Find the range of the function f of x equals the integral from 0 to x of the square root of the quantity 36 minus t squared dt

Question

 

Find the range of the function  .

A)[-6, 6]

B)[0, 6]

C)[0, 9π]

D)[0, 18π]

Answer 1

The graph of y = √(36 - t²) is the upper semicircle with center (0,0) and radius 6, where -6 ≤ t ≤ 6.

So, ∫_{(from 0 to x)} √(36 - t²)dt is the area of the portion of the right half of the semicircle (defined above) that lies between t = 0 and t = x. When x = 0, the value of the integral is also 0. When x = 6, the value of the integral is the area of the quarter circle, which is 36π/4 = 9π. So, the range is [0, 9π].