Answer:
[tex]\displaystyle A = \int_0^{\pi/6}\left(\frac{}{}4\sin(x) - 2\frac{}{}\right)dx + \int_{\pi/6}^{\pi/2}\left(\frac{}{}2-4\sin(x)\frac{}{}\right)dx[/tex]
Step-by-step explanation:
We are being asked to first graph the sine waves:
- y = 3 sin(x)
- y = 2 - sin(x)
and highlight the region between them from x = 0 to x = π/2.
To graph these, we can identify the following properties:
- The first equation is an unshifted sine wave with an unmodified period (2π) and amplitude 3.
- The second equation we can rewrite as:
y = -sin(x) + 2
From here, we can see that it is a flipped sine wave (negative amplitude) that is vertically shifted up 2 units.
See the attached image for a graph.
We can set up the integrand(s) for the integral(s) that models the area between the curves over the given interval [0, π/2] by finding the positive difference between the sine functions. From the graph, we can see that:
- over [0, π/6], the first function is larger than the second; therefore we should subtract the second from the first
- over [π/6, π/2], the second function is larger than the first; therefore we should subtract the first from the second
In equation form, this is:
[tex]\displaystyle A = \int_0^{\pi/6}\left(\frac{}{}3\sin(x) - (2-\sin(x))\frac{}{}\right)dx + \int_{\pi/6}^{\pi/2}\left(\frac{}{}2-\sin(x) - 3\sin(x)\frac{}{}\right)dx[/tex]
We can simplify both integrands by combining like terms:
[tex]\boxed{\displaystyle A = \int_0^{\pi/6}\left(\frac{}{}4\sin(x) - 2\frac{}{}\right)dx + \int_{\pi/6}^{\pi/2}\left(\frac{}{}2-4\sin(x)\frac{}{}\right)dx}[/tex]
