= Solution
For $b>0$, split the interval into $n$ equal widths $\Delta x=b/n$. Since $x^2$ increases on this interval, the left and right rectangles give lower and upper <Riemann sums>:
$$
L_n=\frac{b^3}{n^3}\sum_{k=0}^{n-1}k^2,
\qquad
U_n=\frac{b^3}{n^3}\sum_{k=1}^{n}k^2.
$$
Using the finite sum of squares,
$$
U_n=\frac{b^3n(n+1)(2n+1)}{6n^3}\longrightarrow\frac{b^3}{3}.
$$
The rectangle bounds differ by $U_n-L_n=b^3/n\to0$. The <squeeze theorem> therefore gives, directly from elementary areas,
$$
\boxed{\int_0^b x^2dx=\frac{b^3}{3}.}
$$
For $b=0$ the result is zero. If negative $b$ is allowed, reverse the <integration> limits and use the reflection $x=-u$; the same signed formula holds. No <antiderivative> is needed for this first-principles calculation.
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