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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate the following integrals as limit of sum:
$\int\limits^2_{1}\big(\text{x}^2-1\big)\text{dx}$
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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, $\text{a}=1,\text{ b}=2,\text{ f(x)}=\text{x}^2-1,\text{ h}=\frac{2-1}{\text{n}}=\frac{1}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{1}\big(\text{x}^2-1\big)\text{ dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(1-1)+(\text{h}^2-1)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2-1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{n}-1+\text{h}^2\big\{1^2+2^2+3^2\ ....\ +(\text{n}-1)^2\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[\text{n}-1+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{1}{\text{n}}\bigg[\text{n}-1+\frac{(\text{n}-1)(2\text{n}-1)}{6\text{n}}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\bigg\{1-\frac{1}{\text{n}}+\frac{1}{6}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=1+\frac{1}{3}$
$=\frac{4}{3}$

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