 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2012/p/p2/b/solution

For one interval, the trapezoidal rule gives

I_{1} = \frac{1}{2} {0^{4} + 1^{4}} = \frac{1}{2} .

For two intervals of width

h = 1/2

, the composite trapezoidal rule gives half weights at the endpoints and full weight at the internal node:

I_{1/2} = \frac{1}{2} {\frac{1}{2} 0^{4} + (\frac{1}{2})^{4} + \frac{1}{2} 1^{4}} = \frac{9}{32} .

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