 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2/6/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

For

x > 0

, the kernel is smooth between the two positive endpoints. The Leibniz integral rule gives both endpoint terms and the parameter derivative:

I^{'} (x) = \frac{3 c o s ( x ^{4} ) - 2 c o s ( x ^{3} )}{x} - \int_{x^{2}}^{x^{3}} sin (x t) d t .

The remaining integral equals

[cos (x^{3}) - cos (x^{4})] / x

, so

I^{'} (x) = \frac{4 cos ( x ^{4} ) - 3 cos ( x ^{3} )}{x}, x > 0.

The substitution

u = x t

gives the same result directly from limits

x^{3}, x^{4}

. Negative

x

would make the original interval cross the nonintegrable

1/ t

singularity, so it is not part of this ordinary-integral domain.

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