Express the limit as a definite integral on the given interval.
\[
\begin{array}{l}
\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[6\left(x_{i}^{*}\right)^{3}-5 x_{i}^{*}\right] \Delta x_{,} \quad[2,5] \\
\int_{2}() d x
\end{array}
\]

Step 1 ：Given the limit \(\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[6\left(x_{i}^{*}\right)^{3}-5 x_{i}^{*}\right] \Delta x_{,} \quad[2,5]\), we need to express this limit as a definite integral on the interval [2,5].

Step 2 ：We can write this as \(\int_{2}^{5} (6x^3 - 5x) dx\).

Step 3 ：To calculate this definite integral, we can use the Fundamental Theorem of Calculus, which states that the definite integral of a function from a to b is equal to the antiderivative of the function evaluated at b minus the antiderivative of the function evaluated at a.

Step 4 ：The antiderivative of \(f(x) = 6x^3 - 5x\) is \(F(x) = \frac{3}{2}x^4 - \frac{5}{2}x^2\).

Step 5 ：We need to evaluate \(F(5) - F(2)\).

Step 6 ：After calculating, we find that the definite integral of the function \(f(x) = 6x^3 - 5x\) over the interval [2,5] is 861.

Step 7 ：So, the limit of the sum as a definite integral on the interval [2,5] is \(\boxed{861}\).