Answer
$c=\frac{3}{\sqrt[3] 4}$
Work Step by Step
Find the value of c guaranteed by the Mean Value Theorem
$f(x)=x^3$, [0,3]
Set up in the form
$f(c)(b-a) = \int^b_a f(x)dx$
$3x^3=\int_0^3 x^3dx$, Integrate
$3x^3= [\frac{1}{4}x^4]_0^3$
$3x^3=\frac{81}{4}$
$x^3=\frac{27}{4}$
$c=\frac{3}{\sqrt[3] 4}$
