$f(x)$is continuous in $\left[ 2,5 \right]$

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

Here $h=3/n$

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

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

$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{3}{n} \right)\sum _{r=0}^{n-1}3\left( \frac{9{{r}^{2}}}{{{n}^{2}}}+4+\frac{12r}{n} \right)-5$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{3}{n}\left( \frac{27(n-1)(n)(2n-1)}{6{{n}^{2}}}+12n+\frac{18(n-1)}{n}-5n \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{3}{n}\left( \frac{27({{n}^{2}}-n)(2n-1)}{6{{n}^{2}}}+12n+\frac{18n(n-1)}{n}-5n \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{3}{n}\left( \frac{\left( 54{{n}^{3}}-81{{n}^{2}}+27n \right)+\left( 42{{n}^{3}} \right)\left( 108{{n}^{3}}-108{{n}^{2}} \right)}{6{{n}^{2}}} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{3}{n}\left( \frac{204{{n}^{3}}-189{{n}^{2}}+27n}{6{{n}^{2}}} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{612{{n}^{3}}-567{{n}^{2}}+27n}{6{{n}^{3}}} \right)$

$=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{612}{6} \right)-\left( \frac{567}{6n} \right)+\left( \frac{27}{6{{n}^{2}}} \right)$

$=102$