In your second approach, you actually made the same error that you made in your first approach. I am with you up to this point:

$\displaystyle A_n=(b-a)\left(a(a+3)+\frac{(b-a)(2a+3)(n-1)}{2n}+\frac{(b-a)^2(n-1)(2n-1)}{6n^2} \right)$

Now, when you took the limit as $\displaystyle n\to\infty$ you used a denominator of 6 for the last term within the parentheses, when it should be a 3, do you see why?

I will let you rework this on your own once you understand why you need a 3 there. Everything else you did is correct except for that.

edit: If you have a bunch of quadratic functions, you may want to generalize further and use $\displaystyle f(x)=ax^2+bx+c$ on $\displaystyle [d,e]$.