A stick having a length of $1$ is broken randomly at a point. Expectation of length of smaller part.
Source: brainstellar.com
My approach: Consider the stick is broken at $x$, then expectation of length of smaller part = $x*P(x<1/2) + (1-x)*P(x>1/2)$ = $x/2$ + $(1-x)/2$ = $1/2$. However, answer given is $1/4$.
Actually, this is not a duplicate question, because the OP is asking where he went wrong.
In the 1/2 the time that $~x < (1/2),~$ you will need to take the average value of $~x~$ in this interval. The average value of $~x~$ will be $~(1/4),~$ as indicated by the formula
$$\frac{\int_0^{1/2} x ~dx}{1/2 - 0} = \frac{x^2/2 ~|_{x=0}^{x=1/2}}{1/2} = 2 \times \frac{1/4}{2} = 1/4.$$
That is, the average value of an integrable function $~f(x),~$ from $~a~$ to $~b,~$ where $~a < b~$ is
$$\frac{\int_a^b f(x) ~dx}{b-a}.$$