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].
Answer:8x+11
Step-by-step explanation:2 x+ 6x=8x +11
add 2x to 6x and add 11 to it .
Answer:
No
Step-by-step explanation:
VX = 4 spaces
UW = 3 spaces
I’m pretty sure that in slope intercept form it will be y-10=(0/1)(x-8)
