Find the area under the curves

[tennisgirl]

Hello everyone,

I have really been struggling with the following problem. Your help is really appreciated.
Thank you

(Problem)

Find the area of the region between the curves:

y

=​

sin​

(x)​

​

y

=​

sin​

(2x)​

​

x

=​

0​

​

and

x

=​

2

​

​

​

 

[galactus]

From \(\displaystyle [0,\frac{\pi}{2}]\), the curves intersect at \(\displaystyle \frac{\pi}{3}\)

So, it may help to see it if you break it up:

\(\displaystyle \displaystyle \int_{0}^{\frac{\pi}{3}}[sin(2x)-sin(x)]dx+\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}[sin(x)-sin(2x)]dx\)

Graph them and you can see where the regions lie. This is one way to look at it.