Mathematics, 09.01.2020 22:31 ironclone13
Find the area of the following region y=sin x and y=sin 2x from x=0 x=pi
Answers: 1
![\bf \begin{cases} y=sin(x)\\ y=sin(2x)\\ \textendash\textendash\textendash\textendash\textendash\\ [0,\pi ] \end{cases} \\ \quad \\ \textit{at which points, do they intersect each other? well}\\ ----------------------------\\ sin(x)=sin(2x)\implies sin(x)-sin(2x)=0 \\ \quad \\ sin(x)-2sin(x)cos(x)=0\impliedby \textit{common factor} \\ \quad \\ sin(x)[1=2cos(x)]=0\\ \quad \\ \begin{cases} sin(x)=0\to &x=0\\ 1-2cos(x)=0\to \frac{1}{2}=cos(x)\to &\frac{\pi }{3}=x \end{cases}\\ ----------------------------\\ ](/tpl/images/0448/8852/cd37e.png)
![\bf \textit{notice the [ceiling] function, on the 1st interval, is sin(2x), thus} \\ \quad \\ \int\limits_{0}^{\frac{\pi }{3}}([sin(2x)]-[sin(x)])dx\implies \left[ \cfrac{-cos(2x)}{2}-[-cos(x)] \right]_{0}^{\frac{\pi }{3}}\\ ----------------------------\\ \textit{notice the [ceiling] function, on the 2nd interval, is sin(x), thus} \\ \quad \\ \int\limits_{\frac{\pi }{3}}^{\pi }([sin(x)]-[sin(2x)])dx\implies \left[ -cos(x)-\left( \cfrac{-cos(2x)}{2} \right) \right]_{\frac{\pi }{3}}^\pi](/tpl/images/0448/8852/988f5.png)
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Find the area of the following region y=sin x and y=sin 2x from x=0 x=pi...
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