Evaluating this Riemann sum $$\lim_{x\rightarrow\infty}\sum_{k=1}^{n}\frac{6(k+1)^2}{n^3}\sqrt{1+\frac{2(k+1)^3}{n^3}}$$
How do we determine if it is a Right/Left or midpoint Riemann Sum?
How do we find the values a and b?
Is $$f(x)=\frac{3x^2}{2}\sqrt{1+\frac{x^3}{4}}$$ possible? if $$x=\frac{2(k-1)}{n}$$
But then again how do I find the interval for which I have to integrate?

$$
\sum_{k=1}^{n}\frac{6\left(k+1\right)^2}{n^3}\sqrt{1+\frac{2\left(k+1\right)^3}{n^3}}=\sum_{k=2}^{n+1}6\frac{k^2}{n^3}\sqrt{1+2\frac{k^3}{n^3}}
$$
You need to have only an expression with $\frac{k}{n}$ to idenfity with $f\left(\frac{k}{n}\right)$. So you write it as
$$
\frac{1}{n}\sum_{k=2}^{n+1}6\frac{k^2}{n^2}\sqrt{1+2\frac{k^3}{n^3}}
$$
Can you identify $a$, $b$ and $f$ now ? (notice also that's it starts at $k=2$)
