Evaluate the following integrals as limit of sum: ^3_0(x+4)dx - Vidyadip

Question

Evaluate the following integrals as limit of sum:
$\int\limits^3_0(\text{x}+4)\text{dx}$

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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, $\text{a}=0,\text{ b}=3,\text{ f(x)}=\text{x}+4,\text{ h}=\frac{3-0}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_0(\text{x}+4)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}(0+(\text{n}-1)\text{h})\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+4)+(\text{h}+4)+\ ....\ +((\text{n}-1)\text{h}+4)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}(1+2+\ ....\ +(\text{n}-1))\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{3}{\text{n}}\Big[4\text{n}+\frac{3}{\text{n}}\times\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\Big[12+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=12+\frac{9}{2}$
$=\frac{33}{2}$

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