I doubt if this interval is true and satisfy this $$ x_{i-1}^{2} < \xi_{i}^{2} - \dfrac{1}{3}(x_{i}^{2} + x_{i}x_{i-1} + x_{i-1}^{2}) < x_{i}^{2} < 2\delta^{2}.$$
here is example
$$\int_{a}^{b} x^{2} \,dx= \dfrac{1}{3}(b^{3} - a^{3}) = \dfrac{1}{3}(b-a)(b^{2} + ba + a^{2})$$
and my proof:
Let $\epsilon > 0$ and $C > 0$. Take $\delta = \sqrt{\dfrac{\epsilon}{2C}}$. Let $D' = \lbrace(\xi_{i})[x_{i}-x_{i-1}] \rbrace_{i=1}^{n}$ be $\delta$-fine $C$-division of $[a,b]$. Observe that for each $i \in \lbrace1, 2, \dotsc ,n\rbrace$
$$ x_{i-1}^{2} < \xi_{i}^{2} - \dfrac{1}{3}(x_{i}^{2} + x_{i}x_{i-1} + x_{i-1}^{2}) < x_{i}^{2} < 2\delta^{2}.
$$
Also, note that
\begin{align}
b^{3}-a^{3} &= b^{3} + (x_{1}^{3}-x_{1}^{3}) + (x_{2}^{3}-x_{2}^{3}) + \dotsc +(x_{n-1}^{2}-x_{n-1}^{2}) - a^{2} \\
&= (b^{3}-x_{n-1}^{3}) + (x_{n-2}^{3} - x_{n-1}^{3}) + \dotsc + (x_{n-2}^{3} - a^{3}) \\
&= \sum_{i=1}^{n} (x_{i}^{3}-x_{i-1}^{3})
\end{align}
Now,
\begin{align} &\hphantom{{}={}}\bigg| \sum_{i=1}^{n}f(\xi_{i})(x_{i} - x_{i-1}) - \frac{1}{3}(b^{3} - a^{3}) \bigg| \\ &= \bigg|\sum_{i=1}^{n}\xi^{2}(x_{i} - x_{i-1}) - \frac{1}{3} \sum_{ni=1}^{n}(x_{i}^{3} - x_{i-1}^{3}) \bigg| \\ &= \bigg|\sum_{i=1}^{n}\xi^{2}(x_{i} - x_{i-}) - \frac{1}{3} \sum_{i=1}^{n}(x_{i}-x_{i-1})(x_{i}^{2} + x_{i}x_{i-i} + x_{i-1}^{2}) \bigg| \\ &= \bigg|\sum_{i=1}^{n}(x_{i} - x_{i-1})\Bigl( \xi_{i}^{2} - \frac{1}{3}(x_{i}^{2} + x_{i}x_{i-i} + x_{i-1}^{2}) \Bigr)\bigg| \\ & \leq \sum_{i=1}^{n} \biggl[\big|(x_{i} - x_{i-1})\big| \ \bigg|\xi_{i}^{2} - \frac{1}{3}(x_{i}^{2} + x_{i}x_{i-i} + x_{i-1}^{2} )\bigg|\biggr] \\ &= 2\delta^{2} \sum_{i=1}^{n}\big| (x_{i} - x_{i-1})\big| \\ &\leq 2\delta^{2}\sum_{i=1}^{n}|x_{i} - x_{i-1}| \\ &= 2\delta^{2} C \\ &= 2C\bigg(\sqrt{\frac{\epsilon}{2C}}\bigg)^{2} \\ &= \epsilon \end{align}