Lies Beneath
Find the area of a region that lies beneath the given curve. PLEASE HELP!?
there are two problems i need help with:
1.Find the area of the region bounded by the curves.
y=cosx, y=sin2x, x=0, x= (pie)/2
2. Find the area of the region that lies beneath the given curve Please help, I have no idea how to find the area of these problems and need help.
1. 2. Breaking Benjamin – What Lies Beneath (Lyrics on screen)
y=sinx, 0 is less than or equal to x, and x is less than or equal to (pie)/3; 0
For the first the 2 curves intersect between x = 0 and x = π/2, so you’ll have to split the integral to find the area between them. They intersect at π/6 so that’s where the integral needs to be split and the functions switch. For the first region cos(x) is greater and for the second sin(2x) is greater.
∫[0,π/6] (cos(x) – sin(2x)) dx + ∫[π/6,π/2] (sin(2x) – cos(x)) dx
(sin(x) + (1/2)cos(2x))|[0,π/6] + (-(1/2)cos(2x) – sin(x))|[π/6,π/2]
[(1/2) + (1/2)(1/2) - (0 + (1/2)(1))] + [(-1/2)(-1) - 1 - ((-1/2)(1/2) - (1/2))]
[1/4] + [(1/2) - 1 + (1/4) + (1/2)]
(1/2) + 1 – 1
1/2
…††…†……†…††…†……†…††…†……†…††…†……†…††…†……†…††…†…†
For this one just plug everything into the integral. They’ve given you the function and the limits. This is very trivial and I think you can handle it.
A = ∫[0,π/3] sin(x) dx