 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018/s3/2/b/solution

The probability density is even, so its first moment vanishes. Its second moment follows by integrating the polynomial:

⟨ x^{2} ⟩ = A^{2} [\frac{2 a ^{7}}{3} - \frac{4 a ^{7}}{5} + \frac{2 a ^{7}}{7}] .

Using the normalization from part a gives

⟨ x ⟩ = 0, ⟨ x^{2} ⟩ = \frac{a ^{2}}{7} .

The variance in position is consequently

a^{2} /7

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