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].

p   …   q   …   ¬q   …   p ∨ ¬q   …   (p ∨ ¬q) ⇒ q

T   …   T   …    F   …        T        …          T

T   …   F   …    T   …        T        …          F

F   …   T   …    F   …        F        …          T

F   …   F   …    T   …        T        …          F

Start with the first two columns, taking every possible pair of True/False for p and q.

¬q is just the negation of q, so True becomes False and False becomes True.

p ∨ q is the logical disjunction, or logical "or". It's True if either p or q is True, and False otherwise. So p ∨ ¬q is True only if either p or ¬q is True.

p ⇒ q is the logical implication. It's True only when both p and q are True, or when p is False. So (p ∨ ¬q) ⇒ q is True when both p ∨ ¬q and q are True, or when p ∨ ¬q is False.