We divide [0, 3] into [tex]n[/tex] subintervals,
[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]
so that the right endpoint of each subinterval is given according to the arithmetic sequence,
[tex]r_k=\dfrac{3k}n[/tex]
for [tex]1\le k\le n[/tex].
The Riemann sum is then
[tex]\displaystyle\sum_{k=1}^nf(r_k)\Delta x_k[/tex]
where
[tex]\Delta x_k=r_k-r_{k-1}=\dfrac{3k}n-\dfrac{3(k-1)}n=\dfrac3n[/tex]
With [tex]f(x)=2x^2[/tex], we have
[tex]\displaystyle\frac3n\sum_{k=1}^n2\left(\frac{3k}n\right)^2=\frac{54}{n^3}\sum_{k=1}^nk^2[/tex]
Recall that
[tex]\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6[/tex]
The area under the curve [tex]f(x)[/tex] over the interval [0, 3] is then
[tex]\displaystyle\int_0^32x^2\,\mathrm dx=\lim_{n\to\infty}\frac{54n(n+1)(2n+1)}{6n^3}=\lim_{n\to\infty}9\left(2+\frac3n+\frac1{n^2}\right)=\boxed{18}[/tex]
