Question
Evaluate the following integrals as limit of sum:
$\int\limits^{4}_{0}\big(\text{x}+\text{e}^{2\text{x}}\big)\text{dx}$

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, $\text{a}=0,\text{ b}=4,\text{ f(x)}=\text{x}+\text{e}^{2\text{x}},\text{ h}=\frac{4-0}{\text{n}}=\frac{4}{\text{n}}$
Therefore, $\text{I}=\int\limits^{4}_{0}\big(\text{x}+\text{e}^{2\text{x}}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\Big[(0+\text{e}^0)+(\text{h}+\text{e}^{2\text{h}})+\ .....\ +\Big\{(\text{n}-1)\text{h}+\text{e}^{2(\text{n}-1)\text{h}}\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}\big\{1+2+\ ...+ (\text{n}-1)\text{h}\big\}+\text{e}^0+\text{e}^{2\text{h}}+\text{e}^{4\text{h}}+\ ....+\ \text{e}^{2(\text{n}-1)\text{h}}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}\frac{\text{n}(\text{n}-1)}{2}+\frac{(\text{e}^{2\text{h}})^\text{n}-1}{\text{e}^{2\text{h}}-1}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{16}{\text{n}^2}\times\frac{\text{n}(\text{n}-1)}{2}+\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^8-1}{\frac{\text{e}^{2\text{h}}-1}{\text{h}}}$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{16}{\text{n}^2}\times\frac{\text{n}(\text{n}-1)}{2}+\lim\limits_{\text{h}\rightarrow0}\frac{\text{e}^8-1}{\frac{2(\text{e}^{2\text{h}}-1)}{\text{h}}}$
$=8+\frac{\text{e}^8-1}{2}$
$=\frac{15+\text{e}^8}{2}$

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