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$$\lim_{x\rightarrow\infty}\sum_{k=1}^{n}\frac{6(k+1)^2}{n^3}\sqrt{1+\frac{2(k+1)^3}{n^3}}$$

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?

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$$ \sum_{k=1}^{n}\frac{6\left(k+1\right)^2}{n^3}\sqrt{1+\frac{2\left(k+1\right)^3}{n^3}}=\sum_{k=2}^{n+1}6\frac{k^2}{n^3}\sqrt{1+2\frac{k^3}{n^3}} $$ 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 $k=2$)

Atmos
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    So $f(x)=6x^2\sqrt{1+2x^3}$? –  Mar 21 '21 at 09:46
  • And b-a = 1? How does this change when we integrate? –  Mar 21 '21 at 09:47
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    It's more than $b-a=1$, it's $a=0$ and $b=1$, because the formula is $f(a+(b-a)\frac{k}{n})$. And you have the right $f$, just make sure to make your first indices in $k$ start to $k=1$ – Atmos Mar 21 '21 at 09:48
  • You should find $2\sqrt{3}-\frac{2}{3}$ as the limit as $n$ goes to $+\infty$. – Atmos Mar 21 '21 at 10:06