$A$ is on the positive half of $x$-axis and $B$ is on $y=x(x>0)$. $AB=1$. Find the envelope curve of $AB$ and/or the area below the envelope line.
My idea:
$A(a,0)(1\le a\le\sqrt2), B(x,x)(\dfrac{\sqrt2}2\le x\le1)$, so $(x-a)^2+x^2=1,x=\dfrac{a+\sqrt{2-a^2}}2, AB:y=f(x,a)=\dfrac{1+\sqrt{2-a^2}}{1-a^2}(x-a)$.
Envelope curve: $y=g(x)=\max(f,a)$, aka maximum of $f(x,a)$ when $x$ is fixed. Then $1/2+\int_1^{\sqrt2}g(x)\mathrm dx$ is the answer.
However, I have difficulty calculating $g(x)$. I tried triangular commutator, but it didn't work. Who can save me? qwq