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For instance,

$$a \cdot ds=dt$$

I know that one can take the indefinite integral on both sides to get

$$\int a ds = \int 1 dt$$

But how do I take the definite integral of both sides, and exactly what do I need to know to do this? (Specifically, the bounds. How do I know what bounds to use?)

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  • $\begingroup$ You would do these as you would any other integrals. $\endgroup$ – 123 Jan 8 '15 at 3:08
  • $\begingroup$ So, say if I know the bounds with respect to t will be 0 and t then the bounds of ds will be s(0) and s(t)? $\endgroup$ – Jason Jan 8 '15 at 3:10
  • $\begingroup$ Indeed. As Saibal points out in his answer, you simply use initial conditions to establish bounds of integration. $\endgroup$ – 123 Jan 8 '15 at 3:11
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Suppose you have the initial condition $t=t_0\implies s=s_0$. Then you can integrate: $$\int_{s_0}^{s}ads = \int_{t_0}^{t}dt$$ This is equivalent to first evaluating the indefinite integral and then solving for the constant of integration.

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  • $\begingroup$ How could you prove that it's equivalent? Although it's clear to me that the lower bound depends on the initial condition, it's not immediately clear to me why they're equal. $\endgroup$ – nog642 May 15 '18 at 16:11
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    $\begingroup$ Let's assume an antiderivative of a(s) is A(s). In the indefinite integral method, you get A(s) = t + C. Then to evaluate C, you would plug in the initial condition: A(s0) = t0 + C, i.e. C = A(s0) - t0. Substituting this value of C in the first equation, A(s) = t + A(s0) - t0. Rearranging A(s) - A(s0) = t - t0, which is exactly what you would get using the definite integral formulation. $\endgroup$ – Saibal May 15 '18 at 20:44

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