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