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

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

For the Leibniz integral rule, the upper endpoint has derivative

2 π x

and the lower endpoint has derivative

1

. The integrand has partial derivative

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}) .

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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