$f\left( x \right)$is continuous in $\left[ 0,2 \right]$

$\int _{a}^{b}f(x)dx=\underset{n\to \infty }{\mathop{\lim }}\,h\sum _{r=0}^{n-1}f(a+rh)$, where $h=(b-a)/n$

Here $h=2/n$

$\int _{0}^{2}\left( {{x}^{2}}+x \right)dx=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2}{n} \right)\sum _{r=0}^{n-1}f\left( \frac{2r}{n} \right)$

\[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2}{n} \right)\sum _{r=0}^{n-1}\left( {{\left( \frac{2r}{n} \right)}^{2}}+\left( \frac{2r}{n} \right) \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2}{n} \right)\sum _{r=0}^{n-1}\left( \frac{4{{r}^{2}}}{{{n}^{2}}}+\frac{2r}{n} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{4(n-1)(n)(2n-1)}{6{{n}^{2}}}+\frac{n(n-1)}{n} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{4\left( {{n}^{2}}-n \right)(2n-1)}{6{{n}^{2}}}+\frac{n(n-1)}{n} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{4\left( 2{{n}^{3}}-2{{n}^{2}}-{{n}^{2}}+n \right)}{6{{n}^{2}}}+\frac{n\left( n-1 \right)}{n} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{(8{{n}^{3}}-12{{n}^{2}}+4n)+\left( 6{{n}^{3}}-6{{n}^{2}} \right)}{6{{n}^{2}}} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{14{{n}^{3}}-18{{n}^{2}}+4n}{6{{n}^{2}}} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{28{{n}^{3}}-36{{n}^{2}}+8n}{6{{n}^{3}}} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{28}{6} \right)-\left( \frac{36}{6n} \right)+\left( \frac{8}{6{{n}^{2}}} \right)\]

\[=14/3\]