 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/discrete-mathematics-continuous-mathematics-and-probability/5/d/solution

For

h = 1/ N

T_{N} (f) = \frac{1}{N} (\frac{f ( 0 )}{2} + \sum_{m = 1}^{N - 1} f (m / N) + \frac{f ( 1 )}{2}) .

With

f (x) = 1/ (4 - x^{2})

this becomes

T_{N} = \frac{7}{24 N} + \sum_{m = 1}^{N - 1} \frac{N}{4 N ^{2} - m ^{2}} .

Write the requested right Riemann sum as

S_{N}

S_{N} - T_{N} = \frac{f ( 1 ) - f ( 0 )}{2 N} = \frac{1}{24 N} .

The integrand is continuous on

[0, 1]

, so the trapezoidal values converge to its definite integral; the endpoint correction tends to zero. Therefore

N \to \infty lim m = 1 \sum N \frac{N}{4 N ^{2} - m ^{2}} = \frac{1}{4} lo g 3.

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