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Definite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals as limit of sum:
$\int\limits^{4}_{1}\big(3\text{x}^2+2\text{x}\big)\text{dx}$
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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, $\text{a}=1,\text{ b}=4,\text{ f(x)}=3\text{x}^2+2\text{x},\text{ h}=\frac{4-1}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^{4}_{1}\big(3\text{x}^2+2\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})+\ ....\ +\text{f}\big(1+(\text{n}-1)\text{h}\big)\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\big(3.1^2+2\times1)+(3+(1+\text{h})^2+2(1+\text{h})\big)+\\\ ....+\ \Big\{3\big(1+(\text{n}-1)\text{h}\big)^2+2\big(1+(\text{n}-1)\text{h}\big)\Big\}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[3\Big\{1^2+\big(1+\text{h}^2+(1+2\text{h})\big)^2+\ ....\ +\big(1+(\text{n}-1)\text{h}^2\big)\Big\}\\+2\Big\{1+(1+\text{h})+\ ...+\ \big(1+(\text{n}-1)\text{h}\big)\Big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[3\text{n}+3\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}+6\text{h}\big\{1+2+\ ...+\\\ (\text{n}-1)\text{h}\big\}+2\text{n}+2\text{h}\big\{1+2+\ .....\ +(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+3\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+8\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{3}{\text{n}}\Big[5\text{n}+\frac{9(\text{n}-1)(2\text{n}-1)}{2\text{n}}+12\text{n}-12\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}3\Big[17-\frac{12}{\text{n}}+\frac{9}{2}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\Big]$
$=51+27$
$=78$

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