Evaluate the following integrals as limit of sum: ^ 2 _ 0 (3x^2-2…

Question

Evaluate the following integrals as limit of sum:
$\int\limits^{2}_{0}\big(3\text{x}^2-2\big)\text{dx}$

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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, $\text{a}=0,\text{ b}=2,\text{ f(x)}=3\text{x}^2-2,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{2}_{0}\big(3\text{x}^2-2\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big\{0+(\text{n}-1)\text{h}\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0-2)+(3\text{h}^2-2)+\ ....\ +\{3(\text{n}-1)^2\text{h}^2-2\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-2\text{n}+3\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[-2\text{n}+3\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[-2\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{\text{n}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{-2+2\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=-4+8$
$=4$

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