Evaluate the following integrals as limit of sum: ^3_ 2 (2x^2+1 )… - Vidyadip

Question

Evaluate the following integrals as limit of sum:
$\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$

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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, $\text{a}=2,\text{ b}=3,\text{ f(x)}=2\text{x}^2+1,\text{ h}=\frac{3-2}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^3_{2}\big(2\text{x}^2+1\big)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(2)+\text{f}(2+\text{h})+\ ....\ +\text{f}\big\{2+(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2(2.2)^2+1+\big\{2(2+\text{h})^2+1\big\}+\\\ ....\ +\big\{2((2+\text{n}-1)\text{h})^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+2\Big\{2^2+(2+\text{h})^2+\ .....\big((2+\text{n}-1\big)\text{h}\big)^2\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+8\text{n}+2\text{h}^2\Big\{1^2+2^2+3^3+\ .....\ +(\text{n}-1)^2\Big\}\\+8\text{h}\big\{1+2+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\text{h}\bigg[9\text{n}+\text{h}^2\frac{2\text{n}(\text{n}-1)(2\text{n}-1)}{6}+8\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[9\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{3\text{n}}+4\text{n}-4\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg\{13+\frac{1}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{4}{\text{n}}\bigg\}$
$=13+\frac{2}{3}$
$=\frac{41}{3}$

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