If $a$ is picked randomly in the range $(\frac{1}{4}, \frac{3}{4})$ and $b$ is chosen such that $$\int_{a}^{b} \frac{1}{x^2}dx = 1,$$ compute the expected value of $(b − a)$.
https://www.stanfordmathtournament.com/pdfs/smt2020/team-solutions.pdf
I have soved it differently given as an answer.
let me know if there is any flaw in deriving like you see in my answer.
I have solved it differently.
$\int_{a}^{b} \frac{1}{x^2}dx = 1$
$[\frac{1}{a} - \frac{1}{b}] = 1$
$b = \frac{a}{1-a}$
pdf of a => is 2
pdf of b is derived as below using CDF method:
$P_B(B\leq b) => P(\frac{a}{1-a}\leq b)$
$P(a \leq \frac{b}{1+b})$
$ =\frac{\frac{b}{1+b}-\frac{1}{4}}{{\frac{3}{4}} - \frac{1}{4}}$
$P(B\leq b)=2\frac{b}{1+b} - \frac{1}{2}$
Now differentiate to get the pdf
interval for b transforms to $(\frac{1}{3}, 3)$
Thus $P(B=b) = \frac{2}{(1+b)^2}$
$E(B) = \int_{\frac{1}{3}}^{3} \frac{2b}{(1+b)^2}$
Put $u = (1+b)$ => $db = du$ and $b = u-1$
Now $E(b) = 2 \int_{\frac{4}{3}}^{4} (\frac{(u-1)}{u^2}du$
$E(b) = 2 \int_{\frac{4}{3}}^{4} \frac{1}{u}du - 2 \int_{\frac{4}{3}}^{4}\frac{1}{u^2}du$
$E(b) = 2\ln(3) - 1$
$ = \ln(9) - 1$
$E(a) = \frac{\frac{1}{4}+\frac{3}{4}}{2} = \frac{1}{2}$
Thus $E(b - a) = E(b) - E(a)$
$E(b-a) = \ln(9) - 1 - \frac{1}{2}$
$E(b-a) = \ln(9) - \frac{3}{2}$