# Comparing Reimann to this Integral

• May 26th 2009, 04:11 PM
icedragontt
Comparing Reimann to this Integral
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
• May 26th 2009, 07:30 PM
Media_Man
S= $\int_0^1 xdx=\lim_{n\rightarrow\infty} \sum_{k=0}^n f(x_k)\Delta$ , where $n\Delta=b-a=1$ and $x_k=k\Delta$.

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