Comparing Reimann to this Integral

**icedragontt** · May 26th 2009, 04:11 PM

Given:
n
$Sigma$ k/(n^2)
k = 1

show that
limit as n-> infinity Sn = integral from 0 -> 1 on (xdx)
by comparing the Reimann sum for the integral to the
series.

can someone walk me through this problem?, thanks

**Media_Man** · May 26th 2009, 07:30 PM

S= $\int_0^1 xdx=\lim_{n\rightarrow\infty} \sum_{k=0}^n f(x_k)\Delta$ , where $n\Delta=b-a=1$ and $x_k=k\Delta$ .

$S=\lim_{n\rightarrow\infty} \sum_{k=0}^n k\Delta^2=\lim_{n\rightarrow\infty} \sum_{k=0}^n \frac k{n^2}$

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