Answer
$2\sqrt {3}$
Work Step by Step
Find the values of c that are guaranteed by the mean value theorem
$\int ^6_0 \frac{x^2}{4} dx=\frac{x^2}{4}(6-0)$, Integrate
$[\frac{1}{12}x^3]^6_0 =\frac{3}{2}x^2$
$(18-0) = \frac{3}{2}x^2$
$12=x^2$
$2\sqrt 3= x$
$-2\sqrt 3$ is not on the interval.
