Asked by Henry

Find the range of the function ∫sqrt(4-t^2) dt where a=0 =x

[0, 4π]
[0, 2π]
[-4, 0]
[0, 4]

Answers

Answered by Steve
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.
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