Question
Evaluate the following integrals as limit of sum:
$\int\limits^2_{0}\big(\text{x}^2+1\big)\text{dx}$
$\int\limits^2_{0}\big(\text{x}^2+1\big)\text{dx}$
✓
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, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+1,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\big(\text{x}^2+1\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+1)+(\text{h}^2+1)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{1+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=2+\frac{8}{3}$
$=\frac{14}{3}$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=0,\text{ b}=2,\text{ f(x)}=\text{x}^2+1,\text{ h}=\frac{2-0}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^2_{0}\big(\text{x}^2+1\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(0+1)+(\text{h}^2+1)+\ ....\ +\big\{(\text{n}-1)^2\text{h}^2+1\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\big\{1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[\text{n}+\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}\frac{2}{\text{n}}\Big[\text{n}+\frac{2(\text{n}-1)(2\text{n}-1)}{3\text{n}}\Big]$
$=\lim\limits_{\text{n}\rightarrow\infty}2\bigg\{1+\frac{2}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)\bigg\}$
$=2+\frac{8}{3}$
$=\frac{14}{3}$
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