 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/i/solution

For the Leibniz integral rule, the upper endpoint has derivative

2 π x

and the lower endpoint has derivative

1

4 x t + sin t

. Therefore

h^{'} (x) = (2 π x^{4} + x sin (π x^{2})) 2 π x - (2 x^{3} + x sin x) + \int_{x}^{π x^{2}} (4 x t + sin t) d t .

The last integral is

[2 x t^{2} - cos t]_{x}^{π x^{2}} = 2 π^{2} x^{5} - 2 x^{3} - cos (π x^{2}) + cos x

. Combining terms gives

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

 New to topics? Read the docs here!