这道题就成这个问题

x1 and x2 follows uniform distribution U[0,1]
what is the probability that |x1-x2|>1/3?

case a:
0<=x1<1/3
then as long as x2<=x1+(1/3), |x1-x2|>1/3
so the probability is x1+(1/3)

case b:
1/3<=x1<2/3
then as long as x1-(1/3)<=x2<=x1+(1/3)
so the probability is 2/3

case c:
2/3<=x1<=1
similar to case a
probability is (4/3)-x1

then integrate all the probability, we get 5/9

if we use area method, consider (x1, x2) as a point in square [0，1]^2
then |x1-x2|<1/3 is bounded by two lines, x1-x2<1/3 and x2-x1<1/3

by subtracting the two triangles, the area between the lines is 5/9

done.

\int_{0}^{1}\int_{max(0, x-1/3)}^{min(1, x+1/3)}f(x)*f(y)dydx

华新不支持latex encoding，大家写出来看，_{}表示subscript，^{}表示superscript