Find the range of the function ∫sqrt(4-t^2) dt where a=0 =x
[0, 4π]
[0, 2π]
[-4, 0]
[0, 4]
If f(x) = ∫[0,x] √(4-t^2) dt
f(x) = 1/2 √(4-t^2) + 2 arcsin(x/2)
clearly, the domain is [-2,2]
f(-2) = -π
f(2) = π
Those choices don't seem to work. Better check my math.
To find the range of the function ∫sqrt(4-t^2) dt, we need to evaluate the definite integral over the given interval [a, x].
First, let's find the antiderivative of the function sqrt(4-t^2). The function inside the square root is in the form of a squared term subtracted from a constant squared. This suggests that the integral involves trigonometric functions.
By using the trigonometric substitution t = 2sin(u), where -π/2 ≤ u ≤ π/2, we can simplify the problem. Taking the derivative of t = 2sin(u), we get dt = 2cos(u) du. Substituting this into the integral, we have:
∫sqrt(4-t^2) dt = ∫sqrt(4-(2sin(u))^2) * 2cos(u) du = ∫2cos^2(u) du.
Now, using the double angle identity for cosine, cos^2(u) = (1 + cos(2u))/2, we can rewrite the integral as:
∫2cos^2(u) du = ∫(1 + cos(2u)) du = u + (sin(2u))/2 + C,
where C is the constant of integration.
Now, to find the range of this function over the interval [a, x], we substitute the values of a=0 and x into the antiderivative:
f(x) = x + (sin(2x))/2 + C,
f(0) = 0 + (sin(2*0))/2 + C = 0 + (sin(0))/2 + C = 0 + 0/2 + C = C,
f(x) = x + (sin(2x))/2 + C.
The range of the function ∫sqrt(4-t^2) dt over the interval [0, x] is now dependent on the value of C, the constant of integration. Since C can be any constant, the range of the function is:
[-∞, +∞].
Therefore, none of the given options [0, 4π], [0, 2π], [-4, 0], or [0, 4] are the correct ranges for the function.