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].
Hello!
Answer:
36.225
Step-by-step explanation:
Hope this helps!
What do u want me to solve there is no photo
Answer:
x=(7,0),(-1,0) y=(0,-7)
Step-by-step explanation:
Recall that the area under a curve and above the x axis can be computed by the definite integral. If we have two curves
<span> y = f(x) and y = g(x)</span>
such that
<span> f(x) > g(x)
</span>
then the area between them bounded by the horizontal lines x = a and x = b is
To remember this formula we write
