Evaluate the following integrals as limit of sum: ^1_ -1 (x+3)dx - Vidyadip

Question

Evaluate the following integrals as limit of sum:
$\int\limits^1_{-1}(\text{x}+3)\text{dx}$

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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, $\text{a}=-1,\text{ b}=1,\text{ f(x)}=\text{x}+3,\text{ h}=\frac{1+1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^1_{-1}(\text{x}+3)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(-1)+\text{f}(-1+\text{h})+\\\ ....\ +\text{f}\big\{-1+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(-1+3)+(-1+\text{h}+3)+\\ ....\ +\{-1+(\text{n}-1)\text{h}+3\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\text{n}-1\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(3-\frac{1}{\text{n}}\Big)$
$=6$

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