Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 10 / b / ii

Assume $h$ is continuously differentiable on the interval concerned. For fixed $x>0$, take $u=h(t)$ and $dv=(x-t{)}^{-1/2}dt$, so $v=-2\sqrt{x-t}$. Integration by parts, interpreted first with an endpoint cutoff and then by a limit, givesThe upper endpoint term vanishes, and the lower endpoint supplies the explicit square-root term.