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

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, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\text{e}^{\text{x}},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^\text{b}_{\text{a}}\text{e}^{\text{x}}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{e}^\text{a}+\text{e}^{\text{a}+\text{h}}+\ .....\ +\text{e}^{\{\text{a}+(\text{n}-1)\text{h}\}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{e}^{\text{a}}\bigg\{\frac{(\text{e}^{\text{h}})^{\text{n}}-1}{\text{e}^{\text{h}}-1}\bigg\}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Big[\text{e}^\text{a}\frac{\text{e}^{\text{b}-\text{a}}-1}{\text{e}^{\text{h}}-1}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}=\Bigg[\frac{\text{e}^\text{b}-\text{e}^{\text{a}}}{\frac{\text{e}^\text{h}-1}{\text{h}}}\Bigg]$
$=\frac{\text{e}^\text{b}-\text{e}^\text{a}}{1}$
$=\text{e}^{\text{b}}-\text{e}^\text{a}$

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