 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/4/a/solution

Let

h = (b - a) / n

x_{j} = a + jh

, and

m_{j} = a + (j + \frac{1}{2}) h

M_{n} = h \sum_{j = 0}^{n - 1} f (m_{j}), T_{n} = h [\frac{f ( a ) + f ( b )}{2} + \sum_{j = 1}^{n - 1} f (x_{j})] .

2 n

strips, the spacing is

h /2

. Its odd-indexed nodes are the

m_{j}

and its even interior nodes are the

x_{j}

. Consequently

S_{2 n} = \frac{h}{6} [f (a) + f (b) + 4 j = 0 \sum n - 1 f (m_{j}) + 2 j = 1 \sum n - 1 f (x_{j})] = \frac{2}{3} M_{n} + \frac{1}{3} T_{n} .

This is an identity between the quadrature rules, not an approximation involving an omitted quadrature error.

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