 Solution

ID: 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/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}

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