Originally my remark was intended as a comment on Christian Blatter's answer, but it was too long so I have posted it here as a separate answer.
To put his answer in more easily understood terms for a calculus student, the problem with the way the original question is posed is that it assumes the average is taken uniformly over the $y$-values of the parabolic section corresponding to $x \in [0,2]$. But because the parabola is curved, you can see from the diagram that for "small" $y$, a strip of differential width $dy$ will enclose in some sense "more" points on the parabola than a strip of the same width located higher up on the $y$-axis. Thus, the "average," if taken with respect to measure $dy$, is going to be smaller than if taken with respect to the arc length measure $ds$ of the parabola, which one can argue is the more "natural" measure because it gives equal weight to all points on the parabolic arc, whereas the former clearly does not--it gives less weight to points near the bottom of the arc compared to the top.
As you can see from the animation, $dy$ is represented by the width of the green strip. When the strip is near the bottom of the parabolic arc, it encloses a longer portion of the arc (thicker red segment) than it does when the strip is near the top. So if you integrate the horizontal distance with respect to the height of the bar $y$, you are not giving all of the points on the arc equal weight.
By integrating with respect to the arc length, you guarantee that the differential segment (thick red) in the animation above is the same length along the entire curve, giving each point equal weight in the average distance calculation.