Posted by Anonymous on Friday, January 25, 2013 at 2:41am.
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.
Related Questions
Calculus - 1. Find the area of the region between the curves y=sin(x pi/2) and y...
Calculus (Area Between Curves) - Find the area of the region IN THE FIRST ...
Calculus (Area Between Curves) - Find the area of the region bounded by the ...
Calculus (Area Between Curves) - Find the area of the region IN THE FIRST ...
calculus - Sketch the region enclosed by the given curves. Decide whether to ...
Calculus - Find the area of the region enclosed by the given curves: y=e^6x, y=...
Calculus II - Find the area of the region between y = x sin x and y = x for 0 &...
Calculus AB - Let R be the region bounded by the graphs of y=sin(pi x) and y=(x^...
calculus - Sketch the region bounded by the curves y = x^2, y = x^4. 1) Find ...
Calculus (Area Between Curves) - Find the area of the region bounded by the ...
For Further Reading
