Question
Evaluate the following integrals as limit of sum:
$\int\limits^{\text{b}}_{\text{a}}\text{x}\text{ dx}$

Answer

$\int\limits^{\text{b}}_\text{a}\text{f(x)}\text{dx}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\text{f}(\text{a}+2\text{h})\ +\\ ....\ +\text{f}(\text{a}+(\text{n}-1)\text{h})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\text{x},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^{\text{b}}_{\text{a}}\text{x}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{a}+(\text{a}+\text{h})+(\text{a}+2\text{h})+\ ....+\ \big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{na}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{na}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{b}-\text{a}}{\text{n}}\Big[\text{na}\frac{[\text{b}-\text{a}](\text{n}-1)}{2}\Big]$
$=(\text{b}-\text{a})\text{a}+\frac{(\text{b}-\text{a})^2}{2}$
$=\frac{2\text{ab}-2\text{a}^2+\text{b}^2+\text{a}^2-2\text{ab}}{2}$
$=\frac{\text{b}^2-\text{a}^2}{2}$

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