Find the area under y=4sin(x) and above y=4cos(x) for π2≤x≤3π2. (Note that this area may not be defined over the entire interval.)
First, we need to find the points where the two curves intersect within the given interval:
4sin(x) = 4cos(x)
tan(x) = 1
x = π/4 or x = 5π/4
Therefore, we have two segments to consider: π/2 ≤ x ≤ π/4 and 5π/4 ≤ x ≤ 3π/2.
For the first segment, we have:
Area = ∫π/44cos(x) dx - ∫π/24sin(x) dx
Area = [4sin(x)]π/4π/2 - [4cos(x)]π/4π/2
Area = 2(2) - 2(√2)
Area = 4 - 2√2
For the second segment, we have:
Area = ∫5π/43π/24sin(x) dx - ∫5π/24cos(x) dx
Area = [4sin(x)]5π/43π/2 - [4cos(x)]5π/43π/2
Area = 2(√2) - 2(2)
Area = -4 + 2√2
Therefore, the total area under y=4sin(x) and above y=4cos(x) for π/2≤x≤3π/2 is:
Area = (4 - 2√2) + (-4 + 2√2)
Area = 0
So the area is zero, which makes sense since the two curves are equal at x = π/4 and x = 5π/4, and the area above and below these points cancels out.