If f(x) = ∫[0,x] √(4-t^2) dt
f(x) = 1/2 √(4-t^2) + 2 arcsin(x/2)
clearly, the domain is [-2,2]
f(-2) = -π
f(2) = π
Those choices don't seem to work. Better check my math.
Find the range of the function ∫sqrt(4-t^2) dt where a=0 =x
[0, 4π]
[0, 2π]
[-4, 0]
[0, 4]
1 answer