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