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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
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, $\text{I}=\int\limits^4_{1}\big(\text{x}^2-\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(1-1)+\big\{(1+\text{h}^2)-(1+\text{h})\big\}+\\\ ....\ +\big\{(1+(\text{n}-1)\text{h})^2-(1+(\text{n}-1)\text{h})\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{h}^2\big\{1^2+2^2+3^2\ ....\ +\$\text{n}-1)^2\big\}+1+2\text{h}\big\{1+2+\ ....+\$\text{n}-1)\big\}-\text{n}-\text{h}\big\{1+2+\ .....\ +(\text{n}-1)\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+\text{h}\frac{(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{3}{\text{n}}\bigg[\frac{9(\text{n}-1)(2\text{n}-1)}{6\text{n}}+\frac{3(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}3\bigg[\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)+\frac{3}{2}\Big(1-\frac{1}{\text{n}}\Big)\bigg]$
$=9+\frac{9}{2}$
$=\frac{27}{3}$

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