Question

Find the area of the following region y=sin x and y=sin 2x from x=0 x=pi

Answer 1

if you graph the function, as you should first
you'll notice, there are two intervals with areas in them

where do they intersect? well, at sin(x)=sin(2x)

so.. you use the 1st interval, keeping in mind which function
is the [ceiling] and which one is the [floor]

and you do the same for the 2nd interval
so
`\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 (1)/(2)=cos(x)\to &(\pi )/(3)=x \end{cases}\\ ----------------------------\\`

`\bf \textit{notice the [ceiling] function, on the 1st interval, is sin(2x), thus} \\ \quad \\ \int\limits_(0)^{(\pi )/(3)}([sin(2x)]-[sin(x)])dx\implies \left[ \cfrac{-cos(2x)}{2}-[-cos(x)] \right]_(0)^{(\pi )/(3)}\\ ----------------------------\\ \textit{notice the [ceiling] function, on the 2nd interval, is sin(x), thus} \\ \quad \\ \int\limits_{(\pi )/(3)}^(\pi )([sin(x)]-[sin(2x)])dx\implies \left[ -cos(x)-\left( \cfrac{-cos(2x)}{2} \right) \right]_{(\pi )/(3)}^\pi`