Distance between Random Points on the X and Y Axes

Suppose that point $A$ is given by $(0,y)$ and point $B$ is given by $(x,0)$ , where $0≤y≤1$ and $0≤x≤1$ .

Now, the square of the distance between them is $x^2+y^2$ .

Average all possible values of $x^2+y^2$ in $0≤y≤1$ and $0≤x≤1$ using limits and sigma notation: $\displaystyle\lim_{N\to\infty}\frac{1}{N^2}\sum_{a=0}^{N}\sum_{b=0}^{N}\frac{a^2}{N^2}+\frac{b^2}{N^2}$ .

Rewrite using integral notation by substituting $x=\frac{a}{N}$ , $y=\frac{b}{N}$ , $dx=dy=\frac{1}{N}$ : $\displaystyle\int_{0}^{1}\int_{0}^{1}(x^2+y^2)\ dx\ dy$

This double integral can be integrated through iterated integration: $=\displaystyle\int_{0}^{1}[\frac{x^3}{3}+xy^2]_{x=0}^{x=1}\ dy$ $=\displaystyle\int_{0}^{1}(\frac{1}{3}+y^2)\ dy$ $=[(\frac{y}{3}+\frac{y^3}{3})]_{y=0}^{y=1}$ $=\frac{2}{3}$

Thus, the answer is $3\cdot2+4\cdot3=18$ .

We are interested in computing $E[x^2+y^2]$ , and since $x, y$ are independent of each other we have $E[x^2+y^2] = E[x^2] + E[y^2]$ . This leads us to the following integration:

$E[x^2+y^2] = E[x^2] + E[y^2] = \int_{0}^{1} x^2 \cdot (\frac{1}{1-0}) dx + \int_{0}^{1} y^2 \cdot (\frac{1}{1-0}) dy = \frac{x^3}{3}|_{0}^{1} + \frac{y^3}{3}|_{0}^{1} = \frac{2}{3} = \frac{p}{q}$

and $3p+4q = 3(2) + 4(3) = \boxed{18}.$