 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/4/a/ii/solution

For the composite trapezoidal rule, the mesh width is

h = 1/ N

and the nodes are

x_{n} = n / N

. The endpoint weights are halves and the interior weights are ones, so

T_{N} = \frac{1}{N} [\frac{3 ( 0 ) ^{2}}{2} + n = 1 \sum N - 1 3 (\frac{n}{N})^{2} + \frac{3 ( 1 ) ^{2}}{2}] = \frac{3}{N ^{3}} n = 1 \sum N - 1 n^{2} + \frac{3}{2 N} .

This formula is an approximation to the definite integral, not an exact equality with it for finite

N

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