 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/6/solution

u = 3 - 2 x

, so

d x = - d u /2

and

x = (3 - u) /2

- \frac{1}{4} \int (3 u^{1/2} - u^{3/2}) d u = - \frac{1}{2} u^{3/2} + \frac{1}{10} u^{5/2} + C .

Thus, with an arbitrary constant of integration,

\int x 3 - 2 x d x = - \frac{x + 1}{5} (3 - 2 x)^{3/2} + C .

This real antiderivative applies for

x < 3/2

and extends continuously to the stated endpoint. Its left derivative there agrees with the limiting integrand, zero.

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