Evaluate the following integrals as limit of sum: ^2_ 1 x^2 dx - Vidyadip

Question

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

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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, $\text{a}=1,\text{ b}=2,\text{ f(x)}=\text{x}^2,\text{ h}=\frac{2-1}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{1}\text{x}^2\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big(1+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(1)+(\text{h}+1)^2+\ ....\ +\big((\text{n}-1)\text{h}+1\big)^2\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\\+2\text{h}\big\{1+2+3+\ ....\ +(\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+2\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\text{n}+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}+\text{n}-1\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{2+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{1}{\text{n}}\bigg\}$
$=2+\frac{1}{3}$
$=\frac{7}{3}$

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