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Splitting up [0, 3] into [tex]n[/tex] equally-spaced subintervals of length [tex]\Delta x=\frac{3-0}n = \frac3n[/tex] gives the partition
[tex]\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right][/tex]
where the right endpoint of the [tex]i[/tex]-th subinterval is given by the sequence
[tex]r_i = \dfrac{3i}n[/tex]
for [tex]i\in\{1,2,3,\ldots,n\}[/tex].
Then the definite integral is given by the infinite Riemann sum
[tex]\displaystyle \int_0^3 2x^2 \, dx = \lim_{n\to\infty} \sum_{i=1}^n 2{r_i}^2 \Delta x \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac6n \sum_{i=1}^n \left(\frac{3i}n\right)^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3} \sum_{i=1}^n i^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3}\cdot\frac{n(n+1)(2n+1)}6 = \boxed{18}[/tex]