 Solution

ID: 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/solution

Assume

h

is continuously differentiable on the interval concerned. For fixed

x > 0

, take

u = h (t)

and

d v = (x - t)^{- 1/2} d t

, so

v = - 2 x - t

. Integration by parts, interpreted first with an endpoint cutoff and then by a limit, gives

g (x) = [- 2 h (t) x - t]_{t = 0}^{t = x} + 2 \int_{0}^{x} h^{'} (t) x - t d t = 2 h (0) x + 2 \int_{0}^{x} h^{'} (t) x - t d t .

The upper endpoint term vanishes, and the lower endpoint supplies the explicit square-root term.

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