Sure.

BTW, looking again at my Mathematica session, there's still a small glitch. I told Mathematica that [$]y_1 < y_2[$]. But, indeed since [$]\dot{\xi} < 0[$] for this problem, actually I should have said to Mathematica that [$]y_2 < y_1[$]. That makes both the l.h.s. and r.h.s. positive for [$]t>0[$] (since [$]b(x)<0[$]). However, after making this change in the Assumptions, and rerunning: Mathematica gets the same expression for integral[y1,y2]. In other words, the answer is still the same, confirming Paul's answer.