Work done for conservative forces is path independent Proof So I’m looking at the proof for work that is path independent.
There is a line were the integral
Partial derivative V dr from r1 to r2 becomes
Partial derivative V r’ dt from t1 to t2
I’m a bit confused at how this step was done if anyone could explain it. It would be really helpful also I hope I was clear enough. I don’t know how to to do integrals on this forum
 A: Let ${\bf \vec F}=-\nabla U$ be the conservative force acting on a particle over a path $C: {\bf \vec r}(t), t_1 \leq t \leq t_2 $.
Then, the work by ${\bf \vec F}$ is $$W=\int_C {\bf \vec F} \cdot \mathrm d {\bf \vec r}.$$
Since ${\bf \vec F}$ is conservative, we have that,
$$W=-\int_C \nabla U \cdot \mathrm d{\bf \vec r}.$$
Since ${\bf \vec v}=\dfrac{\mathrm d\bf \vec r}{\mathrm dt}$, (or $\delta {\bf \vec r}={\bf \vec v} \ \delta t$), we have that,
$$W=-\int_{t_1}^{t_2} \nabla U \cdot {\bf \vec v} \ \mathrm dt.$$
Your question was about the integration limits. In our first integral, we are integrating over the path $C$. However, in the integral above, we changed our limits because we are integrating with respect to time.
By the chain rule, we know that $\dfrac{\mathrm d}{\mathrm dt} U\left( {\bf \vec r}(t) \right)=\nabla U\cdot {\bf \vec v}$, so,
$$W=-\left[ U\left( {\bf \vec r}(t_1)\right) - U\left( {\bf \vec r}(t_2)\right)\right],$$ which becomes $$W=U\left( {\bf \vec r}_1\right)-U\left( {\bf \vec r}_2\right). \blacksquare$$
