Evaluate the following definite integral as limit of sum: _ 1 ^ 4… - Vidyadip

Question

Evaluate the following definite integral as limit of sum:
$\int\limits_{1}^{4}(\text{x}^{2}-\text{x)}\ \text{dx}$

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Answer

$\text{we}\ \text{know}\ \text{thet}\ \int\limits_{\text{a}}^{\text{b}}\ \text{f}\ \text{(x)}\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\big[\text{f} \ \text{(a)}+\text{f}\ \text{(a}+\text{h)}+\text{f}(\text{a}+2\text{h)}+.......+\text{f}\text{(a}+\text{(n}-1)\text{h)}\big]$
$\text{where}\ \text{nh}=\text{b}-\text{a}$
$\text{Here},\ \ \text{a}=1,\text{b}=4,\text{nh}=3\ \text{and}\ \text{f}\ \text{(x)}=\text{x}^{2}-\text{x}$
$\therefore \ \ \int\limits_{1}^{4}\text{(x}^{2}-\text{x)}\ \text{dx}=\lim\limits_{\text{h}\rightarrow0}\ \text{h}\bigg[0+\text{h}+\text{h}^{2}+2\text{h}+4\text{h}^{2}+....+\text{(n}-1)\text{h}+\text{(n}-1)^{2}\text{h}^ {2}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0} \ \text{h}\bigg[\text{h}(1+2+3+....+\text{(n}-1)+\text{h}^{2}(1^{2}+2^{2}+.....\text{(n}-1)^{2}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[\text{nh}\frac{\text{(nh}-\text{h)}}{2}+\frac{\text{nh}\text{(nh}-\text{h)}(2\text{nh}-\text{h)}}{6}\bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\bigg[3\frac{(3-\text{h)}}{2}+\frac{3(3-\text{h)}(2.3-\text{h)}}{6}\bigg]$
$=\bigg[3\frac{(3-0)}{2}+\frac{3(3-0)(6-0)}{6}\bigg]=\bigg[\frac{9}{2}+9\bigg]=\frac{27}{9}$

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