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September 22, 2014

September 22, 2014

Posted by **Anonymous** on Friday, January 25, 2013 at 2:41am.

2. Find the area of the region between the curves y=sin(x), y=sin(2x), x=0, and x=pi/2.

- Calculus -
**Reiny**, Friday, January 25, 2013 at 11:42amI will do the harder of the two

2.

If you make a sketch you will see that the curves intersect in your domain 0 ≤ x ≤ π/2

sin2x = sinx

2sinxcosx - sinx = 0

sinx(2cosx - 1) = 0

sinx = 0 or cosx = 1/2

x = 0 , your left domain, or

x = π/3

so we have to do this in two parts

area = ∫(sin2x - sinx) dx from 0 to π/3 + ∫(sinx - sin2x) dx from π/3 to π/2

= [(-1/2)cos2x + cosx] form 0 to π/3 + [-cosx + 1/2)cos2x ] from π/3 to π/2

= (1/4 + 1/2 - (-1/2) + 1)) + (0 + 0 -(-1/2 +(1/2)(-1/2))

= .... you do the arithmetic

and please check my arithmetic above, should have written it out on paper first.

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