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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals as limit of sum:$\int\limits^{3}_{0}\big(2\text{x}^2+3\text{x}+5\big)\text{dx}$
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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, $\text{a}=0,\text{ b}=3,\text{ f(x)}=2\text{x}^2+3\text{x}+5,\text{ h}=\frac{3-0}{\text{n}}=\frac{3}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{0}\big(2\text{x}^2+3\text{x}+5\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[(0+0+5)+(2\text{h}^2+3\text{h}+5)+\\\ ....\ +\big\{2(\text{n}-1)^2\text{h}^2+3(\text{n}-5)\text{h}+1\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[5\text{n}+2\text{h}^2\big(1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big)+\\3\text{h}\big\{1+2+\ ...\ +(\text{n}-1)\text{h}\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\bigg[5\text{n}+2\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+3\text{h}\frac{\text{n}(\text{n}-1)}{2}\bigg]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{3}{\text{n}}\Big[\text{n}+\frac{3\text{n}(\text{n}-1)(2\text{n}-1)}{\text{n}}+\frac{9(\text{n}-1)}{2}\Big]$
$=15+18+\frac{27}{2}$
$=\frac{93}{3}$

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