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

Question

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

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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, $\text{a}=2,\text{ b}=3,\text{ f(x)}=\text{x}^2,\text{ h}=\frac{3-2}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{2}\text{x}^2\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(2)+\text{f}(2+\text{h})\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2^2+(2+\text{h})^2+\ ....+\ \big\{2(\text{n}-1)\text{h}\big\}^2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+4\text{h}\big\}\\+4\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\text{h}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[4\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(\text{n}-2)}{6}+4\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{1}{\text{n}}\Big[4\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}+2\text{n}-2\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}5\Big[6+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{2}{\text{n}}\Big]$
$=6+\frac{1}{3}$
$=\frac{19}{3}$

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