Final answer:
To find the range of the given function, we evaluate the integral of sqrt(16-t^2) from -4 to x. The range is given by the antiderivative of the function evaluated at x.
Step-by-step explanation:
To find the range of the given function, we need to evaluate the integral from -4 to x of sqrt(16-t^2) dt. Let's start by integrating the function:
∫ sqrt(16-t^2) dt
To evaluate this integral, we can use the trigonometric substitution t = 4sin(theta). Then, dt = 4cos(theta) d(theta). The integral becomes:
∫ 4cos^2(theta) d(theta)
We can simplify the integral using the identity cos^2(theta) = (1 + cos(2theta))/2. The integral becomes:
∫ 4(1 + cos(2theta))/2 d(theta)
Now we can integrate each term separately:
= 2∫ (1 + cos(2theta)) d(theta)
Simplifying further:
= 2(theta + (1/2)sin(2theta)) + C
Substituting back:
= 2arcsin(t/4) + (1/2)tsqrt(16-t^2) + C
This is the antiderivative of the function. To find the range, we evaluate the antiderivative at x:
Range = 2arcsin(x/4) + (1/2)xsqrt(16-x^2) + C