$f\left( x \right)$is continuous in $\left[ 1,3 \right]$

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

$\int _{1}^{3}\left( 2{{x}^{2}}+5x \right)dx=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2}{n} \right)\sum _{r=0}^{n-1}f\left( \left( 1+\frac{2r}{n} \right) \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2}{n} \right)\sum _{r=0}^{n-1}\left( 2{{\left( 1+\frac{2r}{n} \right)}^{2}}+5\left( 1+\frac{2r}{n} \right) \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2}{n} \right)\sum _{r=0}^{n-1}\left( 2+\frac{8{{r}^{2}}}{{{n}^{2}}}+\frac{8r}{n}+5+\frac{10r}{n} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{2}{n} \right)\sum _{r=0}^{n-1}\left( 7+\frac{8{{r}^{2}}}{{{n}^{2}}}+\frac{18r}{n} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{8(n-1)(n)(2n-1)}{6{{n}^{2}}}+7n+\frac{9n(n-1)}{n} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{8({{n}^{2}}-n)(2n-1)}{6{{n}^{2}}}+7n+\frac{9n(n-1)}{n} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{8\left( 2{{n}^{3}}-2{{n}^{2}}-{{n}^{2}}+n \right)}{6{{n}^{2}}}+7n+\frac{9n(n-1)}{n} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{(16{{n}^{3}}-24{{n}^{2}}+8n)+\left( 54{{n}^{3}}-{{54}^{2}} \right)+(42{{n}^{3}})}{6{{n}^{2}}} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{2}{n}\left( \frac{112{{n}^{3}}-78{{n}^{2}}+8n}{6{{n}^{2}}} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{224{{n}^{3}}-156{{n}^{2}}+8n}{6{{n}^{3}}} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{224}{6} \right)-\left( \frac{156}{6n} \right)+\left( \frac{8}{6{{n}^{2}}} \right)=112/3$