just in case there's interest, as I don't like muddling in obscurity
here's the solution:as I have given in JB
in order 'integrate' a function of the form [ax^n] one applies the following formula:
∫ax^n dx =ax^n+1/(n+1) so ∫ax dx = ax^2/2 and ∫ax^5 dx = ax^6/6 ...and so on....
for a definite integral between two bounds for example between 2 and 0 and where 2 = b and 0 =a you have to replace those X's with the numbers always subtracting the [b-a]
so for this integral equation ∫axdx =AS^2/2 - [A0^2/2=0]
and since S^2 =S*S we can say that ∫axdx between the bounds{s,0}
= 1/2 ASS 