Originally Posted by

**SlipEternal** Let's find out. First, let's figure out where it crosses the $x$-axis:

$0 = 4x-x^3 = x(4-x^2) = x(2+x)(2-x)$

So, it crosses the axis at $x=-2, x=0, x=2$. We only care about positive $x$-values, and $y(1) >0, y(3)<0$. Hence it is in the 1st quadrant only when $0 < x < 2$. So, we use the average value formula:

$\displaystyle \begin{align*}y_{\text{avg}} & = \dfrac{1}{b-a}\int_a^b ydx \\ & = \dfrac{1}{2-0}\int_0^2(4x-x^3)dx \\ & = \dfrac{1}{2}\left[2x^2-\dfrac{x^4}{4}\right]_0^2 \\ & = 2\end{align*}$

So, the average height is 2, and 8/4 = 2, so that is the correct answer.