Evaluate the following integrals as limit of sum: ^3_1(3x-2)dx - Vidyadip

Question

Evaluate the following integrals as limit of sum:
$\int\limits^3_1(3\text{x}-2)\text{dx}$

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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, $\text{a}=1,\text{ b}=3,\text{ f(x)}=3\text{x}-2,\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_1(3\text{x}-2)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big(1+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(3-2)+(3+3\text{h}-2)+(3+6\text{h}-2)\ +\\ ....\ +\big(3+(\text{n}-1)\text{h}+3-2\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+3\text{h}\big(1+2+3+\ ....\ + (\text{n}-1)\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+3\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+3\text{n}-3\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\Big(4-\frac{3}{\text{n}}\Big)$
$=8$

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