AP Calculus: Range of a Function

Question

Find the range of the function F(x) = Int[-4,x] √(16-t^2)dt

Answer 1

y = √(16-t²) is defined only when -4 ≤ t ≤ 4

 

The graph on the domain [-4,4] is the upper half of the circle centered at (0,0) with radius 4.  

 

F(x) = ∫_{(from -4 to x)} √(16-t²)dt

 

       = area of the portion of the upper half of the circle with center (0,0) and radius 4 that lies between t =            -4 and t = x. 

 

The area of the upper half of the above mentioned circle is (1/2)π(4)² = 8π.

 

So, 0 ≤ F(x) ≤ 8π.

 

That is, the range of F(x) is [0, 8π].