$I = \int\limits_0^4 {\left( {x + {e^{2x}}} \right)dx}$

Let $f\left( x \right) = x + {e^{2x}}$ $\Rightarrow$

$\int\limits_0^4 {f\left( x \right)dx} = \int\limits_0^4 {\left( {x + {e^{2x}}} \right)dx}$

Where $a = 0,b = 4$and $nh = b - a = 4 - 0 = 4$

By definition
$\int\limits_a^b {f\left( x \right)} = \mathop {\lim }\limits_{h \to 0} h\left[ {f\left( a \right) + f\left( {a + h} \right) + f\left( {a + 2h} \right) + ... + f\left( {a + \left( {n - 1} \right)h} \right)} \right]$

$\therefore$ $\int\limits_0^4 {\left( {x + {e^{2x}}} \right)dx}$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {f\left( 0 \right) + f\left( h \right) + f\left( {2h} \right) + .... + f\left( {\left( {n - 1} \right)h} \right)} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {\left\{ {0 + {e^0}} \right\} + \left\{ {h + {e^{2h}}} \right\} + \left\{ {2h + {e^{4h}}} \right\} + .... + \left\{ {\left( {n - 1} \right)h + {e^{2\left( {n - 1} \right)h}}} \right\}} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {h\left( {1 + 2 + 3 + ... + \left( {n - 1} \right)} \right) + {e^{2h}}\left( {1 + {e^{2h}} + {e^{4h}} + ... + {e^{2\left( {n - 2} \right)h}}} \right) + 1} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {h \cdot \cfrac{{\left( {n - 1} \right)\left( {1 + \left( {n - 1} \right)} \right)}}{2} + {e^{2h}} \cdot \cfrac{{1 \cdot \left( {{e^{2\left( {n - 1} \right)h}} - 1} \right)}}{{{e^{2h}} - 1}} + 1} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {\cfrac{{\left( {nh - h} \right)\left( {nh} \right)}}{2} + \cfrac{1}{2} \cdot \cfrac{{{e^{2h}}\left( {{e^{2\left( {n - 1} \right)h}} - 1} \right)}}{{\cfrac{{{e^{2h}} - 1}}{{2h}}}} + h} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {\cfrac{{\left( {4 - h} \right)4}}{2} + \cfrac{1}{2} \cdot \cfrac{{{e^{2h}}}}{{\cfrac{{{e^{2h}} - 1}}{{2h}}}}\left( {{e^{8 - 2h}} - 1} \right) + h} \right]$

$= \cfrac{{4 \times 4}}{2} + \cfrac{1}{2} \cdot \cfrac{{{e^0}}}{{\log e}}\left( {{e^8} - 1} \right) + 0 = 8 + \cfrac{1}{2}\left( {{e^8} - 1} \right) = \cfrac{{16 + {e^8} - 1}}{2} = \cfrac{{{e^8} + 15}}{2}$