\(\text{using }n \text{ subintervals we get}\\ x_k = \dfrac{(3-0)k}{n} = \dfrac{3k}{n}\\ \Delta = x_1- x_0 = \dfrac 3 n\)
\(\text{the left hand Riemann sum is} \\ \dfrac 3 n\sum \limits_{k=0}^{n-1}~\dfrac{3k}{n} + \dfrac{9k^2}{2n^2} = \dfrac{9}{4 n^2}-\dfrac{45}{4 n}+9\\ \text{you need to show this last equality of course}\)
\(\text{Pretty clearly}\\ \lim \limits_{n\to \infty}~\dfrac{9}{4n^2}-\dfrac{45}{4n} + 9 = 9\\ \text{which is the value of the definite integral of }y \text{ over }[0,3]\)
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