Expected distance of a dart to the center of board I have a dart board of radius R. You hit a dart. What is the expected distance of your hit with respect to the center of the board?
Attempt:
Let $x$= distance away from the center. 
$E[X]=\int_{x=0}^{R}1-P(X\leq x) dx$
And $P(X\leq x)=x^2\pi/(\pi R^2)$
Do you guys think my approach is correct?
 A: Your approach is correct, although I would suggest using
the notation more carefully. In particular, if you are taking the
integral of something with multiple terms, put parentheses around
the expression to more clearly indicate you mean the integral of the
entire expression:
$$E[X]=\int_0^R (1-P(X\leq x)) \,dx.$$
Note that it is not necessary to write $x=0$ at the lower end of
the integral sign; the symbol $dx$ tells us that the integration is over
the variable $x.$
Comparing  your formula with this other answer,
keeping in mind that $F_X(x)$ in that other answer is the
same as your $P(X\leq x),$ we see that your formula is almost the same.
In fact, a standard formula for the expectation of
a random variable, when the variable takes only positive values, is
$$E[X]=\int_0^\infty (1-P(X\leq x)) \,dx.$$
In the problem you set out to solve,
$1 - P(X \leq x) = 0$ whenever $x \geq R,$ so
$$\int_R^\infty (1-P(X\leq x)) \,dx = 0$$
and therefore
$$\int_0^\infty (1-P(X\leq x)) \,dx
=\int_0^R  (1-P(X\leq x)) \,dx.$$
So you are justified in taking the integral only from $0$ to $R.$
