 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/undated/prelims/mathematics-iii-resit/2/c/solution

L = \int 1 + (y^{'})^{2} d x

. Since

y^{'} = 2 x

L = \int_{0}^{1} 1 + 4 x^{2} d x .

2 x = sinh t

: then

d x = \frac{1}{2} cosh t d t

and

1 + 4 x^{2} = cosh t

cosh^{2} t = (1 + cosh 2 t) /2

yields

L = [\frac{x}{2} 1 + 4 x^{2} + \frac{1}{4} arsinh (2 x)]_{0}^{1} = \frac{5}{2} + \frac{1}{4} ln (2 + 5) .

Here

arsinh t = ln (t + 1 + t^{2})

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