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Could someone explain step by step how to solve this Riemann sum: $$\frac{1}{n} \sum_{i=0}^n\left(\frac{k}{n}\right)^2\frac{1}{8}\arcsin\frac{k}{n}\frac{1}{2} $$ k/n inside the sum is all squared. I don't know what to do with that 1/2 and 1/8. I thought the integral would be something like $$ x^2\arcsin\frac{x}{2}$$

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  • $\begingroup$ Should the sum be over $k$? $\endgroup$ – SEWillB May 12 '17 at 11:13
  • $\begingroup$ yes I left it like that because I copied from the java help page,my bad $\endgroup$ – Lola May 12 '17 at 11:16
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We have $\frac{1}{n} \sum_{i=0}^n\left(\frac{k}{n}\right)^2\frac18\arcsin\frac{k}{n}\frac{1}{2} \to \frac{1}{8} \int_0^1x^2\arcsin\frac{x}{2} dx $ for $n \to \infty$.

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