Question
Evaluate the following integrals as limit of sum:
$\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
$\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
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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, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\cos\text{x},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\cos(\text{a})+\cos(\text{a}+\text{h})+\ ....+\ \cos\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\cos\big\{\text{a}+(\text{n}-1)\frac{\text{h}}{2}\big\}\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}2\cos\Big(\text{a}+\frac{\text{b}-\text{a}}{2}-\frac{\text{h}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)\Bigg]$ (Using nh = b - a)
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\cos\Big(\frac{\text{a}+\text{b}}{2}-\frac{\text{h}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)$
$=2\cos\Big(\frac{\text{a}+\text{b}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)$
$=\sin\text{b}-\sin\text{a}$ $\Big[\text{Since},2\cos\text{A}\sin\text{B}=\sin(\text{A}+\text{B})-\sin(\text{A}-\text{B})\Big]$
Where, $\text{h}=\frac{\text{b}-\text{a}}{\text{n}}$
Here, $\text{a}=\text{a},\text{ b}=\text{b},\text{ f(x)}=\cos\text{x},\text{ h}=\frac{\text{b}-\text{a}}{\text{n}}$
Therefore, $\text{I}=\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\text{f}(\text{a})+\text{f}(\text{a}+\text{h})+\ ....\ +\text{f}\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\big[\cos(\text{a})+\cos(\text{a}+\text{h})+\ ....+\ \cos\big\{\text{a}+(\text{n}-1)\text{h}\big\}\big]$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}\Bigg[\frac{\cos\big\{\text{a}+(\text{n}-1)\frac{\text{h}}{2}\big\}\sin\frac{\text{nh}}{2}}{\sin\frac{\text{h}}{2}}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\Bigg[\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}2\cos\Big(\text{a}+\frac{\text{b}-\text{a}}{2}-\frac{\text{h}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)\Bigg]$ (Using nh = b - a)
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{\text{h}}{2}}{\sin\frac{\text{h}}{2}}\times\lim\limits_{\text{h}\rightarrow0}2\cos\Big(\frac{\text{a}+\text{b}}{2}-\frac{\text{h}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)$
$=2\cos\Big(\frac{\text{a}+\text{b}}{2}\Big)\sin\Big(\frac{\text{b}-\text{a}}{2}\Big)$
$=\sin\text{b}-\sin\text{a}$ $\Big[\text{Since},2\cos\text{A}\sin\text{B}=\sin(\text{A}+\text{B})-\sin(\text{A}-\text{B})\Big]$
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