dot{v}]$$
$$=\frac{1}{2}m[2(\dot{v} \cdot v)] = m\dot{v} \cdot v = F \cdot v$$
Now for the integration by parts, bring the dt to the right and integrate,
$$dT = F \cdot v \ dt$$
$$T = \int\limits_{t_0}^{t_1} F \cdot v \ dt$$
Recall $v \ dt = dr$,
$$T = \int\limits_{t_0}^{t_1} F \cdot dr$$
$$=\int\limits_{t_0}^{t_1} m \dot{v} \cdot dr$$
$$=m\int\limits_{t_0}^{t_1} \frac{dv}{dt} dr$$
$$=m\int\limits_{t_0}^{t_1} \frac{dv}{dr}\frac{dr}{dt} dr$$
$$=m\int\limits_{v_0}^{v_1}dv \ \frac{dr}{dt} $$
It's at this point where you must change the bound variables because $dv$ is what you're integrating with respect to. Thinking about it in the $u$ and $du$ calculus, here, I used $v$ instead of $u$. 
Recall $\frac{dr}{dt} = \dot{v}$,
$$T = m \int\limits_{v_0}^{v_1} \dot{v} \ dv$$
$$= m[\frac{v_1^2}{2} - \frac{v_0^2}{2}]$$
A: B.C.'s answer is interesting. I never would have thought of using integration by parts to answer this question.
Here's an alternative way to derive it just using the chain rule.
$$T = \frac{1}{2}m{v^2}$$
Assuming the mass is constant:
$$\frac{{dT}}{{dx}} = \frac{m}{2}\frac{d}{{dx}}\left( {{v^2}} \right)$$
$$\frac{d}{{dx}}\left( {{v^2}} \right) = \frac{{d\left( {{v^2}} \right)}}{{dv}}\frac{{dv}}{{dx}} = 2\underbrace v_{\frac{{dx}}{{dt}}}\frac{{dv}}{{dx}} = 2\frac{{dx}}{{dt}}\frac{{dv}}{{dx}} = 2\frac{{dv}}{{dt}} = 2a$$
So from the previous two equations we get
$$\frac{{dT}}{{dx}} = \frac{m}{2}\frac{d}{{dx}}\left( {{v^2}} \right) = \frac{m}{2} \cdot 2a = ma = F$$
And finally:
$$\int_{{x_0}}^{{x_1}} {Fdx}  = \int_{{x_0}}^{{x_1}} {\frac{{dT}}{{dx}}dx = T\left( {v\left( {{x_1}} \right)} \right)}  - T\left( {v\left( {{x_0}} \right)} \right) = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_0^2$$
So basically... the reason we are able to substitute $v_0$ and $v_1$ for $x_0$ and $x_1$ as our limits of integration is because 
$$F = \frac{{dT\left( {v\left( x \right)} \right)}}{{dx}} = \frac{{dT}}{{dv}}\frac{{dv}}{{dx}}$$
For more info on integration by substitution, plus some practice problems, check out these notes
