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.
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].
To find the range of the integral from 0 to x of sqrt(36-t^2), let's break it down step by step.
First, let's draw the graph of the function sqrt(36-t^2) on the coordinate plane. This function represents the upper half of a circle with a radius of 6 units centered at the origin.
Now, to find the range of the integral from 0 to x, we are looking for the area under the curve bounded by the x-axis and the curve itself within the interval [0, x]. You correctly mentioned that the area of the whole circle is 36π, so dividing it by 2 gives us the area of the upper half-circle, which is 18π.
Since we only want the range from 0 to x, we need to evaluate the integral at x and subtract the area of the portion of the curve below the x-axis.
To find the range, we calculate the integral from 0 to x of sqrt(36-t^2) and subtract it from half the area of the full circle. The result will give us the area of the portion of the curve between 0 and x.
Now, to determine whether it should be [0,18π] or [0,9π], we need to consider the negative values of x. Since the function is symmetric about the y-axis (due to the square root), the range of the integral from 0 to -x will be the same as the range from 0 to x. This means that the answer will always be positive, and we don't need to worry about negative values.
So, the correct answer is [0,18π] because we want the range of the integral from 0 to x, which includes the full upper half of the circle.
To summarize, the range of the integral from 0 to x of sqrt(36-t^2) is [0,18π].