Find the area of the region of

y= cos x and y=sin2x, x= 0, x=pi/2

i sketched the curve and y=cos x is on top, so wouldn't it be

\int cosx-sin2x from pi/2 to 0?

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- May 22nd 2011, 03:23 PMdorkymichelleFinding area between curves?
Find the area of the region of

y= cos x and y=sin2x, x= 0, x=pi/2

i sketched the curve and y=cos x is on top, so wouldn't it be

\int cosx-sin2x from pi/2 to 0? - May 22nd 2011, 03:43 PMMondreus
Are you sure that $\displaystyle \cos x \geq \sin{2x}$ on $\displaystyle x\in[0,\pi/2]$?

What about $\displaystyle x=\pi/4$? You need to break up the integral into intervals where one function is greater than or equal to the other. - May 22nd 2011, 03:48 PMdorkymichelle
so would I have to split it into 2 integrals, one where cos x is bigger than sin2x, and the other where sin2x is bigger than cos x?

- May 22nd 2011, 03:51 PMMondreus
Yes. In order to do that you need to solve $\displaystyle \cos x = \sin{2x}\Leftrightarrow \cos x = 2\cos x \sin x$