$\int\limits_0^5 {\left( {x + 1} \right)\,dx}$

Let $f\left( x \right) = x + 1$ $\Rightarrow$ $\int\limits_0^5 {\left( f \right)x\,dx} = \int\limits_0^5 {\left( {x + 1} \right)\,dx}$

Where, $a = 0,b = 5,nh = b - a = 5 - 0 = 5$

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

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

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

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

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

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

$\therefore$ $\int\limits_0^5 {\left( {x + 1} \right)dx} = \mathop {\lim }\limits_{h \to 0} h\left[ {1 + \left( {1 + h} \right) + \left( {1 + 2h} \right) + .... + \left( {1 + \left( {n - 1} \right)h} \right)} \right]$

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

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

$= \mathop {\lim }\limits_{h \to 0} \left( {5 + \cfrac{{\left( {5 - h} \right)5}}{2}} \right) = 5 + \cfrac{{25}}{2} = \cfrac{{35}}{2}$

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