 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/20/b/ii/solution

Holding

x

fixed when taking the antiderivative with respect to

t

gives

h (x) = [x^{2} t^{2} - x cos t]_{x}^{π x^{2}} = π^{2} x^{6} - x^{4} - x cos (π x^{2}) + x cos x .

Use the product rule and the chain rule to differentiate this explicit expression:

h^{'} (x) = 6 π^{2} x^{5} - 4 x^{3} - cos (π x^{2}) + 2 π x^{2} sin (π x^{2}) + cos x - x sin x .

This agrees term by term with part (i).

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