= Solution
Write $d=b-a$ and $h=d/n$. The required <Riemann sum> is
$$
S_n=h\sum_{k=1}^n(a+kh)^2
=h\left[na^2+2ah\sum_{k=1}^nk+h^2\sum_{k=1}^nk^2\right].
$$
Using $\sum k=n(n+1)/2$ and the given sum of squares,
$$
\boxed{S_n=a^2d+ad^2\frac{n+1}{n}+d^3\frac{(n+1)(2n+1)}{6n^2}.}
$$
An equivalent expression separates its <limit> and finite-$n$ corrections:
$$
S_n=\frac{b^3-a^3}{3}+\frac{(b-a)(b^2-a^2)}{2n}+\frac{(b-a)^3}{6n^2}.
$$
Consequently
$$
\boxed{\lim_{n\to\infty}S_n=\frac{b^3-a^3}{3}=\int_a^bx^2\,dx.}
$$
The function is continuous, so the <Riemann sums> converge to its <definite integral>, the desired <area> under the <parabola>. Both corrections are positive when $0<a<b$, agreeing with the geometric fact that these right-endpoint rectangles overestimate an increasing curve.
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