Find the range of the function F(x) = the integral from 0 to x of the square root of 4-t^2 dt

Note that √(4 - t²) is defined only as long as 4 - t² ≥ 0, or -2 ≤ t ≤ 2. Then the real integral exists only if -2 ≤ x ≤ 2. (Otherwise we deal with complex numbers.)

If x = 2, then the integral corresponds to the area of a quarter-circle with radius 2. This means that the integral has a maximum value of 1/4 • π • 2² = π.

On the opposite end, if x = -2, then the integral has the same value, but the integral from 0 to -2 is equal to the negative integral from -2 to 0. So the minimum value is -π.

For all x in between, we observe that the integrand is continuous over the rest of its domain, so F(x) is continuous.

Then the range of F(x) is the interval [-π, π].

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step-by-step explanation:

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