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].
6 penalties/8 penalties
16 min/21.3
Answer:
The value is 9 tenths :)
Step-by-step explanation:
