Consider a $1$ meter stick and suppose you break it into two pieces $X$ meters from the end, where $X \sim \operatorname{Unif}(0,1)$. What is the expected length of the longer piece (after it is broken)?
So the integral would be from $x$ to $1$, but what would be the integrand?
Thanks so much.
On
You can do it by integration.
$\begin{align} E[\max\{X, 1-X\}] & = \int_0^{1}\max\{x, 1-x\}\operatorname d x \\ & = \int_0^{1/2} 1-x\operatorname d x + \int_{1/2}^1 x\operatorname d x \\ & = \big[x-x^2/2\big]_{x=0}^{x=1/2} + \big[x^2/2\big]_{x=1/2}^{x=1} \\ & = 3/4 \end{align}$
Alternatively: The break occurs uniformly, half the time it will be nearer one end and half the time it will be nearer the other, so we can say that the break results in the longer piece's length having a uniform distribution between $(1/2..1)$, and thus the mean longer-piece-length is $3/4$ (as expected).
The length of the longer piece is $\max\{X,1-X\}$. So your expectation would be $\int_0^1 \max\{x,1-x\}\,dx$.