I assume you meant
f(x) = ∫[0,x] √(36-t^2) dt
You do know that f(x) is only defined on the interval [-6,6].
However. the integral if from 0 to x, so the maximum area is only a quarter circle.
That means that
∫[0,0] √(36-t^2) dt = 0
∫[0,6] √(36-t^2) dt = 9π
Thus, the range is [0,0π].
How do I find the range of the integral from 0 to x of sqrt(36-t^x)?
I know you use the formula for the area of a circle to get the area under the curve so A=pi(6)^2. The area of the whole circle would be 36pi, but since it's a semicircle you divide it in half to 18pi. Do I divide it in half again since it only wants the range from 0 to x?
[0,18pi] and [0,9pi] are the answers I'm stuck between.
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