Answer :
[tex]f(x)=\displaystyle\int_{-6}^x\sqrt{36-t^2}\,\mathrm dt[/tex]
The integrand is defined for [tex]36-t^2\ge0[/tex], or [tex]-6\le t\le6[/tex], so the domain should be the same, [tex]-6\le x\le6[/tex].
When [tex]x=-6[/tex], the integral is 0.
The integrand is non-negative for all [tex]x[/tex] in the domain, which means the value of [tex]f(x)[/tex] increases monotonically over this domain. When [tex]x=6[/tex], the integral gives the area of the semicircle centered at the origin with radius 6, which is [tex]\dfrac\pi26^2=18\pi[/tex], so the range is [tex]\boxed{0\le f(x)\le 18\pi}[/tex].