Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 5 / b / ii

Table of contents
- Solution ii

Solution

h = 1/4

, the composite trapezoidal rule gives half weight to the two endpoints:

T_{4} = \frac{1}{4} [\frac{0 ^{2} + 1 ^{2}}{2} + (\frac{1}{4})^{2} + (\frac{1}{2})^{2} + (\frac{3}{4})^{2}] = \frac{1}{4} (\frac{1}{2} + \frac{1 + 4 + 9}{16}) .

Therefore

T_{4} = \frac{11}{32} = 0.34375.

It exceeds the exact definite integral by

1/96

, as expected because

x^{2}

is a convex function and each trapezoid lies above the curve.

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