Help with Calc please!

Question

Find the range of the function .

Options:

1)	[-6, 6]
2)	[0, 6]
3)	[0, 9π]
4)	[0, 18π]

Answer 1

f(x) = ∫₀^xsqrt(36 - t²)dt is area of curve trapezoid bounded by x = 0, x = x and arks of circle x = sqrt(1 - t²).

Minimum of this area is 0(when x = 0) and maximum is area of right quoter(in first quadrant of circle x² + t² = 6²., anothr words, f_max = π·6²/4 = 9π. So range of this function is [0, 9π].

Answer: 3)

Answer 2

the general formula for the definite integral of square root of a^2 - x^2 is

(x/2)sqr(a^2-x^2) + (a/2)sin^-1(x/a)

evaluate it at the lower limit x=0 and a=6, f(x) = 0

that eliminates 1) which doesn't have 0 for the lower limit

maybe the best or most intuitive way to see what's going on is

graph it. f(x) is a semicircle, the top half of a circle centered at the origin

the integral is the area from x=0 to x=x where x varies from 0 to 6

and y varies from 0 to 6, but the area varies up to the area of a full

semi-circle pi(r)^2/2 where r=6

so maximum range value is pi(36)/2 = 18pi

4) is correct