Question

Find the range of the function f of x equals the integral from negative 6 to x of the square root of the quantity 36 minus t squared

Answer 1

$f(x)=\displaystyle\int_{-6}^x\sqrt{36-t^2}\,\mathrm dt$

The integrand is defined for $36-t^2\ge0$, or $-6\le t\le6$, so the domain should be the same, $-6\le x\le6$.

When $x=-6$, the integral is 0.

The integrand is non-negative for all $x$ in the domain, which means the value of $f(x)$ increases monotonically over this domain. When $x=6$, the integral gives the area of the semicircle centered at the origin with radius 6, which is $\dfrac\pi26^2=18\pi$, so the range is $\boxed{0\le f(x)\le 18\pi}$.