$f\left( x \right)$is continuous in

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

Here $h=3/n$

\[\int _{1}^{4}(3{{x}^{2}}+2x)dx=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{3}{n} \right)\sum _{r=0}^{n-1}f\left( \left( 1+\frac{3r}{n} \right) \right)\]

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

\[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{3}{n} \right)\sum _{r=0}^{n-1}3\left( \frac{9{{r}^{2}}}{{{n}^{2}}}+1+\frac{6r}{n} \right)+2\left( 1+\frac{3r}{n} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{3}{n}\left( \frac{27(n-1)(n)(2n-1)}{6{{n}^{2}}}+3n+\frac{9n(n-1)}{n}+2n+\frac{3n(n-1)}{n} \right)\]

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

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

\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{3}{n}\left( \frac{\left( 54{{n}^{3}}-81{{n}^{2}}+27n \right)+\left( 30{{n}^{3}} \right)+\left( 72{{n}^{3}}-72{{n}^{2}} \right)}{6{{n}^{2}}} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{3}{n}\left( \frac{156{{n}^{3}}-153{{n}^{2}}+27n}{6{{n}^{2}}} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{468{{n}^{3}}-459{{n}^{2}}+81n}{6{{n}^{3}}} \right)\]

\[=\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{468}{6} \right)-\left( \frac{459}{6n} \right)+\left( \frac{81}{6{{n}^{2}}} \right)\]

\[=78\]