Evaluate the following integrals as limit of sum: ^ 4 _ 1 (x^2-x … - Vidyadip

Question

Evaluate the following integrals as limit of sum:$\int\limits^{4}_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$

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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, $\text{a}=1,\text{ b}=4,\text{ f(x)}=\text{x}^2-\text{x},\text{ h}=\frac{4-1}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^{4}_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(1-1)+(1+\text{h})^2-(1+\text{h})+\\\ ....\ +\big\{(\text{n}-1)\text{h}+1\big\}^2-\big\{(\text{n}-1)\text{h}+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+4\text{h}\big\}\\-\text{h}\big\{1+2+\ ....+\ (\text{n}-1)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}-\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{3}{\text{n}}\Big[\frac{3(\text{n}-1)(2\text{n}-1)}{2\text{n}}+\frac{3(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}3\Big[\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big]$
$=9+\frac{9}{3}$
$=\frac{38}{3}$

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