Evaluate the following integrals as limit of sum: ^2_ 0 e^ x dx - Vidyadip

Question

Evaluate the following integrals as limit of sum:
$\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$

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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, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{e}^{\text{x}},\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\text{e}^{\text{x}}\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{e}^0+\text{e}^\text{h}+\text{e}^{2\text{h}}+\ .....\ +\text{e}^{(\text{n}-1)\text{h}}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\frac{(\text{e}^{\text{h}})^{\text{n}}-1}{\text{e}^{\text{h}}-1}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\text{e}^2-1}{\frac{\text{e}^{\text{h}}-1}{\text{h}}}\Bigg]$
$=\frac{\text{e}^2-1}{1}$
$=\text{e}^2-1$

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