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?)

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.