Q: How to properly generalize a definite integral? I know, I know.  On the can, this problem seems simple.  Just take $\int_a^bf(x)\mathrm{d}x$ and write is as $\int f(x)\mathrm{d}x$.  However, when I tried to do that on an Engineering Dynamics problem, it didn't work very well.  
We have a motorcycle going around a path with an acceleration of $-0.001s\space \mathrm{m/s}$.  The distance of the path happens to be $150\pi$.  The initial speed is 25.  Since we know $a\mathrm{d}s=v\mathrm{d}v$,  That readily lends itself to this integral equation:
$$-\int_0^{150\pi}0.001s\mathrm{d}s=\int_{25}^vv\mathrm{d}s$$  This give the correct answer as provided in the book.  However, I also attempted the following combination to get the change in $v$:
$$\int_0^s0.001s\mathrm{d}s=\int_0^vv\mathrm{d}s$$which solves to $v=\sqrt{0.001}\times s$.  This renders a likely solution of a change of $14\space\mathrm{m/s}$ in $v$ -the actual change is closer to 5.  This is obviously wrong.I know why the first integral equation does work, but would someone be so kind as to tell me why the second one fails?  
What is wrong with assuming $0$ rather than using, say, $v_1$ and $v_2$?
I need to understand my mistake so I don't make it in the future!  Thanks!

A: $$v_1 - v_0 = \int_{v_0}^{v_1} dv = \int_{s_0}^{s_1} \frac{dv}{ds} ds = \int_{s_0}^{s_1} \frac{dv}{dt} \frac{dt}{ds} ds = \int_{s_0}^{s_1} \frac{a}{v} ds$$
Does that answer your question?
