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enter image description here

I know that the answer is C. But how exactly would you go about solving this problem? Is there a specific formula I should be using? Because I can't seem to find any related ones.

[EDIT] Okay. I just found an explanation for the problem. But it still doesn't make sense how they got (x,x^2) and (sqrt(y),y) and why they changed the interval from [0,2] to [0,4]... Can someone please explain what they did in a clearer way?

enter image description here

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  • $\begingroup$ how do you know the answer is C? $\endgroup$ – JonMark Perry Nov 26 '15 at 7:44
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    $\begingroup$ The ordinary interpretation of average distance would give answer $1$. $\endgroup$ – André Nicolas Nov 26 '15 at 7:49
  • $\begingroup$ @AndréNicolas: See my answer for another interpretation $\ldots$ $\endgroup$ – Christian Blatter Nov 26 '15 at 9:06
  • $\begingroup$ It’s becoming clear from the comments that this problem is ill-posed without more context. What does it mean by “average distance.” $\endgroup$ – amd Nov 26 '15 at 9:43
  • $\begingroup$ I've added another picture. Maybe it'll help? $\endgroup$ – Ordinary Owl Nov 26 '15 at 18:56
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In order to compute an average of a function over some set $S$ we have to have a measure on $S$. The natural measure on the graph in question is not ${\rm d}y$ (as suggested by the source), but arc length measure ${\rm d}s$. It follows that the average distance $\bar D$ in question is given by $$\bar D={\int_0^2(2-x)\sqrt{1+4x^2}\>dx\over\int_0^2\sqrt{1+4x^2}\>dx}={1 + 7 \sqrt{17} + 6 \>{\rm arsinh}\> 4\over 12 \sqrt{17} + 3 \>{\rm arsinh}\> 4}$$ (as computed by Mathematica), with a numerical value $0.760921$.

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  • $\begingroup$ this result matches this simulation: Mean[EuclideanDistance[#, {2, #[[2]]}] & /@ RandomPoint[ImplicitRegion[x^2 == y && 0 <= x <= 2, {x, y}], 10^6]] $\endgroup$ – martin Nov 26 '15 at 9:14
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    $\begingroup$ In other words, none of the above ;) $\endgroup$ – amd Nov 26 '15 at 9:41
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The answer is C. This is because we can see the lines from $x=2$ to the curve $y=x^2$ as horizontal lines starting from $x=2$ and ending at $y=x^2$.

So the total sum of these lines becomes the area under the curve, namely $\frac83$. Our range however is NOT the difference between the limits used to calculate the integral, which would be $2$, but $4$ because we divide by the range of the function, namely $2^2-0^2=4$.

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  • $\begingroup$ Okay. I understand how you got 4. But how does it become 4/3? $\endgroup$ – Ordinary Owl Nov 26 '15 at 18:57
  • $\begingroup$ @OrdinaryOwl you divide the total area, i.e., the sum of the lines, $\frac83$ by the length of the interval, $4$. $\endgroup$ – amd Nov 26 '15 at 20:03
  • $\begingroup$ the integral is horizontal not vertical, but the area under the graph doesn't change so you can use the original one, but you need to change the interval. it's as @amd; says $\endgroup$ – JonMark Perry Nov 26 '15 at 20:14
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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.

enter image description here

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.

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  • $\begingroup$ The easiest way to see the difference is to consider any horizontal segment (on which $dy=0$). $\endgroup$ – A.S. Nov 26 '15 at 20:02

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