Evaluate the following integrals as limit of sum: ^ 2 _ 0 dx - Vidyadip

Question

Evaluate the following integrals as limit of sum:
$\int\limits^{\frac{\pi}{2}}_{0}\cos\text{x dx}$

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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, $\text{a}=0,\text{ b}=\frac{\pi}{2},\text{ f(x)}=\cos\text{x},\text{ h}=\frac{\frac{\pi}{2}-0}{\text{n}}=\frac{\pi}{2\text{n}}$
Therefore, $\text{I}=\int\limits^{\frac{\pi}{2}}_{0}\cos\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(0)+\text{f}(0+\text{h})+\ ....\ +\text{f}\big(0+(\text{n}-1)\text{h}\big)\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\cos0+\cos\text{h}+\cos2\text{h}+\ ....+\ \cos(\text{n}-1)\text{h}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\cos\big((\text{n}-1)\frac{\text{h}}{2}\big)\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\cos\big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}}{\sin\frac{\text{h}}{2}}\Bigg]$ $\Big(\text{Using, nh}=\frac{\pi}{2}\Big)$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times2\cos\Big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\cos\Big(\frac{\pi}{4}-\frac{\text{h}}{2}\Big)\sin\frac{\pi}{4}$
$=2\cos\frac{\pi}{4}\sin\frac{\pi}{4}$
$=2\times\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$
$=1$

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