Question
Evaluate the following integrals as limit of sum:
$\int\limits^{3}_{1}\big(2\text{x}^2+5\text{x}\big)\text{dx}$
$\int\limits^{3}_{1}\big(2\text{x}^2+5\text{x}\big)\text{dx}$
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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, $\text{a}=1,\text{ b}=3,\text{ f(x)}=2\text{x}^2+5\text{x},\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{1}\big(2\text{x}^2+5\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2+5)+\big\{2(1+\text{h})^2+5(1+\text{h})\big\}+\ \\....+\ \big\{2(1+(\text{n}-1)\text{h}^2+5(1+(\text{n}-1)\text{h})\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\Big\{1^2+(1+\text{h}^2)+\ ....+\ \big\{1+(\text{n}-1)\text{h}\big\}^2\Big\}+\\5\big\{1+(1+\text{h})+(1+2\text{h}+\ ....+\ (1+(\text{n}+1)\text{h}))\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+2\text{h}^2(1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+\\4\text{h}\big\{1+2+\ ....+ (\text{n}-1)\big\}+5\text{n}+5\text{h}\big\{1+2+\ ...+\ (\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[7\text{n}+2\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+9\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{2}{\text{n}}\Big[7\text{n}+\frac{4(\text{n}-1)(2\text{n}-1)}{3\text{n}}+9\text{n}-9\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}2\Big[16+\frac{4}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{9}{\text{n}}\Big]$
$=32+\frac{16}{3}$
$=\frac{112}{3}$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=1,\text{ b}=3,\text{ f(x)}=2\text{x}^2+5\text{x},\text{ h}=\frac{3-1}{\text{n}}=\frac{2}{\text{n}}$
Therefore, $\text{I}=\int\limits^{3}_{1}\big(2\text{x}^2+5\text{x}\big)\text{dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(1)+\text{f}(1+\text{h})\ ....\ +\text{f}\big\{1+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[(2+5)+\big\{2(1+\text{h})^2+5(1+\text{h})\big\}+\ \\....+\ \big\{2(1+(\text{n}-1)\text{h}^2+5(1+(\text{n}-1)\text{h})\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\Big\{1^2+(1+\text{h}^2)+\ ....+\ \big\{1+(\text{n}-1)\text{h}\big\}^2\Big\}+\\5\big\{1+(1+\text{h})+(1+2\text{h}+\ ....+\ (1+(\text{n}+1)\text{h}))\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[2\text{n}+2\text{h}^2(1^2+2^2+3^2+\ ....\ +(\text{n}-1)^2+\\4\text{h}\big\{1+2+\ ....+ (\text{n}-1)\big\}+5\text{n}+5\text{h}\big\{1+2+\ ...+\ (\text{n}-1)\big\}\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Big[7\text{n}+2\text{h}^2\frac{\text{n}(\text{n}-1)(2\text{n}-1)}{6}+9\text{h}\frac{\text{n}(\text{n}-1)}{2}\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}\frac{2}{\text{n}}\Big[7\text{n}+\frac{4(\text{n}-1)(2\text{n}-1)}{3\text{n}}+9\text{n}-9\Big]$
$=\lim\limits_{\text{h}\rightarrow\infty}2\Big[16+\frac{4}{3}\Big(1-\frac{1}{\text{n}}\Big)\Big(2-\frac{1}{\text{n}}\Big)-\frac{9}{\text{n}}\Big]$
$=32+\frac{16}{3}$
$=\frac{112}{3}$
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