$\int\limits_1^4 {\left( {{x^2} - x} \right)dx}$

Let $f\left( x \right) = {x^2} - x$ $\Rightarrow$
$\int\limits_1^4 {f\left( x \right)dx} = \int\limits_1^4 {\left( {{x^2} - x} \right)dx}$

Where $a = 1,b = 4,nh = 3$

Now, $f\left( a \right) = f\left( 1 \right) = 1 - 1 = 0$

$f\left( {a + h} \right) = f\left( {1 + h} \right) = {\left( {1 + h} \right)^2} - \left( {1 + h} \right) = {h^2} + h$

$f\left( {a + 2h} \right) = f\left( {1 + 2h} \right) = {\left( {1 + 2h} \right)^2} - \left( {1 + 2h} \right) = 4{h^2} + 2h$

$f\left( {a + 3h} \right) = f\left( {1 + 3h} \right) = {\left( {1 + 3h} \right)^2} - \left( {1 + 3h} \right) = 9{h^2} + 3h$

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$f\left( {a + \left( {n - 1} \right)h} \right) = f\left( {1 + \left( {n - 1} \right)h} \right) = {\left( {1 + \left( {n - 1} \right)h} \right)^2} - \left( {1 + \left( {n - 1} \right)h} \right) = {\left( {n - 1} \right)^2}{h^2} + \left( {n - 1} \right)h$

By definition $\int\limits_a^b {f\left( x \right)dx}$

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

$\therefore$ $\int\limits_1^4 {\left( {{x^2} - x} \right)dx}$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {0 + \left( {{h^2} + h} \right) + \left( {4{h^2} + 2h} \right) + \left( {9{h^2} + 3h} \right) + ...... + \left\{ {{{\left( {n - 1} \right)}^2} + \left( {n - 1} \right)h} \right\}} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {{h^2}\left\{ {1 + 4 + 9 + ... + {{\left( {n - 1} \right)}^2}} \right\} + h\left\{ {1 + 2 + 3 + ... + \left( {n - 1} \right)} \right\}} \right]$

$= \mathop {\lim }\limits_{h \to 0} h\left[ {{h^2}\cfrac{{\left( {n - 1} \right)n\left( {2n - 1} \right)}}{6} + \cfrac{{h\left( {n - 1} \right)n}}{2}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {\cfrac{{\left( {nh - h} \right)nh\left( {2hn - h} \right)}}{6} + \cfrac{{\left( {nh - h} \right)nh}}{2}} \right]$

$= \mathop {\lim }\limits_{h \to 0} \left[ {\cfrac{{\left( {3 - h} \right)3\left( {6 - h} \right)}}{6} + \cfrac{{\left( {3 - h} \right)3}}{2}} \right]$

$= \cfrac{{3\left( 3 \right)\left( 6 \right)}}{6} + \cfrac{{3\left( 3 \right)}}{2} = 9 + \cfrac{9}{2} = \cfrac{{27}}{2}$