$a\underset{n}{lim}\frac{1}{n}[(\frac{1}{n}{)}^2+(\frac{2}{n}{)}^2+$cdots$+(\frac{n}{n}{)}^2].$

Proof Let $f(x)=x^2.$ Let $Pbe$ the partition $of0,1$ into $n$ equal parts: $x_i-x_{i-1}=\frac{1}{n}{U}_P(f)={\displaystyle \underset{1}{\overset{n}{}}}su{p}_{[x_{i-1},x_i]}f(x)(x_i-x_{i-1})$

$={\displaystyle \underset{1}{\overset{n}{}}}{x}_i^2(x_i-x_{i-1})$

$={\displaystyle \underset{1}{\overset{n}{}}}(\frac{i}{n}{)}^2\frac{1}{n}($ for $x_i=\frac{i}{n})$

$=\frac{1}{n}{\displaystyle \underset{1}{\overset{n}{}}}(\frac{i}{n}{)}^2$

$=\frac{1}{n}((\frac{1}{n}{)}^2+(\frac{2}{n}{)}^2+$cdots$+(\frac{n}{n}{)}^2)f(x)=x^2$nlongrightarrow${\displaystyle \underset{0}{\overset{1}{}}}{x}^{2}dx\underset{\text{n}\text{l}\text{o}\text{n}\text{g}\text{r}\text{i}\text{g}\text{h}\text{t}\text{a}\text{r}\text{r}\text{o}\text{w}}{lim}\frac{1}{n}((\frac{1}{n}{)}^2+(\frac{2}{n}{)}^2+$cdots$+(\frac{n}{n}{)}^2)={\displaystyle \underset{0}{\overset{1}{}}}{x}^{2}dx=\frac{x^3}{3}{|}_0^1=\frac{1}{3}-\frac{0}{3}=\frac{1}{3}{x}_2,$ where $x_i-x_{i-1}=\frac{1}{n}.$ Consider

$U_P(f)={\displaystyle \underset{1}{\overset{n}{}}}su{p}_{[x_{i-1},x_i]}f(x)(x_i-x_{i-1})$

$={\displaystyle \underset{1}{\overset{n}{}}}{x}_i^2(x_i-x_{i-1})$

$={\displaystyle \underset{1}{\overset{n}{}}}(\frac{i}{n}{)}^2\frac{1}{n}($ for $x_i=\frac{i}{n})$

$=\frac{1}{n}{\displaystyle \underset{1}{\overset{n}{}}}(\frac{i}{n}{)}^2$

$=\frac{1}{n}((\frac{1}{n}{)}^2+(\frac{2}{n}{)}^2+$cdots$+(\frac{n}{n}{)}^2)$

Since $f(x)=x^{2}is$ continuous, $itisR.$ I., and $so$ the $limitin(4)as$ nlongrightarrow$$ equals

${\displaystyle \underset{0}{\overset{1}{}}}{x}^{2}dx.$ Thus

$\underset{\text{n}\text{l}\text{o}\text{n}\text{g}\text{r}\text{i}\text{g}\text{h}\text{t}\text{a}\text{r}\text{r}\text{o}\text{w}}{lim}\frac{1}{n}((\frac{1}{n}{)}^2+(\frac{2}{n}{)}^2+$cdots$+(\frac{n}{n}{)}^2)={\displaystyle \underset{0}{\overset{1}{}}}{x}^{2}dx=\frac{x^3}{3}{|}_0^1=\frac{1}{3}-\frac{0}{3}=\frac{1}{3}0=x_0$

$x_2,$ where $x_i-x_{i-1}=\frac{1}{n}.$ Consider

$U_P(f)={\displaystyle \underset{1}{\overset{n}{}}}su{p}_{[x_{i-1},x_i]}f(x)(x_i-x_{i-1})$

$={\displaystyle \underset{1}{\overset{n}{}}}{x}_i^2(x_i-x_{i-1})$

$={\displaystyle \underset{1}{\overset{n}{}}}(\frac{i}{n}{)}^2\frac{1}{n}($ for $x_i=\frac{i}{n})$

$=\frac{1}{n}{\displaystyle \underset{1}{\overset{n}{}}}(\frac{i}{n}{)}^2$

$=\frac{1}{n}((\frac{1}{n}{)}^2+(\frac{2}{n}{)}^2+$cdots$+(\frac{n}{n}{)}^2)$

Since $f(x)=x^{2}is$ continuous, $itisR.$ I., and $so$ the $limitin(4)as$ nlongrightarrow$$ equals

${\displaystyle \underset{0}{\overset{1}{}}}{x}^{2}dx.$ Thus

$\underset{\text{n}\text{l}\text{o}\text{n}\text{g}\text{r}\text{i}\text{g}\text{h}\text{t}\text{a}\text{r}\text{r}\text{o}\text{w}}{lim}\frac{1}{n}((\frac{1}{n}{)}^2+(\frac{2}{n}{)}^2+$cdots$+(\frac{n}{n}{)}^2)={\displaystyle \underset{0}{\overset{1}{}}}{x}^{2}dx=\frac{x^3}{3}{|}_0^1=\frac{1}{3}-\frac{0}{3}=\frac{1}{3}$