Q:

What does the area under an acceleration-displacement curve represent? 

Considering the equation where,
$$
\frac {1}{2} \left (v^2_f - v^2_i \right) = \int_0^s ads\,
$$
What does the left-hand side of the equation actually represent?  Is there an intuitive explanation similar to how the area under an a-t curve yields velocity and the area under a v-t curve yields displacement? 

A:

As Newton says $F=ma$ we  have $a=F/m$. Now force times distance is work and, for a body  experiencing only the force $F$,  the amount of work done 
$$
W=\int_{\rm start}^{\rm end}  F\,ds
$$
 is the change in kinetic energy
$$W= \Delta {\rm K.E.} =1/2 mv_{\rm end}^2 - 1/2 mv_{\rm start}^2$$.

